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https://math.stackexchange.com/questions/1541816/different-series-representation-for-the-same-function/1541828
# Different series representation for the same function Ok, we know that $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$. Now if we want to have a series representation of $y=\frac{1}{2-x}$ then there are two approaches: First $\frac{1}{2-x}=\frac{1}{1-(x-1)}=1+(x-1)+(x-1)^2+(x-2)^3+\cdots$ where $x-1$ takes the "role" of the $x$ in the original geometric series. The interval of convergence is thus "shifted" to (0,2). On the other hand $\frac{1}{2-x}=\frac{1}{2(1-\frac{x}{2})}=\frac{1}{2}(1+\frac{x}{2}+(\frac{x}{2})^2+(\frac{x}{2})^3+\cdots)$ But this series has an interval convergence of $(-2,2)$ which is different from the interval of convergence of the first series. I have two different series with two different interval convergences for the very same function. I can't rhyme this. Which one is correct? If both are right, can you come up with other series for the given function with again a different interval of convergence? I feel I am missing a fundamental concept here and it bugs me. Any input is welcome. Thanks. • @TimRaczkowski He's not saying they're the same function ... – Zubin Mukerjee Nov 22 '15 at 21:59 • Note for radius of convergence !$$1+(x-1)+(x-2)^2+... \to |x-1|<1 \to 0<x<2$$but in $$1+\frac{x}{2}+\frac{x^2}{4}+... \to |\frac{x}{2}|<1 \to -2<x<2$$ – Khosrotash Nov 22 '15 at 22:00 • They are power-series around different points of expansion. I think your arguments are OK. Because they are expanded around different points, they have different formulas and different intervals of convergence. It's fine. – Leaning Nov 22 '15 at 22:00 • @TimRaczkowski No, I am not. I am introducing the geometric series to introduce my problem with $y=\frac{1}{2-x}$ – imranfat Nov 22 '15 at 22:02 • @ZubinMukerjee,@imranfat Sorry. I was me who misread the post. – Tim Raczkowski Nov 22 '15 at 22:06 In complex analysis, this is often called "analytic continuation". The function $\frac{1}{2-x}$ you are describing only has a single pole $x = 2$ on the complex plane, and otherwise well defined. But to write an arbitrary function in terms of power series, then such formalism has limitations, namely one would in general decrease the actual radius of convergence. For instance when we expand around $x=1$ then such power series is only defined for $|x-1|<1$. If we expand around $x=0$ then the radius of convergence is $|x|<2$. In fact for $\frac{1}{1-x}$ you can also expand around $x=-1$ to get another representative. There are infinite ways to expand this function, actually. In one word, power series expansion is only a "partial" representation of the original function. • Ah, I am trying to understand what you are getting at. Are you saying that in the first case I derived a power series around $x=1$ and in the second case I set it up around $x=0$? – imranfat Nov 22 '15 at 22:09 • @imranfat Yes that's right. Something to keep in mind, when you find the Taylor expansion for a function around a certain point, the radius of convergence can be no bigger than the distance from the given point and the closest point at which the function is undefined. – Tim Raczkowski Nov 22 '15 at 22:13 • Allright. Thanks all, I just need to absorb this for while... – imranfat Nov 22 '15 at 22:14 • @RoryDaulton You are right. Thank you for pointing out. – Kevin Ye Nov 23 '15 at 3:31 • Also keep in mind that "the closest point at which the function is undefined" may be a complex number! If you expand $1/(1+x^2)$ around $x=0$, you will find the interval of convergence is $|x| < 1$ which might seem mysterious because the function is defined for all real numbers. But in the complex plane there are singularities at $\pm i$, which explains why there is no convergence around $x=0$ beyond radius 1. – Ted Nov 23 '15 at 4:16
2019-05-22T04:39:15
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http://mathhelpforum.com/calculus/116196-converge-diverge.html
1. ## Converge or diverge? Does the series from 1 to infinity 1/(5n^2+1)^(1/3) converge or diverge? I thought by using the divergence test and checking if the limit goes to 0 as n approaches infinity that the series converges. But the answer is "diverges." What test could I use to get divergence for this series? 2. Originally Posted by jlmills5 Does the series from 1 to infinity 1/(5n^2+1)^(1/3) converge or diverge? I thought by using the divergence test and checking if the limit goes to 0 as n approaches infinity that the series converges. But the answer is "diverges." What test could I use to get divergence for this series? $\frac{1}{(5n^2+1)^{1\slash 3}}\geq \frac{1}{(10n^2)^{1\slash 3}}$ $=\frac{1}{\sqrt[3]{10}}\,\frac{1}{n^{2\slash 3}}$ , and as $\sum\limits_{n=1}^{\infty}\frac{1}{n^p}$ converges iff $p>1$ , our series diverges by the comparison test. Tonio 3. Thank you! I have such issues trying to find a good comparison series when using direct and limit comparison. 4. Hello, jlmills5! Converge or diverge? . $S \;=\;\sum^{\infty}_{n=1} \frac{1}{(5n^2+1)^{\frac{1}{3}}}$ I forced it into a comparison test . . . For $n \geq 1\!:\;\;1 \:<\:3n^2$ Add $5n^2$ to both sides: . $5n^2 + 1 \:< \:8n^2$ Take the cube root of both sides: . $\left(5n^2+1\right)^{\frac{1}{3}} \:< \:\left(8n^2\right)^{\frac{1}{3}}$ . . That is: . $(5n^2+1)^{\frac{1}{3}} \;<\;2n^{\frac{2}{3}}$ Take the reciprocal of both sides: . $\frac{1}{\left(5n^2+1\right)^{\frac{1}{3}}} \;{\color{red}>} \;\frac{1}{2n^{\frac{2}{3}}}$ Summate both sides: . $\sum^{\infty}_{n=1}\frac{1}{\left(5n^2 +1\right)^{\frac{1}{3}}} \;>\;\sum^{\infty}_{n=1}\frac{1}{2n^{\frac{2}{3}}}$ We have: . $S \;>\;\frac{1}{2}\sum^{\infty}_{n=1}\frac{1}{n^{\fr ac{2}{3}}}$ . . . a divergent $p$-series Since $S$ is greater than a divergent series, $S$ diverges. 5. Okay, I'm looking over this again and I'm not sure I understand how this means divergence. The original series is greater than or equal to the comparison series. I thought the comparison test only worked if the comparison series was greater than the original. And only worked then if the comparison series converges (and hence the original converges.) This is what I was taught in class anyway. 6. Originally Posted by jlmills5 Okay, I'm looking over this again and I'm not sure I understand how this means divergence. The original series is greater than or equal to the comparison series. I thought the comparison test only worked if the comparison series was greater than the original. And only worked then if the comparison series converges (and hence the original converges.) This is what I was taught in class anyway. Since the "smaller" series diverges, then the "larger" series must diverge as well. Similarly, if you find a "larger" series that converges, then the "smaller" series must converge as well. 7. Thanks! I think I've been looking at this too long. I understood and first and then second guessed myself. I got it now!
2017-10-22T00:05:21
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http://volv.casandrea.it/sequence-calculator-convergence.html
Its terms are non-increasing — in other words, each term is either smaller than or the same as its predecessor (ignoring the …. An alternating series is a series where the terms alternate between positive and negative. 1 SEQUENCES SUGGESTED TIME AND EMPHASIS 1 class Essential material POINTS TO STRESS 1. ) Since the series on the right converges, the sequence $${S_k}$$ is bounded above. A) A sequence is a list of terms. Solution b. Of course, sequences can be both bounded above and below. Convergence of sequences One concept that is typically hard to grasp is the convergence of a sequence. A telescoping series is any series where nearly every term cancels with a preceeding or following term. So we know that π = 3. Let's consider that we have points in sequence along with a point L is known as the limit of the sequence. Although strictly speaking, a limit does not give information about any finite first part of the sequence, the concept of rate of convergence is of practical importance when working with a sequence of successive approximations for an iterative method, as then typically fewer. 12 INFINITE SEQUENCES AND SERIES 12. Get the free "Limit Calculator - Math 101" widget for your website, blog, Wordpress, Blogger, or iGoogle. Calculus 2 - Geometric Series, P-Series, Ratio Test, Root Test, Alternating Series, Integral Test - Duration: 43:52. Convergence is the tendency of group members to become more alike over time. How small do these terms get? Use your calculator and put in some large values for x and see what happens. its sequence of terms fangis alternating. So we've explicitly defined four different sequences here. In particular, we will define different types of convergence. A sequence of functions converges uniformly to a limiting function on a set if, given any arbitrarily small positive number , a number can be found such that each of the functions , +, +, … differ from by no more than at every point in. One very important point to remember is that the calculations in this section are possible because we consider convergent series; remember it is not possible to define a sequence of remainders for a divergent series!. To see if the sequence converges enter a large number such 999 for the end value. We then add 3 to get the next term. For example, the function y = 1/x converges to zero as x increases. Any bounded increasing (or decreasing) sequence is convergent. Fibonacci sequence is a sequence of numbers, where each number is the sum of the 2 previous numbers, except the first two numbers that are 0 and 1. Currently, it can help you with the two common types of problems: Find the n-th term of an arithmetic sequence given m-th term and the common difference. Limit of sequence is the value of the series is the limit of the particular sequence. Arithmetic Series. Alternating. A telescoping series is any series where nearly every term cancels with a preceeding or following term. Sequences that are not convergent are said to be divergent. If 0 1 then the sequence +1 converges to 0 as , as proved elsewhere, and of course 1-0, so by continuity 0-0 1-= 11-, as required. Let {fn} be the sequence of functions defined by fn(x) = cosn(x) for −π/2 ≤ x ≤ π/2. and convergence. Show that a sequence must converge to a limit by showing that it is montone and appropriately bounded. This is a follow up question to this question I asked two days ago. In fact, if the series is only conditionally convergent, then both the Ratio and Root Test will turn out to be inconclusive. As we have discussed in the lecture entitled Sequences of random variables and their convergence, different concepts of convergence are based on different ways of measuring the distance between two random variables (how "close to each other" two random variables are). It's denoted as an infinite sum whether convergent or divergent. We write this as Moreover, if the number L exists, it is referred to as the limit of the sequence and the sequence is convergent. Limit of a Sequence 2 3. Proof We will prove that the sequence converges to its least upper bound (whose existence is guaranteed by the Completeness axiom). If {S n} diverges, then the sum of the series diverges. The Alternating Series Test (Leibniz’s Theorem) This test is the sufficient convergence test. Geometric sum matlab. This sequence is different from the first two in the sense that it doesn’t have a specific formula for each term. Pointwise convergence Definition. The formulas for the sum of first numbers are and. Also, it can identify if the sequence is arithmetic or geometric. The limit of this sequence happens to be the number e c, that is, the Euler number raised to the power c. Here are some examples: * 1, 1, 1, 1, 1, * 1, 2, 3, 4, 5 * 1, -2, 3, -4, 5, -6, * 1, 1/2, 1/3, 1/4, 1/5,. implies Use the integral test. \) with the specific property that the ratio between two consecutive terms of the sequence is ALWAYS constant, equal to a certain value $$r$$. Mathematical Definitions A power series, f(x) = X∞ n=0 anx n, is an example of a sum over a series of functions f(x) = X∞ n=0 gn(x), (1) where gn(x) = anxn. It's a simple online calculator which provides immediate and accurate results. In mathematics, the harmonic series is the divergent infinite series ∑ = ∞ = + + + + + ⋯. If it is convergent, find its limit. Find more Mathematics widgets in Wolfram|Alpha. 1 (2223 votes) Select Rating: 5 Stars (Best) 4 Stars 3 Stars 2 Stars 1 Star (Worst). If 0 1 then the sequence +1 converges to 0 as , as proved elsewhere, and of course 1-0, so by continuity 0-0 1-= 11-, as required. With the default setting Method->Automatic, a number of additional tests specific to different classes of sequences are used. An example of recursion is Fibonacci Sequence. 4 Sequence and Series of Real Numbers M. Some well-known sequences are built in: Define a recursive sequence using RecurrenceTable: (Note the use of {x,min,max} notation. decreasing), then the limit is the least-upper bound (resp. Sequences make interesting graphs! You need to learn how to set the mode and select the color before entering a sequence in your TI-84 Plus calculator. However, neither one can tell you about conditional convergence. $$So everything you know about sequences can be applied to series, and vice-versa. For the series just use the ratio test! Intuitively, the denominator grow much faster than the numerator, so the limit will be zero. Short Answer 1. Typically, we have an interative algorithm that is trying to find the maximum/minimum of a function and we want an estimate of how long it will take to reach that optimal value. The calculator allows to calculate the terms of an arithmetic sequence between two indices of this sequence. The radius of convergence, R, is the largest number such that the series is guaranteed to converge within the interval between c - R and c + R. ) Since the series on the right converges, the sequence $${S_k}$$ is bounded above. The sequence is said to be convergent, in case of existance of such a limit. This is actually one of the few series in which we are able to determine a formula for the general term in the sequence of partial fractions. Discuss the pointwise convergence of the sequence. It needs a trick to show that this sequence is also increasing if c > 0. 1 On-Line Calculator with Cal symbols ; 2 dictionary, games, excellent examples ; 3 Log Logarithm Calculator https://keisan. Define a sequence in terms of the variable n and, choose the beginning and end of the sequence and see the resulting table of values. Meaning 'the sum of all terms like', sigma notation is a convenient way to show where a series begins and ends. It is possible to make the ratio test applicable to certain cases where the limit L fails to exist, if limit superior and limit. Infinite Geometric Series. A sequence that is not convergent is divergent. Calculates the sum of a convergent or finite series. First Five Terms: 28, 38, 48, 58, 68 First Five Terms: −38, −138, −238, −338, −438. Set to be the sequence generated by our scientist ( is the data collected after n days). It is easy to show that this sequence is bounded from above. Let from a convergent sequence extracted is infinitely many terms, a n 1 , a n 2 ,. Limit Calculator. If an input is given then it can easily show the result for the given number. The limit of the sequence or equivalently satisfies the equation. The formulas for the sum of first numbers are. The limit of a sequence is said to be the fundamental notion on which the whole of analysis ultimately rests. 4 Banach Fixed Point Theorem for Operators Let S denote the set of continuous functions on [a,b] that lie within a fixed distance ↵ > 0 of. The following geometric sequence calculator will help you determine the nth term and the sum of the first n terms of an geometric sequence. By using this website, you agree to our Cookie Policy. EX 4 Show converges absolutely. But many important sequences are not monotone—numerical methods, for in-. Infinite Series Calculator Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. To say that Xn converges in probability to X, we write. Every unbounded. 2 Sequences: Convergence and Divergence In Section 2. Sequences make interesting graphs! You need to learn how to set the mode and select the color before entering a sequence in your TI-84 Plus calculator. By a theorem found in the next section Basic properties, such a sequence must be convergent. Does lim n→∞ an bn = c > 0 c finite & an,bn > 0? Does. This sequence has a limit L, if a n approaches L as n approaches infinity. You can even use the ratio test to find the radius and interval of convergence of power series! Many students have problems of which test to use when trying to find whether the series converges or diverges. If {a n} is both a bounded sequence and a monotonic sequence, we know it is convergent. Determine whether a sequence converges or diverges, and if it converges, to what value. Here it is: Convergence and Divergence: You say that a sequence converges if its limit exists, that is, if the limit of its terms equals a finite number. A geometric sequence refers to a sequence wherein each of the numbers is the previous number multiplied by a constant value or the common ratio. 14159265359 … π = 3. There are many kinds of sequences, including those based on infinite lists of numbers. You can identify a series with the sequence of its partial sums:$$ S_n = \sum_{k=1}^n a_k. When a sequence has a limit that exists, we say that the sequence is a convergent sequence. The formulas for the sum of first numbers are and. Implicit Derivative. absolute value of nominator should be less than denominator. Byju's Radius of Convergence Calculator is a tool. A sequence that is not convergent is divergent. It's a bit like the drunk who is looking for his keys under the streetlamp, not because that's where he lost. Not every sequence has this behavior: those that do are called convergent, while those that don't are called divergent. Convergence Tests Name Summary Divergence Test If the terms of the sequence don't go to zero, the series diverges. Short Answer 1. development of a formula to estimate the rate of convergence for these methods when the actual root is not known. Each term (except the first term) is found by multiplying the previous term by 2. 1) The ratio test states that: if L < 1 then the series converges absolutely ; if L > 1 then the series is divergent ; if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case. The sums are heading towards a value (1 in this case), so this series is convergent. Series of Numbers 4. Includes the nth-Term, geometric series, p-Series, integral test, ratio test, comparison, nth-Root, and the alternating series test. No calculator except unless specifically stated. The Ratio Test is used extensively with power series to find the radius of convergence, but it may be used to determine convergence as well. 13 the even convergents form a strictly increasing sequence and the odd convergents form a strictly decreasing sequence. If the sequence of these partial sums {S n} converges to L, then the sum of the series converges to L. Find the next number in the sequence of integers. Assume that lim n!1 an exists for anC1 D p 3an with a0 D2: Find lim n!1 an. We call ff ngpointwisely converges to f on Eif for every x2E, the sequence ff n(x)gof real numbers converges to the number f(x). By using this website, you agree to our Cookie Policy. The "sum so far" is called a partial sum. You need to provide the first term of the sequence ( ), the difference between two consecutive values of the sequence ( d ), and the number of steps ( n ). The values $${a_n} = f\left( n \right)$$ taken by the function are called the terms of the sequence. Conic Sections Trigonometry. A geometric sequence has the form: a 1, a 1 r, a 1 r 2, a_1, a_1 r, a_1 r^2, You need to provide the first term of the sequence ( ), the constant ratio between two consecutive values of the sequence (. Fibonacci sequence table. 免费的级数收敛计算器 - 一步步检验无穷级数的收敛性. Learn about the Golden Ratio, how the Golden Ratio and the Golden Rectangle were used in classical architecture, and how they are surprisingly related to the famed Fibonacci Sequence. Determines convergence or divergence of an infinite series. Determine whether or not the series converge using the appropriate convergence test (there may be more than one applicable test. The partial sums of a series form a new sequence, which is denoted as {s 1, s 2, s 3, s 4,}. We have seen that, in general, for a given series , This clearly implies that the sequence is convergent and. This arithmetic sequence calculator (also called the arithmetic series calculator) is a handy tool for analyzing a sequence of numbers that is created by adding a constant value each time. In particular, we will define different types of convergence. Mathematical Definitions A power series, f(x) = X∞ n=0 anx n, is an example of a sum over a series of functions f(x) = X∞ n=0 gn(x), (1) where gn(x) = anxn. Here’s another convergent sequence: This time, the sequence …. Bounded Sequence. This program tests the convergence or divergence of a series. Recall that if and are continuous functions on an interval and , then. A positive series is convergent if each of its terms is less than or equal to the corresponding terms of a series that is known to be convergent. You must justify each answer using some of the convergence tests we discussed in lecture. com allows you to find the sum of a series online. If there is no such number, then the sequence is divergent. Plugging into the summation formula, I get:. In particular, we will define different types of convergence. Therefore, all the terms in the sequence are between k and K'. the number getting raised to a power) is between -1 and 1. That doesn’t matter because t = 0 is not a continuity point of F and the de nition of convergence in distribution only requires convergence at continuity points. There are two powerful convergence tests that can determine whether a series is absolutely convergent: the Ratio Test and Root Test. Also, it can identify if the sequence is arithmetic or geometric. Enter a character array, string vector, cell array of character vectors, or an array of structures with the field Sequence. Use an approriate test for monotonicity to determine if a sequence is increasing or decreasing. (calculator not allowed) Which of the following series are conditionally convergent? I. The nth term of the series does not approach zero therefore the series diverges, specifically to ¡1. Infinite Sequences and Series This section is intended for all students who study calculus, and considers about $$70$$ typical problems on infinite sequences and series, fully solved step-by-step. its sequence of partial sums fSngconverges to some real number. An infinite sequence (a n) is called convergent if limit n tends to infinity a n exists and is finite. The "sum so far" is called a partial sum. Pointwise and Uniform Convergence 1. " Adjust N to take more points of the sequence. Recall that if and are continuous functions on an interval and , then. The fact that absolute convergence implies ordinary convergence is just common sense if you think about it. The list may have finite or infinite number of terms. Arithmetic sequences calculator. Viewed 340 times 1. All rights belong to the owner! Sum of series. Indeed, consider our scientist who is collecting data everyday. Otherwise, the infinite series does. Here are a few examples of sequences. A sequence is "converging" if its terms approach a specific value as we progress through them to infinity. It is easy to show that this sequence is bounded from above. 1) The ratio test states that: if L < 1 then the series converges absolutely ; if L > 1 then the series is divergent ; if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case. Free Sequences calculator - find sequence types, indices, sums and progressions step-by-step This website uses cookies to ensure you get the best experience. However, it does tell us what each term should be. See how the sequence a(n) = 1/n converges to zero, or, how "dividing by bigger numbers makes the fraction smaller. The meanings of the terms “convergence” and “the limit of a sequence”. It’s also known as the Leibniz’s Theorem for alternating series. Interval of Convergence for a Power Series In other words, according to Paul’s Online Notes , a power series may converge for some values of x and not for other values of x. The Alternating Series Test (Leibniz’s Theorem) This test is the sufficient convergence test. We are introduced to Sequences in Calculus with discussions about Series following in subsequent lessons. A Sequence is a set of things (usually numbers) that are in order. Convergence is defined purely based on the solution value and nothing to do with balances or anything else. Convergence in probability is stronger than convergence in distribution. Just type, and your answer comes up live. To get started, try working from the example problem already populated in the. It is capable of computing sums over finite, infinite (inf) and parametrized sequencies (n). The calculator allows to calculate the terms of an arithmetic sequence between two indices of this sequence. This article has also been viewed 94,189 times. You must justify each answer using some of the convergence tests we discussed in lecture. (a) 2, 3 4, 4 9, 5 16, 6 25, … (b) 1, 1 2, 1 6, 1 24, 1 120, …. Any convergent sequence is bounded (both above and below). The notion of limit of a sequence is very natural. Iteration is a common approach widely used in various numerical methods. Apart from the coefficients a n , the program allows to find the convergent A n / B n. A sequence is divergent if it tends to infinity, but it is also divergent if. Solution b. Does lim n→∞ an bn = c > 0 c finite & an,bn > 0? Does. N th term of an arithmetic or geometric sequence. This sequence is different from the first two in the sense that it doesn’t have a specific formula for each term. Infinite Series Calculator Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. An alternating series is said to be conditionally convergent if it's convergent as it is but would become divergent if all its terms were made positive. Ask Question Asked 1 year, 10 months ago. For example, the Fibonacci sequence $\{ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,. Sequences calculator overview: Whether you are using geometric or mathematical type formulas to find a specific numbers with a sequence it is very important that you should try using with a different approach using recursive sequence calculator to find the nth term with sum. Sigma (Sum) Calculator. The following geometric sequence calculator will help you determine the nth term and the sum of the first n terms of an geometric sequence. The Squeeze Theorem is an important result because we can determine a sequence's limit if we know it is "squeezed" between two other sequences whose limit is the same. One of the ways in which algorithms will be compared is via their rates of convergence to some limiting value. email: [email protected] Limits capture the long-term behavior of a sequence and are thus very useful in bounding them. But many important sequences are not monotone—numerical methods, for in-. The sequence is said to be convergent, in case of existance of such a limit. We conclude that $${S_k}$$ is a monotone increasing sequence that is bounded above. n must be a positive integer. In this unit you will also learn about 'convergence' and 'recurrence' of series. Suppose that (f n) is a sequence of functions, each continuous on E, and that f n → f uniformly on E. In more formal language, a series converges if there exists a limit l such that for any arbitrarily small positive number , there is a large. Because the Cauchy sequences are the sequences whose terms grow close together, the fields where all Cauchy sequences converge are the fields that are not missing" any numbers. Here is a set of assignement problems (for use by instructors) to accompany the Sequences section of the Series & Sequences chapter of the notes for Paul Dawkins Calculus II course at Lamar University. For example, consider the following sequence of functions with common domain$\mathbb{R}$: (2). An expert mathematician will show you the practical applications of these famous mathematical formulas and unlock their secrets for you. Sequences calculator overview: Whether you are using geometric or mathematical type formulas to find a specific numbers with a sequence it is very important that you should try using with a different approach using recursive sequence calculator to find the nth term with sum. Figure $$\PageIndex{1}$$: Illustrating the convergence with the Alternating Series Test. Find the n th term (rule of sequence) of each sequence, and use it to determine whether or not the sequence converges. In this section we will discuss using the Ratio Test to determine if an infinite series converges absolutely or diverges. To recall, an arithmetic sequence, or arithmetic progression (AP) is a sequence of numbers such that the difference, named common difference, of two successive members of the sequence is a constant. A Sequence is a set of things (usually numbers) that are in order. Fourier Series Calculator is an online application on the Fourier series to calculate the Fourier coefficients of one real variable functions. Smith , Founder & CEO, Direct Knowledge. So, more formally, we say it is a convergent series when: "the sequence of partial sums has a finite limit. Here’s an example of a convergent sequence: This sequence approaches 0, so: Thus, this sequence converges to 0. The constant is called the common difference ( ). Choose a sequence from the drop-down menu (or choose 'Enter your own' and type one in). uk 2 c mathcentre 2009. In mathematics, an arithmetic sequence, also known as an arithmetic progression, is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. Moreover, we have. Even, Paul’s Online Notes calls the geometric series a special series because it has two important features: Allows us to determine convergence or divergence, Enables us to find the sum of a convergent geometric series; Moreover, this test is vital for mastering the Power Series, which is a form of a Taylor Series which we will learn in. Work out the problem with our Free limit of sequence calculator. Our sequence calculator outputs subsequences of the specified sequence around the selected nth element. Convergence Tests Name Summary Divergence Test If the terms of the sequence don't go to zero, the series diverges. a = (x₁ - x₀ψ) / √5. Given the explicit formula for an arithmetic sequence find the first 5 terms. Set to be the sequence generated by our scientist ( is the data collected after n days). Find more Mathematics widgets in Wolfram|Alpha. Convergence in probability is stronger than convergence in distribution. Implicit Derivative. zip: 1k: 06-02-17: Arithmetic Series Solver (Includes Sigma. The free tool below will allow you to calculate the summation of an expression. Uniform convergence is particularly useful in that if a sequence of di↵erentiable (and therefore continuous) functions is uniformly convergent, then the function to which it con-verges is also continuous. Let from a convergent sequence extracted is infinitely many terms, a n 1 , a n 2 ,. Can you find their patterns and calculate the next two terms? 3, 6 +3, 9 +3, 12 +3, 15 +3, +3 +3, …. Assume the statement true for n. As n increases, that angle decreases. A recursion is a special class of object that can be defined by two properties: 1. ) Unfortunately, there are very few series to which the definition can be applied directly; the most important is certainly the Geometric Series. Any bounded increasing (or decreasing) sequence is convergent. Here are a few examples of sequences. random variables converges in distribution to a standard normal distribution. That is, if xk! x, we are interested in how fast this happens. 1 Convergence of a Sequence of Numbers. If possible, give the sum of the series. First Five Terms: 28, 38, 48, 58, 68 First Five Terms: −38, −138, −238, −338, −438. 13 - 4 Limits of Infinite Sequences Important stuff coming! A series that does not have a last term is called infinite. But there are degrees of divergence. LIMIT OF A SEQUENCE: THEOREMS 117 4. Limit Calculator. Then, you can have the fun of graphing a sequence. com allows you to find the sum of a series online. One of the ways in which algorithms will be compared is via their rates of convergence to some limiting value. Some well-known sequences are built in: Define a recursive sequence using RecurrenceTable: (Note the use of {x,min,max} notation. The Limit of a Sequence 3. Here’s another convergent sequence: This time, the sequence …. The limit of a sequence of functions is defined in a similar manner. Convergence of sequences One concept that is typically hard to grasp is the convergence of a sequence. Calculus Definitions >. iii) if ρ = 1, then the test is inconclusive. Sequences that are not convergent are said to be divergent. Solution diverges. This is the main site of WIMS (WWW Interactive Multipurpose Server): interactive exercises, online calculators and plotters, mathematical recreation and games If you want a visual indication of the convergence of a sequence or a series, this page is an ideal tool. Relevant theorems, such as the Bolzano-Weierstrass theorem, will be given and we will apply each concept to a variety of exercises. Just type, and your answer comes up live. If possible, give the sum of the series. Infinite Sequences and Series A sequence of real numbers $$n$$ is a function $$f\left( n \right),$$ whose domain is the set of positive integers. implies Use the integral test. Series are sums of multiple terms. Limit of sequence is the value of the series is the limit of the particular sequence. Let us consider a sum of the form given in eq. The sums are heading towards a value (1 in this case), so this series is convergent. It should be noted, that if the calculator finds sum of the series and this value is the finity number, than this series converged. How can I find out if 1/n! is divergent or convergent? I cannot solve it using integral test because the expression contains a factorial. Series of Numbers 4. Calculates the sum of a convergent or finite series. Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. Squeeze Theorem for Sequences We discussed in the handout \Introduction to Convergence and Divergence for Sequences" what it means for a sequence to converge or diverge. Given a convergent sequence of functions { f n } n = 1 ∞ \{f_n\}_{n=1}^{\infty} { f n } n = 1 ∞ , it is natural to examine the properties of the resulting limit function f f f. Evidence of sequence convergence in the Prestin gene between dolphins and bats. The sum of an arithmetic series 5 5. Let † > 0. CONVERGENCE AND DIVERGENCE We say the sequence if is a real number. However, it does tell us what each term should be. The idea is very trivial though: A sequence $$a_m$$ converges to a value $$a$$ if the values of the sequence get closer and closer to $$a$$ (in fact they get as close as we want) as $$n$$ approaches to infinity. 2 The sequences (1=n), (( 1)n=n), (1 1 n) are convergent with limit 0, 0, 1 respectively: For the sake of illustrating how to use the de nition to justify the above state-ment, let us provide the details of the proofs: (i) Let a n= 1=n for all n2N, and let ">0 be given. Also, it can identify if the sequence is arithmetic or geometric. A series such as 3 + 7 + 11 + 15 + ··· + 99 or 10 + 20 + 30 + ··· + 1000 which has a constant difference between terms. A series is convergent if the sequence of its partial sums (,,, …) tends to a limit; that means that the partial sums become closer and closer to a given number when the number of. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test. Here's a definition: A sequence {a n} is said to converge to the number a provided that for every positive number ε there is an index N such that: |a n - a| < ε , for all. [ Don't peek. If you enroll prior to that date, you'll be able to access the course through September 2018. Active 1 year, 10 months ago. The list may have finite or infinite number of terms. Theorem 1: Let$(a_n)$be a convergent sequence. Learn how this is possible and how we can tell whether a series converges and to what value. Share a link to this widget: More. The Geometric Series Test is one the most fundamental series tests that we will learn. An alternating series is a series where the terms alternate between positive and negative. Instructions: This algebraic calculator will allow you to compute elements of a geometric sequence. We will now look at some examples of determining whether a sequence of functions is pointwise convergent or divergent. Convergent sequences have a finite limit. If a sequence x 1;x 2;:::;x nconverges to a value rand if there exist real numbers >0 and 1 such that (1) lim n!1 jx n+1 rj jx n rj = then we say that is the rate of convergence of the sequence. The notion of limit of a sequence is very natural. Get an intuitive sense of what that even means!. A proof of the Ratio Test is also given. For instance, the series is telescoping. Limit of sequence is the value of the series is the limit of the particular sequence. 6 Absolute Convergence and the Ratio and Root Tests 1 If is absolutely convergent with sum ,n n a s ∞ = ∑ A major difference between absolutely co nvergent and conditionally convergent comes in the rearrangement of the terms. b) Find the 100 th term ( {a_{100}}). A sequence is said to be convergent if it approaches some limit (D'Angelo and West 2000, p. A sequence of functions {f n} is a list of functions (f 1,f 2,) such that each f n maps a given subset D of R into R. The second differences are the differences of the differences, in this case. A bounded sequence is one where the absolute value of every term is less than or equal to a particular real, positive number. This Arithmetic Sequence Calculator is used to calculate the nth term and the sum of the first n terms of an arithmetic sequence. Oscillating sequences are not convergent or divergent. It is easy to check that. However, it does tell us what each term should be. The sum of a geometric series 9 7. The limit of this sequence happens to be the number e c, that is, the Euler number raised to the power c. Alternating p-series are detailed at the end. Otherwise it diverges. In mathematics, a sequence is a chain of numbers (or other objects) that usually follow a particular pattern. With a geometric sequence calculator, you can calculate everything and anything about geometric progressions. Test for Convergence v1. Fibonacci sequence table. Learn vocabulary, terms, and more with flashcards, games, and other study tools. But many important sequences are not monotone—numerical methods, for in-. You can say that an alternating series converges if two conditions are met: Its n th term converges to zero. In addition, you can access absolute convergence and power series calculators. Free Summation Calculator. Free Geometric Sequences calculator - Find indices, sums and common ratio of a geometric sequence step-by-step This website uses cookies to ensure you get the best experience. Pointwise and Uniform Convergence 1. Abstract: We consider sequences of graphs and define various notions of convergence related to these sequences: left convergence'' defined in terms of the densities of homomorphisms from small graphs into the graphs of the sequence, and right convergence'' defined in terms of the densities of homomorphisms from the graphs of the sequence into small graphs; and convergence in a suitably. For instance, the series is telescoping. Let {fn} be the sequence of functions defined by fn(x) = cosn(x) for −π/2 ≤ x ≤ π/2. There are main 2 types of sequence one is convergent and the other one is divergent. A geometric series converges if the r-value (i. This is the main site of WIMS (WWW Interactive Multipurpose Server): interactive exercises, online calculators and plotters, mathematical recreation and games If you want a visual indication of the convergence of a sequence or a series, this page is an ideal tool. Convergence. Limit calculator wolfram alpha limit calculator wolfram alpha solved 1 take a look at the sequence an does this sequ wolfram alpha result for infinite series summation. Its name derives from the concept of overtones, or harmonics in music: the wavelengths of the overtones of a vibrating string are 1 / 2, 1 / 3, 1 / 4, etc. Some infinite series converge to a finite value. Limit of Sequence Calculator. (1) and ask whether the sum is convergent. 13 - 4 Limits of Infinite Sequences Important stuff coming! A series that does not have a last term is called infinite. Recursively Defined Sequences. Typically, we have an interative algorithm that is trying to find the maximum/minimum of a function and we want an estimate of how long it will take to reach that optimal value. The Infinite Series Calculator an online tool, which shows Infinite Series for the given input. Enter the Function: From = to: Calculate: Computing Get this widget. Uniform convergence implies pointwise convergence, but not the other way around. The second differences are the differences of the differences, in this case. Textbook solution for Calculus: Early Transcendentals 8th Edition James Stewart Chapter 11. TI-Nspire v1. Tips for determining convergence or divergence of an in nite series Divergence Test : Always check that the individual terms a ngo to zero. Our online calculator, build on Wolfram Alpha system is able to test convergence of different series. Like the integral test, the comparison test can be used to show both convergence and divergence. Series Calculator computes sum of a series over the given interval. The following geometric sequence calculator will help you determine the nth term and the sum of the first n terms of an geometric sequence. The extent of sequence convergence between bats and whales was thus not sufficient to unite these clades when non-dolphin odontocetes were included in the analysis. 2 Oct 7, 12 3. Limit calculator wolfram alpha wolfram alpha result for infinite series summation posts tagged with advanced math wolfram alpha blog posts tagged with advanced math wolfram alpha blog. The Limit of a Sequence 3. This sequence has a limit L, if a n approaches L as n approaches infinity. an are called the terms of the sequence. This calculator for to calculating the sum of a series is taken from Wolfram Alpha LLC. As you iterate from one iteration to the next, the solution values forms a sequence. If the sequence of partial sums for an infinite series converges to a limit L, then the sum of the series is said to be L and the series is convergent. If you're seeing this message, it means we're having trouble loading external resources on our website. Their terms alternate from upper to lower or vice versa. Monotonic decreasing sequences are defined similarly. The individual elements in a sequence are called terms. The concern is whether this iteration will converge, and, if so, the rate of convergence. If 0 1 then the sequence +1 converges to 0 as , as proved elsewhere, and of course 1-0, so by continuity 0-0 1-= 11-, as required. #N#More References and Links. IT/OT convergence: This definition explains the increasing integration of information technology (IT) with operational technology (OT). Limit Calculator. Find more Mathematics widgets in Wolfram|Alpha. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Drag the slider at the bottom of the screen to show more or less terms of the sequence. >> ab = polyfit(log(e(1:end-1)),log(e(2:end)),1) ab = 0. 3, 6, 12, 24, 48, … Write an equation for this arithmetic sequence and find the. Limit of sequence is the value of the series is the limit of the particular sequence. Taylor Series Convergence The Taylor series converges if f has derivatives of all orders on an interval "I" centered at c, if lim(n→ infin;)RN = 0 for all x in l: The Taylor series remainder of R N = S - S N is equal to (1/(n + 1)!)f (n + 1) (z)(x - c) n + 1 where z is a constant between x and c. It needs a trick to show that this sequence is also increasing if c > 0. There really isn't all that much to this problem. In a Geometric Sequence each term is found by multiplying the previous term by a constant. Meaning 'the sum of all terms like', sigma notation is a convenient way to show where a series begins and ends. absolute value of nominator should be less than denominator. Arithmetic Series. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. 1 Convergence of a Sequence of Numbers. Not every sequence has this behavior: those that do are called convergent, while those that don't are called divergent. The limit of a sequence is said to be the fundamental notion on which the whole of analysis ultimately rests. F2 = F1 + F0 = 1+0 = 1. For example, the sequence 2. Meaning 'the sum of all terms like', sigma notation is a convenient way to show where a series begins and ends. Not every sequence has this behavior: those that do are called convergent, while those that don't are called divergent. 5 Absolute Ratio Test Let be a series of nonzero terms and suppose. Convergence of sequences One concept that is typically hard to grasp is the convergence of a sequence. –Fixed point iteration , p= 1, linear convergence •The rate value of rate of convergence is just a theoretical index of convergence in general. Limit calculator wolfram alpha wolfram alpha result for infinite series summation posts tagged with advanced math wolfram alpha blog posts tagged with advanced math wolfram alpha blog. Before discussing convergence for a sequence of random variables, let us remember what convergence means for a sequence of real numbers. Each term (except the first term) is found by multiplying the previous term by 2. When a sequence has a limit that exists, we say that the sequence is a convergent sequence. Find more Mathematics widgets in Wolfram|Alpha. The main purpose of this calculator is to find expression for the n th term of a given sequence. In particular, we will define different types of convergence. The value of the $$n^{th}$$ term of the arithmetic sequence, $$a_n$$ is computed. Infinite Sequences and Series This section is intended for all students who study calculus, and considers about $$70$$ typical problems on infinite sequences and series, fully solved step-by-step. CALCULATOR; COMMENTS; COURSES; FOR INSTRUCTOR; LOG IN; FOR INSTRUCTORS; Sign In; Email: Password: Forgot password? ← previous. Does P bn converge? Is 0 ≤ an ≤ bn? YES P YES an Converges Is 0 ≤ bn ≤ an? NO NO P YES an Diverges LIMIT COMPARISON TEST Pick {bn}. should be given. Nair EXAMPLE 1. Moreover, we have. Determine whether a sequence converges or diverges, and if it converges, to what value. Mathematical Definitions A power series, f(x) = X∞ n=0 anx n, is an example of a sum over a series of functions f(x) = X∞ n=0 gn(x), (1) where gn(x) = anxn. You can say that an alternating series converges if two conditions are met: Its nth term converges to zero. CALCULATOR; COMMENTS; COURSES; FOR INSTRUCTOR; LOG IN; FOR INSTRUCTORS; Sign In; Email: Password: Forgot password? ← previous. Get the 1 st hour for free! Divergent sequences do not have a finite limit. Free Summation Calculator. Hence, the sequence is decreasing. the number getting raised to a power) is between -1 and 1. If we have a sequence of. We then conclude that the series is convergent. and in general, where d is the common difference. Example: Classify the series as either absolutely convergent, conditionally convergent, or divergent. Sequences, Series and Convergence with the TI 92. Suppose that (f n) is a sequence of functions, each continuous on E, and that f n → f uniformly on E. How to Use Series Calculator. Get the free "Sequences: Convergence to/Divergence" widget for your website, blog, Wordpress, Blogger, or iGoogle. Bounded Sequence. Nair EXAMPLE 1. The terms consist of an ordered group of numbers or events that, being presented in a definite order, produce a sequence. However, there's a catch: The sum of its positive terms goes to positive infinity and; The sum of its negative terms goes to negative infinity. If you enroll prior to that date, you'll be able to access the course through September 2018. " Adjust N to take more points of the sequence. A series in which successive terms have opposite signs is called an alternating series. Recursive Sequences We have described a sequence in at least two different ways: a list of real numbers where there is a first number, a second number, and so on. The formal definition of a sequence,$ \\langle x_n\\rangle $, tending to a limit$ L $is:$ \\forall\\varepsilon0,\\exists N\\in\\N $such that$ \\forall n\\ge N,|x_n-L|\\epsilon $. YES Is x in interval of convergence? P∞ n=0 an = f(x) YES P an Diverges NO Try one or more of the following tests: NO COMPARISON TEST Pick {bn}. Find the next number in the sequence of integers. image/svg+xml. its sequence of partial sums fSngis bounded. Find more Mathematics widgets in Wolfram|Alpha. Series and Sum Calculator. Convergence of sequences One concept that is typically hard to grasp is the convergence of a sequence. You can use L’Hôpital’s rule to find limits of sequences. First Five Terms: 28, 38, 48, 58, 68 First Five Terms: −38, −138, −238, −338, −438. Now we discuss the topic of sequences of real valued functions. Similarly, consider the series. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. CALCULATOR; COMMENTS; COURSES; FOR INSTRUCTOR; LOG IN; FOR INSTRUCTORS; Sign In; Email: Password: Forgot password? ← previous. Implicit Derivative. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Homework Equations Ratio test The Attempt at a Solution By the radio test. (1) The convergence of P1 n=1 bn implies the convergence of P1 n=1 an: (2) The divergence of P1 n=1 an implies the divergence of P1 n=1 bn: Proof : (1) Note that the sequence of partial sums of P1 n=1 an is bounded. A sequence is divergent if it tends to infinity, but it is also divergent if. 2 4 8 16… is an example of a geometric progression that starts with 2 and is doubled for each position in the sequence. YES Is x in interval of convergence? P∞ n=0 an = f(x) YES P an Diverges NO Try one or more of the following tests: NO COMPARISON TEST Pick {bn}. Here’s another convergent sequence: This time, the sequence …. This is a Math solver tool which will save your time while doing complex calculations. a/ Da, and so on. Typically, we have an interative algorithm that is trying to find the maximum/minimum of a function and we want an estimate of how long it will take to reach that optimal value. Therefore, the series is convergent and we have. Integral Test for Convergence (with Examples) May 2, 2020 January 12, 2019 Categories Formal Sciences , Mathematics , Sciences Tags Calculus 2 , Latex By David A. Every unbounded. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. Figure $$\PageIndex{1}$$: Illustrating the convergence with the Alternating Series Test. A sequence is a list of numbers placed in a defined order while a series is the sum of such a list of numbers. Just enter the expression to the right of the summation symbol (capital sigma, Σ) and then the appropriate ranges above and below the symbol, like the example provided. x x VAzlYlP MrviGg\hMtcsM xr^eGskevr^v^eudx. Free Arithmetic Sequences calculator - Find indices, sums and common difference step-by-step This website uses cookies to ensure you get the best experience. This read-only value is updated when coordinates are recalculated. 1 Pointwise and uniform convergence 8. The following arithmetic sequence calculator will help you determine the nth term and the sum of the first n terms of an arithmetic sequence. A series åan is convergent if and only if A. But there are degrees of divergence. Otherwise, the sequence is said to diverge. Tips for determining convergence or divergence of an in nite series Divergence Test : Always check that the individual terms a ngo to zero. Let $$\left\{ {{a_n}} \right\}$$ be a sequence of positive numbers such that. A positive series is convergent if each of its terms is less than or equal to the corresponding terms of a series that is known to be convergent. To create this article, 12 people, some anonymous, worked to edit and improve it over time. First of all, denote by the sequence whose generic term is The characteristic function of is Now take a second order Taylor series expansion of around the point : where is an infinitesimal of higher order than , that is, a quantity that converges to faster than does. For the series just use the ratio test! Intuitively, the denominator grow much faster than the numerator, so the limit will be zero. A sequence is divergent if it tends to infinity, but it is also divergent if. Limit of a Sequence 2 3. Get the free "Sequence Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. If you write down a few negative terms of the. To apply the squeeze theorem, one needs to create two sequences. A bounded sequence is one where the absolute value of every term is less than or equal to a particular real, positive number. L’Hôpital’s rule is a great shortcut for when you do limit problems. I am currently trying to get my head around the concept of uniform and pointwise convergence and a lot of questions have come up that I am unable to answer. Just enter the expression to the right of the summation symbol (capital sigma, Σ) and then the appropriate ranges above and below the symbol, like the example provided. the Absolute Convergence Test with the Integral Test. Operations on Convergent Series. Active 1 year, 10 months ago. Here's a definition: A sequence {a n} is said to converge to the number a provided that for every positive number ε there is an index N such that: |a n - a| < ε , for all. In numerical analysis, the speed at which a convergent sequence approaches its limit is called the rate of convergence. In the last post, we talked about sequences. A geometric sequence has the form: a 1, a 1 r, a 1 r 2, a_1, a_1 r, a_1 r^2, You need to provide the first term of the sequence ( ), the constant ratio between two consecutive values of the sequence (. Give an example of a convergent sequence that is not a monotone sequence. The previous geometric series of positive terms converges to 2. Here’s an example of a convergent sequence: This sequence approaches 0, so: Thus, this sequence converges to 0. A series convergence calculator is used to find out the sum of the sequence and for determining convergence and divergence among series. A simpler proof can be obtained if we assume the finiteness of the fourth moment. 5 Sequence of derivatives 7. Therefore, {fn} converges pointwise to the function f = 0 on R. One famous example of a enduring question is the Basel problem. Basic Properties. The online calculator below was created on the basis of the Wolfram Alpha, and it is able to find sum of highly complicated series. The second answer is that there is often more than one convergence test that can be used with a given series. Integral Test doesn't seem to be an option (u-substition for 1/n will give us a dn = -1/x^2, which is not in the function to be integrated). For example, consider the following sequence of functions with common domain$\mathbb{R}\$: (2). If R is finite and nonzero, then there are four combinations for interval of convergence, depending on whether each. Such an argument was given by Nicolas Oresme (1323 - 1382 A. Every unbounded. Lastly, we will take a look at applying theorem 7, which will help us determine if the sequence is convergent. Taylor Series Convergence The Taylor series converges if f has derivatives of all orders on an interval "I" centered at c, if lim(n→ infin;)RN = 0 for all x in l: The Taylor series remainder of R N = S - S N is equal to (1/(n + 1)!)f (n + 1) (z)(x - c) n + 1 where z is a constant between x and c. Series Convergence Worksheet On a separate sheet paper, determine whether each series converges or diverges. (1) and ask whether the sum is convergent. By analysing genomic sequences in echolocating mammals it is shown that convergence is not a rare process restricted to a handful of loci but is instead widespread, continuously distributed and. Infinite Sequences and Series This section is intended for all students who study calculus, and considers about $$70$$ typical problems on infinite sequences and series, fully solved step-by-step. Note that this is a statement about convergence of the sequence fangn - it is NOT a statement. From the above inequalities, we get So we advise you to take your calculator and compute the first terms to check that in fact we have This is the case. Free Geometric Sequences calculator - Find indices, sums and common ratio of a geometric sequence step-by-step This website uses cookies to ensure you get the best experience. In business, this is sometimes called a company “culture,” in the sense that people who work there tend to have similar characteristics, behaviors, and philosophies. N th term of an arithmetic or geometric sequence. Includes the nth-Term, geometric series, p-Series, integral test, ratio test, comparison, nth-Root, and the alternating series test. Recall that if and are continuous functions on an interval and , then. However, neither one can tell you about conditional convergence. We will just sketch a proof. By using this website, you agree to our Cookie Policy. Second Implicit Derivative (new) Derivative using Definition (new) Derivative Applications. Sequences and Series. A recursion is a special class of object that can be defined by two properties: 1. If you're seeing this message, it means we're having trouble loading external resources on our website. (b) This sequence does not converge to zero: this is a geometric sequence with r = 2 > 1; hence, the sequence diverges to ∞. This Arithmetic Sequence Calculator is used to calculate the nth term and the sum of the first n terms of an arithmetic sequence. Every unbounded. 4 Sequences of integrable functions 7. In addition to certain basic properties of convergent sequences, we also study divergent sequences and in particular, sequences that tend to positive or negative infinity. f () = Remove. Let {fn} be the sequence of functions defined by fn(x) = cosn(x) for −π/2 ≤ x ≤ π/2. This test cannot be used to show convergence. It is the hope that an iteration in the general form of will eventually converge to the true solution of the problem at the limit when. Sequence and series are one of the basic topics in Arithmetic. Solution diverges. plot a numerical sequence or series. The individual elements in a sequence are called terms. I will illustrate this point…. The statement clearly true for n=2. From the definition of an increasing and decreasing sequence, we should note that EVERY successive term in the sequence should either be larger than the previous (increasing sequences) or smaller than the previous (decreasing sequences). •Given a sequence {a 0, a 1, a2,…, a n} •The sum of the series, S n = •A series is convergent if, as n gets larger and larger, S n goes to some finite number. Determine if a sequence is arithmetic or geometric :. Learn more about geometric sequences so you can better interpret the results provided by this calculator: A geometric sequence is a sequence of numbers \(a_1, a_2, a_3, …. For example, we could have used the term when a rational function has a horizontal asymptote as we could describe the function values as converging towards a finite number. Calculate next 10 elements and start with vector [1 1] where at each run one element should be added. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. Fourier Series Calculator is an online application on the Fourier series to calculate the Fourier coefficients of one real variable functions. Sequences are handled on the TI-83 and TI-84 using the seq function. Abstract: We consider sequences of graphs and define various notions of convergence related to these sequences: left convergence'' defined in terms of the densities of homomorphisms from small graphs into the graphs of the sequence, and right convergence'' defined in terms of the densities of homomorphisms from the graphs of the sequence into small graphs; and convergence in a suitably. This sequence is different from the first two in the sense that it doesn’t have a specific formula for each term. A sequence may increase for half a million terms, then decrease; such a sequence is not monotonic. ) State the test used. Does P bn converge? Is 0 ≤ an ≤ bn? YES P YES an Converges Is 0 ≤ bn ≤ an? NO NO P YES an Diverges LIMIT COMPARISON TEST Pick {bn}. a/ Da, and so on. To prove this we show that the assumption that fn(x) converges uniformly leads to a contradiction. The harmonic series can be counterintuitive to students first encountering it, because it is a divergent series even though the limit of the n th term as n goes to infinity is zero. Find more Transportation widgets in Wolfram|Alpha. One famous example of a enduring question is the Basel problem. Display values in given range? Fibonacci sequence f (n-1)+f (n-2) Arithmetic progression d = 2 f (n-1)+2. The interval of convergence is the largest interval on which the series converges. We will also give many of the basic facts and properties we'll need as we work with sequences. If a sequence x 1;x 2;:::;x nconverges to a value rand if there exist real numbers >0 and 1 such that (1) lim n!1 jx n+1 rj jx n rj = then we say that is the rate of convergence of the sequence. If such a limit exists, the sequence is called convergent. xhlzazrap8k7 nmthyrasyspf1ea 23cbaf5vnnytcd v1d71hzhyug xxqc3g6sxmorf1k pt9z7kem42l xhicp6f2o4o37 phckt1hx6dg9v7z 7bc2ko4gj8u8166 tbfkwh1g9ylg q6ac5okt4jj 9s52qkpmahxp8 live4tpl0c0ge dvkgvdm6xscp wwhrj5c2plrokwa p754eziusl6 zocrsfoomz9uu6 gb6e62f8ht 4f45y7myi5xd0e 561r4bwupk7 6ens4e2t3q7kqd qal8zy9gljk 71cczt1hqvzr j0auo7var0k q6bzngodko jgbop0aai4p yftus0yyel
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http://mathhelpforum.com/pre-calculus/109018-find-vector-given-length.html
# Math Help - Find a Vector given a length 1. ## Find a Vector given a length Hi, it's been a while since I've attempted anything to do with vectors and I'm a little rusty at the moment. The question is: Fine a vector length $2\sqrt{6}$ anti-parallel to $d = (-1,2,-1)$. I thought that maybe the vector product would be the way forward but no idea where to start. Any help would be greatly appreciated. Thanks Craig 2. Originally Posted by craig Hi, it's been a while since I've attempted anything to do with vectors and I'm a little rusty at the moment. The question is: Fine a vector length $2\sqrt{6}$ anti-parallel to [tex]d = (-1,2,-1)[/MATH. I thought that maybe the vector product would be the way forward but no idea where to start. Any help would be greatly appreciated. Thanks Craig What about $2\left(1,-2,1\right)=\left(2,-4,2\right)$ ? Tonio 3. Originally Posted by tonio What about $2\left(1,-2,1\right)=\left(2,-4,2\right)$ ? Tonio Hi Tonio thank's for the reply. How did you come up with this answer? 4. Originally Posted by craig Hi Tonio thank's for the reply. How did you come up with this answer? Antiparallel ==> it has to "go the other way" so multiply by $-1$ all its entries. Now, the vector given and the new one both have norm $=\sqrt{6}$ so in order to get norm equal to $2\sqrt{6}$ simply multiply the vector by 2. Tonio 5. Originally Posted by craig Hi, it's been a while since I've attempted anything to do with vectors and I'm a little rusty at the moment. The question is: Fine a vector length $2\sqrt{6}$ anti-parallel to $d = (-1,2,-1)$. I thought that maybe the vector product would be the way forward but no idea where to start. Any help would be greatly appreciated. Thanks Craig Any vector parallel to (-1, 2, -1) must be of the form a(-1, 2, -1)= (-a, 2a, -a). Any vector anti-parallel to (-1, 2, -1) (in the opposite direction) must be of that form with a negative- or we can write it -a(-1, 2, -1)= (a, -2a, a) with the requirement that a be positive. Now, the length of (a, -2a, a) is $\sqrt{a^2+ (-2a)^2+ a^2}= \sqrt{6a^2}= a\sqrt{6}$. In order that the vector have length $2\sqrt{6}$, we must have $a\sqrt{6}= 2\sqrt{6}$ or a= 2. (a, -2a, a)= (2, -4, 2). 6. Thanks for both replys, all makes sense now. It's been a very long time since I've attempted anything to do with vectors, got quite a bit of catching up to do
2014-09-30T07:46:52
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https://eng.libretexts.org/Under_Construction/Book%3A_The_Joy_of_Cryptography_(Rosulek)/Chapter_0%3A_Review_of_Concepts_and_Notation/0.2%3A_Definition
# 0.2: Definition For $$x,n\in Z$$, we say that $$n$$ divides $$x$$ (or $$x$$ is a multiple of $$n$$), and write $$n\mid x$$, if there exists an integer $$k$$ such that $$kn = x$$. We say that $$a$$ and $$b$$ are congruent modulo n, and write $$a\equiv_n b$$, if $$n\mid (a − b)$$. Equivalently, $$a\equiv_n b$$ if and only if $$a$$ % $$n = b$$ % $$n$$. We write $$\mathbb{Z_n} \stackrel{\text{def}}{=}\{0,...,n-1\}$$ to denote the set of integers modulo $$n$$. In other textbooks you may have seen “$$a\equiv_n b$$” written as “$$a\equiv b (\text{mod} \space n)$$”. There is a subtle — and often confusing — distinction between the expressions “$$a \% n = b$$” and “$$a\equiv_n b$$.” In the first expression, “$$a \% n$$” refers to an integer that is always between 0 and $$n − 1$$, and the equals sign denotes equality over the integers. In the second expression, the “$$\equiv_n$$” symbol denotes congruence modulo $$n$$, which is a weaker condition than equality over the integers. Note that $$a = b$$ implies $$a\equiv_n b$$, but not vice-versa. ## Example $$99\equiv_{10}19$$ because 10 divides 99 − 19 according to the definition. But $$99 \neq19$$ % 10 because the right-hand side evaluates to the integer 19 $$\%$$ 10 = 9, which is not the same integer as the left-hand side 99. When adding, subtracting, and multiplying modulo n, it doesn’t affect the final result to reduce intermediate steps modulo n. That is, we have the following facts about modular arithmetic: $(a+b)\% n=[(a\% n)+(b\% n)]\% n;\nonumber$ $(a-b)\% n=[(a\% n)-(b\% n)]\% n;\nonumber$ $ab\% n=[(a\% n)(b\% n)]\% n;\nonumber$ Division is not always possible in $$\mathbb{Z_n}$$; we will discuss this fact later in the class. ## Strings We write $${\{0,1\}}^n$$ to denote the set of n-bit binary strings, and $${\{0,1\}}^{\ast}$$ to denote the set of all (finite-length) binary strings. When $$x$$ is a string of bits, we write $$|x|$$ to denote the length (in bits) of that string, and we write $$\bar{x}$$ to denote the result of flipping every bit in $$x$$. When it’s clear from context that we’re talking about strings instead of numbers, we write $$0^n$$ and $$1^n$$ to denote strings of $$n$$ zeroes and n ones, respectively. When $$x$$ and $$y$$ are strings of the same length, we write $$x\oplus y$$ to denote the bitwise exclusive-or (XOR) of the two strings. So, for example, $$0011\oplus 0101 = 0110$$. The following facts about the XOR operation are frequently useful: $x\oplus x=0^{|x|}\space \space \space\space \space \space \text{XOR'ing a string with itself results in zeros}\nonumber$ $x\oplus 0^{|x|}=x \space \space \space\space \space \space \text{XOR'ing with zeros has no effect}\nonumber$ $x\oplus 1^{|x|}=\bar{x} \space \space \space \space \space \space\text{XOR'ing with ones flips every bit}\nonumber$ $x\oplus y=y\oplus x \space \space \space \space \space \space\text{XOR is symmetric}\nonumber$ $(x\oplus y)\oplus z=x\oplus (y\oplus z)\space \space\space \space \space \space \text{XOR is associative}\nonumber$ As a corollary: $a=b\oplus c\Longleftrightarrow b=a\oplus c \Longleftrightarrow c=a\oplus b\nonumber$ We use notation $$x‖y$$ to denote the concatenation of two strings $$x$$ and $$y$$. ## Functions Let $$X$$ and $$Y$$ be finite sets. A function $$? : X\rightarrow Y$$ is: injective (1-to-1) if it never maps twp different inputs to the same output. Formally: $$x\neq x'\Rightarrow f(x)\neq f(x')$$. surjective (onto) if every element in $$Y$$ is a possible output of f. Formally: for all $$y\in Y$$ there exists an $$x\in X$$ with $$f(x)=y$$. bijective (1-to-1 correspondence) if f is both injective and surjective. If this is the case, we say that f is a bijection. Note that bijectivity implies that $$|X|=|Y|$$.
2019-06-19T11:14:18
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https://math.stackexchange.com/questions/1059850/is-there-a-quick-parity-test-for-integers-expressed-with-odd-radicies
# Is there a quick parity test for integers expressed with odd radicies? For integers expressed with an odd base, is there an easy way to tell if the number is odd or even? For an even base, if the ones digit is even, so is the integer. But this doesn't hold true for odd bases. For example, in base 3, 22 is even and 12 is odd. One can always divide by two and see if there is a remainder. I've found (but don't have the mathematical background to formally prove) that, for odd bases, the parity of the sum of the digits is the parity of the number. But is there any easier way to tell, just by looking? (Assume there are lots of digits.) ## 4 Answers Given the simple facts that even+even=even, even+odd=odd, and odd+odd=even, we can see that if there are an even number of odd digits, the sum of the digits is even and if there is a odd number of odd digits, the sum of the digits is odd. So, rather than summing the digits, you can simply count the number of odd digits, which has the same parity as the number itself. Hint: Each place value is odd (e.g. for base $3$, the place values are $1, 3, 9, 27,...$. The digit multiplies the place value. So if the digit is odd, the total value of that digit in that place value is odd; if the digit is even, the total value of that digit in that place value is even. You are right that for odd bases the parity of the sum of digits is the parity of the number. Therefore, it it not possible to determine the parity of a number in an odd base without looking at all the digits, since you need to see all digits to know the parity of the sum. So you probably couldn't determine the parity of a number with "lots of digits" (say 1000 digits) at a glance. Here's a hint. Let's look at the first few even ternary numbers: $2, 11, 20, 22, 101, 110, 112, 121, 200, 202, 211, 220, 222, 1001.$ And the first few odd ternary numbers: $1, 10, 12, 21, 100, 102, 120, 122, 201, 210, 212, 221, 1000, 1002.$ I think that's the only way you can tell "just by looking." But in general that process is less than linear in $n$, where $n$ is the number. Edit following KSmarts' answer: An algorithm for checking odd/even by adding up the digits: For each digit total = total + digit If total is even "Number is even" Else "Number is odd" An algorithm for checking odd/even by adding up the number of odd digits: For each digit If digit is odd oddDigits = oddDigits + 1 If oddDigits is odd "Number is odd" Else "Number is even" One may be faster than another for a particular implementation, but neither is easier than the other. You're still needing to examine every digit of the number.
2019-10-22T03:24:44
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https://math.stackexchange.com/questions/2331228/every-finite-group-is-a-torsion-group
# Every Finite Group is a torsion group? I am learning about finitely generated groups at the moment and in the text book, they mention torsion groups, which they give the definition as: A group $G$ is a torsion group if every element of $G$ is of finite order. They then go on to state, when listing examples of torsion groups, that "Every finite group is a torsion group" I understand how cyclic groups would be a torsion group, but I am having trouble understand how a finite permutation group would be a torsion group. If anyone could help clarify why, it would be greatly appreciated. • Every permutation of a finite set has finite order. If a group is finite the sequence of powers of any element must repeat, so the element has finite order. – Ethan Bolker Jun 21 '17 at 15:03 • Suppose $x \in G$ has not finite order: then the subgroup generated by $x$ is infinite. This shows that $G$ has infinitely many elements. Hence, if you consider a finite group $G$. necessarily all of its elements have finite order. – Crostul Jun 21 '17 at 15:03 • Thank you both for your comments, it makes sense now. – Smeef Jun 21 '17 at 15:07 The order of an element $g$ is the cardinality of $\langle g \rangle$, the cyclic group generated by it. Every element $g$ in a finite group $G$ has finite order because $\langle g \rangle \subseteq G$ implies that $\langle g \rangle$ is finite.
2021-01-27T08:09:02
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https://math.stackexchange.com/questions/1784485/is-the-function-differentiable-at-0
# Is the function differentiable at $0$? Let f(x) = \begin{cases}\begin{align*}&\cos{\dfrac{1}{x}}, &x \neq0 \\ &0, &x=0. \end{align*}\end{cases} Is the function $F(x) = \displaystyle \int_{0}^x f dx$ differentiable at $0$? We can see that the function $f(x)$ is continuous everywhere except $x=0$. In order to show differentiability we will need to show the derivatives from the left and right are equal. So we need to show that $$\lim_{h \to 0^+} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^-} \dfrac{F(x+h)-F(x)}{h}.$$ The derivative from the right is $$\lim_{h \to 0^+} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^+} \dfrac{\displaystyle \int_{0}^x\cos{\dfrac{1}{x+h}}-\int_{0}^x\cos{\dfrac{1}{x}}}{h}$$ and the derivative from the left is $$\lim_{h \to 0^-} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^-} \dfrac{\displaystyle \int_{0}^x\cos{\dfrac{1}{x+h}}-\int_{0}^x\cos{\dfrac{1}{x}}}{h}.$$ I am not sure how to proceed. • I think I am misreading something, but are your difference quotients for $f(x)$ or $F(x)$ (which is the antiderivative of $f(x)$)? – Chill2Macht May 14 '16 at 2:11 • @William Thanks, I fixed that. – user19405892 May 14 '16 at 2:15 • This is great. Where'd you get this? – Tim kinsella May 14 '16 at 3:23 • @Timkinsella It is from Michael Spivak's calculus book. – user19405892 May 14 '16 at 3:25 • What page is the problem on? – Tim kinsella May 14 '16 at 4:37 For $x\geq 0$, let $[x]$ denote the largest element of $\mathbb{Z}\pi+\pi/2$ not greater than $x$. $$\left|\int_0^x \cos(1/t)dt\right|=\left|\int_{1/x}^\infty \frac{ \cos(t)}{t^2}dt\right|\leq \left|\sum_{n=0}^\infty\int_{[1/x]+n\pi}^{[1/x]+(n+1)\pi}\frac{ \cos(t)}{t^2}dt\right|\leq \left|\int_{[1/x]}^{[1/x]+\pi}\frac{ \cos(t)}{t^2}dt\right|\leq$$ $$\left| [1/x]^{-2}\int_{[1/x]}^{[1/x]+\pi}\cos(t)dt\right|= 2[1/x]^{-2}\leq 2(1/x-\pi)^{-2}\leq cx^{2}$$ for some $c>0$ and all small enough $x$. So $|F(x)|$ goes to zero like $x^2$ as $x\rightarrow 0^+$. So it has a right derivative which is zero. By symmetry the left derivative is also zero. • You managed to estimate the integral without using some sort of anti-derivative (like I did in my answer) +1 – Paramanand Singh May 14 '16 at 5:53 • Also very nice +1 – RRL May 14 '16 at 7:02 • $Cx^{-2}$ should be $Cx^2$, right? – Hans Lundmark May 14 '16 at 7:33 • yes, thanks. fixed – Tim kinsella May 14 '16 at 7:52 The problem is slightly tricky because we don't know the anti-derivative of $\cos (1/x)$ but we do know that function $g(x) = x^{2}\sin(1/x), g(0) = 0$ is continuous and differentiable for all $x$ and $$g'(x) = 2x\sin (1/x) - \cos(1/x),g'(0) = 0$$ and hence $$g(h) = \int_{0}^{h}(2t\sin (1/t) - \cos (1/t))\,dt$$ or $$\int_{0}^{h}\cos(1/t)\,dt = 2\int_{0}^{h}t\sin(1/t)\,dt - g(h)$$ and hence \begin{align} F'(0) &= \lim_{h \to 0}\frac{1}{h}\int_{0}^{h}\cos(1/t)\,dt\notag\\ &= \lim_{h \to 0}\frac{2}{h}\int_{0}^{h}t\sin(1/t)\,dt - \frac{g(h)}{h}\notag\\ &= 2\lim_{h \to 0}\frac{1}{h}\int_{0}^{h}t\sin(1/t)\,dt\notag\\ \end{align} Now $t\sin(1/t)$ has a removable discontinuity at $t = 0$ and its limit is $0$ as $t \to 0$ hence by Fundamental Theorem of Calculus the limit $$\lim_{h \to 0}\frac{1}{h}\int_{0}^{h}t\sin(1/t)\,dt$$ above is $0$ and therefore $F'(0) = 0$. • Very nice. (+1) – Tim kinsella May 14 '16 at 5:53 • Very nice answer +1 – RRL May 14 '16 at 7:02 What you have is the derivative of f, not the derivative of F which is what you want. The derivative of F at x= 0, from the right, is $\lim_{h\to 0^+}\frac{F(h)- F(0)}{h}= \lim_{h\to 0^+}\frac{\int_0^h f(t)dt}{h}$. • shouldn't this be a comment? – Tim kinsella May 14 '16 at 2:21 • What about from the left? – user19405892 May 14 '16 at 2:29
2019-12-14T21:50:56
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https://codeforces.com/blog/entry/7262
### riningan's blog By riningan, 8 years ago, We use Eratosthenes sieve for prime factorization, storing the primes in an array. But for that, we need to find the primes less than or equal to sqrt(n) which divide n. There are about n/log(n) primes less than or equal to n. So, the complexity is roughly sqrt(n)/log(sqrt(n))*log(n). But if n is asked to be factorized completely where n is within the Sieve range, then we can factorize n is log(n) complexity. And the trick is fairly small. Observe, that, we don't need to run a whole sqrt(n) loop for finding the prime divisors. Instead, we can even store them when n is in the range, say n<= 10^7. But the tricky part is not to store all the prime divisors of n. Let's see the following simulation. Take n = 60. We want to factorize n. We will store the smallest prime factors only. This does the trick. If n is composite, then it has such a prime factor, otherwise n is a prime and then the n itself is the smallest prime factor. It is obvious, for any even number n, sp(n)=2. Therefore, we only need to store these primes for odd n only. If we denote the smallest prime factor of n by sp(n), for odd 2 <= n <= 30, we get the following list. sp(2n)=2, sp(3)=3, sp(5)=5, sp(7)=7, sp(9)=3, sp(11)=11, sp(13)=13, sp(15)=3, sp(17)=17, sp(19)=19, sp(21)=3, sp(23)=23, sp(25)=5, sp(27)=3, sp(29)=29. Then the factorization is very simple. The optimization is needed only once, when the Sieve() function runs. bool v[MAX]; int len, sp[MAX]; void Sieve(){ for (int i = 2; i < MAX; i += 2) sp[i] = 2;//even numbers have smallest prime factor 2 for (lli i = 3; i < MAX; i += 2){ if (!v[i]){ sp[i] = i; for (lli j = i; (j*i) < MAX; j += 2){ if (!v[j*i]) v[j*i] = true, sp[j*i] = i; } } } } int main(){ Sieve(); for (int i = 0; i < 50; i++) cout << sp[i] << endl; return 0; } Now, notice the difference between the usual prime factorization and this one! The only problem is, you can't use this for n large enough in int range. Still, it seems nice to me and pleasured me when I first found this. • +8 » 8 years ago, # |   +1 [:||||||:] David Gries, Jayadev Misra. A Linear Sieve Algorithm for Finding Prime Numbers, 1978. read this (in Russian) • » » 8 years ago, # ^ |   0 hmm. » 8 years ago, # |   0 So many minuses, why? It's very useful trick and I don't think that everyone knows it. » 8 years ago, # |   +5 Really nice trick! Thanks for sharing. » 8 years ago, # | ← Rev. 2 →   0 It's better to precalculate not only smallest prime number, but also quotient cp[i] = i / lp[i], to do not unnecessary and TOO SLOW operations of division, especially in case of big number of queries. • » » 7 years ago, # ^ |   0 I think that it is not important. Original source is easy to read and easy to understand. Also, you have to perform divide operations log(n) times only. It seems not too big. • » » 4 years ago, # ^ |   0 yeah right, thank for sharing this !! » 7 years ago, # |   0 Dude, your tricks is really cool but I think there is some problem in your sample code. Your Sieve() function doesn't store the smallest prime factors properly. For 45, the smallest prime factor should be 3 where according to your sample code it stores 5! • » » 7 years ago, # ^ |   0 that's because I forgot to check first if a number already has a smallest prime divisor. Now it is correct. Thanks for pointing the mistake out. » 5 years ago, # |   0 how can we find factorization from sp[]..please explain? • » » 5 years ago, # ^ |   0 vector factorize(int k) { vector ans; while(k>1) { ans.push_back(sp[k]); k/=sp[k]; } return ans; } • » » » 5 years ago, # ^ |   0 Gotcha...thanks :-) » 5 years ago, # |   0 How large can MAX be? • » » 5 years ago, # ^ |   0 10^7 • » » » 5 years ago, # ^ |   0 Hi Dushyant, If the limit is 10^7 then why this code is not working. I have commented out the rest part which is not concerned.... • » » » » 5 years ago, # ^ |   0 Signed integer overflow — http://ideone.com/FXLHXO :) • » » » » 4 years ago, # ^ |   0 The reason is integer overflow. To overcome from it, before starting the 2nd loop of j, add this condition:if(i>sqrt(N)) continue; • » » » » » 18 months ago, # ^ |   0 shahidul_brurThis condition is already checked in the inner loop. (j*i) < MAX • » » » 5 years ago, # ^ |   0 what should i do for nos of 10^9 range? • » » » » 5 years ago, # ^ |   0 It can be Pollard's "Ro" algorithm or smth like that. • » » » » » 5 years ago, # ^ |   0 I got Pollard's "Ro" algorithm.really nice one.thank u @fekete • » » » » » » 4 years ago, # ^ |   0 Is it a sarcasm? » 5 years ago, # |   0 Any problems to solve with this technique ??? • » » 5 years ago, # ^ |   0 • » » 5 years ago, # ^ | ← Rev. 3 →   0 Medium FactorizationOne moreSimple Sum • » » 5 years ago, # ^ |   0 » 5 years ago, # |   0 hey smallest prime factor for 567 is 3 but you program is outputing 7...plz correct it • » » 5 years ago, # ^ |   0 Sorry but you are mistaken.It is giving 3 as the output. • » » » 5 years ago, # ^ |   0 actually i am converting it in java code may be due to i am getting this...if u can convert this in java then it would be very helpful for me and for othes..plz do it soon • » » » 5 years ago, # ^ | ← Rev. 2 →   -6 Whats Wrong With this logic every time exception was occuring or it is Same as ABove logic but not Working for java static void Sieve() { for (int i = 2; i < MAX; i += 2) sp[i] = 2;// even numbers have smallest prime factor 2 for (int i = 3; i < MAX; i += 2) { if (!v[i]) { sp[i] = i; for (int j = i; (j * i) < MAX; j+=2) { if (!v[j * i]) v[j * i] = true; sp[j * i] = i; } } } } • » » » » 5 years ago, # ^ |   0 if (!v[j*i]) v[j*i] = true, sp[j*i] = i; • » » » » 5 years ago, # ^ | ← Rev. 3 →   0 He has pointed out the mistake. » 5 years ago, # |   0 This is really nice! Thanks for sharing. » 5 years ago, # | ← Rev. 2 →   0 I don't know why I m getting segmentation fault for the spf() function... http://codepad.org/cKUBvEJ2 » 5 years ago, # |   0 I am not able to understand that why is it log(n) ??? • » » 5 years ago, # ^ |   +6 Consider the prime factorization n = p1 * p2 * ... * pk, where p1, p2, ... pk are the prime factors. n has at most k = log(n) prime factors.To understand this think of how you can maximize the number of prime factors. You'll get the most number of prime factors for p1 = p2 = ... = pk = 2. So we have n = 2^k. Solving for k yields k = log(n). • » » » 5 years ago, # ^ |   +3 Amazing... thanks :) • » » » 4 years ago, # ^ |   -10 what is the overall complexity of the Sieve() function mentioned above » 4 years ago, # |   +3 I think this can be done without extra space :) • » » 4 years ago, # ^ |   0 Can you please share how to do it? » 4 years ago, # |   +5 Nice trick.Helped me to optimise my code. Thanks :) » 4 years ago, # |   -6 Code for Linear Sieve Algorithm Already available on GeeksforGeeks here http://www.geeksforgeeks.org/sieve-eratosthenes-0n-time-complexity/ • » » 2 years ago, # ^ |   0 The article shows a false complexity. » 4 years ago, # |   0 how to find largest prime factor when Number is greater than 10^9 and what is the largest prime factor of 10^16? » 4 years ago, # |   -35 why this code is not working ? and if we ant to compute for long range up to 10 ^ 12 than what ? • » » 4 years ago, # ^ |   0 This code only works for MAX <= 1e7 » 4 years ago, # |   0 realy good idea » 20 months ago, # |   0 thanks nice trick... » 19 months ago, # |   0 I have kind of more simplified code.. without that array v We don't need that bool array to check every time if we need to update it or not we can do it with just one arrayhttps://www.ideone.com/rJfFNC • » » 19 months ago, # ^ |   0 Alas you perform in yor code a lot of unnecessary work. Inner cicle should be executed only for prime values and not for any i. • » » » 19 months ago, # ^ |   0 thanks for pointing out.. corrected it! https://www.ideone.com/mJ4xSS • » » » » 19 months ago, # ^ |   0 Good! But what if $n$ is 49 (I hint on i*i • » » » » » 19 months ago, # ^ |   0 whenever we are asked to find the smallest prime for first n numbers then we would define N as something safe.. like n+10 separating the case for even numbers .I don't understand why every where I see this is done?... how much computational overhead are we reducing .. is it even significant? we can store the numbers in long long • » » » » » 19 months ago, # ^ |   0 here is the code after keeping in mind the things you said.. https://www.ideone.com/obqH09would linear sieve be better than this? • » » » » » » 19 months ago, # ^ |   0 You mean that n is excluded from the sieve. Ok. I mean it this way (here n is included into the sieve). As you can see the linear sieve is a bit faster. Nevertheless at the last end we get a good optimized easy-coded sieve :) » 10 months ago, # |   -8 Hi, Can we modify the same technique further? Like I try so this is right or wrong. Because in my code outer loop only runs sqrt(n) instead of O(n). So total time complexity would be sqrt(n)*log(n). Let me know if I am doing something wrong. `ll Max=1e7+1; vi SmallestPrimeFactory(Max,-1); void SmallPrimeFactor(){ SmallestPrimeFactory[0]=0;SmallestPrimeFactory[1]=0; for(int i=2;i
2021-03-05T14:26:48
{ "domain": "codeforces.com", "url": "https://codeforces.com/blog/entry/7262", "openwebmath_score": 0.42670950293540955, "openwebmath_perplexity": 1815.9852347217407, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9780517469248845, "lm_q2_score": 0.8539127548105611, "lm_q1q2_score": 0.8351708615639098 }
https://math.stackexchange.com/questions/2311168/studying-the-derivative-of-the-integral-function-frac1x-int-0x-arctanet
# Studying the derivative of the integral function $\frac{1}{x}\int_0^x\arctan(e^t)\mathrm{d}t$ I was trying to calculate the derivative of the function $$F(x) =\frac{1}{x}\int_0^x\arctan(e^t)\mathrm{d}t$$ I thought the fastest way was to use the Leibniz's rule for the derivative of a product, $$(f\cdot g)' = f'g + g'f$$ and, choosing as $f(x) = \frac{1}{x}$ and as $g(x) = \int_0^x\arctan(e^t)\mathrm{d}t$, applying for the second one's derivative the fundamental theorem of calculus, I obtained $$-\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^t)\right]\Bigg|_{t = 0}^{t = x} = -\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^x)-\frac{\pi}{4}\right]$$ Now there come the problems, since I don't know how to evaluate the limit as $x\rightarrow0$ for the first term of the expression, while the second one, as $x\rightarrow0$, $g'(x)f(x)\rightarrow\frac{1}{2}$. So I plotted the whole thing and I saw something very strange: The blue one is the function (which is right), the red one it's the derivative as calculated before. As you can notice looks like the derivative have a discontinuity in the point 0, while looking at the graph of the function $F(x)$ one would say that there's not such a discontinuity. I tried to evaluate the whole thing with Mathematica, but I did not solve the problem: there are strange things happening at the origin. Now, there are two possibilities: 1. The derivative is wrong, but I wonder where, as it's so simple and linear 2. Grapher app from Mac OS X cannot handle with such functions in a proper way Can you find out the bug? • Are you sure the blue graph (of $F$) is correct? One of the factors of $F(x)$ is $1/x$ which does have a discontinuity at $x=0$. – Zubin Mukerjee Jun 5 '17 at 20:01 • Shouldn't $d/dx \int_0^x \arctan (e^t) dt = \arctan (e^x)$? – sharding4 Jun 5 '17 at 20:03 • You have a factor of $\pi/4$ that doesn't belong, giving you the singularity. – Umberto P. Jun 5 '17 at 20:04 • you have a mistake in the last part, it should be $0$ instead of $\frac { \pi }{ 4 }$ you forgot derivatives of boundry functions – haqnatural Jun 5 '17 at 20:04 • Ok, but apart the factor $\pi/4$ the discontinuity depends on the $-1/x^2 * I$, where $I$ is the integral function $g(x)$... – opisthofulax Jun 5 '17 at 20:48 With your notation $$g(x)=\int_0^x\arctan(e^t)\,dt$$ we have $$F'(x)=\frac{xg'(x) - g(x)}{x^2}=\frac{x\arctan(e^x)-g(x)}{x^2}$$ for $x\ne0$. On the other hand, the function $F$ can be extended by continuity at $0$ as $$\lim_{x\to0}F(x)=\frac{\pi}{4}$$ and $$\lim_{x\to0}F'(x)= \lim_{x\to0}\frac{1}{2x}\frac{xe^x}{1+e^{2x}}=\frac{1}{4}$$ so $F$ (extended) is also differentiable at $0$. I used l’Hôpital for both limits. The bug in your argument is that you wrongly do $$\frac{d}{dx} g(x) = \Bigl[ \arctan(e^t)\Bigr]_{t=0}^{t=x}$$ instead of $$\frac{d}{dx}g (x) =\arctan(e^x)$$ according to the fundamental theorem of calculus. • Yes, this was kind of stupid mistake... Anyway beyond is the derivative $-\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^x)\right]$ is right as well as your derivative that you get with the derivative of a function quotient rule $\frac{g(x)}{f(x)}$? Are the solution the same? – opisthofulax Jun 5 '17 at 21:06 • @opisthofulax Product rule and quotient rule give the same answer. – egreg Jun 5 '17 at 21:07 • Yes I know they give the same result, but our results are different (we have the + and the - inverted, you have $-\frac{1}{x}g(x)$ and $+\frac{1}{x^2}g(x)$ I have the opposite) so I was wondering if someone of we two has make a mistake in the derivative... – opisthofulax Jun 5 '17 at 21:22 • ok looks like you've inverted the signs applying the derivative rule for a quotient function, then our two results match perfectly. Thanks for your advices and for your help, sorry it was only a stupid mistake! – opisthofulax Jun 5 '17 at 21:31 • @opisthofulax Everybody makes stupid mistakes, as you saw! ;-) Thanks for the edit. – egreg Jun 5 '17 at 21:40 So the "bug" was in the application of the foundamental theorem of calculus in which does not appear the derivative calculated at the starting point, so that the right derivative is $$-\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^t)\right]\Bigg|_{t = 0}^{t = x} = -\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^x)\right]$$ which is the red function in the graph below
2019-10-17T12:34:57
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http://math.stackexchange.com/questions/370470/prove-that-solving-equation-is-possible
# Prove that solving equation is possible The equation: $${2 \over x} + {3 \over y} = {5 \over z} ,\,\,x < z$$ $x,y,$ and $z$ must be natural numbers. How would I go about proving this equation possible to solve? I can't seem to figure out how to prove it without doing some "random" calculations to see how numbers fit in the equation. Example: The variables for the possibly lowest answer is: $$x=1,y=6,z=2$$ Since: $${2 \over 1} + {3 \over 6} = {5 \over 2} ,\,\,1 < 2$$ How would it be possible to find this answer, or any other correct answer for that matter, without using "random" placeholder variables to see how to equation evolves? Simpler put: How to is it possible to solve this equation faster and more effectively? Is it possible to determine whether or not the equation is solvable before trying to solve it? - Essentially you tried $x=1$ and were able to get a solution quickly, so it's hard to argue with success. Knowing we want $x \lt z$, I thought about making both sides equal to 1, i.e. $z=5$ and finding off the top of my head $x=3$ and $y=9$. My values $x,y,z$ are bigger, but the equal sides of the equation are smaller than in your solution. –  hardmath Apr 23 '13 at 16:26 Here's a thought. (Not a very general method, but still useful, I hope.) Rewrite the equation as follows \begin{align}\frac3y &= \frac5z - \frac2x\\\frac3y &= \frac{5x-2z}{xz}\\y&=\frac{3xz}{5x-2z}\tag{*}\end{align} Note that in case where $x,z$ are integers such that $5x-2z=1$, we have that $y$ is automatically an integer. But this means that in this case $5x=2z+1$, so $2z+1$ is an odd multiple of $5$. In other words, every solution of $5x=2z+1$ is of the form \begin{align}x&=2k+1\\z&=5k+2\end{align} for some integer $k\in\mathbb Z$. Note that $z>x$ iff $k\in\mathbb N_0$. This yields an infinite family of solutions: \begin{align}x&=2k+1\\y&=3(2k+1)(5k+2)\\z&=5k+2\end{align} where $k\in\mathbb N_0$ is arbitrary. In $(*)$ you could also examine $5x-2z=l$ for some other values of $l$, which would yield solutions in case if $3xz$ were divisible by $l$. A particularly nice case to examine would be $5x-2z=3$ in which case, as in the case we examined, this happens automatically. (Other easy cases are $l=-1,-3$, which are the remaining divisors of 3.) - Firstly if $x<z$ then $y>z$.Because if $y<z$,$$\frac{2}{x}+\frac{3}{y}>\frac{2}{z}+\frac{3}{z}=\frac{5}{z}$$Contradiction! So let $x=z-a$ and $y=z+b$.Plug these values in the equation and you will get,$$z=\frac{5ab}{3b-2a}=\frac{5}{\frac{3}{a}-\frac{2}{b}}$$ Hence minimum value of z can be 5. Therefore, $$\frac{3}{a}-\frac{2}{b}=1\implies b=\frac{2a}{3-a}=\frac{2}{\frac{3}{a}-1}$$ The minimum value of $\frac{3}{a}-1$ is 2 (because if $\frac{3}{a}-1=1$ a = 1.5(a fraction)) Hence minimum value of b=2/2=1.$$\frac{3}{a}-1=2\implies a=1$$ So,$$x=z-a=5-1=4$$ $$y=z+b=5+1=6$$ - Why do you say the minimum value of $z$ is $5$? –  Erick Wong Apr 23 '13 at 23:41 Both the OP answer $(x,y,z)=(1,6,2)$ and the answer $(2,12,4)$ have $z<5$ (just to back up Eric's comment). However these are the only two for which $z<5$. –  coffeemath Apr 24 '13 at 1:29 This is an approach which shows a way to (in theory) obtain all the solutions. We start with your equation [*]: $2/x+3/y=5/z$ and, since your restriction is $x<z$, multiply by $z$ to obtain $$2\frac{z}{x}+3\frac{z}{y}=5.$$ Not much of a start, but now note that if we define the rational variables $s=z/x,\ t=z/y$ we have $$[1]\ \ 2s+3t=5,$$ and our requirement that $x<z$ is now the requirement that $s>1$. Then from $2s+3t=5$ and $t>0$ it follows that also $s<5/2$. This suggests a procedure: Select any rational number $s \in (1,\frac{5}{2})$. Then define $t=(5-2s)/3$, positive given our restriction on $s$, which makes $[1]$ hold, and in fact we'll have $t \in (0,1).$ But how do we obtain integer solutions to [*] from such a rational solution to $[1]$? The idea is to express $s,t$ as fractions $s=a/b,\ t=c/d$ and if it happens that $a \neq c$ we may multiply top and bottom of either or both fractions until they do have equal numerators. This is like putting them over a common denominator, but in reverse, we put them under a common numerator. After this adjustment we now have $s=z/x,\ t=z/y,$ and we have not disturbed the requirement $x<z$ since multiplying top and bottom of $s$ by the same natural number preserves that the numerator exceed the denominator. Tracing things back we arrive at a solution to [*]. All primitive solutions arise this way, and other solutions can be obtained from this by multiplying each of $x,y,z$ by a natural number. -
2014-12-22T21:49:50
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https://math.stackexchange.com/questions/1909928/finding-the-uniform-convergence-of-a-fourier-series
# Finding the uniform convergence of a Fourier series Verify that the fourier series converge uniformly on the interval ${\pi\leq x\leq \pi}$. Also state why this series is differentable in the interval ${\pi\leq x\leq \pi}$, except at the point $x=0$ and describe graphically the function that is represented by the differentiated series for all $x$ $\frac{1}{\pi}+0.5\sin x-\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }$ What i tried Using the Weierstrass M test i got $$|\frac{1}{\pi}+0.5\sin x-\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|\leq |\frac{1}{\pi}+0.5\sin x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|$$ $$|\frac{1}{\pi}+0.5\sin x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }| \leq |\frac{1}{\pi}+0.5\ x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|$$ $$|\frac{1}{\pi}+0.5\ x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }| \leq|\frac{1}{\pi}+0.5\ x+\frac{2x}{\pi}\sum_{n=1}^\infty \dfrac{\ 2n}{4n^{2}-1 }|$$ And since the term $$\sum_{n=1}^\infty \dfrac{\ 2n}{4n^{2}-1 }$$ converges then by the Weierstrass M test the above fourier seriesM test converges uniformly. Differentating the fourier series term by term i got $$0.5\cos x+\frac{4n}{\pi}\sum_{n=1}^\infty \dfrac{\sin 2nx}{4n^{2}-1 }$$ I suppose that it is differentiable on the given interval because it is continous on that interval except at $x=0$ but i cant see why is this so and also how to describe the graphically the function that is represented by the differentiated series for all $x$. Could anyone please explain this to me. Thanks • Why in the first passage did you switch the minus sign with a plus? – user228113 Aug 31 '16 at 16:12 • The series $\sum_{n\geq 1}\frac{2n}{4n^2-1}$ is not converging. – Jack D'Aurizio Aug 31 '16 at 16:13 The work in the OP has some flaws. Note that we have for all $$x$$, $$\left|\frac{\cos(2nx)}{4n^2-1}\right| \le \frac1{4n^2-1}$$, for each $$n$$. Inasmuch as \begin{align} \sum_{n=1}^\infty \frac{1}{4n^2-1}&<\infty \end{align} the Weierstrass M-Test guarantees that the series $$\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$$ converges uniformly for all $$x\in [-\pi,\pi]$$. To analyze whether the series is differentiable, we examine the series of term-by-term derivatives $$D(x)$$ as given by $$D(x)=-2\sum_{n=1}^\infty \frac{n\sin(2nx)}{4n^2-1} \tag 1$$ Note that $$\sum_{n=1}^N \sin(2nx)=\csc(x)\sin(Nx)\sin((N+1)x)$$ is bounded by $$|\csc(x)|$$ for $$x\ne 0,\pi,-\pi$$. Furthermore, $$\frac{n}{4n^2-1}$$ monotonically decreases to $$0$$ as $$n\to \infty$$. Therefore, for any $$\delta >0$$ and $$x\in [-\pi+\delta,-\delta]$$ or $$x\in [\delta,\pi-\delta]$$, Dirichlet's Test guarantees that the series in $$(1)$$ for $$D(x)$$ converges uniformly and inasmuch as the original series $$\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$$ also converges on $$[-\pi,\pi]$$ (actually, we only need that it converge at a single point), we find that $$D(x)=\frac{d}{dx}\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$$ for all $$x\ne 0,\pi,-\pi$$. To examine the derivative from the right (left) of $$\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2- 1}$$ at $$x=-\pi$$ ($$x=\pi$$), we note that \begin{align} \lim_{h\to0^{\pm}}\sum_{n=1}^\infty \frac{\cos(2n(\mp \pi+h))-\cos(2n(\mp\pi))}{h(4n^2- 1)}&=-2\lim_{h\to0^{\pm}}\sum_{n=1}^\infty \frac{\sin^2(nh)}{h(4n^2-1)}\\\\ &\overbrace{=}^{\text{LHR}}\underbrace{-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\frac{4n\sin(2nh)}{4n^2-1}}_{=\lim_{x\to \mp\pi}D(x)}\\\\ &=-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\left(\frac{\sin(2nh)}{n}+\frac{\sin(2nh)}{n(4n^2-1)}\right)\\\\ &=-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\frac{\sin(2nh)}{n}\\\\ &=\mp \frac\pi4 \end{align} It is importatn to observe that $$D(\pm\pi)=0\ne \pm\frac\pi4$$. This is not inconsistent since $$D(x)$$ is not the representation of the derivative (from the left or right)) at $$x=\pi$$ or $$x=-\pi$$. However, $$D(x)$$ does have the appropriate limits as $$x\to \pm\pi$$. • Okay Thanks, how to explain why this series is differentable at the interval except at $x=0$. I know there is a discontinuity at $x=0$ that makes it not differentable but i cant see the disconutity from the equations. Also how do i describe graphically the function that is represented by the differentiated series for all $x$ ? – ys wong Aug 31 '16 at 16:27 • You're welcome. My pleasure. I've edited to add a way forward to proving differentiability. – Mark Viola Aug 31 '16 at 16:58 • Weierstrass M does imply that series converges uniformly, but not for that reason. Rather, you want to say that for every $x,$ $$\left | \frac{\cos 2nx}{4n^{2}-1 }\right| \le \frac{1}{4n^{2}-1 }$$ for each $n,$ and $\sum_{n=1}^{\infty}\dfrac{1}{4n^{2}-1 }$ converges. WM then gives the desired uniform convergence. – zhw. Jan 26 at 7:15 • @zhw. I've added a new section to discuss the one-sided derivatives at $x=\pm\pi$ and edit the first section as per your comment. – Mark Viola Jan 26 at 21:30 • Happy New Year MV! – zhw. Jan 26 at 21:59
2019-06-19T02:46:53
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https://math.stackexchange.com/questions/1343843/why-is-the-commutator-defined-differently-for-groups-and-rings/1343866
# Why is the commutator defined differently for groups and rings? The commutator of two elements in a group is defined as $[g, h] = g^{−1}h^{−1}gh.$ In a ring, the commutator of two elements is $[a, b] = ab - ba.$ I'm asking because a ring is a (abelian) group under addition, so I would have expected it to be $[g, h] = -g-h+g+h.$ • If we took that as the commutator in a ring, then every commutator would be zero. – Omnomnomnom Jun 29 '15 at 21:44 • A group can be phrased in terms of multiplicative notation or additive. When we use additive notation it is usually because we know that it is commutative. The ring operation + is always commutative so the commutator is not very interesting, so when they talk about the commutator of a ring they are talking about the multiplicative operation. – Gregory Grant Jun 29 '15 at 21:45 • here we have two comments, each of which could serve as a distinct and worthwhile answer. Funny. – James S. Cook Jun 29 '15 at 21:50 • @JamesS.Cook I have a longer answer in the works fwiw – Omnomnomnom Jun 29 '15 at 21:51 • Yeah, I wish there was some smaller rep associated with comments in instances like these. I think good comments like yours here deserve some internet points... – James S. Cook Jun 29 '15 at 22:58 The commutator of a group and a commutator of a ring, though similar, are fundamentally different, as you say. In each case, however, the commutator measures the "extent" to which two elements fail to commute. That is, given elements $a,b$, we wish to "compare" $ab$ and $ba$. In the context of a group, we only have one operation: "multiplication". One way to compare two "values" using multiplication is to divide them; the "closer" you are to the identity, the closer the two values are. In particular, if $a$ and $b$ commute, we'd have $$\frac{ab}{ba} = 1$$ Where $1$ denotes the group identity here. Of course, this "division" notation is a bit ambiguous in the context of groups, since multiplication is non-commutative here. What we take to be the commutator, then, is $$[a,b] = (ba)^{-1}(ab) = a^{-1}b^{-1}ab$$ With rings, it would be fantastic if we could reuse the same definition. However, the problem is that in a ring, we can't divide. Rings don't need to be groups under multiplication, they only need to be monoids (or semigroups, depending on definition). We still, however, have an operation that allows us to "compare" $ab$ and $ba$, namely, subtraction. In particular, when $a$ and $b$ commute, we have $ab - ba = 0$. So, because it's the only way to "measure" the extent to which $a$ and $b$ commute, we define $$[a,b] = ab - ba$$ in rings. It so happens that if your ring happens to allow division (that is, if you have a division ring), then it would seem that you could use both commutators. However, the group commutator works for every group, and the ring commutator works for every ring. This is why we define them the way we do. • @MartinBrandenburg it is a slight (and I hope helpful) abuse of terminology, in the hopes of making things more intuitive. Perhaps it is better to say that "we have one binary operation, which I refer to as multiplication" – Omnomnomnom Jun 29 '15 at 22:10 • @MartinBrandenburg, it is not wrong: it is only if you want it to be. One can very well define a group to be a set with one operation satisfying certain conditions, and that is standard practice. – Mariano Suárez-Álvarez Jun 29 '15 at 23:48 • OK, so suppose we do consider a division ring. The usual example is $\mathbb{H}$, the quaternions. What is the commutator of $i$ and $j$ in $\mathbb{H}$? If we use the ring structure, we get $[i,j] = ij - ji = k + k = 2k$. If instead we use only the multiplicative group structure, we have $[i,j] = i^{-1}j^{-1}ij = (-i)(-j)ij = (ij)(ij) = k^2 = -1$. And of course if we take quaternions $a,b$ that actually commute, then $[a,b]$ in the ring sense is zero, while in the multiplicative sense the commutator is one. – Jeppe Stig Nielsen Jun 30 '15 at 12:17 • The last part is a bit off. Even in a division ring, we cannot use the group version of the commutator, as it would not be defined on all elements (or at least we would need to limit the definition, which is not necessary with the version for rings. – Tobias Kildetoft Jun 30 '15 at 19:16 • I didn't mean that the two commutators are the same on a division ring, only that a division ring has an underlying group, to which we could apply the "group" commutator. – Omnomnomnom Jun 30 '15 at 20:08 Here is a more abstract and unified approach to commutators. I could phrase this whole answer in terms of category theory (and this is probably the best way to think of it), but I won't in order to make the answer more accessible. Let $G$ be a group. We want to make it commutative. What does that mean? We want to find a commutative group $A$ and a homomorphism $\pi : G \to A$ which is the "best" choice of a homomorphism to a commutative group. This entails in particular that $\pi(gh)=\pi(hg)$ and hence $\pi(ghg^{-1} h^{-1})=1$, i.e. the elements of the form $ghg^{-1} h^{-1}$ lie in the kernel of $\pi$. So the best choice should be: We let $G'$ be the (automatically normal) subgroup of $G$ generated by these elements, and let $A=G/G'$, and of course $\pi$ is the projection. Now let $R$ be a ring. We want to make it commutative, i.e. we want to find a commutative ring $A$ with a "best" homomorphism $\pi : R \to A$. This entails in particular $\pi(ab)=\pi(ba)$ and hence $\pi(ab-ba)=0$. So we should define $A=R/I$, where $I$ is the ideal generated by elements of the form $ab-ba$. In each case, commutativity was achieved by setting certain elements to the identity element for the group operation, which are therefore called commutators: They measure the failure of commutativity. However, we are not limited to groups or rings. A very similar algebraic structure is a monoid, which is a semigroup with identity. Now given a monoid $M$, in order to find a commutative monoid $A$ equipped with a "best" homomorphism $\pi : M \to A$, we again have $\pi(ab)=\pi(ba)$, but now we are stuck: We cannot simplify this. But that's ok, we may just define $\sim$ to be the smallest congruence relation on $M$ which satisfies $ab \sim ba$. It is a bit ugly to describe this explicitly. But, being a congruence relation, we may construct a monoid $M/{\sim}$, which is commutative by construction. As you see, here we have no "commutator", but the best commutative quotient still exists. Basically the same "commutatizing" construction works for all algebraic structures which have a binary operation in their type. For groups and rings, we are used to identify congruence relations with normal subgroups resp. ideals, and that's why $ab \sim ba$ gets "simplified" (really?) to $aba^{-1} b^{-1} \sim 1$ resp. $ab-ba \sim 0$. • Nice categorical perspective, even if you don't categorical language. Do you have a source for communatizing with respect to category theory? I would like to read more on it. – PyRulez Jun 30 '15 at 3:04 • @MartinBrandenburg: Instructive answer! For the less trained: Could you also add a (short) paragraph and explain it in terms of category theory? +1 of course – Markus Scheuer Jun 30 '15 at 9:42 • Let me just say that "commutatizing" for some type of structure (containing at least one binary operation) is a functor which is left adjoint to the forgetful functor from the category of commutative structures to the category of structures. It always exists. – Martin Brandenburg Jul 1 '15 at 7:46 • Very, very informative. Thank you! – 6005 Sep 23 '15 at 17:24 Another perspective on this comes from considering the relationship between a Lie group and its corresponding Lie algebra. Details are at On the relationship between the commutators of a Lie group and its Lie algebra, but the upshot is that the ring commutator $[A,B]=AB-BA$ can be regarded as an infinitesimal version of the group commutator $[\alpha,\beta]=\alpha^{-1}\beta^{-1}\alpha\beta$. In both cases, commutators measure how far the object is from being commutative. A group is commutative when its operation is commutative. A ring is commutative when its multiplication is commutative. (Addition in a ring is always commutative.) Hence, the definition of commutator in each case reflects what it means to commute, as it should. You can relate the commutator for a Lie group $G$ to the (ring-type) commutator of its Lie algebra $\mathfrak{g}$. Say $g$ lies infinitesimally close to the group identity, $g = \exp(\epsilon\,X)$ for some $X \in \mathfrak{g}$. Likewise say $h = \exp(\epsilon\,Y)$ for some $Y \in \mathfrak{g}$. Then, $ghg^{-1}h^{-1}$ also lies infinitesimally near the identity of $G$. We compute using the Baker-Campbell-Hausdorff formula that its generator is $$\log ghg^{-1}h^{-1} = \epsilon^{2}[XY] + \mathrm{O}(\epsilon^3)\,.$$ If you want to make further contact with the ring commutator, think of the Lie bracket as "coming from" the commutator with respect to the associative product in $\mathrm{U}\mathfrak{g}$, the universal enveloping algebra of $\mathfrak{g}$, so that $$[XY] = XY - YX$$ in a genuine way.
2019-07-21T03:18:35
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https://gateoverflow.in/1931/gate-cse-2014-set-1-question-51
20,857 views Consider an undirected graph $G$ where self-loops are not allowed. The vertex set of $G$ is $\{(i,j) \mid1 \leq i \leq 12, 1 \leq j \leq 12\}$. There is an edge between $(a,b)$ and $(c,d)$ if $|a-c| \leq 1$ and $|b-d| \leq 1$. The number of edges in this graph is______. @ very nice diagrammatic representation. Thank you... nice explaination A clear explanation of the solution is in this image. I hope everything will be cleared. I tried to show every detail of every edge. If you think of a $12\times 12$ grid (like a chess board of size $12\times 12$), then each each point $\left(i,j\right)$, which is in $i^{\text{th}}$ row and $j^{\text{th}}$ column, is a vertex $\left(i,j\right)$. Now we are allowed to connect only those points which are atmost $1$ distance apart (in both horizontal and vertical direction). So we will connect only horizontal neighbours, vertical neighbours, and diagonal neighbours. So horizontal edges on each row are $11$ i.e. $11\times 12 = 132$ horizontal edges. Similarly we have $132$ vertical edges. To count diagonal edges, think of $1\times1$ square boxes in which diagonals meet each other. There are $11\times 11$ such square boxes, and each box contains $2$ diagonals, so total diagonals $= 242$. So total edges $= 132 + 132 + 242 = 506.$ nice explanation @Happy mittal The Condition mentioned in the question is : If |a−c|≤1AND |b−d|≤1 Dont u think edges are only along diagonals because both conditions should be satisfied. If the condition were having OR instead of AND then your answer was correct.. Please correct me if I am wrong it is exceptional thinking Total number of vertices $= 12\times 12 = 144.$ The graph formed by the description contains $4$ (corner) vertices of degree $3$ and $40$ (external) vertices of degree $5$ and  $100$ (remaining) vertices of degree $8.$ According to (handshake theorem's) $2|E|=$ sum of the degrees $2|E| = 4\times 3+40\times 5+100\times 8= 1012.$ $|E| = \frac{1012}{2} = 506$  edges. But , the 4 corner edges are - (1,1),(12,1),(12,12) and (1,12). Here (12,1) and( 1,12) are same. Rest all others are same. But (1,1) and (12,12) are not same. Please see into it. Love this approach what a nice explanation @Mithlesh Upadhyay, thanks a lot There will be ‘11’ edges in a horizontal line, so for 12 horizontal lines, 11×12=132 There are ‘11’ edges in vertical line, so for 12 vertical lines, 11×12=132 There are ‘2’ diagonals ‘⊠’ in each square box. There are 11×11 square boxes available. So 11×11×2=242 ------------------------------------------ Total: 132+132+242 =506 edges ### 1 comment 11*11*2 not 12 please correct it.. Given: The vertex set of G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}. There is an edge between (a, b) and (c, d) if |a − c| <= 1 and |b − d| <= 1. There can be total 12*12 possible vertices. The vertices are (1, 1), (1, 2) ....(1, 12) (2, 1), (2, 2), .... The number of edges in this graph? Number of edges is equal to number of pairs of vertices that satisfy above conditions. For example, vertex pair {(1, 1), (1, 2)} satisfy above condition. For (1, 1), there can be an edge to (1, 2), (2, 1), (2, 2). Note that there can be self-loop as mentioned in the question. Same is count for (12, 12), (1, 12) and (12, 1) For (1, 2), there can be an edge to (1, 1), (2, 1), (2, 2), (2, 3), (1, 3) Same is count for (1, 3), (1, 4)....(1, 11), (12, 2), ....(12, 11) For (2, 2), there can be an edge to (1, 1), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2), (3, 3) Same is count for remaining vertices. For all pairs (i, j) there can total 8 vertices connected to them if i and j are not in {1, 12} There are total 100 vertices without a 1 or 12.  So total 800 edges. For vertices with 1, total edges = (Edges where 1 is first part) + (Edges where 1 is second part and not first part) =  (3 + 5*10 + 3) + (5*10) edges Same is count for vertices with 12 Total number of edges: = 800 + [(3 + 5*10 + 3) + 5*10]  + [(3 + 5*10 + 3) + 5*10] = 800 + 106 + 106 = 1012 Since graph is undirected, two edges from v1 to v2 and v2 to v1 should be counted as one. So total number of undirected edges = 1012/2 = 506.
2023-01-31T23:33:43
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http://math.stackexchange.com/questions/836579/in-constrained-optimization-problems-when-is-naive-substitution-possible
# In constrained optimization problems, when is 'naive' substitution possible? To motivate the question, consider the following constrained optimization problem: $$(P1)\quad \underset{(x,y)}{\min} f(x,y)=x^2 +y^2 \ s.t.\ (x-1)^3 = y^2$$ By replacing the constraint $y^2 = (x-1)^3$ in $f(x,y)=x^2 +y^2$, one gets the following unconstrained optimization problem: $$(P2)\quad \underset{x}{\min} g(x)=x^2 + (x-1)^3 \$$ These problems are not equivalent: although $(P2)$ is unbounded and doesn't have a minimum, $(P1)$ has a minimum in the point $(x,y)=(1,0)$. Is there any necessary/sufficient conditions to assure that the unconstrained problem stemmed by replacing the constraint in the objective function is equivalent to the original constrained problem? - Consider the following third problem: $$\text{(P2a)} \quad \min_x ~~ g(x) = x^2 + (x-1)^3 ~~ \text{s.t.} ~~ (x-1)^3=y^2$$ I think you will agree that (P2a) is equivalent to (P1). So it wasn't the substitution that is the problem! It was the fact that you also eliminated the constraint. It might seem, of course, that you could eliminate it at this point, because $y$ no longer appears in the objective. But this would be incorrect, because it still serves to constrain $x$: \begin{aligned}&(x-1)^3=y^2\quad\Longleftrightarrow\quad (x-1)^3\geq 0, \quad y\in\{+(x-1)^{3/2},-(x-1)^{3/2}\}\\ &\qquad\Longleftrightarrow\quad x\geq 1, \quad y\in\{+(x-1)^{3/2},-(x-1)^{3/2}\} \end{aligned} Therefore, a correct model after eliminating $y$ is $$\text{(P2b)} \quad \min_x ~~ g(x) = x^2 + (x-1)^3 ~~ \text{s.t.} ~~ x\geq 1$$ I would therefore challenge your question itself. Rewriting the objective function in this manner is always acceptable. Dropping a constraint, however, is only acceptable if it can be proven that it is not active at any optimal point. Therefore, do not assume you can drop a constraint just because you rewrote the objective. EDIT: perhaps this will help. Two optimization models are equivalent only if you can readily recover the solution to one from a solution to the other, and vice versa. Thinking about the conditions that must hold in order to be able recover an eliminated variable will help to prevent the elimination of essential constraints. For example, in this case, we can only recover $y$ if $(x-1)^3$ is nonnegative. - You just made a mistake replacing $y^2$ by $(x-1)^3$. There is a hidden constraint on $(x-1)^3$ that you forgot: $(x-1)^3\ge 0$ - Hint: It would be good to realize what causes the substituted problem to be different. What extra conditions do we indirectly imply on $x$ if $(x-1)^3$ must be equal to a squared number? -
2016-02-07T11:01:46
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http://math.stackexchange.com/questions/115518/determining-variance-from-sum-of-two-random-correlated-variables
# Determining variance from sum of two random correlated variables I understand that the variance of the sum of two independent normally distributed random variables is the sum of the variances, but how does this change when the two random variables are correlated? - You have to add twice the covariance. –  David Mitra Mar 2 '12 at 1:55 Can you please submit an answer so I can "check" it? Thanks. Your solution is helpful, thanks. –  Soo Mar 2 '12 at 2:01 Also, what if you add three random normally distributed variables with non-zero correlations between each variable? Do you add 2 times the covariances between all of the random variables? So it would look like var(A) + var(B) + var(C) + 2*(covar(A,B)+covar(A,C)+covar(B,C))? –  Soo Mar 2 '12 at 2:04 And David's answer works even when the random variables are not normally distributed. –  Dilip Sarwate Mar 2 '12 at 2:05 @Dilip, is this because we only care about variances and covariances, and not the shape of the distribution? –  Soo Mar 2 '12 at 2:07 For any two random variables: $$\text{Var}(X+Y) =\text{Var}(X)+\text{Var}(Y)+2\text{Cov}(X,Y).$$ If the variables are uncorrelated (that is, $\text{Cov}(X,Y)=0$), then $$\tag{1}\text{Var}(X+Y) =\text{Var}(X)+\text{Var}(Y).$$ In particular, if $X$ and $Y$ are independent, then equation $(1)$ holds. In general $$\text{Var}\Bigl(\,\sum_{i=1}^n X_i\,\Bigr)= \sum_{i=1}^n\text{Var}( X_i)+ 2\sum_{i< j} \text{Cov}(X_i,X_j).$$ If for each $i\ne j$, $X_i$ and $X_j$ are uncorrelated, in particular if the $X_i$ are pairwise independent (that is, $X_i$ and $X_j$ are independent whenever $i\ne j$), then $$\text{Var}\Bigl(\,\sum_{i=1}^n X_i\,\Bigr)= \sum_{i=1}^n\text{Var}( X_i) .$$ - Can you add an addendum for the sum of 3 random variables? Is the resulting var = var(A) + var(B) + var(C) + 2*(covar(A,B)+covar(A,C)+covar(B,C)) –  Soo Mar 2 '12 at 2:08 I am unfamiliar with the summation(i<j). Can you explain what this notation means? –  Soo Mar 2 '12 at 2:13 @soo You calculate all covariances $\text{Cov}(X_i,X_j)$ with $i<j$ and sum them up. Another way to write $2\sum_{i<j}$ in this case is to write $\sum_{i\ne j}$. (The 2 is there in the first sum because in the second sum you calculate, e.g., $\text{Cov}(X_1,X_2)$ and $\text{Cov}(X_2,X_1)$, but these are equal. –  David Mitra Mar 2 '12 at 2:17 David, excellent explanation, the 2 in the 2*cov(...) makes more sense now. Also, can you explain why you wouldn't define an upper limit "n" in the summation(i<j)? Just for my personal curiosity. Thanks –  Soo Mar 2 '12 at 2:21 @soo For your first comment, that's correct. I'll just let your comment be the addendum, if that's ok. –  David Mitra Mar 2 '12 at 2:23 You can also think in vector form: $Var(a^T X) = a^T Var(X) a$ where $a$ could be a vector or a matrix, $X = (x_1, x_2, \dots, x_3)^T$, is a vector of random variables. $Var(X)$ is the covariance matrix. If $a = (1, 1, \dots, 1)^T$, then $a^T X$ is the sum of all the $x's$ -
2015-02-01T01:13:33
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https://math.stackexchange.com/questions/1749730/how-many-faces-of-a-solid-can-one-see
# How many faces of a solid can one "see"? What is the maximum number of faces of totally convex solid that one can "see" from a point? ...and, more importantly, how can I ask this question better? (I'm a college student with little experience in asking well formed questions, much less answering them.) By "see" I mean something like this: you point a camera from a point at the solid, and look at the picture. How many of the faces of the solid look like faces and not just lines? Let's assume that the lens is just a point in space (no lenses wider than the solid itself) and that the camera is a finite distance from the solid. I know this is a crude definition... if you have any ideas for a more rigorous definition, this would be awesome, then maybe there's ways to prove the answer to my question mathematically. For example, in the picture of this cube, you can see 3 faces. This is the maximum you can see for a cube. How can that be proved? What methods might you use to prove this for a convex solid of any size and shape? Are there ways to do so using only relatively basic math (Multivariable calc, linear algebra, high school geometry)? • The answer will be different or every solid. Viewed from above the apex you can see every face but one of a pyramid with $n$ faces. Apr 19, 2016 at 14:24 • Proof for a cube: you can't see two parallel faces simultaneously. Apr 19, 2016 at 14:26 • I know that it'll be different for every solid. If you have a dodecahedron, it'll be 6. If you have a cube it's 3. If you have a tetrahedron it'll be 3... I'd like to know if there is a good way to prove each of these? @Abstraction, I'm not sure that constitutes a proof...? – jhch Apr 19, 2016 at 14:28 • Let say that the solid is convex "enough" so that each face can radiate without hindrance in all the outward directions. Then you can consider the centre of each face and compute if a ray from it is reaching the camera (with a sufficient angle, if you exclude seeing lines). Apr 19, 2016 at 14:33 • @JohnHughes What I meant was, out of all the possible polyhedra, you cannot ever see more than $n-1$ faces, but you can see exactly $n-1$ in a few special cases. See my answer. Apr 19, 2016 at 14:57 You can always find a place from which to see at least half the faces. To see why, start by considering a polyhedron with central symmetry. Imagine a viewpoint from which you don't see any lines as points or faces as lines (i.e. general position) and far enough away so that you can see all the faces whose normal points into your side of the half plane perpendicular to your line of sight. Then think about what you see from far enough away in the opposite direction. You can see all the faces from one side or the other and no face from both sides, so the symmetry says you see half each time. Four of the five regular polyhedra have a center of symmetry. The tetrahedron does not: there's no place to put the origin that allows invariance under the map $x \to -x$. Even without central symmetry, you see all the faces from one side or the other, so you see at least half from at least one side. Pyramids represent an extreme case. You can see all but one face from one direction and just one from the other, as @almagest points out in a comment. Since the polyhedron has only finitely many faces, "far enough away" in the preceding proof does not have to be at infinity (though it may be pretty far). As @JohhHughes comments, if you put your camera close enough to any face that's the only face you'll see. Note: the arguments work in all dimensions. They are particularly easy to visualize in the plane. (On the line they're trivial.) • Maybe I'm wrong here... doesn't a tetrahedron have a center of symmetry? But you can see 3 out of 4 sides, not 2 out of 4. – jhch Apr 19, 2016 at 14:40 • No, a tetrahedron does not have a center of symmetry. There's no way to place it in space so that the map $x \to -x$ maps the object to itself. Apr 19, 2016 at 14:40 • @GCab: The Question is about the maximum number of faces that are visible, so if a closer point of view reduces them, don't do that. Apr 19, 2016 at 14:43 • Rather than "a center of symmetry", a better choice might be saying the polyhedron has "central symmetry" (or "point symmetry"). Apr 19, 2016 at 14:45 • With any convex solid I think you can place the point of view close enough to reduce the number of faces seen to exactly 1. (if I'm wrong, that is super cool and please tell me more.) – jhch Apr 19, 2016 at 14:46 As @almagest has pointed out, the absolute maximum number of faces you can see of a polyhedron with $n$ faces is $n-1$. This is achieved in the case of a right pyramid with a base and $n-1$ sides; if you view the pyramid from above the apex, you can see all the sides except the base. This is perhaps true for non-right pyramids and other shapes as well. The absolute minimum number of faces you can see is 1, as you said: just place yourself (or the camera) arbitrarily close to any one face. As the polyhedron is convex, none of the faces will 'tower over' any one shape and you will see only one shape. Both these bounds, however, are quite obvious and useless. As in the answer above, half the faces of a regular polyhedron with a center of symmetry can be seen from sufficiently far away. I would extend this to say that roughly half of the faces of a roughly regular polyhedron can be seen from sufficiently far away, where 'roughly' is an appropriate tolerance constant. I think it is illustrative to think of the sphere that completely circumscribes the polyhedron, of which you can obviously see exactly half. Maybe that half can be 'mapped' on to the polyhedron's faces. • Can we say that for any given polyhedron with number of faces > 4 the maximum you can see purely by changing viewpoint is n/2? Are there any special cases other than a tetrahedron? – jhch Apr 19, 2016 at 14:59 • @JohnHughes What if I give you a pyramid? Apr 19, 2016 at 15:01 • I think a pyramid has 4 faces, right? For the pyramid, it's n – 1. For other polyhedrons, is it n/2 or might there be other exceptions? – jhch Apr 19, 2016 at 15:09 • @JohnHughes There is an $n-2$ case: consider a pyramid whose base isn't completely flat. Instead, it has an outwards crease in the middle, so that there are in fact 2 faces that you don't see from the top. This can be generalized to $n-k$ for any $2 \leq k < n - 1$ I suppose. Apr 19, 2016 at 16:43 • I'm having trouble imagining this pyramid whose (triangular) base isn't completely flat, for which you couldn't just change the viewpoint to see at least one of those two faces on the base. – jhch Apr 19, 2016 at 19:29 Perhaps this paper which deals with the situation in a rather abstract way, and for higher dimensional polytopes might interest some: http://www-math.mit.edu/~rstan/transparencies/vis.pdf Not only for a pyramid, but also for a "hemi-spheric diamond" cut with any number of flat faces, when looked from far enough, will show $n-1$ faces. If you consider regular polyhedron only, than the answer may vary. Coming back to a general method, for a general convex polyhedron and a "common" camera, then consider the camera visual cone (something less than ${2\pi }$ steradians), with the vertex placed at the camera position. It will collect all the rays within that cone. Now consider the polyhedron as light emitting. You may consider that each face radiate from its centre in all the directions radiating outwards. If the polyhedron is convex, each face should freely radiated in the outward hemisphere. Exclude rays at $90°$ from vertical if you do not account for face=line. Check if there is a radiated ray that can reach the camera, and if it is within its view cone. Or invert the ray path. With a proper math description of the polyhedron, the SW task is not too difficult. • Can you clarify what a "hemi-spheric diamond" is? – jhch Apr 19, 2016 at 15:46 • @JohnHughes I mean a polyhedron, with a flat (large) face and with all vertices lying on a dome over that, perimeter (rim) included. View point on the vertical axis, as far as it can view the most inclined faces along the perimeter. Apr 19, 2016 at 19:14 Here are some definitions (in dimention 3, but you can easily generalize): Definition : Given a finite number of points with coordinates $P_1 = (x_1,y_1,z_1), .., P_n = (x_n,y_n,z_n)$, a convex solid is the convex hull of these points, i.e. all the points of the space defined as $\sum_{1\leq i \leq n} \lambda_i P_i$ with $\forall i \in [1,n], \lambda_i \in \mathbb{R}, 0 \leq \lambda_i \leq 1, \sum_{1 \leq i \leq n } \lambda_i = 1$ For instance, the points (0,0,0); (1,0,0); (0,1,0); (1,1,0); (0,0,1); (1,0,1); (0,1,1); (1,1,1) define a cube. Rq : If I add the point (0.5,0.5,0.5), I define the same cube. Definition : An oriented plane is 4 real numbers $a, b, c$ and $d$. Definition : An oriented plane is a face of a solid, iff at least three points defining the solid verify $ax+by+cz = d$ (are in the plane), and all points of the solid verify $ax+by+cz \leq d$. For instance, the plane defined by $a=1, b=0, c=0, d=1$ in the cube described above is a face. The one defined $a=1, b=1, c=0, d=1$ is not, since it fails the second condition of the definition. The one defined by $a=-1, b=0, c=0, d=-1$ is not either. Indeed, the definition specifies $ax+by+cz \leq d$, and it is not the case here. The definition of a face given here enables to control on which "side" of the plane the solid is. Putting a "-" sign in front of all the coefficients defining a face would mean the solid is on the other side of the plane. Definition : A view direction is a vector $(x_v,y_v,z_v)$ with $x_v^2+y_v^2+z_v^2=1$. The direction can be identified at the way you look at. Here, it is assumed you look from a point at an infinite distance. Definition : A face $(a,b,c,d)$ of a solid is visible from a direction $(x_v,y_v,z_v)$ iff $ax_v+by_v+cz_v < 0$. Here is the proof (short version) for the cube (with the points described earlier). All faces of this cube are defined by the following quadruples (and no others) : • (1,0,0,1) • (0,1,0,1) • (0,0,1,1) • (-1,0,0,0) • (0,-1,0,0) • (0,0,-1,0) So, with any viewing direction, one sees at most three faces (the ones corresponding to the sign of your viewing direction). (you see less if one of the viewing direction coordinates is 0, it would mean we see only some of its edges). For a different kind of solid, you need its defining points, and then the proof would basically go this way • first find all faces of this solid • assume there is another face and discover it is one you already have (so you have proven that you have all of them) • Assume you have a viewing direction. Then, a lot of cases to look at, like "if my direction is such that $x_v + ... < ...$" • find, in all the cases, the maximal number of faces If you think there is a leaner/more conventional way to define the problem, do not hesitate to mention it.
2022-06-30T23:55:53
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https://kstc.com/parabolic-listening-ozjqqgn/6tqq5g.php?ca6c6f=minkowski-distance-supremum
m. An object with distance information to be converted to a "dist" object. The Minkowski distance defines a distance between two points in a normed vector space. Here I demonstrate the distance matrix computations using the R function dist(). p p=2, the distance measure is the Euclidean measure. Similarity measure 1. is a numerical measure of how alike two data objects are. Minkowski Distance. As mentioned above, we can manipulate the value of p and calculate the distance in three different ways-p = 1, Manhattan Distance . 2. higher when objects are more alike. < It means, the distance be equal zero when they are identical otherwise they are greater in there. This problem has been solved! let p = 1.5 let z = generate matrix minkowski distance y1 y2 y3 y4 print z The following output is generated Minkowski Distance – It is a metric intended for real-valued vector spaces. The Minkowski distance or Minkowski metric is a metric in a normed vector space which can be considered as a generalization of both the Euclidean distance and the Manhattan distance. Minkowski distance is typically used with being 1 or 2, which correspond to the Manhattan distance and the Euclidean distance, respectively. Click to see full answer Herein, how do you calculate Minkowski distance? Maximum distance between two components of x and y (supremum norm) manhattan: Absolute distance between the two vectors (1 norm aka L_1). The Euclidean Distance tool is used frequently as a stand-alone tool for applications, such as finding the nearest hospital for an emergency helicopter flight. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . Additionally, how do you calculate Supremum distance? Asked By: Jianyun Norton | Last Updated: 24th February, 2020. In the limit that p --> +infinity , the distance is known as the Chebyshev distance. Which approach can be used to calculate dissimilarity of objects in clustering? Furthermore, how do you calculate Supremum distance? skip 25 read iris.dat y1 y2 y3 y4 skip 0 . Alternatively, this tool can be used when creating a suitability map, when data representing the distance from a certain object is needed. Does Hermione die in Harry Potter and the cursed child? pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. 3. groups of data that are very close (clusters) Dissimilarity measure 1. is a num… Although theoretically infinite measures exist by varying the order of the equation just three have gained importance. p It always gives the shortest distance between the two points, It may give a longer distance between the two points. p = ∞, the distance measure is the Chebyshev measure. Kruskal 1964) is a generalised metric that includes others as special cases of the generalised form. {\displaystyle p<1} What's the difference between Koolaburra by UGG and UGG? What is data governance in data warehouse? The definition was slightly modified and renamed in (Rosenfeld, 1985): same supremum for the distance H to exist; this is a serious drawback of this definition. {\displaystyle p} The traditional Minkowski distances are induced by the corresponding Minkowski norms in real-valued vector spaces. MINKOWSKI DISTANCE. m: An object with distance information to be converted to a "dist" object. However, a metric can be obtained for these values by simply removing the exponent of In R, dist() function can get the distance. More than 50 million people use GitHub to discover, fork, and contribute to over 100 million projects. p Show transcribed image text . The first property is called positivity. ... Euclidean distance (L 2 norm) r = ∞. The supremum distance (also referred to as L max, L ∞ norm and as the Chebyshev distance) is a generalization of the Minkowski distance for h → ∞. Minkowski Distance p1 p2 p3. p See the answer. This is the maximum difference between any component of the vectors. This difference is the supremum distance, defined more formally as: Here (theta) gives the angle between two vectors … Use the online Minkowski distance program below for your tool of exploration. 3. often falls in the range [0,1] Similarity might be used to identify 1. duplicate data that may have differences due to typos. (where In a set of real numbers the completeness axiom is valid: Every non-empty set of real numbers which is bounded from above has a supremum. In the limiting case of p The Minkowski distance or Minkowski metric is a metric in a normed vector space which can be considered as a generalization of both the Euclidean distance and the Manhattan distance. The power of the Minkowski distance. I am using scipy distances to get these distances. The second property called symmetry means the distance between I and J, distance between J and I should be identical. What is the difference between Euclidean distance and Manhattan distance? is an integer) between two points. The Chebyshev distance is the limiting case of the order-Minkowski distance, when reaches infinity. if p = 1, its called Manhattan Distance ; if p = 2, its called Euclidean Distance; if p = infinite, its called Supremum Distance Question: Question One Calculate The Similarity Measures Using Euclidean, Minkowski, And Supremum Distances Of The Following Points 60 Points 20 10 30 20 50 30 40 30 20 10 0 10 20 Euclidean Distance Nkowski Distance Supremum Distance. Previous question Next question Get more help from Chegg . The power of the Minkowski distance. Maximum distance between two components of \(x\) and \(y\) (supremum norm) manhattan: Absolute distance between the two vectors (1 … Mainly, Minkowski distance is applied in machine learning to find out distance similarity. Exercise1! where r is a parameter, n is the number of dimensions (attributes) and x k and y k are, respectively, the k-th attributes (components) or data objects x and y. To compute it, we find the attribute f that gives the maximum difference in values between the two objects. See the applications of Minkowshi distance and its visualization using an unit circle. When p = 1, Minkowski distance is same as the Manhattan distance. When p = 1, Minkowski distance is same as the Manhattan distance. p > 1 b) Euclideandistance! E.g. 1 Then the third one called triangular inequality means for the distance between i and j. , but the point (0,1) is at a distance 1 from both of these points. Minkowski Distance. The Minkowski distance is computed using Equation (2.18). {\displaystyle 2^{1/p}>2} {\displaystyle p} , the Minkowski distance is a metric as a result of the Minkowski inequality. / Minkowski distance Objective. We can manipulate the above formula by substituting ‘p’ to calculate the distance between two data points in different ways. Supremum Distance p1 p2 p3. 2 Minkowski distance is a metric in a normed vector space. if p = 1, its called Manhattan Distance ; if p = 2, its called Euclidean Distance; if p = infinite, its called Supremum Distance; I want to know what value of 'p' should I put to get the supremum distance or there is any other formulae or library I … Supremum distance Let's use the same two objects, x1 = (1, 2) and x2 = (3, 5), as in Figure 2.23. The scipy function for Minkowski distance is: distance.minkowski(a, b, p=?) TITLE Minkowski Distance with P = 1.5 (IRIS.DAT) Y1LABEL Minkowski Distance MINKOWSKI DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns .
2021-08-04T21:46:30
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https://taptapdesign.com/9iu2mi/cardinality-of-functions-from-n-to-n-2277b1
The set of even integers and the set of odd integers 8. Definition13.1settlestheissue. rationals is the same as the cardinality of the natural numbers. What's the cardinality of all ordered pairs (n,x) with n in N and x in R? The cardinality of N is aleph-nought, and its power set, 2^aleph nought. Relevance. We quantify the cardinality of the set $\{\lfloor X/n \rfloor\}_{n=1}^X$. There are many easy bijections between them. Julien. ... 11. In a function from X to Y, every element of X must be mapped to an element of Y. Special properties If X is finite, then there is a unique natural n for which there is a one to one correspondence from [n] → X. Relations. We discuss restricting the set to those elements that are prime, semiprime or similar. I understand that |N|=|C|, so there exists a bijection bewteen N and C, but there is some gap in my understanding as to why |R\N| = |R\C|. R and (p 2;1) 4. (Of course, for 4 Cardinality of Sets Now a finite set is one that has no elements at all or that can be put into one-to-one correspondence with a set of the form {1, 2, . A minimum cardinality of 0 indicates that the relationship is optional. 0 0. Now see if … But if you mean 2^N, where N is the cardinality of the natural numbers, then 2^N cardinality is the next higher level of infinity. . , n} for any positive integer n. The next result will not come as a surprise. (a₁ ≠ a₂ → f(a₁) ≠ f(a₂)) N N and f(n;m) 2N N: n mg. (Hint: draw “graphs” of both sets. In this article, we are discussing how to find number of functions from one set to another. In counting, as it is learned in childhood, the set {1, 2, 3, . A function with this property is called an injection. First, if $$|A| = |B|$$, there can be lots of bijective functions from A to B. Describe your bijection with a formula (not as a table). In the video in Figure 9.1.1 we give a intuitive introduction and a formal definition of cardinality. Becausethebijection f :N!Z matches up Nwith Z,itfollowsthat jj˘j.Wesummarizethiswithatheorem. The functions f : f0;1g!N are in one-to-one correspondence with N N (map f to the tuple (a 1;a 2) with a 1 = f(1), a 2 = f(2)). . Show that the two given sets have equal cardinality by describing a bijection from one to the other. Cardinality of an infinite set is not affected by the removal of a countable subset, provided that the. , n} for some positive integer n. By contrast, an infinite set is a nonempty set that cannot be put into one-to-one correspondence with {1, 2, . 8. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides In 1:n, 1 is the minimum cardinality, and n is the maximum cardinality. Example. find the set number of possible functions from - 31967941 adgamerstar adgamerstar 2 hours ago Math Secondary School A.1. Define by . Every subset of a … An interesting example of an uncountable set is the set of all in nite binary strings. Theorem13.1 Thereexistsabijection f :N!Z.Therefore jNj˘jZ. It is a consequence of Theorems 8.13 and 8.14. Note that A^B, for set A and B, represents the set of all functions from B to A. Sometimes it is called "aleph one". Onto/surjective functions - if co domain of f = range of f i.e if for each - If everything gets mapped to at least once, it’s onto One to one/ injective - If some x’s mapped to same y, not one to one. Fix a positive integer X. 1 Functions, relations, and in nite cardinality 1.True/false. b) the set of all functions from N to {0,1} is uncountable. We introduce the terminology for speaking about the number of elements in a set, called the cardinality of the set. . (a)The relation is an equivalence relation Solution False. An example: The set of integers $$\mathbb{Z}$$ and its subset, set of even integers $$E = \{\ldots -4, … f0;1g. Theorem 8.16. More details can be found below. If A has cardinality n 2 N, then for all x 2 A, A \{x} is finite and has cardinality n1. A function f from A to B is called onto, or surjective, if and only if for every element b ∈ B there is an element a ∈ A with f(a) For each of the following statements, indicate whether the statement is true or false. Lv 7. Solution: UNCOUNTABLE. Set of polynomial functions from R to R. 15. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. find the set number of possible functions from the set A of cardinality to a set B of cardinality n 1 See answer adgamerstar is waiting … SetswithEqualCardinalities 219 N because Z has all the negative integers as well as the positive ones. For example, let A = { -2, 0, 3, 7, 9, 11, 13 } Here, n(A) stands for cardinality of the set A We only need to find one of them in order to conclude \(|A| = |B|$$. The set of all functions f : N ! (hint: consider the proof of the cardinality of the set of all functions mapping [0, 1] into [0, 1] is 2^c) Set of linear functions from R to R. 14. Note that since , m is even, so m is divisible by 2 and is actually a positive integer.. What is the cardinality of the set of all functions from N to {1,2}? 2. This corresponds to showing that there is a one-to-one function f: A !B and a one-to-one function g: B !A. This function has an inverse given by . The 3 years ago. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. An infinite set A A A is called countably infinite (or countable) if it has the same cardinality as N \mathbb{N} N. In other words, there is a bijection A → N A \to \mathbb{N} A → N. An infinite set A A A is called uncountably infinite (or uncountable) if it is not countable. Cardinality To show equal cardinality, show it’s a bijection. Homework Equations The Attempt at a Solution I know the cardinality of the set of all functions coincides with the respective power set (I think) so 2^n where n is the size of the set. Cardinality of a set is a measure of the number of elements in the set. Subsets of Infinite Sets. Since the latter set is countable, as a Cartesian product of countable sets, the given set is countable as well. Theorem. All such functions can be written f(m,n), such that f(m,n)(0)=m and f(m,n)(1)=n. De nition 3.8 A set F is uncountable if it has cardinality strictly greater than the cardinality of N. In the spirit of De nition 3.5, this means that Fis uncountable if an injective function from N to Fexists, but no such bijective function exists. Is the set of all functions from N to {0,1}countable or uncountable?N is the set … Theorem. That is, we can use functions to establish the relative size of sets. Let S be the set of all functions from N to N. Prove that the cardinality of S equals c, that is the cardinality of S is the same as the cardinality of real number. Give a one or two sentence explanation for your answer. The existence of these two one-to-one functions implies that there is a bijection h: A !B, thus showing that A and B have the same cardinality. Show that (the cardinality of the natural numbers set) |N| = |NxNxN|. Theorem 8.15. Thus the function $$f(n) = -n… Section 9.1 Definition of Cardinality. Surely a set must be as least as large as any of its subsets, in terms of cardinality. Set of continuous functions from R to R. Show that the cardinality of P(X) (the power set of X) is equal to the cardinality of the set of all functions from X into {0,1}. The proof is not complicated, but is not immediate either. Injective Functions A function f: A → B is called injective (or one-to-one) if each element of the codomain has at most one element of the domain that maps to it. Answer the following by establishing a 1-1 correspondence with aset of known cardinality. This will be an upper bound on the cardinality that you're looking for. ∀a₂ ∈ A. . Set of functions from N to R. 12. Theorem \(\PageIndex{1}$$ An infinite set and one of its proper subsets could have the same cardinality. The number n above is called the cardinality of X, it is denoted by card(X). If there is a one to one correspondence from [m] to [n], then m = n. Corollary. A relationship with cardinality specified as 1:1 to 1:n is commonly referred to as 1 to n when focusing on the maximum cardinalities. It’s at least the continuum because there is a 1–1 function from the real numbers to bases. View textbook-part4.pdf from ECE 108 at University of Waterloo. . Z and S= fx2R: sinx= 1g 10. f0;1g N and Z 14. A.1. {0,1}^N denote the set of all functions from N to {0,1} Answer Save. Cantor had many great insights, but perhaps the greatest was that counting is a process, and we can understand infinites by using them to count each other. It is intutively believable, but I … , n} is used as a typical set that contains n elements.In mathematics and computer science, it has become more common to start counting with zero instead of with one, so we define the following sets to use as our basis for counting: Formally, f: A → B is an injection if this statement is true: ∀a₁ ∈ A. Second, as bijective functions play such a big role here, we use the word bijection to mean bijective function. It’s the continuum, the cardinality of the real numbers. show that the cardinality of A and B are the same we can show that jAj•jBj and jBj•jAj. It's cardinality is that of N^2, which is that of N, and so is countable. 46 CHAPTER 3. SETS, FUNCTIONS AND CARDINALITY Cardinality of sets The cardinality of a … Set of functions from R to N. 13. Prove that the set of natural numbers has the same cardinality as the set of positive even integers. . a) the set of all functions from {0,1} to N is countable. 2 Answers. 3.6.1: Cardinality Last updated; Save as PDF Page ID 10902; No headers. Discrete Mathematics - Cardinality 17-3 Properties of Functions A function f is said to be one-to-one, or injective, if and only if f(a) = f(b) implies a = b. It is well-known that the number of surjections from a set of size n to a set of size m is quite a bit harder to calculate than the number of functions or the number of injections. … In 0:1, 0 is the minimum cardinality, and 1 is the maximum cardinality. Here's the proof that f … Functions and relative cardinality. Need to find one of them in order to conclude \ ( \PageIndex { 1,,... Graphs ” of both sets has all the negative integers as well as the set of linear functions R... Be lots of bijective functions play such a big role here, we use the bijection... A! B and a one-to-one function f: a! B and a one-to-one function g:!... Your bijection with a formula ( not as a cardinality of functions from n to n product of countable sets, the given set not. { 1, 2, 3, it ’ s the continuum there!, show it ’ s at least the continuum because there is a one-to-one function g B! F … it ’ s a bijection an infinite set is countable $\ \lfloor... Functions play such a big role here, we use the word bijection mean... A surprise, indicate whether the statement is true or false a ) the set of all from! Set is countable ) ≠ f ( a₁ ) ≠ f ( ). Those elements that are prime, semiprime or similar cardinality as the positive ones can be lots of functions... One to one correspondence from [ m ] to [ N ], then m = n..... A function with this property is called an injection of X, it is learned childhood. Is actually a positive integer note that A^B, for set a and B, represents the of. 1G N and f ( a₂ ) is optional and Z 14 elements respectively ≠ f ( N ; )! In terms of cardinality, so m is even, so m is divisible by and. Classes ( Injective, surjective, bijective cardinality of functions from n to n of functions, you can this... A … 1 functions, relations, and so is countable introduction and one-to-one. F ( N ; m ) 2N N: N, 1 is the minimum cardinality of set... Bijection with a formula ( not as a Cartesian product of countable sets, the given set the. Numbers has the same cardinality as the set B ) the set of even integers 2 and actually... Positive even integers ; 1g N and f ( N ; m ) 2N N: N mg. Hint... R. 14 but is not affected by the removal of a set must be as least as large any! To B element of X must be as least as large as any its... A^B, for set a and B, represents the set of all functions from - adgamerstar! And the set show that ( the cardinality of the set to another: Let X and Y are sets! Bijective functions from R to R. 15 set is countable as well as set. 'Re looking for quantify the cardinality of 0 indicates that the relationship is optional consequence of Theorems and. From [ m ] to [ N ], then m = n. Corollary Z up. Is that of N^2, which is that of N is countable are prime, semiprime or similar describe bijection! That is, we use the word bijection to mean bijective function to mean bijective function every of! As the set of all functions from R to R. 15 the number elements. Since the latter set is not immediate either theorem \ ( |A| = |B|\,... N and Z 14 X and Y are two sets having m N! Numbers to bases not come as a surprise terms of cardinality fx2R: sinx= 1g 10. f0 1g.$ \ { \lfloor X/n \rfloor\ } _ { n=1 } ^X $,,. That is, we can use functions to establish the relative size of sets of a countable subset provided. Continuum, the set of positive even integers of both sets by 2 and is actually a positive n.. N to { 1,2 } be lots of bijective functions play such a big role here, we the! Set of all in nite binary strings of cardinality of N^2, which that. Of odd integers 8 as bijective functions play such a big role here, we the! Terms of cardinality a and B, represents the set to those elements that prime. Restricting the set$ \ { \lfloor X/n \rfloor\ } _ { }... The minimum cardinality of a set must be mapped to an element of X, is! Use functions to establish the relative size of sets adgamerstar 2 hours ago Math Secondary School A.1 B a... Is the cardinality that you 're looking for prove that the S= fx2R: sinx= 1g 10. f0 ; N! ) ≠ f ( a₁ ≠ a₂ → f ( N ; m ) N. Here 's the proof that f … it ’ s the continuum, the set. Properties First, if \ ( |A| = |B|\ ) mapped to an element of Y at... ) ≠ f ( a₁ ≠ a₂ → f ( a₁ ) ≠ f ( a₁ ) ≠ (... Z matches up Nwith Z, itfollowsthat jj˘j.Wesummarizethiswithatheorem function g: B! a X to Y every. Divisible by 2 and is actually a positive integer n. Section 9.1 Definition of cardinality 2 1. Use the word bijection to mean bijective function, relations, and its power,. Lots of bijective functions play such a big role here, we use the bijection! Is a consequence of Theorems 8.13 and 8.14 the relative size of.. The continuum, the cardinality of the set of polynomial functions from R to R. 15 a one or sentence! Integers as cardinality of functions from n to n as the positive ones draw “ graphs ” of both sets above is an! F0 ; 1g N and Z 14 has all the negative integers as well it cardinality! Since, m is divisible by 2 and is actually a positive integer here the... If this statement is true or false is a consequence of Theorems 8.13 and 8.14 for set a and,... ), there can be lots of bijective functions play such a big here! ) of functions, relations, and N cardinality of functions from n to n the minimum cardinality, and so countable. Show it ’ s a bijection continuum because there is a measure of the set of natural numbers has same! M ] to [ N ], then m = n. Corollary to showing that there a! Positive integer n. Section 9.1 Definition of cardinality for understanding the basics functions... Only need to find one of them in order to conclude \ ( \PageIndex { 1 } \ an! Each of the set ) of functions formula ( not as a surprise Cartesian product of sets! Minimum cardinality of a countable subset, provided that the set $\ \lfloor... Functions, relations, and 1 is the minimum cardinality, show it ’ at... Definition of cardinality of its proper subsets could have the same cardinality as the set$ {! Is countable, as it is learned in childhood, the given set the... Adgamerstar 2 hours cardinality of functions from n to n Math Secondary School A.1, bijective ) of functions from R R.... M = n. Corollary with this property is called an injection if this statement is true or false a ). Is actually a positive integer property is called an injection if this is. About the number of possible functions from { 0,1 } to N aleph-nought! Not complicated, but is not immediate either result will not come as a table ) in counting, a... Math Secondary School A.1 if there is a 1–1 function from X to Y, every element X! For each of the set of polynomial functions from N to { 1,2?! Because Z has all the negative integers as well of the following statements, indicate the! B, represents the set of linear functions from one set to another: Let X and Y are sets! } is uncountable N N and f ( a₂ ) linear functions from to! Is countable as well as the set of all functions from one set to those elements that are prime semiprime... We discuss restricting the set of linear functions from a to B and is actually a positive... Of a countable subset, provided that the set number of functions, you can refer this Classes... 1, 2, 3, least the continuum because there is a one or sentence. B is an equivalence relation Solution false function with this property is called an injection if this statement true! Example of an infinite set and one of its subsets, in terms of cardinality maximum cardinality →! The cardinality of the set of odd integers 8 the set of natural numbers has the same as. The minimum cardinality of an uncountable set is the maximum cardinality { 0,1 } is uncountable a₁ ≠ a₂ f. R to R. 15 and Y are two sets having m and N is aleph-nought, and is! Nwith Z, itfollowsthat jj˘j.Wesummarizethiswithatheorem semiprime or similar 2N N: N, and so is,... A₂ ) bijective functions from B to a is the maximum cardinality to a in the in... There can be lots of bijective functions from - 31967941 adgamerstar adgamerstar 2 hours ago Secondary.: B! a one set to another: Let X and Y are two sets m. A countable subset, provided that the 1 } \ ) an infinite and. _ { n=1 } ^X \$: draw “ graphs ” of both sets for the! Even integers N because Z has all the negative integers as well conclude \ ( |A| = )... For set a and B, represents the set of positive even.! 2 and is actually a positive integer mapped to an element of must... Pearce Grip Pg-39, Can Babies Get The Coronavirus Disease, Hidden Restaurants In Nairobi, Prada Nylon Mini Hobo Bag, Container Chassis Dimensions, How Long Does Waxing Last On Legs, Revita Shampoo Costco, Delta Flynn Toilet Paper Holder, Cow Hide Rug Ikea,
2021-02-27T01:23:20
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https://math.stackexchange.com/questions/3005409/koch-curve-from-cantor-sets-paradox
# Koch curve from Cantor sets (paradox) The Koch curve is normally constructed by taking a line segment, replacing the middle third with two copies of itself forming legs in an equilateral triangle, and then repeating this recursively for every subsegment. See image below. At every step, the length of the curve is multiplied by $$4/3$$ so the final length is infinite. Notice that every line segment undergoes the construction of the Cantor set: Therefore one could consider replacing every line segment with Cantor sets already from the beginning. Thus, start with the Cantor set, take two smaller copies of the Cantor set and make a $$\wedge$$ in the middle opening, and then repeat recursively. In this case the final length will be zero since the Cantor set has length zero. Question: How is this paradox resolved? Will the limit set of my construction be different than the ordinary Koch curve? If so, what points are missing? • Aren't there a countably infinite number of Cantor sets in the Koch snowflake if you look at it this way? Nov 19 '18 at 19:46 • @MichaelSeifert. Yes, but a countable union of sets with zero length (Lebesgue measure) also has zero length. Nov 19 '18 at 19:49 • I'm not entirely sure what the paradox is here. You're just taking something with $0$ length and "adding" it together countably infinite times, right? I'm not sure why anyone would expect it to have positive length, in that case. Am I missing something? Nov 19 '18 at 19:55 • You've not argued that the final Koch snowflake is in fact the union of countably many Cantor sets (and indeed it isn't), so there's no contradiction. Nov 19 '18 at 21:15 • The MSE question 12906 "The staircase paradox, or why $\pi\ne 4$" seems similar. Or else MSE question 1525410 "Length of diagonal compared to the limit of lengths of stair-shaped curves converging to it". Nov 19 '18 at 21:25 Convergence in fractal geometry is typically defined in terms of the Hausdorff metric. Roughly, two sets are "close" with respect to the Hausdorff metric, if every point in one is close to some point of the other. Your collection of Cantor sets is indeed dense in the Koch curve with respect to the Hausdorff metric. The Hausdorff metric, however, doesn't respect length. That is, two sets can be close in the Hausdorff metric, yet their lengths can be very far apart. You've found one example illustrating this but there are others. For example, if $$Q_n = \{k/n:0\leq k \leq n\},$$ then the Hausdorff distance between $$Q_n$$ and the unit interval is less than $$1/n$$. $$Q_n$$ is finite, yet the sequence of $$Q_n$$s converges to a set of positive length. Similarly, you could strengthen your example by using the set of endpoints of the intervals that approximate the Koch curve. Here is a strategy to find points in the Koch curve that do not lie on any of your Cantor sets. First, note that the Koch curve is invariant set of the iterated function system: \begin{align} T_1(x,y) &= \left(\frac{x}{3},\frac{y}{3}\right) \\ T_2(x,y) &= \left(\frac{1}{6} \left(x-\sqrt{3} y+2\right),\frac{1}{6} \left(\sqrt{3} x+y\right)\right) \\ T_3(x,y) &= \left(\frac{1}{6} \left(x+\sqrt{3} y+3\right),\frac{1}{6} \left(-\sqrt{3} x+y+\sqrt{3}\right)\right) \\ T_4(x,y) &= \left( \frac{x}{3},\frac{y+2}{3} \right). \end{align} These functions map the Koch curve onto the four sub-parts below Now, any point On the Koch curve can be realized as the limit of a sequence \begin{align} & T_{i_1}(0,0) \\ & T_{i_1} \circ T_{i_2}(0,0)\\ & \vdots \\ & T_{i_1} \circ T_{i_2} \circ \cdots T_{i_n}(0,0) \\ & \vdots \end{align} where $$(i_1,i_2,i_3,\ldots)$$ is a sequence in $$\{1,2,3,4\}$$. The point lies on one of your Cantor sets precisely when the sequence contains only finitely many 2s and 3s, so that it ends in a string of only 1s and 4s. If, for example, the sequence has only 1s and 4s, and no 2s or 3s, then we get a point in Cantor's ternary set lying on the unit interval. If the sequence start with a 2 and then contains only 1s and 4s, we generate a point in the red Cantor set shown below; this is exactly the image of the ternary Cantor set under the function $$T_2$$. If the sequence starts 3, then 2, then contains only 1s and 4s, we generate a point in the blue Cantor set below; this is exactly the the image of the ternary Cantor set under the function $$T_3 \circ T_2$$. Finally, if we have any other sequence of 1s, 2s, 3s, and 4s, then we generate some other point on the Koch curve that is not in any of your Cantor sets. There are uncountably many such points. I suppose the simplest one to find explicitly corresponds to the sequence containing only 2s, which is exactly the fixed point of $$T_2$$. To find it, we need only solve $$T_2(x,y) = (x,y),$$ which yields $$(5/14,\sqrt{3}/14)$$. That point is shown in red in the sequence of approximations below. If we zoom into the last picture on the red point so that it's centered in a square of side length 0.04, we get the following: Thus, the edges keep jutting out closer to the point but never actually hit it. It's in the limit but not on any of the edges. • Could you give me some points that will be in the Koch curve but not in my construction? Nov 20 '18 at 10:10 • @md2perpe Sure - the point $(5/14,\sqrt{3}/14)$ (along with many other points) is in the Koch curve but not in your construction, as explained in my edit. Nov 20 '18 at 15:48 • So that point is neither in any step of the construction of the Koch curve nor in any step of my construction. Therefore it's not in the $\limsup$ limit of neither sequence. But it is in the Hausdorff limit of both? Nov 20 '18 at 18:58 • That is correct. If you were thinking in terms of the limsup, that might explain your confusion. Nov 20 '18 at 19:15 Not every point in the Koch curve is in one of your Cantor sets! Indeed, this follows immediately from the Baire category theorem: each of your Cantor sets is a closed subset of the Koch curve with empty interior, and the union of countably many such sets will still have empty interior and thus cannot be the whole Koch curve. The proof of the Baire category theorem also gives a recipe for finding points that are not in any of the Cantor sets. Let's say the $$n$$th Cantor set is $$C_n$$ and the Koch curve is $$K$$. Start with a point $$x_0\in K\setminus C_0$$; then there is some closed ball $$B_0$$ around $$x_0$$ which does not intersect $$C_0$$. Then pick a point $$x_1\in B_0\setminus C_1$$. There is again some closed ball $$B_1$$ around $$x_1$$ which does not intersect $$C_1$$. Then pick a point $$x_2\in B_0\cap B_1\setminus C_2$$, and so on. This procedure gives a sequence of points $$(x_n)$$ in $$K$$ and balls $$B_n$$ such that $$x_m\in B_n$$ for all $$m\geq n$$ and $$B_n\cap C_n=\emptyset$$. Since $$K$$ is compact, this sequence $$(x_n)$$ must accumulate at some point $$x\in K$$ (in fact, we can choose the balls $$B_n$$ to have radius converging to $$0$$ and then $$(x_n)$$ will be Cauchy and thus will converge). Then $$x\in B_n$$ for all $$n$$, so $$x$$ cannot be in $$C_n$$ for any $$n$$. Another way to think of it is that the Koch curve is parametrized by a continuous function $$f:[0,1]\to \mathbb{R}^2$$ which is the limit of functions $$f_n:[0,1]\to\mathbb{R}^2$$ parametrizing the approximations to the Koch curve. Each of your Cantor sets corresponds to the image of some Cantor set in $$[0,1]$$ under $$f$$, and that Cantor set in $$[0,1]$$ will have Lebesgue measure $$0$$ if you've chosen your parametrization reasonably. So, the union of all those Cantor sets in $$[0,1]$$ is still a null set (and a set of first category), so most points in $$[0,1]$$ aren't in any of those Cantor sets. Note that you can also think of a point $$x\in[0,1]$$ that is not in any of those Cantor sets as a point such that the sequence $$(f_n(x))$$ never stabilizes but instead keeps moving infinitely often as you build the successive approximations. The value of $$f(x)$$ will then be the limit of this sequence, a point of the Koch curve that was not in any of the approximations but instead is a limit of points in the approximations.
2021-12-03T08:35:20
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https://math.stackexchange.com/questions/68327/what-properties-of-numbers-allow-us-to-remove-parentheses-from-expressions/68332
# What properties of numbers allow us to remove parentheses from expressions? I've seen it asserted in several places (e.g., Spivak's Calculus, p.3) that the fact that "parentheses can be freely rearranged" in expressions involving only addition ($+$) is based solely on (P1) associtivity of addition, $$a+\left(b+c\right) =\left(a+b\right)+c,$$ and can see that this is the case in every example I've tired. However it is also asserted that one can eliminate parenthesis altogether, so that, for example $$a+b+c$$ is identical to the above expressions. I can't see how to prove this with using only P1. The proof would seem to require (P2) additive identity, $$a+0=a,$$ (P3) additive inverse, $$a+\left(-a\right)=0,$$ and (P4) commutativity, $$a+b=b+a.$$ For example, the proof $$\left(a+b\right)+c$$ $$=\left(a+b\right)+c+0$$ $$=0+\left(a+b\right)+c$$ $$=a+\left(-a\right)+\left(a+b\right)+c$$ $$=a+\left(\left(-a\right)+a\right)+b+c$$ $$=a+\left(a+\left(-a\right)\right)+b+c$$ $$=a+0+b+c$$ $$=a+b+c$$ requires P2, P4, P3, P1, P4, P3, and P2. Am I missing something that allows one to conclude that $$\left(a+b\right)+c=a+b+c$$ based solely on P1? Perhaps there something subtle about what parentheses represent that I'm missing. • I always thought brackets simply represent order of operations. So a + b + c is the same as (a + b) + c, if we define addition as left-to-right (and the opposite if we define the other way). – davin Sep 28 '11 at 21:57 • Note that parentheses can only be freely rearranged when the sum involves finitely many terms: $(1 + -1) + (1 + -1) + \ldots \neq 1 + (-1 + 1) + (-1 + 1) + \ldots$ for instance. – user7530 Sep 28 '11 at 22:21 • I think my question wasn't clear. I'm given just P1, as written above. Is that sufficient to conclude $(a+b)+c=a+b+c$ or do I need additional properties? – orome Sep 29 '11 at 1:25 • @davin: Is that my mistake then (and the reason that my question is dumb)? The parentheses aren't just arbitrary symbols that require formal steps to remove (as in my "proof"). They're just what we all "know" they are: a way of specifying the order of operations. – orome Sep 29 '11 at 2:02 HINT $\:$ It is easy: keep pushing ')'s rightward using the rewrite rule $\rm\ (x+y)+z\ \to\:\ x+(y+z)\:.\$ An easy induction shows that this process terminates with the right-associated normal form where all ')'s are at the right end, e.g. $\rm\ a+(b+(c+(d+\:\cdots\:)))\:.\:$ By associativity, the rewrite rule preserves equality, so every possible bracketing of the summands is equal to the normalized bracketing. Thus we can omit the brackets, yielding a well-defined $\rm\:n$-ary addition operation $\rm\ a_1+ a_2+\: \cdots\:+a_n\:.$ However, for nonassociative operations, different bracketings need not yield equal values, so the brackets are required in order to uniquely specify the intended value. It might help to think of the expressions presented as (parse) trees, e.g. below. • So it is the case then that I'm attaching too much significance to the parentheses as symbols, and thus to the formalism if getting rid of them—that their only significance is the obvious and common one of specifying an order, so that once things are in the normal form above, we can simply drop them, based on their effect on the order of operation? If that's the case, I see that only P1 is indeed necessary (and the elaborate "proof" is not). – orome Sep 29 '11 at 1:34 • @rax What I wrote above holds true for any associative operation. None of the other laws are required to show that sums are independent of the bracketing. For example consider, for products $\rm\:((ab)c)d = (ab)(cd) = a(b(cd))\:.$ – Bill Dubuque Sep 29 '11 at 1:58 • Yes, I see that, as for addition. But what I was unsure of was what additional rules or definitions were required to remove the parentheses altogether. If I understand correctly, you've answered that. – orome Sep 29 '11 at 2:05 • @rax Yes, only the associative law is required. – Bill Dubuque Sep 29 '11 at 2:09 • @rax The brackets specify not order but, rather, shape (tree structure). – Bill Dubuque Sep 29 '11 at 3:46 $+$ is initially defined only as a binary operation. Then we can define $a+b+c$ as either $(a+b)+c$ or $a+(b+c)$; associativity says it doesn't matter which one we use, since they are equal. But until that definition is made, the expression $a+b+c$ has no meaning. • So is it simply that? The full definition of associativity is just $$a+\left(b+c\right) =\left(a+b\right)+c=a+b+c$$ and there is, in fact, nothing further to prove? – orome Sep 29 '11 at 1:17 • @rax: Dear raxacoricofallapatorius, Yes, although the two equalities are of a slightly different nature: the first is not true by definition, but is rather an axiom (the associative law). The second is true by definition: it is what "$a+b+c$" means. Regards, – Matt E Sep 29 '11 at 5:18 The point is that no matter how you parenthesize $a+b+c$, the result is the same, so the expression without parentheses is unambiguous. Thus, it doesn’t matter whether you define $a+b+c$ to be $(a+b)+c$ or $a+(b+c)$. (Note that some such definition is ultimately required, since $+$ is originally defined only as a binary operation.) Compare this with $a-b-c$: in general $(a-b)-c\ne a-(b-c)$, so the expression $a-b-c$ is uninterpretable without some convention, e.g., work from left to right. When the operation is associative, no such convention is required.
2019-08-18T13:45:34
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https://math.stackexchange.com/questions/2515694/when-dividing-by-a-fraction-why-can-you-not-take-the-reciprocal-of-term-involvi
# When dividing by a fraction, why can you not take the reciprocal of term involving addition/subtraction? Given something like: $$\frac{a}{\frac{a}{b}}$$ You would multiply the numerator $a$ by the reciprocal of the denominator, $\tfrac ba$ to get: $$a\cdot\frac{b}{a}= \frac{ab}{a}=b$$ Given $$\frac{1}{\frac 1a + \frac 1b}$$ By taking the LCM and adding the denominators you get: $$\frac{1}{\left(\frac{a+b}{ab}\right)}$$ Given the reciprocal division rule in example one: $$\frac{ab}{a+b}$$ Why can you not take the reciprocal of $\frac 1a + \frac 1b$ to begin with? I did this and ended up with: $$1\cdot\left(\frac a1 + \frac b1\right) =a + b$$ However $\frac{1}{1/a + 1/b}$ is not the same as $a+b$ so this is incorrect. I was trying to find a similar example online but I could not, why is this incorrect? Does the rule only work with one fraction as the denominator and not terms linked by addition and/or subtraction? • This is exactly the reason you should be using Mathjax! Nov 11 '17 at 21:54 • Sorry! I am reading the page now! Nov 11 '17 at 21:56 • Cool! I did the first edit for you ... Nov 11 '17 at 21:59 • Thanks, I am working on the rest! Nov 11 '17 at 22:03 • Your edit made the result of the first multiplication $a$ instead of $b$. Nov 11 '17 at 22:09 ## 2 Answers When you took the reciprocal of $\frac 1a+\frac 1b$ to get $a+b$, you implicitly assumed that the reciprocal of a sum is the sum of the reciprocals. This is not true. Note that $\frac 1a$ is $a^{-1}$. Just as the square of a sum is not the sum of the squares (in general), that is, $(x+y)^2\neq x^2+y^2$, so also the reciprocal of a sum is not in general the sum of the reciprocals: $(x+y)^{-1}\neq x^{-1}+y^{-1}$. • Nope, not me. My name is pretty common. Nov 11 '17 at 23:00 The reciprocal of $\frac{1}{a} + \frac{1}{b}$, is by definition $$\frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{1}{\frac{a+b}{ab}} = \frac{ab}{a+b}.$$ So your reasoning fails because $\frac{a}{1} + \frac{b}{1}$ is not the reciprocal of $\frac{1}{a} + \frac{1}{b}$.
2021-09-25T00:39:20
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https://quant.stackexchange.com/questions/57957/calculating-european-call-option-the-bjork-way
# Calculating European call option, the Bjork way We have a 3 period binomial tree with values: 59.65 (C33 = 7.65) 56.24 (C22 = ?) 53.03 53.03 (C32 = 1.03) 50 50 (C21 = ?) 47.14 47.14 (C31 = 0) 44.45 (C20 = ?) 41.91 (C30 = 0) W want to calculate a European call option, no arbitrage, with properties K = 52, u = 1.0606, d = 1/u = 0.943, maturity in 9 weeks, r = 0.001 per week. The value for a call option is given by $$max[S_t -K, 0]$$. We can calculate $$C_2^{2}$$ given the risk neutral formula from the literature (Bjork 3ed, 2.1.4): $$C_2^{2} = \frac{1}{1+R} (q*C_3^{3} + (1-q)*C_3^{2})$$, $$\frac{1}{1+R}$$, given by Bjork Proposition 2.6, but as we have multiple nodes I assume we need to discount it, which gives the formula $$e^{r-(T-t)}= e^{0.053348-(9/52)} = 1.009276$$ $$R = 1.001^{52} = 5.3348pct = 0.053348 ,$$ $$q = \frac{(1+R)-d}{u-d} = \frac{1.009276 - 0.943}{1.0606 - 0.943} = 0.5636$$, if we then plug the values into the formula: $$C_2^{2} = \frac{1}{1+R} (q*C_3^{3} + (1-q)*C_3^{2}) = 1.009276*(0.5636*7.65 + 0.4363*1.03) = 4.8051$$, My questions are: a) is the value of $$C_2^{2}$$ correct? b) is there a faster way to calculate the option value of the tree because this takes a lot of time (yes you can write a program but I am following the theory and I believe I have to learn it by hand as well). • I think you need to fix your discounting in a few places. In general, multiplying by $1+r(T-t)$ is similar to multiplying by $e^{r(T-t)}$. You are mixing those up here. Sep 12 '20 at 22:50 • When I follow the method given by @RRL, it seems that the discount factor is $\frac{1}{(1+r)^9} = \frac{1}{(1+0.001)^9} = 0.9910$, can you point out where the mistakes are? Sep 13 '20 at 7:40 • It depends on if you do exponential or periodically-compounded discounting. I've seen trees built with both, but Bjork is using one so for replication you should use that. Sep 13 '20 at 7:52 • Are you referring to proposition 2.11? Sep 13 '20 at 7:58 Is there a faster way to calculate the option price? With a recombining binomial tree, the terminal asset price has a binomial distribution -- as you might have expected. For a tree with $$n$$ steps, the probability of reaching price $$S_{n,k}$$ where $$k$$ is the number of up moves is $$P_{n,k} = \frac{n!}{k!(n-k)!}q^k(1-q)^{n-k}$$ The option price is the discounted risk-neutral expectation of the payoff, $$C = \frac{1}{(1+r_s)^n}\sum_{k=0}^n\frac{n!}{k!(n-k)!}q^k(1-q)^{n-k} \max(S_{n,k}-K,0),$$ where $$r_s$$ is the interest rate per period associated with a single step. Using this formula avoids working backwards and computing option values at intermediate steps. In this case we have $$n= 3$$ and $$(1+r_s) = (1+0.001)^3$$ (since each step spans 3 weeks). Hence, since $$C_{31} = C_{30} = 0$$, $$C = \frac{1}{(1+r)^9} (1 \cdot q^3 C_{33} + 3 \cdot q^2(1-q) C_{32})$$ (Note that the coefficient $$1$$ for the first term arises because there is one path through the tree reaching the node $$(3,3)$$ and the coefficient $$3$$ for the second term arises because there are three paths through the tree reaching the node $$(3,2)$$.) • Many many thanks: if we apply your formula $C = \frac{1}{(1+r)^9} (1 \cdot q^3 C_{33} + 3 \cdot q^2(1-q) C_{32})$ = $\frac{1}{(1+0.001)^9} (1 \cdot 0.5636^3 (7.65) + 3 \cdot 0.5636^2(0.4363)(1.03))$ = $0.9910 \cdot ((0.1790 \cdot 7.65) + (3 \cdot (0.3176 \cdot 0.4363)1.03)) = 1.4918$ as the option price. Is it possible that you check the value of the price? Also because @kurtosis pointed out that my discounting is not correct Sep 13 '20 at 7:34 • You’re welcome. First, to get the risk neutral probability you have the correct formula $q = (1+R-d)/(u-d)$ but the $R$ here must be the interest rate for the single step period. It seems you are not given a continuously compounded rate but rather a weekly compounded rate $r = 0.001$. Since a single step is 3 weeks we have $R = (1+r)^3 -1$. – RRL Sep 13 '20 at 14:08 • Look at the Wikipedia entry for compound interest and in particular the example when a monthly rate is specified. There is a distinction between nominal and effective annual rate. If I am told that the interest rate per week is $0.001$ then I would assume this is a rate that is compounded at a weekly frequency. The nominal annual rate here is $52\times 0.001$. – RRL Sep 13 '20 at 14:18 • The problem in the notes says time to maturity is 9 weeks, rfi = 0.1% per week thus a 10000 means 10 usd every week. If I follow your line of thinking, this means that we have indeed $r = 0.001$, you say $R$ must be the interest rate for the single step period. We have 9 weeks left, and 3 steps is indeed 3 periods; $R = (1+r)^3 -1= (1.001^3 -1) = 0.003$, plug into q, gives us (1.003 - 0.943)/(1.0606−0.943) = 0.5102. If we plug this in your formula we get 0.9910⋅((0.1328⋅7.65)+(3⋅(0.2603⋅0.4898)1.03))= 3.46 which seems a bit high.. Sep 13 '20 at 14:55 • @wecandothis: I get $q= 0.5108$ and $q^3C_{33} + 3q^2(1-q)C_{32} = 1.4140$. Dividing by $(1+r)^9 = 1.009036$ I get $C = 1.4013$. This is the same as the result obtained by working backwards through the tree. – RRL Sep 15 '20 at 0:40
2021-10-16T02:00:26
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https://math.stackexchange.com/questions/3730907/z-frac43-2-how-to-know-which-complex-roots-to-keep-from-this-equat
# $z^{\frac{4}{3}} = -2$ ; How to know which complex roots to keep from this equation So I recently came upon the following complex algebra problem: $$z^{\frac{4}{3}} = -2$$ So, to solve it I have to find the z values that solve the following: $$z = (-2)^{\frac{3}{4}}$$ To do this I express -2 in exponential form: $$z = (2e^{i(\pi + 2\pi n)})^{\frac{3}{4}}$$ Then, I solve for that trying for $$n=0,1,2,3$$ and I come up with 4 roots: $$z_1 = 2^{\frac{3}{4}}e^{i\frac{\pi}{4}}$$ $$z_2 = 2^{\frac{3}{4}}e^{i\frac{3\pi}{4}}$$ $$z_3 = 2^{\frac{3}{4}}e^{i\frac{5\pi}{4}}$$ $$z_4 = 2^{\frac{3}{4}}e^{i\frac{7\pi}{4}}$$ However, if I try to check these solutions for the original problem, only $$z_2$$ and $$z_3$$ succeed, while $$z_1$$ and $$z_4$$ do not solve the initial equation. Even plugging the original equation into Wolfram, gives me just those two roots. I have been thinking about this over and over and don't understand where I'm going wrong or what is it that I'm failing to consider. Does anybody have any idea of where I'm going wrong? • Btw, Mathjax works in titles too – sai-kartik Jun 23 '20 at 2:57 • Thanks! I didn't know that – Alejandro Flores Jun 23 '20 at 2:59 • I would have said to keep the roots with principle arguements but looks like it doesn't match with what you're getting.. – sai-kartik Jun 23 '20 at 3:02 • From this chatroom,you can see that -π to π is the definition of the principle argument, and it helps to keep it in this range, cus then you're gonna easily avoid repeated solutions – sai-kartik Jun 23 '20 at 3:59 • Also you can convert 5π/4 and 7π/4 further to make it fall in the principle argument range – sai-kartik Jun 23 '20 at 4:00 I presume you're treating $$z^{4/3}$$ as a multivalued function, and you're allowing any $$z$$ such that any branch of $$z^{4/3}$$ is $$2$$. By definition, $$z^{4/3} = \exp((4/3) \log(z))$$ where $$\log(z)$$ is any branch of the logarithm of $$z$$. If $$\text{Log}(z)$$ is the principal branch (with imaginary part in $$(-\pi, \pi]$$), the other branches of $$\log(z)$$ are $$\text{Log}(z) + 2 \pi i n$$ for arbitrary integers $$n$$, and the corresponding branches of $$z^{4/3}$$ are $$\exp((4/3) \text{Log}(z) + (8 \pi i n/3))$$. There are three possibilities, corresponding to the values of $$n \mod 3$$. Now this is supposed to be $$-2 = 2 \exp(\pi i)$$. Thus for $$n \equiv 0 \mod 3$$, $$2 = \exp((4/3) \text{Log}(z) - \pi i)$$ where $$\text{Im}((4/3) \text{Log}(z) - \pi i) = 0$$ and $$\text{Re}((4/3) \text{Log}(z) = \text{Log}(2)$$. We get either $$\text{Log}(z) = (3/4) \text{Log}(2) + 3 \pi i/4$$, i.e. $$z = 2^{3/4} e^{3 \pi i/4}$$, or $$\text{Log}(z) = (3/4) \text{Log}(2) - 3 \pi i/4$$, i.e. $$z = 2^{3/4} e^{-3\pi i/4}$$. (this $$2^{3/4}$$ being the real $$3/4$$ power). For $$n \equiv 1 \mod 3$$, $$2 = \exp((4/3) \text{Log}(z) + 5 \pi i/3)= \exp((4/3) \text{Log}(z) - \pi i/3$$ where $$\text{Im}((4/3) \text{Log}(z) - \pi i/3 = 0$$. We get $$\text{Log}(z) = (3/4) \text{Log}(2) + \pi i/4$$, or $$z = 2^{3/4} e^{\pi i/4}$$. For $$n \equiv 2 \mod 3$$, $$2 = \exp((4/3) \text{Log}(z) + 13 \pi i/3) = \exp((4/3) \text{Log}(z) + \pi i/3$$ where $$\text{Im}((4/3) \text{Log}(z) + \pi i/3 = 0$$. We get $$\text{Log}(z) = (3/4) \text{Log}(2) - \pi i/4$$, or $$z = 2^{3/4} e^{-\pi i/4}$$. So there are indeed four solutions. However, if you try to verify these with Mathematica or most other computer algebra systems, they won't all work, as they like to use the principal branch rather than multivalued functions. • If $\operatorname {Log}$ is the principal branch, then $$\operatorname {Re}(4 \operatorname {Log}(z)/3) = \operatorname {Log} 2 \land \operatorname {Im}(4 \operatorname {Log}(z)/3 - \pi i) = 0$$ (the case $n \equiv 0 \bmod 3$) has only one solution. The second equation should be $$\operatorname {Im}(4 \operatorname {Log}(z)/3 - \pi i) = 2 \pi k.$$ – Maxim Jul 12 '20 at 18:33 Please note that the complex function $$f(z)=z^{\frac{1}{n}}$$, $$n \in \mathbb{N}, \, n \ge 2$$ is a multivalued function. Writing the function in polar form,$$z=re^{i \theta } \qquad \rightarrow \qquad f(z)=(re^{i \theta })^{\frac{1}{n}}=r^{\frac{1}{n}}e^{i \frac{\theta }{n}},$$we can easily conclude that a point with arguments $$\theta$$, $$\theta + 2\pi$$, ..., $$\theta + 2(n-1)\pi$$ in the domain plane corresponds to $$n$$ distinct points with the arguments $$\frac{\theta }{n}$$, $$\frac{\theta }{n}+\frac{2\pi }{n}$$, ..., $$\frac{\theta }{n}+\frac{2(n-1)\pi }{n}$$ in the image plane. In other words, this function is a one-to-$$\bf{n}$$ correspondence. With a similar argument, one can show that the function $$f(z)=z^{\frac{4}{3}}$$ is a one-to-three correspondence. You have solved $$z^{\frac{4}{3}}=-2$$ correctly. However, please note that to check the solutions for the original problem, you should use the same representation of points you reached by solving the problem, that is,$$z_1 = 2^{\frac{3}{4}}e^{i\frac{3\pi}{4}}$$$$z_2 = 2^{\frac{3}{4}}e^{i\frac{9\pi}{4}}$$$$z_3 =2^{\frac{3}{4}}e^{i\frac{15\pi}{4}}$$$$z_4 =2^{\frac{3}{4}}e^{i\frac{21\pi}{4}},$$which are clearly satisfy the original problem. Otherwise, you may get other values of $$z^{\frac{4}{3}}$$ not satisfying the original problem.
2021-06-20T03:49:56
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https://math.stackexchange.com/questions/1443756/positive-borel-measures-mu-n-mu-on-mathbbr-with-mu-n-to-mu-weak
# Positive Borel measures $\mu_n,\mu$ on $\mathbb{R}$ with $\mu_n \to \mu$ weak-*. Show $\limsup \mu_n(K) \leq \mu(K)$ for $K$ compact This is a problem that I've seen a few times on UCLA's analysis quals that I've been trying to solve, and I have a few questions. Suppose $\mu_n,\mu$ are positive finite Borel measures on $\mathbb{R}$ such that $$\int f\,d\mu_n \to \int f \,d\mu \qquad \forall f \in C_c(\mathbb{R}).$$ Show that $\limsup_{n\to\infty} \mu_n(K) \leq \mu(K)$ for all compact subsets $K\subset \mathbb{R}$. Here's my attempt at a solution. Let $\Lambda: C_c(\mathbb{R}) \to \mathbb{R}$ be the functional given by $\Lambda(f) =\int f \,d\mu$. We have that $|\Lambda(f)| \leq ||f||_\infty \mu(\mathbb{R})$, which shows that $\Lambda$ is a bounded linear functional. By the Riesz Representation theorem, $\Lambda(f) = \int f \,d\tilde\mu$ where $\tilde \mu$ is a regular complex Borel measure. Question 1: Can we assume/prove that $\mu = \tilde \mu$? Suppose we can. Let $K\subset \mathbb{R}$ be a compact set. Fix $\epsilon>0$. As $\mu$ is regular, we can find an open set $U\supset K$ with $\mu(U\setminus K)<\epsilon$. Then, by Urysohn's lemma there exists $g$ continuous with $g(x)=0$ for $x\notin U$, $g(x)=1$ for $x\in K$, and $0\leq g(x)\leq 1$ for $x \in U\setminus K$. Clearly, $g$ has compact support. We have that $$\int g \,d\mu_n \to \int_{U\setminus K} g \,d\mu + \int_K g\,d\mu \leq \mu(K) + \epsilon.$$ On the other hand, $$\limsup_{n\to \infty} \mu_n(K)\leq \limsup_{n\to \infty}\left(\int_K g\,d\mu_n + \int_{U\setminus K} g\,d\mu_n \right)\leq \epsilon + \mu(K).$$ The first inequality follows since $\int_{U\setminus K}g\,d\mu_n \geq 0$, and the second follows by the above limit. As $\epsilon$ is arbitrary, we get the desired result. Assuming the first question I made above, is this a valid solution? And, if question 1 is not true, what are some different methods for proving this? Also, my apologies if this question has been asked before; I couldn't seem to find it anywhere on the site. Edit: I was able to solve the problem without needing to know whether or not Question 1 is true. Regularity of $\mu$ may be already true, but my answer below shows how to get the open set $U$ in my argument above without knowing that $\mu$ is regular. • Where do you use Q1 in your proof? – Giovanni Sep 20 '15 at 17:09 • I use it to assume that $\mu$ is regular, in order to choose $U$ add I have. – Andrew Sep 20 '15 at 17:10 • It is a Borel measure by assumption, this should imply that it is regular, doesn't it? – Giovanni Sep 20 '15 at 17:11 • Maybe some assume Borel means regular by comvention, but I do not. It may be that this problem assumes this convention, and I just don't know. – Andrew Sep 20 '15 at 17:14 • Ok, here is the thing: you can prove that any finite Borel measure on a compact space is regular. You can use this in this context extending this result by sigma finiteness to $\mathbb{R}$ (which can be expressed as union of compact sets), hence you can assume that $\mu$ is regular. – Giovanni Sep 20 '15 at 17:22 Define $f_m\in C_c(\mathbb{R})$ by $f_m(x)=\left(1-m d(x,K)\right)_+$ where $d(x,K)=\inf(d(x,k): k\in K)$ is the distance of $x$ from $K$. Then $f_m\geq 1_K$ so that $$\int f_m \,d\mu_n\geq \int 1_K \,d\mu_n.\tag1$$ Taking the $\limsup_n$ in (1) gives $$\int f_m \,d\mu\geq \limsup_n \mu_n(K).\tag2$$ Letting $m\to\infty$ in (2) gives the result. • Definitely not the first proof I would have thought of, but it is very nice! A quick question: those $f_m$ are dominated by $f_1$, correct? Then dominated convergence lets you compute the $m\to \infty$ limit? – Andrew Sep 20 '15 at 23:21 • @Andrew Correct. $(f_m)$ is a decreasing sequence of functions. – user940 Sep 21 '15 at 2:25 We claim that $K=\bigcap_{n=1}^\infty V_n$ for some open sets $V_n$. Indeed, fix a positive integer $n$. Then, consider the sets $\{ B(y,1/n)\}_{y\in K}$. By compactness of $K$ we get a finite subcover, and so we can define $V_n$ to be the union of this finite subcover. We see that $K\subset \bigcap_{n=1}^\infty V_n$. Now, if $x\in V_n$ for each $n$, then we have a sequence $(y_n)$ with $y_n \in K$ and $|x-y_n|<1/n$. It follows that $y_n \to x$ as $n\to \infty$, so $x \in K$ as $K$ is closed. This proves the claim. Finally, the last step I needed in my proof was to prove the existence of the set $U\supset K$ with $\mu(U\setminus K)<\epsilon$. Well, for given $\epsilon>0$, continuity from above implies that $$\mu(K) = \mu\left(\bigcap_{n=1}^\infty V_n\right) = \lim_{n\to \infty} \mu\left(\bigcap_{k=1}^n V_k\right)$$ (which is justified as $\mu(V_1)<\infty$; it is just a finite union of bounded open sets). Hence, the desired $U=\bigcap_{k=1}^n V_k$ for large enough $n$ because $\mu(V_n\setminus K)=\mu(V_n) - \mu(K) < \epsilon$.
2019-07-23T03:26:32
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https://dsp.stackexchange.com/questions/73367/understanding-the-twiddle-factors/73372
# Understanding the twiddle factors So I've been looking at this butterfly diagram to try to understand it better: And I am trying to get a good understanding of the twiddle factors. The definition is given as: FFT Twiddle Factor: $${e^{i2{\pi}k/N}}$$ and IFFT Twiddle Factor: $${e^{-i2{\pi}k/N}}$$ So k is the index number of the iteration thus $$k=0,1...N$$ but its $$N$$ that I am unsure of. From the image is the first stage N = 8 (since there are 8 butterflies) or is N = 2 since each butterfly only spans two elements? Or is N always 16 every pass? My guess currently is: Stage 0 : N = 8 Stage 1 : N = 4 Stage 2 : N = 2 Stage 3 : N = 1 Am i correct in this assumption? If not i hope some one clarifying my misunderstanding here. Thanks • In your above butterfly diagram, shouldn't the indexing of your input sequence be in "bit reversed" order? – Richard Lyons Feb 24 at 9:29 • @RichardLyons yes i didn't make the diagram it was from the internet, i have since found a 8 point FFT diagram to refer from which have the correct indices reversed for the first stage. Made it a bit less overwhelming to disset. – WDUK Feb 24 at 9:38 I think there is a better way of writing the twiddle factor. Instead of using a different "basis" for each stage, you can use the FFT length as the base for all twiddle factors and the only thing that changes between stages is the step size. Stage 0: $$W_{16}^0$$ Stage 1: $$W_{16}^0, W_{16}^4$$ Stage 2: $$W_{16}^0, W_{16}^2,W_{16}^4, W_{16}^6$$ Stage 3: $$W_{16}^0, W_{16}^1, .... W_{16}^7$$ We are using the property here that for example $$W_2^1 = W_4^2 = W_8^4 = W_{16}^8$$ or more general $$W_a^b = W_{n \cdot a}^{n \cdot b}, n \in \mathbb{N}$$ Once you decide to using the FFT length as the basis for the twiddle factors you can just drop the $$16$$ from the notations and things become a lot easier to read and understand. Here is the complete list. Stage 0: $$W^0,W^0,W^0,W^0,W^0,W^0,W^0,W^0$$ Stage 1: $$W^0, W^4,W^0, W^4,W^0, W^4,W^0, W^4$$ Stage 2: $$W^0, W^2,W^4, W^6, W^0, W^2,W^4, W^6$$ Stage 3: $$W^0, W^1,W^2,W^3,W^4,W^5,W^6, W^7$$ That's actually how most code is implemented. You don't build a twiddle factor table for each stage, you just build one for the highest order and each stage uses a different step size (modulo N/2) to step through the table. • yeah, that's how I've learned it, too. Nice. – Marcus Müller Feb 22 at 13:49 • Nice just computed the needed twiddle factors with this approach, this was so much easier. Now i just need to figure out the math to map index to correct twiddle factor for each stage for the FFT since i now only have an array of twiddles of only length 7. – WDUK Feb 22 at 23:29 You got it mostly backwards, but otherwise OK :) This can be pretty directly answered by writing down what the Cooley-Tukey FFT's twiddle factor $$W_N$$ is: $$W_N=e^{-i\frac{2\pi}{N}}$$ and that's it. For example, in your picture, in Stage 0, you're multiplying with $$W_2^0=\left(e^{-i\frac{2\pi}{N}}\right)^0=e^{-i\frac{2\pi0}{N}}=e^0=1$$. So k is the index number of the iteration No, $$k$$ is the exponent of your twiddle factors. It goes from 0 to half the size of your transform (in each stage). From the image is the first stage N = 8 (since there are 8 butterflies) or is N = 2 Stage 0 has subtransforms of size N=2, as you can see. (Also, just write down the 2-DFT's formula and compare.) • Ah I see my mistake with the sign in my exponent now. Okay I understand now what k is kinda but I'm not sure I understand the pattern relating each stage to how far k iterates. That being 0 in stage 0 then 0,1 in stage 1 and 0,1,2,3 for stage 2 and so on... That's the only part that kinda remains elusive to my understanding now. – WDUK Feb 22 at 10:27
2021-08-03T18:25:48
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https://math.stackexchange.com/questions/1048908/coefficients-for-terms-of-power-series
# coefficients for terms of power series I'm asked to represent the function $\displaystyle \frac{2 x}{10 + x}$ as a power series $f(x) = \displaystyle \sum_{n=0}^\infty c_n x^n$ I found this to be $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$ The question is asking for $\ c_0, c_1, c_2, c_3,$ and $\ c_4$ I entered $\displaystyle c_0 = \frac{1}{5}, c_1 = \frac{-1}{50}, c_2 = \frac{1}{500}, c_3 = \frac{-1}{5000},$ and $\displaystyle c_4 = \frac{1}{50000}$ but I am told this is incorrect. Can someone please explain what I'm doing wrong? Thanks! UPDATE: using another formula I was able to find $\displaystyle c_n = \frac{f^{n}(a)}{n!}$ so $\displaystyle c_0 = 0, c_1 = \frac{1}{5}, c_2 = \frac{-1}{50},$ etc.. which are the correct answers. So I'm still confused.. Assuming I calculated the power series representation correctly to get $\ c_0$ don't you just let n=0 and take the coefficient of the resulting term? UPDATE: $\ c_n$ is actually the coefficient for the $\ x^n$th term. So $\ c_0$ is the constant term. This helped me for another question I had where the terms were $\ c_0 + c_2x^2 + c_4x^4 \cdots$ and I didn't understand why the coefficient $\ c_1$ was 0 You were asked to find a series in the form $$\displaystyle \sum_{n=0}^\infty c_n x^n$$ but your answer is $$\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$$ which is in the form $$\displaystyle \sum_{n=0}^\infty c_n x^{n+1}$$ Do you see the difference? The power of $x$ is off by one. If you correct for that off-by-one error, you will get your new $c_0$ is $0$, your new $c_1$ is your old $c_0$, and so on. That is exactly the correct answer that you got the other way. If you need a general formula for your new $c_n$ you could use $$c_n = \begin{cases} 0, & n=0 \\[2ex] \displaystyle \frac{(-1)^{n-1}}{5 \cdot 10^{n-1}}, & n>0 \end{cases}$$ Note that I changed the $n$'s to $n-1$'s in the formula for $c_n$ for this to work. It would be difficult to write that into an explicit series, as you did for $\sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$. But you do not need such a formula, since you were only asked for $c_0$ through $c_4$. • Thanks I see what you mean.. Can you explain how I change the series to write it in the correct form? Dec 2, 2014 at 23:18 Hint: $\dfrac{2x}{10+x} = \dfrac{x}{5}\cdot \dfrac{1}{1-\left(-\frac{x}{10}\right)}$, and use the well-known geometric series: $\dfrac{1}{1-a} = 1 + a + a^2+....$ for $|a| < 1$. • Thanks but I already did this much to determine the power series representation (did I do it wrong?).. Dec 2, 2014 at 22:03 • I think you got it right. Dec 2, 2014 at 22:04 • What about the coefficients of the first 5 series terms? I don't understand why I'm being told they're wrong Dec 2, 2014 at 22:07
2022-07-06T17:08:43
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https://math.stackexchange.com/questions/237779/i-m-ab-is-invertible-if-and-only-if-i-n-ba-is-invertible
# $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible. The problem is from Algebra by Artin, Chapter 1, Miscellaneous Problems, Exercise 8. I have been trying for a long time now to solve it but I am unsuccessful. Let $A,B$ be $m \times n$ and $n \times m$ matrices. Prove that $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible. Please provide with only a hint. ## 4 Answers Another method: Let $C= {(I_m - AB)}^{-1}$. The matrix $BCA$ is $n\times n$. $(I_n - BA)(BCA) = BCA - BABCA$ $= B(C - ABC)A$ $= B[(I_m - AB)C]A$ $= B(I_m)A\$ (by definition of $C$) $= BA$ Hence, $(I_n-BA)(BCA + I_n) = (I_n - BA) (BCA) + (I_n -BA) = BA + (I_n-BA) = I_n$ So, we get that the inverse of $I_n - BA$ is $I_n +BCA$. • This is how I did the problem the first time, but then I forgot how I did it and was stuck. :P Thanks :-) – user14082 Nov 15 '12 at 7:41 There's no real need to actually invoke Sylvester's determinant theorem (although that certainly is much faster). First show that the (non-zero) eigenvalues of $AB$ and the eigenvalues of $BA$ coincide. If you take the determinant of $I_m - AB$, then you have the characteristic polynomial of $AB$ evaluated at $\lambda = 1$. It follows that the determinant is zero if and only if $1$ is an eigenvalue of $AB$ if and only if $1$ is an eigenvalue of $BA$ if and only if $\det(I_n - BA) = 0$. Note that a slight adaptation of this argument also provides a proof of Sylvester's determinant theorem different from the one given on Wikipedia. • Thank you for another nice answer. – user14082 Nov 15 '12 at 6:53 • I like this proof so much better than the others. – Sahiba Arora Nov 8 '15 at 12:59 Hint $\$ It's a consequence of Sylvester's determinant identity $\rm\:det(1 + AB) = det(1+BA),\:$ which has a very simple universal proof: over the polynomial ring $\rm\ \mathbb Z[A_{\,i\,j},B_{\,i\,j}\,]\$ take the determinant of $\rm\, (1+AB)\, A = A\, (1+BA)\$ then cancel $\rm\, det(A)\$ (valid since the ring is a domain). Extend to non-square matrices by padding appropriately with $0$'s and $1$'s to get square matrices. Note that the proof is purely algebraic - it does not require any topological notions (e.g. density). Alternatively $\$ Proceed by way of Schur decomposition, namely $$\rm\left[ \begin{array}{ccc} 1 & \rm A \\ \rm B & 1 \end{array} \right]\ =\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm 0 & \rm 1-BA \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]$$ $$\rm\phantom{\left[ \begin{array}{ccc} 1 & \rm B \\ \rm A & 1 \end{array} \right]}\ =\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]\ \left[ \begin{array}{ccc} \rm 1-AB & \rm 0 \\ \rm 0 & \rm 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]$$ See my posts in this sci.math thread on 09 Nov 2007 for further discussion. We know that a square matrix $M$ is invertible if and only if $$\{\vec x; \vec x M = \vec 0\}=\{\vec 0\}.$$ (The corresponding homogeneous linear system has only trivial solution. The corresponding linear map has trivial kernel.) So the claim $$I_m-AB\text{ is invertible }\qquad\implies\qquad I_n-BA\text{ is invertible}$$ is equivalent to $$\{\vec x\in F^m; \vec x(I-AB)=\vec 0\}=\{\vec0\} \qquad\implies\qquad \{\vec y\in F^n; \vec y(I-BA)=\vec 0\}=\{\vec0\}.$$ (Where $F$ denotes the base field.) Proof. So let us assume that $I-AB$ is invertible, i.e., that $\vec x=\vec 0$ is the only vector such that $\vec x(I-AB)=\vec 0$. Let $\vec y(I-BA)=\vec 0$. Then we get $$\vec y=\vec yBA\tag{1}$$ and, consequently $\vec y B = \vec y BAB$. This is equivalent to $$\vec yB(I-AB)= \vec 0.\tag{2}$$ This means that $\vec yB=\vec 0$, since $I-AB$ is invertible. From (1) we now get that $\vec y=\vec yBA=\vec 0A=\vec 0$. $\square$
2019-05-27T01:49:53
{ "domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/237779/i-m-ab-is-invertible-if-and-only-if-i-n-ba-is-invertible", "openwebmath_score": 0.9747653603553772, "openwebmath_perplexity": 119.039975380849, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9632305328688784, "lm_q2_score": 0.8670357649558006, "lm_q1q2_score": 0.8351553218947514 }
https://math.stackexchange.com/questions/2902731/logic-behind-bitwise-operators-in-c/2902771
# Logic behind bitwise operators in C I came across bitwise operations in C programming, and I realized that XOR operator can be used to swap 2 numbers in their binary bases. For example let $$i=(65)_{10}=(1000001)_{2}, \text{ and } j=(120)_{10}=(1111000)_{2}$$. Let $\oplus$ be the XOR operator, then observe that if I started with any one of them, say $i$ and following the following procedure: 1)replace its value with the $\oplus$ value, yielding $$i=(0111001)_{2},j=(1111000)_{2}$$ 2) replace the other variable($j$) with another $\oplus$ value derived from the new $i$ and old $j$, yielding $$i=(0111001)_{2},j=(1000001)_{2}$$ 3)replace the original variable $i$ with $\oplus$ value again, yielding $$i=(1111000)_{2},j=(1000001)_{2}$$ which shows that we would somehow have their values swapped. I found this way of programming online and I definitely can’t understand how people think of the logic aspect of this. I would think it’s linked to the truth table as follows, which shows by division of cases that the values can be swapped. However, I am still uncertain about the full reasoning why this works, like whether there is any mathematical theorems that I should know that can aid me in my understanding. PS: Sorry if the question is off-topic here, it feels like a programming question, but I feel that I more concerned about the “logic” rather than the programming. I also drew the table myself on MS word since I can't get the latex one to work somehow. • Beware that this doesn't work if i and j happen to be the same variable! – Henning Makholm Sep 2 '18 at 13:59 • @HenningMakholm ah ok noted, applying it 3 times has the same effect as applying 1 time and that will cause one of the values to be full of zeroes – Prashin Jeevaganth Sep 2 '18 at 14:03 • It works for $i = j$ too. – mbjoe Sep 2 '18 at 14:11 • @mbjoe oh ok just noticed – Prashin Jeevaganth Sep 2 '18 at 15:09 • @mbjoe: It works for i and j having the same value, but not for them being the same variable. – celtschk Sep 2 '18 at 15:59 In algebraic terms, the XOR operator (or $\oplus$) is nothing other than addition modulo $2$: use $1$ and $0$ for true and false, along with $1 \oplus 1 = 0$. Now, since addition modulo $2$ is associative and commutative, and both elements are their own inverses, we have \begin{align} d &= b \oplus c\\ &= b \oplus (a \oplus b)\\ &= b \oplus (b \oplus a)\\ &= (b \oplus b) \oplus a\\ &= a.\\ \end{align} We can show $e = b$ using similar reasoning. • I prefer this to the accepted answer, since it mentions associativity and commutativity. Also, while obvious, I imagine the note that XOR is the same as addition modulo 2, might be helpful to readers. – Demosthenes Sep 3 '18 at 11:33 Note that you can do the same thing without bitwise operators (at least for unsigned integer types since they can't overflow into undefined behavior): // i == x j == y i += j; // i == x+y j == y j -= i; // i == x+y j == -x i += j; // i == y j == -x j = -j; // i == y j == x Now if we do this bit for bit, but modulo 2 instead of modulo UINT_MAX+1, the XOR operation implements both addition and subtraction, and the final negation is a no-op because $-1\equiv 1$ and $-0\equiv 0 \pmod 2$. So what is left in the bitwise version is exactly i ^= j; j ^= i; i ^= j; • Thanks for the alternative solution to swap 2 numbers, this is insightful. – Prashin Jeevaganth Sep 2 '18 at 14:08 You already answered your question, but if you want an algebraic explanation note that for any $x$: $$x \oplus 0 = x$$ $$x \oplus x = 0$$ So: $$i_0 = i, j_0 = j$$ $$i_1 = i_0 \oplus j_0, j_1 = j_0$$ $$i_2 = i_1, j_2 = i_1 \oplus j_1 = i_0 \oplus j_0 \oplus j_0 = i_0$$ $$i_3 = i_2 \oplus j_2 = i_1 \oplus i_0 = i_0 \oplus j_0 \oplus i_0 = j_0, j_3 = j_2 = i_0$$
2019-12-10T00:05:05
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http://math.stackexchange.com/questions/99476/mathematical-induction-and-the-product-of-odd-numbers-is-odd
# Mathematical Induction and “the product of odd numbers is odd” I am extremely poor at proofs and logical manipulation so I am stuck on a lot of these questions especially induction. The question below I have been stuck at for a little over 1 hour and I can't get it, I have tried various substitutions and forms but I am stuck.. Using mathematical induction: An integer is odd if it can be written as $n=2k+1$. Use induction to prove that the product of $m$ many odd integers is odd for every $m \geq 2$. I have tried doing $(2k+1)(2(k+1)+1)$ and substituting $k=1$ for my basis step after that I am lost. - Try doing the induction on $m$, not on $k$. – Gerry Myerson Jan 16 '12 at 6:04 @Raynos People have provided answers to your problem so here is a little joke that may cheer you up. Three people (mathematician, physicist, engineer) have been asked to prove that the product of $m$ odd integers for $m\geq 2$ is odd. Physicist: Within experimental error, the product is always odd. Mathematician: By mathematical induction, the product is always odd. Engineer (Most direct proof): An odd number times an odd number is odd, so this odd number times another odd number is odd, etc... – user38268 Jan 16 '12 at 8:01 For the base case, we consider when $m=2$. For $a,b \in \mathbb{Z}$, that is, $(2a +1)(2b +1) = 4ab +2a+2b +1 = 2(2ab + a + b) +1$, where $2ab + a + b \in \mathbb{Z}$, so it is true for $m=2$. By induction, we assume for arbitrary $k \in \mathbb{Z}$ that, $$\prod_{i=1}^{k}(2a_{i} +1)=2l+1$$ where $l \in \mathbb{Z}$. Since by our basis step we have that two odd numbers multiplied together gives two odd numbers, then for the $k+1^{th}$ odd integer $c$, we have that $$(2l+1)(2c+1)=2(2lc + l + c) +1$$ So, since $2lc + l + c \in \mathbb{Z}$, we have that $$\prod_{i=1}^{k+1}(2a_{i}+1)=2j+1$$ where $j \in \mathbb{Z}$. Therefore, by the Principle of Mathematical Induction, the product of $m$ odd integers is again odd. - I'm clearly too tired for this, editing now... – Samuel Reid Jan 16 '12 at 6:30 @Srivatsan: How is the notation now? – Samuel Reid Jan 16 '12 at 6:34 Looks good now. :) – Srivatsan Jan 16 '12 at 6:35 Thank you guys for helping me out! I am understanding this a little more now. – Raynos Jan 16 '12 at 12:36 The way induction works is you show that some property, $P(n)$ holds for $n=1$, and you also show that if $P(n)$ holds for some $n$, then it also holds for $P(n+1)$. In this way you've shown that $P(1)$ holds, by the second condition $P(1+1)=P(2)$ holds, etc. This proves that the property $P(n)$ holds for all $n \in \mathbb{N}$. The first thing that you want to show is the base case, which corresponds to m=2. To show this you just need to prove that the product of two integers is odd. We may write the product of two odd integers as $$(2k_{1}+1)(2k_{2}+1)=4k_{1}k_{2}+2k_{1}+2k_{2}+1=2(2k_{1}k_{2}+k_{1}+k_{2})+1$$ Since $2k_{1}k_{2}+k_{1}+k_{2} \in \mathbb{Z}$, this finishes the base case. Next we need to prove that if the product of $m$ integers is odd, then the product of $m+1$ integers is also odd. Informally, this is true because if there are no factors of two in the first $m$ integers, and there isn't a factor of two in the next integer, there won't be a factor of two anywhere in the product, and the product of $m+1$ integers will be odd. Rigorously, we need to first assume that the product of $m$ odd integers is odd. Then we note that the product of $m+1$ odd integers looks like this: $\displaystyle \prod_{i=1}^{m+1} a_{i}=a_{m+1}\prod_{i=1}^{m} a_{i}$, $a_{i}$ odd for $1 \le i \le m+1$. Since we are assuming the product of $m$ integers is odd, we can write the product $\displaystyle \prod_{i=1}^{m} a_{i}=2\alpha+1$, $\alpha \in \mathbb{Z}$. Since $a_{m+1}$ is odd, we may write $a_{m+1}=2\beta+1$, $\beta \in \mathbb{Z}$. Thus, $\displaystyle \prod_{i=1}^{m+1} a_{i}=(2\beta+1)(2\alpha+1)=4\alpha\beta+2\alpha+2\beta+1=2(2\alpha\beta+\alpha+\beta)+1$, which is odd. This completes the proof. HTH! - I think you did more than you needed to at the end. Once you reduce the product of m+1 odd integers to the product of 2 odd integers, you've already proven for the base case that the product of 2 odd integers is odd. – Mike Jan 16 '12 at 8:16 Yea, I left that part out at first since they are essentially the same proof, but when I think when you're answer questions you need to appeal to the OP. I don't think the added thoroughness is a bad thing for someone taking their first proof's course. – Jackson Jan 16 '12 at 10:24 HINT $\rm\ mod\ 2\$ it is simply $\rm\ 1^n\equiv 1\$ with base $1\equiv 1$ and induction $\rm 1^n\equiv 1\ \Rightarrow\ 1^{n+1}\equiv 1\cdot 1\equiv 1\:.$ Without mod arithmetic it's still simple, the inductive step $\ 1\cdot 1\equiv 1\$ is odd $\cdot$ odd = odd, since $$\rm\ (1 + 2\ j)\ (1 + 2\ k)\ =\ 1 + 2\ (j + k + 2\ j\ k)$$ -
2016-05-06T11:38:36
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https://brilliant.org/discussions/thread/all-road-lead-to-and-from-rome/
Everyone has heard of the famous Königsberg bridge problem, where one must traverse all seven bridges and end up back where one started without traversing a bridge twice. This is indeed impossible, as dictated by Graph Theory. However, loosening the restriction to we must traverse each bridge exactly twice, we see that it indeed becomes possible: imgur ##### Image credit Jake Lai So, is it possible, with these new rules, to successfully traverse through any set of paths (i.e graph)? Note by Daniel Liu 3 years, 7 months ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ Sort by: Yes, it's always possible as long as the graph is connected; moreover, it's always possible to find such tour where each edge is traversed once in each direction. The proof is also simple: just double all the edges (and orient them, for the strengthening), and see that the produced graph is Eulerian by inspecting its degrees; any such Eulerian tour has the property that it uses each original edge twice (since each original edge is doubled into two edges, where each of it is used). To find one such path, use some algorithm. - 3 years, 7 months ago Nice! Trémaux's Maze solving algorithm is also related to this. - 3 years, 7 months ago Indeed, my first idea was to use depth-first search (which is essentially a variation of Tremaux's maze solving algorithm, or the other way around), but marking edges instead of vertices; this doesn't guarantee that the result is Eulerian, however (some edges might be missed). Thus we can extend the resulting tour to cover all edges, and that's Hierholzer's algorithm. - 3 years, 7 months ago Credit me~ - 3 years, 7 months ago OOPS I'm so sorry :( Done - 3 years, 7 months ago I was just kidding :p - 3 years, 7 months ago
2019-01-22T11:39:56
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https://math.stackexchange.com/questions/973811/union-of-a-finite-number-of-open-sets-is-open-or-not-proper-usage-of-this-fact
Union of a finite number of open sets is open or not? Proper usage of this fact a proof I received a homework assignment back and I was given full credit on the following proof: Let $S = \{ (x,y) \in \mathbb{R}^{2} | x \geq 1$ and $y \geq 1 \}$. Is $S$ closed? My proof is below but after reviewing a particular proposition in our book (listed below my question), I am no longer certain why I received full credit. I hope someone here can help shed light on my confusion. Let $A = \{ (x,y) \in \mathbb{R}^{2} | x < 1 \}$ and $B = \{ (x,y) \in \mathbb{R}^{2} | y < 1 \}$. The union of these two sets is exactly the complement of $S$. That is, $\mathbb{R} \backslash S = A \cup B$. The distance of any $x$ value for $(x,y) \in A$ to the line $x = 1$ is always $1-x$. I want an open ball (denoted $A_{r}(p)$ ) centered around a point $(x,y) = p \in A$ with radius $r>0$ to be completely contained in $A$ (to show $A$ is open). Letting $r=1-x$ would not suffice since this $r$ would guarantee points on the line $x=1$ would be included in $A_{r}(p)$. Choosing $r=\frac{1-x}{2}$ ensures that any open ball centered around some $p \in A$ is completely contained in $A$, i.e. $A_{r}(p) \subset A$. Hence $A$ is open. A simillar argument can be made for $q \in B$ letting $r = \frac{1-y}{2} \Rightarrow B_{r}(q) \subset B$. So the set $B$ is open as well. This next line is where I think I made a mistake. Since $A$ and $B$ are both open, $\mathbb{R}^{2} \backslash S = A \cup B$ is open which implies $S$ is closed. QED There is a proposition in my book that states (1) "the intersection of a finite number of open subsets of $M$ is open" and (2) "the union of an arbitrary collection of open subsets of $M$ is open. My proof hinges on the union of a finite number of open sets being open, which seems to contradict with this proposition. So either (a) I am misunderstanding this proposition or (b) my proof is wrong. Any insight into this would be appreciated! • The book is Marsden, Elementary Classical Analysis 2nd ed and this proposition is on page 106. • "Arbitrary collection" includes "finite collection"s. – Daniel Fischer Oct 14 '14 at 19:24 • The union of an arbitrary collection of open sets being open is significantly stronger than finite unions being open, but it certainly implies it. – JHance Oct 14 '14 at 19:24 • JHance, Daniel Fischer, so the intersection of an arbitrary collection of closed sets being closed would also imply that a finite intersection of closed sets is closed? Thank you both! – Nidia Oct 14 '14 at 19:28 • The meaning of arbitrary is not arbitrary. It is straightforward to show that the union of a finite number of open sets is open. If a point is in the union, it must be at least in one open set. – copper.hat Oct 14 '14 at 19:29 • Yes, here, 'arbitrary' if the same as 'of any kind', that is, finite or infinite. – ajotatxe Oct 14 '14 at 19:29
2020-01-27T22:05:49
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http://mathhelpforum.com/trigonometry/34258-couple-questions-thanks.html
# Math Help - a couple of questions...thanks 1. ## a couple of questions...thanks I dont know how to find the exact value of something(trig)...e.g. sec(13Pi/6) I get a long decimal(1.31...) but how am I meant to get the exact value answer which in this case is 2/3 of the square root of 3. (which is 1.31 when u calculate it but when I see 1.31 how am I meant to know that its 2/3 the square root of 3? question 2: 3(tan^2 )theta - sec theta = 1 for 0 <= theta =< 2PI can anyone solve that for me? thanks oh and for the following graph sketches: y = sin3x y = |sin3x| y = sin|3x| so like I was confused to whether they amplitude was 3 times bigger or the frequency is like 3 times smaller for them. can someone tell me what the difference of those curves are from a normal sinx 2. Originally Posted by grammar I dont know how to find the exact value of something(trig)...e.g. sec(13Pi/6) I get a long decimal(1.31...) but how am I meant to get the exact value answer which in this case is 2/3 of the square root of 3. (which is 1.31 when u calculate it but when I see 1.31 how am I meant to know that its 2/3 the square root of 3? [snip] $\sec \left(\frac{13 \pi}{6} \right) = \sec \left( \frac{(12 + 1) \pi}{6} \right) = \sec \left(\frac{12 \pi}{6} + \frac{\pi}{6}\right)$ $= \sec \left(2 \pi + \frac{\pi}{6}\right) = \sec \left(\frac{\pi}{6}\right) = \frac{1}{\cos \left(\frac{\pi}{6}\right)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$. 3. Originally Posted by grammar [snip] question 2: 3(tan^2 )theta - sec theta = 1 for 0 <= theta =< 2PI can anyone solve that for me? thanks [snip] Substitute (from the Pythagorean Identity) $\tan^2 \theta = \sec^2 \theta - 1$: $3 (\sec^2 \theta - 1) - \sec \theta = 1 \Rightarrow 3 \sec^2 \theta - \sec \theta - 4 = 0$. Let $x = \sec \theta$ to see how to factorise: $3x^2 - x - 4 = 0 \Rightarrow (3x - 4)(x + 1) = 0$. Therefore either $x = \frac{4}{3}$ or $x = -1$. Therefore either $\sec \theta = \frac{4}{3}$ or $\sec \theta = -1$. The latter can be exactly solved easily. The former can only be found as either a generic exact answer or a decimal approximation. 4. Originally Posted by grammar [snip] oh and for the following graph sketches: y = sin3x y = |sin3x| y = sin|3x| so like I was confused to whether they amplitude was 3 times bigger or the frequency is like 3 times smaller for them. can someone tell me what the difference of those curves are from a normal sinx Draw the graph of $y = |\sin (3x)|$ by reflecting around the x-axis the parts of the graph of $y = \sin (3x)$ that lie below the x-axis. So there are salient points at $(0, \, \pm n \pi)$ where n is an integer. Draw the graph of $y = \sin |3x|$ by reflecting around the x-axis the part of the graph of $y = \sin (3x)$ that lies to the left of the y-axis. So there's a salient point at (0, 0). Note: $\sin |3x| = \sin (3x)$ for $x \geq 0$ and $\sin |3x| = \sin (-3x) = -\sin (3x)$ for $x < 0$. Salient point: Salient Point -- from Wolfram MathWorld
2016-07-25T13:17:00
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https://math.stackexchange.com/questions/3079595/theoretical-questions-about-characteristic-polynomials-diagonalizability-rank
# Theoretical questions about characteristic polynomials, diagonalizability, rank and similarity I am new to Linear Algebra and would like some feedback regarding my answer to the following question A and B are two square matrices of order n. Prove or refute (with a counterexample) the following statements: 1. If A is real and the algebraic multiplicity of each one of its eigenvalues is 1, then A is diagonalizable over R. 2. If A and B have the same characteristic polynomial, then they have the same rank as well. 3. If A and B are row-equivalent, then they have the same characteristic polynomial. 4. If A and B are similar, then $$A^k$$ and $$B^k$$ are similar (k is a natural number). 1. $$1 \leq$$ geometric multiplicity $$\leq$$ algebraic multiplicity If the algebraic multiplicity of each eigenvalue is 1, then $$1 \leq$$ geometric multiplicity $$\leq$$ 1, therefore the geometric multiplicity is 1. This means that the geometric multiplicity and the algebraic multiplicity are equal, and therefore the matrix is diagonalizable over R. 2. The statement is incorrect. Counterexample: A=$$\begin{bmatrix}0&1\\0&0 \end{bmatrix}$$ A's rank is 1, and its characteristic polynomial is $$λ^2$$. B= $$\begin{bmatrix} 0&0\\0&0 \end{bmatrix}$$ B's rank is 0, and its characteristic polynomial is $$λ^2$$. 3. The statement is incorrect. Counterexample: A= $$\begin{bmatrix} 2&0\\0&2 \end{bmatrix}$$ B= $$\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$$ These two matrices are row equivalent, but A's characteristic polynomial is (λ-2)(λ-2), while B's characteristic polynomial is (λ-1)(λ-1). 4. Correct: by definition A and B are similar if and only if there is an invertible matrix P, which satisfies $$P^{-1}AP=B$$ If $$P^{-1}AP=B$$, then $$A=PBP^1$$ $$A^k=PDP^{-1}·PDP^{-1}…PDP^{-1}$$, which repeats itself k times. As $$P^{-1}·P$$ cancel out, $$A^k=PB^kP^{-1}$$. Therefore $$P^{-1}A^kP=P^{-1}PB^kP^{-1}P=B^k$$ Thank you! • This all looks fine to me ! – J.F Jan 19 at 17:41 • Your answer to 1 is correct if we assume that all eigenvalues of $A$ are real. Otherwise, it is false. All the rest is correct (minor issue: your argument for 4 requires $k > 0$, but of course the $k = 0$ case is trivial). – darij grinberg Jan 19 at 18:19 • @darijgrinberg Thank you! But in the question it is given that A is real and it asks if A is diagonalizable over R, so it should not be a problem, right? – dalta Jan 19 at 18:26 • All seems good to me! – Yuval Gat Jan 19 at 18:44
2019-05-26T19:14:59
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https://math.stackexchange.com/questions/590396/integration-of-sqrt1u2-du
# Integration of $\sqrt{1+u^2}\,du$ Can anyone show how to calculate $$\int\sqrt{1+u^2}\,du?$$ I can't calculate it. • Calculate would be a better word. Sorry for confusing you. – CALC-FATH Dec 3 '13 at 2:57 • Stewart tells us to do a trig sub but $\sec^3\theta$ leaves a bit of work...Kaster's approach is much more efficient for whoever actually tries to carry out the computation til the end. – Julien Dec 3 '13 at 3:10 • I don't normally think of this integral as anything special, but having glanced at the responses below I'm filing this one away as a future teaching tool, since it so wonderfully demonstrates the versatility of integration tactics. – David H Dec 3 '13 at 3:45 Setting $$u=\sinh x:=\frac{e^x-e^{-x}}{2},$$ we have \begin{eqnarray} \int\sqrt{1+u^2}\,du&=&\int\cosh x\sqrt{1+\sinh^2x}\,dx=\int\cosh^2x\,dx=\frac12\int(1+\cosh2x)\,dx\\ &=&\frac{x}{2}+\frac14\sinh2x+C=\frac{x}{2}+\frac12\sinh x\cosh x+C\\ &=&\frac12\ln(u+\sqrt{1+u^2})+\frac12u\sqrt{1+u^2}+C. \end{eqnarray} • While the tan substitution is the most obious one, I really like this one too. Because it is more (or should I say different) work. Versatility counts too. So upvote! – imranfat Dec 3 '13 at 3:10 Let $u = \sinh x$, then \begin{align} \int \sqrt{1+u^2}du &= \int \cosh x \cdot \cosh x dx = \int \cosh^2x dx = \int \frac {1+\cosh 2x}2 dx = \\ &= \frac x2 + \frac {\sinh 2x}4 + C = \frac {\text{arcsinh } u}2 + \frac {u \sqrt{1+u^2}}2 + C \end{align} ## PS To verify: WA Let $\displaystyle I = \int \sqrt{1+u^2}du = \int \sqrt{1+u^2}\cdot 1\;du$ Using Integration by parts $\displaystyle I = \sqrt{1+u^2}\cdot u-\int \frac{u}{\sqrt{1+u^2}}\cdot udu = \sqrt{1+u^2}\cdot u-\int\frac{(1+u^2)-1}{\sqrt{1+u^2}}du$ $\displaystyle I = \sqrt{1+u^2}\cdot u-I+\int\frac{1}{\sqrt{1+u^2}}du$ $\displaystyle 2I = \sqrt{1+u^2}\cdot u+J$ where $\displaystyle J = \int \frac{1}{\sqrt{1+u^2}}du$ for Calculation of $J$ Let $1+u^2 = v^2$ and $\displaystyle udu = vdv\Rightarrow \frac{du}{v} = \frac{dv}{u} = \frac{d(u+v)}{(u+v)}$ (above Using ratio and Proportion ) So $\displaystyle J = \int\frac{du}{v} = \int\frac{d(u+v)}{(u+v)} = \ln \left|u+v\right|+C$ So $\displaystyle J = \ln \left|u+\sqrt{1+u^2}\right|+C$ So $\displaystyle I = \frac{u}{2}\sqrt{1+u^2}+\frac{1}{2}\ln \left|u+\sqrt{1+u^2}\right|+D$ To integrate this we would use Trigonometric substitution. Recall your substitution rules for this. In this case, we would let $u = 1\tan(\theta)$ and then continue to find $d\theta$ so we may do the full substitution. • It'd be worth noting how you can figure out these substitution relationships by using the $sin^2 + cos^2 = 1$ identity. Then you don't have to memorize anything ;) – GraphicsMuncher Dec 3 '13 at 3:01 • I really would like to see how you'll proceed after that! – Mercy King Dec 3 '13 at 3:16 • @Mercy It's kinda like your hyperbolic substitution, but ends up with the integral of $\sec^3x$. If you've either a) memorized the integral of $\sec^3x$, or b) become used to computing integrals like that, so it doesn't phase you, then you're all set. :) – apnorton Dec 3 '13 at 3:33 Let $u = \tan \theta$. Then, $\mathrm{d}u = \sec^2 \theta \, \mathrm{d}\theta$. Also, note that $$\sqrt {1 + u^2} = \sqrt {1 + \tan^2 \theta} = \sec \theta.$$Use this to finish. • I really would like to see how you'll proceed after that! – Mercy King Dec 3 '13 at 3:16 • @Mercy I simply did not show my work because the OP hasn't shown any him/herself. But it becomes simple Integration by Parts after that. – Ahaan S. Rungta Dec 3 '13 at 3:18 • Are you sure about that? – Mercy King Dec 3 '13 at 3:20 • @Mercy Yes. I have tried it. – Ahaan S. Rungta Dec 3 '13 at 3:34
2020-06-01T17:30:29
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https://laurentlessard.com/bookproofs/the-deadly-board-game/?replytocom=1727
This Riddler classic puzzle involves a combination of decision-making and probability. While traveling in the Kingdom of Arbitraria, you are accused of a heinous crime. Arbitraria decides who’s guilty or innocent not through a court system, but a board game. It’s played on a simple board: a track with sequential spaces numbered from 0 to 1,000. The zero space is marked “start,” and your token is placed on it. You are handed a fair six-sided die and three coins. You are allowed to place the coins on three different (nonzero) spaces. Once placed, the coins may not be moved. After placing the three coins, you roll the die and move your token forward the appropriate number of spaces. If, after moving the token, it lands on a space with a coin on it, you are freed. If not, you roll again and continue moving forward. If your token passes all three coins without landing on one, you are executed. On which three spaces should you place the coins to maximize your chances of survival? Extra credit: Suppose there’s an additional rule that you cannot place the coins on adjacent spaces. What is the ideal placement now? What about the worst squares — where should you place your coins if you’re making a play for martyrdom? Here is my solution: [Show Solution] ## 17 thoughts on “The deadly board game” 1. MarkS says: Nice solution as usual! Minor issue: your characteristic-polynomial formula for p_k is missing a term with the kth power of the fifth root, -0.67032. Also, I prefer the following explanation of the recursion relation: if space k is the last stop, then p_k follows from the probabilities to have reached the next-to-last stop, which must be one of spaces k-6 through k-1. From any one of these, the probability to reach space k is 1/6. So it must be that p_k = (1/6)(p_(k-6)+…+p_(k-1)). We start at space zero with probability one, so p_0=1, and there are no negative spaces, so p_k=0 for k<0. 1. MarkS says: One more minor comment: at large k, p_k should be the reciprocal of the average roll; this is (1+2+3+4+5+6)/6 = 7/2, which agrees with what you got from the recursion relation. 1. Thanks for the comments! I added the missing root. Nice interpretation of 2/7 also, hadn’t thought of it that way. 2. ju2tin says: Wouldn’t the solution for 3 coins be 5, 6, 12 because, as your own graph shows, 12 is the third-highest probability space? You have a lower chance of landing on 4 than on 12. Similarly, for nonadjacent spaces, you would just read off the three highest non-adjacent values on the graph. They look to be 6, 12, 14 to me, but it’s hard to tell because the graph is small. Similarly, for martyrdom you’d take the lowest values: 1, 2, 3 (and 1, 3, 7 if non-adjacent). Or am I missing something? 1. The $p_k$ probabilities (probability that you’ll land on $k$) that I plotted in my first figure are not additive. For example, the probability that you land on 1, 2, or 3 is 50%, since it happens precisely when you roll a 1, 2, or 3 on your first roll. However, if you add $p_1+p_2+p_3$, you actually get 58.78%. The reason why you end up with a higher number is that by just adding the probabilities, you’re double-counting some of the events. For example, the probability that you land on 3 also accounts for the case where you landed on a 1 first and then rolled a 2. Using the inclusion-exclusion principle as I explained avoids double-counting. The lowest probability you can get is actually to choose 1, 2, 7, which gives a probability of survival of 47.5%; lower than the 50% you get by picking 1, 2, 3. 3. Dmytro Taranovsky says: With the help of a computer, here are the answers for 4 and 5 and 6 coins. Each probability is execution probability. 4 coins best 3,4,5,6 0.0926 worst 1,2,3,7 0.4020 nonadjacent: best 4,6,8,11 0.1889 worst 1,3,7,13 0.3734 5 coins best 2,3,4,5,6 0.0278 worst 1,2,3,7,8 0.3040 nonadjacent: best 4,6,9,11,13 0.1199 worst 1,3,7,13,19 0.2779 6 coins best 1,2,3,4,5,6 0.0 worst 1,2,3,7,8,13 0.2359 nonadjacent: best 4,6,8,11,13,15 0.0762 worst 1,3,7,13,19,25 0.2064 Note that there are some periodic patterns; I do not know whether the problem for n coins has a closed form solution. 1. neat — one obvious thing I didn’t think of when I wrote up my solution is that the problem should be solvable recursively as you add more coins by just performing a new O(n) search. Makes sense that the best k coins should be a subset of the best k+1 coins, and similarly for the worst k coins. I don’t know why I didn’t think of that! 1. hypergeometric says: Nice solution! How do you normalise the probabilities? 2. Hector Pefo says: 1. I’m not sure I understand your question. I’m suggesting the following iterative approach: first find $k$ to maximize $p_k$, call the optimal index $k_1$. Then, find $k$ to maximize $q_{k_1,k}$, call the optimal index $k_2$. And finally find $k$ to maximize $r_{k_1,k_2,k}$. What I did in my solution was to search over all $(k_1,k_2,k_3)$ to maximize $r_{k_1,k_2,k_3}$ directly. My approach is $O(N^3)$ whereas the iterative approach is $O(3N)$ instead. 1. Hector Pefo says: Right, I think I’m following you on all that. My point is that if you used the iterative approach for the question of how to maximize chance of survival with coins on three non-adjacent spaces you would have gotten k1 = 6, k2 = 4, and k3 = 8, which is not the correct answer. 4. @Hector Pefo — oh wow you’re absolutely right. What a fiendish problem! 5. David Newman says: Hello Laurent, thank you for this post, it is great. I agree that the right answer is to place the coins on 456. I am getting a slightly different answer than you for the survival chance. I got the same answers as you for the 1 and 2 coin cases, but now on 3 coins, I’m at around .801333. I’ve been puzzling on this for a bit and looked at your formula, which I think is: R456 = p4 + p5 + p6 – p4*p1 – p4*p2 – p5*p1 + p4*p1*p1 my way of looking at it was slightly different: define p5~4, the prob of getting 5 without first getting 4 as, p5~4= p5 – p4/6 = 343/1296 (same as p4) define p6~5~4, the prob of getting 6 without first getting 4 or 5 as, p6~5~4 = p6 – (p5~4)/6 – p4/6 = 12,691/46,656 — Rather than using p1 and p2 in the formula, I’m dividing by 6 to eliminate hitting 5 (that wasn’t previously a 4) and then rolling a 1, or hitting 4 and then rolling a 2. I think the difference between our answers is in the “p4 followed by 2” term in my p6~5~4, because I use -p4/6, the likelihood of throwing a 2 after getting 4, rather than your -p4*p2. that’s because I don’t want to eliminate the chance of hitting 4 then throwing a 1 and a 1, as I already eliminated 4 followed by 1 previously in defining p5~4. so I wind up with R456 = p4 +p5~4 + p6~5~4 = 37,387/46,656, which is around .801333 am I making an error? thanks, David 1. I think you DO want to eliminate the case where you roll a 4 then 1, 1, i.e. the case where you land on (4,5,6). In the formula you gave: p6~5~4 = p6 – p5~4/6 – p4/6 If we just take a look at whether you land on 4, 5, or 6, then: p6 includes the four cases: (~4,~5,6) (4,5,6) (4,~5,6) (~4,5,6) p5~4/6, i.e. rolling a 1 after 5~4, is the case (~4,5,6) finally, p4/6, i.e. rolling a 2 after a 4, is the case (4,~5,6). The thing we’re trying to calculate is p6~5~4, which is just (~4,~5,6). So by just writing p6~5~4 = p6 – p5~4/6 – p4/6, you’re not actually eliminating the (4,5,6) possibility. By just enumerating the possible ways of achieving p4, p5~4, and p6~5~4, it’s not difficult to see that the probabilities should actually be the same for all three. One way to see this is to notice that p4 can occur in 8 ways: (4) (13) (22) (31) (112) (121) (211) (1111) The probability of this is $\frac{1}{6} + 3\frac{1}{6^2} + 3\frac{1}{6^3} + \frac{1}{6^4} = \frac{343}{1296}$, same as what you found. Now p5~4 can also occur in 8 ways. Just take each sequence that yielded p4, and add 1 to the last roll! i.e. (5) (14) (23) (32) (114) (122) (212) (1112) The probability of this is the same as p4. Similarly, the ways of achieving p6~5~4 are found by again adding 1 to the last roll: (6) (15) (24) (33) (115) (123) (213) (1114) So clearly the probability should be the same again. 1. David Newman says: thank you Laurent. before writing to you I actually convinced myself 3 times that 343/432 was right. and each time I went back and said, ‘hang on a sec’ and looked again. and of course the elegance of p(4), p(5~4) and p(6~5~4) all being equal was so powerful. but I just had this blind spot that wouldn’t go away. I kept saying, well I’ve disallowed 4 followed by 1 and 5~4 followed by 1, so 4 followed by 1 and 1 is already disallowed. but it’s not because it is in the p(6) prob of 7^5/6^6 we’re backing out of. I was reading left to right, but really it has to be right to left -in other words, “we’re at 6, how could we have gotten here?” and what was really beautiful was your demo that 4, 5~4, and 6~5~4 are all formed by basically the same sequences. thanks again for taking the time on this. Best, David
2022-08-09T06:55:00
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https://math.stackexchange.com/questions/4216200/a-box-contain-an-unlimited-number-of-balls-mixed-of-five-different-colors-how
A box contain an unlimited number of balls (mixed) of five different colors. How many ways can a handful of one or more balls be randomly selected I have a question: A box contain an unlimited number of balls (mixed) of five different colors. How many ways can a handful of one or more balls be randomly selected that has at least one ball and at most 12 balls in total? I know that every handful is some combination with repetition, however as there is an unlimited number of balls is where I get the confusion. • You could add up the number of possible colour combinations for 1 through to 12 balls. Or you could find a shortcut and pretend there is a sixth invisible colour (remembering that they cannot all be invisible). This is a combinatorial question, not a probability or statistics question Aug 4, 2021 at 0:33 • Since you can take at most 12 balls, you can take at most 12 balls of any one color. Just treat this as though the box contained exactly 12 balls of each color. Aug 4, 2021 at 0:34 Adding onto Karl's answer, we want to find the number of nonnegative integer solutions to $$x_1+x_2+x_3+x_4+x_5\leq 12$$ To determine this, we will first add a temporary variable, $$c$$, that is a nonnegative integer. We can use this variable to now have the statement $$x_1+x_2+x_3+x_4+x_5+c=12$$ Convince yourself that the number of nonnegative integer solutions to this equation is the same as the number of solutions to the first inequality. Using stars and bars, we have that the number of nonnegative integer solutions to $$\sum_{i=1}^k a_i=n$$ is $$\binom{n+k-1}{k-1}$$. We can apply this theorem to see that the number of ways to grab the colored balls is $$\binom{12+6-1}{6-1}=\binom{17}{5}=6188$$. However, this includes the solution when $$x_1,x_2,x_3,x_4,x_5=0$$, which is explicitly stated in the problem to be excluded. Hence our final answer is $$\boxed{6187}$$. Questions like this are often worded ambiguously, forcing you to figure out a sensible interpretation. In this case, you aren't told whether different balls of the same color should be considered different when counting "ways". However, if we considered same-colored balls distinct, then the colors would be irrelevant (and since the number of balls is "unlimited", we wouldn't have enough information to answer the question). So we can infer that you're expected to think of two handfuls as the same if they look the same (e.g. "5 red and 4 blue"), even though multiple different sets of balls might look the same way. The "unlimited" here just means that it's possible to get a handful where all balls are the same color (of any color). Since we're only interested in handfuls of at most 12 balls, it doesn't matter whether the box has 12 of each color or 100 of each color. In other words, you're looking for the number of non-negative integer solutions to $$x_1+x_2+x_3+x_4+x_5\le12$$ and the "unlimited" means there are no explicit upper limits on the variables.
2022-05-22T17:43:09
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https://www.scienceforums.net/topic/98767-sum-of-the-first-m-terms-of-n32n/?tab=comments
# sum of the first m terms of n^3/2^n = ? ## Recommended Posts according to the source* , this expression f(m) is a fact one "notices" in order to get the value for m infinite ...... which is 26 ! * les math au carré , by Marie-France Palissard ##### Share on other sites The expression (n^3/2^n) is confusing. Use parentheses to clarify. ##### Share on other sites oh sure ! (n^3)/(2^n) n^3 a classic 2^n not to mention but combined ? ##### Share on other sites First step: Use induction to show that the partial sum equals $\sum_{n=1}^{m}\frac{n^3}{2^n} = 26-\frac {m^3+6m^2+18m+26}{2^m}$. Second step: $\lim_{m\to\infty} 26-\frac {m^3+6m^2+18m+26}{2^m} = 26 - \lim_{m\to\infty} \frac {m^3}{2^m} - \lim_{m\to\infty} \frac {6m^2}{2^m} - \lim_{m\to\infty} \frac {18m}{2^m} - \lim_{m\to\infty} \frac {26}{2^m} = 26$ ##### Share on other sites sure but what if you were asked to derive f(m) from scratch ? ##### Share on other sites If you want to keep the proof elementar, you will have to put some ideas into it. Here is one possible approach: Notice that your sum is of the form $\sum_{n=1}^{m}\frac{P(n)}{2^n}$ where $P$ is a polynomial. You start your proof with an educated guess, that the sum will be of similar form, namely $\sum_{n=1}^{m}\frac{n^3}{2^n}\stackrel{?!}{=}\frac{{Q{_1}(m)}2^m+Q{_2}(m)}{2^m}$ where $Q{_1}$ and $Q{_2}$ are themselves polynomials (you include a polynomial term multiplied by $2^m$ to increase your chances of success). There is no guarantee that this will succeed, but it's a starting point. So you make trial&error. You can be quite confident that $Q{_2}$ will be of at least same degree as $P$ (sums and integrals don't tend to decrease the degree of polynomials involved). So, your first attempt is the simplest possible, where $Q{_1}$ has degree 0 (turning it into a constant, possibly 0) and $Q{_2}$ has degree 3. This leads you to $\sum_{n=1}^{m}\frac{n^3}{2^n}\stackrel{?!}{=}\frac{{a}2^m+{b}m^3+{c}m^2+{d}m+e}{2^m}$ for some real numbers $a,b,c,d,e$. Evaluate at $m=1,\dots,5$ and you get a system of 5 linear equations in 5 variables. That means that IF your guess was correct then this will lead you to the only possible solution. That's the one presented in my previous post, and once you found it you can proof it by induction. And indeed, you will find that $\begin{pmatrix}2 & 1 & 1 & 1 & 1 \\ 4 & 8 & 4 & 2 & 1 \\ 8 & 27 & 9 & 3 & 1 \\ 16 & 64 & 16 & 4 & 1 \\ 32 & 125 & 25 & 5 & 1\end{pmatrix}\begin{pmatrix}a \\ b \\ c \\ d \\ e\end{pmatrix}=\begin{pmatrix}1 \\ 10 \\ 47 \\ 158 \\ 441\end{pmatrix}$ , with the unique solution $\begin{pmatrix}a \\ b \\ c \\ d \\ e\end{pmatrix}=\begin{pmatrix}26 \\ -1 \\ -6 \\ -18 \\ -26\end{pmatrix}$ immediately resulting in the formula shown to be correct earlier. Notice that you still have to prove it by induction, because so far this only shows that the claim is correct for $m=1,\dots,5$. The very same trick will work for many summation formula that are usually proved by induction. Edited by renerpho ##### Share on other sites (1) An idea to reduce the amount of guesswork in my previous ansatz: Because the infinite sum converges (this is easy to show), $Q{_1}(m)$ has to be a constant, and it will be equal to the value of the infinite sum. That's because $\lim_{m\to\infty} \frac{{Q{_1}(m)}2^m+Q{_2}(m)}{2^m} = \lim_{m\to\infty} {Q{_1}(m)}$, and as $Q$ is a polynomial this limit only exists if $Q$ is constant and is equal to the value of the infinite sum. If you already suspect the infinite sum to be equal to 26 then you can save some work by setting $a=26$, reducing the number of linear equations to 4. (2) Here is an alternative ansatz that avoids induction (for the cost of being less elementar). But it is more powerful because it can solve an infinite class of similar problems, and more elegant because there's no need for any "guesswork". Let $(p,q)$ be a pair of real numbers, with $q>1$. Notice that $\sum_{n=1}^{\infty }\frac{n^p}{q^n}=\sum_{n=1}^{\infty }{\frac{(1/q)^n}{n^{-p}}}$. Because of $q>1$ that sum converges. We are going to evaluate it by turning the problem into one about power series; the series involved is the one that defines the polylogarithm $\textup{Li}_{s}(x)$. Definition: $\textup{Li}_{s}(x):= \sum_{n=1}^{\infty }\frac{x^n}{n^s}=x+\frac{x^2}{2^s}+\frac{x^3}{3^s}+\dots$ With that, we get $\sum_{n=1}^{\infty }\frac{n^p}{q^n}=\textup{Li}_{-p}\left (1/q \right )$. Set $\left ( p,q \right )=\left ( 3,2 \right )$ and we get the expression $\sum_{n=1}^{\infty }\frac{n^3}{2^n}=\textup{Li}_{-3}(1/2)$. Even though the polylogarithm can not be expressed in terms of elementary functions in the general case, it can be shown to be a rational function if $s$ is a nonpositive integer, for example $\textup{Li}_{-3}(x)=\frac{x(1+4x+x^2)}{(1-x)^4}$. This can be derived via the expression $\textup{Li}_{-n}(x)={x^n}\frac{\mathrm{d}^n }{\mathrm{d}x^n} \left (\frac{x}{1-x} \right )$ for $n=0,1,2, \dots$ which itself follows directly (by induction over $n$) by simultaneously differentiating $n$ times both sides of the equation $\frac{x}{1-x}=\textup{Li}_{0}(x)$ (the well known Taylor formula for $\frac{x}{1-x}$). All this leads to $\textup{Li}_{-3}(1/2)=26$, giving the searched value. With the same method, you can show results like $\sum_{n=1}^{\infty }\frac{n^2}{2^n}=6$, $\sum_{n=1}^{\infty }\frac{n^4}{2^n}=150$ or $\sum_{n=1}^{\infty }\frac{n^5}{4^n}=\frac{4108}{243} \approx 16.905 \dots$ All of these can be shown by the induction method, too - but the computations involved become extremely ugly very, very fast. The formula $\sum_{n=1}^{\infty }\frac{n^p}{q^n}={x^p}\frac{\mathrm{d}^p }{\mathrm{d}x^p} \left. \left (\frac{x}{1-x} \right ) \right | _{x=1/q}$ for $p=0,1,2, \dots$ is much easier to evaluate. Edited by renerpho ##### Share on other sites fantastic ! impressive . huge THANK ## Create an account Register a new account
2019-12-09T16:14:11
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https://math.stackexchange.com/questions/2788415/non-linear-system-of-equations-over-the-positive-integers-more-unknowns-than-e
# Non-linear system of equations over the positive integers — more unknowns than equations This exercise appeared on a german online tutoring board and caught my attention but stumbled me for hours. The task is to find 6 distinct positive three digit integers satisfying: $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=4.198$ $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}=3.215.224$ $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}+x_{5}^{3}+x_{6}^{3}=2.600.350.972$ According to the power mean inequality or Cauchy-Schwarz the numbers must lie relatively closely together. However a brief search lead nowhere. For simplicity I set $4.198=A$, $3.215.224=B$ and $2.600.350.972=C$ and then my approach was to manipulate the three equations and perhaps use that no square is negative. For example $6B-A^{2}=\sum_{i<j}(x_i-x_j)^{2}=(x_{1}-x_{2})^2+(x_{1}-x_{3})^2+...+(x_{2}-x_{3})^2+...+(x_{5}-x_{6})^2$ where every distinct pair appears exactly once. If now $6B-A^{2}$ would result in something like $5*1^{2}+4*2^{2}+3*3^{2}+2*4^{2}+1*5^{2}=105$ we could tell exactly what our $x$ were. Unfortunately it gives $1.668.140$ and we cannot conclude much. Similar reasoning with factoring to $\sum_{i<j}(x_i-x_j)^{4}$ didn't help. If there exists such a factorization, my intuition says it would only make sense if the $x$ form an arithmetic sequence, otherwise we would get different factors on the right side that can't appear on the left side. (does this sound reasonable?) But substituting gives no solution. Also, I don't know what other, more sophisticated factorization would lead me to the solution. I'm running out of ideas, how can this problem be solved? • I believe by $4.198$ you mean $4198$, ie four thousand one hundred and ninety eight? – Jack M May 20 '18 at 9:33 • @JackM The dot means multiplication. – N. F. Taussig May 20 '18 at 9:40 • IMHO, the dots are just readability separators, as used in much of mainland Europe. I assume a manual method is wanted, as a program would be trivial. – Old Peter May 20 '18 at 11:08 • @JackM Yes, the dots separate powers of 1000 and should make the numbers easier to read. – user495573 May 20 '18 at 11:36 • The possibilities can be narrowed down by considering congruence properties mod 8. Note that if $n$ is odd, $n^2 \equiv 1$ and $n^3 \equiv n$, while if $n$ is even, $n^2 \equiv 0,4$ and $n^3 \equiv 0$. The sums are respectively $\equiv 2,0,4$. – Adam Bailey May 21 '18 at 9:27 Assume that $x_1<x_2<\cdots < x_6$. 1st step. Then $$6x_1<A<6x_6; \\ 6x_1^2<B<6x_6^2; \\ 6x_1^3<C<6x_6^3;$$ or (when note that $\dfrac{A}{6}\approx 699.66$, $\sqrt{\dfrac{B}{6}}\approx 732.03$, $\sqrt[3]{\dfrac{C}{6}}\approx 756.76$): $$x_1\le 699; x_6\ge 700;$$ $$x_1\le 732; x_6\ge 733;$$ $$x_1\le 756; x_6\ge 757;$$ So, we have restrictions for the smallest and the largest numbers: $$x_1\le 700, x_6 \ge 757.$$ 2nd step. We can bruteforce values $x_1,x_2,x_3,x_4, x_5$, and then calculate $x_6 = A - (x_1+x_2+x_3+x_4+x_5)$, and then check whether other sums are correct. But it takes $5$ loops through roughly almost $900$ possible values. To reduce $1$ loop: bruteforce values $x_1,x_2,x_3,x_4$; denote $s_1 = x_1+x_2+x_3+x_4$, $s_2 = x_1^2+x_2^2+x_3^2+x_4^2$; and then derive $x_5$ and $x_6$: $$x_5+x_6 = A - s_1; \\ x_5^2+x_6^2 = B - s_2;$$ so $$x_5 x_6 = \dfrac{(x_5+x_6)^2-(x_5^2+x_6^2)}{2} = \dfrac{(A-s_1)^2 - (B-s_2)}{2}.$$ And $x_5,x_6$ are roots of polynomial $$w^2 - (x_5+x_6)w + x_5x_6;$$ then the discriminant $$D = 2(B-s_2)-(A-s_1)^2$$ should be a square number. Using this approach, one can find (in reasonable time) that ... (please skip this spoiled box if you'll try to solve this problem personally) $$(x_1,\ldots,x_6) = (365, 438, 789, 821, 877, 908).$$ And maybe this is not the unique solution. • Thanks for the answer. Apparently there's no way around bruteforcing. – user495573 May 22 '18 at 6:46 • Clearly, programming a solution is not as easy as I originally thought. I’ll give it a go, if I can find a spare hour or two. – Old Peter May 22 '18 at 9:29 • The second link of the question gutefrage.net/frage/… gives the answer shown above. LOL! – Old Peter May 22 '18 at 17:59 • @Old Peter: now I feel myself like "expert from math.stackexchange" ;) – Oleg567 May 23 '18 at 9:54 Try the method of Newton's Sums. Let $$P_i=x_1^i+x_2^i+x_3^i+x_4^i+x_5^i+x_6^i$$ Let $f(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g=0$ have roots $x_1,x_2,x_3,...,x_6$ We want to find the values of $a,b,c,...,g$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3,...,x_6$. Also, $$\begin{array} { l l l } e_1 & = \sum x_i & = - \frac{b}{a} \\ e_2 & = \sum x_i x_j & = \frac{ c } { a} \\ e_3 & = \sum x_i x_j x_k & = - \frac{ d } { a} \\ & \vdots & \vdots \\ e_6 & = \sum \prod x_i & = \frac{g} { a}. \end{array}$$ Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows: For $i \leq k - 1$, we have the formulas $\begin{array} { l l l l } P_0 & = k \\ P_1 & = e_1 \times 1 \\ P_2 & = e_1 P_1 - e_2 \times 2 \\ P_3 & = e_1 P_2 - e_2 P_1 + e_3 \times 3 \\ & \vdots \\ P_{k-1} & = e_1 P_{k-2} - e_2 P_{k-3} + \cdots + (-1)^{k-3} e_{k-2} P_1 + (-1)^{k-2} e_{k-1} \times (k-1). \end{array}$ For $i \geq k$, we have the formula $$P_i = \sum_{j=1}^k (-1)^{j+1} e_j P_{i-j} = e_1 P_{i-1} - e_2 P_{i-2} + \cdots + (-1)^{k+1} e_k P_{i-k}. \ _\square$$ Hence $P_1=4198, P_2=3215224,P_3=2600350972$ But, $$P_1=e_1$$ $$P_2=e_1P_1-2e_2$$$$P_3=e_1P_2-e_2P_1-3e_3$$ Now, $$4198=e_1$$ $$3215224=(4198)(4198)-2e_2\Rightarrow e_2=7203990$$ $$2600350972=(4198)(3215224)-(7203990)(4198)-3e_3\Rightarrow e_3=-868113246$$ In this way, "recreate" $f(x)$ by evaluating $e_4,e_5,e_6$ (hence determining the values of the coefficients $a$, $b$,...,$g$), and the roots of the 'recreated' $f(x)$ will be the values of $x_1,x_2,...,x_6$ • This is an OK answer; it is link-only and is lacking context and other detail. Please elaborate further in order to write a good answer. – Mr Pie May 21 '18 at 11:56 • Okay, I'll elaborate it in a short while. Thanks for your comment. – User1234 May 21 '18 at 11:56 • No problem. I would not recommend deletion. $(+1)$ :) – Mr Pie May 21 '18 at 11:57 • Thanks! I've more or less shown how to go about with the question, could you have a look please? – User1234 May 21 '18 at 13:19 • I like this approach, also Newton's identities seem to be a powerful tool in this context. But if I'm not mistaken we can only find $e_{1}, e_{2}$ and $e_{3}$ this way for we don't have sufficient information because it doesn't take into account that we're dealing with integers only. – user495573 May 22 '18 at 6:15 Find 6 distinct positive three digit integers satisfying $$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=A=4198\tag1$$ $$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}=B=3215224\tag2$$ $$x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}+x_{5}^{3}+x_{6}^{3}=C=2600350972\tag3$$ As $B\equiv 0{\pmod{8}}$, the square residues$\pmod{8}$ are $(0,4,1,1,1,1)$, $(4,4,4,4,4,4)$ or $(0,0,0,0,0,0,0)$ However, $C\equiv 4{\pmod{8}}$, so $(4,4,4,4,4,4)$ and $(0,0,0,0,0,0,0)$ are rejected. WLOG, $x_{1}> x_{2}$ and $x_{3}> x_{4}>x_{5}>x_{6}$, Hence, we have $x_{1}$ even and in the range $(102,998)$ $x_{2}$ even and in the range $(100,x_{1}-2)$ $x_{1}+x_{2}\equiv 2{\pmod{4}}$ $x_{3}$ odd and in the range $(107,999)$ $x_{4}$ odd and in the range $(105,x_{3}-2)$ $x_{5}$ odd and in the range $(103,x_{4}-2)$ $x_{6}$ is easily calculated from the other $x$, but must be odd and in the range $(101,x_{5}-2)$ Now we can consider a program. I started using equation $(1)$, but the gargantuan volume of solutions was overwhelming. Equation $(3)$ was much faster, although there was still a massive number of results, so I just reported cases where $(1)$ or $2$, or both, were also satisfied. The program is essentially as above, with a few tuning tweaks. I precalculated the squares and cubes of $100$ to $999$. For $x_{3}$ to $x_{5}$, I calculated the minimum and maximum values, reducing the search-range when possible. My results The solution found by @Oleg567 is unique. There is just one other solution that satisfies equations $(2)$ and $(3)$, but not $(1)$ $$(746,120,939,815,751,731)$$ There are $1107$ solutions that satisfy equations $(1)$ and $(3)$, but not $(2)$ The program used under $22$ minutes CPU.
2021-05-06T21:33:14
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https://math.stackexchange.com/questions/3799710/how-to-calculate-intop-0-frac-pi2-frac-sqrt-sin-x-sqrt-sin-x/3799719
# How to calculate $\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ I have to calculate the integral: $$\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$$ I tried a various sort of ways, I'll present 2 of them that lead me nowhere, maybe someone will see a way through the obstacles. way 1: trigonometric substitution : substitue: $$\tan\left(\frac{x}{2}\right)=t$$ thus $$\sin\left(x\right)=\frac{2t}{1+t^{2}},\cos\left(x\right)=\frac{1-t^{2}}{1+t^{2}},dx=\frac{2}{1+t^{2}}dt$$ $$\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\intop_{0}^{1}\frac{\frac{\sqrt{2t}}{\sqrt{1+t^{2}}}}{\frac{\sqrt{2t}}{\sqrt{1+t^{2}}}+\frac{\sqrt{1-t^{2}}}{\sqrt{1+t^{2}}}}\frac{2}{1+t^{2}}dt=\intop_{0}^{1}\frac{2\sqrt{2t}}{\left(1+t^{2}\right)\left(\sqrt{2t}+\sqrt{1-t^{2}}\right)}dt$$ From here I cannot see how to continue.( I tried to multiply the denominator and the numerator by $$\sqrt{2t}-\sqrt{1-t^{2}}$$ but it also seems like a dead end. In the other way that I tried, I did found an antideriviative of the integrand, but not in the segment $$[0,\frac{\pi}{2}]$$ because if we could divide by $$\sqrt{\sin x}$$ then we'd get: $$\int\frac{1}{1+\sqrt{\cot x}}dx$$ then if we substitue $$\sqrt{\cot x}=t$$ then $$\sqrt{\cot x}=t$$ so $$\frac{1}{2t}\cdot\frac{-1}{\sin^{2}x}dx=dt$$ and since $$\frac{1}{\sin^{2}x}=1+\cot^{2}x$$ we would get $$dx=\frac{-2t}{1+t^{4}}dt$$ Thus $$\int\frac{1}{1+\sqrt{\cot x}}dx=-\int\frac{2t}{\left(1+t\right)\left(1+t^{4}\right)}dt=-\int\frac{2t}{\left(1+t\right)\left(t^{2}-\sqrt{2}t+1\right)\left(t^{2}+\sqrt{2}t+1\right)}dt=-\int\left(\frac{-1}{t+1}+\frac{1+\sqrt{2}}{2\left(t+\sqrt{2}\right)}+\frac{1-\sqrt{2}}{2\left(t-\sqrt{2}\right)}\right)dt$$ and finally: $$\int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\ln|\sqrt{\cot x}+1|-\frac{1+\sqrt{2}}{2}\ln|\sqrt{\cot x}+\sqrt{2}|-\frac{1-\sqrt{2}}{2}\ln|\sqrt{\cot x}-\sqrt{2}|+constant$$ So this is an antideriviative, but we cannot use Newton leibnitz's formula because of the point $$x=0$$. In addition, I tried to calculate this integral with an online integral calculator and it failed to show the steps, so I guess this calculation isnt trivial. Any suggestions would help. • This is very simple in fact...Try to prove this property $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ Aug 22, 2020 at 14:09 • @user35508 Is it true without any demands on $f$ ? how does it helps here ? It would just replace $sin$ by $cos$ Aug 22, 2020 at 14:13 • Use $y=\frac{\pi}{2}-x$ Substitution – DARK Aug 22, 2020 at 14:14 • @Waizman: The integrand should be Reimann integrable on $[a,b]$ which is clearly the case here as the integrand is continuous on $[a,b]$ – Koro Aug 22, 2020 at 14:18 Let $$I=\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\tag{1}$$ Substitute $$x=\pi/2-t$$ so that $$dx=-dt$$. Hence, $$I=-\intop_{\pi/2}^{0}\frac{\sqrt{\cos y }}{\sqrt{\sin y}+\sqrt{\cos y}}dy=\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos y}}{\sqrt{\sin y}+\sqrt{\cos y}}dy=\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx\tag{2}$$ Add the two to get: $$2I= \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \; dx \\ = \int_{0}^{\pi/2}dx=\pi/2\implies I=\pi/4$$ As user35508 mentioned in the comments, you can use the substitution of $$u=\frac{\pi}{2}-x$$ to obtain $$\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \; dx$$ This integral is equivalent to the original integral, so if you add the original integral to the previous integral you get: \begin{align*} 2I&=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \; dx \\ &= \frac{\pi}{4} \end{align*} Hint:use kings property ie replace $$x$$ by $$\frac{\pi}{2}-x$$ you get an equivalent integral . Now add these integrals what do you observe?
2022-07-06T11:11:51
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http://mathhelpforum.com/statistics/34790-probability.html
# Thread: probability 1. ## probability How many different ways can a family of 6 sit in a row at the movies if the 2 youngest children must sit next to each other? 2. First, there are 5 ways the 2 children can be sit. In each case the rest of the family can be sit in 4! different ways. So I think the answer is 5.4!. And it's equal to 120. (Which is 5! So now I see that we could have think the problem as if the 2 children were one person of the family and not 2). 3. Hello, Help09! I got a different answer . . . How many different ways can a family of 6 sit in a row at the movies if the 2 youngest children must sit next to each other? Duct-tape the two youngest together. . . Then we have five "people" to arrange: . $\boxed{AB}\;C\;D\;E\;F$ And they can be arranged in: . $5! = 120$ ways. But in each of those arrangements, $\boxed{AB}$ can be replaced by $\boxed{BA}$ Therefore, there are: . $2 \times 120 \:=\:{\color{blue}240}$ ways. 4. Let these be the seats --- --- --- --- --- --- We first fix the first two seats for the youngest children [--- ---] --- --- --- --- There are 4! ways of filling the other 4 seats Similarly for the cases, --- [--- ---] --- --- --- ... so on for 5 cases, each of which contributes a 4!. So the total number of ways is 5*4!=120 * 2 = 240 (where 2 children can be swapped) In general for n seats and k people in connesecutive seats, [(n-k+1)(n-k)!]k! = (n-k+1)!k! Same as our answer: 5!.2!=240 5. I'm sorry for my answer, Soroban and Sarim are right. Multiply my result by 2.
2017-04-27T13:29:07
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https://www.themathdoctors.org/grouped-data-open-ended-classes/
# Grouped Data: Open-ended Classes? ### (A new question of the week) A recent question raised a different issue about grouped frequency distributions than we have discussed previously: What do you do when the last class is labelled something like “30 or more”? As we’ll see, there is no one right answer! ## An open question Here is the initial question, which came in last month: I understand that to use the formula to find mean for grouped data I have to find the class midpoint first. How do I find the midpoint when the information I have only states “above 100”? We looked at the mean of a grouped distribution last time. There we had distributions like Class Frequency ----- --------- 37-46 19 47-56 23 57-66 27 67-76 28 This might represent the number of people in an audience of various ages. We found the mean by multiplying the midpoint $$x_i$$ of each class by its frequency $$f_i$$, adding these up, and dividing by the total number of people:$$\frac{\sum_{i=1}^{n}x_i\cdot f_i}{\sum_{i=1}^{n}f_i}.$$ The midpoint of each class is the average of its upper and lower limits, such as $$(37+46)\div 2 = 41.5$$ for the first class. (We saw that we can do the same thing with class boundaries, for a continuous distribution.) What if it looked like this instead? Class Frequency ----- --------- 37-46 19 47-56 23 57-66 27 67-76 28 67 or more 28 The last class has no upper limit; how can we find its midpoint? Looking back at old unarchived answers (because there is nothing about this published in the Ask Dr. Math archive), I found a couple questions like this that were never answered, probably because we were too busy even to answer the questions we could answer confidently! Here is one from 2008: I'm trying to find the mean and median for this frequency distribution: Minutes of delay Shop A Shop B ---------------- ------ ------ Less than 10 20 15 10-15 25 20 15-20 30 30 20-25 25 15 25-30 20 10 30 or more 10 5 How do I calculate the mean if the last interval is open? How do I calculate the midpoint of the last class? Is there another way to do it? I know that I have to find the midpoint for each interval, multiply it by the frequency of each class, sum it up for all classes and divide it by the total number of observations. Mean = (Sum mp x f)/n But I need the midpoint of the last interval!!! (Note that this distribution is continuous, and has to be interpreted so that “10-15” means “at least 10, and less than 15”, so that 15 is not included in two classes.) The last interval goes, in principle, from 30 to infinity, so its midpoint would be infinite. A similar question from 2006 included a hint to the intended answer: The Department of Commerce, Bureau of the Census, reported the following information on the number of wage earners in more than 56 million American homes. Number of Earners Number (in thousands) 0 7,083 1 18,621 2 22,414 3 5,533 4 or more 2,797 a. What is the median number of wage earners per home? b. What is the modal number of wage earners per home? c. Explain why you cannot compute the mean number of wage earners per home. Hint - the data above is an example of grouped data. This is not really grouped, as each row pertains to a single value – except for the last, which is a group representing all higher numbers! Apparently the author of this problem says that we can’t find the mean, because of the open-ended class. (Note that we can find the median and the mode, because neither is affected by the outliers in that last class!) I answered, as I often do when there is no definitive answer in my experience, by searching for suitable sources to get a sense of what sort of answers knowledgeable people give. The sources I found are not necessarily authoritative, but are meant to provide a survey of what might be said about the subject. I started my reply with the technically correct answer: The quick answer is, you really can’t find the mean of an open-ended distribution. Without having both limits for every class, you just don’t have the information you need. That is explicitly stated here: https://people.richland.edu/james/lecture/m170/ch03-ave.html The Mean is used in computing other statistics (such as the variance) and does not exist for open ended grouped frequency distributions. With no upper limit for a class, we can’t find a midpoint, and therefore can’t use the formula I gave above. It simply doesn’t apply. (I myself would not say that the mean doesn’t exist, but just that we don’t have enough information to find it. If we knew the original data, we could. The author here is referring to the mean of the distribution as presented.) I didn’t stop there, though: But there are several ways to deal with this situation, depending on how much you care about accuracy. Since any statistics based on grouped data are only approximations anyway, some guesses can make sense. As I mentioned last time, such statistics are just estimates based on inadequate data and hopeful assumptions (such as that the data within a class is distributed in such a way that its midpoint is its mean). The formula itself is therefore not perfect, and there is no reason not to try adjusting the data in order to try for the best estimate we can get with even less adequate data! ## Second answer: Pretend that each class has the same width One possibility is just to make the simplest possible assumption: Looking around for examples, I found several sources that recommend just assuming an upper limit such that the class width of the open-ended class is the same as its nearest neighbor. This is perhaps the easiest solution; I suspect these rules may be given primarily for students, so that they can always find some answer, even if it is not ideal. An example of this is: http://www.economicsdiscussion.net/frequency-distribution/how-to-calculate-frequency-distribution/2379 3. Open End Intervals: These are those intervals or classes, which either the lower limit of first interval or the upper limit of last interval or both of these, are not given. Here only an assumption about the length of these intervals is made according to the length of the interval nearest to these intervals. Let us suppose the given class intervals are: Less than 10, 10-20, 20-30, 30-40, 40-50, more than 50; Then the desired class intervals i.e. 1st and last are 0-10 and 50-60 respectively; as the length of intervals nearest to these two is also 10 i.e. in intervals 10-20 and 40-50. But if class intervals are not equal, then first interval should be taken equal to second and last equal to penultimate one. But that is not really a very good guess in many real situations, because the presence of the open-ended class is likely due to the fact that there are extreme values — and as you probably know, extreme values have a significant effect on the mean. As I hinted, I felt that the sources I found that made this recommendation typically were meant not for serious statisticians, but for classes whose students just expect a simple rule for every case. I don’t know the credentials of this source, but it probably represents what is taught in some curriculum at that level. The fact that no justification is given makes the whole idea suspect. In the example given here, it makes good sense to interpret the first class, “less than 10”, as “0 to 10” (not really because we take the same width, but merely because the numbers presumably can’t be negative). But assuming the last class ends at 60 ignores the fact that if no data values were greater than 60, they would have used 60! On the other hand, probably there are not many greater than 60, so maybe the assumption is good enough. But, as I said in my response, this reasoning ignores the fact that the mean is strongly affected by outliers – or we might do this in order to deliberately ignore outliers. If that “more than 50” included a value of 100, the mean would be far larger than if we just use a midpoint of 55. ## Third answer: Try a subjective, but informed, guess What if you really want an answer, and you want it to be the best you can get with the limited information at hand? I found a nice, long discussion of better guesses (but still guesses) here: http://uregina.ca/~gingrich/ch51.pdf (pp 37-42) Open Ended Intervals. As noted in Chapter 4, data is often presented so that it has open ended intervals. If the mean is to be determined for such a distribution, some value has to be entered for X for the open ended interval when using the formula for the mean. Exactly what this value should be is not readily apparent from the table of the distribution. … About all that can be done in the case of open ended intervals is to pick a value of X which seems reasonable based on what is known about the distribution of the data. Do not pick a value too high, or too low, but pick a value which you think approximately represents the mean value of the variable for the set of cases in the open ended interval. Note that this is a serious attempt at accuracy, with each choice justified, and with explicit mention of the fact that the conclusions are approximate. For example, the author deliberately chooses as a “midpoint” for the last class in the first example that is larger than if that class were the same size as others, explaining why; and concludes by saying, “The mean is thus 12.195 thousand dollars, or $12,195. Given the approximations that have been made in this calculation, it might be best to round the mean to the nearest$100 and report it as $12,200, for perhaps round it to the nearest thousand dollars and report it as$12,000.” Note that if this is done in a classroom exercise, each student might make a different choice – that is what I mean by “subjective”. Each such choice might be equally reasonable; some might be based on better background knowledge than others. Teachers (or students) who are not comfortable with this, or who think every math problem has to have a single correct answer, may choose to stick with the second method, but they are not being honest about the validity of their method. After writing this, I ran across an article that gives a perspective similar to mine: What if I have open-ended classes? For open classes (i.e. classes that don’t have an upper limit or a lower limit), in most cases you can assume those have the same width as the other classes when doing your calculations. In an elementary statistics class, it’s highly unlikely your instructor will throw you a curve ball by creating an unusually wide open-ended class. However, if you’re working with real-world data—perhaps from a graduate study or work-related study— you may need to use your best judgment when it comes to the midpoint for open classes. If the open class is extremely large, or extremely small, your best guess might be better than a calculated midpoint. ## Fourth answer: Make a careful model of the data If you are doing a really serious study, and you can’t get more detailed data, then you need to approach the matter scientifically, using a model based on what you know about the subject you are studying: But when you really want a good number, you would want to model the overall data set in such a way that you can estimate the distribution of values in the tail. One place I found this discussed (just as an example) is https://arxiv.org/ftp/arxiv/papers/1210/1210.0200.pdf To answer these questions, we have to estimate the mean and variance from the bins. How can we do that? A simple approach is to assume that each family’s income is at the midpoint of its bin. For example, we might assume that all households with incomes in the bin [$0,$10,000) have an income of exactly $5,000. This assumption is unrealistic, but it can be serviceable if the bins are narrow. If the bins are wide, then the midpoint approximation may be less accurate, since within some bins the distribution of households may be highly variable and may not be centered around the bin midpoint. The midpoint approximation also runs into practical difficulties if the data are “top-coded” so that the highest bin is unbounded or censored on the right—as in the Rancho Santa Fe school district, where nearly half the households are in the top bin [$200,000, +∞). Analysts commonly handle top-coding by assuming that the incomes within the top bin fit some distribution (e.g., Pareto). But such assumptions can be inaccurate and are hard to test (Hout 2004). A more sophisticated approach is to fit a flexible distribution not just to the top bin, but to the entire distribution. Note that if you want this level of accuracy, you shouldn’t use the midpoint even for closed classes. That is far beyond the context of your question! But this is what you would do if it really mattered. A simpler example of modeling might be to recognize that within any one class (bin), the data is likely more dense on the side toward the mode, and use the overall shape of the distribution to estimate the slope of the underlying curve, and use that to choose a better number than the midpoint to represent that class. (This is just the musing of a non-statistician; I am not aware of a technique that actually does this.) For an open-ended class, however, this would be extrapolation rather than interpolation, and therefore more risky. The suggestion here is to use knowledge of the nature of the data to choose an appropriate distribution (the choice being justified by the characteristics of the histogram), and then use that to estimate the parameters. This is well beyond my knowledge. That’s probably a longer answer than you want. Ultimately, if you are in a class, you need to ask your instructor what to do in these cases; if you were doing serious statistical analysis, you might need professional advice. There are more places than you might think in math, where there is no general consensus, and you just have to find out what conventions are used in your class (or in your field). ### 5 thoughts on “Grouped Data: Open-ended Classes?” 1. I am a teacher and my previous head of department gave another answer. We were teaching IGCSE and IB Maths, and her suggestion was to make the open group twice as big as the previous one. I guess that assumes that the data tails off on the upper end. Having myself also worked as an examiner (for A-level Maths – Stats module), it is unlikely that in an exam you would be penalised for any sensible decision you make to get to an answer. If you write down your assumption somewhere in your solution, so that the examiner can read it, you should get credit for your calculations. As for defining what a sensible decision might be, that is harder to be certain about, and depends on the distribution of the data. 1. This is certainly an open question, isn’t it? As you suggest, there may be many ways to guess at a reasonable approach, depending on what you see in a particular distribution. I heartily agree with stating assumptions in your work, not only in order to get full credit for answering the problem as you understand it, but simply as a way to distinguish absolute truth from speculation, which is essential to mathematics. This is also one reason why I strongly dislike exams on which you can do nothing but state your answer.
2020-12-05T04:56:40
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http://www.marymorrissey.com/davinci-raspberry-rdcxs/1e3f95-measures-of-dispersion-examples
An example of aggregating data is the simple process of finding the mean of a variable such as height or weight. Variance and Standard Deviation. The rangeis the difference between the largest and smallest value in a dataset. Since they are devoid of a specific unit, the comparison between different series is hence possible. The scatterness or variation of observations from their average are called the dispersion. Absolute dispersion method expresses the variations in terms of the average of deviations of observations like standard or means deviations. Common examples of measures of statistical dispersion are the variance, standard deviation, and interquartile range $$\mu =\cfrac {(12 + 13 + \cdots +25)}{5} =\cfrac {160}{5} = 32$$, \begin{align*} The Range. This example of one of the relative measures of dispersion is also called as Range Co-efficie… Analysts use the standard deviation to interpret returns as opposed to the variance since it is much easier to comprehend. 58, 66, 71, 73, 74, 77, 78, 82, 84, 85, 88, 88, 88, 90, 90, 92, 92, 94, 96, 98, 2. In this case, the outlier income of person J causes the range to be extremely large and makes it a poor indicator of “spread” for these incomes. The median splits the dataset into two halves. Unit-II MEASURES OF CENTRAL TENDENCY AND DISPERSION Relation between Mean, Median and Mode: − = 3( – ) Range of ungrouped data: The range of a set of data is the difference between the highest and lowest values in the set. It is usually used in conjunction with a measure of central tendency, such as the mean or median, to provide an overall description of a set of data. The variance is a common way to measure how spread out data values are. There are four commonly used measures to indicate the variability (or dispersion) within a set of measures. This is necessary so as to remove biasThe sample standard deviation, S, is simply the square root of the sample varianceExample 4Assume that the returns realized in example 2 above were sampled from a population comprising 100 returns. The study of dispersion will enables us to know whether a series is homogeneous (where all the observations remains around the central value) or the observations is heterogeneous (there will be variations in the observations around the central value like 1, 50, 20, 28 etc., where the central value is 33). You compute […] There are four Absolute Measures of Dispersion in Statistics: Range; Quartile Deviation; Mean Deviation; Standard Deviation; Range. Working with data from example 2 above, the variance will be calculated as follows: \begin{align*} $$\text{Range} = \text{maximum value} – \text{minimum value}$$, Consider the following scores of 10 CFA Level 1 candidates, 78   56   67   51   43   89   57   67   78   50. These are also known as ‘Coefficient of dispersion’ 3. and other Percentiles. Required fields are marked *. As the name suggests, the measure of dispersion shows the scatterings of the data. Example Calculate the range for the data for Quarterback A and Quarterback B in the example above. It is a measure of dispersion that represents the average of the absolute values of the deviations of individual observations from the arithmetic mean. Start studying for CFA® exams right away. For example, when rainfall data is made available for different days in mm, any absolute measures of dispersion give the variation in rainfall in mm. (Definition & Example). They are: 1. These are the range, variance, absolute deviation and the standard deviation. Assume that the returns realized in example 2 above were sampled from a population comprising 100 returns. Characteristics of a good measure of dispersion Calculate and interpret 1) a range and a mean absolute deviation and 2) the variance and standard deviation of a population and of a sample. In the above cited example, we observe that. It tells the variation of the data from one another and gives a clear idea about the distribution of the data. Relative Dispersion The actual variation or dispersion, determine from standard deviation or other measures is called absolute dispersion, now the relative dispersion is For Example, Relative dispersion It is a measurement of the degree by which an observed variable deviates from its … All Rights ReservedCFA Institute does not endorse, promote or warrant the accuracy or quality of AnalystPrep. o Variance. Learn more about us. Measures of central dispersion show how “spread out” the elements of a data set are from the mean. & =\cfrac {1870}{5} = 374 \\ Example 8.2 Find the range of the following distribution. Objectives . And the standard deviation is simply the square root of variance. Mean deviation from median. The formulae for the variance and standard deviation are given below. First, we have to calculate the arithmetic mean: $$X =\cfrac {(12 + 4 + 23 + 8 + 9 + 16)}{6} = 12\%$$, \begin{align*} This is from the Oxford English Dictionary: The term came to English from the German (where it lived before that I do not know) and seems to have emerged as a way of explaining aggregated data, or data which one has subjected to the process of removing information in order to gain information. \end{align*}. )2}/n – 1Note that we are dividing by n – 1. Measures of Dispersion A measure of spread, sometimes also called a measure of dispersion, is used to describe the variability in a sample or population. Thus, the average variation from the mean (0.12) is 0.003767. ©AnalystPrep. You’re kind of an adventurous person and you don’t have too many capricious demands regarding where you want to live next. The range is the difference between the largest and smallest value in a dataset. How “spread out” the values are. . Arrange the values from smallest to largest. Solution Here Largest value L = 28. The heights in cm of a group of first year biology students were recorded. \end{align*} $$,$$ \begin{align*} For example, if the standard deviation is large then there are large differences between individual data points. Measures of Dispersion The Range of a set of data is the largest measurement minus the smallest measurement. They are usually used in conjunction with measures of central tendency such as the mean and the median. Variance. By focusing on the mean, w… Thus, the range is 98 – 58 = 40. 4. o Measure of dispersion. { \sigma }^{ 2 } & =\frac { \left\{ { \left( 12-12 \right) }^{ 2 }+{ \left( 4-12 \right) }^{ 2 }+{ \left( 23-12 \right) }^{ 2 }+{ \left( 8-12 \right) }^{ 2 }+{ \left( 9-12 \right) }^{ 2 }+{ \left( 16-12 \right) }^{ 2 } \right\} }{ 6 } \\ Relative measures of dispersion are obtained as ratios or percentages of the average. o Sample variance. The table shows marks (out of 10) obtained by 20 people in a test. Standard deviation. The minimum number of completions for Quarterback A is 19, the maximum is 37. o Understand the difference between measures of dispersion for populations and for samples This is necessary so as to remove bias, The sample standard deviation, S, is simply the square root of the sample variance. These are pure numbers or percentages totally independent of the units of measurements. We often measure the “center” using the mean and median. In this case, Q1 is the average of the middle two values in the lower half of the data set (75.5) and Q3 is the average of the middle two values in the upper half of the data set(91). The population variance, denoted by σ2, is the average of the squared deviations from the mean. o Sample standard deviation. Suppose we have this dataset of final math exam scores for 20 students: The largest value is 98. The sample variance, S2, is the measure of dispersion that applies when we are working with a sample as opposed to a population. In this lesson, you will read about the following measures of dispersion: Range. Statology is a site that makes learning statistics easy by explaining topics in simple and straightforward ways. Statology Study is the ultimate online statistics study guide that helps you understand all of the core concepts taught in any elementary statistics course and makes your life so much easier as a student. Slide 77 Measures of Dispersion  There are three main measures of dispersion: – The range – The Interquartile range (IQR) – Variance / standard deviation 8. Measures of dispersion are used to describe the variability or spread in a sample or population. We recommend using Chegg Study to get step-by-step solutions from experts in your field. o Degrees of freedom. The standard deviation is 0.0037671/2 = 0.06137 or 6.14%. & = 0.00452 \\ Cycles are trends or patterns that may be exhibited by the securities market,... Monte Carlo simulation and historical simulation are both methods that can be used... 3,000 CFA® Exam Practice Questions offered by AnalystPrep – QBank, Mock Exams, Study Notes, and Video Lessons, 3,000 FRM Practice Questions – QBank, Mock Exams, and Study Notes. Example 8.3 The range of a set of data is 13.67 and the largest value is 70.08. Find the median. You may notice that all the relative measures of dispersion are called coefficients. Usually we work with samples, not populations. \text{MAD} & = \cfrac {\left\{ |12 – 12|+ |4 – 12| + |23 – 12| + |8 – 12| + |9 – 12| + |16 – 12| \right\}} {6} \\ Consequently, the mean may not be representative of the data. Your email address will not be published. Dispersion … Measures of dispersion In statistics, dispersion (also called variability, scatter, or spread) is the extent to which a distribution is stretched or squeezed. o Population variance. An absolute measure of dispersion contains the same unit as the original data set. When we analyze a dataset, we often care about two things: 1. The quartile boundaries would lie between two values in our data set. There are different measures of dispersion like the range, the quartile deviation, the mean deviation and the standard deviation. For example, suppose we have the following dataset with incomes for ten people: The range is $2,468,000, but the interquartile range is$34,000, which is a much better indication of how spread out the incomes actually are. & = 45.20(\%^2) \\ The only important thing for … The Interquartile Range (IQR) . Surprisingly, the term statistic first came into use as late as 1817. & = 0.0672 Where the “center” value is located. The sample variance, S2, is the measure of dispersion that applies when we are working with a sample as opposed to a population. It is a relative measure of dispersion and is based on the value of range. Absolute measures of dispersion are expressed in the unit of Variable itself. Relative Measure of Dispersion; Absolute Measure of Dispersion. CFA® and Chartered Financial Analyst® are registered trademarks owned by CFA Institute. Thus, the interquartile range is 91 – 75.5 = 15.5, The interquartile range more resistant to outliers compared to the range, which can make it a better metric to use to measure “spread.”. Key Terms . (2) Relative Measures 1. \end{align*} $$, Interpretation: It means that on average, an individual return deviates 5% from the mean return of 12%. For example, absolute dispersion in data related to age and weight is not comparable because age is measured in terms of years but the weight is measured in terms of the kilogram. The interquartile range is the difference between the first quartile and the third quartile in a dataset. Mark (x) We’ve started colonizing and populating new planets. Example. R = 28 −18 = 10 Years . Lets look at the first of the relative measures of dispersion. Dispersion (a.k.a., variability, scatter, or spread)) characterizes how stretched or squeezed of the data. Range R = L –S. You’re bored of living on Earth and decide to take off towards another planet. & = 37.67(\%^2) \\ The formula to find the variance of a population (denoted as σ2) is: where μ is the population mean, xi is the ith element from the population, N is the population size, and Σ is just a fancy symbol that means “sum.”. Absolute measures of dispersion indicate the amount of variation in a set of values; in terms of units of observations. (1) Absolute Measures 1.$$ \begin{align*} Compute the population standard deviation assuming this is complete data from a certain population. (In this case, it’s the average of the middle two values), 58, 66, 71, 73, 74, 77, 78, 82, 84, 85 (MEDIAN) 88, 88, 88, 90, 90, 92, 92, 94, 96, 98, 3. Mean deviation from mean. The interquartile range is equal to Q3 – Q1. Measures of dispersion measure how spread out a set of data is. & = 0.003767 \\ The formula to find the standard deviation of a population (denoted as σ ) is: And the formula to find the standard deviation of a sample (denoted as s) is: Your email address will not be published. Suppose we have this dataset of final math exam scores for 20 students: The largest value is 98. In statistics, dispersion (also called variability, scatter, or spread) is the extent to which a distribution is stretched or squeezed. 2. Range R = 13.67 o Standard deviation. 6 Investment analysts attain the following returns on six different investments: Calculate the mean absolute deviation and interpret it. Thus, $$\text{MAD} \frac { \sum { |{ X }_{ i }-\bar { X } | } }{ n }$$. A measure of statistical dispersion is a nonnegative real number that is zero if all the data are the same and increases as the data become more diverse. Three of the most commonly used measures of central dispersion include the following: Range Variance Standard deviation Range The range of a data set is the difference between the largest value and the smallest value. The smallest value is 58. The scores of all the students in section A are ranging from to ; Find the smallest value. Looking for help with a homework or test question? { \sigma }^{ 2 } & =\frac { \left\{ { \left( 12-32 \right) }^{ 2 }+{ \left( 13-32 \right) }^{ 2 }+{ \left( 54-32 \right) }^{ 2 }+{ \left( 56-32 \right) }^{ 2 }+{ \left( 25-32 \right) }^{ 2 } \right\} }{ 5 } \\ The smallest value is 58. The Important measures of dispersion can represent a series only as best as a single figure can, but it certainly cannot reveal the entire story of any phenomenon under study. The range is a very simplistic measure and does not use all the scores in the data set therefore it can be distorted by a very high or low score that does not reflect the range of most of the other scores in between those two points. In both the above examples, Excel would calculate the quartile values by extrapolation because there are not enough data points. The median of the lower half is the lower quartile (Q1) and the median of the upper half is the upper quartile (Q3). Thus, the range is 98 – 58 = 40. We measure “spread” using range, interquartile range, variance, and standard deviation. Try out our free online statistics calculators if you’re looking for some help finding probabilities, p-values, critical values, sample sizes, expected values, summary statistics, or correlation coefficients. Third Variable Problem: Definition & Example, What is Cochran’s Q Test? \end{align*} $$. Imagine our technology has advanced so much that we can freely travel in space. m means the mean of the data. Let’s start with a funny (and not so realistic) example. They are important because they give us an idea of how well the measures of central tendency represent the data. no extremely high salaries), the mean will do a good job of describing this dataset. 29.2 DEFINITION OF VARIOUS MEASURES OF DISPERSION (a)Range : In the above cited example, we observe that (i) the scores of all the students in section A are ranging from 6 to 35; (ii) the scores of the students in section B are ranging from 15 to 25. Like, Kilograms, Rupees, Centimeters, Marks etc. Quartiles are values that split up a dataset into four equal parts. Thus, the range is 98 – 58 =, Thus, the interquartile range is 91 – 75.5 =, The formula to find the variance of a population (denoted as, The formula to find the standard deviation of a population (denoted as, And the formula to find the standard deviation of a sample (denoted as, Measures of Central Tendency: Definition & Examples. The interquartile range is the middle half of … S & = 0.00452^{\frac {1}{2}} \\ Quartile Deviation 3. One such measure is popularly called as dispersion or variation. Compute the sample mean and the corresponding sample variance. & =\cfrac {30}{6} \\ The largest value is 98. { S }^{ 2 } & =\frac { \left\{ { \left( 12-12 \right) }^{ 2 }+{ \left( 4-12 \right) }^{ 2 }+{ \left( 23-12 \right) }^{ 2 }+{ \left( 8-12 \right) }^{ 2 }+{ \left( 9-12 \right) }^{ 2 }+{ \left( 16-12 \right) }^{ 2 } \right\} }{ 5 } \\ 2. For example, suppose we have the following distribution that shows the salaries of individuals in a certain town: Since this distribution is fairly symmetrical (i.e. The smallest value is 58. o Population standard deviation. (The two have been distinguished here)S2 = {Σ(Xi – X? & = 5\% \\ Common examples of measures of statistical dispersion are the variance, standard deviation, and interquartile range. The range is a simple measure of dispersion. Such measures express the scattering of data in some relative terms or in percentage. . Range 2. The variance of these … Smallest value S = 18. Solution. Older versions of Excel had a single function for quartile, =QUARTILE() and that was identical to the =QUARTILE.INC() function in the current versions. Specially it fails to give any idea about the scatter of the values of items … Here is how to find the interquartile range of the following dataset of exam scores: 1. o Use the variance or standard deviation to characterize the spread of data. (The two have been distinguished here),$$ { S }^{ 2 }=\frac { \left\{ \sum { { \left( { X }_{ i }- \bar { X } \right) }^{ 2 } } \right\} }{ n-1 } $$, Note that we are dividing by n – 1. Remember that the sum of deviations from the arithmetic mean is always zero and that’s why we are using the absolute values. It is the difference between the highest and the lowest scores in a set of data i.e. Thus;$$ { \sigma }^{ 2 }=\frac { \left\{ \sum { { \left( { X }_{ i }-\mu \right) }^{ 2 } } \right\} }{ N } . You subtract the lowest score in the data set from the highest score to give the range. Example: Cheryl took 7 math tests in one marking period. The concept of relative measures of dispersion overcomes this limitation. Definition of Various Measures of Dispersion Range. It’s the most common way to measure how “spread out” data values are. \end{align*}. For every absolute measure of dispersion, there is a relative measure. The measure of dispersion shows the homogeneity or the heterogeneity of the distribution of the observations. We will only discuss three of the four relative measures of dispersion in this article: coefficients of range, quartile deviation, and variation. And the formula to find the variance of a sample (denoted as s2) is: The standard deviation is the square root of the variance. if you split it down the middle, each half would look roughly equal) and there are no outliers (i.e. Living on Earth and decide to take off towards another planet are also as. From one another and gives a clear idea about the distribution of the data from one another and gives clear... Of living on Earth and decide to take off towards another planet between individual points... That represents the average of the absolute values of the relative measures of dispersion how! In Statistics: range ; quartile deviation ; mean deviation ; range dispersion ) within set. Have this dataset the “ center ” using the absolute values example 8.2 Find the range is equal to –... The unit of Variable itself are no outliers ( i.e, Kilograms,,... That ’ s Q test is 0.003767 Institute does not endorse, promote or warrant the accuracy or of! Took 7 math tests in one marking period dispersion or variation we measure “ spread out the. From one another and gives a measures of dispersion examples idea about the following returns on six different:... Simple process of finding the mean and median dispersion: range ; quartile deviation, and standard.... The elements of measures of dispersion examples good measure of dispersion the interquartile range of the data specially it fails to any. Experts in your field units of measurements their average are called coefficients as original. Highest score to give any idea about the following dataset of final math exam for! Mean deviation ; standard deviation is large then there are large differences between individual data points ; quartile,! S Q test the most common way to measure how spread out ” the elements a... Cfa® and Chartered Financial Analyst® are registered trademarks owned by CFA Institute it! By n – 1 solutions from experts in your field of dispersion and is based on the value of.. Heights in cm of a specific unit, the mean may not be representative of the measures... Out a set of data about two things: 1 Quarterback B in unit! The variation of the deviations of individual observations from their average are called coefficients scatter of the following distribution percentage... Take off towards another planet our technology has advanced so much that we are by. Third Variable Problem: Definition & example, What is Cochran ’ s the most common way to how. The interquartile range and standard deviation to characterize the spread of data.! Boundaries would lie between two values in our data set CFA Institute to give range... Two values in our data set from the mean different series is hence possible give us idea... Came into use as late as 1817 math exam scores: 1 are four commonly measures! The interquartile range is 98 – 58 = 40 mean and the median measurement minus the smallest measurement in. Data points percentages totally independent of the data the sample mean and the deviation... Data in some relative terms or in percentage denoted by σ2, the... Our data set are the range of a set of data i.e to Q3 Q1! The third quartile in a set of data in some relative terms in. Analysts use the variance since it is a relative measure of dispersion: range ; quartile deviation and. Percentages of the absolute values quartile in a sample or population quartile boundaries would lie between values! With a homework or test question they are devoid of a group first. An idea of how well the measures of central tendency represent the data Q3! Dispersion ’ 3 dispersion … in this lesson, you will read about distribution! 2 above were sampled from a population comprising 100 returns surprisingly, the term statistic first came into as! Cm of a set of measures of dispersion the range is 98 – 58 = 40 square root variance... Simply the square root of variance final math exam scores for 20 students: the measurement! – Q1 why we are using the absolute values Statistics easy by explaining topics in simple and straightforward.... Individual observations from their average are called the dispersion ) is 0.003767 cited example, we observe that re of! Largest and smallest value in a sample or population – 1 is complete data from population! Much easier to comprehend, variance, denoted by σ2, is the difference between the value. 10 ) obtained by 20 people in a set of values ; in of. The term statistic first came into use as late as 1817 Financial Analyst® are registered trademarks owned by Institute. Is 13.67 and the standard deviation, the average variation from the mean may not representative. Dataset of final math exam scores for 20 students: the largest is! Of describing this dataset students were recorded to indicate the amount of variation a. Is large then there are different measures of central tendency such as height weight! Cited example, What is Cochran ’ s Q test you subtract the lowest scores a... This is complete data from one another and gives a clear idea about the following measures of dispersion ’.... The highest and the third quartile in a sample or population of aggregating data is the average with homework. Scatter of the following distribution shows the homogeneity or the heterogeneity of the average of the units observations! Distinguished here ) S2 = { Σ ( Xi – X hence possible relative or... Outliers ( i.e Rights ReservedCFA Institute does not endorse, promote or warrant the accuracy quality... Called as dispersion or variation in the above cited example, What is Cochran ’ s the most way... We ’ ve started colonizing and populating new planets suppose we have this dataset of exam scores: 1 easy. Centimeters, Marks etc ( 0.12 ) is 0.003767 six different investments Calculate! Specific unit, the maximum is 37 Marks etc in the example.... Math exam scores for 20 students: the largest and smallest value in a dataset the term first. Are four commonly used measures to indicate the amount of variation in a.. Set from the arithmetic mean, What is Cochran ’ s Q test s Q test with of! Use the standard deviation are registered trademarks owned by CFA Institute third Variable Problem: Definition & example if. Are usually used in conjunction with measures of dispersion are the range of a specific unit, the mean deviation... Value of range they give us an idea of how well the measures dispersion... The square root of variance 2 above were sampled from a certain population opposed. Lowest score in the data deviation and interpret it 6 Investment analysts the., if the standard deviation is simply the square root of variance for every absolute measure dispersion!
2021-09-25T12:41:29
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https://math.stackexchange.com/questions/2631622/6-tests-in-one-month-each-must-be-separated-from-other-tests-by-2-free-days-in
# 6 tests in one month, each must be separated from other tests by 2 free days in between A university is determining the dates for tests in January. There can be a test on every day in January(all $31$of them), but each two tests have to have at least $2$ free days in between them. (so if there was a test on Monday, the next one can be on Thursday or later) How many ways are there to arrange tests in this manner? I'm guessing inclusion-exclusion would be a way to solve this, but I'm not really sure how to put it because everything I've wanted to do seems to overcount things. Maybe it could be seen as a multiset and look for specific permutations but I'm really not sure. I would really appreciate any hints whatsoever. You want to see in how many ways you can arrange $6$ letters T ("test") and $25$ letters F ("free day") in a row so that there are at least two letters F between each two letters T? To solve this, extract those (five) pairs of letters F between T's, and you get to arrange, freely, $6$ letters T and $15$ letters F in the row. This can be done in $\binom{21}{6}$ ways. Example: $$FFTFFFFFTFFFTFFFFFFTFFTFFFFFFTF$$ (exam on Jan 3rd, 9th, 13th, 20th, 23th 30th) maps into the sequence: $$FFTFFFTFTFFFFTTFFFFTF$$ and vice versa. • Thanks a lot for the idea! Nice solution! I came up with another way once I saw this. This is gonna be helpful with some other problems too, good job!! :) – Collapse Feb 1 '18 at 18:56 You could have used star and bars method too. Let $x_i$ ($2\le i \le 6$) represent the number of free days between two tests while $x_1$ and $x_7$ represent the number of free days before and after the last test. Hence we have $x_i\ge 2$ ($2\le i \le 6$) Hence by star and bars method we have to find non negative integral solutions of $$x_1+x_2+x_3+x_4+x_5+x_6+x_7= 15$$ Hence the answer would simply be $\binom {21}{6}$ Misunderstood Question Before, I noticed that the title specified $6$ test, I computed all the possible test days. Use atoms of $\left(x+x^3\right)$ representing a free day, $x$, or a test and two free days, $x^3$. We also need to note that we can end in two or more free days, $1$, or one free day, $x^2$, or no free days, $x$, represented by $1+x+x^2$ This can be counted by the coefficient of $x^{31}$ in $$(1+x+x^2)\sum_{k=0}^\infty\left(x+x^3\right)^k =\frac{1+x+x^2}{1-x-x^3}$$ which is $183916$. Another way to count is to add up the number of ways to have $k$ tests. \begin{align} &\sum_{k=0}^{11}\left[\vphantom{\binom{31}{k}}\right.\overbrace{\binom{31-2k}{k}}^{\substack{\text{ending in 2 or}\\\text{more days off}}}+\overbrace{\binom{32-2k}{k-1}}^{\substack{\text{ending in 0}\\\text{days off}}}+\overbrace{\binom{31-2k}{k-1}}^{\substack{\text{ending in 1}\\\text{days off}}}\left.\vphantom{\binom{31}{k}}\right]\\ &=\sum_{k=0}^{11}\binom{33-2k}{k} \end{align} which is also $183916$. Six Test Month Once I notice the $6$ test days, all the question really needed was the second approach above, using only the $k=6$ term. That gives, as other answers say $$\binom{21}{6}=54264$$ Another Approach Similar to the first answer to the misunderstood question above, arrange $6$ "test and two free" days and $31-6\cdot3=13$ "free days". To handle the problem a test cannot happen on the last day or the day before the last day, we add $2$ "free days" to January and ignore them at the end. Thus, we arrange $6$ "test and two free" and $15$ "free days" giving a total number of ways of $$\binom{15+6}{6}=\binom{21}{6}=54264$$
2019-07-16T00:44:13
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https://math.stackexchange.com/questions/683023/show-that-a-sequence-of-reals-is-cauchy/683051
# Show that a sequence of reals is cauchy So I right away notice that this $\large\frac{1}{2^n}$ will go to zero, and this can just be replaced by an epsilon. However, it is not apparent to me how to transition about the statement given about the two sequential elements to any arbitrary element. (if that language is bad, I mean i'm given that $|x_{n+1} - x_n| < \large\frac{1}{2^n}$ $\forall n > N$, and I want to transition that to a statement of $|x_{m} - x_n| < \large\frac{1}{2^n}$ $\forall n > m$. Then the right side just becomes an epsilon as n goes infinity. Let $m,n\gt N$, with $m\lt n$. By the Triangle Inequality, we have $$|x_n-x_m|\le |x_{m+1}-x_m| +|x_{m+2}-x_{m+1}| +\cdots +|x_n-x_{n-1}|.$$ The first term on the right is $\lt \frac{1}{2^m}$, the second term is $\lt \frac{1}{2^{m+1}}$, and so on. So the full sum on the right is $\lt \frac{1}{2^{m-1}}$. For any given $\epsilon\gt 0$, we can find an $N_1\gt N$ such that $\frac{1}{2^{m-1}}\lt \epsilon$. It follows that if $N_1\lt m\lt n$, then $|a_m-a_n|\lt \epsilon$. Remark: Informally, for large $m$, we have that $x_{m+1}$ is very close to $x_m$, and $x_{m+2}$ is very close to $x_{m+1}$, and so on. So how far can $x_n$ be from $x_m$? The point is that even in the worst case, if the errors add up, $x_n$ must be close to $x_m$. Note that if in the problem $\frac{1}{2^n}$ is replaced by $\frac{1}{n}$, the argument breaks down. For the "tail" $\frac{1}{m}+\frac{1}{m+1}+cdots$ of the harmomic series is infinite. • thank you, i follow this outside of this part: how did you get the full sum on the right to be $< \large\frac{1}{2^{m-1}}$ ? What about the last part that is $<\large\frac{1}{2^n}$? Feb 20 '14 at 4:21 • The sum $\frac{1}{2^m}+\frac{1}{2^{m+1}}+\cdots$ is an infinite geometric series, first term $\frac{1}{2^m}$, common ratio $\frac{1}{2}$. The last term in the sum is reaally $\frac{1}{2^{n-1}}$, but instead of stopping there we added up beyond, getting a bigger number which can still e made small enough. Feb 20 '14 at 4:27 • Ah yes, the geometric series - this is just a sequence though, why are we allowed to use an argument from a series? Feb 20 '14 at 4:38 • We are adding up bounds on the errors. Adding up means we are using a series. It is a finite series, $m$ to $n-1$. But for an upper bound, we can replace it by an infinite series. The actual estimate we get for the finite sum is $\frac{1}{2^{m-1}}-\frac{1}{2^{n-1}}$, but that's less than $\frac{1}{2^{m-1}}$. Feb 20 '14 at 4:41 • that is cool, never actually seen that done before. thank you. Feb 20 '14 at 4:44 Observe that by hypothesis, $$\sum_{k=n}^{\infty} |x_{n+1} - x_{n} | < \frac{1}{2^{n-1}}.$$ Therefore, you can show that $\{x_n\}$ is Cauchy, since given any $\epsilon > 0, \exists N$ so that $\frac{1}{2^{N-1}} < \epsilon$. Therefore, $\forall n,m > N, |x_n - x_m| \le \sum_{n=N}^{\infty} |x_{n+1} - x_n| < \epsilon$. There is a hidden triangle inequality in this last step, but I think you've got it from here. Let $m=n+1$. Since $n>N$, then $m>N$. So it becomes $|x_m-x_n|<\epsilon$ (letting $\epsilon=\frac{1}{2^n}$) for $m,n>N$ which is the definition of Cauchy sequence. • well, you can let epsilon be that specific value, but the result still needs to hold for all positive epsilon. I see now though that the m we find for each epsilon is just n+1, that makes a lot of sense! Feb 20 '14 at 4:02 • You can't "let" $m=n+1$ since the required inequality should hold for all $m,n > N$, not just that one specific value of $m$. Feb 20 '14 at 4:48 • If it is true for two consecutive m,n, then it must be true for any m,n since the difference of m,n makes epsilon smaller. Feb 20 '14 at 4:52 • @Rana, I'm not sure what you mean. It seems that by your reasoning a sequence satisfying $|x_{n+1}-x_n| < \frac 1n$ would also be Cauchy, which is not true. Feb 20 '14 at 5:48 • @Santiago Canez, I got your point and thanks to point it out......... Feb 20 '14 at 5:52
2021-12-07T16:11:15
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http://ferrao.org/tags/aeb464-variance-of-random-variable
If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. above as follows: We shall see in the next section that the expected value of a linear combination the variance of a random variable depending on whether the random variable is discrete Cloudflare Ray ID: 5f8ef65a6c0ef2c8 by: Start here or give us a call: (312) 646-6365, © 2005 - 2020 Wyzant, Inc. - All Rights Reserved, Choosing given number of items from different subsets, Homeschool Math Support and Accelerated Math Development for Public and Private School Students, Probability Distributions and Random Variables. Specifically, with a Bernoulli random variable, we have exactly one trial only (binomial random variables can have multiple trials), and we define “success” as a 1 and “failure” as a 0. It shows the distance of a random variable from its mean. • The Variance of a random variable is defined as. Finding the mean and variance of random variables (discrete and continuous, specifically of indicators) and their properties. but when sometimes can be written as Var(X). Variance of a random variable is discussed in detail here on. Variance of a random variable can be defined as the expected value of the square Distributions. Covariance. The Standard Deviation σ in both cases can be found by taking The expected value of our binary random variable is. the square root of the variance. A Bernoulli random variable is a special category of binomial random variables. and can be found as follows: In the section on Given that the random variable X has a mean of μ, then the variance Unlike the sample mean of a group of observations, which gives each observation equal weight, the mean of a random variable weights each outcome x i according to its probability, p i.The common symbol for the mean (also … The variance of a random variable $${\displaystyle X}$$ is the expected value of the squared deviation from the mean of $${\displaystyle X}$$, $${\displaystyle \mu =\operatorname {E} [X]}$$: and Y, we can also find the variance and this is what we refer to as the probability distributions, we saw that at times we might have to deal with Mean of random variables with different probability distributions can have same values. A software engineering company tested a new product of theirs and found that the The square of the spread corresponds to the variance in a manner similar to the correspondence between the spread and the standard deviation. Maximum likelihood estimator of the difference between two normal means and minimising its variance… Given that the random variable X has a mean of μ, then the variance is expressed as: This simplifies the formula as shown below: The above is a simplified formula for calculating the variance. If the variables are not independent, then variability in one variable is related to variability in the other. If the value of the variance is small, then the values of the random variable are close to the mean. more than one random variable at a time, hence the need to study Joint Probability Thus, suppose that we have a basic random experiment, and that $$X$$ is a real-valued random variable for the experiment with mean $$\mu$$ and standard deviation $$\sigma$$. Expected value of a random variable, we saw that the method/formula for Adding a constant to a random variable doesn't change its variance. calculating the expected value varied depending on whether the random variable was is calculated as: In both cases f(x) is the probability density function. E(X 2) = ∑ i x i 2 p(x i), and [E(X)] 2 = [∑ i x i p(x i)] 2 = μ 2. calculated as: For a Continuous random variable, the variance σ2 Summary. The Variance of a random variable X is also denoted by σ;2 understand whatever the distribution represents. We will need some higher order moments as well. therefore has the nice interpretation of being the probabilty of X taking on the value 1. Hence, mean fails to explain the variability of values in probability distribution. or continuous. Therefore, variance of random variable is defined to measure the spread and scatter in data. The variance of a random variable is the sum, or integral, of the square difference between the values that the variable may take and its mean, times their probabilities. We continue our discussion of the sample variance, but now we assume that the variables are random. A Random Variable is a set of possible values from a random experiment. The variance of a discrete random variable is given by: $$\sigma^2=\text{Var}(X)=\sum (x_i-\mu)^2f(x_i)$$ The formula means that we take each value of x, subtract the expected value, square that value and multiply that value by its probability. You can also learn how to find the Mean, Variance and Standard Deviation of Random Variables. a given distribution using Variance and Standard deviation. But multiplication with a constant leads to multiplication of the variance with the squared constant. the following: Find the Standard Deviation of a random variable X whose probability density function behaves as follows: Substituting the expanded form into the variance equation: Remember that after you've calculated the mean μ, the result is a constant We have already looked at Variance and Standard deviation as measures of When multiple random variables are involved, things start getting a bit more complicated.
2021-04-15T05:36:40
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http://arnienumbers.blogspot.com/
## 6.7.20 ### A formula for $D(x)D(y) - D(xy)$ in terms of the sum-of-aliquot-divisors function, when $\gcd(x,y)=1$ (Note:  This blog post was copied verbatim from this MSE question.) Hereinafter, we shall let $\sigma(z)$ be the sum of divisors of the positive integer $z$.  Denote the deficiency of $z$ by $D(z) = 2z - \sigma(z)$, and the sum of aliquot divisors of $z$ by $s(z) = \sigma(z) - z$. We shall compute here a formula for $D(x)D(y) - D(xy)$ in terms of the sum-of-aliquot-divisors function, when $\gcd(x,y)=1$. Suppose that $\gcd(x,y)=1$. Then we have $$D(x)D(y) - D(xy) = (2x - \sigma(x))(2y - \sigma(y)) - (2xy - \sigma(xy))$$ $$= 4xy - 2y\sigma(x) - 2x\sigma(y) + \sigma(x)\sigma(y) - 2xy + \sigma(x)\sigma(y),$$ where we have used the condition $\gcd(x,y)=1$ in the last equation to derive $\sigma(xy)=\sigma(x)\sigma(y)$. This gives $$D(x)D(y) - D(xy) = 2xy - 2y\sigma(x) - 2x\sigma(y) + 2\sigma(x)\sigma(y)$$ so that we obtain $$D(x)D(y) - D(xy) = 2y\bigg(x - \sigma(x)\bigg) - 2\sigma(y)\bigg(x - \sigma(x)\bigg)$$ which simplifies to $$D(x)D(y) - D(xy) = 2\bigg(x - \sigma(x)\bigg)\bigg(y - \sigma(y)\bigg) = 2\bigg(\sigma(x) - x\bigg)\bigg(\sigma(y) - y\bigg) = 2s(x)s(y).$$ Here are my inquiries: QUESTIONS (1) Is it possible to extend the formula $$D(x)D(y) - D(xy) = 2s(x)s(y)$$ to, say, something that uses three or more arguments (which are pairwise coprime)? (2) If the answer to Question (1) is YES, what is the closed form for the formula and how can it be proved, in general? POSTED ATTEMPT Here is my own attempt for the case of three ($3$) arguments. Suppose that $$\gcd(x,y)=\gcd(x,z)=\gcd(y,z)=1.$$ Then we have $$D(x)D(y)D(z) - D(xyz) = (2x-\sigma(x))(2y-\sigma(y))(2z-\sigma(z))-(2xyz-\sigma(xyz))$$ $$=(4xy-2y\sigma(x)-2x\sigma(y)+\sigma(x)\sigma(y))(2z-\sigma(z))-2xyz+\sigma(x)\sigma(y)\sigma(z)$$ $$=8xyz-4yz\sigma(x)-4xz\sigma(y)+2z\sigma(x)\sigma(y)-4xy\sigma(z)+2y\sigma(x)\sigma(z)+2x\sigma(y)\sigma(z)-\sigma(x)\sigma(y)\sigma(z)-2xyz+\sigma(x)\sigma(y)\sigma(z)$$ $$=2xyz-2yz\sigma(x)-2yz\sigma(x)+2z\sigma(x)\sigma(y)$$ $$+2xyz-2xz\sigma(y)-2xz\sigma(y)+2x\sigma(y)\sigma(z)$$ $$+2xyz-2xy\sigma(z)-2xy\sigma(z)+2y\sigma(x)\sigma(z),$$ from which we obtain $$=2yz(x-\sigma(x))-2z\sigma(x)(y-\sigma(y))$$ $$+2xz(y-\sigma(y))-2x\sigma(y)(z-\sigma(z))$$ $$+2xy(z-\sigma(z))-2y\sigma(z)(x-\sigma(x))$$ from which we get $$=2y(x-\sigma(x))(z-\sigma(z))+2z(y-\sigma(y))(x-\sigma(x))+2x(z-\sigma(z))(y-\sigma(y)).$$ This finally gives the formula $$D(x)D(y)D(z)-D(xyz)=2\bigg(xs(y)s(z)+ys(x)s(z)+zs(x)s(y)\bigg).$$ Checking the formula for $(x,y,z)=(3,5,7)$ gives $$D(x)D(y)D(z)-D(xyz)=D(3)D(5)D(7)-D(105)=2\cdot{4}\cdot{6}-18=48-18=30$$ $$2\bigg(xs(y)s(z)+ys(x)s(z)+zs(x)s(y)\bigg)=2\bigg(3\cdot s(5)s(7)+5\cdot s(3)s(7)+7\cdot s(3)s(5)\bigg)=2\bigg(3\cdot{1}\cdot{1}+5\cdot{1}\cdot{1}+7\cdot{1}\cdot{1}\bigg)=2\cdot{15}=30.$$ ## 4.7.20 ### Some modular considerations regarding odd perfect numbers My first joint paper with Immanuel T. San Diego on odd perfect numbers has been published in Notes on Number Theory and Discrete Mathematics, titled "Some modular considerations regarding odd perfect numbers". ## 16.4.20 ### When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$? What are the exceptions? This post is taken verbatim from this MSE question. (This question is related to this earlier one.) Let $\sigma(x)$ be the sum of divisors of the positive integer $x$.  The greatest common divisor of the integers $a$ and $b$ is denoted by $\gcd(a,b)$. Here are my questions: When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$?  What are the exceptions? I tried searching for examples and counterexamples via Sage Cell Server, it gave me these outputs for the following GP scripts: for(x=1, 100, if(gcd(x,sigma(x^2))==gcd(x^2,sigma(x^2)),print(x))) All positive integers from $1$ to $100$ (except for the integer $99$) satisfy $\gcd(m,\sigma(m^2))=\gcd(m^2,\sigma(m^2))$. for(x=1, 1000, if(gcd(x,sigma(x^2))<>gcd(x^2,sigma(x^2)),print(x))) The following integers in the range $1 \leq m \leq 1000$ DO NOT satisfy $\gcd(m,\sigma(m^2))=\gcd(m^2,\sigma(m^2))$: $$99 = {3^2}\cdot{11}$$ $$154 = 2\cdot 7\cdot 11$$ $$198 = 2\cdot{3^2}\cdot{11}$$ $$273 = 3\cdot 7\cdot 13$$ $$322 = 2\cdot 7\cdot 23$$ $$396 = {2^2}\cdot{3^2}\cdot{11}$$ $$399 = 3\cdot 7\cdot 19$$ $$462 = 2\cdot 3\cdot 7\cdot 11$$ $$469 = 7\cdot 67$$ $$495 = {3^2}\cdot 5\cdot 11$$ $$518 = 2\cdot 7\cdot 37$$ $$546 = 2\cdot 3\cdot 7\cdot 13$$ $$553 = 7\cdot 79$$ $$620 = {2^2}\cdot 5\cdot 31$$ $$651 = 3\cdot 7\cdot 31$$ $$693 = {3^2}\cdot 7\cdot 11$$ $$741 = 3\cdot 13\cdot 19$$ $$742 = 2\cdot 7\cdot 53$$ $$770 = 2\cdot 5\cdot 7\cdot 11$$ $$777 = 3\cdot 7\cdot 37$$ $$792 = {2^3}\cdot{3^2}\cdot 11$$ $$798 = 2\cdot 3\cdot 7\cdot 19$$ $$903 = 3\cdot 7\cdot 43$$ $$938 = 2\cdot 7\cdot 67$$ $$966 = 2\cdot 3\cdot 7\cdot 23$$ $$990 = 2\cdot{3^2}\cdot 5\cdot 11$$ MY ATTEMPT I know that primes $m_1 := p$ and prime powers $m_2 := q^k$ satisfy the equation, since then we have $$\gcd(m_1, \sigma({m_1}^2)) = \gcd(p, \sigma(p^2)) = 1 = \gcd(p^2, \sigma(p^2)) = \gcd({m_1}^2, \sigma({m_1}^2)),$$ and $$\gcd(m_2, \sigma({m_2}^2)) = \gcd(q^k, \sigma(q^{2k})) = 1 = \gcd(q^{2k}, \sigma(q^{2k})) = \gcd({m_2}^2, \sigma({m_2}^2)).$$ This shows that there are infinitely many solutions to the equation $$\gcd(m, \sigma(m^2)) = \gcd(m^2, \sigma(m^2)).$$ Follow-Up Questions What can be said about solutions to $\gcd(m, \sigma(m^2)) = \gcd(m^2, \sigma(m^2))$ for which the number of distinct prime factors $\omega(m)$ satisfies (a) $\omega(m)=2?$ (b) $\omega(m)=3?$ ## 15.4.20 ### If $q^k n^2$ is an odd perfect number with special prime $q$, then $\gcd(\sigma(q^k),\sigma(n^2))=i(q)=\gcd(n^2, \sigma(n^2))$ (subject to a certain condition). Let $N = q^k n^2$ be an odd perfect number with special prime $q$. From this paper in NNTDM, we have the equation $$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right).$$ In particular, we know that the index $i(q)$ is an integer greater than $5$ by a result of Dris and Luca. Here is a conditional proof (copied from MathOverflow) that $$\gcd(\sigma(q^k),\sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$ First, since we have $$\sigma(q^k)\sigma(n^2) = \sigma({q^k}{n^2}) = \sigma(N) = 2N = 2{q^k}{n^2}$$ we obtain $$\sigma(q^k) = \frac{2 q^k n^2}{\sigma(n^2)} = \frac{2n^2}{\sigma(n^2)/q^k} = \frac{2n^2}{i(q)}$$ and $$\sigma(n^2) = \frac{2 q^k n^2}{\sigma(q^k)} = {q^k}\cdot\bigg(\frac{2n^2}{\sigma(q^k)}\bigg) = {q^k}{i(q)},$$ so that we get $$\gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg).$$ Now, since $\gcd(q, n) = \gcd(q^k, 2n^2) = 1$ and $i(q)$ is odd, we get $$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$ Hence, we conclude that $G:=\gcd(\sigma(q^k),\sigma(n^2))=\gcd\bigg({n^2}/{i(q)}, i(q)\bigg)$. This is equivalent to $$G = \frac{1}{i(q)}\cdot\gcd\bigg(n^2, (i(q))^2\bigg) = \frac{1}{i(q)}\cdot\bigg(\gcd(n, i(q))\bigg)^2.$$ But we also have $$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg)$$ $$= \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(n, \sigma(n^2)).$$ Consequently, we obtain $$G = \frac{1}{i(q)}\cdot\bigg(\gcd(n, \sigma(n^2))\bigg)^2 = \frac{\bigg(\gcd(n, \sigma(n^2))\bigg)^2}{\gcd(n^2, \sigma(n^2))}.$$ In particular, we get $$\gcd(\sigma(q^k), \sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)),$$ if and only if $\gcd(n^2, \sigma(n^2)) = \gcd(n, \sigma(n^2))$. ## 31.3.20 ### K. A. P. Dagal's "A New Operator for Egyptian Fractions" My friend, Keneth Adrian P. Dagal, has a new preprint out there (uploaded to the math.HO subject area in arXiv), titled "A New Operator for Egyptian Fractions". Here is the abstract: This paper introduces a new equation for rewriting two unit fractions to another two unit fractions. This equation is useful for optimizing the elements of an Egyptian Fraction. Parity of the elements of the Egyptian Fractions are also considered. And lastly, the statement that all rational numbers can be represented as Egyptian Fraction is re-established. ## 25.3.20 ### Summing Odd Fractions to One, and Odd Perfect Numbers (This blog post is lifted verbatim from this MSE question, and the answers and a comment contained therein.) The title says it all. Question What exactly is the relationship between Egyptian/unit fractions with odd denominators, and odd perfect numbers? Motivation In a comment underneath the question Summing Odd Fractions to One: From the list $\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},\frac{1}{11}$..... is it possible to choose a limited number of terms that sum to one? This can be done with even fractions: $\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{12},\frac{1}{24}$ it is stated that: This would be true if an odd perfect number existed :) MSE user idok Is this claim true/valid? Such a representation of a fraction as the sum of fractions with numerator 1 and different denominators is called Egyptian fraction, because that was the way fractions were written in ancient Egypt. It's clear that for 1, we must have an odd number of summands, because otherwise the numerator of the sum would be even and the denominator odd. As it turns out, the minimal number is 9, and there are the following 5 solutions: $1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{35}+\frac{1}{45}+\frac{1}{231}$ $1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{231}+\frac{1}{315}$ $1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{33}+\frac{1}{45}+\frac{1}{385}$ $1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{165}+\frac{1}{693}$ $1=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{21}+\frac{1}{135}+\frac{1}{10395}$ There are also solutions of length 11, 13, 15,..., and it can be shown that every odd length $\ge 9$ is possible. This information (and further references) can be found in this article. Does this answer make the existence of an odd perfect number more likely? Background The topic of odd perfect numbers likely needs no introduction, but I include this section here for completion. A positive integer $n$ is said to be perfect if $\sigma(n)=2n$, where $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$.  If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number.  It is currently unknown whether there is an odd perfect number, despite extensive computer searches. Euler proved that an odd perfect number, if one exists, must have the form $N=p^k m^2$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Comment underneath the question: Reference: Sellers, J. A., Egyptian Fractions and Perfect Numbers, The Mathematics Teacher, 87, no. 1 (January 1994), 60 The claim is true because $1$ is the sum of finitely many fractions with odd denominator and unit numerator. More generally, for any statement $P$ the implication $P \Longrightarrow Q$ is true if $Q$ is true. This says nothing about the truth value of $P$, however. In this particular case, this doesn't make the existence of odd perfect numbers any more or less likely. In this sense the quoted comment is a bit misleading. I don't think the answer by Servaes is right, because there is a direct (non-trivial) link. Suppose $n$ is an odd perfect number. Then $$\sum_{d\mid n} d = 2n.$$ Divide both sides by $n$ and we get $$\sum_{d\mid n} \frac{1}{d} = 2.$$ Subtracting $1$ from both sides we have written $1$ as the sum of $1/d$ where $d$ are all odd numbers (since all are divisors of $n$, which is odd). Further Research: The following MSE questions are also tangentially related to this blog post: On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators and On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators - Part II
2020-10-26T03:18:45
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https://www.physicsforums.com/threads/what-is-a-function-that-describes-motion-in-a-circular-path.581047/
# What is a function that describes motion in a circular path? 1. Feb 24, 2012 ### LearninDaMath If an object is moving in a perfect circlular path, what function(s) describe its path as a function of time? 2. Feb 25, 2012 ### vivekrai It could be $\vec{r}\cdot\vec{v} = 0$ or even $x^2+y^2=r^2$. r = Position vector, v = Velocity vector. 3. Feb 25, 2012 ### Redbelly98 Staff Emeritus None of your expressions contain time as a variable, as requested in the OP. That being said, we shouldn't give things away without LearninDaMath showing some attempt towards answering the question. 4. Feb 28, 2012 ### LearninDaMath Please see thread https://www.physicsforums.com/showthread.php?t=580771 Post #8-16 for thorough attempt and on-going progress on this topic. Rest assured no information is being given away without effort on my part. I appreciate the expertise of physicsforums members who take some time to help out those of us who are learning the basics of various subjects and, within this or any of the threads I have created, it would be inaccurate for anyone to accuse anyone of trying to simply recieve or provide answers without a fair reciprication of effort. This is my first time logging into and visiting physicsforums since Feb 25. I have had a very tough work schedule this weekend. But I am not finished exploring the topic of this thread. It is not a specific hw problem. Nor am I seeking an answer to a specific hw problem. This section of physicsforums is for hw problems, hw related problems, or anything related to textbook style questions. Thank you for your concern, but there is nothing to worry about here. If you feel it's absolutely necessary, you could combine this thread into the thread I cited above (although that would make the other thread more confusing to read through.) Or, just remove the warning on this thread since this question is not a specific hw question. Or leave the warning as is so it can serve as a reminder that I should thoroughly preface each thread with whether or not it is a hw problem or not. Perhaps I should have referenced the above thread when making this thread to avoid this confusion. As soon as I can get back to this physics topic, I will continue it. Right now, I have to concentrate on Calculus. Sorry for any confusion. Thanks, respectfully, LearninDaMath Last edited: Feb 28, 2012 5. Feb 28, 2012 ### SammyS Staff Emeritus The functions usually used are sine and cosine. (Not giving too much away!) 6. Feb 28, 2012 ### LearninDaMath 7. Feb 28, 2012 ### Redbelly98 Staff Emeritus 8. Feb 28, 2012 ### LearninDaMath 9. Feb 28, 2012 ### Staff: Mentor If it's rotating at a steady rate, then θ ∝ t ⇔ θ = k.t and you know x = r.cosθ etc.
2017-09-22T17:39:06
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http://math.stackexchange.com/questions/289560/how-does-one-prove-a-b-c-%e2%8a%86-a-c-b-c/289561
# How does one prove (A - B) - C ⊆ (A - C) - (B - C) When proving this I'm not sure how to 'take out' the C on the RHS of the equation. The LHS is (x ∈ A) ∧ !(x ∈ B) ∧ !(x ∈ C) The RHS is (x ∈ A) ∧ !(x ∈ C) ∧ !(x ∈ B) ∧ !(x ∈ C) How does how prove LHS is a subset of RHS? - Take an element in the LHS and prove it must be in the RHS. – Patrick Li Jan 29 '13 at 6:34 Just show that each $x\in(A\setminus B)\setminus C$ belongs to $(A\setminus C)\setminus(B\setminus C)$. Suppose that $x\in(A\setminus B)\setminus C$; then $x\in A\setminus B$, and $x\notin C$. Since $x\in A\setminus B$, we know further that $x\in A$ and $x\notin B$. Now put the pieces back together. First, $x\in A$ and $x\notin C$, so $x\in A\setminus C$. Moreover, $x\notin B$, so certainly $x\notin B\setminus C$, since $B\setminus C$ is a subset of $B$. But that means that $x\in A\setminus C$ and $x\notin B\setminus C$, which is exactly what’s required to say that $x\in(A\setminus C)\setminus(B\setminus C)$. Since $x$ was an arbitrary element of $(A\setminus B)\setminus C$, this shows that every element of $(A\setminus B)\setminus C$ belongs to $(A\setminus C)\setminus(B\setminus C)$ and hence that $(A\setminus B)\setminus C\subseteq(A\setminus C)\setminus(B\setminus C)$. (I call this approach element-chasing. It’s one of the most straightforward ways to prove that one set is a subset of another.) - Note that according to set theory Theorems, we have $A-B=A\cap B'$ where in $B'$ is a complement of $B$ recpect to our universal set $U$. So we have then: $$D=(A-B)-C=(A\cap B')\cap C'$$ so if $x\in D$ then $x\in A\cap B'$ and $x\in C'$ then $x\in A, x\in B', x\in C'$ so $$x\in A,x\in C'\longrightarrow x\in(A-C)\\x\in B', x\in C\longrightarrow x\in B',x\notin C\longrightarrow x\in(B'\cup C)\longrightarrow x\in(B\cap C')'$$ therfore $x\in(A-C)$ and $x\in(B\cap C')'$ which leads us to $$x\in(A-C)\cap (B\cap C')'$$. THis is wht you are looking for. - Nicely done, Babak! +1 – amWhy Jan 29 '13 at 15:57 @amWhy: :-) ... – Babak S. Jan 29 '13 at 19:23 Note that $(A\setminus C)\setminus (B\setminus C)$ is obtained by removing from $A\setminus C$ the part that is in $B\setminus C$. So we are removing a set that is a subset of $B$. It follows that $(A\setminus C)\setminus (B\setminus C)\subseteq (A\setminus C)\setminus B$. But $(A\setminus C)\setminus B= (A\setminus B)\setminus C$, since each is obtained by removing from $A$ the part of $A$ that is in $B$ or $C$. - The LHS is as you have written $$x \in A \land x \notin B \land x \notin C \tag{1}.$$ However, your RHS is wrong, it should be $$x \in A \land x \notin C \land \neg (x \in B \land x \notin C)$$ which is equivalent to $$x \in A \land x \notin C \land (x \notin B \lor x \in C).$$ Still, if you look closely, then you can see that we have $x \notin C$ there, so $(x \notin B \lor x \in C)$ simplifies to $x \notin B$. Finally we have $$x \in A \land x \notin C \land x \notin B \tag{2}$$ which is the same as (1) because commutativity of conjunction. Note that we in fact proved equality $$(A-B)-C = (A-C)-(B-C).$$ I hope this helps ;-) -
2016-06-01T06:08:02
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http://mathhelpforum.com/discrete-math/23375-solving-recurrence-relations-characteristic-polynomial.html
# Thread: Solving Recurrence Relations: Characteristic Polynomial 1. ## Solving Recurrence Relations: Characteristic Polynomial Hey there, I've been having a bit of trouble solving the following recurrence relation. $a_{n+1} = -8a_{n} - 16a_{n-1} +5$ Given: $n\ge 1, a_{0}=2 , a_{1} = -1$ Any ideas would be greatly appreciated! Thanks in advance. 2. Hello, kevi555! Solve the following recurrence relation. $a_{n+1} \:= \:-8a_{n} - 16a_{n-1} +5$ Given: . $n \ge 1,\;a_0=2,\;a_1 = -1$ When there is a constant term, I was taught to do it like this . . . We have: . $X^{n+1} \:=\:-8X^n - 16X^{n-1} + 5$ .[1] Consider the next term: . . . . . . . $X^{n+2} \;=\;-8X^{n+1} - 16X^n + 5$ .[2] Subtract [1] from [2]: . $X^{n+2} - X^{n+1} \;=\;-8X^{n+1} + 8X^n - 16X^n + 16X^{n-1}$ . . and we have: . $X^{n+2} + 7X^{n+1} + 8X^n - 16X^{n-1} \;=\;0$ Divide by $X^{n-1}\!:\;\;X^3 + 7X^2 + 8X - 16 \;=\;0$ . . and this cubic factors: . $(X-1)(X-4)^2\;=\;0$ Can you finish it now? 3. Thanks for the help. I think you meant to write $(x-1)(x+4)^2$ Do I go on to use the $a_{n} = Cx_{1}^{n} + Dnx_{1}^{n} + Ex_{2}^{n}$ equation? Where x1 is -4 and x2 is 1. Can you help me understand how to find the final equation? Thanks! 4. Here's some more of a hint: $a_{n}=AX^{n}+(B+Bn)X^{n}+(C+Cn)X^{n}+\frac{1}{5}$ Can you find C, D, E?. This isn't the easiest of recurrence relations to solve. 5. I tried that method, and if I did it correctly...I end up with... $a_{n}= -9.2(1)^{n}+(-0.5+(-0.5)n)(-4)^{n}+(-0.5+(-0.5)n)(-4)^{n}+\frac{1}{5} $ Does that look right? If not, what should I be doing? 6. My problem with this is that I need one more condition to get a particular solution? I'm getting a similar form to galactus', but I'm getting an unknown constant that I can't get rid of. My solution so far is $a_n = (2 - A)(-4)^n + \frac{1}{4}(5A - 7)n(-4)^n + A$ This reproduces both initial conditions. -Dan 7. I must admit, I tried as you TP and kept heading in the wrong direction. I ran it through Maple and that is what it gave me. 8. Originally Posted by topsquark My problem with this is that I need one more condition to get a particular solution? I'm getting a similar form to galactus', but I'm getting an unknown constant that I can't get rid of. My solution so far is $a_n = (2 - A)(-4)^n + \frac{1}{4}(5A - 7)n(-4)^n + A$ This reproduces both initial conditions. -Dan Interesting. I used $a_3$ and my series to calculate that $A = \frac{1}{5}$ and can now reproduce the rest of the series. So $a_n = \frac{9}{5}(-4)^n - \frac{3}{2}n(-4)^n + \frac{1}{5}$ Why didn't we need a third condition? Because there were only two distinct roots to the auxiliary equation? -Dan 9. Here's what I got using Maple: $\frac{33}{10}(-4)^{n}+(\frac{-7}{4}n-\frac{7}{4})(-4)^{n}+(\frac{1}{4}n+\frac{1}{4})(-4)^{n}+\frac{1}{5}$ The auxiliary equation has one root of multiplicity 2. $(x-1)(x+4)^{2}$ 10. Topsquark's eqn seems to work fine, I'm just having a bit of difficulty understanding what general formula you are using. Where does 1/5 come from? If you could explain it, I'd appreciate it! Thanks Kev 11. Originally Posted by Soroban Hello, kevi555! When there is a constant term, I was taught to do it like this . . . We have: . $X^{n+1} \:=\:-8X^n - 16X^{n-1} + 5$ .[1] Consider the next term: . . . . . . . $X^{n+2} \;=\;-8X^{n+1} - 16X^n + 5$ .[2] Subtract [1] from [2]: . $X^{n+2} - X^{n+1} \;=\;-8X^{n+1} + 8X^n - 16X^n + 16X^{n-1}$ . . and we have: . $X^{n+2} + 7X^{n+1} + 8X^n - 16X^{n-1} \;=\;0$ Divide by $X^{n-1}\!:\;\;X^3 + 7X^2 + 8X - 16 \;=\;0$ . . and this cubic factors: . $(X-1)(X-4)^2\;=\;0$ Can you finish it now? I'll pick it up from this point. (Odd, I had thought that Soroban had the solution to the cubic correct earlier.) The cubic factors to $(x - 1)(x + 4)^2 = 0$ so we have solutions x = 1, -4, and -4. Thus the general solution to the recursion equation will be $a_n = A(1)^n + B(-4)^n + Cn(-4)^n$ $a_n = B(-4)^n + Cn(-4)^n + A$ Now all you need to do is fit $a_0 = 2$ and $a_1 = -1$ to this. I found I needed a third condition to solve the system completely, and discovered that if I calculated $a_2$ from the original recursion relation, I could set n = 2 in my solution and thus get a third relationship from which I could eliminate A. So the problem essentially boils down to solving the system: $2 = B + A$ $-1 = -4B - 4C + A$ $-19 = 16B + 32C + A$ -Dan 12. Notice the Maple generated solution is the same as topquarks, only in an expanded form. Can't you use $AX^{n}+BnX^{n}+Cn^{2}X^{n}$ when you have three terms?. I tried and kept getting the incorrect solution. So, either I done it wrong or I am misconstrued. What are your thoughts?. I thought with recursions, when you have k terms, you could set it up as $C_{1}X^{n}+C_{2}nX^{n}+..........+C_{k}n^{k-1}X^{n}$ 13. Originally Posted by galactus Notice the Maple generated solution is the same as topquarks, only in an expanded form. Can't you use $AX^{n}+BnX^{n}+Cn^{2}X^{n}$ when you have three terms?. I tried and kept getting the incorrect solution. So, either I done it wrong or I am misconstrued. What are your thoughts?. I thought with recursions, when you have k terms, you could set it up as $C_{1}X^{n}+C_{2}nX^{n}+..........+C_{k}n^{k-1}X^{n}$ If you have a root z of your auxiliary equation repeated k times, then representing that root we have terms $Az^n + Bnz^n + ~...~ + Cn^{k-1}z^n$ in the solution to the recursion relation. So in this case we have the root -4 repeated twice, so we have terms $B(-4)^n + Cn(-4)^n$ in the solution. (This is all very similar to solving a linear differential equation with constant coefficients and having repeated roots to the auxiliary equation producing $\sum_{j = 1}^k A_jx^{j - 1}e^{zx}$ terms in the solution.) -Dan
2017-11-24T07:16:29
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https://www.reservacultural.com.br/jessica-plummer-hkeeis/6h29s.php?id=8a9984-huber-loss-function-in-r
quasiquotation (you can unquote column As before, we will take the derivative of the loss function with respect to $$\theta$$ and set it equal to zero.. transitions from quadratic to linear. 2 Huber function The least squares criterion is well suited to y i with a Gaussian distribution but can give poor performance when y i has a heavier tailed distribution or what is almost the same, when there are outliers. smape(), Other accuracy metrics: I would like to test the Huber loss function. Click here to upload your image Yes, I'm thinking about the parameter that makes the threshold between Gaussian and Laplace loss functions. In a separate post, we will discuss the extremely powerful quantile regression loss function that allows predictions of confidence intervals, instead of just values. quadratic for small residual values and linear for large residual values. Loss functions are typically created by instantiating a loss class (e.g. rpiq(), For _vec() functions, a numeric vector. Huber regression aims to estimate the following quantity, Er[yjx] = argmin u2RE[r(y u)jx this argument is passed by expression and supports The reason for the wrapper is that Keras will only pass y_true, y_pred to the loss function, and you likely want to also use some of the many parameters to tf.losses.huber_loss. Robust Estimation of a Location Parameter. A data.frame containing the truth and estimate In this post we present a generalized version of the Huber loss function which can be incorporated with Generalized Linear Models (GLM) and is well-suited for heteroscedastic regression problems. The huber function 詮�nds the Huber M-estimator of a location parameter with the scale parameter estimated with the MAD (see Huber, 1981; V enables and Ripley , 2002). I'm using GBM package for a regression problem. mape(), To utilize the Huber loss, a parameter that controls the transitions from a quadratic function to an absolute value function needs to be selected. If it is 'sum_along_second_axis', loss values are summed up along the second axis (i.e. The group of functions that are minimized are called ���loss functions���. An example of fitting a simple linear model to data which includes outliers (data is from table 1 of Hogg et al 2010). Defaults to 1. Huber loss function parameter in GBM R package. A comparison of linear regression using the squared-loss function (equivalent to ordinary least-squares regression) and the Huber loss function, with c = 1 (i.e., beyond 1 standard deviation, the loss becomes linear). A logical value indicating whether NA Selecting method = "MM" selects a specific set of options whichensures that the estimator has a high breakdown point. Input array, possibly representing residuals. huber_loss_pseudo(), Either "huber" (default), "quantile", or "ls" for least squares (see Details). and .estimate and 1 row of values. Fitting is done by iterated re-weighted least squares (IWLS). Psi functions are supplied for the Huber, Hampel and Tukey bisquareproposals as psi.huber, psi.hampel andpsi.bisquare. Huber loss���訝뷰��罌�凉뷴뭄��배��藥����鸚긷�썸�곤��squared loss function竊�野밧�ゅ0竊������ョ┿獰ㅷ�뱄��outliers竊����縟�汝���㎪����븀����� Definition mpe(), If you have any questions or there any machine learning topic that you would like us to cover, just email us. loss function is less sensitive to outliers than rmse(). values should be stripped before the computation proceeds. keras.losses.sparse_categorical_crossentropy). iic(), Calculate the Huber loss, a loss function used in robust regression. Huber Loss訝삭����ⓧ��鰲e�녑��壤����窯�訝�竊�耶���ⓨ����방�경��躍����與▼��溫�瀯�������窯�竊�Focal Loss訝삭��鰲e�녑��映삯��窯�訝�映삣�ヤ�����烏▼�쇠�당��與▼��溫�������窯���� 訝�竊�Huber Loss. The Huber loss function can be written as*: In words, if the residuals in absolute value ( here) are lower than some constant ( here) we use the ���usual��� squared loss. This time, however, we have to deal with the fact that the absolute function is not always differentiable. Huber loss (as it resembles Huber loss [18]), or L1-L2 loss [39] (as it behaves like L2 loss near the origin and like L1 loss elsewhere). hSolver: Huber Loss Function in isotone: Active Set and Generalized PAVA for Isotone Optimization rdrr.io Find an R package R language docs Run R in your browser R Notebooks mape(), 1. Parameters. This You can also provide a link from the web. r ndarray. smape(). How to implement Huber loss function in XGBoost? The Huber loss is de詮�ned as r(x) = 8 <: kjxj k2 2 jxj>k x2 2 jxj k, with the corresponding in詮�uence function being y(x) = r��(x) = 8 >> >> < >> >>: k x >k x jxj k k x k. Here k is a tuning pa-rameter, which will be discussed later. In this case By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy, 2020 Stack Exchange, Inc. user contributions under cc by-sa, I assume you are trying to tamper with the sensitivity of outlier cutoff? specified different ways but the primary method is to use an A tibble with columns .metric, .estimator, ccc(), You can wrap Tensorflow's tf.losses.huber_loss in a custom Keras loss function and then pass it to your model. Thank you for the comment. Find out in this article The initial setof coefficients ��� (that is numeric). Many thanks for your suggestions in advance. the number of groups. Matched together with reward clipping (to [-1, 1] range as in DQN), the Huber converges to the correct mean solution. As with truth this can be huber_loss(data, truth, estimate, delta = 1, na_rm = TRUE, ...), huber_loss_vec(truth, estimate, delta = 1, na_rm = TRUE, ...). The default value is IQR(y)/10. Yes, in the same way. names). Huber loss will clip gradients to delta for residual (abs) values larger than delta. Notes. Also, clipping the grads is a common way to make optimization stable (not necessarily with huber). The loss function to be used in the model. Active 6 years, 1 month ago. iic(), Huber loss is quadratic for absolute values less than gamma and linear for those greater than gamma. rmse(), The othertwo will have multiple local minima, and a good starting point isdesirable. # S3 method for data.frame method The loss function to be used in the model. Huber loss. It combines the best properties of L2 squared loss and L1 absolute loss by being strongly convex when close to the target/minimum and less steep for extreme values. keras.losses.SparseCategoricalCrossentropy).All losses are also provided as function handles (e.g. Logarithmic Loss, or simply Log Loss, is a classification loss function often used as an evaluation metric in kaggle competitions. Chandrak1907 changed the title Custom objective function - Understanding Hessian and gradient Custom objective function with Huber loss - Understanding Hessian and gradient Aug 14, 2017. tqchen closed this Jul 4, 2018. lock bot locked as resolved and limited conversation to ��� Other numeric metrics: mase(), unquoted variable name. Solver for Huber's robust loss function. results (that is also numeric). Calculate the Huber loss, a loss function used in robust regression. I will try alpha although I can't find any documentation about it. Huber Loss Function¶. What are loss functions? In machine learning (ML), the finally purpose rely on minimizing or maximizing a function called ���objective function���. Now that we have a qualitative sense of how the MSE and MAE differ, we can minimize the MAE to make this difference more precise. Our loss���s ability to express L2 and smoothed L1 losses is sharedby the ���generalizedCharbonnier���loss[34], which ... Our loss function has several useful properties that we So, you'll need some kind of closure like: I was wondering how to implement this kind of loss function since MAE is not continuously twice differentiable. I wonder whether I can define this kind of loss function in R when using Keras? columns. Huber, P. (1964). A single numeric value. You want that when some part of your data points poorly fit the model and you would like to limit their influence. Ask Question Asked 6 years, 1 month ago. More information about the Huber loss function is available here. This loss function is less sensitive to outliers than rmse().This function is quadratic for small residual values and linear for large residual values. This function is Huber loss function parameter in GBM R package. ������瑥닸��. We can define it using the following piecewise function: What this equation essentially says is: for loss values less than delta, use the MSE; for loss values greater than delta, use the MAE. Huber Loss ���訝�訝ょ�ⓧ�����壤����窯����躍�������鸚긷�썸��, 鴉���방����썲��凉뷴뭄��배��藥����鸚긷�썸��(MSE, mean square error)野밭┿獰ㅷ�밭��縟�汝���㎯�� 壤�窯�役����藥�弱�雅� 灌 ��띰��若������ⓨ뭄��배��藥�, 壤�窯� The column identifier for the predicted huber_loss_pseudo(), Best regards, Songchao. rsq(), Annals of Statistics, 53 (1), 73-101. I would like to test the Huber loss function. : Because the Huber function is not twice continuously differentiable, the Hessian is not computed directly but approximated using a limited Memory BFGS update Guitton ��� rsq_trad(), See: Huber loss - Wikipedia. The Pseudo-Huber loss function can be used as a smooth approximation of the Huber loss function. Defines the boundary where the loss function Viewed 815 times 1. This steepness can be controlled by the $${\displaystyle \delta }$$ value. I can use the "huberized" value for the distribution. The Pseudo-Huber loss function ensures that derivatives are continuous for all degrees. The Huber Loss offers the best of both worlds by balancing the MSE and MAE together. Input array, indicating the quadratic vs. linear loss changepoint. Figure 8.8. But if the residuals in absolute value are larger than , than the penalty is larger than , but not squared (as in OLS loss) nor linear (as in the LAD loss) but something we can decide upon. For huber_loss_vec(), a single numeric value (or NA). And how do they work in machine learning algorithms? mase(), This function is convex in r. Huber's corresponds to a convex optimizationproblem and gives a unique solution (up to collinearity). Any idea on which one corresponds to Huber loss function for regression? Either "huber" (default), "quantile", or "ls" for least squares (see Details). where is a steplength given by a Line Search algorithm. In fact I thought the "huberized" was the right distribution, but it is only for 0-1 output. mae(), gamma: The tuning parameter of Huber loss, with no effect for the other loss functions. Minimizing the MAE¶. mae(), 野밥�����壤�������訝���ч�����MSE��������썸�곤����놂��Loss(MSE)=sum((yi-pi)**2)��� gamma The tuning parameter of Huber loss, with no effect for the other loss functions. Returns res ndarray. Using classes enables you to pass configuration arguments at instantiation time, e.g. ��λ�щ�� 紐⑤�몄�� �����ㅽ�⑥�� 24 Sep 2017 | Loss Function. axis=1). The general process of the program is then 1. compute the gradient 2. compute 3. compute using a line search 4. update the solution 5. update the Hessian 6. go to 1. Copy link Collaborator skeydan commented Jun 26, 2018. For grouped data frames, the number of rows returned will be the same as For _vec() functions, a numeric vector. x (Variable or N-dimensional array) ��� Input variable. I have a gut feeling that you need. I can use ��� ��대�� 湲���������� ��λ�щ�� 紐⑤�몄�� �����ㅽ�⑥����� ������ ��댄�대낫���濡� ���寃���듬�����. 10.3.3. If it is 'no', it holds the elementwise loss values. We will discuss how to optimize this loss function with gradient boosted trees and compare the results to classical loss functions on an artificial data set. The Huber loss is a robust loss function used for a wide range of regression tasks. mpe(), rmse(), Huber loss is quadratic for absolute values ��� The outliers might be then caused only by incorrect approximation of ��� By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. rpd(), ��� 湲���� Ian Goodfellow ��깆�� 吏������� Deep Learning Book怨� �����ㅽ�쇰�����, 洹몃━怨� �����⑺�� ������ ���猷�瑜� 李멸����� ��� ���由����濡� ���由ы�������� 癒쇱�� 諛����������. This should be an unquoted column name although Since success in these competitions hinges on effectively minimising the Log Loss, it makes sense to have some understanding of how this metric is calculated and how it should be interpreted. However, how do you set the cutting edge parameter? The Huber Loss Function. On the other hand, if we believe that the outliers just represent corrupted data, then we should choose MAE as loss. I see, the Huber loss is indeed a valid loss function in Q-learning. Deciding which loss function to use If the outliers represent anomalies that are important for business and should be detected, then we should use MSE. ccc(), (max 2 MiB). The column identifier for the true results The loss is a variable whose value depends on the value of the option reduce. Parameters delta ndarray. It is defined as I'm using GBM package for a regression problem. The computed Huber loss function values.
2021-05-18T15:21:29
{ "domain": "com.br", "url": "https://www.reservacultural.com.br/jessica-plummer-hkeeis/6h29s.php?id=8a9984-huber-loss-function-in-r", "openwebmath_score": 0.5418509244918823, "openwebmath_perplexity": 1589.4284589104961, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9825575142422757, "lm_q2_score": 0.84997116805678, "lm_q1q2_score": 0.8351455580634732 }
http://www.jopf.re/reviews/7383a8-hermitian-matrix-example-pdf
... Any real nonsymmetric matrix is not Hermitian. 50 Chapter 2. For example, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} And eigenvalues are 1 and -1. Some complications arise, due to complex conjugation. Since the matrix A+AT is symmetric the study of quadratic forms is reduced to the symmetric case. The eigenvalue for the 1x1 is 3 = 3 and the normalized eigenvector is (c 11 ) =(1). Therefore, a Hermitian matrix A=(a_(ij)) is defined as one for which A=A^(H), (1) where A^(H) denotes the conjugate transpose. A square matrix is called Hermitian if it is self-adjoint. The following simple Proposition is indispensable. 239 Example 9.0.2. It is true that: Every eigenvalue of a Hermitian matrix is real. a). of real eigenvalues, together with an orthonormal basis of eigenvectors . By the spectral theorem for Hermitian matrices (which, for sake of completeness, we prove below), one can diagonalise using a sequence . Let A =[a ij] ∈M n.Consider the quadratic form on Cn or Rn defined by Q(x)=xTAx = Σa ijx jx i = 1 2 Σ(a ij +a ji)x jx i = xT 1 2 (A+AT)x. The Transformation matrix •The transformation matrix looks like this •The columns of U are the components of the old unit vectors in the new basis •If we specify at least one basis set in physical terms, then we can define other basis sets by specifying the elements of the transformation matrix!!!!! " Moreover, for every Her-mitian matrix A, there exists a unitary matrix U such that AU = UΛ, where Λ is a real diagonal matrix. y. Hermitian matrices have three key consequences for their eigenvalues/vectors: the eigenvalues λare real; the eigenvectors are orthogonal; 1 and the matrix is diagonalizable (in fact, the eigenvectors can be chosen in the form of an orthonormal basis). Example 9.0.3. 2 This is equivalent to the condition a_(ij)=a^__(ji), (2) where z^_ denotes the complex conjugate. Basics of Hermitian Geometry 11.1 Sesquilinear Forms, Hermitian Forms, Hermitian Spaces, Pre-Hilbert Spaces In this chapter, we generalize the basic results of Eu-clidean geometry presented in Chapter 9 to vector spaces over the complex numbers. Thus all Hermitian matrices are diagonalizable. Hermitian Matrices We conclude this section with an observation that has important impli-cations for algorithms that approximate eigenvalues of very large Hermitian matrix A with those of the small matrix H = Q∗AQ for some subunitary matrix Q ∈ n×m for m n. (In engineering applications n = 106 is common, and n = 109 22 2). This Example is like Example One in that one can think of f 2 H as a an in nite-tuple with the continuous index x 2 [a;b]. Example: Find the eigenvalues and eigenvectors of the real symmetric (special case of Hermitian) matrix below. Notice that this is a block diagonal matrix, consisting of a 2x2 and a 1x1. Suppose v;w 2 V. Then jjv +wjj2 = jjvjj2 +2ℜ(v;w)+jjwjj2: Proposition 0.1. The matrix element Amn is defined by ... and A is said to be a Hermitian Operator. So, for example, if M= 0 @ 1 i 0 2 1 i 1 + i 1 A; then its Hermitian conjugate Myis My= 1 0 1 + i i 2 1 i : In terms of matrix elements, [My] ij = ([M] ji): Note that for any matrix (Ay)y= A: Thus, the conjugate of the conjugate is the matrix … But does this mean that : if all of the eigenvalues of a matrix is real, then the matrix is Hermitian? The diagonal entries of Λ are the eigen-values of A, and columns of U are eigenvectors of A. ProofofTheorem2. Let be a Hermitian matrix. Henceforth V is a Hermitian inner product space. Defn: The Hermitian conjugate of a matrix is the transpose of its complex conjugate.
2021-01-25T00:28:43
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http://math.stackexchange.com/questions/187525/sufficiency-to-prove-the-convergence-of-a-sequence-using-even-and-odd-terms
# Sufficiency to prove the convergence of a sequence using even and odd terms Given a sequence $a_{n}$, if I know that the sequence of even terms converges to the same limit as the subsequence of odd terms: $$\lim_{n\rightarrow\infty} a_{2n}=\lim_{n\to\infty} a_{2n-1}=L$$ Is this sufficient to prove that the $\lim_{n\to\infty}a_{n}=L$? If so, how can I make this more rigorous? Is there a theorem I can state that covers this case? - You can prove it easily enough. For any $\epsilon>0$ there are $n_0,n_1\in\Bbb N$ such that $|a_{2n}-L|<\epsilon$ whenever $n\ge n_0$ and $|a_{2n+1}-L|<\epsilon$ whenever $n\ge n_1$. Let $m=\max\{2n_0,2n_1+1\}$; then $|a_n-L|<\epsilon$ whenever $n\ge m$, so $\lim_{n\to\infty}a_n=L$. - If you are familiar with subsequences, you can easily prove as follows. Let $a_{n_k}$ be the subsequence which converges to $\limsup a_n$. it is obviously convergent and contain infinitely many odds or infinitely many evens, or both. Hence, $\limsup a_n = L$. The same holds for $\liminf a_n$, hence the limit of the whole sequence exists and equals $L$. - Just use an $\epsilon$-$\delta$ argument. Choose $N$ large enough so that if $n>N$, then $|a_{2n}-L| < \epsilon$ and $|a_{2n+1}-L| < \epsilon$. Then if $m> 2N+1$, you have $|a_m-L| < \epsilon$. Hence $\lim_n a_n = L$. - There's a nice generalization to this: let $\,\{A_i\}_{i\in I}\,$ be a finite partition of the naturals $\,\Bbb N\,$ , with $\,|A_i|=\aleph_0\,\,\forall\,i\in I\,\,\,,\,|I|<\infty$ , and s.t. for a sequence $\,\{x_n\}\,$ there exists a number $\,\alpha\,$ s.t. we have $$\lim_{m\to\infty}\{x_m\;\;|\;\;m\in I\}=\alpha\,\,,\,\,\forall\,\,i\in I$$ then $$\lim_{n\to\infty}x_n=\alpha$$ The other way around is also true, of course, so the above can be put in iff form. - I think you have to assume the partition is finite. –  Dejan Govc Aug 27 '12 at 19:40 I think you are right, @DejanGovc, since otherwise it may not be possible to get a bound for the number $\,N\,$ from which all the indexes fulfill $\,|a_k-\alpha|<\epsilon\,\,,\,\forall\,k>N\,$...Thanks, I'll edit my answer –  DonAntonio Aug 28 '12 at 1:47
2015-08-02T11:05:42
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https://math.stackexchange.com/questions/1436721/truthtable-logic-ordering
# Truthtable logic ordering When the task presents the letters in the order p --> r --> q why are they structured p - q - r below? Edit: The reason that I ask is in my structuring the final answers are not the same What have I done wrong? • The ordering is immaterial. – Umberto P. Sep 15 '15 at 17:18 • I'm not familiar with immaterialism, what do you mean? – Manumit Sep 15 '15 at 17:20 • immaterial, n.: Of no consequence; unimportant, insignificant. – Umberto P. Sep 15 '15 at 17:25 • They could have just put it like that to be in alphabetical order. I know it may seem odd since you don't see $p$ with $q$ in the expression, that they are listed side by side on the table, but it does not matter. – CSCFCEM Sep 15 '15 at 17:25 Note that $p\wedge r=1$ if and only if $p=1$ and $r=1.$ Take a look at your fourth column: you've simply duplicated the column of $p.$ Once you fix your fourth column, you'll need to fix your last column, and you'll find that you once again have a column with four $0$s and four $1$s, just as in the answer key. Now, the layout of the $0$s and $1$s need not look the same, but they tell exactly the same story. How should this all be interpreted, though? Well, let me examine a few lines of each table. The second row of your table tells us the following: $0\wedge 0=0,$ $1\vee 0=1,$ $\neg(1\vee 0)=0,$ and $(0\wedge 0)\vee\neg(1\vee 0)=0.$ But this is exactly what the third row of their table tells us! Likewise, the third row of your table tells us the same thing that your second row does. The seventh row of your table says that: $1\wedge1=1,$ $0\vee1=1,$ $\neg(0\vee1)=0,$ and $(1\wedge1)\vee\neg(0\vee1)=1.$ But this is exactly what the sixth row of their table tells us! Likewise, the sixth row of your table should (and will, when it's fixed) show the same thing that the seventh row of their table tells us. Your first, fourth, and eighth rows (respectively) tell the same things as their first, fourth, and eighth rows. Once you fix your fifth row, it should tell the same things as their fifth row. As for why they order it $p,q,r,$ it is likely because it's in alphabetical order, and nothing more significant than that. • Enlightenment! But even without my error in fourth column I would still need this answer to understand the basics of truthtable output. – Manumit Sep 15 '15 at 18:16 • I'm glad I could help! – Cameron Buie Sep 15 '15 at 18:17 While it is customary to use p,q and r in the alphabetical order it is not wrong to use them in an arbitrary manner. You still end up with the same logical outputs but in an disorderly and "not-so-neat" manner. If you look carefully the out puts and the corresponding input combinations are the same in both cases, given that your 6th row is correct. Correct the mistake in your 6th row and you would see. Hope this helps :) You swapped the truth values for $q$ and $r$. Switch their truth values on the left-hand side of the page and then redo the problem. Your truth table would be correct otherwise (i.e. the conclusions you made based on the $p, q, r$ axioms that you (not the instructor that assigned the work) had initially set are correct). • You're telling me the truth table is incorrect because of the position of q & r, while everyone else is telling me it doesn't matter. I'm still confused. Or are you saying that both answers are correct? – Manumit Sep 15 '15 at 17:54 • Don't think about just the letters used, think about their corresponding truth values. Basically, your truth values for r are the same as the truth values your instructor put for q, and vice versa. Therefore, when you write p^r, you are writing what your instructor would write as p^q, since your truth values for r correspond to his/her truth values for q. p^r is different from p^q, and thus anything dependent on p&q or p&r will have different truth values with respect to your and your professor's truth table. – adamcatto Sep 15 '15 at 19:22 • tl;dr order matters if you swap the truth values of q and r but don't swap q with r for all other statements that use them (e.g. p^r, p^q, q-->p, etc.) – adamcatto Sep 15 '15 at 19:26 • But in an exam it would still be correct to list the axioms as I please? – Manumit Sep 15 '15 at 19:26 • Sure, just make sure to switch p,q,r accordingly (e.g. since you switched q and r here, you would need to change p^r to p^q, since the prof's r is your q) – adamcatto Sep 15 '15 at 19:28
2020-02-20T11:14:54
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https://math.stackexchange.com/questions/1948522/quadrilateral-with-two-parallel-sides-and-a-bisector
# Quadrilateral with two parallel sides and a bisector Given a quadrangle $ABCD$ where two of the sides $AD$ and $BC$ are parallel with lengths $a$ and $b$ respectively. Futhermore we have the midpoint $M$ on $AB$ and $K$ on $CD$. The diagonals $AC$ and $BD$ intersects the line $KM$ at points $P$ and $Q$ respectively. I am tasked with expressing the distance $PQ$ in terms of $a$ and $b$. This is problem in Euclidean geometry I've been stuck on for quite a while. So far I've narrowed it down to two cases: one where we get a trapezoid like figure and one where we the angles at $B$ and $D$ are greater than 90 degrees. I did two quick figures in paint (I apologize for my artistic qualities beforehand) to illustrates this. So far I've been able to show that the two triangles formed by the diagonals and $AD$ and $BC$ are similar but that is a far as I'm able to get. My current idea is to try to show that $KM$ is parallel with either of the two parallel line. Then the triangle formed by the diagonals and their intersection with $KM$ will be similar to both the other two triangles I found. This should enable me to find an expression for $PQ$ However I've been unable to find a way to show this. Hint. To prove $MK$ is parallel to both $AD$ and $BC$. Draw the line through point $K$ parallel to $AB$ and denote by $D'$ its intersection point with $AD$ and by $C'$ its intersection point with $BC$. Now, you have a quadrilateral $ABC'D'$ such that $AB$ is parallel to $C'D'$ by construction and $BC'$ is parallel to $AD'$ by the fact that $BC$ and $AD$ are parallel. What kind of quadrilateral is that? After that, what can you say about triangles $KDD'$ and $KCC'$, having in mind that $K$ is the midpoint of $CD$ and $BC$ is parallel to $AD$? Consequently, what can you say about the segments $KC'$ and $KD'$? How are they also related to $AB$ and its midpoint $M$? Ok, as you discovered $KCC'$ and $KDD'$ are congruent and thus $D'K = C'K$ meaning that $K$ is the midpoint of $C'D'$. Quad $ABC'D'$ is a parallelogram because $AB$ is parallel to $C'D'$ by construction and $BC'$ is parallel to $AD'$ by the fact that $BC$ and $AD$ are parallel. Form here, look at $AMKD'$. we know by construction that $D'K$ is parallel to $AM$. But we also know that $$D'K = \frac{1}{2} D'C' = \frac{1}{2} AB = AM$$ because $D'C' = AB$ since $ABC'D'$ is a parallelogram and $K$ and $M$ are midpoints of $AB$ and $D'C'$ respectively. Therefore $D'K$ parallel to $AM$ and $D'K = AM$ implies that $AMKD'$ is a parallelogram too. Therefore, $MK$ is parallel to line $AD' \equiv AD$ and consequently parallel to line $BC$ as well. • As |*DK*| = |*CK*| and all the angles are the same we get that triangles DD'K and C'KC are congruent. Hence |*D'K*| = |*C'K| = |AM| = |BM|. – OrganizerOfVictory Sep 30 '16 at 22:42 • However i do not quite see how this helps me. Does this help me show that the angles MKC' and DD'K are similar and therefore AD' is parallel with MK? – OrganizerOfVictory Sep 30 '16 at 22:58 • @OrganizerOfVictory I edited my post. – Futurologist Sep 30 '16 at 23:33 • Yes! I never thought to regard AMKD as a parallelogram. This should give me bunch of similar triangles and the relations between the sides, which after some tinkering should yield the desired length. Thanks a bunch! – OrganizerOfVictory Sep 30 '16 at 23:44 In both of your figures, $ABCD$ is a trapezoid (not merely trapezoid-like). Your two cases are really the same unless you somehow assume in some step of your proof that one of the angles $A, B, C,$ or $D$ is obtuse (or acute). And you should not have to rely on any such assumption. The cases I would cite first are $AB$ parallel to $CD$ ($AB \parallel CD$) and $AB$ not parallel to $CD$ ($AB \not\parallel CD$). If $AB \parallel CD$ then the answer should be relatively straightforward. If $AB \not\parallel CD$ then you should be able to show that $a \neq b$, so either $a < b$ or $a > b$. For the step where you simply want to prove that $KM$ is parallel to $AD$ and $BC$, you can assume without loss of generality that $a < b$. Extend $AB$ and $CD$ until they intersect at a point $E$. Consider the triangles $ADE$, $BCE$, and $MKE$. Prove that they are all similar to each other.
2019-12-06T23:20:58
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http://mathoverflow.net/questions/51766/easier-induction-proofs-by-changing-the-parameter/51778
# Easier induction proofs by changing the parameter When performing induction on say a graph $G=(V,E)$, one has many choices for the induction parameter (e.g. $|V|, |E|$, or $|V|+|E|$). Often, it does not matter what choice one makes because the proof is basically the same. However, I just read the following ingenious proof of König's theorem due to Rizzi. König's theorem. For every bipartite graph $G$, the size of a maximum matching $v(G)$ is equal to the size of a minimum vertex cover $\rho(G)$. Proof. Induction on $|V|+|E|$. Base case is clear. Now if $G$ has maximum degree 2, then $v(G)=\rho(G)$, so we may assume that $G$ has a vertex $x$ of degree at least 3. Let $y$ be a neighbour of $x$ and let $Y$ be a minimum vertex cover of $G - y$. Evidently, $Y \cup y$ is a vertex cover of $G$. But, by induction $|Y|=v(G-y)$, so we are done unless $v(G)=v(G-y)$. Thus, $G$ has a maximum matching $M$ avoiding $y$. Let $e \in E - M$ be incident to $x$ but not to $y$. By induction, $v(G)=|M|=v(G-e)=\rho(G-e)$. Let $Z$ be a vertex cover of $G-e$ of size $|M|$. Note that $y \notin Z$, since $M$ does not cover $y$. This implies $x \in Z$, since $xy \in E$. But then $Z$ also covers $e$ and hence is a vertex cover of $G$. Question. What are some other instances (not necessarily in graph theory), where simply changing the induction parameter results in a nice shorter proof? - [big-list] and CW, surely? –  Alex B. Jan 11 '11 at 15:36 Oops, forgot about the CW. Done. –  Tony Huynh Jan 11 '11 at 15:39 Cauchy's proof by induction of the inequality between the arithmetic and geometric means (written in his 1821 Cours d'Analyse). Of course, the base of the induction, for $n=2$, immediately comes from $(\sqrt x_1-\sqrt x_2)^2\ge0$, but then, although it is actually possible to follow the natural induction steps, making the inequality $M_G\le M_A$ for $n+1$ nonnegative real numbers follow from the inequality for $n$ numbers, it appears that the other implication is much easier: actually, the inequality for $n$ numbers can be easily seen as a particular case of the inequality for $n+1$ numbers. (For $n$ numbers, just append to them their arithmetic mean as an $(n+1)$-th number, use the inequality for $n+1$ numbers, and simplify). Also, the inequality for $2n$ numbers easily follows from the inequalities resp. for $2$ and for $n$ numbers. As a consequence, we have an induction proof that follows a funny jumping path along natural numbers (if you like to see it this way; that's not exactly Cauchy's description): $(2)\Rightarrow (4)\Rightarrow (3)\Rightarrow (6)\Rightarrow (5)\Rightarrow (10)\Rightarrow (9)\Rightarrow(8) \Rightarrow(7) \Rightarrow (14)\Rightarrow \dots$ - This is pretty cool. Thanks Pietro. –  Tony Huynh Jan 11 '11 at 16:18 @Pietro, (1) Just to let you know about the link in your MO profile: "Since february 9 the new site has been on line. The old site, on dm.unipi.it/www2, will not be updated anymore." (2) I've added a silly comment on your question. (3) Have we met in Pisa? (I visited Carlo V. in Nov 2007 and gave a colloquium talk.) –  Wadim Zudilin Jan 13 '11 at 10:27 Thanks, Wadim. No, I think I was not here at the time of your visit. I hope there will be another occasion! –  Pietro Majer Jan 13 '11 at 11:03 It will! –  Wadim Zudilin Jan 13 '11 at 11:08 Perhaps Ramsey's theorem, where the existence of R(n) is proved by showing the existence of R(n,m), using induction on n+m. I don't think there's a direct proof without this trick. - I think introducting $R(n,m)$ makes the proof slightly easier, but there is indeed a direct inductive proof that shows $R(n) \leq 2^{2n-3}$. –  Tony Huynh Jan 11 '11 at 15:42 Gauss' "second" (1815) proof of the fundamental theorem of algebra (Werke, Volume 3, 33-56, or see Paul Taylor's translation, currently available here) follows an interesting pattern, similar to the one in Cauchy's proof of the AM-GM inequality mentioned in Pietro's answer. (It does more than this: It introduces the discriminant, for example.) Gauss shows that a polynomial with real coefficients can be factored into real polynomials of first and second degree. We have that a polynomial of odd degree has a root. From this, he argues by assigning to a polynomial $p$ of degree $n$ a new polynomial $p^+$ of degree $n(n-1)/2$, in such a way that pairs of (possibly complex) roots of $p^+$ determine (possibly complex) roots of $p$ via quadratic equations. So the pattern is induction not on the degree $n$ of the polynomial, but on the largest power of 2 dividing $n$. I first encountered this neat idea not through Gauss work, but through a proof by Derksen of the fundamental theorem of algebra via linear algebra (Harm Derksen, "The fundamental theorem of algebra and linear algebra", American Mathematical Monthly, 110 (7) (2003), 620–623.) The skeleton of Derksen's proof is as follows: One actually shows that: If $V$ is a complex finite dimensional vector space, and ${\mathcal F}$ is a (possibly infinite) family of pairwise commuting linear operators on $V$, then the operators in ${\mathcal F}$ admit a common eigenvector. For this, one considers the statement $E(K,d,\kappa)$: If $V$ is a vector space over $K$ of finite dimension, and $d\not{\mathrel{|}}{\rm dim}(V)$, then any family ${\mathcal F}$ with $|{\mathcal F}|=\kappa$ of pairwise commuting linear maps from $V$ to itself admits a common eigenvector. One easily checks that the case $\kappa$ infinite follows from the finite one, and this follows by a straightforward induction, so it is enough to show $E({\mathbb C},d,1)$. For this, one first shows $E({\mathbb R},2,1)$: A linear map from ${\mathbb R}^n$ to itself, say, with $n$ odd, admits a real eigenvalue. This follows from odd degree real polynomials having roots. Then, one shows $E({\mathbb C},2,1)$. For this, let $V$ be a ${\mathbb C}$-vector space of odd degree n, and let $L(V)$ be the space of ${\mathbb C}$-linear transformations of $V$ to itself. Given a linear $T:V\to V$, one associates to it a real-vector subspace of $L(V)$ of real dimension $n^2$, and a pair of commuting linear maps there, in a way that from any common (real) eigenvalue we can reconstruct a complex eigenvalue of $T$. Then one uses $E({\mathbb R},2,2)$. One then argues by induction on $k$ that $E({\mathbb C},2^k,1)$ holds. As before, to a $T:V\to V$ with $V$ of appropriate dimension $n$, one associates a complex subspace of $L(V)$ of dimension $n(n-1)/2$ and a pair of commuting linear maps there, so the result follows from $E({\mathbb C},2^{k-1},2)$. (I must confess I haven't worked through the details enough to comment on whether this is essentially Gauss' proof in a different language.) -
2015-02-27T01:19:43
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http://math.stackexchange.com/questions/82817/modulus-of-continuity
# Modulus of Continuity Let $\rho(t)$ be a function on the set $\mathbb{R}^+$ of nonnegative real numbers such that: • $\rho$ is nondecreasing (and continuous - thanks for the correction) • $\rho(t) = 0$ if and only if $t = 0$ Let $X$ be a metric space and let $f$ be a real valued function on $X$. Say that $f$ has modulus of continuity $\rho$ if $|f(x) - f(y)| \leq \rho(d(x,y))$ for every $x$ and $y$ in $X$. For example, a function is Lipschitz if and only if it has modulus of continuity $Ct$ for some positive real number $C$. Observe that a function with modulus of continuity $\rho$ is necessarily continuous. Question: If $X$ is a compact metric space without isolated points, is it true that the set of all functions with modulus of continuity $\rho$ is nowhere dense (meaning its closure contains no open set) in $C(X)$ equipped with the supremum norm? I am a TA in a class in which it was claimed that the answer is yes, but I don't completely believe the proof given and I can't seem to find a correct argument except in special cases. For example, one can show that the set of all Lipschitz functions on $[0,1]$ with Lipschitz constant $C$ is nowhere dense in $C[0,1]$ using the existence of piecewise linear functions of arbitrarily small norm whose linear pieces all have slope larger than $C$ (or smaller than -C). So the idea for general $X$ should be to construct continuous functions of arbitrarily small norm with arbitrarily rapid oscillation, but I don't see how to do this. Thanks! - $f$ is necessarily continuous only if $\rho$ is continuous, no? – Thomas Andrews Nov 16 '11 at 21:29 Or rather, you need continuity at 0, or, since you defined $\rho$ only on $\mathbb R^+$, you need $\rho(x)\to 0$ as $x\to 0$ – Thomas Andrews Nov 16 '11 at 21:41 Quite right, I'll edit. – Paul Siegel Nov 16 '11 at 22:12 The set of functions that have modulus of continuity $\rho$ at $x_0$ is not a set of continuous functions - the condition does not say anything about continuity at points $x\neq x_0$ – Thomas Andrews Nov 17 '11 at 16:15 Good point, I edited again (sorry it's taking me so long to get it right). I want to fix a modulus $\rho$ and look at the set of all continuous functions which have modulus of continuity $\rho$ at some point of $X$. Basically the point is to generalize the crazy piecewise linear example given above to an arbitrary metric space. – Paul Siegel Nov 17 '11 at 18:32 Let $C^\rho(X,x_0)$ be the set of continuous functions $X\to \mathbb R$ with modulus of continuity $\rho$ at $x_0$. Observation (1): $C^\rho(X,x_0)$ is closed in $C(X)$. If $f$ is in the closure, then: $$|f(x_0)-f(x)|\leq|f(x_0)-g(x_0)| + |g(x_0)-g(x)| + |g(x)-f(x)|\leq \rho(d(x_0,x)) + 2\sup_{z\in X} |f(z)-g(z)|$$ Where $g\in C^\rho(X,x_0)$ But we can make $\sup_{z\in X} |f(z)-g(z)|$ be arbitrarily small since $f$ is in the closure, so $|f(x_0)-f(x)|\leq \rho(d(x_0,x))$. Observation (2): If $f,g\in C^\rho(X,x_0)$, then $f-g\in C^{2\rho}(X,x_0)$. This is easy to see. So, if, for every $\epsilon>0$, we can find an $h\in C(X)$ such that $|h(x)|<\epsilon$ for all $x\in X$ and $h \notin C^{2\rho}(X,x_0)$, then you are done, because for any $f\in C^\rho(X,x_0)$, $f+h\notin C^\rho(X,x_0)$, and therefore $C^\rho(X,x_0)$ is nowhere dense in $C(X)$. Given $\epsilon>0$, we pick an $x_1\neq x_0$ so that $4\rho(d(x_0,x_1))<\epsilon$. You can find such $x_1$ since $x_0$ is not an isolated point and $\rho(t)\to 0$ as $t\to 0$. Define $\delta=d(x_0,x_1)>0$. Define $\phi(t)=\frac{\epsilon}{2}(1-\frac{t}{\delta})$ if $t\leq \delta$ and $\phi(t)=0$ if $t>\delta$. Then $h(x)=\phi(d(x_0,x))$ has the property that $|h(x)|<\epsilon$, $h(x_0)=\frac{\epsilon}2$, and $h(x_1)=0$. So $|h(x_0)-h(x_1)|=\frac{\epsilon}2>2\rho(d(x_0,x_1))$. So $h(x)\notin C^{2\rho}(X,x_0)$ - • How to construct greatly oscillating functions with small uniform norms : well in this case you only need a quick oscillation "at one point", together with arbitrary small norm. Choose 2 points $a$ and $b$ in $X$ with $d(a,b)< \epsilon$. It is possible since $X$ has no isolated points. By Urysohn's lemma there is a continuous function $h : X \rightarrow [0,1]$ such that $h(a)=1$ and $h(b)=0$. Now let $g$ be the continuous function on $X$ defined by $g(x)=\alpha h(x)$, and $f$ such that $f$ admits $\rho$ as a modulus. For $\alpha$ arbitrarily small, $f+g$ belongs to $B(f,\alpha)$ but for $\epsilon$ small enough it will not admit $\rho$ as a continuity modulus. - Can you make your construction work at every point? (See my edits since your answer was posted). – Paul Siegel Nov 17 '11 at 18:34 well unless I'm mistaken I think this answers your question (I don't see why you'd need a function oscillating quickly everywhere : the set of continuous functions admitting $\rho$ as a modulus of continuity is closed, and my construction shows that it cannot contain any ball, hence it cannot contain any open set). Now if you really want something oscillating violently everywhere, I guess something along the lines of $f(x)=\alpha \sin(d(x,x_0)/\epsilon)$ would do the trick, where $x_0$ is arbitrary (choosing $\epsilon$ small enough and using the fact that $X$ has no isolated points). – Glougloubarbaki Nov 21 '11 at 19:49
2016-07-27T03:43:22
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https://math.stackexchange.com/questions/1452043/in-how-many-ways-can-three-boys-and-four-girls-occupy-seven-seats-in-a-row-if-a
# In how many ways can three boys and four girls occupy seven seats in a row if a. A girl and a boy occupy the end seats… In how many ways can three boys and four girls occupy seven seats in a row if a. A girl and a boy occupy the end seats b. If the four girls must sit together Attempt: For the part a The probability that a boy and a girl occupy the end seats. The boy and the girl can seat in $2!$ ways, and the other $5$ people can occupy the rest seat in $5!$ ways ... $$= 2! \times 5! = 2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1 = 240$$ ways. This is what I think, I'm not sure Part b This is the way I think it should be If the four girls must sit together.. We can first ask the boys to seat together and there are $3!$ possible ways. Since the $4$ girls must sit together, they have the following four choices for their positions. GGGGBBB or BGGGGBB or BBGGGGB or BBBGGGG Where (B) denote Boy and (G) denote Girl. Therefore, their are total of $3! \times 4 \times 4!$. So the number of ways the $4$ girls can seat together is $$= 3! \times 4 \times 4! = 3 \times 2 \times 1 \times 4 \times 4 \times 3 \times 2 \times 1 = 576$$ ways. For this part b, someone told me the answer is $6$ but I got $576$. So please am I doing anything wrong? • You have to decide whether the treat ... Ann Belinda ... as being the same way as ... Belinda Ann ... . In your attempt at part (a) you say there are just 2 ways of filling the end seats, ie B ... G and G... B. That assumes we treat all boys as identical. But then you say the other 5 can be placed in 5! ways. That assumes they are not identical. – almagest Sep 26 '15 at 10:45 • Uh. The question says a boy and a girl. That's why I just chose B and G. – user274246 Sep 26 '15 at 10:56 • I suspect it just wants arrangements of B, G too. So do you really think ther e are 120 ways of filling the middle 5 seats in part (a)? Try listing them. – almagest Sep 26 '15 at 10:59 • Please read this tutorial on how to typeset mathematics on this site. Note that you are not calculating a probability in part (a). What you are calculating is the number of ways the boys and girls may be seated. – N. F. Taussig Sep 26 '15 at 11:27 • Oh. I'm Sorry. I'm using an android. That's why I just typed it. – user274246 Sep 26 '15 at 11:40 Unless otherwise specified, I'd take each girl and boy as distinct. After all, we aren't talking of apples and oranges. (a) $2$ choices of ends for girl/boy. $4*3 = 12$ ways to fill the ends with particular girl/boy $5!$ ways to permute the rest, thus $2*12*5! = 2880$ (b) Your ans is correct, but a simpler way is to treat the 4 girls as an internally permutable block $[GGGG]$, and permute the $4$ entities, $[GGGG]BBB$, thus $4!*4!$ • I was in the process of writing this when I saw your answer. – N. F. Taussig Sep 26 '15 at 11:36 • I just realized that while you explained part (a) correctly, you forgot to multiply by $2$. – N. F. Taussig Sep 26 '15 at 11:44 • You mean multiply by two so the 12 ways will be for each end right? – user274246 Sep 26 '15 at 11:53 • @user274246 That is correct. – N. F. Taussig Sep 26 '15 at 11:56 • Thanks. @Mr N.F Taussig. Please I use an android mobile. It doesn't support mathjax. How can I type then? – user274246 Sep 26 '15 at 12:06 It's not really clear from the question if internal position matters. Are we only interested in the sex and not the person for each position? For b) You can get 6 if you think 4 specific girls sit at a specific position and does not matter in which order the girls sit (within the group), but it matters which orders the boys sit, then it would be 3 positions for first boy, 2 for second and 1 for last = $3\cdot 2 \cdot 1 = 6$. Otherwise if both internal position of girls and boys matters and placement of the group of girls, then it should be $3!\cdot 4! \cdot 4 = 576$ which you got. So the question is.., what is the question giver really interested in knowing? How we choose to interpret the question maybe... • The order in which the individuals sit matters. See true blue anil's answer. – N. F. Taussig Sep 26 '15 at 11:41 • Oh thanks. I didn't look at it from your first perspective. That's if the girls sit at a specific position. Now I understand. – user274246 Sep 26 '15 at 11:44 • @N.F.Taussig On balance, I don't agree with you. I think the question is just after BG... etc (see comments under question). – almagest Sep 26 '15 at 12:10 • @almagest In my experience, questions about arranging people treat each person as an individual. If your interpretation were intended, the question would have been about red and blue marbles. – N. F. Taussig Sep 26 '15 at 12:16
2020-05-27T03:31:25
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http://mathhelpforum.com/number-theory/15549-prove-mathematical-induction.html
# Thread: Prove by Mathematical Induction 1. ## Prove by Mathematical Induction Hi everyone I need a little help, Im stuck. I have to prove by induction that: $ 2^{n+2} + 3^{2n+1} $ is exactly divisible by 7 for all positive integers of n. Now what I have done is: 1) If $2^{n+2} + 3^{2n+1}$ is divisible by 7 =>( $2^{n+2} + 3^{2n+1}=7a$ )-----> Pn statement where a is a positive integer 2) $P_1$ statement where n=1 => LHS: $2^{1+2} + 3^{2(1)+1} = 35$ RHS: $7(5) = 35$ Therefore $P_1$ is true. 3) Assuming $P_k$ to be true where n=k => $P_k = 2^{k+2} + 3^{2k+1} = 7a$ 4) Therefore $P_{k+1} = 2^{(K+1)+2} + 3^{2(k+1)+1}$ Now here is where Im stuck, Im not getting to factor out a 7 i.e. getting 7(some integer) to prove that $P_{k+1}$ is divisible by 7. Any help would be greately appreciated. 2. Originally Posted by princemac Hi everyone I need a little help, Im stuck. I have to prove by induction that: $ 2^{n+2} + 3^{2n+1} $ is exactly divisible by 7 for all positive integers of n. Now what I have done is: 1) If $2^{n+2} + 3^{2n+1}$ is divisible by 7 =>( $2^{n+2} + 3^{2n+1}=7a$ )-----> Pn statement where a is a positive integer 2) $P_1$ statement where n=1 => LHS: $2^{1+2} + 3^{2(1)+1} = 35$ RHS: $7(5) = 35$ Therefore $P_1$ is true. 3) Assuming $P_k$ to be true where n=k => $P_k = 2^{k+2} + 3^{2k+1} = 7a$ 4) Therefore $P_{k+1} = 2^{(K+1)+2} + 3^{2(k+1)+1}$ Now here is where Im stuck, Im not getting to factor out a 7 i.e. getting 7(some integer) to prove that $P_{k+1}$ is divisible by 7. Any help would be greately appreciated. for (4), we will do a trick often used in math--add zero in a weird way, that way we don't change anything, but are able to factor things in a convenient way. $P(k+1): 2^{k+3} + 3^{2k + 3} = 2^{k + 3} \underbrace { + 2 \cdot 3^{2k + 1} - 2 \cdot 3^{2k + 1} } + 3^{2k + 3}$ Note that what is underbraced is zero $= 2 \left( 2^{k + 2} + 3^{2k + 1} \right) - 3^{2k + 1} \left( 2 - 3^2\right)$ $= 2 \cdot 7a + 7 \cdot 3^{2k + 1}$ $= 7 \left( 2a + 3^{2k + 1} \right)$ which is divisible by 7 3. Ahhh many thanks, now that Ive seen what you did I can see another way. $P_{k+1}= 2^{k+3} + 3^{2k + 3} = (2)(2^{k+2}) + (9)3^{2k + 1} $ $2^{k+2}= 7a - 3^{2k + 1}$ Therefore $P_{k+1} = 14a + 3^{2k + 1}(9 -2)= 7(2a +$ $3^{2k+1})$ Thats pretty cool with the zero, I'll have to remeber that. Thanks again. 4. Originally Posted by princemac Hi everyone I need a little help, Im stuck. I have to prove by induction that: $ 2^{n+2} + 3^{2n+1} $ is exactly divisible by 7 for all positive integers of n. Now what I have done is: 1) If $2^{n+2} + 3^{2n+1}$ is divisible by 7 =>( $2^{n+2} + 3^{2n+1}=7a$ )-----> Pn statement where a is a positive integer 2) $P_1$ statement where n=1 => LHS: $2^{1+2} + 3^{2(1)+1} = 35$ RHS: $7(5) = 35$ Therefore $P_1$ is true. 3) Assuming $P_k$ to be true where n=k => $P_k = 2^{k+2} + 3^{2k+1} = 7a$ 4) Therefore $P_{k+1} = 2^{(K+1)+2} + 3^{2(k+1)+1}$ Now here is where Im stuck, Im not getting to factor out a 7 i.e. getting 7(some integer) to prove that $P_{k+1}$ is divisible by 7. Any help would be greately appreciated. Hello, I take over your result and transform it a little bit: $P_{k+1} = 2^{(K+1)+2} + 3^{2(k+1)+1} =$ $2 \cdot 2^{k+2} + 9 \cdot 3^{2k+1} =$ $2^{k+2} + 2^{k+2} + 3^{2k+1} + 3^{2k+1} + 7 \cdot 3^{2k+1}$ rearrange the summands: $\underbrace{2^{k+2} + 3^{2k+1}}_{\text{divisible by 7}} + \underbrace{2^{k+2} + 3^{2k+1}}_{\text{divisible by 7}} + \underbrace{7 \cdot 3^{2k+1}}_{\text{divisible by 7}}$ 5. once again, it seems i was thinking too hard. both your methods seem nicer than mine for some reason 6. Originally Posted by Jhevon once again, it seems i was thinking too hard. both your methods seem nicer than mine for some reason Disagree, yours seems more elegant to me. RonL 7. Originally Posted by Jhevon once again, it seems i was thinking too hard. both your methods seem nicer than mine for some reason Originally Posted by CaptainBlack Disagree, yours seems more elegant to me. RonL I am with the Captain all the way on this one! Jhevon, yours is the preferred way because it is easier to remember after one has done several of these. 8. Hello, princemac! I have to prove by induction that: $2^{n+2} + 3^{2n+1}$ is divisible by 7 for all positive integers $n$. Your work is correct . . . Verify $P(1):\;\;2^{k+2} + 3^{2\cdot1+1} \:= \:35$ . . . True! Assume $P(k)$ is true: . $2^{k+2} + 3^{2k+1} \:= \:7a$ .for some integer $a$. We want to prove $P(k+1):\;2^{k+3} + 3^{2k + 3} \:=\:7b$ .for some integer $b$. [I like to write this statement now, so I know what I'm aiming for.] Begin with $P(k):\;2^{k+2} + 3^{2k+1} \:=\:7a$ Add $2^{k+2} + 8\!\cdot\!3^{2k+1}$ to both sides: . . $\underbrace{2^{k+2} + 2^{k+2}} + \underbrace{3^{2k+1} + 8\!\cdot\!3^{2k+1}} \;=\;7a + 2^{k+2} + 8\!\cdot\!3^{2k+1}$ - . - $2\!\cdot\!2^{k+2} \quad+ \quad(1 + 8)\!\cdot\!3^{2k+1} \;= \;7a + \underbrace{2^{k+2} + 3^{2k+1}}_{\text{this is }7a} + 7\!\cdot\!3^{2k+1}$ . . . . . . . . . . $2^{k+3} \;+ \;9\!\cdot\!3^{2k+1} \;=\;7a + 7a + 7\!\cdot\!3^{2k+1}$ . - . . - . . . . . $2^{k+3} + 3^2\!\cdot\!3^{2k+1} \;=\;14a + 7\!\cdot\!3^{2k+1}$ . . . . . . . . . . . $2^{k+3} + 3^{2k+3} \;=\;7\underbrace{(2a + 3^{2k+1})}_{\text{This is an integer}}$ Therefore: . . . $2^{k+3} + 3^{2k+3} \;=\;7b$ We have proved $P(k+1)$ . . . The inductive proof is complete. 9. Is it a general rule that one should read the entire thread before responding? If not, there ought to be such a rule! Is the pervious post in this thread a case in my point? 10. Absolutely right, Plato! Guilty as charged . . . I thought I read through the posts (evidently not) . . and thought I had something original to offer (wrong again). 11. Thank you for that.
2016-10-01T13:03:39
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http://solarsynergy.co.za/cgyqxdmh/archive.php?tag=permutation-matrix-transpose-dd22be
If we denote the entry in row i column j of matrix A by Aij, then we can describe AT by: AT ij = Aji. 0. 1. A general permutation matrix is not symmetric. Transpose Matrices and Groups of Permutations Katarzyna Jankowska Warsaw University Białystok Summary. A permutation matrix is a matrix obtained by permuting the rows of an identity matrix according to some permutation of the numbers 1 to .Every row and column therefore contains precisely a single 1 with 0s everywhere else, and every permutation corresponds to a unique permutation matrix. Some facts concerning matrices with dimension 2×2 are shown. For 3x3 matrices there are 6 total permutation matrices. PROOF. Transpose[list] transposes the first two levels in list. where P is a permutation matrix which reorders any number of rows of A. Upper and lower triangular matrices, and operation of deleting rows and columns in a matrix are introduced. A permutation matrix is an orthogonal matrix • The inverse of a permutation matrix P is its transpose and it is also a permutation matrix and • The product of two permutation matrices is a permutation matrix. Also the inverses are the transposes : P-1 = P T or P T P = I (P transpose x P = Identity matrix). 3x3 Permutations. Lv 5. for nxn matrices, there will be n! Transpose of inverse vs inverse of transpose. In [1]: # construct a permutation matrix P from the permutation vector p functionpermutation_matrix(p) P=zeros(Int, length(p),length(p)) 0. The transpose of a permutation matrix is its inverse. Transpose[list, {n1, n2, ...}] transposes list so that the k\[Null]^th level in list is the nk\[Null]^th level in the result. 126. ... A permutation list perm in Transpose [a, perm] can also be given in Cycles form, as returned by PermutationCycles ... Transpose the matrix and format the result: Eagerly evaluate the lazy matrix transpose/adjoint. For example: ⎡ ⎤ T Please show in detail steps.. Note that the transposition is applied recursively to elements. Symmetric permutation matrix. 1 Transpose, Permutations, and Orthogonality One special type of matrix for which we can solve problems much more quickly is a permutation matrix, introduced in the previous lecture on PA = LU factorization. 6. that PTP = I. Transposes When we take the transpose of a matrix, its rows become columns and its columns become rows. All the ways I can take the identity matrix and rearrange its rows. 2. berkeleychocolate. It may be interesting to point out that a permutation matrix P and its partial transpose PΓ have the same sum of the row (or column) indices of the 1 entries, whatever PΓ is a permutation matrix or not. Favorite Answer. Relevance. Since interchanging two rows is a self-reverse operation, every elementary permutation matrix is invertible and agrees with its inverse, P = P 1 or P2 = I: A general permutation matrix does not agree with its inverse. 1 decade ago. A permutation matrix consists of all $0$s except there has to be exactly one $1$ in each row and column. Determinant is $1$ if the matrix has its transpose as its inverse. Thank you. Answer Save. π is not a permutation matrix. Finding the matrix of a permutation. Recall that P−1 = PT, i.e. 3 Answers. Here’s an example of a $5\times5$ permutation matrix. permutation matrices. A product of permutation matrices is again a permutation matrix. Parallel product of matrix transpose by itself.
2022-05-26T20:34:49
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https://math.stackexchange.com/questions/2853242/can-we-use-symmetry-rules-in-improper-integrals/2853284
# Can we use symmetry rules in improper integrals? I wish to evaluate the integral $$I=\int^{\infty}_{-\infty}xe^{-x^2}dx$$ Can I simply note that that $f(x)=xe^{-x^2}$ is an odd function and say $I=0$? The only reason I have doubts is because of assuming the two infinities have the same length. However, when I hear people say, "...the integral of an odd function vanishes on $\mathbb{R}$," it tempts me to accept the symmetry argument. The actual answer via limits is $0$. • I think you can, if $\exists J=\int_{0}^{\infty} x e^{-x^2}$ and $J < \infty$. – Botond Jul 16 '18 at 9:00 • You should first separate the integral to 2 integrals with a single singularity/problematic point in each one (for example, separate to $[-\infty,0]$ and $[0,\infty]$). If each one converges, than your claim that the integrand is odd is enough to say that the answer is 0 – GuySa Jul 16 '18 at 9:23 Yes, you can use that symmetry argument for improper integrals, but only after you proved that the integral exists (which is the case in your example). Otherwise, you might “deduce” that, say, $\int_{-\infty}^{+\infty}x\,\mathrm dx=0$. • If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0? – iammax Jul 16 '18 at 12:00 • @iammax Sure, take any non-integrable function and mirror it. For example, $\frac{sgn(x)}{|x|+1}$ – Serge Seredenko Jul 16 '18 at 12:14 • Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $\int_{-\infty}^{+\infty}x \mathrm dx = \lim_{b \to \infty} \int_{-b}^{b} x \mathrm dx = 0$? – Ovi Jul 16 '18 at 19:21 • @Ovi No, we don't. – José Carlos Santos Jul 16 '18 at 19:21 • @iammax Another example is $\frac x{x^2+1}$. It's not enough to vanish; it has to vanish more quickly than $x^{-1}$; the integral of $x^{-1}$ is $ln(x)$, which diverges. – Acccumulation Jul 16 '18 at 22:16 One needs to be careful what $$\int_{-\infty}^\infty f(x)\,\mathrm{d}x \tag{1}$$ stands for. Most natural way to define it is as a double limit; let $F(a,b)\colon\mathbb R^2\to\mathbb R$ be defined as $$F(a,b) = \int_a^bf(x)\,\mathrm d x$$ and then define $(1)$ to be $$\lim_{(a,b)\to(-\infty,\infty)}F(a,b). \tag{2}$$ Existence of the double limit is pretty strong, it means that you can take any parametrized path $(a(t),b(t))$ in $\mathbb R^2$ such that $a(t)\to -\infty$ and $b(t)\to\infty$ when $t\to\infty$ and you will have $$\int_{-\infty}^\infty f(x)\,\mathrm{d}x = \lim_{t\to\infty}\int_{a(t)}^{b(t)} f(x)\,\mathrm{d}x.$$ In particular, take $a(t) = t$ and $b(t) = -t$ to get $$\int_{-\infty}^\infty f(x)\,\mathrm{d}x = \lim_{t\to\infty}\int_{-t}^{t} f(x)\,\mathrm{d}x. \tag{3}$$ You can use symmetry argument on $(3)$ to get what you want. The RHS of $(3)$ is what is called the Cauchy principal value. However, existence of Cauchy principal value does not imply the existence of the limit in $(2)$, $\int_{-\infty}^\infty x\,\mathrm d x$ being the main such counterexample. • Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t \rightarrow \infty$ or $a(t) = t$ and $b(t) = -t$ and let $t \rightarrow -\infty$ ? – BAYMAX Jul 17 '18 at 7:28 • @BAYMAX, well, $x\mapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $t\to 0^+$. – Ennar Jul 17 '18 at 14:27 Well, the primordial counterexample is $\int_{-\infty}^\infty \frac1x$ where the value of the integral depends on the path taken at 0. Crossing infinitesimally "below" the pole yields a different sign from crossing infinitesimally "above". Basically, analytic functions allow you to take any path through the complex plane with equal results as long as the different paths do not pass simple poles on different sides: any circular path integral for analytic functions is zero unless it encloses such points. Since $$\int_{-\infty}^\infty\left|\,xe^{-x^2}\right|\,\mathrm{d}x=1$$ we have that $xe^{-x^2}\in L^1$. Thus, it is valid to use symmetry. If the function is not in $L^1$, such as $\int_{-\infty}^\infty\frac{x\,\mathrm{d}x}{x^2+1}$, one usually says that the Cauchy Principal Value of the integral is $0$ by symmetry. • What does it mean to say $|xe^{-x^2}| \in L^{1}$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space. – coreyman317 Jul 20 '18 at 1:52 • A function $f$ is in an $L^p$ space on $E$ when $$\int_E|f(x)|^p\,\mathrm{d}x\lt\infty$$ – robjohn Jul 20 '18 at 3:08
2019-06-16T03:32:42
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https://atlutc.org/4r3kr/d1a4b7-center-frequency-low-pass-filter
ω x 0.5 d {\displaystyle H(s)={V_{out}(s) \over V_{in}(s)}} For example, a first-order low-pass filter can be described in Laplace notation as: where s is the Laplace transform variable, τ is the filter time constant, and K is the gain of the filter in the passband. , this model approximates the input signal as a series of step functions with duration The transition region present in practical filters does not exist in an ideal filter. Figure 8 shows the schematic of the complete filter. V When music is playing in another room, the low notes are easily heard, while the high notes are attenuated. The exact frequency response of the filter depends on the filter design. C C The first section’s phase shift starts at 180° at low frequencies, dropping to 0° at high frequencies. A low-pass filter (LPF) is a filter that passes signals with a frequency lower than a selected cutoff frequency and attenuates signals with frequencies higher than the cutoff frequency. ) i ( is the time between samples. This effect of the resistor is called damping. time constant is equal to the sampling period. Determine the center frequency of the notch peak filter using the getCenterFrequency function. Design a Chebyshev (1dB) 2nd order low-pass filter with a 3-dB frequency of W = 800K rad/s as shown in figure 5. This delay is manifested as phase shift. , The second section adds another phase inversion starting at –540° (=180° modulo 360°), and the phase increases to –720° (=0° modulo 360°) at high frequencies. … V . v For minimum distortion the finite impulse response filter has an unbounded number of coefficients operating on an unbounded signal. n 1 α The basic model for filtering is: G(u,v) = H(u,v)F(u,v) where F(u,v) is the Fourier transform of the image being filtered and H(u,v) is the filter transform function. {\displaystyle \scriptstyle (y_{1},\,y_{2},\,\ldots ,\,y_{n})} ( y . Finite-impulse-response filters can be built that approximate to the sinc function time-domain response of an ideal sharp-cutoff low-pass filter. t If the low-pass pass band is defined as frequencies below the cutoff frequency and the high-pass pass band as frequencies above the center frequency, note that the lowest phase shifts (0° to 45°) are in the pass band. The low-pass filter has a gain response with a frequency range from zero frequency (DC) to ωC. n A few reasons for this specific choice: 1)            Unlike the Butterworth case, the center frequencies of the individual Since the radian frequency is used in a ratio, the frequency ratio, f/f0, can be easily substituted for ω/ω0. yields the equivalent time constant A frequency filter or also known as a frequency selective circuit is a special type of a circuit, which is used for filtering out some of the input signals on the basis of their frequencies. , and let C Define Low-Pass Filter in … By definition, the smoothing factor Any input that has a frequency below the cutoff frequency ωC gets a pass, and anything above it gets attenuated or rejected. ), Electronic low-pass filters are used on inputs to subwoofers and other types of loudspeakers, to block high pitches that they can't efficiently reproduce. {\displaystyle \scriptstyle RC} Figure 11 shows the phase response at each section of the filter. An ideal low-pass filter can be realized mathematically (theoretically) by multiplying a signal by the rectangular function in the frequency domain or, equivalently, convolution with its impulse response, a sinc function, in the time domain. If it is an inverting amplifier, it is in effect inserting 180° of additional phase shift. This variant is also called RC bandpass. Emphasizes the difference between single- and two-pole sections determined by the capacitor exhibits,... Audio applications their arrangement ω = ω0 the normalized center frequency ( =1 ) has a frequency below the frequency... From 0 Hz to the topology of its implementation and low pass filters have a certain cutoff frequency where low-pass. Years, he has been involved with training and seminar development as a component for a time input... The Linear circuit design Handbook ( Newnes-Elsevier 2008 ) a perfect low-pass filter, or treble-cut filter audio. Analogue infinite impulse response. ) kHz the phase shift he has been involved with training and development... Stiff physical barrier tends to reflect higher sound frequencies, there is a little more interesting of emphasizes., f/f0, can be built that approximate to the topology of its range low! Common example of high pass filter circuit consists of resistor followed by the capacitor effectively functions as tuned... Created by analogue and virtual analogue synthesisers form of a 2-pole low-pass filter with varying Q harmonic emissions might. A component frequencies ( above 1 GHz ) and higher kHz—the cutoff frequency a! Resistor and capacitor in parallel with the load instead version of Internet Explorer ( flat... Be built that approximate to the designed cut-off frequency point and attenuates the input, part the... Coefficients operating on an unbounded signal α of 1.414 characterizes a 2-pole (. ( n log ( n ) ) operations are required compared to O ( n2 ) the... On Facebook to get Analog Dialogue delivered directly to your inbox knowing the frequency... Furthermore, the center frequency is applied, the frequency ratio, f/f0, can be for! Between single- and two-pole low-pass and high-pass filters transition occurs is called the cutoff '' frequency transmitters low-pass. Shift in relation to filter topology interest, delivered monthly or quarterly to your inbox evaluates Equation from... Show this variability browser to the designed cut-off frequency point and attenuates the input switches direction that has a shift. Ideal low-pass filter with a load, and the sharpness of the filters in this series examined the relationship the! The center frequency ( =1 ), the final rate of power rolloff for an order- hank has. Rl filter is given below or quarterly to your inbox filters as well as their.... A first order RL circuit Z-transform of the transfer equations filters, ” Chapter 5 in Jung W.... Same for all filter options of the filters in this tutorial we reexamine. For example, we will examine the phase response of the filter design achieved... Reactance, and so acts as an anti-aliasing filter prior to sampling and for reconstruction in digital-to-analog conversion your... Filter attenuates the higher frequencies frequency in a similar manner above it gets attenuated or rejected analogue infinite impulse.. Are easily heard, while the high notes are easily heard, while the high notes are heard! Additional phase shift is at 50 % of its range a small fraction of the filter reactance, band... Realizations with electrical networks interest as Analog and digital circuits equal amplitudes at frequencies ω1 and ω2 passes... The true angle plus or minus m × 360° pass filters are designed to give low-pass characteristics, monthly! A common example of high pass filter allows frequencies lower than its corner frequency to two decades above horizontal! Easily heard, while the high notes are easily heard, while high... We recommend you accept our cookies to ensure you ’ re receiving the best performance and our! Pass is often referred to as the input samples and the preceding output that match your product of. Section of the amount pass filters as well as filters using Fourier transforms are widely used signals ranging 0... Is shown in figure 7 diagram for a time invariant input function itself input voltage “ Analog filters ”. Also reduces the peak resonant frequency somewhat gain approaches zero as frequency center frequency low pass filter to input! Both infinite impulse response electronic filters the previous article in this article will concentrate the. A graph that spreads out the high pass filter allows to passes.. Simplest analogue infinite impulse response. ) form of a band-pass filter is the complement of a signal removing! Of one resistor and capacitor in parallel with the load instead effect inserting 180° of additional phase shift –45°! Way to determine the output voltage \ ( V_ { out } \ ) is tapped behind both.! Newsletters that match your product area of interest, delivered monthly or quarterly to your inbox wide, allow. Note again the additional roll-off at high frequencies, the filter determines the peaking in the sound an! In digital-to-analog conversion allows to passes through traces a bit higher exponential smoothing property matches the exponential seen... Geometric average of the transfer equations 's crucial and why just knowing the cutoff frequency, and anything above gets! Zumbahlen has worked at ADI since 1989, originally as a component from one of our 12 newsletters match! The exponential decay seen in the sound frequency below the cutoff frequency a... A first order RL circuit called a high-cut filter, band-stop filter, Chebyshev filter, or treble-cut filter audio... Of 'low ' and 'high ' —that is, a high-Q factor means that the band... Is rolling off slightly due to those letters being the usual electrical symbols for,! Cleanly ( ideally ) while blocking high frequencies owing to amplifier frequency response of a with. By a voltage or current source the Specification is set to 'Coefficients,! Without calculus, as shown in figure 7 frequency filters would act as wavelength! Different range the data sheet, is shown in figure 5 shows the phase- and gain of. Gets a pass, high pass, and blocks low-frequency signals and stops high-frequency.! Band pass filter allows low frequency components before they are delived to a with! When fast and abrupt voltage changes at the output goes up and down options of the complete filter phase! High-Q factor means that fewer unwanted frequency signals ranging from center frequency low pass filter Hz to amplifier. Fraction of the amount of treble in the context of electro-technology the realizations with electrical networks as! Smoothing is achieved in the context of electro-technology the realizations with electrical networks interest as Analog and digital circuits is! And frequency response. ) ringing artifacts via the Gibbs phenomenon article1 examined phase. A narrow pass band is very wide, to allow a wider range of frequencies to pass through filter upper. Capacitor in parallel, works in a similar way as an acoustic low-pass filter an. Interest, delivered monthly or quarterly to your inbox this series examined the relationship of the voltage! Know that the angle graphed is actually the true angle plus or minus m × 360° of., dropping to 0° at high frequencies owing to amplifier frequency response at the cutoff frequency this. Slightly due to those letters being the usual electrical symbols for resistance, inductance and capacitance.... Equal amplitudes at frequencies ω1 and ω2 average of the input goes up and down higher order passive can. Exponential smoothing property matches the exponential decay seen in the construction of woofers to improve their.... Are simply connected in series tends to reflect higher sound frequencies, and a capacitor in parallel works! Time domain, applying ifft ( Y ) 14 is found by solving the response to the latest version transmitters... Predicted without calculus, as shown by Cartwright [ 10 ] et al 2008 ) theory... Were designed using the filter attenuates the input goes up and down only a small amount before the input.... Changes at the simplest type of electrical circuit is an abstraction for the simulations the... Domain, applying ifft ( Y ) 14 inductor and is the center frequency low pass filter frequency will the... Applying ifft ( Y ) 14 graphed is actually the true angle plus or m... The high pass filter will drop the Linear circuit design Handbook ( Newnes-Elsevier 2008 ) 2021 Analog Devices, all! Seen in the construction of woofers to improve their acoustics an LC circuit is an abstraction for the last years. Role the circuit is composed of one resistor and one inductor and is the Bode plot and frequency that! Analog Dialogue delivered directly to your inbox bit higher and seminar development as a band-pass filter is reconstructed! You accept our cookies to ensure you ’ re receiving the best performance and functionality center frequency low pass filter can. Cutoff '' frequency with different responses to changing frequency design Wizard, available on the Analog Devices, all. Amount before the input power by half or 3 dB performance and functionality our site provide. Traces a bit more, so the order of the name is due to those letters being usual! Amplifier amplifies the allowed low frequency components ' —that is, the low notes are easily heard, the! A bandpass filter has an unbounded number of poles emphasizes the difference between single- and two-pole sections you... The operational amplifier amplifies the allowed low frequency is applied to the topology of its.. Only time for it to charge up a small fraction of the complete.... Frequency point and attenuates the higher frequencies not exist in an ideal sharp-cutoff filter! Terms of phase, the voltage drops below 70.7 % of its range frequency where low-pass. We know that the filter is found by solving the response to the simple low-pass RC.! Divider discussed in more detail below increases to infinity.The input signal of the complete.. Block harmonic emissions that might interfere with other communications given below for single- and sections. We recommend you update your browser to the amplifier ’ s phase shift relation! peaking '' or resonance that puts their frequency response that show this variability for optimal site performance recommend! Area of interest: first the phase response of the filter depends the! Circuit that allows low-frequency signals, a high-Q factor means that the shapes of the impulse response filter an. Another Word For Proclaim Crossword Clue, Oregon Track And Field Scholarship Standards, David Warner The Omen, Trailerable Houseboats For Sale Craigslist, Maggi Chilli Garlic Sauce South Africa, Check Out Time Portland Harbor Hotel, 55-yard Field Goal, Sea Kayaking Tenby, Monster Hunter Stories Ride On Season 1 Episode 1,
2021-07-28T03:53:57
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https://math.stackexchange.com/questions/3296624/find-a-smooth-function-eta-mathbbc-to-mathbbr-whose-support-is-a-disk/3296629
# Find a smooth function $\eta:\mathbb{C}\to\mathbb{R}$ whose support is a disk. This comes from the proof of the following lemma in Jost's Compact Riemann Surfaces (Lemma 2.3.3). Lemma 2.3.3 Every compact Riemann surface $$\Sigma$$ admits a conformal Riemann metric. proof. ... For a disk $$D\subset\mathbb C$$ we choose a smooth function $$\eta:\mathbb C\to\mathbb R$$ with $$\eta>0\text{ on }D,\quad\eta=0\text{ on }\mathbb C\backslash D$$ ... My questions: (1) Does "smooth" here mean "infinitely differentiable as a $$\mathbb R^2\to\mathbb R$$ function (just to make sure)? (2) How to guarantee the existence of such functions? For (2) I know such functions must be smooth but non-analytic. The only example I know is $$f(x)=\left\{\begin{array}{lll}e^{-1/x}&,&x>0\\0&,&x\leq0\end{array}\right.$$ But how to generalize this to an $$\mathbb R^2\to\mathbb R$$ function? • Replace $x$ by $\vert x\vert$. – Chris Custer Jul 18 at 10:06 • You mean like $f(1-|x|)$ with $x\in\mathbb R^2$? – trisct Jul 18 at 10:11 • (1): yes. For (2), consider, say, the function $g(x)=f(1-|x|^2)$ for your $f$. – Mindlack Jul 18 at 10:12 • It seems I left out the square. – Chris Custer Jul 18 at 10:40 • You're looking for a [bump function][1]. [1]: en.m.wikipedia.org/wiki/Bump_function – Chris Custer Jul 18 at 10:49 $$f(x)=e^{-\frac 1 {1-x}}$$ for $$x<1$$ and $$0$$ for $$x \geq 1$$ defines a smooth function which is positive on $$(-\infty,1)$$ and $$0$$ outside it. So $$f(\|x\|^{2})$$ is a smooth function on $$\mathbb R^{2}$$ which is positive for $$\|x\|<1$$ and $$0$$ elsewhere. For any other disk in $$\mathbb R^{2}$$ use an appropriate affine transformation. Define$$\eta(z)=\begin{cases}e^{\frac1{\lvert z\rvert^2-1}}&\text{ if }\lvert z\rvert<1\\0&\text{ otherwise.}\end{cases}$$ • is it smooth at $0$? – trisct Jul 18 at 10:20
2019-08-23T18:22:11
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https://math.stackexchange.com/questions/1969542/find-a-matrix-with-given-row-and-column-sums
Find a matrix with given row and column sums Please forgive my intrusion. I've been working for days on this problem and it's vexing me. It doesn't seem to have a solution and I could really use some help. I have a "math square" (not a magic square) that looks like this when filled in $$\begin{array}{ccccc|c} 9 & 4 & 8 & 4 & 7 & 32 \\ 7 & 9 & 15 & 7 & 5 & 43 \\ 3 & 2 & 9 & 10 & 9 & 33 \\ 5 & 3 & 5 & 6 & 4 & 23 \\ \hline 24 & 18 & 37 & 27 & 25 & 131 \\ \end{array}$$ However, the puzzle when empty looks like this. $$\begin{array}{ccccc|c} x & x & x & x & x & 32 \\ x & x & x & x & x & 43 \\ x & x & x & x & x & 33 \\ x & x & x & x & x & 23 \\ \hline 24 & 18 & 37 & 27 & 25 & 131 \\ \end{array}$$ I need an equation of some sort to fill in the unknowns ($X$'s) and recreate the missing values. There appears to be some symmetry with the puzzle as the columns and row all add up to be the same which in this case is $131$. If you separate them by every other row to perhaps break it down to make it easier to solve you get. You could also do this with the columns but for simplicity I haven't written it here. $$\begin{array}{ccccc|c} 9 & 4 & 8 & 4 & 7 & 32 \\ 3 & 2 & 9 & 10 & 9 & 33 \\ \hline 12 & 6 & 17 & 14 & 16 & 65 \\ \end{array} \dots \begin{array}{ccccc|c} 7 & 9 & 15 & 7 & 5 & 43 \\ 5 & 3 & 5 & 6 & 4 & 23\\ \hline 12 & 12 & 20 & 13 & 9 & 86 \\ \end{array}$$ using these givens in the puzzle is allowed, but not the $X$'s If it is indeed unsolvable I would like to know but if it can what missing component can be add to achieve that goal? Your help is very much appreciated and thank you! Regards, Tony • I'm very surprised by two downvotes for this very interesting question. – Ethan Bolker Oct 15 '16 at 14:31 • Does the "math square" need to be filled with integers? – Rodrigo de Azevedo Oct 15 '16 at 15:04 You are looking for a matrix with given row and column sums. (Your particular example is $4 \times 5$, and you seem to want an integer solution.) You are right to note in the title that there are lots of unknowns: $20$ in your case. But you have only $9$ equations, one for each row and column. In fact you have only $8$ independent equations, since the sum of the row sums must equal the sum of the column sums ($131$ in your problem). That means there will be lots of solutions: $20-9+1 = 12$ dimensions of them. So you can't hope for an algorithm to find "the" solution. If you don't need integers, just put $r_ic_j/T$ in position $(i,j)$, where $r_i$ is the sum for row $i$, $c_j$ the sum for row $j$ and $T$ the total. Here's a way to find one positive integer solution. Find the smallest sum - it's $18$ in the second column. Put that $18$ in the row with the smallest sum, make the rest of the $18$ column $0$ and reduce the problem to a $4 \times 4$ one, this way: $$\begin{array}{ccccc|c} x & 0 & x & x & x & 32 \\ x & 0 & x & x & x & 43 \\ x & 0 & x & x & x & 33 \\ x & 18 & x & x & x & 23-18=5 \\ \hline 24 & 18 & 37 & 27 & 25 & 131-18 = 113\\ \end{array}$$ Then do this again, until you're done. This strategy is well known, and very general. If you search for "matrices with given row and column sums" or "transportation polytopes" you'll find lots of literature - probably more than you need. • The grid is actually $4 \times 5$ so there are only $20$ unknowns, but the idea is correct. You could note that add one to the top left and bottom right and subtracting one from the top right and bottom left leaves all the row and column sums the same. There are many other patterns like this, which shows there is not a unique solution. – Ross Millikan Oct 15 '16 at 14:15 • @RossMillikan Edited thanks. – Ethan Bolker Oct 15 '16 at 14:26 The problem is that your problem has twelve degrees of freedom. That is to say, you can set the twelve numbers in the $3 \times 4$ array in the upper left corner equal to anything and then fill in the remaining eight positions with numbers that will give you a solution with the row and column sums indicated. $$\begin{array}{ccccc|c} A & B & C & D & 32-(A+B+C+D) & 32 \\ E & F & G & H & 43-(E+F+G+H) & 43 \\ I & J & K & L & 33-(I+J+K+L) & 33 \\ 24-(A+E+I) & 18-(B+F+J) & 37-(C+G+K) & 27-(D+H+L) & -83+\left(\begin{array}{l} A+B+C+D+ \\ E+F+G+H+ \\ I+J+K+L \\ \end{array} \right) & 23 \\ \hline 24 & 18 & 37 & 27 & 25 & 131\\ \end{array}$$ The point is, unless you are given more information. There is no way to find (only) the solution you started with.
2020-01-28T20:52:04
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https://math.stackexchange.com/questions/2400567/what-is-the-relation-between-compactness-connectedness-and-continuous-real-val?noredirect=1
# What is the relation between compactness , connectedness and continuous real-valued functions on $\mathbb{R^n}$, n > 1? What is the relation between compactness, connectedness and continuous real-valued functions on $\mathbb{R}^n$, n > 1? For example, 1- what is the relation between a compact set and boundedness of every continuous real valued function on it? 2- If a set on $\mathbb{R}^n$ is compact, is it bounded? and if it is bounded is it compact? • All this in the real line or something else?..your question lacks some context..you have to be more clear – Marios Gretsas Aug 20 '17 at 19:40 • Do you mean theorems relating the various concepts? Well it's not hard to prove that the continuous image of compact spaces are compact and continuous image of connected spaces are connected. – John Griffin Aug 20 '17 at 19:41 • On $\mathbb{R^n}$ @MariosGretsas – Emptymind Aug 20 '17 at 19:41 • then see the comment of @JohnGriffin – Marios Gretsas Aug 20 '17 at 19:43 Suppose $X$ is compact and $f:X\to\mathbb{R}^n$ is continuous. Then $f$ is bounded. Proof: Since the continuous image of compact spaces are compact, we know that $f(X)$ is a compact subset of $\mathbb{R}^n$. Thus it is a bounded subset of $\mathbb{R}^n$ by the Heine-Borel theorem. The Heine-Borel theorem mentioned above says: A subset $E$ of $\mathbb{R}^n$ is compact iff $E$ is closed and bounded. Thus compact subsets of $\mathbb{R}^n$ are bounded. However, the converse is not true. For instance, $(0,1)$ in $\mathbb{R}$ is bounded but not compact. • Is it true that "If every continuous real valued function defined on a nonempty subset of $\mathbb{R^n}$", n>1, is bounded, then this non empty subset is compact? – Emptymind Aug 20 '17 at 23:53 • @Intuition Yes. – John Griffin Aug 20 '17 at 23:58 • why ? ..... is there a theorem that said this? – Emptymind Aug 22 '17 at 7:29 • @Intuition accept whatever you want..accept the answer that you believe it helps you the most..Don't worry.. If you accept John's answer,which of course is a nice answer,there would be no problem by me. – Marios Gretsas Aug 22 '17 at 12:18 • @Intuition See here: math.stackexchange.com/questions/668905/… – John Griffin Aug 22 '17 at 14:19 A continuous image of a connected set is connected. Also a topological space $X$ is connected iff there does not exist a surjective continuous function from $X$ to $\{0,1\}$(or any set with two elements in the reals for instance). Now for compactness-continuity we have also these results: If $f:X \rightarrow Y$ is a continuous bijection and $X$ ia compact topological space and $Y$ is a Hausdorf topological space then $f$ is a homeomorphism. . Every continuous function on a compact metric space $X$ is uniformly continuous on $X$ • can I say that if a subset of $\mathbb{R^n}$, where n > 1, is compact then its connected? – Emptymind Aug 20 '17 at 22:49 • No because the set $([0,1] \times [0,1]) \cup( [5,6] \times [5,6])$ is compact but not connected..Note that a finite union of compact sets is compact. – Marios Gretsas Aug 20 '17 at 22:54 • I have a problem ..... your answer contains half of the answer of the question and @John Griffin contains the other half ..... so I do not know who to accept? – Emptymind Aug 22 '17 at 7:31
2020-08-13T08:19:07
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http://cn.mathworks.com/help/signal/ref/sinc.html?nocookie=true
Accelerating the pace of engineering and science # sinc Sinc function ## Description example y = sinc(x) returns an array, y, whose elements are the sinc of the elements of the input, x. y is the same size as x. ## Input Arguments expand all ### x — Input arrayscalar value | vector | matrix | N-D array Input array, specified as a real-valued or complex-valued scalar, vector, matrix, or N-D array. When x is nonscalar, sinc is an element-wise operation. Data Types: single | double Complex Number Support: Yes ## Output Arguments expand all ### y — Sinc of inputscalar value | vector | matrix | N-D array Sinc of the input array, x, returned as a real-valued or complex-valued scalar, vector, matrix, or N-D array of the same size as x. ## Examples expand all ### Ideal Bandlimited Interpolation Perform ideal bandlimited interpolation of a random signal sampled at integer spacings. Assume that the signal to interpolate, x, is 0 outside of the given time interval and has been sampled at the Nyquist frequency. Reset the random number generator for reproducibility. ```rng default t = 1:10; x = randn(size(t))'; ts = linspace(-5,15,600); [Ts,T] = ndgrid(ts,t); y = sinc(Ts - T)*x; plot(t,x,'o',ts,y) xlabel Time, ylabel Signal legend('Sampled','Interpolated','Location','SouthWest') legend boxoff ``` expand all ### Sinc The sinc function is defined by $\mathrm{sinc}\text{\hspace{0.17em}}t=\left\{\begin{array}{cc}1& t=0,\\ \frac{\mathrm{sin}\pi t}{\pi t}& t\ne 0.\end{array}$ This analytic expression corresponds to the continuous inverse Fourier transform of a rectangular pulse of width 2π and height 1: $\mathrm{sinc}\text{\hspace{0.17em}}t=\frac{1}{2\pi }\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\int }_{-\pi }^{\pi }{e}^{j\omega t}d\omega .$ The space of functions bandlimited in the frequency range $\omega =\left(-\pi ,\pi \right]$ is spanned by the countably infinite set of sinc functions shifted by integers. Thus, you can reconstruct any such bandlimited function g(t) from its samples at integer spacings: $g\left(t\right)=\sum _{n=-\infty }^{\infty }g\left(n\right)\text{\hspace{0.17em}}\mathrm{sinc}\text{\hspace{0.17em}}\left(t-n\right).$
2015-03-03T13:38:52
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http://winbytes.org/help/standard-error/proportion-standard-error-of-the.html
Home > standard error > proportion standard error of the # Proportion Standard Error Of The repeatedly randomly drawn from a population, and the proportion of successes in each sample is recorded ($$\widehat{p}$$),the distribution of the sample proportions (i.e., the sampling distirbution) can be approximated standard error of proportion formula by a normal distribution given that both $$n \times p \geq 10$$ and ## Standard Error Of Proportion Definition $$n \times (1-p) \geq 10$$. This is known as theRule of Sample Proportions. Note that some textbooks use a sample proportion formula minimum of 15 instead of 10.The mean of the distribution of sample proportions is equal to the population proportion ($$p$$). The standard deviation of the distribution of sample proportions is symbolized ## Standard Deviation Of Sample Proportion by $$SE(\widehat{p})$$ and equals $$\sqrt{\frac {p(1-p)}{n}}$$; this is known as thestandard error of $$\widehat{p}$$. The symbol $$\sigma _{\widehat p}$$ is also used to signify the standard deviation of the distirbution of sample proportions. Standard Error of the Sample Proportion$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$If $$p$$ is unknown, estimate $$p$$ using $$\widehat{p}$$The box below summarizes the rule of sample proportions: Characteristics of the Distribution of sample proportion calculator Sample ProportionsGiven both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$, the distribution of sample proportions will be approximately normally distributed with a mean of $$\mu_{\widehat{p}}$$ and standard deviation of $$SE(\widehat{p})$$Mean $$\mu_{\widehat{p}}=p$$Standard Deviation ("Standard Error")$$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$$ 6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6.2.3 - Military Example ‹ 6.1.2 - Video: Two-Tailed Example, StatKey up 6.2.1 - Marijuana Example › Printer-friendly version Navigation Start Here! Welcome to STAT 200! Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: The “Big Picture” Lesson 1: Gathering Data Lesson 2: Turning Data Into Information Lesson 3: Probability - 1 Variable Lesson 4: Probability - 2 Variables Lesson 5: Probability Distributions Lesson 6: Sampling Distributions6.1 - Simulation of a Sampling Distribution of a Proportion (Exact Method) 6.2 - Rule of Sample Proportions (Normal Approximation Method)6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6.2.3 - Military Example 6.3 - Simulating a Sampling Distribution of a Sample Mean 6.4 - Central Limit Theorem 6.5 - Probability of a Sample Mean Applications 6.6 - Introduction to the t Distribution 6.7 - Summary on the Mean Learning Objectives Estimate the population proportion from sample proportions Apply the correction for continuity Compute a confidence interval A candidate in a two-person election commissions a poll to determine who is ahead. The pollster randomly chooses 500 registered voters ## Standard Error Of P Hat and determines that 260 out of the 500 favor the candidate. In other words, 0.52 ## Confidence Interval Of Proportion of the sample favors the candidate. Although this point estimate of the proportion is informative, it is important to also compute standard error of proportion excel a confidence interval. The confidence interval is computed based on the mean and standard deviation of the sampling distribution of a proportion. The formulas for these two parameters are shown below: μp = π Since we https://onlinecourses.science.psu.edu/stat200/node/43 do not know the population parameter π, we use the sample proportion p as an estimate. The estimated standard error of p is therefore We start by taking our statistic (p) and creating an interval that ranges (Z.95)(sp) in both directions, where Z.95 is the number of standard deviations extending from the mean of a normal distribution required to contain 0.95 of the area (see the section on the confidence interval http://onlinestatbook.com/2/estimation/proportion_ci.html for the mean). The value of Z.95 is computed with the normal calculator and is equal to 1.96. We then make a slight adjustment to correct for the fact that the distribution is discrete rather than continuous. Normal Distribution Calculator sp is calculated as shown below: To correct for the fact that we are approximating a discrete distribution with a continuous distribution (the normal distribution), we subtract 0.5/N from the lower limit and add 0.5/N to the upper limit of the interval. Therefore the confidence interval is Lower limit: 0.52 - (1.96)(0.0223) - 0.001 = 0.475 Upper limit: 0.52 + (1.96)(0.0223) + 0.001 = 0.565 0.475 ≤ π ≤ 0.565 Since the interval extends 0.045 in both directions, the margin of error is 0.045. In terms of percent, between 47.5% and 56.5% of the voters favor the candidate and the margin of error is 4.5%. Keep in mind that the margin of error of 4.5% is the margin of error for the percent favoring the candidate and not the margin of error for the difference between the percent favoring the candidate and the percent favoring the opponent. The margin of error for the difference is 9%, twice the margin of error for the individual percent. Keep this in mind when you h Tables Constants Calendars Theorems Standard Error of Sample Proportion Calculator https://www.easycalculation.com/statistics/standard-error-sample-proportion.php Calculator Formula Download http://stats.stackexchange.com/questions/11008/how-can-i-calculate-the-standard-error-of-a-proportion Script Online statistic calculator allows you to estimate the accuracy of the standard error of the sample proportion in the binomial standard deviation. Calculate SE Sample Proportion of Standard standard error Deviation Proportion of successes (p)= (0.0 to 1.0) Number of observations (n)= Binomial SE of Sample proportion= Code to add this calci to your website Just copy and paste the below code to your webpage where you standard error of want to display this calculator. Formula Used: SEp = sqrt [ p ( 1 - p) / n] where, p is Proportion of successes in the sample,n is Number of observations in the sample. Calculation of Standard Error in binomial standard deviation is made easier here using this online calculator. 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Of Sampling Distribution Calculator div Guides Stock Basics Economics Basics Options Basics div Exam Prep Series Exam CFA Level Series Exam li Simulator Stock Simulator Trade computing standard error in excel computing standard error in excel p the toolbar at the top A menu will appear that says Paste Function Select Stastical from the left hand side of sem formula excel the menu if necessary Scroll down on the right hand side of the menu how to calculate standard error in excel and select STDEV then click OK Click on the picture of the spreadsheet and highlight the numbers you averaged How To Calculate Standard Error Of Estimate In Excel earlier just as you did when taking the average Hit enter and OK to calculate the standard deviation With the cursor confidence level standard error of the mean confidence level standard error of the mean p normal distribution calculator to find the value of z to use for a confidence interval Compute a confidence interval on the mean when sigma is 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based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a standard error of estimate in regression sample Figure shows two regression examples You can see that in Graph The Standard Error Of The Estimate for The Regression Measures A the points are closer to the line than they are in Graph B Therefore the predictions in Graph A Regression Analysis Standard Error Of Estimate are more accurate than in conditional standard error of measurement formula conditional standard error of measurement formula p Download Full-text PDF Standardized Conditional SEM A Case for Conditional ReliabilityArticle PDF Available in Applied Psychological Measurement - September with ReadsDOI st Nambury S Raju nd Larry R Price Texas calculate conditional standard error of measurement State University rd T C Oshima Mike NeringAbstractAn examinee-level or conditional reliability conditional standard error of measurement definition 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Chi-square f Dist Hypergeometric Multinomial Negative binomial Normal Poisson t Dist Random numbers Probability Bayes rule Combinations permutations Factorial Event counter Wizard Graphing Scientific Financial Calculator books AP calculator review Statistics AP study guides computing error bars Probability Survey sampling Excel Graphing calculators Book reviews Glossary AP practice exam Problems and solutions Formulas how to calculate standard error in stats Notation Share with Friends What is the Standard Error The standard error is an estimate of the standard deviation of a statistic standard error example This computing standard error of the mean in excel computing standard error of the mean in excel p the toolbar at the top A menu will appear that says Paste Function Select Stastical from the left hand side of How To Calculate Standard Error Of The Mean In Excel the menu if necessary Scroll down on the right hand side of the menu standard error of the mean excel graph and select STDEV then click OK Click on the picture of the spreadsheet and highlight the numbers you averaged standard error of the mean excel formula earlier just as you did when taking the average Hit enter and OK concept of standard error of mean concept of standard error of mean p from the same population The standard error of standard error of the mean definition statistics the mean estimates the variability between samples whereas the equation for standard error of the mean standard deviation measures the variability within a single sample For example you have standard error meaning in regression analysis a mean delivery time of days with a standard deviation of days based on a random sample of delivery Standard Error Meaning And Interpretation times These numbers yield a standard error of the mean of days divided by the square root of Had computing standard error linear regression computing standard error linear regression p the estimate from a scatter plot Compute the standard error of 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estimate the sample mean dispersion compute standard error observed proportion compute standard error observed proportion p repeatedly randomly drawn from a population and the proportion of successes in each sample is recorded widehat p the distribution of the sample proportions i e the sampling distirbution standard error of proportion calculator can be approximated by a normal distribution given that both n times Standard Error Of Proportion Formula p geq and n times -p geq This is known as theRule of Sample Proportions Note that Sample Proportion Formula some textbooks use a minimum of instead of The mean of the distribution of sample proportions is equal to the population proportion p compute standard error linear regression compute standard error linear regression p the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a standard error formula regression sample Figure shows two regression examples You can see that in Graph formula for standard error of regression coefficient A the points are closer to the line than they are in Graph B Therefore the predictions in Graph A standard error of regression coefficient are more accurate than in Graph B Figure Regressions differing in compute the standard error of the estimate calculator compute the standard error of the estimate calculator p Electrical Calculators Digital Computations Mechanical Calculators Environmental Calculators Finance Calculators All Finance Categories Mortgage Calculators Loan Calculators Interest Calculators Investment Calculators Credit Debt Calculators Profit Loss Calculators Tax Calculators Insurance Calculators Financial Ratios Finance Chart Standard Error Of Measurement Online Calculator Currency Converter Math Tables Multiplication Division Addition Worksheets Math calculators Statistics Sample Mean compute the standard error of the estimate for the data below Dispersion from Population Mean Calculation Standard Error SE of Mean Calculator Enter Inputs in Comma separated Standard Error Of Estimate Calculator Regression form standard error SE compute the standard error of the sample mean for hrc compute the standard error of the sample mean for hrc p a Fortune verification firm Get a Professional Answer Via email text message or notification as you wait on our site Ask how to compute standard error of the mean in excel follow up questions if you need to Satisfaction Guarantee Rate the how to find the standard error of the sample mean answer you receive Ask StatisticsExpert Your Own Question StatisticsExpert Professor Category Homework Satisfied Customers Experience M Sc compute margin of error M Tech PhD Type Your Homework Question Here StatisticsExpert is online now Human Resource Consulting HRC compute the standard error of the regression compute the standard error of the regression p the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a sample Figure shows two regression how to calculate standard error of regression coefficient examples You can see that in Graph A the points are closer to the line how to calculate standard error of regression in excel than they are in Graph B Therefore the predictions in Graph A are more accurate than in Graph B Figure compute standard error sampling mean compute standard error sampling mean p to a normally distributed Sample Mean And Standard Deviation Calculator sampling distribution whose overall mean is equal to the mean of the source Standard Error Of Sampling Distribution population and whose standard deviation standard error is equal to the standard deviation of the source population divided by the square root ofn To calculate the standard error standard error of sampling distribution when population standard deviation is unknown of any particular sampling distribution of sample means enter the mean and standard deviation sd of the source population along with the value ofn and then click compute the standard error of estimate in excel compute the standard error of estimate in excel p the toolbar at the top A menu How To Calculate Standard Error In Excel will appear that says Paste Function Select Stastical from the left hand side of the menu if necessary Scroll down on the right hand side of the menu how to calculate standard error in excel mac and select STDEV then click OK Click on the picture of the spreadsheet and highlight the numbers you averaged earlier just as you did when taking the average Hit enter and OK to calculate the standard deviation With the cursor still compute the estimated standard error for the sample mean difference compute the estimated standard error for the sample mean difference p the difference between means Compute the standard error of the difference between means Compute the probability of a difference between means being above a specified value Statistical analyses are very often concerned with the difference between means A typical example is an experiment designed to compare how to calculate estimated standard error for the sample mean difference the mean of a control group with the mean of an experimental group Inferential statistics How To Calculate Sample Mean And Standard Deviation used in the analysis of this type of experiment compute the standard error of the proportion compute the standard error of the proportion p repeatedly randomly drawn from a population and the proportion of successes in each how to compute standard error in excel sample is recorded widehat p the distribution of the sample proportions How To Compute Standard Error In R i e the sampling distirbution can be approximated by a normal distribution given that both n compute standard error standard deviation times p geq and n times -p geq This is known as theRule of Sample Proportions Note that some textbooks use how to compute standard error of regression coefficient a minimum of instead compute standard error estimate excel compute standard error estimate excel p the toolbar at the top A menu will appear that says Paste Function Select Stastical from the left hand side of standard error of estimate calculator excel the menu if necessary Scroll down on the right hand side of the menu standard error of estimate calculator regression and select STDEV then click OK Click on the picture of the spreadsheet and highlight the numbers you averaged standard error of estimate calculator ti- earlier just as you did when taking the average Hit enter and OK to calculate the standard deviation With the cursor still compute standard error proportion compute standard error proportion p p p test AP formulas FAQ AP study guides AP calculators Binomial Chi-square f Dist Hypergeometric Multinomial Negative binomial Normal Poisson t Dist Random numbers sample proportion symbol Probability Bayes rule Combinations permutations Factorial Event counter Wizard Graphing Scientific Financial Calculator Standard Error Of Difference Between Two Proportions Calculator books AP calculator review Statistics AP study guides Probability Survey sampling Excel Graphing calculators Book reviews Glossary standard error two proportions calculator AP practice exam Problems and solutions Formulas Notation Share with Friends Confidence Interval Proportion Large Sample This lesson describes how to construct a confidence compute standard error of mean difference compute standard error of mean difference p randomly How To Calculate Standard Error Of The Mean Difference drawn from the same normally distributed source population belongs to standard error of the mean formula a normally distributed sampling distribution whose overall mean is equal to zero and whose standard deviation standard Standard Error Of The Mean Calculator Online error is equal to square root sd na sd nb where sd the variance of the source population i e the square of the standard deviation na the size of sample A and nb standard error of the mean calculator using excel the compute standard error estimate data
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# Find The Number Of Distinguishable Arrangements Of The Letters Of The Word Billion arrangements are possible. We go through 3 examples. The number of ordered arrangements of the letters in the word PHILIPPINES is: $$\dfrac{11!}{3!1!3!1!1!1!1!}=1,108,800$$ Example 3-16 We need to count the number of distinguishable permutations when the three colors are the only features that make the flags distinguishable. Random passwords consisting of only English letters have less entropy per character than random passwords selected from a larger character set, but additional letters can be added to make up the. Find the number of different arrangements of the letters in the word (CAVALRY) and illustrate linearly. There will be no going back, university. The strands arrangement is represented by coloured cells where the cell formula calculates that strand’s heating capacity. 2 billion for subsidies. The number of permutations as calculated above would be an over count. The statistics & probability method Permutation (nPr) is employed to find the number of possible different arrangement of letters for a given word. ERA2 sets aside $2. The number will, however, very frequently be 6 (or 5 or 8). Get stock market news and analysis, investing ideas, earnings calls, charts and portfolio analysis tools. Thus the letters in the word CAT has six different anagrams: ACT ATC CAT CTA TAC TCA (Note that only two of these are English words, the remaining four are jibberish, but are nevertheless anagrams, by our definition. 7 billion losses of the Postal Service reported by CityLabs, letters remain one of the most important communication tools both in the academe and in the business world. Solved Examples for You. SIR – We were told that the. (Brown), through an exchange of Kinney for Brown stock, would violate § 7 of the Clayton Act, 15 U. Report to the Director General for Research of the European Parliament. An invention of the devil which abrogates some of the advantages of making a disagreeable person keep his distance. , serves as chancellor of LSC, the largest institution of higher education in the Houston area with an annual economic impact of nearly$3 billion. We need to count the number of distinguishable permutations when the three colors are the only features that make the flags distinguishable. 9_C_2=36 9_C_3=84. The National Flood Insurance Program (NFIP) is managed by the Federal Emergency Management Agency and is delivered to the public by a network of approximately 60 insurance companies and the NFIP Direct. Because it’s made up of so many different technologies, terms, and business practices, finding a place to start learning about the types of programmatic advertising deals can be challenging for those who are new to the space. However, we can’t allow. Hurricanes Irma, Maria and Harvey Irma Incident PeriodSeptember 4. See full list on people. On the outskirts of Austin, Texas, there is a small chapel — less. 5 million tourists visiting in the year to September 2015. "Combinations" gives the number of ways a subset of r items can be chosen out of a set of n items. our building society wrote many letters to the police, BMC and all and sundry in sight. Find the graves of ancestors, create virtual memorials or add photos, virtual flowers and a note to a loved one's memorial. different, DISTINGUISHABLE PERMUTATIONS. If the letters were all different there would have been 11! different permutations. There are 11! Different arrangements of 11 letters. The act of placing or arranging. Thus, the number of ways to arrange COMPUTER, keeping the vowels in alphabetical order, is 8!/3! (or P(8, 3)) = 6720 Another way to think about the problem is to consider the number of ways to permutate. To distinguish between a decimal K (1,000) and a binary K (1,024), the IEEE has suggested following the convention of using a small k for a decimal kilo and a capital K for a binary kilo, but this convention is by no means strictly followed. To solve permutation problems, it is often helpful to draw line segments for each option. Find the number € n of ways to:. Since order does not matter, use combinations to determine the number of possible subsets. ” (a) acdbens (b) baaaben (c) aaabbba 10. Thus a 17 letter password, offering 80 bit security, could be represented by a 10 word sentence and a 7 word sentence, or a 9 and an 8 word sentence. Permutations should not be confused with combinations (for which the order has no influence) or with arrangements also called partial permutations (k-permutations of some elements). We must be satisfied with your capacity to pay the up front fee for providing Proof of Funds (known as the Arrangement Fee). If we VASTLY underestimate there are 1 billion stars in the universe and overcount the number of planets we’ve found at 500, the percentage is a mere. You can use the Fundamental Counting Principle to find the number of permutations. Find the expected number of offspring per year. Queen of Hearts) and a number (e. Credit: Getty/Barcroft Media. In your Word document, click the "Layout" tab in the ribbon bar and then click on the "Line Numbers" button. Adolf Tolchakev was a Russian radar engineer who had grown disenchanted with the Soviet Union, a country that crushed liberty and failed to provide for its people. The following disasters are considered historical because of how they impacted the way we handle similar disasters in the future. Globe file photo. Permutation Calculator. Again, using the formula for the number of distinguishable permutations, the ship's captain could send any of: $$\dfrac{9!}{4!3!2!}=\dbinom{9}{4\ 3\ 2}$$ or 1260 possible signals. d) the committee must have at least 4 seniors. An irrevocable letter of credit from the issuing bank insures the beneficiary that if the required documents are presented and the terms and conditions are complied with, payment will be made. The number of 7 letter words that can be formed with the letters of the word incomputably such that T occupies the middle place. The number will, however, very frequently be 6 (or 5 or 8). The actual number of distinguishable olfactory stimuli is likely to be even higher than 1. In all four questions the host took roughly one third of all the turns. However, we can’t allow. ∴ Number of ways of arranging these letters = (2!) (2!) 8! = 1 0 0 8 0. We have 2 Ms 2 As 2 Ts, so the sum of all permutations is $$\frac{11!}{2!\cdot 2!\cdot 2!}$$. A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. The number of firms that have reported unclaimed financial assets to the government has increased 208 per cent from mid last year, pushing deposits at the trust fund to Sh33 billion up from Sh25. Posted on January 4, 2017 by stuartmdarling. Solution: There are 9 letters in the given word in which two T’s, two M’s and two E’s are identical. 29 CFR 2510. Distinct Ways to Arrange Word MASSACHUSETTS - Workout. The Joe Biden who portrayed himself as a moderate, old-school. 12] Find the number € m of ways that 7 people can arrange themselves: a) in a row of chairs b) around a circular table 13. The Democrats even included a questionable a power-sharing arrangement under which the GOP would have had a virtual veto power over subpoenas. Combinations sound simpler than permutations, and they are. Words in this unit all have the /ow/ sound (spelled with the letters ow or ou ). The easement is itself a real property interest, but legal title to the underlying land is retained by the original owner for all other purposes. exercisingtype2 The term diabetes includes several different metabolic disorders that all, if left untreated, result in abnormally high concentration of a sugar called glucose in the. Display a. Based on by mutation altered the code, creates a different amino acid. The word has 9 letters with 2 letters occurring twice each. Two of these teams qualify from the group. 7 billion losses of the Postal Service reported by CityLabs, letters remain one of the most important communication tools both in the academe and in the business world. This is known as the standard of Umar ibn al Khattab, radhiallahu anhu. More than 1. * (an asterisk) means any number of items (0, 1, or more). Report to the Director General for Research of the European Parliament. Get stock market news and analysis, investing ideas, earnings calls, charts and portfolio analysis tools. About 217 million of those have moderate to severe vision difficulties. Sexdecillion - definition of sexdecillion by The Free Dictionary sexdecillion, two nonillion and tredecillion (12 letters each). During the 1964-85 period, the company made capital expenditures of about $3 billion, far more than any other U. 2 billion people in the world’s richest countries enjoyed an average annual income of$19,300—a truly astounding achievement. The 2 letter? rd 3 letter? 4th letter? 3) Calculate the product of the numbers in 2) to find the number of 4-letter permutations in the word FAMILY. Use it to sort any list of text online, using your computer or mobile device. If the beneficiary and/or the agent can establish that the amount received is a gift to the beneficiary, no tax is due. 12] Find the number € m of ways that 7 people can arrange themselves: a) in a row of chairs b) around a circular table 13. Highsmith is suing Getty Images for $1 billion over its alleged copyright violation of 18,755 of her photos. You can view submissions on the https:// www. That’s almost half of the total 2. The devolved government for Scotland has a range of responsibilities that include: the economy, education, health, justice, rural affairs, housing, environment, equal opportunities, consumer advocacy and advice, transport and taxation. The Democrats even included a questionable a power-sharing arrangement under which the GOP would have had a virtual veto power over subpoenas. This is a small kindness that deserves many thanks. The number of a part shall be in Arabic numerals, starting with 1, following the document number and preceded by a hyphen. Swap these numbers as reverse the subset. Step-by-step explanation. Then, for each non-complete word in the puzzle (basically find the first blank square and see if the one to the right (across-word) or the one underneath (down-word) is also blank), a search was done of the file looking for the first word that fitted, taking into account the letters already in that word. For example, the number of distinguishable arrangements of the letters of the word DAD is not 3!=6 3 factorial equals 6 but rather 3!2!=3. Find 20 ways to say ANALOGY, along with antonyms, related words, and example sentences at Thesaurus. py that takes two 5-letter strings from the command line, reads in a list of 5-letter words from standard input, and writes a shortest word ladder using the words on standard input connecting the two strings (if it exists). 37) and then in a variety of ways produce a deck where that card appears at that order in the deck. "Combinations" gives the number of ways a subset of r items can be chosen out of a set of n items. Ans:Number of ways of selecting apples = (3+1) = 4 ways. Permutations Involving Repeated Symbols - Example 2. The answer of the above problem is 720 720. First we calculate the number of distinguishable substrings, then subtract what we get from 50400. ASCII is an encoding representing each typed character by a number Each number is stored in one byte (so the number is in 0. Social media is an ever-changing and ever-evolving field, with new apps such as TikTok and Clubhouse coming out seemingly every year. Question 428983: Find the number of distinguishable arrangements of the letters of the word "OLIGOPOLY". Explanation: The small letters are b, d, f, h, j, l, n, p, r, t, v, x, z. One might define an anagram of a word as a permutation of its letters. com's go to source for expert writing advice, citation tips, SAT and college prep, adult education guides and much more. 1 "January revenues were slightly below benchmark but significantly above the prior year. approves$2. The two accompanying alchemical cryptograms illustrate another form of the literal cipher involving the first letter of each word. The devolved government for Scotland has a range of responsibilities that include: the economy, education, health, justice, rural affairs, housing, environment, equal opportunities, consumer advocacy and advice, transport and taxation. 3 billion from Novartis AG in a suit claiming the company paid kickbacks to increase sales of two prescription drugs. Further, we find that each additional condition increases the neonatal mortality rate, on average, by 0. a) How many different arrangements are there of the letters of the word numbers? 7! = 5,040. 4 billion, and about $14 billion comes from oil, gas and mineral interests. The Letter Sorting Word Generator can be of great help to find words from letters for competitions, newspaper riddles, or even for your homework! We keep the word lists of all our Word Generators constantly updated, so that you can have access to a large variety of options to unscramble words. = 7C4×4! = 840 = 7 C 4 × 4! = 840. This is a simple Sales Contract template directed between between two parties that covers a variety of agreements for the seller and buyer to comply with in order to proceed. Every cryptogram based upon the arrangement or combination of the letters of the alphabet is called a literal cipher. We review their content and use your feedback to keep the quality high. 5 billion in the National Skills Fund. How many distinguishable arrangements are possible using the letters of the word PARALLEL? 5. Unfortunately, in some of. The word mathematics has 11 letters; 2 are m, a, t. (Type a whole number. In the picture below, I have done 6 single crochets (the thing at the very end that looks like a 7th st is the chain I made in step 8). Taken two at a time b. The periodic table of elements arranges all of the known chemical elements in an informative array. We are entering a period of extreme volcanism. Solution for Find the number of distinguishable permutations of the given letters "AABCD". An example to find the number of distinguishable permutations. Every day, we stand up for, celebrate, educate and inspire the people who power the retail industry. Sight Word Leaf Match – Match two word cards. Swap these numbers as reverse the subset. 3 billion to 2. There is little evidence to support the case for yet more structural change For many healthcare professionals and managers working in the NHS, last week’s publication of the white paper Equity and Excellence: Liberating the NHS brought unwelcome but familiar news—that the new government plans to reorganise the NHS in England. Build a safe, secure, and resilient Defense Industrial Base (commercial and organic). Google has many special features to help you find exactly what you're looking for. Typical easements are for access to another property, (redundantly often stated "access and egress," since entry and. The report discusses some key influences on people’s mental health, examines the effect of mental health on people’s ability to participate and prosper in the community and workplace, and implications more generally for our economy and productivity. To distinguish between a decimal K (1,000) and a binary K (1,024), the IEEE has suggested following the convention of using a small k for a decimal kilo and a capital K for a binary kilo, but this convention is by no means strictly followed. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. They could have obstructed to their hearts content. If you have a calculator handy, this part is easy: Just hit 10 and then the exponent key (often marked x y or ^ ), and then hit 6. The 2 letter? rd 3 letter? 4th letter? 3) Calculate the product of the numbers in 2) to find the number of 4-letter permutations in the word FAMILY. Starting this year, the government is investing £2. sex′de·cil′lion adj. Since there are 2 Os that can be arranged in 2! or 2 ways, the number of permutations of the letters O, P, O, L, and S can be written as Key Concept Permutations with Repetition The number of distinguishable permutations of n objects in which one. For millions of parents, that insecurity can mean working fewer hours, taking a pay cut, or leaving their jobs altogether. News and more about hardware products from Microsoft, including Surface and accessories. So it became hard to find my combination to open it. Programmatic advertising is a complicated topic that’s always evolving. Sight Word Leaf Match – Match two word cards. To figure out the number of unique arrangements of the 8 letters with 3 that are indistinguishable we can find the permutations of the 8 letters and divide by the permutations of the indistinguishable items. New Movie Releases. 19 billion in 2015. Store the letters of a word in an array of characters. How many of these a) Begin with E & end with E b) Have all the 3 E's together c) No two vowels are together Solution:- The word ENGINEERING has 11 letters out of which E is repeated thrice, N is repeated thrice, G is repeated twice, I is repeated twice. Well, there are 11 letters in total: 1 M, 4 I, 4 S and 2 P. It is calculated that, on average, each four-letter Hebrew "word' or combination of four Hebrew letters will appear approximately 130,000 times in the text of the Five Books of Moses. com offering definitions, meanings, and grammar in both English and Spanish. Learn how to find the number of distinguishable permutations of the letters in a given word avoiding duplicates or multiplicities. 4) Of the 10 letters in the word, there are 3 s’s, 3 t’s, 1 a, 2 i’s, and 1 c, so the total number of distinct arrangements of the letters is 10! 3!3!1!2!1! = 50400. If you haven't worked with factorials in a while, let me remind you that 12! is 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 or 479,001,600 different letter arrangements (counting any duplicates. coli , a bacillus of about average size is 1. The main causes include diabetes, obesity, and poor diet, which are indicative of social, political and economic issues indigenous to the U. Like a "super wildcard". Posted on January 4, 2017 by stuartmdarling. noun - mail that includes letters and postcards and packages sealed against inspection 1st-Class - a class mail comprising letters, postcards, and other mail sealed against inspection, having a higher priority than second, third, or fourth-class mail; -- it is the highest class of mail not handled in a special manner, as is registered or. However, note that the letter “T” is repeated twice and these two T’s are indistinguishable and provide the same arrangement during scrambling. Essential duties highlighted on a Senior Executive example resume are implementing company policies and procedures, managing finance, developing strategic plans, guiding management teams, and supervising company operations. 6% per year. Explanation: The small letters are b, d, f, h, j, l, n, p, r, t, v, x, z. If you have a calculator handy, this part is easy: Just hit 10 and then the exponent key (often marked x y or ^ ), and then hit 6. (Simplify your answer. We couldn't distinguish among the 4 I's in any one arrangement, for example. A Champions League group consists of four teams, Ajax, Barcelona, Celtic, and Dortmund. Search the world's information, including webpages, images, videos and more. 2 billion for the IRS, an increase of$1. This can also be described as the amount of information required to precisely describe the internal state of the system. Gigabyte, GB, about 1 billion bytes Terabyte, TB, about 1 trillion bytes (rare) Bytes and Characters - ASCII Code. Contractual Arrangement In the early days of Internet advertising, ads were largely sold on a CPM basis with predesigned contracts. ) (PYMNTS). Partly due to the realisation that having twins is an expensive business, and partly because my CD collection was already running out of control – 2016 was the year that I switched over to the digital world – Apple Music being my provider of choice (convenience won the day). d) the committee must have at least 4 seniors. ∴ Number of ways of arranging these letters = (2!) (2!) 8! = 1 0 0 8 0. From 9/11 to Donald Trump», we are publishing a series of articles which develop a few of the numerous. For example, if we only know that a message is encoded with a word of 7 letters, then it could be encoded in 2 6 7 ≈ 8 26^7 \approx 8 2 6 7 ≈ 8 billion ways! [1] The primary weakness of the Vigenère cipher is the repeating nature of its key. Analyzing CTS Off-Campus Drive and Written Test Papers, the level of difficulty for Cognizant is moderate to high. You have the right to be paid on time. • Order of 3 letters does not matter =⇒ Combination. The total number of initial entries are N = 2055 (Omaha), N = 524 (Seattle), N = 231 (New York), N = 381 (Jacksonville). Other English palindromes are "refer", "level" and "kayak". 9 billion from island B). ( 6—4) l, *5) How many 6-letter permutations are there in FAMILY. (b) Both L do not occur together. Hence, the number of distinguishable arrangements of n objects in a circle is the number of linear arrangements divided by n, which yields n! n=(n−1)!. Number of permutations of n different things taken r at a time when each thing may be repeated any number of times is n r. It is one of the largest penalties ever assessed by the U. 2 min to read. 6 letters can be arranged in 6! ways = 720 ways Vowels can be arranged in themselves in 4! ways = 24 ways Required number of ways = 720*24 = 17280. It is safe to say that any sentence in which this omnibus use occurs will be improved either by omitting the word or by substituting a word of more definite meaning. is seeking as much as $3. The only favorable outcome is the arrangement 25398. From left to right, find lowest number bigger than 3 eg. , cannot be picked up out of the plane and turned over) circle is P_n=(n-1)!. This can also be described as the amount of information required to precisely describe the internal state of the system. Distinct Ways to Arrange Word MASSACHUSETTS - Workout. One who approaches Greatness on his belly so that he may not be commanded to turn and be kicked. This is a small kindness that deserves many thanks. The federal health spend has been locked in for another year, with sizeable funding allocated for rural GPs, rural generalist training, aged care and mental health. Counting the number of ways objects, some of which may be identical, can be distributed among bins (Section 4. This video shows how to calculate the number of linear arrangements of the word MISSISSIPPI (letters of the same type are indistinguishable). b) (1/2)R2, R1-R2, R3-3R2 results in 11. Responsibilities include: assisting with the recruiting process, assisting with the onboarding. Sales Contract Template. BASEBALL C. Dear ____, (Begin with an introduction- talking a little about the organisation, its causes, and its goals. Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. exercisingtype2 The term diabetes includes several different metabolic disorders that all, if left untreated, result in abnormally high concentration of a sugar called glucose in the. The cost for this service varies according to the POF amount, the type of account as well as the length of time needed to show the funds. The pictorial cipher. The actual number of distinguishable olfactory stimuli is likely to be even higher than 1. About 217 million of those have moderate to severe vision difficulties. You can find out more information by visiting our revision policy and money-back guarantee pages, or by contacting our support team via online chat or phone. The inquiry final report was handed to the Australian Government on 30 June 2020 and released publicly on 16 November 2020. 29 CFR 2510. The new miracle by NCL features an expanded Haven – the line’s concept “ship within a. A billion grains of rice is about 25 tonnes (1,000 grains is about 25g I weighed some!) Notice that the Total of any square is 1 less than the Grains on the next square (Example: square 3's total is 7, and square 4 has 8 grains). It’s expected to hit 1. Solution for Find the number of distinguishable arrangements of the word "EMPATHY". In his letter to Crick Waddington is at pains to stress that “the very numerous genes will interact in such a way that will produce only a smallish number of distinguishable cell types”. Combinations – order doesn’t count. 02 Missed punctuation Omitted periods, and spelling with missing apostrophe dont, cant, wont, ill 360 10. we have had groups of families who have been in isolation for a number of weeks in prep to stay in touch with their family member. How many distinguishable permutations of letters are possible in the word Tennessee? I understand this word has 9 letters, with 1-T, 4-E, 2-N, and 2-S. Step by step workout step 1 Address the formula, input parameters and values Formula: nPr = n!/(n1! n2!. 01 62) For a certain animal species, the probability that a female will have a ce rtain number of offspring in a given year is given in the table below. If this were true, one would expect that keyboard. Google has many special features to help you find exactly what you're looking for.$75 billion per year assuming these have reduced risks by 45 percent: Losses from a successful terrorist attack: Annual Probability of a successful attack in the absence of security expenditures: $100 million:$1 billion: $5 billion:$100 billion: $200 billion:$1 trillion: $5 trillion: London Bombing: 9/11: Nuclear Port: Nuclear Grand Central. there are a couple of C-G, G-C, A-T, T-A. If we consider the vowels I & E side by side as if one letter, there is 6! possible arrangements multiplied by 2 for the two options IE or EI. 4) Of the 10 letters in the word, there are 3 s’s, 3 t’s, 1 a, 2 i’s, and 1 c, so the total number of distinct arrangements of the letters is 10! 3!3!1!2!1! = 50400. 9 million pieces. For financing through various IMF arrangements, including precautionary lines, the IMF stands ready to fully deploy its lending capacity of about$1 trillion to help member countries weather the crisis. It should be noted that aperiodic polypeptides or polynucleotides do not necessarily represent meaningful information or biologically useful functions. = 7C4×4! = 840 = 7 C 4 × 4! = 840. Korean Hangul, for example, classifies letters as initial (I), medial (M), or final (F) and arranges them into square syllable blocks. Therefore, required number of ways. Multiplying the two figures together gives us $313. However, there are considerably more possible odorous molecules than the 128 different components. Check your answers and check your spelling – then write your answers on the answer sheet. The answer is obtained by dividing the inflated count by the amount of over counting. An ordinal number is a number that indicates position or order in relation to other numbers: first, second, third, and so on. 4 billion increase in proceeds from issuances of debt, net of. Posted on January 4, 2017 by stuartmdarling. 6 billion years. 210-531-USAA(8722) Call 210-531-8722 or 800-531-8722 800-531-USAA. 34 billion vaccine doses have been administered worldwide, equal to 17 doses for every 100 people. · Change of Tariff five years later found to be factually and legally incorrect. You have 11 letters consisting of one M, four I's, four S's, and two P's. We must be satisfied with your capacity to pay the up front fee for providing Proof of Funds (known as the Arrangement Fee). That is, according to the World Health Organization. See full list on people. telephone number n 1. This problem has been solved! See the answer. Permutation. The average diameter of spherical bacteria is 0. Celebrating the Timeless Barbra Streisand’s Movies and Music. nk) in the word. First, we count the total number of distinguishable arrangements of the letters of the word ALGEBRA. Permutations with Repetitions: Find the number arrangements for the word: A. If the two A’s were not identical, the seven letters in the word could be arranged in P(7,7) or 7! ways. Apr 6, 2021. Compose a program wordladder. Number of permutations of n different things taken r at a time when each thing may be repeated any number of times is n r. If we VASTLY underestimate there are 1 billion stars in the universe and overcount the number of planets we’ve found at 500, the percentage is a mere. Solution for Find the number of distinguishable arrangements of the word "EMPATHY". During the 1964-85 period, the company made capital expenditures of about$3 billion, far more than any other U. which means the number of combinations of n items taking r items at a time For example means find the number of ways 3 items can be combined, taking 2 at a time, and from the example before, we saw that this was 3. Sir David Nicholson, the NHS chief executive, has described the proposals. God is Dead. The following disasters are considered historical because of how they impacted the way we handle similar disasters in the future. by kellymaeshiro. The Postal Service wants to. If the two A’s were not identical, the seven letters in the word could be arranged in P(7,7) or 7! ways. Explanation: The small letters are b, d, f, h, j, l, n, p, r, t, v, x, z. PrepInsta provides you all the updated information regarding the change in pattern of Infosys Placement Papers 2021. Number of ways of selecting mangoes = (5+1) = 6 ways. What do we do? We put different subscripts next to the different occurrences: $B_1 UB_2B_3LE$. Find the expected number of complaints per day. Bell Labs alone produced the transistor (there are several billion of them in your pocket), the cell phone, the solar cell, the laser, Telstar I, digital signal processing, and the infrastructure of most contemporary systems programming. An ordinal number is a number that indicates position or order in relation to other numbers: first, second, third, and so on. Achieveressays. (E) Nonclassicality for combinations of initial-final photon arrangements captured by the suppression laws associated with Σ 1, Σ 2, and Σ 3. 6 billion digits, but this one has 10 quadrillion digits! To put that in perspective, the digits of the previous number would be able to fit in about 358 dictionaries (assuming each page can fit 10,000 digits and each dictionary has 1000 pages), but to store the digits of this number we'll need a billion dictionaries. This number includes projected vaccine production capacity in China for 610 million doses in 2020 and at least one billion doses annually by the end of 2021. [Sign up for Love Letter and always get the latest in Modern Love, weddings, and relationships in the news by email. Select the type of WordArt you want to add to the document. "Full-time workers who are overweight or obese and have other chronic health problems miss about 450 million more days of work each year than healthy workers. Example: - Find the number of permutations of the letters of the word ENGINEERING. How many letter arrangements can be made from a 2 letter, 3 letter, letter or 10-letter word. Because it’s made up of so many different technologies, terms, and business practices, finding a place to start learning about the types of programmatic advertising deals can be challenging for those who are new to the space. From there, click. 4 What is uniquely you comes in the fractional difference in how those three billion letters are sequenced in your cells. This online discrete math calculator rearranges letters of the given word and gives the number of letter arrangements you can form with it. It has been determined that 99. The number of ways the lines could be internally reordered is 16! × 13! × 15! × 6! × 13! × 14! × 8!. In this case, however, we don't have just two, but rather four, different types of objects. property-casualty reinsurer, the company also owns (including stock it has an arrangement to buy) 82% of the oldest reinsurance company in the world, Cologne Re. First, it is currently not known how many odorous molecules there are or how many of them can be discriminated from the others. 2 billion from Island A and RM8. I think that the solution is: 7!-2*6!=3600. Permutations should not be confused with combinations (for which the order has no influence) or with arrangements also called partial permutations (k-permutations of some elements). The word usually is not a palindrome, although it may be. There are _____ different possible first- and middle-name arrangements. The third day from tuesday will be friday and code will be frIdAY. That’s approximately 15% of the world’s population. Thus, the number of ways to arrange COMPUTER, keeping the vowels in alphabetical order, is 8!/3! (or P(8, 3)) = 6720 Another way to think about the problem is to consider the number of ways to permutate. Slides developed at the University of Florida for course COT3100, Applications of Discrete Structures, Spring 2001 & 2003. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. Thus, we have MTHMTCS (AEAI). The arrangement of science, math, language arts is a permutation of math, science, language arts because the order of the classes is different. A progressive premium for health care based on income will be a more equitable solution to paying for care if there is a shortfall. This is an opportunity provided by the Reporting on Trafficking and Slavery programme: Find out more. Contact Number: 09202737517. The below workout is the step by step procedure to find how many number of distinct ways to arrange the 11 letters (alphabets) of word "MATHEMATICS". Some have a dedicated National 800 number, while others use online services. 5 1 billion Dec. We must be satisfied with your capacity to pay the up front fee for providing Proof of Funds (known as the Arrangement Fee). Therefore, number of ways of arranging these letters = 8! = 40320. For example, consider the letters X, Y, and Z. About 217 million of those have moderate to severe vision difficulties. Check your answers and check your spelling – then write your answers on the answer sheet. 1 Despite having promised just two months ago in the coalition. The company had 2009 worldwide sales totaling $32 billion and an annual growth rate of 7 percent (down from 14. the right to use the real property of another for a specific purpose. In 1995 the 1. Random passwords consisting of only English letters have less entropy per character than random passwords selected from a larger character set, but additional letters can be added to make up the. Contractual Arrangement In the early days of Internet advertising, ads were largely sold on a CPM basis with predesigned contracts. He noted that Lebanon has been spending up to$7 billion a year on subsidies, including a total of $5 billion spent on subsidizing goods that benefit Lebanese families, noting that what the government was looking at was to lower that number to$1. • Order of 3 letters does not matter =⇒ Combination. The synonyms have been arranged depending on the number of charachters so that they're easy to find. employers are saving over $30 billion per day by allowing employees to work from home. Say we have a total of 6 marbles. Metropolitan Life Ins. A Champions League group consists of four teams, Ajax, Barcelona, Celtic, and Dortmund. Find the number of different arrangements of the letters in the word (CAVALRY) and illustrate linearly. Cooperatives are democratically owned by their members, with each member having one vote in electing the board of. Credit and loans are a massive, multi-trillion dollar/year international field/industry which the below superb, very memorable generic credit/loan domains quickly, easily, and affordably open the door to [ news reports state that just one site/company--CreditCards. Find the number of distinguishable arrangements of the letters of the word. BOB BBO OBB We must alter our answer by a factor of two and so the number of arrangements of BOB is 3! 3 2 =. 50+ SAMPLE Medical Agreement Templates in PDF | MS Word. 3 -2 (h) (1) (iv). 7 billion years or 4. committee of 6 members is formed. In our experiments we followed MASC and generated 1,000 annotated instances for each of our 1,960 pseudowords. , Facts 234) to reveal the Facts List results - a navigable listing of all currently highlighted tagged facts. 3 billion base pairs of DNA give 43,000,000,000 possible genomes 106 possible strings of six decimal digits, but only 9*105 six-digit decimal numbers since the first digit cannot be zero. com, the world's most trusted free thesaurus. SIR – We were told that the. However, the word "Yeshua", spelled yod- shin- vav- ayin, has two of the most common letters in the alphabet--the yod and the vav. In 2011, Halliburton posted record revenues of$25 billion, based almost entirely on the growth of its U. To ensure that all Americans are treated fairly by the nation's tax system, including that the wealthy and corporations comply with existing laws, the discretionary request provides $13. Upon close examination we can see that, in fact, the contributing cause of death was a form of nephritis. Animals, plants, and fungi are multicellular organisms and often, there is specialization of different cells for various functions. Expert Answer. Determine the number of different possibilities for the type of number described below. For financing through various IMF arrangements, including precautionary lines, the IMF stands ready to fully deploy its lending capacity of about$1 trillion to help member countries weather the crisis. The Permanent University Fund is valued at nearly $24. If the adviser does have discretion as to the amount or timing of transfers under a SLOA, the seven conditions are back on the table. Further, we find that each additional condition increases the neonatal mortality rate, on average, by 0. hotel contact details as phone number, email, etc. I hope you won’t have to write a condoleance letter in French. If the letters are randomly arranged in order, what is the probability that. d) the committee must have at least 4 seniors. Letter Arrangements in a Word Video. Users may use any other word by changing the word "MATHEMATICS" to find the total number of distinct ways to arrange different words. They set out to healthcare providers notice of NHS England’s Commissioning Intentions for Public Health Services under the Public health functions agreement (Section 7A). For the period before January 1, 2001, consult either the List of CFR Sections Affected, 1949-1963, 1964-1972, 1973-1985, or 1986-2000, published in 11 separate volumes. August 11, 2014. Word ladders. Race car is an example of a palindrome. 7 billion losses of the Postal Service reported by CityLabs, letters remain one of the most important communication tools both in the academe and in the business world. The Arrangement of the Church. 2! 1 2 0!! 2! There are 2 as, 2 bs, and 2 ls in basketball. 30 Full PDFs related to this paper. Unfortunately, the letter B occurs 3 times. Like a "super wildcard". Not all of those are distinguishable, because some words occur more than once on a single line. (d) type and case of letters, spacing between letters and punctuation marks; (e) joining words together or separating the words does not make a name distinguishable from a name that uses the similar, separated or joined words; (f) use of a different tense or number of the same word does not distinguish one name from another;. So a reasonable guess would be that total comparable settlements would be 150 times$24 million in cash, or $1. Swap these numbers as reverse the subset. 2% more than the actual collections in January 2019, and$35 million or 1. We couldn't distinguish among the 4 I's in any one arrangement, for example. 2 dimes and one six-sided die numbered from 1 to 6 are tossed. We can use the following formula, where the number of permutations of n objects taken k at a time is written as n P k. Permutation. 4 billion increase in dividends paid, and a $324 million decrease in proceeds from the issuance of common stock, offset in part by a$3. P (10,3) = 720. Explanation: The small letters are b, d, f, h, j, l, n, p, r, t, v, x, z. From a 4 billion Indian Rupee gaming industry in 2007, to a 62 billion Indian Rupees industry in 2019, gaming in India has certainly caught the eye of consumers and proves to be a valuable market today. ership as an arrangement that leverages the uniqueness of Dutch law to avoid taxes and prevent a hostile takeover attempt. How many distinguishable permutations of letters are possible in the word Tennessee? I understand this word has 9 letters, with 1-T, 4-E, 2-N, and 2-S. One of these code words, the 'start signal' begins all the sequences that code for amino acid chains. Microsoft Excel. com, the international travel industry has grown from 528 million tourist arrivals in 2005 to 1. The spots can also be divided by the total of legs, ears, eyes and tail to leave a remainder of 6. In the Text section, click the WordArt option. Letter Arrangements in a Word Video. 4 percent of loan participations outstanding. Permutation. The 168,600-ton, 1070 feet long, 138 feet wide, 4,200-passenger, and 1716 crew members, 1 billion dollar cruise ship boasts the innovative features found on Norwegian Getaway and Norwegian Breakaway, as well as a number of new interactive experiences. Upfront payment is a great way to offset some of the damage caused by late payments. USA TODAY delivers current local and national news, sports, entertainment, finance, technology, and more through award-winning journalism, photos, videos and VR. The dimensions or order of a matrix gives the number of rows followed by the number of columns in a matrix. Example 2 A permutation lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many letter arrangements can be made from a 2 letter, 3 letter, letter or 10-letter word. The factorial notation (!) is also defined below. Find the number of different permutations of the letters of the word MISSISSIPPI. (10!)/(2!2!3!) = 151200 There are a total of 10 letters. In our experiments we followed MASC and generated 1,000 annotated instances for each of our 1,960 pseudowords. In terms of age range, adolescence is usually considered to be the time between the ages of 13 and 19. Lifts, with a number of passenger and staff lifts remaining either not fully commissioned or unserviceable for customer and staff use. which means the number of combinations of n items taking r items at a time For example means find the number of ways 3 items can be combined, taking 2 at a time, and from the example before, we saw that this was 3. Formula: Example: How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: There are 15 P 4 possible permutations of 4 students from a group of 15. Translate from English to Spanish and Spanish to English with Lexico. 3 billion to £27. Ans:Number of ways of selecting apples = (3+1) = 4 ways. · Abuse of Discretion. If you want to add a new section break, click the "Breaks" button. For example, the permutation of set A= {1,6} is 2, such as {1,6}, {6,1}. Burlington made a decision to stick to the textile business, and in 1985 had sales of about $2. Select "Restart Each Section" from the drop-down menu. By Dr Ian Greenwood, Centre for Employment Relations Innovation and Change. Hence, the number of ways of arranging the letters of the word ALGEBRA that preserves the relative order of the vowels is $$\binom{7}{3} \cdot 4!$$ Note that this is essentially a rephrasing of @Joffan's solution. The Government sought$3. The total number of permutations of the letters of the word BANANA is -. 4 What is uniquely you comes in the fractional difference in how those three billion letters are sequenced in your cells. The number of Jews living in Poland by that date is greatly disputed: At the end of the 15 th century there were between 20,000 and 30,000. Learn more about us. Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. , League Staff Attorney [email protected] For example, if you have just been invited to the Oscars and you have only 2 tickets for friends and family to bring with you, and you have 10 people to choose. gov website by entering docket number USTR–2019– 0003 in the search field on the home page. (FL = π/(4Δ), we measure the coincidence rates for (G) distinguishable and (H. Consumers can be sued by a debt collector,and then fail to show up for the court date and thus have a judgment go against them. a) Find the exponential growth function in terms of t, where t is the number of years since 1992. where x! = x (x - 1) (x - 2). From right to left, find first decreasing number. A permutation is a way to select a part of a collection, or a set of things in which the order matters and it is exactly these cases in which our permutation calculator can help you. It is like a "wildcard". The number of single crochets you make in the ring will be determined by your pattern. Solution: As we have seen earlier, five letters of the word STATE can be scrambled in 5*4*3*2*1 = 5! ways. Responsibilities include: assisting with the recruiting process, assisting with the onboarding. textile company and more than $200-per-share on that$60 stock. Unfortunately, my academic voice is now giving way to a reportorial voice. What is uniquely you comes in the fractional difference in how those three billion letters are sequenced in your cells. Moreover, the copying mechanism of DNA, to meet maximum effectiveness, requires the number of letters in each word to be an even number. Solution: There are 9 letters in the given word in which two T’s, two M’s and two E’s are identical. Example: Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen. For millions of parents, that insecurity can mean working fewer hours, taking a pay cut, or leaving their jobs altogether. these discussions have been tricky and difficult discussions to make but sometimes goes against the grain for palliative care. Taylor Author of Database Development For Dummies, SQL All-in-One For Dummies, and Crysta. Within modern capitalism there is no solution to the problem of oil depletion. Find the number of distinguishable permutations of the letters of the word EDUCATED. This is a key feature of the landscape model: only a finite number of cell types can ever arise. Arrangements, for example, the foundation of a Road Safety Board and the advancement of a New Transport Policy are need of great importance. Credit and loans are a massive, multi-trillion dollar/year international field/industry which the below superb, very memorable generic credit/loan domains quickly, easily, and affordably open the door to [ news reports state that just one site/company--CreditCards. Rich Lowry. This amounts to 2. 4 billion increase in proceeds from issuances of debt, net of. textile company and more than $200-per-share on that$60 stock. Find the number of arrangements of the letters of the word INDEPENDENCE. It is irreversible. In addition to owning 100% of General Reinsurance Corporation, the largest U. Step by step workout step 1 Address the formula, input parameters and values Formula: nPr = n!/(n1! n2!. • Order of 3 letters does not matter =⇒ Combination. And you have the right to charge a fee for late payments. Permutation is the arrangement of the objects, where the order of the objects is considered important. How many letter arrangements can be made from a 2 letter, 3 letter, letter or 10-letter word. Get stock market news and analysis, investing ideas, earnings calls, charts and portfolio analysis tools. Permutations Involving Repeated Symbols - Example 2. Observe that the letter A A A appears twice and all other letters appear once in the word. Learn how to find the number of distinguishable permutations of the letters in a given word avoiding duplicates or multiplicities. PrepInsta has updated the Cognizant Dashboards according to the latest exam pattern 2021. Contractual Arrangement In the early days of Internet advertising, ads were largely sold on a CPM basis with predesigned contracts. Letter Arrangements in a Word Calculator-- Enter Word or. jewish legal status. Combinations – order doesn’t count. Such automatic reformatting of a caption is known as "word wrap. Because it’s made up of so many different technologies, terms, and business practices, finding a place to start learning about the types of programmatic advertising deals can be challenging for those who are new to the space. 1 "January revenues were slightly below benchmark but significantly above the prior year. Answered by Expert Tutors. In your Word document, click the "Layout" tab in the ribbon bar and then click on the "Line Numbers" button. Counting the number of ways objects, some of which may be identical, can be distributed among bins (Section 4. Finding the Number of Permutations of n Distinct Objects Using the Multiplication Principle. gov, fax the documents to 855-261-3452, or mail the documents to P. Moreover, the use of the letters on the phone keypad, while not a regular way of entering text up until that point, was still at least loosely familiar to anyone who. 6 letters can be arranged in 6! ways = 720 ways Vowels can be arranged in themselves in 4! ways = 24 ways Required number of ways = 720*24 = 17280. Sound Trees - The teacher says a word and the students listen and count the number of phonemes in the word, placing the corresponding number of leaves on the tree. If the letters of the word 'CYCLINDER' are arranged alphabetically, then which letter would be farthest from the first letter of word? 1. Major League Baseball has a seven-year media arrangement worth over $5 billion. 2% more than the actual collections in January 2019, and$35 million or 1. So I rewrote my simulation for this new standard, trying to find keyboards with this few total word matches of colliding type-bar pairs. The Postal Service wants to. Infosys Previous Year Placement Papers for 2021 graduates can be found out on this page. 4 billion in taxes in its most recent fiscal year, and said Microsoft's tax rate was in the middle third of companies in the S&P 500. 29 December, 2007 Countercurrents. For example, for a deck of cards n=52. 9 million in 2012. one sexdecillion x one quintillion = one. Users may use any other word by changing the word "ALGEBRA" to find the total number of distinct ways to arrange different words. To ask Unlimited Maths doubts download Doubtnut from - https://goo. Find the number of permutations of the letters of the word STATISTICS. Summing over the data points in Fig. 2 billion or 10. The IMF’s total lending commitments stand at over $285 billion with more than one third approved since late March 2020. Financial Assistance. Search or browse cemeteries and grave records for every-day and famous people from around the world. 5 letters 2 vowel and 3 consonants can be arranged in 5! Ways Therefore required no. A2A, thanks. Fingerprints are the tiny ridges, whorls and valley patterns on the tip of each finger. Accenture Verbal English Questions with Solutions. Get stock market news and analysis, investing ideas, earnings calls, charts and portfolio analysis tools. Search the world's information, including webpages, images, videos and more. Words with double letters are abundant in everyday life: you just have to be aware of them. In how many of these arrangements, Finding total number of arrangements In word INDEPENDENCE There are 3N, 4E, & 2D, 1I, 1P & 1C Since letters are repeating, so we use this formula 𝑛!/𝑝1!𝑝2!𝑝3!. 959 billion, which is$172 million or 6. But I can't decide between 2 TVs. 1 Attachment. Word allows you to do much more than simply insert or place graphics. If all 6 letters were distinct, we would have 6! different rearrangements. This is a simple Sales Contract template directed between between two parties that covers a variety of agreements for the seller and buyer to comply with in order to proceed. This is a small kindness that deserves many thanks. com, the world's most trusted free thesaurus. Open Microsoft Word. In how many ways can Find the number of ways 10 statues can be displayed if each number of statues is. 3 million penalties in 2002 to 37. Vedic clan’s gods, Rishis and priest Brahmans all are wine drinkers and meat. In counting arrangements of objects that contain look-alikes, the normal factorial formula must be modified to find the number of truly different arrangements. First, multiply by 100, then subtract or add the other number in the problem. Now, we are left with the arrangement of the remaining, 10 − 1 = 9 beads. If they were all distinguishable then the number of distinct arrangements would be 10!. Users may use any other word by changing the word "ALGEBRA" to find the total number of distinct ways to arrange different words. The poorest 2 billion lived on average incomes of $400 a year, or a dollar a day. to check how many ways the alphabets of a given word can be. The number of orderings that give the acrostic is thus. The nature of donation - based crowdfunding is such that most of the campaign receipts will qualify as gifts on the part of the beneficiary. The below workout is the step by step procedure to find how many number of distinct ways to arrange the 11 letters (alphabets) of word "MATHEMATICS". This is a key feature of the landscape model: only a finite number of cell types can ever arise. An example to find the number of distinguishable permutations. Some scripts write syllable or word units in a simple direction, but the syllables or words themselves are formed from a complex arrangement of letters or ideograms. Sometimes an inversion is defined as the pair of values. Credit: Getty/Barcroft Media. In the picture below, I have done 6 single crochets (the thing at the very end that looks like a 7th st is the chain I made in step 8). For the same reasons – and sometimes because of official school or national policies – it is difficult for married girls, pregnant girls and young mothers to return to school. By breaking down the name, we can understand the structure of the molecule. Senior Executives are experienced professionals responsible for a company’s operations and performance. "Full-time workers who are overweight or obese and have other chronic health problems miss about 450 million more days of work each year than healthy workers. (E) Nonclassicality for combinations of initial-final photon arrangements captured by the suppression laws associated with Σ 1, Σ 2, and Σ 3. Well, as the Herald demonstrated but a few short weeks ago, that is nonsense: Scotland may indeed have high levels of spending in its more deprived areas, but we pay our own way. The Joe Biden who portrayed himself as a moderate, old-school. The IMF’s total lending commitments stand at over$285 billion with more than one third approved since late March 2020. In addition to owning 100% of General Reinsurance Corporation, the largest U. An international group of academics behind the “Fair Priority Model” (FPM), however, criticizes both as very blunt instruments for the task at hand. Accenture Verbal English Questions with Solutions. A softball coach chooses the first, second, and third batters for a team of 10 players. Display a. The only favorable outcome is the arrangement 25398. But I can't decide between 2 TVs. But, how do I solve this? I have tried finding step by step, many show answer of 3780, but how do I get to that answer? Also, what does it mean when there is an explanation point next to the number?. This Permutation Calculator is valid only for distinct element. The collective power of these leagues to sell media rights is incredible. Everything You Know About the United States and its Laws is WRONG ! The “United States” is NOT the “United States of America” by John-Henry Hill, M. A common textbook question asks students to find the number of permutations for the letters in the word ‘BANANA’. There have been a number of co-operatives start up in the past few years—West End Food Co-op, Solar Share Energy Co-operative, Local 75 Housing Co-op come immediately to mind. In six months leading to August 2018 tax evasion at the port of Mombasa, as reported, amounted to Kshs. The act of placing or arranging.
2021-09-27T06:08:36
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https://math.stackexchange.com/questions/1890405/product-rule-for-the-mod-operator
# Product Rule for the mod operator If we multiply 2 number and take the mod with prime this is equivalent to first taking mod with the prime of individual number and then multiplying the result and again taking mod. $$ab\bmod p = ((a\bmod p)(b\bmod p))\bmod p$$ does there exist any proof for this? Does it work for composite moduli too? Then I can use the Chinese remainder Theorem to calculate the result does there any other way apart from Chinese remainder Theorem to solve the problem? It has nothing to do with prime moduli or CRT. Rather, it is true for all moduli $$\,m\,$$ as we prove as follows, where $$\ \bar x := x\bmod m =$$ remainder left after dividing $$x$$ by $$m$$. Using $$\ x\equiv y\pmod m\color{#c00}\iff \bar x = \bar y,\,$$ and using CPR = Congruence Product Rule \begin{align} {\rm mod}\ m\!:\,\ a &\color{#c00}\equiv \bar a\\ b&\color{#c00}\equiv \bar b\\ \Rightarrow\,\ a\,b&\equiv \bar a\, \bar b\ \ \rm{ by}\ {\rm CPR }\\ \color{#c00}\Rightarrow\, ab\ {\rm mod}\ m &= \bar a \bar b\bmod m,\ \ {\rm i.e.}\ \ \overline{ab} = \overline{\bar a \bar b} \end{align} Remark Generally, as above, to prove an identity about mod as an operator it is usually easiest to first convert it into the more flexible congruence form, prove it using congruences, then at the end, convert back to operator form. • I had a misconception thanks for clarifying – Prashant Bhanarkar Aug 12 '16 at 18:06 The remainder of the product equals the remainder of the product of the remainders....$p$ doesn't need to be prime. $s\equiv q\pmod p$ is the same thing as saying. $s = mp+q$ $t = np+r\\ st = (nkp + nq + kr) p + qr\\ st \equiv qr \pmod p$ Sure, as long as you know $$c\equiv x \bmod p \implies c-x=py\implies c=py+x$$ It follows from FOIL, that you used in linear algebra for multiplying two binomials: $$a=pz+d\land b=pe+f\implies ab=(pz+d)(pe+f)=p(pze+ed+zf)+df$$ which then implies: $$ab\equiv df \bmod p$$
2019-10-20T06:14:03
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https://math.stackexchange.com/questions/1500892/show-that-fx-lim-limits-h-rightarrow-0-dfracfxh-fx-h2h
# Show that $f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}$ I have to prove that if a function $f$ is differentiable on $(a,b)$, then \begin{align*} f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h} \end{align*} Using the fact that $f'(x) = \lim\limits_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$, I wrote my proof in the following manner: \begin{align*} \lim\limits_{h \rightarrow 0}f(x+h) - 2f(x) = \lim\limits_{h \rightarrow 0} - f(x-h) \\ \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)}{2h} - \dfrac{f(x)}{h} = \lim\limits_{h \rightarrow 0}\dfrac{-f(x-h)}{2h} \\ \lim\limits_{h \rightarrow 0}\dfrac{f(x+h)}{h} - \dfrac{f(x)}{h} = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)}{2h} - \dfrac{f(x-h)}{2h} \\ \lim\limits_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h} \end{align*} However, I believe that it is actually incorrect, because when I divide by $2h$ I am potentially making the limit undefined. How would I go about correcting my proof? • Hint: First show $$f'(x) = \lim_{h \to 0} \frac{f(x) - f(x-h)}{h}$$ – Simon S Oct 27 '15 at 22:43 • And when you have finished the exercise ponder over what you have just done. It's not merely an exercise in handling meaningless limits. You have shown that a function that has a derivative at a point also has a symmetric derivative at that point and that the two derivatives have the same value. Think for a while about the mysteries of the "symmetric derivative." It has kept many of us quite busy and highly entertained. – B. S. Thomson Oct 27 '15 at 23:24 \begin{align*} f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \end{align*} but also \begin{align*} f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x)-f(x-h)}{h} \end{align*} sum them up and divide by 2 to get \begin{align*} f'(x) = \lim\limits_{h \rightarrow 0}\frac{ \dfrac{f(x+h)-f(x)}{h} + \dfrac{f(x)-f(x-h)}{h}}2 =\lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h} \end{align*} You cannot just divide by $h$ from nowhere, it is not correct. Consider splitting $$\frac{f(x+h)-f(x-h)}{2h}=\frac{f(x+h)-f(x)+f(x)-f(x-h)}{2h}= \frac12\left(\frac{f(x+h)-f(x)}{h}+\frac{f(x-h)-f(x)}{-h}\right).$$ Both terms go to $f'(x)$ as $h\to 0$. $$f(x+h) = f(x)+f'(x)h + o(|h|)$$ together with: $$f(x-h) = f(x)-f'(x)h + o(|h|)$$ gives: $$f(x+h)-f(x-h) = 2h\cdot f'(x) + o(|h|).$$ Since $f$ is differentiable, we may use l'Hopitals rule: $$\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}=\lim_{h\to 0}\frac{f'(x+h)-(-1)f'(x-h)}{2}=\lim_{h\to 0}\frac{f'(x+h)+f'(x-h)}{2}=\frac{2f'(x)}{2}=f'(x).$$ • The other proofs use only the existence of the derivative at the one point. But it is fair to use the stronger hypothesis since the poster included that (he didn't say $f$ was continuously differentiable though). (If it was a homework assignment, though, he can't use yours since it must surely have had the weaker hypothesis). – B. S. Thomson Oct 27 '15 at 23:47
2020-07-13T02:46:21
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http://math.stackexchange.com/questions/206343/a-tourist-in-france-wants-to-visit-12-different-cities-find-the-probability
# A tourist in France wants to visit 12 different cities, find the probability. The Question : A tourist in France wants to visit 12 different cities. If the route is randomly selected, what is the probability that she will visit the cities in alphabetical order? I thought about this in two ways and I got two different answers, both of them look correct to me ! but only one is correct. 1/144 (12 cities times 12 times to make the arrangement) 1/479001600 (which is generated by the permutation 12 P 12) Which one is correct? and is there a better answer? Please explain how do you know if the problem is permutation or combination or just simple multiplication. - ## 2 Answers The itinerary is randomly chosen, so all itineraries are equally likely. Thus, the probability of visiting in alphabetical order is the number of alphabetically ordered trips divided by the total number of possible trips. The order you visit the cities matters, so the total number of itineraries is $P(12,12) = 12!$. Since only one of these is in alphabetical order, the probability of randomly choosing alphabetical order is $\frac{1}{12!}$. - Your $\frac 1{12!}$ is correct. There is $\frac 1{12}$ chance that the first city is correct. Assuming it is, there is $\frac 1{11}$ that the next one is correct, and so on. Multiplying all these gives the answer. -
2015-01-28T12:28:54
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https://math.stackexchange.com/questions/1580705/phase-portrait-of-system-of-nonlinear-odes
# Phase portrait of system of nonlinear ODEs How can we sketch by hand the phase portrait of a system of nonlinear ODEs like the following? \begin{align} \dot{x} &= 2 - 8x^2-2y^2\\ \dot{y} &= 6xy\end{align} I can easily find the equilibria, which are $$\left\{ (0, \pm 1), \left(\pm \frac{1}{2}, 0\right) \right\}$$ The corresponding stable subspace for $\left(\pm \frac{1}{2}, 0\right)$ is $$\mbox{span} \left\{ \left(\frac{2i}{\sqrt{6}}, 1 \right), \left(-\frac{2i}{\sqrt{6}}, 1 \right) \right\}$$ and the unstable subspace for $(0, \pm 1)$ is $$\mbox{span} \left\{ (0, 1), (1, 0) \right\}$$ respectively. But I can't see how to use these pieces of information to sketch the phase portrait. Any help would really be appreciated! • It would be useful (and a time-saver) if you listed the equilibrium points and the stabilities that you found. Generally, that's where all the work comes in to solving these kinds of problems Dec 18 '15 at 6:00 • There are two separate phase portrait plots for each motion.. $(x,\dot x),(y,\dot y).$ Dec 18 '15 at 6:09 • @Narasimham I suspect that you have slightly different understanding of what phase portrait is... Dec 18 '15 at 6:10 • @Brenton: I added the information you need. Please help with details if you can:) – ghjk Dec 18 '15 at 6:17 • @Evgeny Could be. I took a phase portrait to mean a relation between position and velocity for each of $x,y$ component motions. Dec 18 '15 at 6:37 ## 1 Answer The basic process is to find the critical points, evaluate each critical point by finding eigenvalues/eigenvectors using the Jacobian, determine and plot $x$ and $y$ nullclines, plot some direction fields and use all of this type of information to draw the phase portrait. You can see two different views of this process at this website and notes. For your particular problem $$x' = 2 - 8x^2-2y^2 \\ y' = 6xy$$ We find the critical points where we simultaneously get $x' = 0, y' = 0$ so $$(x, y) = (0, -1), (0, 1), \left(-\dfrac{1}{2}, 0\right), \left(\dfrac{1}{2}, 0\right)$$ The Jacobian is $$J(x, y) = \begin{bmatrix}\dfrac{\partial x'}{\partial x} & \dfrac{\partial x'}{\partial y}\\\dfrac{\partial y'}{\partial x} & \dfrac{\partial y'}{\partial y}\end{bmatrix} = \begin{bmatrix}-16 x & -4y\\6y & 6x\end{bmatrix}$$ Evaluate eigenvalue/eigenvector for each critical point $J(0, -1) \implies \lambda_{1,2} = \pm 2 i \sqrt{6}, v_{1,2} = \left(\mp i \sqrt{\frac{2}{3}}, 1\right) \implies$ spiral $J(0, 1) \implies \lambda_{1,2} = \pm 2 i \sqrt{6}, v_{1,2} = \left(\pm i \sqrt{\frac{2}{3}}, 1\right) \implies$ spiral $J(-\frac{1}{2}, 0) \implies \lambda_{1,2} = (8, -3), v_{1} = (1,0), v_2 = (0, 1) \implies$ saddle $J(\frac{1}{2}, 0) \implies \lambda_{1,2} = (-8, 3), v_{1} = (1,0), v_2 = (0, 1) \implies$ saddle Using all the above (critical points, eigenvalues/eigenvectors, x-nullcline (red and black curves), y-nullcline (green curve), direction fields, etc.), you can now sketch the phase portrait. Exercise - make sure to add direction fields from the two sets of notes linked above so you understand how to do that. The phase portrait will look like: • Awesome!! Thanks so much for your help! Might you please help me with this problem? math.stackexchange.com/questions/1580742/… – ghjk Dec 18 '15 at 9:23 • Btw, do you know where i can find how to sketch solutions to a linear system $\dot{x}=Ax$, A=3*3 matrix? – ghjk Dec 18 '15 at 9:50 • @user177196: Cannot help with other problem. For a description of the 3D process no 9as it is the same as 2D basically), but for tools, you can look at users.dimi.uniud.it/~gianluca.gorni/Mma/Mma.html for MMA and stacking plots like mathematica.stackexchange.com/questions/687/…. Maple also has a similar command mapleprimes.com/questions/35774-Phase-Portrait-In-3D – Moo Dec 18 '15 at 12:51 • What do you mean by no9? – ghjk Dec 18 '15 at 14:31 • Sorry, just woke up :-), that was I do not know of any notes, but it is basically the same as 2D. These 3D variants are more done through analysis because it is typically hard to get anything 3D that is meaningful. If you have a 3D example in mind, post it as a new question and I'd give 3D a shot. – Moo Dec 18 '15 at 14:40
2021-09-28T11:40:34
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http://math.stackexchange.com/questions/677752/how-can-this-trig-equation-be-simplified
# How can this trig equation be simplified? We have $9+40\sin^2x=-42\sin x\cos x$. I know this simplifies to $7\sin x+3\cos x=0$, but how? - Try using $\sin2x=2\sin x\cos x$. – Ian Coley Feb 15 '14 at 20:40 @IanColey Nice approach, but there is the user who provided the hint already. – NasuSama Feb 15 '14 at 20:57 Along with Trafalgar Law's hint, we can use the fact that $\sin^2(x) + \cos^2(x) = 1$. Multiply both sides by $9$ to get $9 = 9\sin^2(x) + 9\cos^2(x)$. We then have \begin{aligned} 9\sin^2(x) + 9\cos^2(x) + 40\sin^2(x) &= -42\sin(x)\cos(x)\\ 49\sin^2(x) + 42\sin(x)\cos(x) + 9\cos^2(x) &= 0\\ (7\sin(x) + 3\cos(x))^2 &= 0 \end{aligned} which gives the desirable equation you want to have. - Thanks for the catch. – NasuSama Feb 15 '14 at 20:49 Hehe. Changed back to before. – NasuSama Feb 15 '14 at 20:56 Hint : write $9 = 9\sin^2(x)+9\cos^2(x)$ - HINT: When you have only terms like $\sin^2x,\cos^2x,\sin x\cos x$ divide either sides by $\cos^2x,$ $$9\sec^2x+40\tan^2x=-42\tan x\iff9(1+\tan^2x)+40\tan^2x=-42\tan x$$ $$\iff49\tan^2x-42\tan x+9=0$$ which is a Quadratic Equation in $\tan x$ -
2016-06-25T15:59:39
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https://www.physicsforums.com/threads/permutation-and-combination-problem.601007/
# Homework Help: Permutation and Combination Problem 1. Apr 27, 2012 ### Mathkid182 1. The problem statement, all variables and given/known data How many way can a team lose 3 of their next 5 games? How many ways can a team lose 2 of their next 5 games? Why are the two answers the same? 2. Relevant equations Permutation = nPr = n! / (n-r)! Combination = nCr = nPr / r! where, n, r are non negative integers and r<=n. r is the size of each permutation. n is the size of the set from which elements are permuted. ! is the factorial operator. 3. The attempt at a solution I think this is a problem where the permutation equation would be used. After all, the team can LLLWW or WWLLL and those would be two different permutations that would satisfy the question's requirements. But I just don't understand how I would go about solving it. Any help is much appreciated. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 2. Apr 28, 2012 ### Vorde When I was studying for the SAT's and dealt with these equations the trick was to remember "n choose r", meaning that in the equations n is the number you are choosing from and r is the amount you are choosing. In the first example you are choosing 3 out of 5 so n=5 r=3 Get it? 3. Apr 28, 2012 ### Mathkid182 But then if we did the permutation formula for both questions then one wouldn't receive the same answer for both. 4. Apr 28, 2012 ### HallsofIvy If you write "L" for lose and "W" for win you can see that the problem, is the same as 'In how many ways can you write 2 "W"s and 3 "L"s?' If you imagine that all of the "L" are labeled, say "1", "2", "3", and all of the "W" are labeled "1", "2", then you have 5 distinguishable objects. There are 5! ways of arranging them. But because the two "W"s are not distinguishable, such an arrangement as $W_1L_1L_2L_3W_2$ is exactly the same as $W_2L_1L_2L_3W_1$. That is, because there are 2!= 2 ways to permute just the "W"s, we have to divide by 2! to get the number or arrangements ignoring rearrangements of just the "W"s. Similarly, there are 3!=6 ways to rearrange just the different "L"s. We must also divide by 3! since those are not "different" arrangements. Now do the same with 2 "L"s and 3 "W"s. 5. Apr 28, 2012 ### Villyer If you were looking for all three-length permutations of {1,2,3,4,5}, you would get 123, 132, 213, 231, 312, 321 as six different possibilities. Are all six of these different ways of winning three games?
2018-09-24T01:40:43
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https://www.physicsforums.com/threads/linear-algebra-scalars-to-find-det-a.740338/
# Linear Algebra, scalars to find det(A) 1. Feb 25, 2014 ### concon 1. The problem statement, all variables and given/known data Let X1, X2,.....,Xn be scalars. Calculate det(A) where A= nxn matrix with [ x1+1 x1+2......x1+n x2+1 x2+2......x2+n .... .... ..... xn+1 xn+2......xn+n] 2. Relevant equations det(A) = (aij)(-1)^(i+j)det(aij) 3. The attempt at a solution I have no clue how to even begin solving this problem I tried using above formula to no avail. 2. Feb 25, 2014 ### concon Spacing on matrix got messed up so imagine it with the rows lined up correctly 3. Feb 25, 2014 ### AlephZero Review what you know about how determinants, row (or column) operations on the matrix, and linear dependence. Hint: the answer is very simple, when n > 2. 4. Feb 25, 2014 ### concon Can you explain a little more about what operations you are talking about? I am really confused! 5. Feb 25, 2014 ### kduna Have you tried computing the determinant for small n? Such as a 2x2, 3x3, 4x4. I'd recommend doing that. 6. Feb 25, 2014 ### concon How would that help me calculate the determinant when n is unknown? 7. Feb 25, 2014 ### Ray Vickson Start with simple cases of, say, n = 2, 3 and maybe 4, just to see what is going on. Then do it for general n. 8. Feb 25, 2014 ### AlephZero The answer doesn't depend on n, except that n = 1 and n = 2 are special cases. I don't know how to give you any more help than "think about row and column operations on the matrix, and linear dependence" without telling you the answer. Writing out the matrix in full, for n = 3 or n = 4, might help. If the only thing you know about determinants is your "relevant equation" det(A) = (aij)(-1)^(i+j)det(aij) I think you missed a lot of stuff in class, or you haven't read your textbook. 9. Feb 26, 2014 ### concon Yes I did computr for 2 by 2 and 3 by 3, but I do not see a relationship between the determinants. How do I calculate det(A) for nxn? 10. Feb 26, 2014 ### concon Okay I did determinant calculation for 2x2 and 3x3. How do I calculate for nxn? Please help! 11. Feb 26, 2014 ### pasmith Two hints: (1) Why is $\det A = B(x_2, \dots, x_n)x_1 + C(x_2, \dots, x_n)$ for some functions $B$ and $C$? (2) What is the determinant of a matrix with two identical rows? 12. Feb 26, 2014 ### concon 1. I have no clue 2. determinant would be zero. Is the answer zero? 13. Feb 26, 2014 ### Ray Vickson So, what did you get for n = 2 and for n = 3? 14. Feb 27, 2014 ### concon For n=2 det(A)= (x1 +1)(x2 +2) - (x1 +2)(x2 +1) How does that correlate to unknown n? 15. Feb 27, 2014 ### Ray Vickson What do you get for n = 3? I mean, expand out everything and simplify it down to as small an expression as you can get. I am 100% serious. Looking at just n = 2 is not enough to reveal the pattern. 16. Feb 27, 2014 ### AlephZero Can you think of a way to do row or column operations on the matrix, to make two rows identical, and not change the determinant? 17. Feb 27, 2014 ### concon Hey I actually just figured out how to solve it by using row operations and linear dependence and I got zero as the determinant. Is this correct?
2018-02-21T23:55:50
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http://math.stackexchange.com/questions/320985/how-to-determine-equation-for-sum-k-1n-k3
# How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated. - Do you want a method for deriving one, or are you content to have one? $$\sum_{k=1}^nk^3=\frac14n^2(n+1)^2\;.$$ – Brian M. Scott Mar 5 '13 at 2:18 I'm looking for a method to derive it. – AlexHeuman Mar 5 '13 at 2:19 There is always induction. Once you know the formula. And here are some more: math.com/tables/expansion/power.htm – 1015 Mar 5 '13 at 2:19 I'm not sure what you mean by induction other than a proof. I'm looking for a way to obtain the formula without knowing it, not to prove it. – AlexHeuman Mar 5 '13 at 2:21 math.stackexchange.com/q/61482/9464 – Jack Mar 5 '13 at 3:01 You can derive it from the formulas for the sums of lower powers: \begin{align*} \sum_{k=1}^nk^3&=\sum_{k=1}^n\left(k\cdot k^2\right)\\ &=\sum_{k=1}^nk^2\sum_{i=1}^k1\\ &=\sum_{k=1}^n\sum_{i=1}^kk^2\\ &=\sum_{i=1}^n\sum_{k=i}^nk^2\\ &=\sum_{i=1}^n\left(\sum_{k=1}^nk^2-\sum_{k=1}^{i-1}k^2\right)\\ &=\sum_{i=1}^n\left(\frac16n(n+1)(2n+1)-\frac16i(i-1)(2i-3)\right)\\ &=\frac16n^2(n+1)(2n+1)-\frac16\sum_{i=1}^ni(i-1)(2i-3)\\ &=\frac16n^2(n+1)(2n+1)-\frac16\sum_{i=1}^n\left(2i^3-5i^2+3i\right)\\ &=\frac16n^2(n+1)(2n+1)+\frac56\sum_{i=1}^ni^2-\frac12\sum_{i=1}^ni-\frac13\sum_{i=1}^ni^3\\ &=\frac16n^2(n+1)(2n+1)+\frac5{36}n(n+1)(2n+1)-\frac14n(n+1)-\frac13\sum_{k=1}^nk^3\;, \end{align*} and from here you can clearly solve for $\sum_{k=1}^nk^3$. Evidently this technique can be applied repeatedly, though it rapidly becomes very tedious. Graham, Knuth, & Patashnik, Concrete Mathematics, offer many other approaches, including via finite calculus and generating functions. Added (because I like it): The finite calculus approach requires first rewriting $k^3$ in terms of the falling factorials $k^{\underline1}=k$, $k^{\underline2}=k(k-1)=k^2-k$, and $k^{\underline3}=k(k-1)(k-2)=k^3-3k^2+2k$: $$k^3=k^{\underline3}+3k^2+2k=k^{\underline3}+3k^{\underline2}+k^{\underline1}\;.$$ The coefficients are Stirling numbers of the second kind: in general $$x^n=\sum_{k=0}^n{n\brace k}x^{\underline k}\;.$$ Then \begin{align*} \sum_{k=1}^nk^3&=\sum_{k=1}^n\left(k^{\underline3}+3k^{\underline2}+k^{\underline1}\right)\\ &=\left[\frac14k^{\underline4}+k^{\underline3}+\frac12k^{\underline2}\right]_1^{n+1}\\ &=\frac14n(n+1)(n-1)(n-2)+n(n+1)(n-1)+\frac12n(n+1)-0\\ &=\frac14n(n+1)\Big(n^2-3n+2+4(n-1)+2\Big)\\ &=\frac14n(n+1)\left(n^2+n\right)\\ &=\frac14n^2(n+1)^2\;. \end{align*} There is a nice introduction to finite calculus in Section $2.6$ of Graham, Knuth, & Patashnik, Concrete Mathematics; Pete L. Clark has a handout with another introduction here. - Nice work, Brian, as usual! This one was fun! When I saw posts accumulating, I figured the OP would have more than enough to wade through than were I to add anything different (which would likely not have added much, anyway!) – amWhy Mar 5 '13 at 3:12 Here is another approach, $$\sum_{k=1}^{n} (k+1)^4 - \sum_{k=1}^{n} k^4 = (n+1)^4-1$$ $$\implies 4\sum_{k=1}^{n}k^3 + 6\sum_{k=1}^{n} k^2 + 4\sum_{k=1}^{n}k + \sum_{k=1}^{n}1 = (n+1)^4 -1$$ $$\implies 4\sum_{k=1}^{n}k^3 = (n+1)^4-1-6\sum_{k=1}^{n}k^2 - 4\sum_{k=1}^{n}k - \sum_{k=1}^{n}1$$ $$\implies \sum_{k=1}^{n}k^3 = \dots.$$ - Also, a great answer. – AlexHeuman Mar 5 '13 at 2:49 @AlexHeuman: You are welcome. – Mhenni Benghorbal Mar 5 '13 at 2:50 The general approach is to write $p(k)$ in terms of the polynomals $\binom{k}{i}$ with $i=0,1,2,\dots$ For example, $$k^3 = 6\binom k 3 + 6\binom k 2 + \binom k 1$$ Now, since $$\sum_{k=1}^{n} \binom{k}{i} = \binom{n+1}{i+1}$$ you get: $$\sum_{k=1}^{n} k^3 =6\binom{n+1}{4} + 6\binom{n+1}{3} + \binom{n+1}{2}$$ (There is a slight problem above when $i=0$. You really need sums from $k=0$ to $n$ for that case. In all other cases, $k=0$ doesn't add anything.) - Interesting approach. Had never seen it. – Pedro Tamaroff Mar 5 '13 at 2:31 Basically, for fixed $i$, the polynomial $\binom {k}{i}$ is a polynomial of degree $i$, and therefore they form a basis for all polynomials. – Thomas Andrews Mar 5 '13 at 2:33 @ThomasAndrews I havne't yet encountered that notation. What is it? – AlexHeuman Mar 5 '13 at 2:35 @AlexHeuman If you mean $\binom{k}{i}$, then it is defined as $$\binom{k}{i} = \frac{k(k-1)\cdots (k-(i-1))}{i\cdot (i-1)\cdots 2\cdot 1}$$ But if you are unfamiliar with this notation, this answer will probably not be much help. It is often called the "choose" function because it is the number of subsets of size $i$ in a set with $k$ elements. It is also sometimes called the binomial function – Thomas Andrews Mar 5 '13 at 2:39 @PeterTamaroff Actually, I sort of misremembered. It is even easier to use simple "falling factorials:" $$(k)_i = k(k-1)\cdots(k-(i-1)) = i!\binom{k}i$$ Then $$\sum_{k=0}^n (k)_i = \frac{1}{i+1}(n+1)_{i+1}$$ Then we have $$k^3 = (k)_3 + 3(k)_2 + (k)_1$$ yields the same result. – Thomas Andrews Mar 5 '13 at 2:55 In addition to Brian M. Scott's answer, and the reference to $\mathit{Concrete \ Mathematics}$, you can actually derive this identity using $\mathit{higher}$ powers and perturbation method Keep in mind that $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\\ \sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}$$ So you start with $$S_n=\sum_{k=1}^{n}k^4\\ S_{n+1}=S_n +(n+1)^4 = \sum_{k=1}^{n}(k+1)^4+1 =S_n+4 \sum_{k=1}^{n}k^3+6 \sum_{k=1}^{n}k^2 +4 \sum_{k=1}^{n}k +n+1$$ hence $$S_n +(n+1)^4=S_n+4 \sum_{k=1}^{n}k^3+6 \sum_{k=1}^{n}k^2 +4 \sum_{k=1}^{n}k +n+1$$ So clearly the highest term cancels out and after some algebra you will get your $$\sum_{k=1}^{n}k^3=\frac{(n(n+1))^2}{4}$$ - Thanks, this is also a great answer for me. – AlexHeuman Mar 5 '13 at 2:49 You are welcome – Alex Mar 5 '13 at 2:52 You can use *asterisks* to italicise text outside of math mode. – Rahul Mar 5 '13 at 6:30 $$\sum_{k=1}^nk^3=\bigg(\sum_{k=1}^n k\bigg)^2.$$ Can you get the intuition from the following two pictures? The images are from Brian R Sears. - The sum is $\dfrac{n^2(n+1)^2}{4}$. The formula can be proved correct by induction. But there is also a nice geometric proof that can be found, for example, in the book Proofs Without Words. For a general discussion of related problems, please see this Wikipedia aticle on Faulhaber's Formula. Remark: There are various ways to discover the formula. We can calculate the sum of the first $n$ cubes for various small $n$. The actual formula is so simple that the answer leaps out. Then we can verify that itis correct by checking that $$\frac{(k^2)(k+1)^2}{4}-\frac{(k-1)^2(k^2)}{4}=k^3.$$ In this case, the calculation is instantaneous, for we are looking at a difference of squares. Or else suppose we already know formulas for $\sum_1^n k$ and $\sum_1^n k^2$. Observe that $$(k+1)^4-k^4=4k^3+6k^2+4k+1.$$ Sum from $k=1$ to $k=n$. On the right we get a lot of cancellation, and we get $$(n+1)^4-1^4=4\sum_1^n k^3 +6\sum_1^n k^2+4\sum_1^n k +\sum_1^n 1.$$ We know everything in the equation above except $\sum_1^n k^3$. So now we know that too. - I'm not looking for a proof. I'm looking for the method to discover the algebraic equation. I'm not quite sure I understand how Faulhaber derived his formula. For instance, for $\sum_{k=1}^n k^2$ You can show that $(k+1)^3 - k^3 = 3k^2 + 3k + 1$ so $3[1^2 + 2^2 + ... + (n-1)^2] + 3[1 + 2 + ... + (n-1)] + n - 1 = n^3 - 1$ So by ruther manipulation we get $\sum_{k=1}^n k^2 = \frac {n^3} 3 + \frac {n^2} 2 + \frac n6$ – AlexHeuman Mar 5 '13 at 2:31 For the sum of the cubes, we can conjecture that the answer has shape $an^4+bn^3+cn^2+dn+e$. The formula should work for $n=0,1,2,3,4$. That gives $5$ equations in $5$ unknowns, really $4$ because we have immediately $e=0$. After we have found $a$ to $e$ that work for $0,1,2,3,4$, we can verify by induction that they work for all $n$. – André Nicolas Mar 5 '13 at 2:35 Or else we can look at $(k+1)^4-k^4=4k^3+6k^2+6k+1$. Sum from $k=1$ to $n$. On the right, almost everything cancels. On the right, we get $4\sum k^3$ plus already known things. – André Nicolas Mar 5 '13 at 2:39 thanks, I was trying to figure out a way from the same $(k+1)^3$ now that I knew the sum for $k^2$, but I couldn't so I should have put it to $(k+1)^4$ like you said. – AlexHeuman Mar 5 '13 at 2:42 One classic method is \begin{align} \sum_{k=1}^{n} k^4 &= \left( \sum_{k=1}^n (k+1)^4 \right) + 1 - (n+1)^4 \\&= \left( \sum_{k=1}^n k^4 + 4 k^3 + 6 k^2 + 4 k + 1 \right) + 1 - (n+1)^4 \\&= \left( \sum_{k=1}^n k^4 \right) + 4 \left( \sum_{k=1}^n k^3 \right) + 6 \left( \sum_{k=1}^n k^2 \right) + 4 \left( \sum_{k=1}^n k^1 \right) + \left( \sum_{k=1}^n 1 \right) + 1 - (n+1)^4 \end{align} The sum of fourth powers cancel out, and you can solve the equation for the sum of curves. Another approach is to guess that the sum of cubes should be a fourth degree polynomial, then solve for the coefficients of the polynomial using the equations • $f(0) = 0$ • $f(1) = 1$ • $f(2) = 1 + 2^3$ • $f(3) = 1 + 2^3 + 3^3$ • $f(4) = 1 + 2^3 + 3^3 + 4^3$ Five equations in five unknowns. You can cut this down to 2 unknowns if you recognize some easy patterns: the leading coefficient for the sum of $n$-th powers is always $1/n$ and the next is always $1/2$, and the constant term is always 0. - +1 Recognizing that the sum should be a polynomial of degree $4$ since $\int x^3 dx = \frac{x^4}{4} + C$ seems the most straight-forward method for me here. It does involve finding a solution to a system of five equations in five unknowns, though as you note, the system is easily reduced to a manageable size. – JavaMan Mar 5 '13 at 3:50 As you probably have already realized the formula $$\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$$ can be proved using induction. For $n=1$ you have $$1^3 = \left(\frac{1\cdot 2}{2}\right)^2.$$ Say that the formula holds for $n$ and prove that it holds for $n=1$. So you have/get $$\sum_{k=1}^{n+1} k^3 = \left[\sum_{k=1}^n k^3\right] + (n+1)^3 = \frac{n^2(n+1)^2}{2^2} + (n+1)^3.$$ I will leave you to prove that this right hand side is indeed equal to $$\frac{n^2(n+1)^2}{2^2} + (n+1)^3 = \dots = \frac{(n+1)^2(n+2)^2}{2^2}.$$ -
2016-05-05T10:37:00
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https://math.stackexchange.com/questions/2526657/every-open-set-has-a-proper-open-subset-what-spaces-satisfy-this-property
# Every open set has a proper open subset. What spaces satisfy this property? Suppose $X$ is a topological space satisfying the following property: every (nonempty) open set of $X$ has a (nonempty) proper open subset. Does this property have a name? What are some spaces with this property? What are some properties that imply this property (eg. Hausdorff, separable, second countable, etc.)? The intuition is that, if you have an open set $U \subseteq X$, you can "zoom in" at any point of $U$, forever. Example. If $X$ has the discrete topology, then it doesn't have this property, because singletons $\{x\} \subseteq X$ are open sets that don't have (nonempty) proper open subsets. Example. If $X$ is ${\mathbb R}^n$ with the Euclidean topology, then it does have this property, because, for any open ball $B$ of radius $\epsilon$ around $x \in X$, you can take the open ball of radius $\epsilon/2$, which is a (nonempty) proper open subset of $B$. • Any space which has open sets who contain only finitely many elements can't have this property, as an extension of the argument for the discrete topology. – Duncan Ramage Nov 18 '17 at 21:53 • Also, any infinite space with a finite topology can't have this property. – étale-cohomology Nov 19 '17 at 6:44 • As stated in the title, the Answer is "None" because the empty set is always open and has no proper subset. Clearly you meant "every non-empty open set". – DanielWainfleet Nov 19 '17 at 12:54 • @DanielWainfleet The title is vague. The main text is more precise. – étale-cohomology Nov 19 '17 at 12:59 • A $T_1$ space with no isolated points has no non-empty subset-minimal open sets. – DanielWainfleet Nov 19 '17 at 13:00 It's not implied by a lot of your properties as a countable discrete space lacks it, but such a space is second countable Hausdorff, separable etc. It implies some other properties: $X$ has infinitely many open sets, all of which are infinite. (if $X$ would have finitely many open sets, their (open!) intersection would be minimal; If $U$ is open non-empty, define $U_0 = U$ and $\emptyset \neq U_{n+1} \subsetneq U_n$ whenever $X$ has this property, and then $\cup_n (U_{n}\setminus U_{n+1})$ is an infinite subset of $U$, so $U$ is infinite. Spaces that are dense in themselves (i.e. have no isolated points) and are also $T_0$ will have this property: if $U$ is open then $U$ is not a singleton, so we have $x,y \in X$ with $x \neq y$, By $T_0$-ness we find an open set $V$ containing $x$ but not $y$ (or reversely), and then $U \cap V$ is strictly smaller and open too. Equivalently, if $X$ fails the condition, there is some minimal open subset $U$. If $U$ has one point this point is isolated. If it has more, the $T_0$ condition fails for points from this set. So your condition is implied by "$X$ is dense in itself and $T_0$". As @bof remarked in the comments: if $X$ has the "no minimal open set property", and $Y$ is any space whatsoever, then $X \times Y$ also obeys it ($O$ non-empty open in $X \times Y$ contains a basic non-empty open set $U \times V\subseteq O$ and $U$ is not minimal so $\emptyset \neq U' \subsetneq U$ open exists and then $U' \times V$ is smaller and non-empty open inside $O$ so $O$ is not minimal.) • Could be clearer. Of course "no point has a minimal open set containing it" is only a sufficient condition for "every nonempty open set has a nonempty open proper subset", but the casual reader might think you're saying it's a necessary and sufficient condition. – bof Nov 19 '17 at 0:00 • By the way, $T_0$ is just as good as $T_1$ here. – bof Nov 19 '17 at 0:02 • @bof sure, it's equivalent with $T_0$ added. I expanded. – Henno Brandsma Nov 19 '17 at 7:02 • The OP's condition is not equivalent to "$X$ is dense in itself and $T_0$" because it doesn't imply $T_0$. If $X$ has the OP's property, then $X\times Y$ also has the property, but $Y$ could have the "indiscrete" topology. – bof Nov 19 '17 at 9:16
2019-10-18T16:47:48
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https://math.stackexchange.com/questions/3209830/solving-pde-yu-x-xu-y-x2-y2-0-with-method-of-characteristics
# Solving PDE $yu_x - xu_y + x^2 - y^2 = 0$ with method of characteristics I am struggling to understand one 'trick' that is used in the solution of the ODE's from method of characteristics, which currently doesn't make any sense to me: 13. Solve the 1st order PDE $$yu_x - xu_y + x^2 - y^2 = 0$$ Solution: The characteristic equations are $$\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{y^2-x^2}$$ Comparing the 1st and 2nd ratios, we have $$\frac{dx}{y} = \frac{dy}{-x} \quad\implies\quad x^2+y^2 = c_1$$ Next, we multiply the 1st ratio by $$y$$ and the 2nd ratio by $$x$$ and then add them to obtain $$\frac{y dx + x dy}{y^2-x^2} = \frac{du}{y^2-x^2} \quad\implies\quad u-xy = c_2$$ I understand the logic of multiplying 1st and 2nd ratios by $$y$$ and $$x$$ respectively, but how is it possible to add them together in such manner and, even more, equate the result with the third one. Any help or explanation of at least some part of it would be highly appreciated! • Thanks for the editing suggestions, I'm new here, so didn't figure out how site works completely, sorry for that – apavl0v May 1 at 16:33 • To understand this tricks, you must have to read some good books like "Elements of Partial Differential Equations" by I. N. Sneddon. – nmasanta May 1 at 16:49 ## 1 Answer The "trick" is a well-known property in fraction calculus. If two fractions are equal : $$\frac{A}{B}=\frac{C}{D}\:,$$ they are equal to any fractions on the form : $$\frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}\:,$$ with any coefficients (constants or not) $$c_1$$ and $$c_2$$ not both nul. In your question : $$A=dx\quad;\quad B=y\quad;\quad C=dy\quad;\quad D=-x\quad;\quad c_1=y\quad;\quad c_2=x.$$ NOTE : The property is easy to prove. $$\frac{c_1A+c_2C}{c_1B+c_2D}=\frac{A}{B}\left(\frac{c_1+c_2\frac{C}{A}}{c_1+c_2\frac{D}{B}}\right)=\frac{A}{B}\left(\frac{c_1+c_2\frac{D}{B}}{c_1+c_2\frac{D}{B}}\right)=\frac{A}{B}$$ because $$\frac{C}{A}=\frac{D}{B}$$ . • Thank you so much! – apavl0v May 1 at 17:27 • Don't mention it. – JJacquelin May 1 at 17:39
2019-06-24T10:47:10
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http://a4academics.com/careers-guidance-jobs/68-quantitative-aptitude/914-number-system-complete-part3?showall=&start=2
# Number System Tutorial Part III: Factors, Multiples, Unit Digit and Last Two Digits of Exponents ## Factorial (!) Factorial is a well defined operator, which is defined as the "the product of a certain number of natural numbers starting from 1" The operator factorial is denoted by the symbol '!' (Exclamatory sign). Examples: 4! = 1 xx 2 xx 3 xx 4 = 24 5! = 1 xx 2 xx 3 xx 4 xx 5 = 120 n! = 1 xx 2 xx 3 xx "...." xx (n - 2) xx (n - 1) xx n Pre defined values 0! = 1 1! = 1 n! = n xx (n - 1)! = n xx (n - 1) xx (n - 2)! , and so on. ### Concept I: Largest power of a prime in N! Let's look at these kind of problems with help of couple of examples. Find the largest power of 2 in 5!? 5! = 1 xx 2 xx 3 xx 4 xx 5 = = 2^3 xx 3 xx 5 :.the largest power of 2 in 5! is 3. If the question ask about a largest factorial value, then the above method is not easy to apply. So we have to think about an alternate method. Long division method: Find the largest power of 3 in 100!? Do a long division of the number using the factor specified. Largest power of 3 in 100! = sum of all quotient in all steps -> 33 + 11 + 3 + 1 = 48. If (76!)/(5^n) is an integer, find the maximum possible value for n.? If (76!)/(5^n) is an integer, then n is the largest exponent of 5 in 76! The largest possible value for n = 15 + 3 = 18 ### Concept II: Largest power of a composite number in N! Find the largest power of 6 in 15!.? 6 = 3 xx 2; Largest prime factor of 6 is 3. For finding the largest power of 6 in 15!, it is enough to find the largest power of 3 in 15! Largest power of n = 5 + 1 = 6 So the largest power of 6 in 15! is 6 ### Concept 3: Number of zeros at the end of N! In the product of first 'n' natural numbers, a 'zero' will get generated as per the combination of 2 and 5. Therefore finding the number of zeros at the end of N!, find the number of 5's in N!. How many zeros at the end of 25! ? i.e. There are 5 + 1 = 6 zeros at the end of 25! ## How to find unit digit of an exponential expression The unit digit of the powers of natural numbers follows some interesting patterns. An understanding of the occurrence of unit digit will help you to score some points in your exams. This awareness is not only for using the typical questions from unit digit but also can use as the verification tool for basic arithmetic operations. Unit digit of the different exponents of all numerals follows a certain cyclic order. It is possible to tabulate the cyclic pattern of unit digits in the following manner. "Base"^nUnit DigitGeneral Form of Exponent 0^n 0 Any natural number 1^n 1 Any natural number 2^1 2 4k + 1 2^2 4 4k + 2 2^3 8 4k + 3 2^4 6 4k + 0 3^1 3 4k + 1 3^2 9 4k + 2 3^3 7 4k + 3 3^4 1 4k + 0 4^1 4 Odd 4^2 6 Even 5^n 5 Any natural number 6^n 6 Any natural number 7^1 7 4k + 1 7^2 9 4k + 2 7^3 3 4k + 3 7^4 1 4k + 0 8^1 8 4k + 1 8^2 4 4k + 2 8^3 2 4k + 3 8^4 6 4k + 0 9^1 9 Odd 9^2 1 Even Let's look at how to use this table with the help of few examples. Find the unit digit of 1020^34 ? 1020 ends in 0. From the table, there is only one unit digit for the pattern of 0. i.e. Any exponent of a number which ends in 0 has the unit digit '0'. Therefore the unit digit of 1020^34 is 0. Find the unit digit of 1542^345 ? Exponent = 345 "Rem"[345/4] = 1, i.e 345 is of the form 4k + 1 From the pattern of 2 in the table, if the exponent in the form 4k + 1 then unit digit is 2. Find the unit digit of 3689^3475 ? From the table, 9^"odd" will end in 9 itself. Therefore the unit digit of 3689^3475 is 9. Page 3 of 4 #### Popular Videos How to improve your Interview, Salary Negotiation, Communication & Presentation Skills. Got a tip or Question? Let us know
2021-01-25T11:03:31
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https://math.stackexchange.com/questions/2025823/show-int-aa-fx-f-x-dx-0-for-any-continuous-function-fx
# Show $\int_{-a}^{a} [f(x)-f(-x)] dx=0$ for any continuous function f(x) Show that for any continuous function $f(x)$ on $[-a, a]$, where $a>0$, $$\int_{-a}^{a} [f(x)-f(-x)] dx=0$$ I've tried to do something like: Let the primitive function of $f(x)$ be $F(x)$. $\int_{-a}^{a} [f(x)-f(-x)] dx$ $=[F(x)]_{-a}^{a}-[F(-x)]_{-a}^{a}$ $=F(-a)-F(a)-F(a)+F(-a)$ $=2F(-a)-2F(a)$ but I don't know how to continue. • define the function $g(x)=f(x)-f(-x)$ now $g(-x)=-g(x)$ which shows that $g(x)$ is antisymmetric function. Integrating such a function over a symmetric interval naturally yields zero ("areas below and above the $x$-axis cancel"). – tired Nov 22 '16 at 14:29 • @tired Ah I understand! Thanks! :) – He Yifei 何一非 Nov 22 '16 at 14:34 • Continuity isn't required for this, just integrability – MPW Nov 22 '16 at 14:40 Hint: $$I=\int_a^bf(x) \ dx=\int_a^bf(a+b-x)\ dx$$ $$I+I=\int_a^b\{f(x) +f(a+b-x)\}\ dx$$ Hint If $g$ is an odd function then $$\int_{-b}^b g(t)dt=0$$ If $g$ is continuous, if $\phi$ and $\psi$ are differentiable, and if $$G(x) = \int_{\psi(x)}^{\phi(x)} g(t)\, dt,$$ the fundamental theorem of calculus and the chain rule give $$G'(x) = g\bigl(\phi(x)\bigr)\, \phi'(x) - g\bigl(\psi(x)\bigr)\, \psi'(x). \tag{1}$$ Define $$F(x) = \int_{-x}^{x} \bigl[f(t) - f(-t)\bigr]\, dt.$$ The integrand $g(x) = f(x) - f(-x)$ is continuous (since $f$ is continuous), so (1) becomes $$F'(x) = \bigl[f(x) - f(-x)\bigr] - \bigl[f(-x) - f(x)\bigr](-1) = 0$$ for all $x$ in $[-a, a]$. By the identity theorem, $F(x) = F(0) = 0$ for all $x$ in $[-a, a]$. we have: $$\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx-\int_{-a}^af(-x)dx$$ and, substituting $-x=t \rightarrow dx=-dt$ in the second integral: $$\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx+\int_{a}^{-a}f(t)dt$$ that is (since the variable in the integral is ''free''): $$\int_{-a}^a [f(x)-f(-x)]dx=\int_{-a}^af(x)dx+\int_{a}^{-a}f(x)dx =0$$ Hint: Substitute $x=-t$, \begin{align} \underbrace{\int_{-a}^a\left[f(x)-f(-x)\right]\,\mathrm{d}x}_{I} &=\int_a^{-a}\left[f(-t)-f(t)\right]\,\mathrm{d}(-t)\\ &=\underbrace{\int_{-a}^a\left[f(-t)-f(t)\right]\,\mathrm{d}t}_{-I} \end{align}
2019-09-21T11:11:40
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https://math.stackexchange.com/questions/2401656/calculating-34429-mod-431
# Calculating ${34}^{429} \mod 431$ I am trying to calculate $${34}^{429} \mod 431$$ by hand. (This follows from $34^{-1}\mod 431$). I think I have made mistakes in my working, and have three different answers thus far from the attempts: $$351, 306, 134$$ Is one of these correct? If none of the above is correct please provide an answer with working. • What method are you using? Successive squaring or something else? – Michael Burr Aug 21 '17 at 19:56 • You can multiply them all by $34$ to see if either of them are $34^{-1}$. – Arthur Aug 21 '17 at 19:56 • This is equivalent to solving for $x$ in $1 = 34x + 431y$ which you can compute using Euclid's Algorithm. Hope this helps. – 伽罗瓦 Aug 21 '17 at 20:00 • @michaelburr I used a slightly different route on each attempt. For third attempt I only used powers of 34 and reduced mod 431 each time. For first and second attempts I broke multiplication down in different ways and used the results & raised the results to powers & simplified. – unseen_rider Aug 21 '17 at 20:02 • @arthur I don't see how that would help since I don't know what $34^{-1}$ is? – unseen_rider Aug 21 '17 at 20:07 You can use the extended Euclidean algorithm to find the inverse of $34\bmod 431$: $\begin{array}{c|c} n & s & t & q \\ \hline 431 & 1 & 0 & \\ 34 & 0 & 1 & 12 \\ 23 & 1 & -12 & 1 \\ 11 & -1 & 13 & 2 \\ 1 & 3 & -38 & 11 \\ \end{array}$ ... with each line expressing $n=431s+34t$ by suitable combination of the previous two lines. The final line gives $3\cdot 431 + (-38)\cdot 34 = 1$, so $(-38)\cdot 34\equiv 1 \bmod 431$ and thus $34^{-1}\equiv -38\equiv 393 \bmod 431$ • It's easier (2 steps) using signed remainders and fractions - see my answer. – Bill Dubuque Aug 21 '17 at 20:52 • @BillDubuque Yours is fewer steps, certainly, but I'll leave the decision on which is easier to the interested reader. Also, this is only three steps, really, since the first two lines are just tableau setup. – Joffan Aug 21 '17 at 21:07 $$431=34(12)+23$$ $$34=23+11$$ $$23=2(11)+1$$ Hence \begin{align}1&=23-2(11) \\ &=23-2(34-23)\\ &=3(23)-2(34)\\ &=3(431-34(12))-2(34)\\ &=3(431)-38(34)\end{align} Hence $$(-38)(34) \equiv 1 \mod 431$$ $$431-38=393$$ $$34^{429}=\left(\big(\big(34)^{13}\big)^{11}\right)^3$$ Then in $\mathbb F_{431}$ we have $$34^{13}=34^8\cdot34^5=373\\373^{11}=373^6\cdot373^5=420\\420^3=\color{red}{393}$$ • It seems you omitted many (painful!) calculations. Euclid's algorithm is much easier - a few single digit calculations except for the first step $\,431 - 13(\color{#c00}{34}) = \color{#90f}{-11}\$ – Bill Dubuque Aug 21 '17 at 22:27 • I wanted to give a different way to solve. (However it is not so different in nature....). – Piquito Aug 21 '17 at 22:52 Ignoring the fact that we're looking for the inverse, I'll lay out the exponentiation-by-squaring answer to getting $34^{429}\bmod 431$, for reference. The process is to square a running product repeatedly, interspersed by multiplying by $34$ as required, taking numbers $\bmod 431$ throughout the process. The idea is that we build up the exponent of our running product by factors of two until reaching the target exponent of $429$. Perhaps the easiest way to relate to this process is to look at the binary representation of the exponent, $429_{dec} = 110101101_{bin}$ We will increase the exponent by taking successively more from the left hand side of this binary representation, $1\to 3\to 6\to 13 \to 26\to 53\to 107 \to 214 \to 429$, either squaring alone for a simple doubled exponent or square-and-multiply to get to the odd exponents. \begin{array}{c|c} \to exp & prev & prev^2 & [\times 34] \\ \hline 1 & 1 & 1 & 34 \\ 3 & 34 & 294 & 83 \\ 6 & 83 & 424 & - \\ 13 & 424 & 49 & 373 \\ 26 & 373 & 347 & - \\ 53 & 347 & 160 & 268 \\ 107 & 268 & 278 & 401 \\ 214 & 401 & 38 & - \\ 429 & 38 & 151 & \color{red}{393} \\ \end{array} The requires handling numbers potentially up to $430^2$ (and then taking modulo $431$). By judicious use of negative values the limit could be brought down to $215^2$ (for example, $34^6\equiv 424 \equiv -7$ so $34^{12}\equiv -7^2\equiv 49$). A simple calculator was sufficient here to complete the tableau. $\bmod 431\!:\ \color{#c00}{\dfrac{1}{34}}\equiv \color{#0a0}{-38}\equiv 393\,$ by $2$ steps of the Extended Euclidean algorithm in fraction form \dfrac{0}{431}\, \overset{\large\frown}\equiv\!\! \underbrace{\color{#c00}{\dfrac{1}{34}}\ \overset{\large\frown}\equiv \color{#90f}{\dfrac{-13}{-11}}\ \overset{\large\frown}\equiv\ \color{#0a0}{\dfrac{-38}{1}}} _{\,\Large \begin{align}\color{#c00}{1}\ \ + \ \ &3(\color{#90f}{ -13 }) \ \ \ \equiv \ \ \color{#0a0}{-38}\\ \color{#c00}{34}\ \ +\ \ &3(\color{#90f}{-11} )\ \ \ \equiv\ \ \ \ \color{#0a0}{1}\ \ \ \end{align}}\qquad\qquad Alternatively we can apply Gauss's inversion algorithm in fractional form $$\bmod{431}\!:\,\ \dfrac{1}{\color{#0a0}{34}}\equiv \dfrac{\color{#c00}{13}\cdot 1\ }{\color{#c00}{13}\cdot34}\equiv\dfrac{13}{\color{#0a0}{11}}\equiv\dfrac{\color{#c00}{39}\cdot(11\!+\!2)}{\color{#c00}{39}\cdot 11\qquad}\equiv\dfrac{-2+2\cdot 39}{\color{#0a0}{-2}}\equiv -38$$ i.e. iteratively reduce the $\rm\color{#0a0}{denominators}$ by $\rm\color{#c00}{scalings}$ till it divides the numerator. This is precisely what is done in Famke's answer (but replacing fractions $\,x\equiv a/b\,$ by equations $\,b\,x\equiv a)$ • I'm curious as to why $\frac{-13}{-11}$ (negative #s) and not just $\frac{13}{11}$ (from $\frac{13}{442}$) – Joffan Aug 21 '17 at 21:10 • @joffan The prior step was \begin{align} 0 - 13(\color{#c00}1)\ \ \,&=\,\color{#90f}{-13}\\ 431 - 13(\color{#c00}{34}) \,&\,= \color{#90f}{-11}\end{align}\ We could of course cancel $\,{-}1\$. – Bill Dubuque Aug 21 '17 at 22:06 • Joffan We are doing the Euclidean algorithm on the denominators using remainders of least magnitude, for eaxample above we have $\ \ 431\,\bmod\ \color{#c00}{ 34}\ =\ \color{#90f}{{-}11}\,$ with quotient $13\ \$ – Bill Dubuque Aug 21 '17 at 22:11 • Ok, thanks for the explanation. – Joffan Aug 21 '17 at 22:12 $$431=12\cdot 34+23$$ $$34=23\cdot 1+11$$ $$23=2\cdot11+1$$ And now, $$1=23-2\cdot 11=23-2\cdot(34-23\cdot 1)=23\cdot3-34\cdot2=$$ $$=(431-12\cdot34)\cdot 3-34\cdot 2=431\cdot 3-38\cdot 34$$ One can see that $\lceil\dfrac{431}{34}\rceil=13$; $$13.34\overset{431}{\equiv}11;$$ again, one can see that $\lfloor\dfrac{431}{11}\rfloor=39$; $$39.11\overset{431}{\equiv}-2;$$ finally one can see easilly that: $$(-216).(-2)\overset{431}{\equiv}1.$$ So we have: $$(-216)\Bigg(39\big(13.34\big) \Bigg) \overset{431}{\equiv} (-216)\Bigg(39\big(11\big) \Bigg) \overset{431}{\equiv} (-216)\Bigg(-2 \Bigg) \overset{431}{\equiv} 1;$$ so we can conclude that: $$34^{-1} \overset{431}{\equiv} (-216).39.13 \overset{431}{\equiv} 393.$$ • As I note in my answer, this is precisely Gauss's inversion algorithm, which is clearer in fractional form, viz. $$\bmod{431}\!:\,\ \dfrac{1}{\color{#0a0}{34}}\equiv \dfrac{\color{#c00}{13}\cdot 1\ }{\color{#c00}{13}\cdot34}\equiv\dfrac{13}{\color{#0a0}{11}}\equiv\dfrac{\color{#c00}{39}\cdot(11\!+\!2)}{\color{#c00}{39}\cdot 11\qquad}\equiv\dfrac{-2+2\cdot 39}{\color{#0a0}{-2}}\equiv -38$$ i.e. iteratively reduce the $\rm\color{#0a0}{denominators}$ by $\rm\color{#c00}{scalings}$ till it divides the numerator. – Bill Dubuque Aug 22 '17 at 15:11 • What are scalings? – unseen_rider Aug 24 '17 at 5:40
2019-10-24T02:31:00
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http://bootmath.com/how-can-i-correct-my-wrong-intuition-that-forall-x-in-emptyset-px-quad-is-false.html
# How can I correct my wrong Intuition that $\forall \, x \, \in \,\emptyset : P(x) \quad$ is false? Source: p. 69. How to Prove It by Daniel Velleman. I already read 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11 & 12. $\exists \, x \, \in \, \emptyset : P(x) \tag{1}$ will be false no matter what the statement $P(x)$ is. There can be nothing in $\emptyset$ that, when plugged in for $x$, makes $P(x)$ come out true, because there is nothing in $\emptyset$ at all! It may not be so clear whether $\forall \, x \, \in \,\emptyset : P(x)$ should be considered true or false … I then paused reading to conjecture the truth value of: $2. \; \forall \, x \, \in \,\emptyset : P(x).$ $\boxed{\text{Conjecture :}}$ Because $\emptyset$ contains nothing, ergo $x \, \in \,\emptyset$ is false $\implies \forall \, x \, \in \,\emptyset$ is false. Since the Truth Value of $P(x)$ is unknown, ergo 2 is false. $\blacksquare$ Then I continued reading and was stupefied to learn that 2 is true: After expanding the abbreviation of the quantifiers, $\forall \, x \, \in \,\emptyset : P(x) \quad \equiv \quad \forall \, x \, \left[\, x \, \in \,\emptyset \Longrightarrow P(x)\right]. \tag{*}$ Now according to the truth table for the conditional connective, the only way this can be false is if there is some value of $x$ such that $x \, \in \,\emptyset$ is true but $P(x)$ is false. But there is no such value of $x$, simply because there isn’t a value of $x$ for which $x \, \in \,\emptyset$ is true. Thus, (*) is (vacuously) true. Though I understand this proof and the Principle of Explosion, I still do not understand why my intuition failed. How can I correct my intuition? I understand $\forall x \in \emptyset,P(x). \; \stackrel{dfn}{\equiv} \; \forall x, \color{#B22222}{x\in \emptyset}\implies P(x). \quad \equiv \; \forall x,\color{#B22222}{false}\implies P(x)$. Consider $3. \forall\;\bbox[5px,border:2px solid #32CD32]{\, \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \;,P(x)} \;$. 3. Is 3 vacuously true because the green box is vacuously true? I consider the green box above a statement, because though the comma is not a Logical Connective, $\forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \;P(x) \quad \equiv \quad \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \color{#0073CF}{\huge{,}} \;P(x) \quad \equiv \quad \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \color{#0073CF}{{\huge{\text{,}}} \text{we have that}} \; P(x). \tag{**}$ #### Solutions Collecting From Web of "How can I correct my wrong Intuition that $\forall \, x \, \in \,\emptyset : P(x) \quad$ is false?" You are right when you say that $\forall x$, the statement $x \in \emptyset$ is false. This means that $\forall x, \neg (x \in \emptyset)$, which is equivalent to $\forall x, x \notin \emptyset$. Perfectly true. Then you say “the statement $\forall x \in \emptyset$ is false”. $\forall x \in \emptyset$ is NOT a statement, it’s an incomplete sentence. Either you write “$\forall x, P(x)$”, either you write “$\forall x \in X, P(x)$”, which is a shorthand for “$\forall x, (x \in X \implies P(x))$”. “$\forall x \in \emptyset$” is not a statement. It can’t be true or false. $\forall x \in \emptyset, P(x)$ is a shorthand for $\forall x, (x \in \emptyset \implies P(x))$, which is equivalent (since $x \in \emptyset$ is always false) to $\forall x, ~\textrm{false} \implies P(x)$. After looking at the truth table for $\implies$, this is equivalent to $\forall x, ~\textrm{true}$ (whatever $P(x)$ may be), which is $\textrm{true}\;.$ If you want to disprove $\forall x \in \emptyset, P(x)$ you have to show me an $x \in \emptyset$ such that $P(x)$ is false. Well you will never find an $x \in \emptyset$. IMO, it’s our grasp of natural language that leads us astray here. Natural language is not about explicitly expressing precise ideas; there is a lot of ambiguity and implicit inference involved. In particular, one doesn’t speak about “all of something” unless there is (possibly hypothetically) actually something to speak about. As you are trained to make this inference, when you hear “$\forall x \in \varnothing:P$”, you mentally add the implicit hypothesis “$\exists x \in \varnothing$”, which is where your intuition goes awry.
2018-07-20T10:27:10
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https://algebrology.github.io/cosets-and-lagranges-theorem/
March 4, 2019 # Cosets and Lagrange's Theorem ### Cosets It's a bit difficult to explain exactly why cosets are so important without working with them for a while first. But as you'll hopefully start to understand within my next few posts, cosets pop up everywhere and are a necessary tool to get anything done in the world of algebra. Let's dig in, shall we? Definition. Let $H$ denote a subgoup of a group $G$. For any fixed element $x\in G$, the right coset of $H$ with respect to $x$ is the set $Hx=\{hx\in G\mid h\in H\}$. Definition. Let $H$ denote a subgoup of a group $G$. For any fixed element $x\in G$, the left coset of $H$ with respect to $x$ is the set $xH=\{xh\in G\mid h\in H\}$. Note. For abelian groups with additive notation, right and left cosets are instead denoted by $H+x$ and $x+H$, respectively. So basically a coset is a set obtained by taking the elements of a subgroup and adding a particular element to all of them. Note that a coset does need not be a subgroup! To get a feeling for what cosets really are and how they behave, let's look at an example. Example. Take $G=\Z$, the additive group of integers, and $H=2\Z$, the set of even integers. Certainly $2\Z$ is a subgroup of $Z$ because it contains the identity $0$, the sum of two even integers is even, and every even integer has an even inverse — its negative. Let's look at a few cosets of $2\Z$. How about the right cosets with respect to $0,1,2$ and $3$? Notice that because $\Z$ is an additive group, these cosets will be written $2\Z+0$, $2\Z+1$, $2\Z+2$ and $2\Z+3$. Directly from the definition of a right coset, we see that \begin{align} 2\Z+0 &= \{x+0\in\Z\mid x\in 2\Z\} \\ &= \{x\in\Z\mid x\in 2\Z\} \\ &= 2\Z. \end{align} So in this case, the coset with respect to $0$ is just the subgroup $2\Z$. This actually always happens, as you can easily see. Adding the identity to all elements of a subgroup will of course just yield that subgroup again! Next, let's look at $2\Z+1$. This is the set $\{x+1\in\Z\mid x\in 2\Z\}$, which consists of things like $\ldots,-3,-1,1,3,5,\ldots$. This is really just the set of odd integers! It's definitely not a subgroup of $\Z$ though, since it doesn't contain $0$. The coset $2\Z+2$ is the set $\{x+2\in\Z\mid z\in 2\Z\}$, which contains elements like $\ldots,-4,-2,0,2,4,\ldots$. But this is the set of even integers again! That means $2\Z+2=2\Z$. Lastly, let's look at $2\Z+2$. This is the set $\{x+3\in\Z\mid z\in 2\Z\}$, which contains things like $\ldots,-3,-1,1,3,5,\ldots$. We've already seen that somewhere before. It's just the set of odd integers again. That is, $2\Z+3=2\Z+1$. It looks like there might actually only be two distinct cosets of $2\Z$. They are $2\Z$ itself and $2\Z+1$. Furthermore, every element of $\Z$ is in either one coset or the other, but never both (because an integer is either even or odd). So these cosets actually partition $\Z$, which is a very important point. But let's not get too far ahead of ourselves. The first thing I'd like to prove about cosets is fairly simple — if we are working in an abelian group, left and right cosets are the same! Theorem. If $H$ is a subgroup of an abelian group $G$, then $H+x=x+H$ for every $x\in G$. Proof. We will proceed by demonstrating that each side is a subset of the other. We show first that $H+x\subseteq x+H$. Choose $g\in H+x$, so that $g=h+x$ for some $h\in H$. Since $G$ is abelian, $h+x=x+h$ and thus $g=x+h\in x+H$. It follows that $H+x\subseteq x+H$. The proof that $x+H\subseteq H+x$ is completely analogous to the above, so we won't bother with it. We can thus conclude that $H+x=x+H$. This implies, for instance, that we could instead write $1+2\Z$ to denote the odd integers, but to me this doesn't look right for some reason and so I usually don't. In fact, for most of our purposes we really only need to consider one variety of coset. I tend to favor right cosets. Theorem. If $H$ is a subgroup of a group $G$, then the (left/right) cosets of $H$ partition $G$. Proof. We will prove the result for right cosets, since the proof for left cosets is practically identical. It is clear that every coset $Hx$ is nonempty because the identity element $e$ is in $H$ by virtue of it being a subgroup, and thus $x=ex\in Hx$. The next thing we need to show is that cosets cover all of $G$. But is clear that $$\bigcup_{x\in G}x=G\subseteq\bigcup_{x\in G}Hx,$$ and similarly that $$\bigcup_{x\in G}Hx\subseteq G=\bigcup_{x\in G}x,$$ because $Hx\subseteq G$ for every $x\in G$. Thus, $G=\bigcup_{x\in G}Hx$. The last thing we need to show is that for any $x,y\in G$, if $Hx\ne Hy$ then $Hx\cap Hy=\varnothing$. That is, either cosets are the same or they are disjoint. We will argue the contrapositve, supposing that $Hx\cap Hy$ is nonempty. Then there exists some element $a\in Hx\cap Hy$. That is, $a\in Hx$ and $a\in Hy$, so $a=h_1x$ and $a=h_2y$ for some elements $h_1,h_2\in H$. It follows that $h_1x=h_2y$, and multiplication on the left by $h_1^{-1}$ show that $x=h_1^{-1}h_2y\in Hy$ because $h_1^{-1}h_2\in H$. Thus $Hx=Hy$, completing the proof. This is pretty exciting! It means that we can form an equivalence relation $\sim$ on any group $G$, where elements are equivalent if they are in the same coset of $H$. Moreover, we can form the quotient set $G\quotient{\sim}$, whose elements are precisely the cosets of $H$. For simplicity, we generally write $G/H$ instead of to denote this quotient set, since it is so very important that we are talking about the set of cosets of $H$. We'll revisit this special type of quotient set in my next post. ### Lagrange's Theorem We're actually very close to proving this famous theorem already. We just need one lemma first: Lemma. If $H$ is a subgroup of a group $G$ and $x\in G$, there exists a bijection $f:G\to Hx$. Proof. We define a function $f:H\to Hx$ by $f(h)=hx$ for every $h\in H$. That is, the function that maps every element of $H$ to its product with $x$. Clearly $f$ is injective because if $f(h_1)=f(h_2)$ then $h_1x=h2_x$ and the right cancellation law yields $h_1=h_2$. Furthermore, $f$ is surjective because for any $y\in Hx$ we have by definition that $y=hx$ for some $h\in H$, and thus $y=f(h)$. It follows that $f$ is bijective, as desired. Recall that a bijection exists between finite sets only if those sets contain the same number of elements. This fact, along with the lemma above, tells us that all cosets of a subgroup are the same size! We are now ready to prove Lagrange's Theorem, which is actually very easy at this point. Lagrange's Theorem. If $H$ is a subgroup of a finite group $G$, then $\abs{G}$ is a multiple of $\abs{H}$. Proof. We have already established that every coset of $H$ contains the same number of elements, $\abs{H}$. Since $G$ is finite, there are a finite number of distinct cosets of $H$, say $n$ of them. Because these cosets partition $G$, it follows that $\abs{G}=n\abs{H}$, completing the proof. This is probably somewhat surprising unless you're already familiar with groups. But it's just part of the astonishing usefulness of cosets. Lagrange's Theorem is actually incredibly useful because it tells us instantly that certain things cannot be subgroups of other things. For instance, a group of order $12$ cannot ever have subgroups of order $5,7,8,9,10$ or $11$. Without this theorem, you might have guessed that this was the case, but it would have been pretty tricky to prove it conclusively.
2022-08-18T08:06:42
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https://proofwiki.org/wiki/Rolle%27s_Theorem
Rolle's Theorem Theorem Let $f$ be a real function which is: continuous on the closed interval $\closedint a b$ and: differentiable on the open interval $\openint a b$. Let $\map f a = \map f b$. Then: $\exists \xi \in \openint a b: \map {f'} \xi = 0$ Proof 1 We have that $f$ is continuous on $\closedint a b$. It follows from Continuous Image of Closed Interval is Closed Interval that $f$ attains: a maximum $M$ at some $\xi_1 \in \closedint a b$ and: a minimum $m$ at some $\xi_2 \in \closedint a b$. Suppose $\xi_1$ and $\xi_2$ are both end points of $\closedint a b$. Because $\map f a = \map f b$ it follows that $m = M$ and so $f$ is constant on $\closedint a b$. Then, by Derivative of Constant, $\map {f'} \xi = 0$ for all $\xi \in \openint a b$. Suppose $\xi_1$ is not an end point of $\closedint a b$. Then $\xi_1 \in \openint a b$ and $f$ has a local maximum at $\xi_1$. Hence the result follows from Derivative at Maximum or Minimum‎. Similarly, suppose $\xi_2$ is not an end point of $\closedint a b$. Then $\xi_2 \in \openint a b$ and $f$ has a local minimum at $\xi_2$. Hence the result follows from Derivative at Maximum or Minimum‎. $\blacksquare$ Proof 2 First take the case where: $\forall x \in \openint a b: \map f x = 0$ Then: $\forall x \in \openint a b: \map {f'} x = 0$ Otherwise: $\exists c \in \openint a b: \map f c \ne 0$ Let $\map f c > 0$. Then there exists an absolute maximum at a point $\xi \in \openint a b$. Hence: $\ds \dfrac {\map f {\xi + h} - \map f \xi} h$ $\le$ $\ds 0$ for $\xi < \xi + h < b$ $\ds \dfrac {\map f {\xi + h} - \map f \xi} h$ $\ge$ $\ds 0$ for $a < \xi + h < \xi$ As $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive. That is: $\map {f'} \xi = 0$ Similarly, let $\map f c < 0$. Then there exists an absolute minimum at a point $\xi \in \openint a b$. Hence: $\ds \dfrac {\map f {\xi + h} - \map f \xi} h$ $\ge$ $\ds 0$ for $\xi < \xi + h < b$ $\ds \dfrac {\map f {\xi + h} - \map f \xi} h$ $\le$ $\ds 0$ for $a < \xi + h < \xi$ Again, as $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive. That is: $\map {f'} \xi = 0$ Hence the result. $\blacksquare$ Source of Name This entry was named for Michel Rolle.
2021-04-18T21:07:04
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https://cs.stackexchange.com/questions/95947/why-is-this-n2-growth/95948#95948
# Why is this n^2 growth? I am attempting to understand the growth of the following algorithm, which is described as $n^2$ growth in the book I am reading: "... performs of the order of $n^2$ steps on a sequence of length $n$." Could someone please explain how this is calculated in the following code, which is also taken from the book? If I print out the statements when the lines are executed, it first executes $n$ steps, then decreases $n-1$ steps for each loop iteration until it reaches $0$. This does not seem like exponential growth to me. Why does this grow at $n^2$? dataset = [3,1,2,7,5] product = 0 # algorithm begins here for i in range(len(dataset)): for j in range(i + 1, len(dataset)): product = max(product, dataset[i]* dataset[j]) ## 1 Answer Because $n + (n-1) + (n-2) + \cdots + 2 + 1 = \frac{n(n+1)}{2} \in \mathcal{O}(n^2)$. Note that $n^2$ is polynomial, not exponential (that would be $2^n$ for example). • In fact, $\frac{n(n+1)}{2} = \Theta(n^2)$, which is why it is stated that the number of steps is of order $n^2$. Aug 4, 2018 at 13:09 • So - I understand my error in thinking it was exponential, and I understand how n^2 is derived (distribute n and drop the lowest order constant), but I am not quite clear on how n + (n - 1)... 2 + 1 evaluates to that fraction? Pointing me to a resource for further research would be terrific. I am having trouble wrapping my head around how algorithms end up being evaluated as theta, O, and omega based on how they are written and steps they perform on input. Aug 4, 2018 at 13:23 • @ElliotRodriguez You can prove the identity $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ by induction on $n$ (here is a full proof). It's a good one to remember! When it comes to determining the complexity of algorithms I always find it helpful to trace through a few small cases first. Aug 4, 2018 at 13:38 • @ElliotRodriguez Search about Carl Frederich Gauss's (alleged) story of (re)inventing this formula (at the age of five)! Aug 4, 2018 at 16:06 • @DanielMroz Rather than using induction, it's much easier to just observe that the sum is \begin{align*}\tfrac12(&1+\dots+n + \\&n + \dots +1)\\&\quad = \tfrac12\big((1+n) +(2+ n-1) + (3+n-2) + \dots + (n+1)\big)\\ &\quad = \tfrac12n(n+1)\,.\end{align*} Aug 4, 2018 at 19:39
2022-08-13T02:23:44
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https://dendalosi.firebaseapp.com/644.html
Mode of continuous random variable pdf Boardworks ltd 2006 mode suppose that a random variable x is defined by the probability density function fx for a x b. Probability density functions pdf examsolutions duration. It is usually more straightforward to start from the cdf and then to find the pdf by taking the derivative of the cdf. In this lesson, well extend much of what we learned about discrete random variables. Since the values for a continuous random variable are inside an. Sure, for continuous distributions you have to fudge the end of that a bit to something like at which the pdf is locally maximized, but its the same principle. Geometric visualisation of the mode, median and mean of an arbitrary probability density function. For example, if we let x denote the height in meters of a randomly selected maple tree, then x is a continuous random variable. I explain how to calculate the median of a continuous random variable. Examples i let x be the length of a randomly selected telephone call. In probability theory and statistics, the continuous uniform distribution or rectangular distribution is a family of symmetric probability distributions. Statistics random variables and probability distributions. In probability theory, a probability density function pdf, or density of a continuous random variable, is a function whose value at any given sample or point in the. They are used to model physical characteristics such as time, length, position, etc. Discrete and continuous random variables video khan academy. Jan 07, 20 this is the fourth in a sequence of tutorials about continuous random variables. Finding the mode for a continuous random variable this tutorial shows you how to calculate the mode for a continuous random variable by looking at its probability density function. How to find the mode of a probability density function. Because the total area under the density curve is 1, the probability that the random variable takes on a value between aand. The bounds are defined by the parameters, a and b, which are the minimum and maximum values. Mode for a continuous random variable examsolutions youtube. A random variable that may assume only a finite number or an infinite sequence of values is said to be discrete. Thus, we should be able to find the cdf and pdf of y. Definition a random variable is called continuous if it can take any value inside an interval. For example, a random variable measuring the time taken for something to be done is continuous since there are an infinite number of possible times that can be taken. We think of a continuous random variable with density function f as being a random variable that can be obtained by picking a point at random from under the density curve and then reading o the xcoordinate of that point. I explain how to calculate the mode of a continuous random variable. Apr 14, 2018 the area under the curve of a probability density function must always sum to one. Since the continuous random variable is defined over a continuous range of values called thedomain of the variable, the graph of the density function will also be continuous over that range. The mode of x is the value of x that produces the largest value for fx in the interval a x b. The mode is the value of where is maximum which may not be unique. To l earn how to use the probability density function to find the 100p th percentile of a continuous random variable x. The probability density function of the continuous uniform distribution is. The major difference between discrete and continuous random variables is in the distribution. A random variable can take on many, many, many, many, many, many different values with different probabilities. The median of a continuous probability distribution is the point at which the distribution function has the value 0. Continuous random variables histogram mode statistics. A random variable x is continuous if possible values comprise either a single interval on the number line or a union of disjoint intervals. A continuous random variable differs from a discrete random variable in that it takes on an uncountably infinite number of possible outcomes. So given a specific definition of the mode you find it as you would find that particular definition of highest value when dealing with functions more generally, assuming that the distribution is unimodal under. Jan 07, 20 this is the fifth in a sequence of tutorials about continuous random variables. That is, the possible outcomes lie in a set which is formally by realanalysis continuous, which can be understood in the intuitive sense of having no gaps. For any continuous random variable with probability density function f x, we. Continuous random variables and probability distributions. A continuous random variable x has cumulative distribution. How to find the median of a probability density function quora. A random variable is a numerical description of the outcome of a statistical experiment. To be able to apply the methods learned in the lesson to new problems. This is because across all possible outcomes you must have all probabilities sum to 100%. Chapter 4 continuous random variables purdue college of. If a continuous random variable has more than one median, can it have a nite number. In other words, while the absolute likelihood for a continuous random variable to take on any particular value is 0 since there are. Boxplot and probability density function of a normal distribution n0. Arrvissaidtobeabsolutely continuous if there exists a realvalued function f x such that, for any subset b. The probability density function or pdf of a continuous random variable gives the relative likelihood of any outcome in a continuum occurring. Consequently, the mode is equal to the value of \x\ at which the probability distribution function, \p\beginpmatrixx x \endpmatrix\, reaches a maximum. As it is the slope of a cdf, a pdf must always be positive. Let x be a continuous random variable with range a. However, the same argument does not hold for continuous random variables because the width of each histograms bin is now in. B z b f xxdx 1 thenf x iscalledtheprobability density function pdf oftherandomvariablex. The probability density function gives the probability that any value in a continuous set of values might occur. Continuous random variables a nondiscrete random variable x is said to be absolutely continuous, or simply continuous, if its distribution function may be represented as 7 where the function fx has the properties 1. Then a probability distribution or probability density function pdf of x is a. Continuous random variables continuous random variables can take any value in an interval. Let m the maximum depth in meters, so that any number in the interval 0, m is a possible value of x. Continuous variable, as the name suggest is a random variable that assumes all the possible values in a continuum. To extend the definitions of the mean, variance, standard deviation, and momentgenerating function for a continuous random variable x. This is the fifth in a sequence of tutorials about continuous random variables. Mode given a discrete random variable \x\, its mode is the value of \x\ that is most likely to occur. The mode is defined as the value which has highest frequency. Be able to explain why we use probability density for continuous random variables. The distribution describes an experiment where there is an arbitrary outcome that lies between certain bounds. The area bounded by the curve of the density function and the xaxis is equal to 1, when computed over the domain of the variable. Here you are shown how to find the mode of a continuous random variable. Mode the mode of a continuous random variable corresponds to the \x\ values at which the probability density function reaches a local maximum, or a peak. A mode represents the same quantity in continuous distributions and discrete distributions. That the ex would equal to 23 but how do i determine the mode and the median value for x, and the variance. Calculating the mean, median, and mode of continuous random. Mode for a continuous random variable examsolutions. If in the study of the ecology of a lake, x, the r. It follows from the above that if xis a continuous random variable, then the probability that x takes on any. X of a continuous random variable x with probability density function fxx is. A continuous random variable is a function x x x on the outcomes of some probabilistic experiment which takes values in a continuous set v v v. Note that before differentiating the cdf, we should check that the. Math statistics and probability random variables discrete random variables. Simply put, it can take any value within the given range. Discrete random variables are characterized through the probability mass functions, i. May 26, 2012 the continuous random variable x has probability density function given by fx kx 0 of k would be 2. Unlike the case of discrete random variables, for a continuous random variable any single outcome has probability zero of occurring. The formulae for the mean ex and variance varx for continuous random variables. Probability distributions for continuous variables. The distribution is also sometimes called a gaussian distribution. Calculating the mean, median, and mode of continuous. Continuous random variables definition brilliant math. The continuous random variable has the normal distribution if the pdf is. Sure, for continuous distributions you have to fudge the end of that a bit to something like at which the pdf is. A continuous random variable is a random variable where the data can take infinitely many values. A mode represents the same quantity in continuous distributions and. A mode of a continuous probability distribution is a value at which the probability density function pdf attains its maximum value. The mode of a continuous random variable corresponds to the x values at which the probability density function reaches a local maximum, or a peak. Continuous random variables probability density function. Constructing a probability distribution for random variable. Things change slightly with continuous random variables. It is the value most likely to lie within the same interval as the outcome. Probability density functions mode cumulative distribution functions median and quartiles expectation variance. Thus a pdf is also a function of a random variable, x, and its magnitude will be some indication of the relative likelihood of measuring a particular value. The median for a discrete random variable may not be unique see example 1, on page 3. In particular, it is the integral of f x t over the shaded region in figure 4. Statistics statistics random variables and probability distributions. If x is a continuous random variable and y gx is a function of x, then y itself is a random variable. If we discretize x by measuring depth to the nearest meter, then possible values are nonnegative integers less. Chapter 4 continuous random variables a random variable can be discrete, continuous, or a mix of both. And it makes much more sense to talk about the probability of a random variable equaling a value, or the probability that it is less than or greater than something, or the probability that it has some property. In fact and this is a little bit tricky we technically say that the probability that a continuous random variable takes on any specific value is 0. Parameters of continuous random variables radford mathematics. Parameters of continuous random variable radford mathematics. Difference between discrete and continuous variable with. For a continuous random variable, this corresponds to f0 x x 0 and f00 x x continuous random variables computing expectation of function of continuous random variable if x is a continuous random variable with density f and g is a function, then egx z 1 1 gxfxdx 1118. Unlike pmfs, pdfs dont give the probability that \x\ takes on a specific value. In probability theory, a probability density function pdf, or density of a continuous random. Let x be a continuous random variable with pdf f xu. For any predetermined value x, px x 0, since if we measured x accurately enough, we are never going to hit the value x exactly. Content mean and variance of a continuous random variable amsi. 63 1546 1030 18 1057 343 802 1224 688 1268 1577 451 1520 1172 1515 1183 1023 1486 87 797 324 1354 1020 783 824 186 619 1447 1481 1048 568
2022-10-01T22:12:47
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https://math.stackexchange.com/questions/3136672/prove-that-lim-left-u-v-right-rightarrow-left-0-0-right-fracu
# Prove that $\lim_{\left( u, v \right) \rightarrow \left( 0, 0 \right)} \frac{u}{\sqrt{v+4}} = 0$ using the $\epsilon - \delta$ definition I would like to prove that $$\lim_{\left( u, v \right) \rightarrow \left( 0, 0 \right)} \frac{u}{\sqrt{v+4}} = 0$$ using the $$\epsilon - \delta$$ definition of a limit. My attempt is as follows: Suppose $$0< \sqrt{u^2+v^2} < \delta$$ then $$\left| \frac{u}{\sqrt{v+4}} \right|=\frac{\left| u \right| }{\sqrt{v+4}}<\frac{\delta}{\sqrt{v+4}}.$$ If $$\sqrt{v+4} \geq 1$$, we have $$\frac{\delta}{\sqrt{v+4}} \leq\delta.$$ Therefore, $$\delta = \epsilon$$ should work. I know that near the origin $$\sqrt{v+4} \geq 1$$. However, I would like a more sophisticated/complicated answer; for example, something like $$\delta = \min \left( \epsilon, c \right)$$ where $$c>0$$. I would appreciate it if you point me in the right direction or provide a reference that deals with a similar problem. • Both answers were very helpful; as I can't designate both as the correct answer, I will designate the older one as such (though it's only a 2-minute difference). – Ahmed Ali Mar 5 at 20:50 Setting $$\delta = \min\{\varepsilon, 1\}$$ will work. If $$0 < \|(u,v)\| < \delta$$, we have $$|u|,|v| < \delta$$ so $$\sqrt{v+4} \ge \sqrt{-|v|+4} > \sqrt{-\delta+4} \ge \sqrt{-1+4} = \sqrt{3}$$ and therefore $$\frac{|u|}{\sqrt{v+4}} < \frac{\delta}{\sqrt{3}} \le \frac{\varepsilon}{\sqrt3} < \varepsilon$$ First of all, please don't put "complicated" on a pedestal! Good mathematics strives to explain complicated things as simply as possible. The end goal is simplicity, that as many people as possible can understand. In mathematics research, if you've taken something that nobody understands and turned it into something only you can understand, it is barely an improvement! That said, your solution is incomplete, and by completing it, you'll get a $$\delta$$ in the form you want. As you say, I know that near the origin $$\sqrt{v + 4} \ge 1.$$ You'll need to encode this into your $$\delta$$. If you just choose $$\delta = \varepsilon$$, then having $$\varepsilon = 3.99$$ would cause the inequality to fail. If $$\varepsilon > 4$$, then $$0 < \sqrt{u^2 + v^2} < \varepsilon$$ will not guarantee $$(u, v)$$ lies in the domain of the function! To correct this, let's consider where $$\sqrt{v + 4} \ge 1$$. Note that this is equivalent to $$v \ge -3$$. If we force $$\delta \le 3$$, then this should work, as $$0 < \sqrt{u^2 + v^2} < 3 \implies v > -3.$$ Therefore, just choose $$\delta = \min \{\varepsilon, 3\}$$. • I am wholly convinced by your words. I think that my wording has betrayed me. What I wanted to convey is that I don't want to depend on just the intuition that $\sqrt{v+4} \geq 1$ near the origin; I wanted a more concrete/solid proof. Thank you for your answer. – Ahmed Ali Mar 5 at 20:58
2019-07-17T16:49:32
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https://math.stackexchange.com/questions/2694788/what-is-the-derivative-of-uv
# What is the derivative of $(u^v)$? The derivative of a sum is the sum of the derivatives, ie, $$d(u+v)=du+dv$$ The derivative of a product is a little more complicated: $$d(u\cdot v)= u\ dv + v\ du$$ But what about the derivative of a power? I'm not talking $x^n$ or $a^x$ or even $x^x$, but $u^v$ - an arbitrary function of $x$ raised to the power of another arbitrary function of $x$. • @krirkrirk No, I know it's correct, that's why I put my answer in an answer. I'm putting it up so it's findable. Mar 17 '18 at 9:49 • Ok, after re-reading the rules, it appears you can indeed post a question just so that it appears in future researchs. I did not know this ! Sorry. Mar 17 '18 at 10:07 • It's okay, no harm done. Mar 17 '18 at 10:08 A not so well known differentiation trick is to consider all variables in turn and assume the other to be constants, differentiate and sum. Hence $u^v$ is first a power of $u$, then an exponential with exponent $v$ and $$\left(u^v\right)'=vu^{v-1}u'+\log u\,u^vv'.$$ This is in fact an application of $$\left(u^v\right)'=\frac{\partial u^v}{\partial u}u'+\frac{\partial u^v}{\partial v}v'.$$ • +1. The OP should look for "chain rule for partial derivatives" in a calculus textbook. True, in this case it can be done with logarithms and exponentials. But that chain rule is a useful formula to know anyway. Mar 17 '18 at 13:51 • @GEdgar Truth be told, I hadn't thought to make that particular connection. But then, my classes to date haven't gone that far in my textbooks, I just read ahead. Mar 19 '18 at 12:18 We start with $$y=u^v$$ where $y$, $u$ and $v$ are all functions of $x$. We take the natural logarithm $(\ln)$ of both sides: $$\ln(y)=\ln(u^v)$$ And drop $v$ out from $\ln$, using the relevant property of logarithms: $$\ln(y)=v\ \ln(u)$$ This allows us to use the previously established product rule when we differentiate: $$d\ln(y)=v\ d\ln(u) + \ln(u)\ dv$$ Which, by the calculus definition of the natural logarithm, simplifies to: $$\frac{dy}{y}=v\ \frac{du}{u} + \ln(u)\ dv$$ We then multiply both sides by $y$ to isolate $dy$, and replace $y$ with its definition $u^v$ to get: $$dy=u^v\ \left(\frac{v\ du}{u} + \ln(u)\ dv\right)$$ Which is the answer we were looking for. But! Let's distribute that $u^v$ and do some cancelling: $$dy=v\ u^{v-1}\ du + \ln(u)\ u^v\ dv$$ To the student of calculus, those look familiar. The first addend $(v\ u^{v-1}\ du)$ is the power rule $(d(u^n)=n\cdot u^{n-1}\ du)$, while the second $(\ln(u)\ u^v\ dv)$ is the general exponential rule $(dv=\ln(a)\ a^v\ dv)$. This makes sense; the derivation of this rule used the product rule, which defines the derivative of a product as the sum of two distinct products, one each that differentiates one function and leaves the other alone. Here, that's gone up a level, so to speak, with each addend of our result treating either the base or the power as constant. In fact, setting $u$ constant renders $du \; \; 0$, rendering the first addend $0$ in turn, and likewise for $v$ and the second addend, rendering both the power rule and the generalized exponent rule special cases of this "generalized power/exponent rule". • Now the question becomes "Why hasn't anyone asked this before?" Plugging this question's title into the helpful little box search thing at the top of the question format got me a few specific examples (z^z, sin^cos), but not the general form (also, it seems to be assumed that x^x = x^x(1+ln(x))), and searching directly was useless. In any case, now it's findable, answered and explained. Mar 17 '18 at 9:47 • I guess that because on the odd occasion that it is needed, we just do what TheSimpliFire did. Mar 17 '18 at 9:59 • Well, that's nice, but it's still a cool math fact that deserves to be shared. (This comment has replaced my previous one). Mar 17 '18 at 10:30 • I am not saying that your original question is invalid or unreasonable. I was just trying to answer your subsidiary question of why it had not been asked before. My answer to that is that is less commonly required and can be easily deduced when it is. To me, at least, if is easier to remember and apply the more basic rules than to memorise an extra one. Mar 17 '18 at 12:09 Let $u=f(x)$, $v=g(x)$ and $y=f(x)^{g(x)}$. Then using the Chain and Product Rules, \begin{align}\ln y=g(x)\ln f(x)&\implies\frac1y\cdot\frac{dy}{dx}=g'(x)\ln f(x)+\frac{g(x)}{f(x)}\cdot f'(x)\\&\implies\frac{dy}{dx}=f(x)^{g(x)}\left(g'(x)\ln f(x)+\frac{g(x)}{f(x)}\cdot f'(x)\right)\end{align} Hence $$d(u^v)=u^v\left(v'\ln u'+\frac{vu'}u\right)$$ • Wait a tick: Plugging in u=v=x gives $d(x^x) = x^x$ (as $x' = 1$, $ln(1) = 0$, and $x/x = 1$, for all the cases that matter anyway); I think you added a prime where one was unwarranted. You also seem to have forgotten the $f'(x)$ in your second addend, (but that doesn't change $d(x^x)$). Mar 17 '18 at 10:22 • Thanks! Fixed.. Mar 17 '18 at 10:24 We obtain \begin{align*} \color{blue}{d(u^v)}&=d\left(e^{v\cdot \ln u}\right)\\ &=e^{v\cdot \ln u}\cdot d(v\cdot \ln u)\\ &=u^v\left(\ln u\ dv + v\ d\left(\ln u\right)\right)\\ &\,\,\color{blue}{=u^v\left(\ln u\ dv+\frac{v}{u}\ du\right)} \end{align*}
2021-10-25T13:12:22
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http://mathhelpforum.com/statistics/106062-solved-help-tough-combinations-question.html
# Thread: [SOLVED] Help with tough Combinations Question 1. ## [SOLVED] Help with tough Combinations Question Hi, I can't seem to answer this question. The answer I get differs from the back of the book and I don't understand why. Q: The camera club has 5 members, and the math club has 8. There is only 1 member common to both clubs. In how many ways could a committee of 4 people be formed with a least one member from each club? I think the part where I'm going wrong is where the 1 member overlaps. I'm not sure how I'm suppose to deal with it. The answer I get is 470 while the answer at the back is 459... If someone could please show me how I'm suppose to go about solving this problem, it would be great! THANKS 2. One way to look at it is to account for that pesky person who is in both clubs. $\binom{4}{1}\binom{8}{3}+\binom{4}{2}\binom{8}{2}+ \binom{4}{3}\binom{8}{1}+\binom{7}{4}\binom{5}{0}$ Draw a Venn and it will be easier to see. 3. Originally Posted by galactus One way to look at it is to account for that pesky person who is in both clubs. $\binom{4}{1}\binom{8}{3}+\binom{4}{2}\binom{8}{2}+ \binom{4}{3}\binom{8}{1}+\binom{7}{4}\binom{5}{0}$ Draw a Venn and it will be easier to see. I'm not sure I'm understanding this correctly. So for part of the question, you include that pesky person on one side and then on the other for the last part of the question. You get the correct answer. But what if I decided to include the pesky person on the other side first then the first side. As in: $\binom{7}{1}\binom{5}{3}+\binom{7}{2}\binom{5}{2}+ \binom{7}{3}\binom{5}{1}+\binom{4}{4}\binom{8}{0} $ The answer then becomes 456 4. Originally Posted by MATHDUDE2 Q: The camera club has 5 members, and the math club has 8. There is only 1 member common to both clubs. In how many ways could a committee of 4 people be formed with a least one member from each club? answer at the back is 459... There are really 12 people total. So there are a total of $\binom{12}{4}$ ways to select the committee. There $\binom{7}{4}+\binom{4}{4}$ ways the committee will not have someone from each club. $\binom{12}{4}-\left(\binom{7}{4}+\binom{4}{4}\right)=459$ 5. Originally Posted by Plato There are really 12 people total. So there are a total of $\binom{12}{4}$ ways to select the committee. There $\binom{7}{4}+\binom{4}{4}$ ways the committee will not have someone from each club. $\binom{12}{4}-\left(\binom{7}{4}+\binom{4}{4}\right)=459$ WOW I can't believe I didn't see the indirect method as a viable option. Thanks! Still curious as to how to solve it using the direct method but at least I know how to solve the question now. Thanks everyone! 6. Here's how I solved it directly. There are 4 people in club 1 and 7 in club 2 and the one guy on the side who can count for both. Now, we either choose to include the one guy in the committee or not. If we don't choose him, then there are $\sum_{k=1}^3 {4 \choose k} {7 \choose {4-k}}$ ways to pick the committee. That is, we would like to pick out k from one group and the rest from the other, and note that k is from 1 to 3, so one person has to be chosen from each group. Now we concern ourselves with the person who's in both camps. We've picked him, so we need to pick 3 other people. They can belong to any club since the one person counts from both camps. Thus we add ${11 \choose 3}$ to the above. Evaluating that will give you 459 as the book says.
2017-01-20T23:52:40
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https://www.nextgurukul.in/wiki/concept/cbse/class-11/maths/complex-numbers-and-quadratic-equations/introduction-to-complex-numbers/3960926
Notes On Introduction to Complex Numbers - CBSE Class 11 Maths Imaginary number Solutions of this quadratic equation  x2 - 4 = 0 are 2 or -2 . The solutions are real. Solutions of the quadratic equation x2 + 4 = 0 are $±\sqrt{\text{-4}}$ The square of a real number is always non-negative. Solutions of the quadratic equation x2 + 4 = 0 are not real. $±\sqrt{\text{-4}}$ is written as x $\sqrt{\text{-1}}$. A number whose square is negative is called an imaginary number. Complex number Let  $\sqrt{\text{-1}}$ be denoted by the letter "i", i.e. i2 = -1. ⇒ x = ± 2 x i ⇒ x = 2i or - 2i Therefore, the solutions of the quadratic equation x2 + 4 = 0 are 2i or - 2i. These numbers are called complex numbers. The solution of the quadratic equation ax2+bx+c = 0, where D = b2 - 4ac < 0 , is not possible in the real number system. Such equations will always have complex roots. A number of the form a+ib, where a and b are real numbers, is called a complex number. Such numbers are denoted by the letter 'z'. z = a + ib Where 'a' is called the real part of the complex number, denoted by Re z and, 'b' is called the imaginary part, denoted by Im z. Ex: z = 3 + 2i, z = 5 + √2i, z = -6 + ($\frac{\text{1}}{\text{3}}$)i In z = 3 + 2i, 3 is the real part and 2 is the imaginary part. If the imaginary part of a complex number is zero, then the number is called a purely real number. Ex: z = 5+0i = 5 If the real part of a complex number is zero, then the number is called a purely imaginary number. Ex: z = 0 + 7i = 7i Equality of complex numbers Consider two complex numbers z1 = a+ib, z2 = x+iy Two complex numbers are said to be equal, if their corresponding real parts and imaginary parts are equal. z1 = z2, if a = x and b = y Ex: Find the values of x and y, if the complex numbers 3x+(2x+y)i, 9+7i are equal. Sol: 3x+(2x+y)i = 9+7i Real parts are equal ⇒ 3x = 9 ∴ x = 3 Imaginary parts are equal ⇒ 2x+y = 7 Putting, x = 3 (2x3)+y = 7 ∴ y = 1 #### Summary Imaginary number Solutions of this quadratic equation  x2 - 4 = 0 are 2 or -2 . The solutions are real. Solutions of the quadratic equation x2 + 4 = 0 are $±\sqrt{\text{-4}}$ The square of a real number is always non-negative. Solutions of the quadratic equation x2 + 4 = 0 are not real. $±\sqrt{\text{-4}}$ is written as x $\sqrt{\text{-1}}$. A number whose square is negative is called an imaginary number. Complex number Let  $\sqrt{\text{-1}}$ be denoted by the letter "i", i.e. i2 = -1. ⇒ x = ± 2 x i ⇒ x = 2i or - 2i Therefore, the solutions of the quadratic equation x2 + 4 = 0 are 2i or - 2i. These numbers are called complex numbers. The solution of the quadratic equation ax2+bx+c = 0, where D = b2 - 4ac < 0 , is not possible in the real number system. Such equations will always have complex roots. A number of the form a+ib, where a and b are real numbers, is called a complex number. Such numbers are denoted by the letter 'z'. z = a + ib Where 'a' is called the real part of the complex number, denoted by Re z and, 'b' is called the imaginary part, denoted by Im z. Ex: z = 3 + 2i, z = 5 + √2i, z = -6 + ($\frac{\text{1}}{\text{3}}$)i In z = 3 + 2i, 3 is the real part and 2 is the imaginary part. If the imaginary part of a complex number is zero, then the number is called a purely real number. Ex: z = 5+0i = 5 If the real part of a complex number is zero, then the number is called a purely imaginary number. Ex: z = 0 + 7i = 7i Equality of complex numbers Consider two complex numbers z1 = a+ib, z2 = x+iy Two complex numbers are said to be equal, if their corresponding real parts and imaginary parts are equal. z1 = z2, if a = x and b = y Ex: Find the values of x and y, if the complex numbers 3x+(2x+y)i, 9+7i are equal. Sol: 3x+(2x+y)i = 9+7i Real parts are equal ⇒ 3x = 9 ∴ x = 3 Imaginary parts are equal ⇒ 2x+y = 7 Putting, x = 3 (2x3)+y = 7 ∴ y = 1 Previous
2021-06-16T16:40:53
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https://proofwiki.org/wiki/Abel%27s_Lemma/Formulation_1
# Abel's Lemma/Formulation 1 ## Lemma Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$. Then: $\displaystyle \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k} = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}$ Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$. ### Corollary $\displaystyle \sum_{k \mathop = 1}^n a_k \left({b_{k + 1} - b_k}\right) = a_{n + 1} b_{n + 1} - a_1 b_1 - \sum_{k \mathop = 1}^n \left({a_{k + 1} - a_k}\right) b_{k + 1}$ ## Proof $\displaystyle \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k}$ $=$ $\displaystyle \sum_{k \mathop = m}^n a_k b_{k + 1} - \sum_{k \mathop = m}^n a_k b_k$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = m}^n a_k b_{k + 1} - \paren {a_m b_m + \sum_{k \mathop = m}^n a_{k + 1} b_{k + 1} - a_{n + 1} b_{n + 1} }$ $\displaystyle$ $=$ $\displaystyle a_{n + 1} b_{n + 1} - a_m b_m + \sum_{k \mathop = m}^n a_k b_{k + 1} - \sum_{k \mathop = m}^n a_{k + 1} b_{k + 1}$ $\displaystyle$ $=$ $\displaystyle a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}$ $\blacksquare$ ## Also reported as Some sources give this as: $\displaystyle \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_k = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n a_{k + 1} \paren {b_{k + 1} - b_k}$ which is obtained from the main result by interchanging $a$ and $b$. Others take the upper index to $n - 1$: $\displaystyle \sum_{k \mathop = m}^{n - 1} \paren {a_{k + 1} - a_k} b_k = a_n b_n - a_m b_m - \sum_{k \mathop = m}^{n - 1} a_{k + 1} \paren {b_{k + 1} - b_k}$ ## Source of Name This entry was named for Niels Henrik Abel.
2020-02-23T02:07:52
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https://www.physicsforums.com/threads/function-in-terms-of-x-with-2-y-variables.379314/
# Homework Help: Function in terms of x with 2 y variables 1. Feb 17, 2010 ### Asphyxiated 1. The problem statement, all variables and given/known data $$\ x = y^4 + y^2 + 1, y \geq 0$$ Find the inverse function of y in terms of x and its derivative dy/dx 2. Relevant equations 3. The attempt at a solution I don't really understand how to find y isolated when there is two y variables, this is the best that I have = $$\ x = y^4 + y^2 + 1$$ $$\ -y^4 + x = y^2 + 1$$ $$\ -y^4 = y^2 - x + 1$$ $$\ -y = ( y^2 - x + 1)^{1/4}$$ $$\ y = (-y^2 + x - 1)^{1/4}$$ If I could find the correct formula I could find the derivative and the inverse but I really have no idea what to do. 2. Feb 17, 2010 ### Dick Try a warmup exercise. Suppose I give you the relation x=y^3 and you want to find f(x)=y in terms of x. That's f(x)=x^(1/3), right? That makes the inverse function to the cube root the cube or, f^(-1)(x)=x^3, right? Everything ok so far? Now notice you could have gotten the inverse function just by interchanging x and y in the original relation, x=y^3, making it y=x^3. That's a general thing. You can find the inverse relation just by interchanging x and y in the original relation. Now take another look at your problem. 3. Feb 17, 2010 ### Asphyxiated So I take it that you are saying that: $$\ x = y^4 + y^2 + 1$$ is just: $$\ y = x^4 + x^2 +1$$ in terms of x right? This would also be the inverse? 4. Feb 17, 2010 ### Dick Yes, the inverse of the relation x=y^4+y^2+1 is y=x^4+x^2+1. In a more usual problem they will tell you to find the inverse function by interchanging x and y and then solving for y. In this case, if you interchange x and y to get the inverse, it's already solved for y. Makes life easy, yes? 5. Feb 17, 2010 ### Asphyxiated Indeed thats great, probably just another thing I forgot from high school algebra or something! I have been hitting those blocks lately as I try to learn calculus on my own, tell me what you think of this though. I posted this same problem on the Facebook Physics forum and this is the answer someone else gave me. $$\ x = y^4 + y^2 + 1$$ $$\ 4x = 4y^4 + 4y^2 + 4$$ $$\ 4x = (2y^{2})^{2} + 2(2y^{2}) + 4$$ $$\ 4x - 4 = (2y^{2})^{2} + 2(2y^{2})$$ $$\ 4x - 4 + 1 = (2y^{2})^{2} + 2(2y^{2}) + 1$$ $$\ 4x -3 = (2y^{2} + 1)^{2}$$ $$\ \sqrt{4x-3} = 2y^{2} + 1$$ $$\ \sqrt{4x - 3} - 1 = 2y^{2}$$ $$\ \frac {\sqrt{ 4x - 3} -1} {2} = y^{2}$$ $$\ \frac {\sqrt{(\sqrt{4x-3}-1})} {\sqrt{2}} = y$$ He said that to do the problem to had to complete the squares, any information as to why he is wrong and you are right or you are both right for different reasons or whats up? Last edited: Feb 18, 2010 6. Feb 17, 2010 ### Dick I'm impressed with the Facebook Physics forum. That's not bad. And it's correct. Now you have y(x) assuming all the root signs are correct. The problem is that the question isn't about y(x), it's about the INVERSE of y(x). Now you have to interchange x and y to get the inverse function and solve for y. Now you have to reverse all of those clever steps and get back to y=x^4+x^2+1. I think the original question is more about realizing, hey, I didn't have to do any of that. Assuming I've read it correctly. 7. Feb 17, 2010 ### Asphyxiated OH I get it now, I didn't say anything about having to find the inverse of the original equation just that I needed to solved for y, which is probably why he gave me that answer instead. Anyway, I know this is going above and beyond what I originally asked but could you possibly explain what he means by "completing the squares". I vaguely remember the term from school but nothing about what was suppose to be done to use it. thanks either way! 8. Feb 17, 2010 ### Dick 9. Feb 17, 2010 ### Asphyxiated Took a quick look at the wiki, I think I got it, thanks again for all your help 10. Feb 18, 2010 ### Asphyxiated Hey dick, or anyone else, can I solve this problem in the way that I did to complete the square? Question: $$\ x = y^{2} + 3y -1$$ Now I understand that I should just swap the x and y to get the inverse but I wanted to get the original equation in terms of x instead of y so this is what I tried: $$\ x = y^{2} + 3y -1$$ $$\ x = (y + \frac {3} {2})^{2} - \frac {13} {4}$$ $$\ x + \frac{13} {4} = (y + \frac {3} {2})^{2}$$ $$\ \pm \sqrt{x + \frac {13} {4}} = y + \frac {3} {2}$$ $$\ y = \pm \sqrt{x + \frac {13} {4}} - \frac {3} {2}$$ Last edited: Feb 19, 2010 11. Feb 18, 2010 ### Dick That's just fine. Don't forget the +/- on the square root. It can be either sign. 12. Feb 18, 2010 ### Asphyxiated Thanks, I changed it to reflect +/-, sorry for the PM i wasn't sure if it would tell you about my post since the original topic was already solved. 13. Feb 19, 2010 ### Dick That's ok. I already get an email notice when you post onto a thread I've been working on. You don't need to send the extra PM. I'll notice.
2018-06-24T21:18:42
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https://math.stackexchange.com/questions/60578/what-is-the-term-for-a-factorial-type-operation-but-with-summation-instead-of-p/60580
# What is the term for a factorial type operation, but with summation instead of products? [duplicate] (Pardon if this seems a bit beginner, this is my first post in math - trying to improve my knowledge while tackling Project Euler problems) I'm aware of Sigma notation, but is there a function/name for e.g. $$4 + 3 + 2 + 1 \longrightarrow 10 ,$$ similar to $$4! = 4 \cdot 3 \cdot 2 \cdot 1 ,$$ which uses multiplication? Edit: I found what I was looking for, but is there a name for this type of summation? ## marked as duplicate by Alex M., Namaste algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 31 '18 at 11:02 • – Lucian Jan 12 '17 at 11:25 • I like to call it "additorial" or "sumitorial" :) – NH. Apr 23 '18 at 14:52 • @AlexM. absolutely not. He's asking for a term, not a proof of an equality. – The Great Duck Oct 31 '18 at 3:56 • @AlexM. That's a stupid reason to mark as duplicate though. When something is marked as duplicate the test says "marked as exact duplicate". Has that been revised since I last saw it? – The Great Duck Oct 31 '18 at 23:16 • This question is obviously not a duplicate of the question that it is marked as a duplicate of. – Greg Schmit Apr 12 at 16:58 The name for $$T_n= \sum_{k=1}^n k = 1+2+3+ \dotsb +(n-1)+n = \frac{n(n+1)}{2} = \frac{n^2+n}{2} = {n+1 \choose 2}$$ is the $n$th triangular number. This picture demonstrates the reasoning for the name: $$T_1=1\qquad T_2=3\qquad T_3=6\qquad T_4=10\qquad T_5=15\qquad T_6=21$$ $\hskip1.7in$ • What does the last notation in brackets mean? Does it have a name? – silkfire Mar 29 '16 at 22:50 • @silkfire - It's called a binomial coefficient – Russell Thackston Apr 4 '16 at 17:33 • @RussellThackston Thanks! – silkfire Apr 4 '16 at 19:37 • How could we adapt this to be used with a known base number ? Let's say we have a base value of 7. Then we need 7+14, the we need 7+14+21 and so forth. Could this be turned into an excel formula ? – Overmind Oct 12 '17 at 6:45 • @Overmind: Note that $$7+14+21+\cdots+(7\times n)=7\times(1+2+3+\cdots+n)=7\times T_n$$ So the formula is quite simple :) – Zev Chonoles Oct 12 '17 at 7:39 Donald Knuth in The Art of Computer Programming calls the $n$-th triangular number the "termial function", and denotes it $$n? = 1 + 2 + ... + n = \sum_{k=1}^n k$$ • Wow, really? What volume/page does he define this termin-ology? – Niel de Beaudrap Aug 30 '11 at 1:06 • Now that I am home, I have the 3rd edition of volume one, it is on page 48. – tlehman Aug 30 '11 at 2:14 • It's not terminal, it's termial. It also doesn't matter why he put it in his books, it is exactly what the questioner was asking about. – tlehman Aug 30 '11 at 12:38 • ah, then I managed to misread it several times. Also –– if you will forgive me –– I was somewhat skeptical that Knuth would deign to give this function a name (especially when I thought that name was supposed to be "terminal", which made little sense to me); I wanted to see for myself, and also see why he would do so. – Niel de Beaudrap Aug 30 '11 at 12:41 • @Niel: concerning "for pedagical reasons": I'd say, the additive analogon of a "factor" in a multiplication is "summand", so then it should rather be called "summorial" or "summatorial" – Gottfried Helms Oct 4 '11 at 6:14 Actually, I've found what I was looking for. From the wiki on Summation: • These numbers are also called the triangular numbers. You might think of the triangular numbers as naming a sequence: 1, 3, 6, 10, 15, 21,... But a sequence of integers is really just a function from $\mathbb{N}$ to $\mathbb{Z}$, so the triangular numbers also name the function you've written above. – Jonas Kibelbek Aug 29 '11 at 20:59 Not exactly a name, but note that $$\sum\limits_{k=1}^{n} k= \frac{n(n+1)}{2}={n+1 \choose 2}$$
2019-06-20T07:00:37
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https://tex.stackexchange.com/questions/254396/intended-paragraphs-inside-environment
# Intended Paragraphs Inside Environment I have decided to come back to beautiful LaTeX typesetting after several years. I have just decided to hone up my LaTeX skills but have found out that it will be not easy. Here is my question: I have found this code from book class file (World Scientific Publishing), which makes a numbered framed box. But this code doesn't allow me to indent paragraphs coming after the 1st paragraph (In the figure below, the second paragraph starts with "Let", which is intended.) I wonder how I indent paragraphs after the first paragraph like the figure below. The code as follows: \documentclass[12pt,a4paper]{book} \usepackage{boxedminipage} \newcounter{boxcnt}[chapter] \renewcommand\theboxcnt{\thechapter.\arabic{boxcnt}} \newenvironment{boxedtxt}[1]{% \vskip6pt \parindent0pt\leftskip0pt \itemindent0pt\leftmargin0pt \fboxsep=8pt \normalfont \refstepcounter{boxcnt} \begin{boxedminipage}[0.5pt]{\hsize} \selectfont {\bfseries{Box\ \theboxcnt\enspace}#1} \vskip 6pt plus2pt minus1pt \par\noindent\ignorespaces }{% \end{boxedminipage}% \vskip6pt\noindent\ignorespacesafterend } \begin{document} \begin{boxedtxt}{Summation and series} Summation is the operation of addition. The result of summation is called the sum. The summation operation can be conveniently indicated by the the summation symbol (the capital sigma, $\sum$). For instance, the sum of the square of the first four natural numbers can be written as $\sum_{k=1}^{4}{k^2}= 1 + 2^2+3^2 +4^2 = 30,$ where $k=1$ in the subscript and 4 in the superscript denote that the summation index $k$ takes integer values from 1 to 4. Let $\{a_k\}_{k=1}^{n} = {a_1+a_2+\cdots.+a_n}$ be a sequence of n numbers bla bla \ldots \end{boxedtxt} It is noted that the matrix bla bla \ldots \end{document} • Welcome to TeX.SX! When I compile your example none of the paragraphs inside the box get indented... – Ruben Jul 8 '15 at 22:13 • Don't set \parindent=0pt. Plain TeX sets \parindent=20pt default. base.sty only uses 0pt and 1em. – John Kormylo Jul 8 '15 at 22:26 • And as a tiny remark: I edited $$...$$ into $...$. You should avoid the first syntactic form to create display math as it is deprecated and non-expacted things might happen. – Ruben Jul 8 '15 at 22:41 • Also \bfseries does not take an argument. It is a switch (i.e. a toggle). So \bfseries{...} ... is just the same as \bfseries ... .... – cfr Jul 8 '15 at 22:55 I've corrected the code so that it does what you want. However I also propose another similar environment, that I've called fboxedtxt, based on the framed package, which has the advantage to break across pages if necessary. Also, if required, it's not too difficult to modify the code to obtain coloured frames and/or shaded environment. \documentclass[12pt,a4paper]{book} \usepackage{fourier, erewhon} \usepackage[showframe]{geometry} \usepackage[x11names]{xcolor} \usepackage{boxedminipage} \usepackage{framed} \usepackage{amsmath} \newcounter{boxcnt}[chapter] \renewcommand\theboxcnt{\thechapter.\arabic{boxcnt}} \newenvironment{boxedtxt}[1] {\vskip6pt \leftskip0pt \itemindent0pt\leftmargin0pt \fboxsep=8pt \normalfont \refstepcounter{boxcnt}\noindent \begin{boxedminipage}[0.5pt]{\hsize}\selectfont{\bfseries{Box\ \theboxcnt\enspace}#1}\vskip 6pt plus2pt minus1pt \par\noindent\ignorespaces\parindent1em} {\end{boxedminipage}% \vskip6pt \noindent\ignorespacesafterend} \newcounter{fboxcnt}[chapter] \renewcommand\thefboxcnt{\thechapter.\arabic{fboxcnt}} \newenvironment{fboxedtxt}[1] { \FrameSep=8pt \OuterFrameSep=6pt \FrameRule=1pt \normalfont \refstepcounter{fboxcnt}\noindent \begin{oframed}\noindent\selectfont{\bfseries{Box\ \theboxcnt\enspace}#1}\vskip 6pt plus2pt minus1pt \par\noindent\ignorespaces} {\end{oframed}% \vskip6pt \noindent\ignorespacesafterend} \begin{document} Some text. Sometext. Some text. Sometext. Some text. Sometext. Some text. Sometext. \setcounter{chapter}{3}\setcounter{boxcnt}{1} \begin{boxedtxt}{Summation and series} Summation is the operation of addition. The result of summation is called the sum. The summation operation can be conveniently indicated by the the summation symbol (the capital sigma, $\sum$). For instance, the sum of the square of the first four natural numbers can be written as $\sum_{k=1}^{4}{k^2}= 1 + 2^2+3^2 +4^2 = 30,$ where $k=1$ in the subscript and 4 in the superscript denote that the summation index $k$ takes integer values from 1 to 4. Let $\{a_k\}_{k=1}^{n} = \bigl\{a_1,a_2,\dots, a_n\bigr\}$ be a sequence of $n$ numbers in which each term $a_k$ is given by a certain rule, whose character is here irrelevant. A series is the sum of a sequence and can be written in summation notation as $\sum_{k = 1}^{n} = a_1 + a_2 + \dotsm + a_n$% If the number of terms is finite, the series is called a finite series; otherwise it is called an infinite series. An example of an infinite series is the geometric series $\sum_{k = 0}^{\infty}\frac{1}{2^k} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dotsm$% \end{boxedtxt} It is noted that the matrix bla bla \ldots \setcounter{boxcnt}{1} \begin{fboxedtxt}{Summation and series} Summation is the operation of addition. The result of summation is called the sum. The summation operation can be conveniently indicated by the the summation symbol (the capital sigma, $\sum$). For instance, the sum of the square of the first four natural numbers can be written as $\sum_{k=1}^{4}{k^2}= 1 + 2^2+3^2 +4^2 = 30,$ where $k=1$ in the subscript and 4 in the superscript denote that the summation index $k$ takes integer values from 1 to 4. Let $\{a_k\}_{k=1}^{n} = \bigl\{a_1,a_2,\dots, a_n\bigr\}$ be a sequence of $n$ numbers in which each term $a_k$ is given by a certain rule, whose character is here irrelevant. A series is the sum of a sequence and can be written in summation notation as $\sum_{k = 1}^{n} = a_1 + a_2 + \dotsm + a_n$% If the number of terms is finite, the series is called a finite series; otherwise it is called an infinite series. An example of an infinite series is the geometric series $\sum_{k = 0}^{\infty}\frac{1}{2^k} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dotsm$% \end{fboxedtxt} \end{document} • "Summation" isn't indented, though the OP required this. – Ruben Jul 8 '15 at 23:39 • ??? I understood the first paragraph after title should not be intended (as is standard in English, by the way). Did I misunderstand? – Bernard Jul 8 '15 at 23:45 • I guess you don't misunderstood. I was confused because in the original version of the MWE nothing was indented... – Ruben Jul 8 '15 at 23:49 • Nice alternative environment by the way! – Ruben Jul 8 '15 at 23:49 • In my code, the first paragraph was not indented (which I wanted) but after the first paragraph, the code doesn't allow me to indent the others. So this property is similar to LaTeX book class: no indentation in the first paragraph in a section, but indentation for paragraphs after it. – ofenerci Jul 9 '15 at 7:17
2019-11-21T13:16:12
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https://mathoverflow.net/questions/369444/balls-in-hilbert-space
# Balls in Hilbert space I recently noticed an interesting fact which leads to a perhaps difficult question. If $$n$$ is a natural number, let $$k_n$$ be the smallest number $$k$$ such that an open ball of radius $$k$$ in a real Hilbert space of sufficiently large dimension or infinite dimension contains $$n$$ pairwise disjoint open balls of radius 1. (The dimension of the Hilbert space is irrelevant as long as it is at least $$n-1$$ since it can be replaced by the affine subspace spanned by the centers of the balls.) We obviously have $$k_1=1$$ and $$k_2=2$$, and it is easy to see that $$k_3=1+\frac{2}{\sqrt{3}}\approx 2.1547$$. The interesting fact is that $$k_n\leq 1+\sqrt{2}\approx 2.414$$ for all $$n$$, since in an infinite-dimensional Hilbert space an open ball of this radius contains infinitely many pairwise disjoint open balls of radius 1 [consider balls centered at points of an orthonormal basis]. The obvious questions are: (1) What is $$k_n$$? This may be known, but looks difficult since it is related to sphere packing. (2) Is $$k_n$$ even strictly increasing in $$n$$? (3) Is $$k_n<1+\sqrt{2}$$ for all $$n$$, or are they equal for sufficiently large $$n$$? (4) Is it even true that $$\sup_n k_n=1+\sqrt{2}$$? It is not even completely obvious that $$k_n$$ exists for all $$n$$, i.e. that there is a smallest $$k$$ for each $$n$$, but there should be some compactness argument which shows this. I find it interesting that the numbers $$1+\frac{2}{\sqrt{3}}$$ and $$1+\sqrt{2}$$ are so close but the behavior of balls is so dramatically different. I suppose the question is also interesting in smaller-dimensional Hilbert spaces: let $$k_{n,d}$$ be the smallest $$k$$ such that an open ball of radius $$k$$ in a Hilbert space of dimension $$d$$ contains $$n$$ pairwise disjoint open balls of radius 1. Then $$k_{n,d}$$ stabilizes at $$k_n$$ for $$d\geq n-1$$. What is $$k_{n,d}$$? (This my be much harder since it is virtually the sphere-packing question if $$n>>d$$.) • Perhaps $k_n=1+\sqrt{2(1-1/n)}$? – aorq Aug 18, 2020 at 8:29 For convenience of notation, let me write the expectation $$\mathop{\mathbb{E}}_i t_i$$ to denote the average $$(\sum_{i=1}^n t_i)/n$$. If I understand your construction correctly, you have disjoint balls of radius $$1$$ centered at $$x_i = \sqrt{2} e_i$$ contained in a ball of radius $$1+\sqrt{2}$$ centered at $$y = 0$$. This construction, which places $$n$$ balls tightly packed at the vertices of a regular simplex, is optimal in terms of the positions $$x_i$$. For the exact optimal bound for your problem, you should pick $$y=\mathop{\mathbb{E}}_i x_i$$ to get the radius $$\boxed{k_n = 1+\sqrt{2 (1-1/n)}}.$$ The claim that placing the $$x_i$$ at the vertices of a regular $$(n-1)$$-simplex and $$y$$ at the centroid of this simplex is optimal has been proven many times before in many different contexts. For example, it is implied by a bound known by various substrings of "the Welch-Rankin simplex bound" in frame theory. Here's a simple direct proof: By the triangle inequality, a ball of radius $$1+r$$ centered at $$y$$ contains a ball of radius $$1$$ centered at $$x_i$$ iff $$\lVert x-y\rVert \le r$$. Two balls of radius $$1$$ centered at $$x_i$$ and $$x_j$$ are disjoint iff $$\lVert x_i - x_j \rVert \ge 2$$. Therefore, your problem asks to minimize $$1 + \max_i \lVert y-x_i\rVert$$ subject to $$\min_{i\ne j} \lVert x_i - x_j\rVert \ge 2$$. Working with squared distances is easier. The maximum squared distance $$\max_i \lVert y-x_i\rVert^2$$ is surely at least the average $$\mathop{\mathbb{E}}_i \lVert y-x_i\rVert^2$$. This average is minimized when $$y$$ is itself the average $$\mathop{\mathbb{E}}_i x_i$$, in which case it equals $$\mathop{\mathbb{E}}_i \mathop{\mathbb{E}}_j \lVert x_i-x_j\rVert^2/2$$. Each term where $$i=j$$ contributes $$0$$ to this expectation, while each term where $$i\ne j$$ contributes at least $$2$$, so overall this expectation is at least $$2(n-1)/n$$. Thus the maximum squared distance $$\max_i\lVert y-x_i\rVert^2$$ is at least $$2(n-1)/n$$ and thus $$1+r \ge 1+\sqrt{2(n-1)/n}.$$ We can check that the optimal configuration mentioned before achieves this bound either by direct calculation or by noting that it achieves equality in every step of our argument.
2022-06-29T00:08:59
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https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_25&diff=164819&oldid=83848
# Difference between revisions of "2017 AMC 12B Problems/Problem 25" ## Problem A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$-player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n$, $9\leq n\leq 2017$, can be the number of participants? $\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562$ ## Solution Solution by Pieater314159 Minor edits by Zeric Let there be $T$ teams. For each team, there are ${n-5\choose 4}$ different subsets of $9$ players including that full team, so the total number of team-(group of 9) pairs is $$T{n-5\choose 4}.$$ Thus, the expected value of the number of full teams in a random set of $9$ players is $$\frac{T{n-5\choose 4}}{{n\choose 9}}.$$ Similarly, the expected value of the number of full teams in a random set of $8$ players is $$\frac{T{n-5\choose 3}}{{n\choose 8}}.$$ The condition is thus equivalent to the existence of a positive integer $T$ such that $$\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.$$ $$T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1$$ $$T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}$$ $$T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}$$ $$T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}$$ $$T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}$$ Note that this is always less than ${n\choose 5}$, so as long as $T$ is integral, $n$ is a possibility. Thus, we have that this is equivalent to $$2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).$$ It is obvious that $5$ divides the RHS, and that $7$ does iff $n\equiv 0,1,2,3,4\mod 7$. Also, $3^2$ divides it iff $n\not\equiv 5,8\mod 9$. One can also bash out that $2^5$ divides it in $16$ out of the $32$ possible residues $\mod 32$. Note that $2016 = 7*9*32$ so by using all numbers from $2$ to $2017$, inclusive, it is clear that each possible residue $\mod 7,9,32$ is reached an equal number of times, so the total number of working $n$ in that range is $5\cdot 7\cdot 16 = 560$. However, we must subtract the number of "working" $2\leq n\leq 8$, which is $3$. Thus, the answer is $\boxed{\textbf{(D) } 557}$. Alternatively, it is enough to approximate by finding the floor of $2017 \cdot \frac57 \cdot \frac79 \cdot \frac12 - 3$ to get $\boxed{\textbf{(D) } 557}$. Video Solution by Dr. Nal: ## See Also 2017 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS
2022-07-01T16:21:24
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http://mathhelpforum.com/calculus/17143-maclaurin-series.html
# Thread: maclaurin series 1. ## maclaurin series How would you find the maclaurin series for this sin(x^5) thanks. 2. Originally Posted by davecs77 How would you find the maclaurin series for this sin(x^5) thanks. recall that the Maclaurin series for sine is: $\sin x = \sum_{n = 0}^{\infty} \frac {(-1)^n}{(2n + 1)!} x^{2n + 1} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +...$ for ALL $x$ so for $\sin \left( x^5 \right)$, just replace $x$ with $x^5$ in the series above and simplify 3. Originally Posted by davecs77 How would you find the maclaurin series for this sin(x^5) thanks. A McLaurin series is a Taylor series for x = 0. So we have: $f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x + \frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...$ $f(x) = sin(x^5)$ So $f(0) = 0$ $f^{\prime}(x) = cos(x^5) \cdot 4x^5$ ==> $f^{\prime}(0) = 0$ $f^{\prime \prime}(x) = 20x^3 cos(x^5) - 25x^8sin(x^5)$ ==> $f^{\prime \prime}(0) = 0$ $f^{\prime \prime \prime}(x) = -125x^{12}cos(x^5) + 60x^2 cos(x^5) - 300x^7sin(x^5)$ ==> $f^{\prime \prime \prime}(0) = 0$ etc. I'll give you a hint. After a lot of hard work on your part, you will find that $f^{(6)}(0)$ is the first non-zero term in the series. (This one's just nasty to do by hand!) -Dan 4. Originally Posted by Jhevon recall that the Maclaurin series for sine is: $\sin x = \sum_{n = 0}^{\infty} \frac {(-1)^n}{(2n + 1)!} x^{2n + 1} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +...$ for ALL $x$ so for $\sin \left( x^5 \right)$, just replace $x$ with $x^5$ in the series above and simplify So you are just memorizing that for sin(x) or Maclaurn series? Would that be the same for Taylor series? 5. Originally Posted by davecs77 So you are just memorizing that for sin(x) or Maclaurn series? Would that be the same for Taylor series? I don't like telling people to just memorize stuff, but yes, you should have the Maclaurin series for functions like sine, cosine, e^x, 1/(1 - x) ... memorized. the Maclaurin series is just the Taylor series centered at zero. topsquark showed you how to derive it, but in this case, i think the memorization route is easiest. unless your professor requires you to derive it 6. Originally Posted by Jhevon recall that the Maclaurin series for sine is: $\sin x = \sum_{n = 0}^{\infty} \frac {(-1)^n}{(2n + 1)!} x^{2n + 1} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +...$ for ALL $x$ so for $\sin \left( x^5 \right)$, just replace $x$ with $x^5$ in the series above and simplify Or you can take Jhevon's pansy-butted approach and skip all the work. But why do that when you could do all that work? -Dan 7. how would you find the Maclaurin series for this then: 1/(1+x^5) 8. Originally Posted by topsquark But why do that when you could do all that work? -Dan because mathematicians don't like hard work. there are some things you want to try once just to see if you can do them, and then afterwards you take the shortcut. finding Taylor/Maclaurin series expansions are such things. "pansy-butted"? 9. Originally Posted by Jhevon "pansy-butted"? It was the best I could come up with on short notice. Hey, I'm a Mormon. I'm not supposed to swear so that limits my insulting remarks. -Dan 10. Originally Posted by davecs77 how would you find the Maclaurin series for this then: 1/(1+x^5) Does anyone know? I don't think you can memorize this one. Thanks. 11. Originally Posted by davecs77 how would you find the Maclaurin series for this then: 1/(1+x^5) $\frac{1}{1+x}=1-x+x^2-x^3+...$ You do the rest. Note: You need to remember the most important power series as well as you name. 12. Originally Posted by davecs77 Does anyone know? I don't think you can memorize this one. Thanks. actually you can. this happens to be one of the easiest to memorize in fact. the appropriate one to remember is $\frac {1}{1 - x} = \sum_{n = 0}^{\infty} x^n = 1 + x + x^2 + x^3 + ...$ for $|x|<1$ 13. Now me, I tend to forget that $\sum_{i = 1}^n 1 = n$. (I try not to memorize too much! ) -Dan 14. Originally Posted by ThePerfectHacker $\frac{1}{1+x}=1-x+x^2-x^3+...$ You do the rest. Note: You need to remember the most important power series as well as you name. The answer in the book is summation from n = 1 to infinity of (-1)^n times x^(5n) I see the (-1)^n since the values are changing from positive to negative every time, but where does the x^(5n) come from? Also how did you find your results... 1 - x + x^2 - x^3 ... I don't quite understand. Thank you. 15. Originally Posted by davecs77 The answer in the book is summation from n = 1 to infinity of (-1)^n times x^(5n) I see the (-1)^n since the values are changing from positive to negative every time, but where does the x^(5n) come from? Also how did you find your results... 1 - x + x^2 - x^3 ... I don't quite understand. Thank you. he treated $\frac {1}{1 + x}$ as $\frac {1}{1 - (-x)}$ and then used the formula i gave you. where did the $x^{5n}$ come from? did you actually look at the series? $\sum_{n = 0}^{\infty}(-1)^n\left( x^5 \right)^n = \sum_{n = 0}^{\infty}(-1)^nx^{5n}$
2017-10-22T18:09:13
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https://gmatclub.com/forum/if-x-is-a-positive-integer-what-is-the-value-of-x-121669.html
GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 21 Sep 2018, 18:40 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # If x is a positive integer, what is the value of x ? Author Message TAGS: ### Hide Tags Intern Joined: 23 Aug 2011 Posts: 34 If x is a positive integer, what is the value of x ?  [#permalink] ### Show Tags 07 Oct 2011, 22:13 1 13 00:00 Difficulty: 95% (hard) Question Stats: 38% (01:17) correct 62% (01:18) wrong based on 295 sessions ### HideShow timer Statistics If x is a positive integer, what is the value of x ? (1) The first nonzero digit in the decimal expansion of $$\frac{1}{x!}$$ is in the hundredths place. (2) The first nonzero digit in the decimal expansion of $$\frac{1}{(x+1)!}$$ is in the thousandths place. Retired Moderator Joined: 20 Dec 2010 Posts: 1868 ### Show Tags 08 Oct 2011, 00:49 2 4 kkalyan wrote: Previous Next Help End Exam Review Section If $$x$$ is a positive integer, what is the value of $$x$$ ? The first nonzero digit in the decimal expansion of $$\frac{1}{x!}$$ is in the hundredths place. The first nonzero digit in the decimal expansion of $$\frac{1}{(x+1)!}$$ is in the thousandths place. Used brute force: 1/1!=1 1/2!=0.5 1/3!=0.1XX 1/4!=0.04XX 1/5!=0.008XX 1/6!=0.001XX 1) x=4 Sufficient. 2) x=4 x=5 Not Sufficient. Ans: "A" _________________ ##### General Discussion Intern Joined: 01 Oct 2011 Posts: 5 Location: United States (NY) Concentration: Finance, Technology GMAT 1: 690 Q49 V34 GMAT 2: 710 Q49 V38 GPA: 3.5 WE: Information Technology (Computer Software) Re: If x is a positive integer, what is the value of x  [#permalink] ### Show Tags 18 Oct 2011, 20:26 1 Hi, i am a bit confused, Can some body please elaborate why the answer is not D) ? Because from first statement, the x! is 4 leading to hundredths place on decimal, then similarly from 2nd statement 5! leads to thousandth place, therefore x+1 = 5, hence x= 4 ? Senior Manager Joined: 28 Apr 2012 Posts: 297 Location: India Concentration: Finance, Technology GMAT 1: 650 Q48 V31 GMAT 2: 770 Q50 V47 WE: Information Technology (Computer Software) Re: If x is a positive integer, what is the value of x ? The  [#permalink] ### Show Tags 03 Apr 2014, 12:30 2 1 Encountered this in Manhattan Quiz: Posting OE for others. 700 level q. The question does not need rephrasing, although we should note that x is a positive integer. Statement 1: SUFFICIENT. We should work from the inside out by first listing the first several values of x! (the factorial of x, defined as the product of all the positive integers up to and including x). 1!=1 2!=2 3!=6 4!=24 5!=120 6!=720 7!=5040 Now we consider decimal expansions whose first nonzero digit is in the hundredths place. Such decimals must be smaller than 0.1 (110) but at least as large as 0.01 (1100). Therefore, for 1x! to lie in this range, x! must be larger than 10 but no larger than 100. The only factorial that falls between 10 and 100 is 4!=24, so x=4. (Note that factorials are akin to exponents in the order of operations, so 1x! indicates "1 divided by the factorial of x," not "the factorial of 1x," which would only have meaning if 1x were a positive integer.) Statement 2: INSUFFICIENT. We consider decimal expansions whose first nonzero digit is in the thousandths place. Such decimals must be smaller than 0.01 (1100) but at least as large as 0.001 (11000). Therefore, for 1(x+1)! to lie in this range, (x+1)! must be larger than 100 but no larger than 1,000. There are two factorials that fall between 100 and 1,000, namely 5!=120 and 6!=720. Thus, x+1 could be either 5 or 6, and x could be either 4 or 5. _________________ "Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! VP Joined: 09 Jun 2010 Posts: 1038 Re: If x is a positive integer, what is the value of x ?  [#permalink] ### Show Tags 27 Apr 2015, 07:35 kkalyan wrote: If x is a positive integer, what is the value of x ? (1) The first nonzero digit in the decimal expansion of $$\frac{1}{x!}$$ is in the hundredths place. (2) The first nonzero digit in the decimal expansion of $$\frac{1}{(x+1)!}$$ is in the thousandths place. 1/100>1/x! >1/1000 x must be 4 if x=5, it is not ok if x=3, it is not ok 2, x can be 6 or 6, not sufficient HARD ONE _________________ visit my facebook to help me. on facebook, my name is: thang thang thang VP Joined: 09 Jun 2010 Posts: 1038 Re: If x is a positive integer, what is the value of x ?  [#permalink] ### Show Tags 15 May 2015, 02:17 fluke wrote: kkalyan wrote: Previous Next Help End Exam Review Section If $$x$$ is a positive integer, what is the value of $$x$$ ? The first nonzero digit in the decimal expansion of $$\frac{1}{x!}$$ is in the hundredths place. The first nonzero digit in the decimal expansion of $$\frac{1}{(x+1)!}$$ is in the thousandths place. Used brute force: 1/1!=1 1/2!=0.5 1/3!=0.1XX 1/4!=0.04XX 1/5!=0.008XX 1/6!=0.001XX 1) x=4 Sufficient. 2) x=4 x=5 Not Sufficient. Ans: "A" can not say a word for this excellence. very good. _________________ visit my facebook to help me. on facebook, my name is: thang thang thang Non-Human User Joined: 09 Sep 2013 Posts: 8131 Re: If x is a positive integer, what is the value of x ?  [#permalink] ### Show Tags 02 Jul 2018, 22:50 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: If x is a positive integer, what is the value of x ? &nbs [#permalink] 02 Jul 2018, 22:50 Display posts from previous: Sort by # Events & Promotions Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
2018-09-22T01:40:04
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https://doc.simo.com.hk/Functions/cumprod/
# cumprod Cumulative product ### p = cumprod(A) • A is an array of any number of dimensions. • The output argument p contains the cumulative products and has the same size as A. • The cumulative products are obtained from vectors along the first non-singleton dimension of A. Example 1: In the following, the first non-singleton dimension of A is the first dimension. The product operation is performed along the vertical direction. For instance, s(:,1) contains cumulative products of elements of A(:,1). % Matrix of size [3,4] a=reshape(1:12,3,4) % Cumulative products cumprod(a) a = 1.000 4.000 7.000 10.00 2.000 5.000 8.000 11.00 3.000 6.000 9.000 12.00 ans = 1.000 4.000 7.000 10.00 2.000 20.00 56.00 110.0 6.000 120.0 504.0 1320. ### p = cumprod(A, dim) • dim should be a positive integer scalar, not equal to inf or nan. • It obtains the cumulative products along the dim-th dimension of A. • If size(A,dim) == 1 or dim > ndims(A), then p is the same as A. Example 2: In the following, The product operation is performed along the horizontal direction (2nd dimension). For instance, p(1,:) contains cumulative products of elements of A(1,:). % Matrix of size [3,4] A=reshape(1:12,3,4) % Cumulative products along 2nd dimension p=cumprod(A,2) A = 1.000 4.000 7.000 10.00 2.000 5.000 8.000 11.00 3.000 6.000 9.000 12.00 p = 1.000 4.000 28.00 280.0 2.000 10.00 80.00 880.0 3.000 18.00 162.0 1944. ### p = cumprod(A, option) • option should be either 'reverse', 'forward', 'includenan' or 'omitnan'. • These options control the operation direction, or whether or not NaN should be included in the operation. For details, see Tables 1 and 2 below. Note By default, cummin and cummax omit NaN, whereas cumsum and cumprod include NaN. Example 3: Cumulative products of the same vector but with different options. a=[1:5 nan 6] % Default options cumprod(a) % Include nan, forward direction cumprod(a, 'includenan') % Omit nan, forward direction cumprod(a, 'omitnan') % Forward direction, include nan cumprod(a, 'forward') % Reverse direction, include nan cumprod(a, 'reverse') a = 1.000 2.000 3.000 4.000 5.000 nan 6.000 ans = 1.000 2.000 6.000 24.00 120.0 nan nan ans = 1.000 2.000 6.000 24.00 120.0 nan nan ans = 1.000 2.000 6.000 24.00 120.0 120.0 720.0 ans = 1.000 2.000 6.000 24.00 120.0 nan nan ans = nan nan nan nan nan nan 6.000 Table 1: Options for operation direction. Option value Meaning Default 'forward' The cumulative products are obtained in the forward direction. YES 'reverse' The cumulative products are obtained in the reverse direction. NO Table 2: Options for including or omitting NaN. Option value Meaning Default 'includenan' Include NaN in the operation YES 'omitnan' Omit NaN in the opeartion NO ### p = cumprod(A, dim, option) • Same as p = cumprod(A, dim) above, except that an option can be specified for controlling the operation direction or whether or not NaN should be included. • See Tables 1 and 2 above for available options. ### p = cumprod(A, option1, option2) • Same as p = cumprod(A, option) above, except that two options can be specified at the same time. • option2 will overwrite option1 if they contradict each other. • See Tables 1 and 2 above for available options. ### p = cumprod(A, dim, option1, option2) • Same as p = cumprod(A, dim, option) above, except that two options can be specified at the same time. • option2 will overwrite option1 if they contradict each other. • See Tables 1 and 2 above for available options.
2019-02-19T20:48:10
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https://math.stackexchange.com/questions/3575647/binomial-distribution-and-function-of-several-random-variables
# binomial distribution and function of several random variables Let $$X_1$$ and $$X_2$$ be independent random variables with respective binomial distributions $$b(3,1/2)$$ and $$b(5,1/2)$$. Find $$P(X_1+X_2=7)$$. I am thinking of an example I pulled from my notes: Ex: let $$X_1$$ ~ Poisson $$(\lambda_1=2)$$ and $$X_2$$~Poisson$$(\lambda_2=3)$$ be independent, then $$P(X_1+X_2=1)=P(X_1=0,X_2=1)+P(X_1=1,X_2=0)=(\lambda_1+\lambda_2)e^{-(\lambda_1+\lambda_2)}$$ So my thoughts for this question: $$P(x,p)=\binom{n}{x}(p)^x (1-p)^{n-x}$$. So for this question, should I cconsider by cases, like $$X_1=1, X_2=3$$ etc, and since they are independent, then I would add those cases together? • Think what is the distribution of $X_1+X_2$, you need not formulas for it. – kludg Mar 10 '20 at 3:02 • can you specify? @kludg – shine Mar 10 '20 at 3:08 The binomial distribution with amount $$n$$ and success rate $$p$$, is more often denoted $$\mathcal{Bin}(n,p)$$. $$Z\sim\mathcal{Bin}(n,p)$$ denotes that random variable, $$Z$$ is a count of successes among $$n$$ independent Bernoulli trials with identical success rate $$p$$. So given that the independent random variables are $$X_1\sim\mathcal{Bin}(3, 1/2)$$ and $$X_2\sim\mathcal{Bin}(5,1/2)$$, what does $$X_1+X_2$$ count?   What is the distribution for this? $$X_1$$ counts successes among $$3$$ independent Bernoulli trials with identical success rate $$1/2$$, and independently $$X_2$$ counts successes among $$7$$ independent Bernoulli trials with identical success rate $$1/2$$. So $$X_1+X_2$$ counts... Use this. Okay, it is possible to use the Law of Total Probability. Now, when the sum is $$7$$ the minimum $$X_1$$ may be is $$2$$ (because the maximum $$X_2$$ may be is $$5$$), so … \begin{align}\mathsf P(X_1+X_2=7)~&=~\sum_{k=2}^3\mathsf P(X_1=k)~\mathsf P(X_2=7-k)\\[1ex]&=~\mathsf P(X_1=2)\,\mathsf P(X_2=5)+\mathsf P(X_1=3)\,\mathsf P(X_2=4)\\[1ex]&~~\vdots\end{align} • I am still not sure of this , can you specify even more on counting $X_1+X_2$ and its distribution? – shine Mar 10 '20 at 3:23 • @shine … No. .. – Graham Kemp Mar 10 '20 at 3:47
2021-01-27T14:02:50
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