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In our question, however, you have already selected a card from the hat and it shows a black face. At first glance it appears that there is a 50/50 chance (ie. probability 1/2) that the other side of the card is black, since there are two cards it might be: the black and the mixed. However, this reasoning fails to exploit all of your information; you know not only that the card on the table has a black face, but also that one of its black faces is facing you.
### Solutions:
#### Intuition
Intuition tells you that you are choosing a card at random. However, you are actually choosing a face at random. There are 6 faces, of which 3 faces are white and 3 faces are black. Two of the 3 black faces belong to the same card. The chance of choosing one of those 2 faces is 2/3. Therefore, the chance of flipping the card over and finding another black face is also 2/3. Another way of thinking about it is that the problem is not about the chance that the other side is black, it’s about the chance that you drew the all black card. If you drew a black face, then it’s twice as likely that that face belongs to the black card than the mixed card.
#### Labels
One solution method is to label the card faces, for example numbers 1 through 6.[2] Label the faces of the black card 1 and 2; label the faces of the mixed card 3 (black) and 4 (white); and label the faces of the white card 5 and 6. The observed black face could be 1, 2, or 3, all equally likely; if it is 1 or 2, the other side is black, and if it is 3, the other side is white. The probability that the other side is black is 2/3.
#### Bayes’ theorem | {
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#### Bayes’ theorem
Given that the shown face is black, the other face is black if and only if the card is the black card. If the black card is drawn, a black face is shown with probability 1. The total probability of seeing a black face is 1/2; the total probability of drawing the black card is 1/3. By Bayes’ theorem,[3] the conditional probability of having drawn the black card, given that a black face is showing, is
$\frac{1\cdot1/3}{1/2}=2/3.$
#### Eliminating the white card
Although the incorrect solution reasons that the white card is eliminated, one can also use that information in a correct solution. Modifying the previous method, given that the white card is not drawn, the probability of seeing a black face is 3/4, and the probability of drawing the black card is 1/2. The conditional probability of having drawn the black card, given that a black face is showing, is
$\frac{1/2}{3/4}=2/3.$
#### Symmetry
The probability (without considering the individual colors) that the hidden color is the same as the displayed color is clearly 2/3, as this holds if and only if the chosen card is black or white, which chooses 2 of the 3 cards. Symmetry suggests that the probability is independent of the color chosen. (This can be formalized, but requires more advanced mathematics than yet discussed.)
#### Experiment
Using specially constructed cards, the choice can be tested a number of times. By constructing a fraction with the denominator being the number of times “B” is on top, and the numerator being the number of times both sides are “B”, the experimenter will probably find the ratio to be near 2/3. | {
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Note the logical fact that the B/B card contributes significantly more (in fact twice) to the number of times “B” is on top. With the card B/W there is always a 50% chance W being on top, thus in 50% of the cases card B/W is drawn, card B/W virtually does not count. Conclusively, the cards B/B and B/W are not of equal chances, because in the 50% of the cases B/W is drawn, this card is simply “disqualified”.
Here's Monty!
First, visit — Monty Hall Problem. Don’t read past ‘Solution.’
Attempt to solve. Write down method of solution.
Carry out ‘Simulation.’
Set Operations (A Wiki Rip)
There are ways to construct new sets from existing ones. Two sets can be “added” together. The union of A and B, denoted by A ∪ B, is the set of all things which are members of eitherA or B.
The union of A and B, or A ∪ B
Examples:
• {1, 2} ∪ {red, white} = {1, 2, red, white}.
• {1, 2, green} ∪ {red, white, green} = {1, 2, red, white, green}.
• {1, 2} ∪ {1, 2} = {1, 2}.
Some basic properties of unions are:
• A ∪ B = B ∪ A.
• A ∪ (B ∪ C) = (A ∪ B) ∪ C.
• A ⊆ (A ∪ B).
• A ∪ A = A.
• A ∪ ∅ = A.
• A ⊆ B if and only if A ∪ B = B.
### Intersections
A new set can also be constructed by determining which members two sets have “in common”. The intersection of A and B, denoted by A ∩ B, is the set of all things which are members of both A and B. If A ∩ B = ∅, then A and B are said to be disjoint.
The intersection of A and B, orA ∩ B.
Examples:
• {1, 2} ∩ {red, white} = ∅.
• {1, 2, green} ∩ {red, white, green} = {green}.
• {1, 2} ∩ {1, 2} = {1, 2}.
Some basic properties of intersections:
• A ∩ B = B ∩ A.
• A ∩ (B ∩ C) = (A ∩ B) ∩ C.
• A ∩ B ⊆ A.
• A ∩ A = A.
• A ∩ ∅ = ∅.
• A ⊆ B if and only if A ∩ B = A.
### Complements | {
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### Complements
Two sets can also be “subtracted”. The relative complement of A in B (also called the set theoretic difference of B and A), denoted by B \A, (or B −A) is the set of all elements which are members of B, but not members of A. Note that it is valid to “subtract” members of a set that are not in the set, such as removing the element green from the set {1, 2, 3}; doing so has no effect.
In certain settings all sets under discussion are considered to be subsets of a given universal set U. In such cases, U \ A is called the absolute complement or simply complement of A, and is denoted by A′.
The relative complement
of A in B.
The complement of A in U.
Examples:
• {1, 2} \ {red, white} = {1, 2}.
• {1, 2, green} \ {red, white, green} = {1, 2}.
• {1, 2} \ {1, 2} = ∅.
• {1, 2, 3, 4} \ {1, 3} = {2, 4}.
• If U is the set of integers, E is the set of even integers, and O is the set of odd integers, then the complement of E in U is O, or equivalently, E′ = O.
Some basic properties of complements:
• A ∪ A′ = U.
• A ∩ A′ = ∅.
• (A′)′ = A.
• A \ A = ∅.
• U′ = ∅ and ∅′ = U.
• A \ B = A ∩ B.
### Cartesian product
A new set can be constructed by associating every element of one set with every element of another set. The Cartesian product of two sets A and B, denoted by A × B is the set of all ordered pairs (ab) such that a is a member of A and b is a member of B.
Examples:
• {1, 2} × {red, white} = {(1, red), (1, white), (2, red), (2, white)}.
• {1, 2, green} × {red, white, green} = {(1, red), (1, white), (1, green), (2, red), (2, white), (2, green), (green, red), (green, white), (green, green)}.
• {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}.
Some basic properties of cartesian products:
• A × ∅ = ∅.
• A × (B ∪ C) = (A × B) ∪ (A × C).
• (A ∪ B) × C = (A × C) ∪ (B × C).
Let A and B be finite sets. Then
• | A × B | = | B × A | = | A | × | B |.
A visual of a Cantor Set
Step One: Gentle intro
“Are rational numbers countable?” | {
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A visual of a Cantor Set
Step One: Gentle intro
“Are rational numbers countable?”
Step Two: The Hotel Infinity
“Related to Cantor’s argument that rational numbers are countable?”
Step Three: Self Similarity
“Can you write out details of constructing the Cantor Set?” | {
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# Why use Taylor series instead of Maclaurin? [duplicate]
I know that Maclaurin series are a special case of Taylor series where we set $a=0$, and it is useful for exponential functions (for example). But my question is: Are Taylor series (evaluate at some other point different to zero) really useful in practice?
• There are many reasons, ranging from solving limits, creating linear approximations and approximating functions in general, studying asymptotic growth (generally through Laurent series), and so forth. This question is very broad as it stands... What do you have in mind? Try to be a bit more specific Sep 18 '16 at 18:08
• my question is principal for approximations. Why do we care to approximate a function at some point different to zero. I mean why evaluate for example f(8) = cos x with a = 5 (or other number different to zero) with taylor formula wikimedia.org/api/rest_v1/media/math/render/svg/… Sep 18 '16 at 18:21
• The simplest answer to that is because you might want to approximate a function elsewhere than around zero. A McLaurin series is usually fine if you want to estimate the value of a function at $x=0.5$, but is often terrible at $x=500.5$. If you want to estimate the function there a Taylor series about $x=500$ might be very useful! Sep 18 '16 at 18:45
• You might then ask why we don't just get the series around $x=500.5$ instead of $x=500$. The key to this is ease of calculation. Imagine you are messing with a square root function and trying to approximate $\sqrt{36.458}$. You'll get a pretty nice estimation if you say it's about $\sqrt{36}=6$. Notice that we are estimating a function by a nearby point not equal to zero.... A Taylor series just makes this more rigorous, algorithmic, and accurate! Sep 18 '16 at 18:49
• thank you @BrevanEllefsen that's a good explication Sep 18 '16 at 19:20 | {
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An important reason is that the function might not be defined at $$0$$. Consider the $$log$$ function.
As mentioned in the comments, you might need to approximate the function in a region not near $$0$$. The Maclaurin series might converge very slowly or not at all. You could look for nearer points st which it was easy to calculate the Taylor coefficients.
Taylor series can be used to approximate function at some point (not always $0$). For example, this is a powerful tool for calculating limits. | {
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# Integrating $\int \frac{1}{\sqrt{x^2+4}}\,dx$ using Trigonometric Substitution
I'm reviewing integration by trigonometric substitution in anticipation of covering it in class next week. I seem to be a bit rusty and keep catching myself making various mistakes. On this particular problem I keep getting the same answer which is very close to being correct. However, I somehow end up dividing by two where I should not. I'm hoping another set of eyes can quickly set me right so I can stop frustrating myself reworking the problem to the same apparently wrong answer repeatedly! Thanks in advance!
$$\int \frac{1}{\sqrt{x^2+4}}\,dx$$
$$ln\lvert x + \sqrt{x^2+4} \rvert + C$$
Somehow I keep getting:
$$ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C$$
Here's my work:
$$\int \frac{1}{\sqrt{x^2+4}}\,dx = \int \frac{1}{\sqrt{4(\frac{1}{4}x^2+1)}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}x^2+1}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{(\frac{1}{2}x)^2+1}}\,dx$$
At this point I substitute as follows:
$$\frac{1}{2}x = \tan\theta$$ $$x = 2\tan\theta$$ $$dx = 2\sec^2\theta$$
So I continue on with:
$$\frac{1}{2}\int\frac{2\sec^2\theta}{\sqrt{\tan^2\theta+1}}\,d\theta = \int\frac{\sec^2\theta}{\sqrt{\sec^2\theta}}\,d\theta = \int\frac{\sec^2\theta}{\sec\theta}\,d\theta = \int\sec\theta\,d\theta = ln\lvert\sec\theta + \tan\theta\rvert + C$$
Finally, to get the answer in terms of x I essentially draw a right triangle and use the fact that $\tan\theta = \frac{x}{2}$. The side opposite $\theta$ I take to be x, the side adjacent $\theta$ is 2, and the hypotenuse is $\sqrt{x^2+4}$. So $\sec\theta = \frac{\sqrt{x^2+4}}{2}$ and $\tan\theta = \frac{x}{2}$.
So, substituting these values back in, as mentioned, I end up with:
$$ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C$$
Can anyone help me see where I'm going wrong or failing to understand something? | {
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Can anyone help me see where I'm going wrong or failing to understand something?
• Both answers are correct. Feb 6, 2015 at 21:58
• Thanks to everyone who answered! I greatly appreciate the help :) Feb 6, 2015 at 22:14
You are correct still. Notice that \begin{align} \ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C &= \ln\Bigg|\frac{\sqrt{x^2+4}+x}{2}\Bigg| + C\\ &= \ln\Bigg|\sqrt{x^2+4}+x\Bigg|-\ln(2) + C\\ &= \ln\Bigg|\sqrt{x^2+4}+x\Bigg|+C' \end{align} where $C'$ is still an arbitrary constant.
$$\ln\left|{\sqrt{x^2+4}\over 2}+{x\over 2}\right|+C=\ln|\sqrt{x^2+4}+x|-\ln2+C.$$ | {
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# How many 4-permutations of the positive integers not exceeding 100 contain three consecutive integers in the correct order
Question:
How many 4-permutations of the positive integers not exceeding $100$ contain three consecutive integers in the correct order
a.) where consecutive means in the usual order of the integers and where these consecutive integers can perhaps be separated by other integers in the permutation?
b.) where consecutive means both that the numbers be consecutive integers and that they be in consecutive integers and that they be in consecutive positions in the permutation?
My Attempt:
3 consecutive integers is, $k, k+1, k+2$, thus $k$ is limited to $98$,
a.) The number of ways we can place and the 2 others is $C(4, 3) = 4$. Since $3$ integers are already taken by $k, k+1, k+2$, the last integer can have 97 integers, thus
$$4*98*97 = 38,024$$
b.) Since $k, k+1, k+2$ have to be consecutive in placement, there's only two possible place for that, thus
$$2*98*97 = 19, 012$$
Problem:
Both of my answers are a bit off compared to the answer key. In the answer key, a.) $37,927$ b.) $18,915$. One might notice that if my answers are subtracted $97$, I get the answer key's answer. So how exactly do you arrive to these numbers? I'm out of ideas.
• Inclusion-exclusion principle? That is, make sure we don't double count the sequences of all 4 integers consecutively. – hardmath Jul 20 '14 at 13:27
• i feel you, bro!! i did the same stupid mistake – Godfather Oct 17 '16 at 10:13 | {
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• We interpret $3$ consecutive as meaning at least $3$ consecutive. You have counted $12,13,14,15$ twice in your way of counting, once as $12,13,14$ followed by $15$, and once as $13,14,15$ preceded by $12$. Every one of the $4$ consecutives has been counted twice. There are $97$ ways to have $4$ consecutives, which is why your answer for b) is greater by $97$ than the official (and correct) answer. To get the right answer, we do as you did, but being aware that one is double-counting, and subtracting $97$ at the end to compensate. (Cont) – André Nicolas Nov 6 '14 at 22:16 | {
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# Master theorem latex
### Master theorem latex | {
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2. MIT One way is to model the algorithm in the form of a recurrence equation and then solve via a number of techniques. The ewcounter command uses a LaTeX internal command, and you can also use it: \@addtoreset{new-name}{master} (but remember that it needs to be between \makeatletter and \makeatother, or in a package of your own). Andrew McGregor Associate Professor If you would like to write your homework solutions in LaTeX, Mergesort, Master Theorem Section 1, 2. It will also automatically number theorems, lemmas, definitions and bibliographic entries. 1, 2. 3 of CLR). Master Theorem 373F19 - Karan Singh 34 •Here’s the master theorem, as it appears in CLRS Useful for analyzing divide-and-conquer running time If you haven’t already seen it, please spend some time understanding it It was typeset using the LaTeX language, with most diagrams done using Tikz. Monday, September 9, 2019. The Master Theorem Consider a function that, for all for all ,satisfies the recurrence relation: with ,integer , real , . 2 Letqbe the product of the ˆrstpnumbers. If , for some constant , and if for some constant and all sufficiently large , then. Seems like reading problem in the morning and thinking about it through out the day helped me a lot. ”, “Theorem (Bob). Master Theorem is used to determine running time of algorithms (divide and So , according to master theorem the runtime of the above algorithm can be 1. I want to cite a theorem using its label, but what should appear is both its theorem number and its name (the text in Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Abstract: We state and prove a quantum-generalization of MacMahon's celebrated Master Theorem, and relate it to a quantum-generalization of the boson-fermion correspondence of Physics. I have been working for the past week on improving the | {
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the boson-fermion correspondence of Physics. I have been working for the past week on improving the proofs of Master theorems, as well generalizing them. A table of contents will automatically be created, complete with clickable links to each section and subsection you create in your presentation. and typed all missing steps in LaTeX. Fix function: In the fix function, compare that node with its children. Cormen, C. Correctness 1. 1. 1: z-r+t=3 Lemma 1. Produce beautiful documents starting from our gallery of LaTeX templates for journals, conferences, theses, reports, CVs and much more. In my opinion, the closest to a "master theorem" is the criterion due to Terry Lyons, according to which a reversible Markov chain on a countable state space (in particular, the simple random walk on a locally finite graph) is transient if and only if there exists a flow of finite energy on the state space. Clicking on it will redirect you to their original webpages. 1/30/19 1 CS4102 Algorithms Spring 2019 Warm up Given any 5 points on the unit square, show there’s always a pair distance ≤" apart 1 1 1 1 1 1 2 1 2 2 If points # LaTeX is a document preparation system used to write scholarly articles, prepare presentations, and many other documents. So I'm curious whether there may be some way to define a new theorem environment that accepts the theorem title as an argument. The textbook Algorithms, 4th Edition by Robert Sedgewick and Kevin Wayne surveys the most important algorithms and data structures in use today. Peace is followed by disturbances; departure of evil men by their return. Proof. 2016. recurrence can not be solve by Master Theorem what is the difference between this two recurrences. Let P be a set of n points in d-dimensional space. Common techniques are master theorem, substitution, recurrence trees, The binary search algorithm can be seen as recurrences of dividing N in half with a comparison. Introduction Code Beamer Features More LATEX Disclaimer #1 I am NOT an | {
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N in half with a comparison. Introduction Code Beamer Features More LATEX Disclaimer #1 I am NOT an expert in LATEX I am NOT an expert in Beamer Disclaimer #2 This talk is designed to introduce you to presentations in LATEX For that purpose, we firstly state and prove a \lambda-extension of Schwinger's Master Theorem (SMT), which turns out to be a useful mathematical tool for us, particularly as a generating function for the unitary-representation functions of the conformal group and for the derivation of the reproducing (Bergman) kernel of L^2_h(D_4,d u_\lambda). A score higher than 100% is Jon Wellner's lecture notes on the master theorem of parametric The Davis-Kahan theorem. 4. Master Theorem Example 3 25 56=3 6 2 +8 Case 1 Θ6BCDEF ≈Θ(6&. Lectures by Walter Lewin. ´) than is the case for an ordinary character. However, a much more powerful and flexible solution is achieved thanks to an auxiliary tool called BibTeX (which comes bundled as standard with LaTeX). Master Theorem: Practice Problems and Solutions. Fact (MacMahon Master theorem or Wronski relation). You can earn more than 100% on most problem sets. I am now quite content with the proofs. The last equation shows that f(n) is not (g +(n)) for any >0, which excludes case (3). See my brief slides on the "school method" for integer division. Extra Credit: This class will have extra credit assignments that can help you boost your exam scores (by up to 10%). If your work relies on important results in your field, including those theorems' proofs in your introduction and background sections makes sense. Master Theorem 373F19 - Nisarg Shah 32 •Here’s the master theorem, as it appears in CLRS Useful for analyzing divide-and-conquer running time If you haven’t already seen it, please spend some time understanding it The report document should be written in Latex and contain: problem description, benchmarking, figures, and discussion. No installation, real-time collaboration, version control, hundreds of LaTeX | {
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and discussion. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. So T (n) = T (n/2) + 1. This is an extremely valuable theorem because it is a vast generalization of Dirichlet's Theorem on primes in an arithmetic progression. 3. Theorem There is no largest prime number. Fri, Sep 27: Prerequisites Review Heaps and Heapsort (continued) Chapter 6 (Heapsort), Section 6. 3rd edition. 1 Oct 2014 And I'm almost certain that everyone reading this blog has heard of the generalization of the master theorem due to Akra and Bazzi. Stein. If you can write basic LATEX, you can easily make a Beamer presentation. 3 (Bulding a Heap), Introduction to Algorithms (3rd Edition) by Cormen et al. Master Theorem 3. If a master-file has been specified, then it will scan that file instead of the current file. If, on the other hand, we know the characteristic function φ and want to find the corresponding distribution function, then one of the following inversion theorems can be used. Q&A for users of TeX, LaTeX, ConTeXt, and related typesetting systems Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A group earns 30% bonus to the entire homework for each solution posted on the class website. Submitted my pdf at around 10 am. The Master Theorem can be applied to any recurrence of the form. It was typeset using the LaTeX language, with most diagrams done using Tikz. Master Theorem. 2: q=r How do I do that? Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I managed to prove that master theorem is applicable, but I don't know how to go about using it. Aug 16, 2016 · To start with, a building can never stand | {
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but I don't know how to go about using it. Aug 16, 2016 · To start with, a building can never stand without a base. As Raphael mentions in his comment, here you are encountering two common versions of the Master Theorem, one of which is stronger and can handle your case. In your algorithm, I believe there is one subproblem of size n/2, so a is 1 and b is 2. Algorithms AppendixII:SolvingRecurrences[Fa’13] Change is certain. It is nearly complete (and over 500 pages total!!), there were a few problems that proved some combination of more difficult and less interesting on the initial pass, so they are not yet completed. Determine a good fancy style for each sentence : box shape; colour; title ( yes/no , style) With "theorem" we can mean any kind of labelled enunciation that we want to look separated from the rest of the text and with sequential numbers next to it. Gauss’s law, also known as Gauss’s flux theorem, is a law relating the distribution of electric charge to the resulting electric field. Homework 1 handed out. Case : If is either of the same order as or exactly logarithmically larger than , then the two terms compound each other, I want to write a LaTeX code to show, Lemma 1 : x+y=z Lemma 1. Algorithm design. Define 1;p to be the first positive eigenvalue of the p-Laplace operator on M, then, for any p 2(1;1), the following sharp estimates holds: 1;p p 1 ˇ p d p; (1. Texts. Prepare your solutions in LaTeX and submit in Canvas, like you would do for a regular assignment, but remember not to put your name on your paper! Solve the following recurrence exactly (in other words, do not use asymptotic notation), where you may assume n is a power of 4. Posted 01-14-2020: In case you want to typeset your homework, Here is a nice Latex Introduction. Case : If is polynomially smaller than , then it is negligible, and the asymptotic behavior is the same as what you would have without it included. If you are using a program with a graphical interface to TeX, you probably | {
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have without it included. If you are using a program with a graphical interface to TeX, you probably need to click on a button. It describes the per-iteration overhead, which may depend on the problem size in each iteration. because like the master theorem it gives you a quick way to generate the desired answer (or at least a guess that you can plug in to the recurrence to check). Pearson Education. The aim of classes is to introduce undergraduate mathematics students to $\LaTeX$ in preparation for writing their second year essay or fourth year project. A lifetime of learning Get started with Brilliant’s course library as a beginner, or dive right into the intermediate and advanced courses for professionals and lifelong learners. In Section 5. Then 练习使用latex嘤嘤嘤 注意 以及 LaTeX will calculate the right numbering for the objects in the document; the marker you have used to label the object will not be shown anywhere in the document. This approach is commonly used for theorems in mathematics, but can be used for anything. In this article, you will learn with the help of examples the DFS algorithm, DFS pseudocode and the code of the depth first search algorithm with implementation in C++, C, Java and Python programs. Fermat's little theorem is the basis for the Fermat primality test and is one of the fundamental results of elementary number theory. Leiserson, R. Working on the O-notation was at first supposed to be a brief side-project to “make things right”, but then ended up taking quite a bit of time. 21 May 2019 Move both the theorem and the proof completely in appendix the-end/raw/ master/demo. J. And typeset documents with LaTeX with just the click of a button. Tardos. 6 Proof of the master theorem 97 5 Probabilistic Analysis and Randomized Algorithms 114 5. Master Theorem · Divide and Conquer Algorithm · Binary Search · Floyd-Warshall Algorithm · Strongly Connected Components · Rabin-Karp Algorithm. By the master theorem, this recurrence has values =, =, =, which is | {
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· Rabin-Karp Algorithm. By the master theorem, this recurrence has values =, =, =, which is a "steady state" tree, and thus we use the steady state case that tells us that T ( n ) = Θ ( n k ⋅ log n ) = Θ ( log n ) . They will make you ♥ Physics. In order to untoggle this functionality resort to the command onfrenchspacing . Abstract. tex simple. : We develop the theory of N-homogeneous algebras in a super setting, with particular emphasis on the Koszul property. Here are the latex sources files of Homework 1. The theorem is named after Pierre de Fermat , who stated it in 1640. Question about Master Theorem was very intuitive. View courses, graduate and undergraduate programs, faculty and research interests, activities, events … Jan 23, 2010 · With " theorem " we can mean any kind of labelled enunciation that we want to look separated from the rest of the text and with sequential numbers next to it. The distance is calculated by taking the square root of the sum of the squared pair-wise distances of every dimension. Master theorems 14. when typesetting in LATEX a master or a doctoral thesis according to current re- quirements of TTU Graduate School mathematics, and sciences – who would like to use LATEX in typesetting their theses. The presentation should be made with about 10-12 slides generated with LaTeX (Beamer). It analyzes the theorem package can be used to customize theorem-like environments. It brings a minimalist and powerful approach to text editing. Master Theorem 2. Each of the exercises below involves a choice among the master theorem templates discussed in lecture. 31 Aug 2016 Master Theorem CSE235 Introduction Pitfalls Examples 4th Condition Master Theorem Slides by Christopher M. PS Contrary to The standard commands of LATEX also work in Beamer. A personal use latex package for highlight the mathematical logic in a text. master template. Once the external bibliography file is imported, the command \cite is used just as in the introductory example | {
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bibliography file is imported, the command \cite is used just as in the introductory example . So as you can see, the recurrence relation is gleaned from looking at the algorithm. The included files must have a file variable TeX-master or tex-main-file pointing to the master So we need to tell RefTeX that theorem and axiom are new labeled Latex-Suite is released under the Vim charityware license. Such recurrences should not constitute occasions for sadness but realities for awareness, so that one may be happy in the interim. If I'm wrong, what did I miss on? The answer is that you didn't miss anything. General results Master Theorem. The \glue" is f(n) = n5=2. BibTeX is a widely used bibliography management tool in LaTeX, with BibTeX the bibliography entries are kept in a separate file and then imported into the main document. the problem size must shrink by a constant factor, the subproblems must all have the same size, Then, once you have the recurrence you can analyze using the Master Theorem. If no case applies, simply state that fact; you are not required to attempt a solution when no master theorem case applies. The Quantum MacMahon Master Theorem By Stavros Garoufalidis, Thang TQ Le, Doron Zeilberger and X [maybe you!] Warning: The proof is incomplete, read below for a chance to win a prize and be co-author. $$T(n) = \sum_i a_i T(n/b_i) + g(n)$$. f(n)=lgn=O(n2−ϵ) for ϵ=1, since lgn=O(n). For the special case where the Hecke operator is the ordinary supersymmetry and typed all missing steps in LaTeX. There are four main functions associated with heapsort 1. 1: As part of your background section. If f(n) = O(nlogb a− ) for some constant > 0, then T(n) = Θ(nlogb a). Homework that is Chapter 7: The Prime Number Theorem for arithmetic progressions RefTeX wraps itself round four LaTeX macros: \label , \ref , \cite , and \index . Since there is always a power of 2 in the range $[\frac {n} {2}, n]$, recurrence master-theorem 追加された 27 9月 2013 〜で 02:06 著者 WSS , それ | {
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2 in the range $[\frac {n} {2}, n]$, recurrence master-theorem 追加された 27 9月 2013 〜で 02:06 著者 WSS , それ マスター定理を使用した次の再帰アルゴリズムの実行時間はどのくらいですか？ The DFS algorithm works as follows: Start by putting any one of the graph's vertices on top of a stack. There are 3 cases: 1. 2 it was indicated that kd-trees can also be used to store sets of points in higher-dimensional space. Results There Is No Largest Prime Number The proof uses . Welcome to the Dartmouth Mathematics Department located in Kemeny Hall. 4 Probabilistic analysis and further uses of indicator random variables 130 •Master Theorem 3. Note that the master theorem does not provide a solution for all f f f. Master Theorem: Practice Problems and Solutions Master Theorem The Master Theorem applies to recurrences of the following form: T(n) = aT(n/b)+f(n) where a ≥ 1 and b > 1 are constants and f(n) is an asymptotically positive function. The broad perspective taken makes it an appropriate introduction to the field. LATEX is preferred but not required. With "theorem" we can mean any kind of labelled enunciation that we want to look separated from the rest of the text and with sequential numbers next to it. 1 The hiring problem 114 5. Recommended for you Latex-Theorem. Show your derivation and then prove that your answer is correct. Dec 19, 2019 · Vim-LaTeX is an extension for Vim that lets you edit and compile LaTex documents. 11 Sep 2019 PDF | We use the method of brackets to evaluate quadratic and quartic type Ramanujan's Master Theorem and the Method of Brackets. T(n) = a · T(n/b) + f(n) that describes the complexity of some recursive functions. Dec 02, 2018 · Jupyter notebook extension which supports (some) LaTeX environments within markdown cells. T(n) = aT(n / b) + O(n d) where a, b, and d are constants. November 10: Midterm this week was quite easy. A score higher than 100% is bonus. . (There are some other formulations, but this above one handles the more common cases). Convergence conditions are also studied. In | {
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but this above one handles the more common cases). Convergence conditions are also studied. In particular, if f f f is smaller or larger than n log b a n^{\log_b{a}} n lo g b a by less than a polynomial factor, then none of the three cases are satisfied. 5. It is recommended that students take both classes. – daleif Aug 10 '17 at 8:28 Also note that your example has no chance to work as \institute and \email is not defined in the article class – daleif Aug 10 '17 at 8:29 latex 定理环境，引理，定义，修改格式 Theorem, lemma, definition 时间 2017-03-11 标签 定理环境 latex 定理格式 定理冒号 定理换行 Jan 24, 2016 · For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. In CLRS terms In the analysis of algorithms, the master theorem provides a cookbook solution in asymptotic The master theorem concerns recurrence relations of the form:. The Master Theorem: Chapter 4 (Divide-and-Conquer), Section 4. If in a min-heap, swap the parent and the smallest child considering that the child is smaller than the parent itself. You may find this template useful. It lets you solve recurrence relations that can help you find the complexity of recursive algorithms. e. For example, there is a book titled "Synopsis of Elementary Results in Pure and Applied Mathematics: Containing Propositions, Formulae, And Methods Of Analysis, With Abridged Demonstrations" (Cambridge Library Collection - Mathematics) by George Shoobridge Carr, which I am told consists of theorem without proof, mostly. Kleinberg, E. Typically, running LaTeX means typing latex <filename> at the command prompt and running BibTeX means typing bibtex <filename>. Master Written (use Latex!) – Submit both zip and pdf ! 21 Dec 2005 The LaTeX template for CS thesis is based on version 2. Comment: AMS-LaTeX, 7 pages, revised and final version. characters directly instead of having to rely on Latex-Suite's method of displaying diacritics. pdf) to see what is possible, or generate it with. If you would like to write your homework solutions in | {
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to see what is possible, or generate it with. If you would like to write your homework solutions in LaTeX, here's a template 3 Sept, Preliminaries, Mergesort, Master Theorem, Slides, Section 1, 2. Week 1. LaTeX provides a command that will let you easily define any theorem-like enunciation. Create a list of that vertex's adjacent nodes. LaTeX. Recursive functions, Master Theorem LaTeX is beautiful and strongly recommended We will post our TeX source files, which you can use as templates. T(n1/2). tex the LATEX master file, to be used as a Note that for LIPIcs all numbered theorem-like environments should use one. Assume A is diagonalizible, then it is can be proved in the eigenbasis and is exactly the previous fact. This article explains how to use the standard verbatim environment as well as the package listings, which provide more advanced code-formatting features. For each, indicate which case applies and specify the asymptotic growth class of the function. 2 of the thesis template for \macs — Master of Applied Computer Science project report; \mcs — Master of diff simple. l) • Case 1: if Q6=7(6BCDU^_‘)for some constant a>0, then 56=Θ(6BCDU^) • Case 2: if Q6=Θ(6BCDU^), then 56=Θ(6BCDU^log6) • Case 3: if Q6=Ω(6BCDU^e‘)for some constant a>0, and if NQ O P ≤;Q(6)for some constant ;<1and all sufficiently large 6, then 56=Θ(Q6) 56=N5 6 S +Q(6) LaTeX Resources. As we will soon see, a polynomial of degree n in the complex number system will have n zeros. Mar 19, 2018 · 1 Answer 1. The current figures use images from external websites. We strongly recommend you use LaTeX to typeset your work so it is easy to read. Since the theorem-like structures use a beamercolorbox, using allowframebreaks option directly won't work in this case. Certainly Data Structures and Algorithms are the base to all programming languages and hence is a must learn… Theorem 5. If you use (b) We cannot apply the master theorem because of the square root, so we draw the recursion tree: T(n). You can | {
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apply the master theorem because of the square root, so we draw the recursion tree: T(n). You can easily create overlays and dynamic effects. T. Again, like I mentioned in class, please remember that the regular expression that is taught in class (CSC236) is different from the regular expression that is presented elsewhere. The extra credit assignments will involve doing machine-checked proofs in Coq , which you will learn about during discussion sections. bst only in that the names of all three authors of Arrow, Hurwicz, and Uzawa (1961) are listed in the first citation to that work, although not in the second, parenthetical, citation. Solutions to problems sets must be typeset using LaTeX . a= 4;b= 2 implies a reference function g(n) = nlog 2 4 = n2. When you run the LaTeX file through LaTeX and BibTeX (instructions below), you'll get output for the body of the document that differs from the output when you use te. Similarly, if x−k is a factor of f (x) , then the remainder of the Division Algorithm f (x) = (x−k)q (x)+r is 0. Theorem. Introduction Code Beamer Features More LATEX Disclaimer #1 I am NOT an expert in LATEX I am NOT an expert in Beamer Disclaimer #2 This talk is designed to introduce you to presentations in LATEX 1/30/19 1 CS4102 Algorithms Spring 2019 Warm up Given any 5 points on the unit square, show there’s always a pair distance ≤" apart 1 1 1 1 1 1 2 1 2 2 If points # Solutions to problems sets must be typeset using LaTeX . Jan 24, 2016 · For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. Miscelanea Defined terms on the margin; Date and time of compilation; Print labels on the margin (equation, section and theorem) Images in separate folders; Color all math; Poster Examples (external link) Bibliography Single (master) BibTeX file for all your projects; Definitions on the margin Writes the term being defined on the margin. - T-F-S/tcolorbox Mason–Stothers theorem (polynomials) Master theorem (analysis of algorithms) | {
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- T-F-S/tcolorbox Mason–Stothers theorem (polynomials) Master theorem (analysis of algorithms) (recurrence relations, asymptotic analysis) Maschke's theorem (group representations) Matiyasevich's theorem (mathematical logic) Max flow min cut theorem (graph theory) Max Noether's theorem (algebraic geometry) Maximal ergodic theorem (ergodic theory) latex 定理环境，引理，定义，修改格式 Theorem, lemma, definition 时间 2017-03-11 标签 定理环境 latex 定理格式 定理冒号 定理换行 Jan 21, 2014 · A Sample Proof Using Mathematical Induction (playing with LaTeX) It’s been a long time since I used LaTeX regularly, and I discovered that I don’t have any leftover files from my days as a math student in Waterloo. Other guides can be found at Wikibooks and NYU. The Master theorem applies if f (n) are exactly of the same order of magnitude, or if their ratio grows at least like n , for some > 0, but there are possible values for a, b and f (n) such that neither case applies. Add the ones which aren't in the visited list to the top of stack. David Wu Guess and Check Method. Dec 08, 2015 · To do this, consider instead the recurrence $T (n) = T (n-1) + T (f (n)) + n$, where $f (n)$ is equal to the greatest power of 2 less than n. An introduction to the lab environment and tools used to write, compile, and run a program in either PHP (for those with prior Java experience) or Java (for those without prior Java experience). LaTeX Templates Theses An academic thesis, also known as a dissertation, is a substantial work produced by a graduate student to communicate their research and earn a degree. Application of Hall's Theorem: every doubly-stochastic matrix is a convex combination of permutation matrices matching markets: market clearing prices always exist (envy free allocation, maximizing total valuation); here are the pdf slides. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. This pair of implications is the Factor Theorem. The | {
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manage projects, and build software together. This pair of implications is the Factor Theorem. The \newtheorem command may have at most one optional argument. [Val12b, main Theorem] Let M be a compact Riemannian manifold with nonnegative Ricci curvature and diameter d, and possibly with convex boundary. Homeworks •Hw1 due Wed, January 30 at 11pm –Start early! –Written (use Latex!) –Submit BOTH pdf and zip! Master Theorem 1. So: Aug 27, 2014 · [back to top] The Euclidean distance is a distance measure between two points or or vectors in a two- or multidimensional (Euclidean) space based on Pythagoras' theorem. T(n) = 2T(bn=4c) + n1 Together, these define the parameters f(n), a, and b in the Master-Theorem equation. 2 Indicator random variables 118 5. This is true even if the proofs are well known. Intro to algorithms. Master Theorem is good to know for programming. CLR calls this the “Master Theorem” (section 4. Together, these define the parameters f(n), a, and b in the Master-Theorem equation. 09. 2 2 : 7 Sept Euler Circle Spring Paper: Čebotarev Density Theorem In this paper, we do exactly what the title implies: prove the Čebotarev Density Theorem. Evaluate, simplify, solve, and plot functions without the need to master a complex syntax. LaTeX is widely used in science and programming has become an important aspect in several areas of science, hence the need for a tool that properly displays code. The law was formulated by Carl Friedrich Gauss (see ) in 1835, but was not published until 1867. Master Theorem 1. CS4102 Algorithms Spring 2019 Warm up Given any 5 points on the unit square, show there’s always a pair distance ≤" apart 1 1 1 All homeworks must be typed, preferably in LaTeX. Mar 04, 2016 · Please, check out some of my other uploads or sub! Open to suggestions for other lack (pvc, vinyl, latex) compilations. 3 Then q+1 is not divisible by any of them. 1/29/19 2 CLRS Readings •Chapter 4 4 Homeworks •Hw1 due Wed, January 30 at 11pm –Start early! –Written | {
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2 CLRS Readings •Chapter 4 4 Homeworks •Hw1 due Wed, January 30 at 11pm –Start early! –Written (use Latex!) –Submit BOTH pdf CS4102 Algorithms Spring 2019 Warm up Given any 5 points on the unit square, show there’s always a pair distance ≤" apart 1 1 1 Luckily, the lemma is nevertheles true for the weaker hypothesis, and surprise!, the theorem also holds with the strong hypthesis replaced by the weaker one. SMT is related to MacMahon's Master Theorem (MMT) and an extension of both in terms of Louck's SU(N) solid harmonics is also provided for completeness. Goodrich & Roberto Tamassia!Algorithms is a course required for all computer science majors, with a strong focus on theoretical topics. Thesis 14. We state and prove a quantum-generalization of MacMahon's celebrated Master Theorem, and relate it to a quantum-generalization of the boson-fermion correspondence of Physics. Comments: AMS-LaTeX, 7 pages, revised and final version. theorem. Course for 3rd year bachelor and master students - Fall 2016. In Mathematics most papers are written in a type setting markup language called LaTeX - which evolved from TeX. CLRS: p93 I've been looking at my algorithms and complexity course's suggested exercises for master theorem and I the only one I can't seem to solve is a case in which k is rational. Since f(n)=g(n) !1 For that purpose, we firstly state and prove a \lambda-extension of Schwinger's Master Theorem (SMT), which turns out to be a useful mathematical tool for us, particularly as a generating function for the unitary-representation functions of the conformal group and for the derivation of the reproducing (Bergman) kernel of L^2_h(D_4,d u_\lambda). Thursday, October 17, 2019 For that purpose, we firstly state and prove a \lambda-extension of Schwinger's Master Theorem (SMT), which turns out to be a useful mathematical tool for us, particularly as a generating function for the unitary-representation functions of the conformal group and for the derivation of the | {
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for the unitary-representation functions of the conformal group and for the derivation of the reproducing (Bergman) kernel of L^2_h(D_4,d u_\lambda). How would I figure out which case of Master's theorem this lie on? Would taking logarithms of both sides for comparison make sense since logarithm is a monotonic function? $$2\cdot \lg(\lg(n)) \quad\text{and} \quad (0. So all is well that ends well. Abstract This article provides useful tools to write a thesis with LATEX. Specifically, this means that. The chngcntr package encapsulates the \@addtoreset command into a command \counterwithin. 3 Randomized algorithms 122? 5. ”, and “Theorem (Will)”. Extracredit will be offered if you write the final project in a Jupyter notebook which includes the LaTeX and Python code. 3 Master theorem. There are two types of class: beginners and intermediate. A LaTeX package to create highly customizable colored boxes. LaTeX provides a convenient way to produce high-quality documents and it is the standard used for typesetting computer science papers. Introducing a NEW addition to our growing library of computer science titles, Algorithm Design and Applications, by Michael T. TODO: Identify all the sentences which require an " encapsulation box" and write a "phylosphical" pseudo-definition of them in the example file. 5) Brilliant guides you through problem solving strategies and challenges you to think outside the box. For that purpose, we firstly state and prove a \lambda-extension of Schwinger's Master Theorem (SMT), which turns out to be a useful mathematical tool for us, particularly as a generating function for the unitary-representation functions of the conformal group and for the derivation of the reproducing (Bergman) kernel of L^2_h(D_4,d u_\lambda). To any Hecke operator on a vector superspace, we associate certain superalgebras and generalizing the ordinary symmetric and Grassmann algebra, respectively. Master theorem states that, For any , If for some constant , then . The | {
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algebra, respectively. Master theorem states that, For any , If for some constant , then . The second equation shows the f(n) is not ( g(n)) and therefore precludes case (2). The solution will tell you the nature of the runtime, i. The various LaTeX processing programs can be found at CTAN. Fortunately, LaTeX has a variety of features that make dealing with references much simpler, including built-in support for citing references. tex 3a4,13 > %Theorem 20 Feb 2013 lipics-sample-article. \newtheorem*{mainA}{Theorem A }. Session 4 Bibliography Management, Theorems, Formatting your CV using LaTex Using Built-in Support If you are writing only one or two documents and aren't planning on writing more on the same subject for a long time, maybe you don't want to waste time creating a database of references you are never going to use. The time for dividing is O(1) and time for recombining is O(1) assuming the analysis is not in terms of bit operations. Lab 01 – Introduction, Gettings Started. 4 Thus q+1 is also prime and greater thanp. Presentations, otherwise known as seminars, talks or lectures, are given to an audience with the purpose of sharing information with a group of people. Also you might want to have a look in the amsthm manual to familiarize yourself with the ewtheorem syntax. The formula in the definition of characteristic function allows us to compute φ when we know the distribution function F (or density f). The master template is not applicable to this recurrence. When I searched on google I found tow questions related with this two recurrence First Question: The formulæ are written using $$\LaTeX$$ and the final render of the page is done by the Javascript library MathJax (currently the simplest way to use $$\LaTeX$$ in HTML). The necessary space above and below the statement of the theorem will automatically be generated by LaTeX. Guide: An introduction to LaTeX can be found here. The master theorem is a formula for solving recurrences of the form T(n) = | {
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LaTeX can be found here. The master theorem is a formula for solving recurrences of the form T(n) = aT(n/b)+f(n), where a ≥ 1 and b > 1 and f(n) is asymptotically Usually this will be the name of the master theorem. {\displaystyle T(n)=\Theta (n^{k}\cdot \log n)=\Theta (\log n). You write text and mathematics in your document without having to know a programming language or exit to an equation editor. An online LaTeX editor that's easy to use. Here's some intuition about the master theorem. It is much better at handling equations than Word type programs. The relation is then solved by either the master theorem or the substitution method or gleaned from drawing a graph of the recursions (recursion tree) and then proved rigorously with the substitution method. Felt so good. In the analysis of algorithms, the master theorem for divide-and-conquer recurrences provides an asymptotic analysis (using Big O notation) for recurrence relations of types that occur in the analysis of many divide and conquer algorithms. 631+\epsilon)\cdot \lg(n)$$ Then figure out what epsilon is Thank you in advance, kpark Sometimes an apparently weaker version of the theorem is equivalent to the full theorem with a short proof (for example, Hilbert's Nullstellensatz). Note that the f(n) function is also a complexity expression. All homeworks must be typed, preferably in LaTeX. If I have a lot of these, it becomes tedious having to always go back to the preamble to define a new theorem environment. Comment: LaTeX, 40 pages, three new Sections and six new references added. We strongly recommend typesetting solutions to the homework assignments using LaTeX. Master theorem solver (JavaScript) In the study of complexity theory in computer science, analyzing the asymptotic run time of a recursive algorithm typically requires you to solve a recurrence relation. T(n) = { LaTeX Original \documentclass{article} \begin{document} {\bf 6 Dec 2015 I think you used that method wrong! As mentioned in master | {
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\begin{document} {\bf 6 Dec 2015 I think you used that method wrong! As mentioned in master theorem case 3. Induction. Homework 5 handed out; here are the tex sources. active oldest votes. See my slides on the homework assignment policy of this class. Rivest, C. We prove that these algebras are N-Koszul. They are free. ) ]: Let. Latex is mandatory for submitting papers to many journals and style files for many of the top journals are available. Also provides support for labels and crossreferences, document wide numbering, bibliography, and more Skip to main content Switch to mobile version Intuitively, an algorithm’s efficiency is a function of the amount of computational resources it requires, measured typically as execution time and the amount of space, or memory, that the algorithm uses. 6. Bourke Instructor: Berthe Y. 5 (The Master Method for Solving Recurrences), Introduction to Algorithms (3rd Edition) by Cormen et al. Submission in LaTeX will increase your chances. Submit both paper copy in class, and digital version by e-mail to the TAs. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Posted 02-06-2020: Here is the Master Theorem. If , then . LaTeX is capable of producing presentation slides using the Beamer class in a simple and easy-to-use way. Bibliography management with Bibtex. 2. Join GitHub today. In your case you should use case 3 which uses Ω-notation which describes 18 Apr 2017 For your first question. The Zero Theorem Latex Slick Shine TV incredible workes Master It's particularly useful when you have recurrences of the form. Theorem 1. The university of closed for Labor Day next Monday, September 4th, so Kyle will hold makeup office hours this Friday, September 1st at the usual time, 3:00pm to 4:00pm. A workaround would be to "simulate" your theorem-like heading with a beamercolorbox and then write | {
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A workaround would be to "simulate" your theorem-like heading with a beamercolorbox and then write the text outside the box with the allowframebreaks option: See how you defined theorem. $\begingroup$ From Wikipedia (in agreement with recall of readings many years ago): Historically, as a book-keeping device of covariant perturbation theory, the graphs were called Feynman–Dyson diagrams or Dyson graphs, because the path integral was unfamiliar when they were introduced, and Freeman Dyson's derivation from old-fashioned perturbation theory was easier to follow for physicists How does python multiply Big Integers? How to solve recurrences without remembering the Master Theorem. It is due Wednesday, September 6th, in class. } Depth first traversal or Depth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Use Vim-LaTeX to bring the best of both the worlds into your document editing processes. 1 Suppose pwere the largest prime number. 5 The master method for solving recurrences 93? 4. Take the top item of the stack and add it to the visited list. [e-studiegids To simplify grading it is preferable that you submit your homework in latex. rather than as “Theorem (Jim). CLRS Readings •Chapter 4 4. There are three reasons to include a proof in your master's thesis - two of them good, and one of them bad. This tells us that k is a zero. It's particularly useful when you have recurrences of the form. det (1 − tM) ( ∑iTrtiSi (M)) = 1 , where Si are symmetric tensor power of M. To reference the number assigned to that result in some other theorem, one gives the command \ref{name}; this will be replaced by the number automatically assigned by LaTeX to that result when the LaTeX program is executed. Instead, LaTeX will replace the string " \ref{ marker } " with the right number that was assigned to the object. Yihang: Apr, 11 R The scribe latex template is available Let A be a matrix. Lecture 4: Karatsuba, Induction, and Master | {
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scribe latex template is available Let A be a matrix. Lecture 4: Karatsuba, Induction, and Master Theorem. 1 •Master Theorem 3. The amount of computational resources can be a complex function of the size and structure of the input set. 6 [The Master Theorem (From page 268 of Goodrich and Tamassia. Instructions for checking out code from GitHub using Eclipse. The Master Theorem applies to recurrences of the following form: T(n) = aT(n/b) + f(n). MIT Instructs LaTex to abstain from inserting more space after a period (´. master theorem latex | {
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# Gather multiple equations with curly bracket
I have two equations, both of which hold under the same condition (x<5):
\begin{align}
f(x) &= \sum_{n} a^i_j \Psi_n(x) &\text{for } x < 5
\\ g(x) &= \sum_{n} b^i_j \Phi_n(x) &\text{for } x < 5
\end{align}
That code results in this output:
Now, what I would like to do is to gather both equations with a curly bracket and this way write the condition only once. I tried this:
\begin{align}
\left.
\begin{array}{rl}
f(x) &= \sum_{n} a^i_j \Psi_n(x)
\\ g(x) &= \sum_{n} b^i_j \Phi_n(x)
\end{array}
\right\} &\text{ for } x < 5
\end{align}
Which results in the following output:
However, this doesn't look good to me: The equations are now too close to each other and the alignment of the left hand side with the equation marks is not convincing.
Do you have a hint how I can resolve that issue?
• You might be interested in rcases from mathtools. – leandriis Dec 18 '20 at 14:37
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{rcases} \begin{aligned} f(x) &= \sum_{n} a^i_j \Psi_n(x) \\ g(x) &= \sum_{n} b^i_j \Phi_n(x) \end{aligned} \end{rcases} x < 5
\end{document}
• Exactly what I needed, thanks. – Amos Egel Dec 18 '20 at 14:49
Another easy solution with empheq enables two equation numbers:
\documentclass{article}
\usepackage{empheq}
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# What is the average of rolling two dice and only taking the value of the higher dice roll?
What is the average result of rolling two dice, and only taking the value of the higher dice roll?
To make sure the situation I am asking about is clear, here is an example: I roll two dice and one comes up as a four and the other a six, the result would just be six.
Would the average dice roll be the same or higher than just rolling one dice?
-
With regard to your final question: what does your intuition tell you? – Benjamin Dickman Oct 29 '12 at 3:51
The number of ways to roll a number $x$ under your definition would be $2(x-1) + 1$.
Therefore the expected value would be $$E[X] = \sum_{x=1}^6\frac{2(x-1)+1}{36}x = \frac{1}{36}\sum_{x=1}^6(2x^2 - x) = \frac{161}{36} \approx 4.47$$ So the average is considerably higher than the average of a single die, being $3.5$.
-
For $k=1,\dots,6$ there are $k^2$ ways to get two numbers less than or equal to $k$. To get two numbers whose maximum is $k$ I must get two numbers that are less than or equal to $k$, but not two numbers that are less than or equal to $k-1$, so there are $k^2-(k-1)^2=k^2-(k^2-2k+1)=2k-1$ ways to get two numbers whose maximum is $k$. Thus, the probability of getting a maximum of $k$ is
$$\frac{2k-1}{36}\;,$$
and the expected value of the maximum is
\begin{align*} \sum_{k=1}^6k\cdot\frac{2k-1}{36}&=\frac1{36}\sum_{k=1}^6\left(2k^2-k\right)\\ &=\frac1{18}\sum_{k=1}^6k^2-\frac1{36}\sum_{k=1}^6k\\ &=\frac{6\cdot7\cdot13}{18\cdot6}-\frac{6\cdot7}{36\cdot2}\\ &=\frac{91}{18}-\frac{21}{36}\\ &=\frac{161}{36}\\ &=4.47\overline{2}\;. \end{align*}
Of course this is larger than the expected value of $\frac72=3.5$ for a single roll of a die: picking the maximum of the two numbers can be expected to bias the result upwards.
-
@Matthew: Thanks for catching the typo. – Brian M. Scott Oct 29 '12 at 4:38 | {
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-
@Matthew: Thanks for catching the typo. – Brian M. Scott Oct 29 '12 at 4:38
This is very much delayed, but consider the case with an $n$-sided die. As has already been observed, the expected value of the maximum of two $n$-sided die is
$${1 \over n^2} \sum_{k=1}^n (2k^2-k)$$
and we can write out this sum explicitly. In particular, we can expand to get
$${1 \over n^2} \left( \left( 2 \sum_{k=1}^n k^2 \right) - \sum_{k=1}^n k \right)$$ and recalling the formulas for those sums, this is
$${1 \over n^2} \left( {2n(n+1)(2n+1) \over 6} - {n(n+1) \over 2} \right)$$
or after some rearrangement
$${(n+1)(4n-1) \over 6n}.$$
In particular this is approximately $2n/3$. This could have been guessed if you know that the expectation of the maximum of two uniform random variables on $[0, 1]$ has the beta distribution $B(2,1)$, which has mean $2/3$.
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# Average Length of the Longest Arc in $S^1$
Suppose that we draw $n$ points $a_i \sim Uniform(S^1)$ for $i = 1 \ldots n$. These points determine a partition or set of disjoint arcs of $S^1$. What is the average length of the longest arc? To even measure the $n$ arc lengths, we need an ordering of the $\{ a_i \}$ , $a_{(1)}, \ldots, a_{(N)}$.
The arc lengths are:
$\begin{gather*} l_1 = |a_{(2)} - a_{(1)}| \\ \vdots \\ l_{n-1} = |a_{(n)} - a_{(n-1)}| \\ l_n = |a_{(1)} - a_{(n)}| \end{gather*}$
At this point we’d still have to order the arc lengths, to get the longest arc. Ultimately, we need the distribution of $l_{(n)}$, $P(l_{(n)} > x)$, to compute anything.
\begin{align*} P(l_{(n)} > x) &= P( l_1 > x, \text{ or } \ldots, \text{ or } l_n > x) \\ &= \sum^{n}_{k} \binom{n}{k} (-1)^{k-1} P( l_1 > x, \text{ and } \ldots, \text{ and } l_k > x), \end{align*} \begin{align*} P( l_1 > x, \text{ and } \ldots, \text{ and } l_k > x) & = \text{probability of } n-1 \text{ cuts in a } 2\pi - kx \text{ long arc}\\ & = \left( \frac{2 \pi - kx}{2\pi}\right)^{n-1} \end{align*}
But why!! The simple graphic below should make this clear. Essentially, we can transform the circle problem to one on a line, after the first cut: the unit circle problem with $n$ points is the same problem with $n-1$ points on a $2 \pi$ long line segment. The first cut is superfluous. From there, the argument boils down to recognizing that each cut is made in segments of total length $2\pi - kx$.
Lemma. A Suppose $X \sim p(x)$ is a non-negative random variable, then $E[X] = \int^{+\infty}_{0} P(X > x) dx.$
proof:
Define $F(x) = \int^{x}_{0} p(x) dx = P(X < x)$. Integrating by parts \begin{align*} \int^{b}_{0} x p(x) dx & = x \Big\lvert^{b}_{0} \int^{b}_{0}p(x)dx - \int^{b}_{0}F(x)dx \\ & = \int^{b}_{0} dx \int^{b}_{0}p(x)dx - \int^{b}_{0}F(x)dx \\ & = F(b)\int^{b}_{0} dx - \int^{b}_{0}F(x)dx \\ & = F(b)\int^{b}_{0} 1 - F(x)dx \\ & = F(b)\int^{b}_{0} P(X > x)dx \\ \end{align*} | {
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Taking the limit as $b \longrightarrow + \infty$ finishes the argument.
q.e.d.
Using the Lemma,
\begin{align*} E[l_{(n)}] &= \int^{+\infty}_{0} P(l_{(n)} > x) dx \\ &= \sum^{n}_{k} \binom{n}{k} (-1)^{k-1} \int^{+\infty}_{0}P(l_k > x)dx \\ &= \sum^{n}_{k} \binom{n}{k} (-1)^{k-1} \int^{+\infty}_{0}\left( \frac{2 \pi - kx}{2\pi}\right)^{n-1}dx \\ \end{align*}
Focusing on the integral, set $u = \frac{2\pi - kx}{2\pi}$,
$\frac{2 \pi - kx}{2\pi} \geq 0 \quad \Longrightarrow \quad \frac{2\pi}{k} \geq x$ \begin{align*} \int^{+\infty}_{0}\left( \frac{2 \pi - kx}{2\pi}\right)^{n-1}dx &= \int^{\frac{2\pi}{k}}_{0}\left( \frac{2 \pi - kx}{2\pi}\right)^{n-1}dx\\ &= \frac{2\pi}{-k}\int^{u(\frac{2\pi}{k})}_{u(0)} u^{n-1}du\\ &= \frac{2\pi}{k}\int^{1}_{0} u^{n-1}du\\ &= \frac{2\pi}{kn} \end{align*}
Substituting back into the formula,
\begin{align*} E[l_{(n)}] &= \frac{2\pi}{n} \sum^{n}_{k} \binom{n}{k} \frac{(-1)^{k-1}}{k} \\ &= \frac{2\pi}{n} \sum^{n}_{k} \frac{1}{k}. \end{align*}
The last step applies a known binomial sum identity.
Awesome! We have an answer! But is it right? Lets check our work, by generating the empirical distribution of max-segment lengths. I wrote up some code to do this.
In [16]: data = array([ MaxArc(10) for x in range(10000)])
In [17]: data.mean()
Out[17]: 1.8404203486925401
In [18]: 2 * pi * harmonic(10)/10
Out[18]: 1.8403250298528779
Both sample mean and true mean are within a few thousands of each other, which I think validates the analysis above. I took $N = 10$, but other choices will yield similar results. | {
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"openwebmath_score": 1.0000098943710327,
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# §4.7 Derivatives and Differential Equations
## §4.7(i) Logarithms
4.7.1 $\frac{\mathrm{d}}{\mathrm{d}z}\ln z=\frac{1}{z},$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $\ln\NVar{z}$: principal branch of logarithm function and $z$: complex variable A&S Ref: 4.1.46 Permalink: http://dlmf.nist.gov/4.7.E1 Encodings: TeX, pMML, png See also: Annotations for 4.7(i), 4.7 and 4
4.7.2 $\frac{\mathrm{d}}{\mathrm{d}z}\operatorname{Ln}z=\frac{1}{z},$
4.7.3 $\frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}\ln z=(-1)^{n-1}(n-1)!z^{-n},$
4.7.4 $\frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}\operatorname{Ln}z=(-1)^{n-1}(n-1)!z% ^{-n}.$
For a nonvanishing analytic function $f(z)$, the general solution of the differential equation
4.7.5 $\frac{\mathrm{d}w}{\mathrm{d}z}=\frac{f^{\prime}(z)}{f(z)}$
is
4.7.6 $w(z)=\operatorname{Ln}\left(f(z)\right)+\hbox{ constant}.$ ⓘ Defines: $w$: solution (locally) Symbols: $\operatorname{Ln}\NVar{z}$: general logarithm function, $z$: complex variable and $f(z)$: non-vanishing analytic function Permalink: http://dlmf.nist.gov/4.7.E6 Encodings: TeX, pMML, png See also: Annotations for 4.7(i), 4.7 and 4
## §4.7(ii) Exponentials and Powers | {
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## §4.7(ii) Exponentials and Powers
4.7.7 $\frac{\mathrm{d}}{\mathrm{d}z}e^{z}=e^{z},$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $\mathrm{e}$: base of exponential function and $z$: complex variable A&S Ref: 4.2.49 Permalink: http://dlmf.nist.gov/4.7.E7 Encodings: TeX, pMML, png See also: Annotations for 4.7(ii), 4.7 and 4
4.7.8 $\frac{\mathrm{d}}{\mathrm{d}z}e^{az}=ae^{az},$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $\mathrm{e}$: base of exponential function, $a$: real or complex constant and $z$: complex variable A&S Ref: 4.2.50 (gives n-th derivative.) Permalink: http://dlmf.nist.gov/4.7.E8 Encodings: TeX, pMML, png See also: Annotations for 4.7(ii), 4.7 and 4
4.7.9 $\frac{\mathrm{d}}{\mathrm{d}z}a^{z}=a^{z}\ln a,$ $a\neq 0$.
When $a^{z}$ is a general power, $\ln a$ is replaced by the branch of $\operatorname{Ln}a$ used in constructing $a^{z}$.
4.7.10 $\frac{\mathrm{d}}{\mathrm{d}z}z^{a}=az^{a-1},$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $a$: real or complex constant and $z$: complex variable A&S Ref: 4.2.52 Permalink: http://dlmf.nist.gov/4.7.E10 Encodings: TeX, pMML, png See also: Annotations for 4.7(ii), 4.7 and 4
4.7.11 $\frac{{\mathrm{d}}^{n}}{{\mathrm{d}z}^{n}}z^{a}=a(a-1)(a-2)\cdots(a-n+1)z^{a-n}.$
The general solution of the differential equation
4.7.12 $\frac{\mathrm{d}w}{\mathrm{d}z}=f(z)w$
is
4.7.13 $w=\exp\left(\int f(z)\mathrm{d}z\right)+{\rm constant}.$
The general solution of the differential equation
4.7.14 $\frac{{\mathrm{d}}^{2}w}{{\mathrm{d}z}^{2}}=aw,$ $a\neq 0$,
is
4.7.15 $w=Ae^{\sqrt{a}z}+Be^{-\sqrt{a}z},$
where $A$ and $B$ are arbitrary constants.
For other differential equations see Kamke (1977, pp. 396–413). | {
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# Problem 4-1¶
Location: assignment-problems/flatten.py
Grading: 1 point for passing test, and then (assuming it passes the test) 1 point for code quality
Write a function flatten which takes a nested dictionary and converts it into a flat dictionary based on the key names. You can assume that the nested dictionary only has one level of nesting, meaning that in the output, each key will have exactly one underscore.
Assert that your function passes the following test:
>>> colors = {
'animal': {
'bumblebee': ['yellow', 'black'],
'elephant': ['gray'],
'fox': ['orange', 'white']
},
'food': {
'apple': ['red', 'green', 'yellow'],
'cheese': ['white', 'orange']
}
}
>>> flatten(colors)
{
'animal_bumblebee': ['yellow', 'black'],
'animal_elephant': ['gray'],
'animal_fox': ['orange', 'white'],
'food_apple': ['red', 'green', 'yellow'],
'food_cheese': ['white', 'orange']
}
# Problem 4-2¶
Location: assignment-problems/convert_to_base_2.py
Grading: 1 point for passing test, and then (assuming it passes the test) 1 point for code quality
Write a function convert_to_base_2 that converts a number from base-10 to base-2. Assert that it passes the following test:
>>> convert_to_base_2(19)
10011
Hint: use $\log_2$ to figure out how many digits there will be in the binary number. Then, fill up the binary number, repeatedly subtracting off the next-largest power of 2 if possible.
# Problem 4-3¶
Location: assignment-problems/linear_encoding_cryptography.py
Grading: for each part, you get 1 point for passing test, and then (assuming it passes the test) 1 point for code quality
In Assignment 1, we encountered the trivial encoding function which maps
• ' ' $\rightarrow 0,$
• 'a' $\rightarrow 1,$
• 'b' $\rightarrow 2,$
and so on.
Using a linear encoding function $s(x) = 2x+3,$ the message 'a cat' can be encoded as follows:
1. Original message: 'a cat'
2. Trivial encoding: [1, 0, 3, 1, 20]
3. Linear encoding: [5, 3, 9, 5, 43] | {
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2. Trivial encoding: [1, 0, 3, 1, 20]
3. Linear encoding: [5, 3, 9, 5, 43]
a. Create a function encode(string,a,b) which encodes a string using the linear encoding function $s(x) = ax+b.$ Assert that your function passes the following test:
>>> get_encoding('a cat', 2, 3)
[5, 3, 9, 5, 43]
b. Create a function decode(numbers,a,b) which attempts to decode a given list of numbers using the linear encoding function $s(x) = ax+b.$
To do this, you should apply the inverse encoding $s^{-1}(x) = \dfrac{x-b}{a},$ to all the numbers in the list and then check if they are all integers in the range from $0$ to $26$ (inclusive). If they are, then return the corresponding letters; if they are not, then return False.
Assert that your function passes the following tests:
>>> decode([5, 3, 9, 5, 43], 2, 3)
'a cat'
for debugging purposes, here's the scratch work for you:
[(5-3)/2, (3-3)/2, (9-3)/2, (5-3)/2, (43-3)/2]
[1, 0, 3, 1, 20]
'a cat'
>>> decode([1, 3, 9, 5, 43], 2, 3)
False
for debugging purposes, here's the scratch work for you:
[(1-3)/2, (3-3)/2, (9-3)/2, (5-3)/2, (43-3)/2]
[-1, 0, 3, 1, 20]
False (because -1 does not correspond to a letter)
>>> decode([5, 3, 9, 5, 44], 2, 3)
False
for debugging purposes, here's the scratch work for you:
[(5-3)/2, (3-3)/2, (9-3)/2, (5-3)/2, (44-3)/2]
[1, 0, 3, 1, 20.5]
False (because 20.5 does not correspond to a letter)
c. Decode the message
[377,
717,
71,
513,
105,
921,
581,
547,
547,
105,
377,
717,
241,
71,
105,
547,
71,
377,
547,
717,
751,
683,
785,
513,
241,
547,
751],
given that it was encoded with a linear encoding function $s(x) = ax+b$ where $a,b \in \{ 0, 1, 2, \ldots, 100 \}.$
You should run through each combination of $a$ and $b,$ try to decode the list of numbers using that combination, and if you get a valid decoding, then print it out. Then, you can visually inspect all the decodings you printed out to find the one that makes sense.
# Problem 3-1¶
Location: assignment-problems/convert_to_base_10.py | {
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# Problem 3-1¶
Location: assignment-problems/convert_to_base_10.py
Grading: 1 point for passing test, and then (assuming it passes the test) 1 point for code quality
Write a function convert_to_base_10 that converts a number from base-2 (binary) to base-10 (decimal). For example, the binary number $10011$ corresponds to the decimal number $$1 \cdot 2^{4} + 0 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 = 19.$$
Assert that your function passes the following test:
>>> convert_to_base_10(10011)
19
# Problem 3-2¶
Location: assignment-problems/make_nested.py
Grading: you get 1 point for passing the test, and then (assuming it passes the test) 1 point for code quality
Write a function make_nested which takes a "flat" dictionary and converts it into a nested dictionary based on underscores in the the key names. You can assume that all keys have exactly one underscore.
Assert that your function passes the following test:
>>> colors = {
'animal_bumblebee': ['yellow', 'black'],
'animal_elephant': ['gray'],
'animal_fox': ['orange', 'white'],
'food_apple': ['red', 'green', 'yellow'],
'food_cheese': ['white', 'orange']
}
>>> make_nested(colors)
{
'animal': {
'bumblebee': ['yellow', 'black'],
'elephant': ['gray'],
'fox': ['orange', 'white']
},
'food': {
'apple': ['red', 'green', 'yellow'],
'cheese': ['white', 'orange']
}
}
# Problem 3-3¶
Location: assignment-problems/stack.py
Grading: you get 0.5 points for passing each test, and then (assuming your code passes all the tests) 2 points for code quality.
Implement a stack. That is, create a class Stack which operates on an attribute data using the following methods:
• push: add a new item on top of the stack
• pop: remove the top (rightmost) item from the stack
• peek: return the top item without modifying the stack
Assert that your class passes the following sequence of 5 tests. (You should write 5 assert statements in total.) | {
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>>> s = Stack()
>>> s.data
[]
>>> s.push('a')
>>> s.push('b')
>>> s.push('c')
>>> s.data
['a', 'b', 'c']
>>> s.pop()
>>> s.data
['a', 'b']
>>> s.peek()
'b'
>>> s.data
['a', 'b']
# Problem 2-1¶
Location: assignment-problems/union_intersection.py
a. (1 point for code quality; 1 point for passing test)
Write a function intersection that computes the intersection of two lists. Assert that it passes the following test:
>>> intersection([1,2,'a','b'], [2,3,'a'])
[2,'a']
b. (1 point for code quality; 1 point for passing test)
Write a function union that computes the union of two lists. Assert that it passes the following test:
>>> union([1,2,'a','b'], [2,3,'a'])
[1,2,3,'a','b']
# Problem 2-2¶
Location: assignment-problems/count_characters.py
(2 points for code quality; 2 points for passing test)
Write a function count_characters that counts the number of each character in a string and returns the counts in a dictionary. Lowercase and uppercase letters should not be treated differently.
Assert that your function passes the following test:
>>> countCharacters('A cat!!!')
{'a': 2, 'c': 1, 't': 1, ' ': 1, '!': 3}
# Problem 2-3¶
Location: assignment-problems/recursive_sequence.py
Consider the sequence defined recursively as
$$a_n = 3a_{n-1} -4, \quad a_1 = 5.$$
a. (1 point for code quality; 1 point for passing test)
Write a function first_n_terms that returns a list of the first $n$ terms of the sequence: $[a_1, a_2, a_3, \ldots, a_{n}]$
Assert that your function passes the following test:
>>> first_n_terms(10)
[5, 11, 29, 83, 245, 731, 2189, 6563, 19685, 59051]
b. (1 point for code quality; 1 point for passing test)
Write a function nth_term that computes the $n$th term of the sequence using recursion. Here's the video that you were asked to watch before class, in case you need to refer back to it: https://www.youtube.com/watch?v=zbfRgC3kukk
Assert that your function passes the following test:
>>> nth_term(10)
59051
# Problem 1-1¶
Getting started... | {
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>>> nth_term(10)
59051
# Problem 1-1¶
Getting started...
2. Create a bash repl named assignment-problems
3. Create a file assignment-problems/test_file.py
5. On repl.it, link assignment-problems to your github and push up your work to github. Name your commit "test commit".
6. After you complete this assignment, again push your work up to github. Name your commit "completed assignment 1".
# Problem 1-2¶
Location: assignment-problems/is_symmetric.py
Note: This problem is worth 1 point for passing both tests, plus another 1 point for code quality (if you pass the tests). So, the rubric is as follows:
• 0/2 points: does not pass both tests
• 1/2 points: passes both tests but code is poor quality
• 2/2 points: passes both tests and code is high quality
Write a function is_symmetric(input_string) that checks if a string reads the same forwards and backwards, and assert that your function passes the following tests:
>>> is_symmetric('racecar')
True
>>> is_symmetric('batman')
False
To be clear -- when you run is_symmetric.py, your code should print the following:
>>> python is_symmetric.py
testing is_symmetric on input 'racecar'...
PASSED
testing is_symmetric on input 'batman'...
PASSED
# Problem 1-3¶
Location: assignment-problems/letters_numbers_conversion.py
a. (1 point for passing test, 1 point for code quality)
Write a function convert_to_numbers(input_string) that converts a string to a list of numbers, where space = 0, a = 1, b = 2, and so on. Then, assert that your function passes the following test:
>>> letters2numbers('a cat')
[1,0,3,1,20]
b. (1 point for code quality, 1 point for passing test)
Write a function convert_to_letters(input_string) that converts a list of numbers to the corresponding string, and assert that your function passes the following test:
>>> convert_to_letters([1,0,3,1,20])
'a cat'
To be clear -- when you run letters_numbers_conversion.py, your code should print the following: | {
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To be clear -- when you run letters_numbers_conversion.py, your code should print the following:
>>> python letters_numbers_conversion.py
testing convert_to_letters on input [1,0,3,1,20]...
PASSED
testing convert_to_numbers on input 'batman'...
PASSED
# Problem 1-4¶
(2 points for passing tests, 2 points for code quality)
Write a function is_prime(n) that checks if an integer input $n > 1$ is prime by checking whether $m | n$ for any integer $m \in \left\{ 2, 3, \ldots, \left\lfloor \dfrac{n}{2} \right\rfloor \right\}.$
• $m|n$ means "$m$ divides $n$"
• $\left\lfloor \dfrac{n}{2} \right\rfloor$ is called the "floor" of $\dfrac{n}{2},$ i.e. the greatest integer that is less than or equal to $\dfrac{n}{2}.$
(Hint: Check for divisibility within a for loop.)
Also, assert that your function passes the following tests:
>>> is_prime(59)
True
>>> is_prime(51)
False
To be clear -- when you run is_prime.py, your code should print the following:
>>> python is_prime.py
testing is_prime on input 59...
PASSED
testing is_prime on input 51...
PASSED
In [ ]: | {
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# How to write $x^m+x^{-m}$ as a polynomial in $x+x^{-1}$.
I did find this question Proving that $x^m+x^{-m}$ is a polynomial in $x+x^{-1}$ of degree $m$. but it only shows by induction that this is possible, not the actual form of the solution. How, for example, would you write $$x^2+x^{-2}$$ in a polynomial of $$x+x^{-1}$$?
• Did you mean $x^2\color{red}+x^{-2}$? Because in that case it's $(x+x^{-1})^2-2$ Oct 25, 2019 at 4:18
• Yes I did, edited. How did you come up with that? Just from squaring $(x+x^-1)$ and subtracting off the extra $2$, or is there a more general way? Oct 25, 2019 at 4:21
• This answer math.stackexchange.com/a/1341510/42969 to the referenced question shows the explicit solution. Oct 25, 2019 at 4:34
It's just a simple system of equations (simple, if the degree is low...)
You want \begin{align} x^2+\frac1{x^2}&=a+b\left(x+\frac1x\right)+c\left(x+\frac1x\right)^2\\ &=a+2c+bx+\frac bx+cx^2+\frac c{x^2}. \end{align} And you see by inspection that $$a+2c=1$$, $$b=0$$, $$c=1$$. So $$a=-2$$. That is \begin{align} x^2+\frac1{x^2}&=-2+\left(x+\frac1x\right)^2\\ \end{align}
• Thanks, I will mark this as the accepted answer since it does generalize for $m>2$. I see it is very simple now, but thank you for the clear explanation. Oct 25, 2019 at 4:27
These are basically Chebyshev polynomials. One can derive an explicit formula for them via generating functions. | {
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Let $$y=x+x^{-1}$$ and consider the formal power series $$F(x,t)=\sum_{n=0}^\infty(x^n+x^{-n})t^n.$$ Then \begin{align} F(x,t)&=\sum_{n=0}^\infty x^nt^n+\sum_{n=0}^\infty x^{-n}t^n =\frac1{1-xt}+\frac1{1-x^{-1}t}=\frac{(1-xt)+(1-x^{-1}t)}{(1-xt)(1-x^{-1}t)}\\ &=\frac{2-(x+x^{-1})t}{1-(x+x^{-1})t+t^2}=\frac{2-yt}{1-yt+t^2}=(2-yt) \frac1{1-(y-t)t}\\ &=(2-yt)\sum_{m=0}^\infty(y-t)^mt^m=(2-yt)\sum_{m=0}^n\sum_{k=0}^{m} (-1)^k\binom{m}{k}y^{m-k}t^{m+k}\\ &=(2-yt)\sum_{m=0}^\infty(y-t)^mt^m=(2-yt)\sum_{n=0}^\infty t^n\sum_{k:0\le k\le n/2}(-1)^k\binom{n-k}{k}y^{n-2k}. \end{align} Comparing coefficients of $$t^n$$ gives $$x^n+x^{-n}=2\sum_{k:0\le k\le n/2}(-1)^k\binom{n-k}{k}y^{n-2k} -\sum_{k:0\le k\le (n-1)/2}(-1)^k\binom{n-1-k}{k}y^{n-2k}.$$ One can simplify this a bit: $$2\binom{n-k}{k}-\binom{n-1-k}{k}=\binom{n-k}{k}+\binom{n-k-1}{k-1} =\binom{n-k}{k}+\frac{k}{n-k}\binom{n-k}{k}$$ to get $$x^n+x^{-n}=\sum_{k:0\le k\le n/2}(-1)^k\frac{n}{n-k} \binom{n-k}{k}(x+x^{-1})^{n-2k}.$$
• Very interesting. I have not studied formal power series in more than one variable before. Oct 25, 2019 at 6:26
In order to get $$x^2$$ and $$x^{-2}$$ terms, square $$(x+x^{-1}$$). The result is $$x^2+2+x^{-2}$$.
So subtract $$2$$ to get what you want: $$x^2+x^{-2}=(x+x^{-1})^2-2$$.
• Thanks, does this generalize to $x^m+x^{-m}$ for $m>2$? Oct 25, 2019 at 4:25
• $(x+x^{-1})^3=x^3+3x+3x^{-1}+x^{-3}$ so $x^3+x^{-3}=(x+x^{-1})^3-3(x+x^{-1})$ Oct 25, 2019 at 4:30
• $(x+x^{-1})^4=x^4+4x^2+6+4x^{-2}+x^{-4}$ so $x^4+x^{-4}=(x+x^{-1})^4-4(x^2+x^{-2})-6=(x+x^{-1})^4-4(x+x^{-1})^2+2$ Oct 25, 2019 at 4:35
• I get the point, you can stop now :) Oct 25, 2019 at 4:39
• :), okay, thanks, will do Oct 25, 2019 at 4:39
The inductive argument does tell you how to do this - and, it's worthwhile to see that induction very often gives you the answers you are after, especially since the fact that inductive proofs can be unrolled is often overlooked. | {
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In the linked answer, it is noted that $$x^{k+1}+x^{-k-1} = (x^k +x^{-k})(x+x^{-1}) - (x^{k-1} + x^{-k+1}).$$ and an inductive argument is built from this, but remember that the inductive hypothesis is just that, for each $$k$$, there exists some polynomial $$P_k$$ such that $$P_k(x+x^{-1})= x^k+x^{-k}$$. If we fill out the proof more completely, the implication here is that $$x^{k+1}+x^{-k-1} = P_k(x+x^{-1})\cdot (x+x^{-1}) - P_{k-1}(x+x^{-1}).$$ And then we see that the left hand side is a polynomial in $$x+x^{-1}$$ as well - but we can be more explicit: Let $$P_{k+1}(z)=zP_k(z)-P_{k-1}(z)$$ where we start the sequence as $$P_0(z)=2$$ and $$P_1(z)=z$$. Then, we have $$P_k(x+x^{-1})=x^k+x^{-k}$$ due to the inductive argument. Note that this sequence is very easy to compute incrementally: $$P_2(z)=z\cdot P_1(z) - P_0(z) = z^2 - 2$$ $$P_3(z)=z\cdot P_2(z) - P_1(z) = z^3 - 3z$$ $$P_4(z)=z\cdot P_3(z) - P_2(z) = z^4 - 4z^2 + 2$$ and so on.
As pointed out in the comments, it is possible to write these terms in a general form, although it's a bit surprising that the form you get is actually a polynomial. I won't go into details, since I'm using the usual tools for solving linear homogenous recurrences:
$$P_k(z) = \frac{\left(z - \sqrt{z^2-4}\right)^k + \left(z + \sqrt{z^2-4}\right)^k}{2^k}$$
Note that the $$\sqrt{z^2-4}$$ terms cancel out due to the symmetry of the sum and this does always leave a polynomial.
(I also think this sequence has a name, but I don't know what it is)
• It might be worthwhile to expand this to actually obtain an explicit formula, which should be easy to do since we already have the recurrence (using e.g. generating functions). Oct 25, 2019 at 4:37
• That is very interesting. I have not thought to do this before. I will have to keep it in mind, thanks! Oct 25, 2019 at 4:41
• @YiFan Thanks, I added it, I'd thought that the explicit formula was a bit worse than it is when I was writing this. Oct 25, 2019 at 4:43 | {
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# Ways of merging two incomparable sorted lists of elements keeping their relative ordering
Suppose that, for a real application, I have ended up with a sorted list A = {$a_1, a_2, ..., a_{|A|}$} of elements of a certain kind (say, Type-A), and another sorted list B = {$b_1, b_2, ..., b_{|B|}$} of elements of a different kind (Type-B), such that Type-A elements are only comparable with Type-A elements, and likewise for Type-B.
At this point I seek to count the following: in how many ways can I merge both lists together, in such a way that the relative ordering of Type-A and Type-B elements, respectively, is preserved? (i.e. that if $P_M(x)$ represents the position of an element of A or B in the merged list, then $P_M(a_i)<P_M(a_j)$ and $P_M(b_i)<P_M(b_j)$ for all $i<j$)
I've tried to figure this out constructively by starting with an empty merged list and inserting elements of A or B one at a time, counting in how many ways each insertion can be done, but since this depends on the placement of previous elements of the same type, I've had little luck so far. I also tried explicitly counting all possibilities for different (small) lengths of A and B, but I've been unable to extract any potential general principle in this way.
The merged list has length $|A|+|B|$. Once you know which $|A|$ positions in it are occupied by the elements of list A, you also know exactly how the whole thing has to be ordered, since the internal orders of the elements of A and the elements of B are already known. Thus, there are $$\binom{|A|+|B|}{|A|}=\binom{|A|+|B|}{|B|}$$ possible merged lists, one for each choice of $|A|$ positions for the elements of list A. | {
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• Great answer! I've made an impl in C++ that generates all possible interleaved combinations of two lists, while maintaining relative order for anyone who is interested: github.com/domfarolino/algorithms/tree/master/src/… Oct 22, 2018 at 20:37
• Can this further generalize to 3 lists? To 4? To N? May 27, 2021 at 15:54
• Indeed, I believe this can further generalize to N lists, by viewing the overall operation as a series of (N-1) merges, so the total number of merged lists is computed as the product of (N-1) Binomial Coeffecients, where the i-th expression computes the selection of the positions of the i-th lists's elements from the available positions left after the first (i-1) lists have already been merged. May 27, 2021 at 21:38
• @BryanPendleton: Yes: it’s a multinomial coefficient, $$\binom{|A_1|+|A_2|+\ldots+|A_N|}{A_1,A_2,\ldots,A_N}\,,$$ which can be expanded as a product of binomial coefficients as shown at the link. May 28, 2021 at 3:57
• Great answer. Since we have to keep the relative ordering. This question is equal to "How many different ways are there to select n positions from m + n positions?“ Sep 17, 2021 at 5:47
$\frac{(m+n)!}{(m)!(n)!}$; where m,n is the length of the arrays. $(m+n)!$ is the total number of ways of arranging without any restriction. The condition given that you can not change the order of each array so you should not change the order of each array , so we need to divide all the permutations of each array!
For merging N sorted lists, here is a good way to see that the solution is $$\frac{(|A_1|+\dots|+|A_N|)!}{|A_1|!\dots |A_N|!}$$ | {
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All the $$(|A_1|+\dots|+|A_N|)$$ elements can be permuted in $$(|A_1|+\dots|+|A_N|)!$$ ways. Among these, any solution which has the ordering of the $$A_1$$ elements different from the given order have to be thrown out. If you keep the positions of everyone except $$A_1$$ fixed, you can generate $$|A_1|!$$ permutations by shuffling $$A_1$$. Only 1 among these $$|A_1|!$$ solutions is valid. Hence the number solutions containing the right order of $$A_1$$ is $$\frac{(|A_1|+\dots|+|A_N|)!}{|A_1|!}$$
Now repeat the argument for $$A_2$$, $$A_2, \dots A_N$$. | {
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# Finding coordinate point if two points and time are known
$A$ is at coordinate $(x_1,y_1)$ at $6:00$ am and is at coordinate $(x_2,y_2)$ at $6:15$ am. How can I know where $A$ will be at $6:05$ am i.e $(x_3,y_3)$? The value for $x_1=392,y_1=455,x_2=512,y_2=452$ are known and is straight line.
I understand there is a formula to calculate this in coordinate mathematics. I will appreciate your answer and if you share the formula or term in mathematics.
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In my answer, I assume $A$ moves at a constant speed. Is that what you need? – Rustyn Sep 24 '13 at 3:45
It takes $A$ $15$ minutes, to go $-3$ units downward and $120$ units rightward. So he's traveling, $$\sqrt{3^2 + 120^2} = 3\sqrt{1601}$$ units per $15$ minutes.
Thus, at $6:05$, he'll have gone $\sqrt{1601}$, units from is starting position, $\left(\tfrac{1}{3}\text{ the distance }\right)$. We know that the point $(x_3,y_3)$ is on the line: $$y-455=\frac{-3}{120}(x-392)$$ Thus, $y_3 = \frac{-3}{120}(x_3-392)+455$
AND: $$\sqrt{(x_3-x_1)^2+(y_3-y_1)^2} = \sqrt{(x_3-392)^2+(y_3-455)^2} = \sqrt{1601}$$ So, putting this all together we have: $$\sqrt{(x_3-392)^2+\left(\tfrac{-3}{120}(x_3-392)+455-455\right)^2} = \sqrt{1601} \Longrightarrow \dots$$ $$x_3=432, y_3 = 454$$ EDIT
As Ross Millikan states, we can avoid the square roots entirely. See his comment below for the simplest solution to this problem.
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You needn't do the square roots. Just compute the $x$ velocity $v_x=\frac {-3}{15}$ and $y$ velocity $v_y=\frac {120}{15}=8$ The position at time $t$ (measured in minutes after $6:00$) is $(x_1+v_xt, y_1+v_yt)$ – Ross Millikan Sep 24 '13 at 3:48
@RossMillikan You are absolutely right. Except for I think you got them switched around? $v_x = 120/15$, $v_y = -3/15$ – Rustyn Sep 24 '13 at 3:49
Is A moving at a constant speed? Use the distance formula: http://en.wikipedia.org/wiki/Distance_formula#Geometry | {
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At a constant speed, the distance traveled is proportional to the time spent traveling because of the distance = rate x time equation.
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# In a Commutative Ring, is Addition Necessarily Commutative?
In A First Course in Abstract Algebra, Fraleigh writes on p. 172 that "a ring in which multiplication is commutative is a commutative ring". Of course, this raises the question: is addition necessarily commutative in a commutative ring? Is it commutative in any ring? Are there examples of rings in which addition is non-commutative?
As far as I could tell, commutativity of the underlying additive operation is not one of the defining properties of a ring. Or is it? I don't think the book has been entirely clear on these issues --- I would be thankful if someone could shed some light on this.
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The definition of a ring $R$ is that it has two binary operations, $+$ which is an abelian group and $\cdot$ which is not necessarily commutative (and need not even be a group, merely associative). Of course, you have your extra distributive properties added as well. – mathematics2x2life Mar 7 '14 at 4:06
Possible duplicate of Why is ring addition commutative? – Najib Idrissi Nov 23 '15 at 18:59
The three basic properties of a ring are;
• The set under addition makes an abelian group,
• Multiplication is associative, and
• Left and right distributive laws hold.
Thus, by definition of "abelian group", the addition must be commutative. Hopes that help.
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Oh. I totally glossed over the fact that the underlying group under addition is abelian. – Newb Mar 7 '14 at 4:11
@Newb A ring with not necessarily commutative addition is sometimes called a near-ring. I don't know a whole lot of study about them (but I'm generally ignorant of that sort of thing). – Mike Miller Mar 7 '14 at 4:25 | {
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In order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other (standard) ring axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right distributive law in different order to the term $\rm\:(1\!+\!1)(x\!+\!y),\:$ viz.
$$\rm (1\!+\!1)(x\!+\!y) = \bigg\lbrace \begin{eqnarray} (1\!+\!1)x\!+\!(1\!+\!1)y\, =\, x \,+\, \color{#C00}{x\!+\!y} \,+\, y\\ \rm 1(x\!+\!y)\!+1(x\!+\!y)\, =\, x\, +\, \color{#0A0}{y\!+\!x}\, +\, y\end{eqnarray}\bigg\rbrace\:\Rightarrow\: \color{#C00}{x\!+\!y}\,=\,\color{#0A0}{y\!+\!x}\ \ by\ \ cancel\ \ x,y$$
Thus commutativity of addition, $\rm\:x+y = y+x,\:$ is implied by these axioms:
$(1)\ \ *\,$ distributes over $\rm\,+\!:\ \ x(y+z)\, =\, xy+xz,\ \ (y+z)x\, =\, yx+zx$
$(2)\ \, +\,$ is cancellative: $\rm\ \ x+y\, =\, x+z\:\Rightarrow\: y=z,\ \ y+x\, =\, z+x\:\Rightarrow\: y=z$
$(3)\ \, +\,$ is associative: $\rm\ \ (x+y)+z\, =\, x+(y+z)$
$(4)\ \ *\,$ has a neutral element $\rm\,1\!:\ \ 1x = x$
Said more structurally, recall that a SemiRing is that generalization of a Ring whose additive structure is relaxed from a commutative Group to merely a SemiGroup, i.e. here the only hypothesis on addition is that it be associative (so in SemiRings, unlike Rings, addition need not be commutative, nor need every element $\rm\,x\,$ have an additive inverse $\rm\,-x).\,$ Now the above result may be stated as follows: a semiring with $\,1\,$ and cancellative addition has commutative addition. Such semirings are simply subsemirings of rings (as is $\rm\:\Bbb N \subset \Bbb Z)\,$ because any commutative cancellative semigroup embeds canonically into a commutative group, its group of differences (in precisely the same way $\rm\,\Bbb Z\,$ is constructed from $\rm\,\Bbb N,\,$ i.e. the additive version of the fraction field construction). | {
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Examples of SemiRings include: $\rm\,\Bbb N;\,$ initial segments of cardinals; distributive lattices (e.g. subsets of a powerset with operations $\cup$ and $\cap$; $\rm\,\Bbb R\,$ with + being min or max, and $*$ being addition; semigroup semirings (e.g. formal power series); formal languages with union, concat; etc. For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.
[1] Gerhard Betsch. On the beginnings and development of near-ring theory. pp. 1-11 in:
Near-rings and near-fields. Proceedings of the conference held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong, Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz. Mathematics and its Applications, 336. Kluwer Academic Publishers Group, Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review
[2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $\$ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel. North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7 Zbl review, AMS review
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I would upvote this answer multiple times if I could. As usual, Bill, your answer is highly informative. – Newb Mar 7 '14 at 7:50
The definition of a ring is that it has two binary operations, $+$ and $\cdot$. The $+$ operation forms an abelian group and $\cdot$ need only be associative. The distributive laws need hold. However, notice that the distributive laws force $+$ to be abelian when $R$ has $1$!
$$(1+1)(x+y)=1(x+y)+1(x+y)=x+y+x+y$$
and
$$(1+1)(x+y)=(1+1)x+(1+1)y=x+x+y+y$$ so that $x+y+x+y=x+x+y+y$ then adding $-x$ and $-y$ ($+$ forms a group so has inverses $-x,-y$) on the left and right yields $y+x=x+y$. | {
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A lot of people don't accept a ring without unity as a ring. – Tim Seguine Mar 7 '14 at 11:18
But they do indeed exist and can be useful, thought they do not appear often. So it is important to know they rings are not required to have such properties. Then one can happily continue working with rings with identity. – mathematics2x2life Mar 7 '14 at 21:16 | {
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# -apc.4.1.2 Find the slope at x=4
• MHB
Gold Member
MHB
$\tiny{apc.4.1.2}$
Find the slope of the tangent line to the graph of
$f(x)=-x^2+4\sqrt{x}$ at $x=4$
$a.\ 8\quad b.\ -10\quad c.\ -9\quad d.\ -5\quad e.\ -7$
$f'(x)=-2x+\frac{2}{\sqrt{x}}$
$m=f'(4)=-2(4)+\frac{2}{\sqrt{4}}=-7$ which is (e)
however not asked for here but I forgot how to find $b$ of $y=-7x+b$
Last edited:
skeeter
$y - f(4) = -7(x-4)$
clean it up ...
Gold Member
MHB
$y - f(4) = -7(x-4)$
clean it up ...
that didn't seem to be a tangent line to f{x}
$-8=-7(4)+b$
$-8+28=20=b$
$y=-7x+20$
Last edited:
skeeter
the equation in my previous post is in point-slope form, $y - y_1 = m(x - x_1)$ where $m = -7, \, x_1 = 4 \text{ and } y_1 = f(4)$
$f(4) = -8$
$y + 8 = -7(x - 4) \implies y = -7x + 20$
HOI
that didn't seem to be a tangent line to f{x}
$-8=-7(4)+b$
$-8+28=20=b$
$y=-7x+20$
Yes, it is. Since $f(x)= -x^2+ 4\sqrt{x}$ so $f(4)= -4^2+4\sqrt{4}= -16+ 8= -8$. That is, the expression skeeter gave, $y- f(4)= -7(x- 4)$ becomes $y- (-8)= y+ 8= -7x+ 28$ so, subtracting 8 from both sides, $y= -7x+ 20$ as you have.
TiffanyBK
Is the option wrong? I used cameramath and the answer is -8
HOI
Is the option wrong? I used cameramath and the answer is -8
View attachment 10874
The answer to what question? You want to find the equation of the tangent line to $f(x)= -x^2+ 4\sqrt{x}$ at x= 4. You have already determined that the derivative of f, so the slope of the tangent line, at x= 4 is -7. Thus you know that the equation of the tangent line is y= -7x+ b for some number, b. So what is the value of y at x= 4? $y= f(4)= -(4^2)+ 4\sqrt{4}= -16+ 4(2)= -8$ -8 is the value of f(4). But that is not necessarily b! Now we know that we must have y= -7(4)+ b= -28+ b= -8. Adding 28 to both sides, b= 20.
The equation of the tangent line at x= 4 is y= -7x+ 20.
Last edited:
Gold Member
MHB
well i think the confusion was f(4) vs f'(4) | {
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# Why is the Monotone Convergence Theorem more famous than it's stronger cousin?
I am reading Stein & Shakarchi. On page 62 we have the Monotone Convergence Theorem:
Suppose $\{f_n\}$ is a sequence of non-negative measurable functions with $f_n\nearrow f.$ Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.
However, just before this theorem we have this much more powerful corrolary of Fatou's lemma:
Suppose $f$ is a non-negative measurable function, and $\{f_n\}$ a sequence of non-negative measurable functions with $f_n(x) \le f(x)$ and $f_n(x) \to f(x)$ for a.e. $x$. Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.
To me, this second corollary seems strictly better than the Monotone Convergence Theorem, yet it is the M.C.T. that has a name and is used often. Am I misunderstanding the theorems, or is there a reason why the M.C.T. is more popular?
Does this corollary have a name?
• Are you sure that corollary is true? I think you would still require integrability of $f$. – Calculon Mar 29 '18 at 9:55
• @Calculon I copied it word for word; I will recheck the proof – Ovi Mar 29 '18 at 9:57
• @Calculon $f$ is required to be non-negative and measurable, so the worst that could happen is that $\int f=+\infty$, in which case the inequality holds trivially. – Arnaud D. Mar 29 '18 at 9:58
• This question seems related : math.stackexchange.com/questions/2468412/… – Arnaud D. Mar 29 '18 at 10:08
• @ArnaudD. Nice catch! – Ovi Mar 30 '18 at 14:57
Some people do call this corollary monotone convergence theorem. Moreover, some instructora introduce this corollary as M.C.T., rather than the origianl version of M.C.T., for instance, Professor Mark J. Schervish and Professor Alessandro Rinaldo at CMU (Lecture notes). It's fine because this corollary is stronger and more practical.
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Let's take a look at the Lebesgue's dominant convergence theorem:
Suppose $\{f_n\}$ and $f$ are measurable functions. If there exists a non-negative function $g$ such that $\int g <\infty$ and $|f_n|\leq g$ a.e., for all $n$. Then $f_n\overset{a.e.}{\rightarrow} f$ or $f_n\overset{\mu}{\rightarrow} f$ implies that $$\lim_{n \to \infty} \int f_n = \int f.$$
Now, compare this with the result you give:
Suppose $f$ is a non-negative measurable function, and $\{f_n\}$ a sequence of non-negative measurable functions with $f_n(x) \le f(x)$ and $f_n(x) \to f(x)$ for a.e. $x$. Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.
We can find that when $\int f <\infty$, the result is implied by the Lebesgue's D.C.T. When $\int f =\infty$, by Fatou's lemma we have
$$\infty = \int f = \int \liminf_{n\rightarrow \infty} f_n \leq \liminf_{n\rightarrow \infty} \int f_n\Rightarrow \int f_n\rightarrow \infty.$$
In summary, this result seems more similar to D.C.T than M.C.T. This is one reason why it cannot take the place of M.C.T.
The other reason is more fundamental. If we take a look at the proof of the Fatou's lemma, we may find that it relies on the monotonicity of $g_{k} = \inf_{n\geq k}f_n$ and $g_k\nearrow \liminf_{n\rightarrow \infty} f_n$. That is to say, the Fatou's lemma is actually implied by the M.C.T.
In other words, M.C.T. is the most fundamental theorem, then the Fatou's lemma and finally the D.C.T. Therefore from the perspective of mathematician, it's definitely better to use the "naive version" (actually the only correct version) of M.C.T., but not the seemingly stronger result you give, which is implied by D.C.T. and Fatou's lemma.
Nevertheless, it is true that the result you give here is much more convenient to use than either M.C.T. or D.C.T. That's why some instructors prefer to introduce this result as a substitution for M.C.T. | {
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• Your statement of Lebesgue's dominant convergence theorem doesn't make sense. (Typos?). What does $g$ do? – DanielWainfleet Mar 30 '18 at 0:42
• Thanks a lot for the reminder! Yes it's a typo haha. @DanielWainfleet Now already fixed. – Wanshan Mar 30 '18 at 0:46
• So just to make sure I understand: the result I posted is in fact implied by D.C.T and Fatou's Lemma? Because the top voted answer in the math overflow question gives an example where D.C.T. doesn't work but this theorem works; so I am thinking that that example does not work specifically for Lebesgue measure? mathoverflow.net/questions/296312/… – Ovi Apr 25 '18 at 16:06
• Thanks for the reference and sorry for the late reply. I think for his example we can still use DCT and Fatou to prove that claim. If $\int_X f d\mu<\infty$, we use DCT, otherwise we can use Fatou following the line in my answer to prove that the right hand side of his equality tends to infinity. He said that equality can be used for the definition of Lebesgue integral, but I think it does not make sense because any of the integral theorem depends on the definition of integral, and we cannot use a higher level result to define a lower level concept. @Ovi – Wanshan May 13 '18 at 21:54 | {
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Question
# Assertion :If the two pair of lines $$\displaystyle x^{2}-2m\:xy-y^{2}=0$$ and $$\displaystyle x^{2}-2n\:xy-y^{2}=0$$ are such that one of them represents the bisector of angle between the other, then $$\displaystyle 1+mn=0$$. Reason: Equation of bisector of angle between the pair of lines $$\displaystyle ax^{2}+2hxy+by^{2}=0$$ is given by $$\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}.$$
A
Both Assertion & Reason are individually true & Reason is correct explanation of Assertion.
B
Both Assertion & Reason are individually true but Reason is not the correct (proper) explanation of Assertion.
C
Assertion is true but Reason is false.
D
Assertion is false but Reason is true.
Solution
## The correct option is A Both Assertion & Reason are individually true & Reason is correct explanation of Assertion.Given $$x^2-2mxy-y^2=0$$The pair if bisector of given pair is $$\dfrac{x^2-y^2}{1-(-1)}=\dfrac{xy}{-m}$$$$\dfrac{x^2-y^2}{2}=\dfrac{xy}{-m}$$ or $$x^2+\dfrac{2xy}{m}-y^2=0$$ which is identical to $$x^2-2nxy-y^2=0$$$$\therefore \dfrac{1}{m}=-n \Rightarrow mn+1=0$$Maths
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# Value of the game from payoff matrix
I am absolutely new to decision theory . I came across this following payoff matrix in the book.(Math. Stats : John E Freund).
Player A
I II
Player B 1 7 -4
2 8 10
The value of the game is given as 8 units.However i . have a question . I agree the optimal strategy for Play B is "2" . But i want to argue that the optimal choice for player A is 2 . This is because in decision theory the assumption is made that "each player must choose a strategy without knowing what the opponent is going to do and that once a player has made a choice it can not be changed" ..Going by this logic, the optimal strategy for player A should be II , because i see in that case , the loss would be less overall for the moves of player B . This , logic if true gives the value of the game to be 10 , which is not correct. I want to know , where i am wrong.
• Is this a zero-sum game? Which player is the minimiser and which is the maximiser? – Theoretical Economist Apr 21 '18 at 2:48
The statement that "each player must choose a strategy without knowing what the opponent is going to do and that once a player has made a choice it can not be changed" just means the choice in each time the game is played are made without information of what the opponent played this time. The players are allowed to examine the whole payoff matrix and think about what the opponent might do. In this game Player B finds that move $2$ dominates move $1$. No matter what A does, B is better off playing $2$ than playing $1$, so he should play $2$ all the time. For A, neither strategy dominates the other, but A "knows" B will play $2$, so by playing $1$ A can lose $8$ instead of $10$. If B were so foolish to play $1$ it is true that A could do better by playing $2$, but we don't count on foolishness.
• Thanks a lot.. Your answer is a perfect clarification. – warrior_monk Apr 21 '18 at 2:55
The basic assumptions in a game theory are: | {
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The basic assumptions in a game theory are:
1) Each player is rational and tries to maximize his/her payoff in the game.
2) Each player knows that his/her opponent is rational and tries to maximize his/her payoff in the game.
You can solve this problem by Reducing by Dominance.
Since the entries in Row 2 are greater (or equal to) the corresponding ones in Row 1 (it implies Row 2 dominates Row 1, i.e. Player 2 is better off with strategy 2), we can eliminate Row 1 to get: $$\begin{array}{cc|cc} &&Player \ 1 \\ &&A&B \\ \hline Player \ 2 & 2 & 8 & 10 \end{array}$$ Since the entry in Column A is less than (or equal to) the corresponding one in Column B (it implies Column A dominates Column B, i.e. the Player 1 is better off with strategy A), we can eliminate Column B to get: $$\begin{array}{cc|c} &&Player \ 1 \\ &&A \\ \hline Player \ 2 & 2 & 8 \end{array}$$ So, the Player 2 has an advantage of $8$ units over Player 1.
• What if you begin with player 1(Player A in original ques)? In that case . "II" is the optimal strategy for him , which is Player 1 – warrior_monk Apr 22 '18 at 4:26
• Option 1: If P1 first chooses the strategy A, then P2 will choose the strategy 2 and P2 will win 8 units. Option 2: If P1 first chooses the strategy B, then P2 will choose the strategy 2 and P2 will win 10 units. Which option is rational (better) for P1? Certainly, Option 1 (because P1 loses less: 8 instead of 10). So, P1 knows that P2 is rational and P2 will maximize his/her payoff. For P2 the strategy 2 is better than strategy 1 (i.e. strategy 2 dominates strategy 1). Now, can you work out the two options when P2 starts and P1 ends and which option is rational for P2? – farruhota Apr 22 '18 at 5:22 | {
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# How many different permutations are there of the sequence of letters in “MISSISSIPPI”?
There are 11 letters in the word.
M - 1 I - 4 S - 4 P - 2
so the number of different permutations is $\dfrac{11!}{1! 4 !4!2!}$
Is this correct solution?
• Yes, this is exactly it. – Jimmy R. Mar 23 '15 at 20:46
Yes! Here's why it's correct:
If you numbered each of the $I$s, $S$s, and $P$s, there would be 11! permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P_1, P_2).$$
But, because you don't care about the numbers of the $P$s, each permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P)$$ is actually double-counted, once with $P_1$ before $P_2$ and once with $P_2$ before $P_1$. For example, in the original 11! permutations $$M,I_1,S_1,S_2,I_2,S_3,S_4,I_3,P_1,P_2,I_1$$ looks the same as $$M,I_1,S_1,S_2,I_2,S_3,S_4,I_3,P_2,P_1,I_1$$ if we ignore the numbers of the $P$s.
So, $\frac{11!}{2!}$ is the number of permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P).$$
For the same reason, we find 24-times the number of permutations of $$(M, I_1, I_2, I_3, I_4, S_1, S_2, S_3, S_4, P, P)$$ as $$(M, I, I, I, I, S_1, S_2, S_3, S_4, P, P),$$ one for each of the $4! = 24$ orderings of $(I_1, I_2, I_3, I_4).$ Thus there are $\frac{11!}{2!4!}$
And again there are 24-times the number of permutations of $$(M, I, I, I, I, S_1, S_2, S_3, S_4, P, P)$$ as $$(M, I, I, I, I, S, S, S, S, P, P),$$ one for each of the $4! = 24$ orderings of $(S_1, S_2, S_3, S_4).$ Thus there are $\frac{11!}{2!4!4!}$ total.
An alternative solution:
• Choose $1$ out of $11$ places for the M
• Choose $4$ out of $10$ remaining places for the I's
• Choose $4$ out of $6$ remaining places for the S's
• Choose $2$ out of $2$ remaining places for the P's
$$\binom{11}{1}\cdot\binom{10}{4}\cdot\binom{6}{4}\cdot\binom{2}{2}$$
Please note that you can apply this process in any order you'd like. | {
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Please note that you can apply this process in any order you'd like.
• I think it might be worth explicitly pointing that the result you get is the same as OP's and the ones on other answers. – Wojowu Mar 23 '15 at 21:32
• @Wojowu: Isn't that obvious from the opening statement an alternative solution? – barak manos Mar 23 '15 at 21:34 | {
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# Thread: prove this identity
1. ## prove this identity
2sin(x)/[(cos(x)+sin(x)] = tan(2x) + 1 - sec(2x)
First, I turn tan(2x) into [sin(2x) / cos(2x)] then I turn sec(2x) into 1/cos(2x)
So, I end up with... [sin(2x) / cos(2x)] + 1 - (1/(cos2x))
I multiple and get the GCD of cos(2x)..
[sin(2x) + cos(2x) - 1] / cos(2x)
Now things get kind of fuzzy.. I probably didn't do this the best way thus far but I can't really advance much further.. I need help, please.
2. Originally Posted by zodiacbrave
2sin(x)/[(cos(x)+sin(x)] = tan(2x) + 1 - sec(2x)
First, I turn tan(2x) into [sin(2x) / cos(2x)] then I turn sec(2x) into 1/cos(2x)
So, I end up with... [sin(2x) / cos(2x)] + 1 - (1/(cos2x))
I multiple and get the GCD of cos(2x)..
[sin(2x) + cos(2x) - 1] / cos(2x)
Now things get kind of fuzzy.. I probably didn't do this the best way thus far but I can't really advance much further.. I need help, please.
Start with right side with the identy
$tan(2x)=\frac{2tanx}{1-tan^2x}$ and $sec(2x)=\frac{1}{cos(2x)}=\frac{1+tan^2x}{1-tan^2x}$
Then the right side reduces to
$\frac{2tanx+2}{1-tan^2x}$
Chage tan to terms of sin and cos and you should be able to get there.
3. Hello, zodiacbrave!
Another approach . . .
Prove: . $\frac {2\sin x}{\cos x+\sin x} \:=\: \tan(2x) + 1 - \sec(2x)$
Multiply the left side by: . $\frac{\cos x - \sin x}{\cos x - \sin x}$
$\frac{2\sin x}{\cos x + \sin x}\cdot\frac{\cos x - \sin x}{\cos x - \sin x}$
. . $=\;\frac{\overbrace{2\sin x\cos x}^{\sin(2x)} - \overbrace{2\sin^2\!x}^{1-\cos(2x)}}{\underbrace{\cos^2\!x - \sin^2\!x}_{\cos(2x)}}$
. . $= \;\frac{\sin(2x) + \cos(2x) - 1}{\cos(2x)}$
. . $=\;\frac{\sin(2x)}{\cos(2x)} + \frac{\cos x}{\cos x} - \frac{1}{\cos(2x)}$
. . $=\;\tan(2x) + 1 - \sec(2x)$
4. Soroban's LATEXing technique is superb and very instructive, thanks MHF expert | {
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# Questions about continuous functions.
Recently when working with my thesis, I've got 2 questions.
1. Let $S_n$ be the set $\{x=(x_1,x_2,\cdots,x_n)\in\mathbb{R}^n\mid x_1+x_2+\cdots+x_n=1~\mbox{and}~0\leq x_i~\mbox{for}~ i=1,2,\cdots,n\}$. Is $S_n$ compact or convex, or closed? Or does it possess any topological property?
2. Let $s$ be a point of the product $S_{n_1}\times S_{n_2}\times\cdots S_{n_k}$, where $S_{n_i}$ is the subset of Euclidean space $\mathbb{R}^{n_i}$ satisfying the description above.
$~~~~~~~~$Let $f$ and $g$ be two continuous functions(linear) from $S_{n_1}\times S_{n_2}\times\cdots S_{n_k}$ to $\mathbb{R}$, and define $\phi:S_{n_1}\times S_{n_2}\times\cdots S_{n_k}\rightarrow\mathbb{R}$ by $\phi(s)=\max\{0,f(s)-g(s)\}$. Then is $\phi$ continuous? Why?
Can anyone help me with these questions?
(Sorry to bring these questions with long long descriptions. :'( )
Added: For the last part, I think my real question is why the function, max, is continuous. Any suggestion? And thanks for other answers and advices. :)
-
The objects $S_n$ are the standard n-Simplices. See en.wikipedia.org/wiki/Simplex#The_standard_simplex for example. The properties you mention can be proven elementary. – canaaerus Feb 9 at 9:05
Yes, $S_n$ is compact, convex and closed : all three. Your map is continuous also, since $\sf max$ is continuous.
To see why $S_n$ is closed, write it as a finite intersection (in fact, $n+1$) of closed sets.
$S_n$ is compact because it is both closed and bounded.
$S_n$ is convex because it is defined by convex conditions.
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$S_n$ is convex because it is defined by convex conditions.
-
Each of your $S_n$ sets is the intersection of the convex and closed positive orthant $\{x \geq 0 \}$ with a level set $\{f=1\}$ of the convex function $f(x) = |x_1+ \dots +x_n|$. Its convexity follows from an application of the triangle inequality: $$\left|\sum \lambda x_i + (1-\lambda)y_i\right| = \left|\lambda \sum x_i + (1-\lambda)\sum y_i\right|\leq \lambda \left|\sum x_i\right| + (1-\lambda) \left|\sum y_i\right|.$$ As level sets of convex functions are also convex and closed, $S_n$ is convex and closed.
They are also compact. To see it you only need to notice that the value of each coordinate of any point in $S_n$ is bounded below by $0$ and above by $1$. Since a closed and bounded set in $\mathbb R^n$ is compact, $S_n$ is compact.
Finally, considering that $f-g$ and $\max\{0,\cdot\}$ are continuous functions, the function $\phi$ defined in your question is a composition of continuous functions, and therefore continuous.
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What is this notation, similar to the binomial coefficient?
I've come accross this notation:
$$\left\{\begin{eqnarray}n\\m\end{eqnarray}\right\}$$
The only other info I have about this notation is that $\left\{\begin{eqnarray}4\\2\end{eqnarray}\right\}=7$
What's the name of this notation and what is it used for?
Thanks
-
I find it slightly disappointing that with the vast knowledge we have here, all three answers do little more than link to wikipedia. – user1729 Jun 18 '14 at 10:33
There is a Stirling numbers, which I added to your post. If you are interested in them then you might want to browse through the questions there. – user1729 Jun 18 '14 at 10:44
Maybe you're using Stirling numbers of the second kind, where $\displaystyle{n\brace k}$ denotes the number of ways to partition a set of $n$ objects into $k$ non-empty subsets.
You can use {n\brace k} to produce ${n\brace k}$ and \displaystyle{n\brace k} to produce $\displaystyle{n\brace k}$.
An example of a 'real world' application of these numbers can be found in this MSE question which asks how many rooks can be placed on a triangular chessboard so that none of them are attacking each other.
-
Any chance of an application? It would make your answer complete...(for example, they can be used to find certain generating sets of certain semigroups). – user1729 Jun 18 '14 at 10:25
Can you elaborate your answer by providing a real life example of its usage? Thanks... – Tunk-Fey Jun 18 '14 at 10:29
Stirling numbers of second kind, the number of partitions of a set of $n$ objects into $k$ non-empty subsets.
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Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 27 Jun 2018, 18:28 7 9 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64 $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
### Show Tags
29 Jun 2018, 06:58
28
9
First, choose a nice, round number, such as 70, within the range of values.
Then calculate the average with the following formula:
Average = Nice number + Average of differences from the nice number
Average = 70 + (4 - 1 - 6 + 9 - 6 + 14 + 7)/7 = 70 + 21/7 =73 | {
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Average = 70 + (4 - 1 - 6 + 9 - 6 + 14 + 7)/7 = 70 + 21/7 =73
Answer: C
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 25 Jun 2018, 05:10 Solution Given: • Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77
To find:
• Smith’s average weekly grocery bill over the 7-week period
Approach and Working:
• Smith’s total bill amount over the 7-week period = (74 + 69 + 64 + 79 + 64 + 84 + 77) = 511
• Therefore, Smith’s average bill amount over the same period = $$\frac{511}{7}$$ = 73
Hence, the correct answer is option C.
Answer: C
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### Show Tags
28 Jun 2018, 18:07
1
Bunuel wrote:
Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
We can determine the average using the formula: average = sum / number:
average = (74 + 69 + 64 + 79 + 64 + 84 + 77)/7 = 511/7 = 73
Answer: C
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 28 Aug 2018, 03:42 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
E. $85 NEW question from GMAT® Official Guide 2019 (PS07369) Easy way out is consider avg 70 - so diff for all terms is 4, -1 , -6, 9 , -6,14, 7 = sum is 21 = divided by 7 = equal to 3 so 70 +3 = 73 avg Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4868 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink]
### Show Tags
10 Sep 2018, 07:34
Bunuel wrote:
Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
NEW question from GMAT® Official Guide 2019
(PS07369)
$$\frac{( 70 + 4 ) + ( 70 - 1 ) + ( 60 + 4 ) + ( 80 - 1 ) + ( 60 + 4 ) + ( 80 + 4 ) + ( 70 + 7 )}{7}$$
= $$\frac{( 70 + 70 + 60 + 80 + 60 + 80 + 70 ) + ( 4 - 1 + 4 - 1 + 4 + 4 + 7 )}{7}$$
= $$\frac{490 + 21}{7}$$
= $$\frac{490}{7} + \frac{21}{7}$$
= $$70 + 3$$
= $$73$$, Answer must be (C)
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### Show Tags
02 Oct 2018, 00:09
1
This solution might be too blunt for some people. Sum the unit digits and you will have 41.
Now there are 7 numbers, which mean if x multiply by 7 must will result a number with unit digit of 1, so the x is 3.
The only answer with 3 as its unit digit is C.
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### Show Tags
27 Jan 2020, 12:45
Notice that all options' values have different unit digit;
To Solve: Add the unit digits of each values(4+9+4+9+4+4+7) = 41 and check the unit digit of output (unit digit 1).
for division by 7, 3 should be unit digit of the quotient.
Only Option C satisfies, it is $73! _________________ Unable to Give Up! Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink] 27 Jan 2020, 12:45
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# New Free Booklet on Applied Category Theory
edited September 2018
Tai-Danae Bradley has a new free "booklet" on applied category theory. It grew out of the workshop Applied Category Theory 2018, and I think it makes a great complement to Fong and Spivak's books Seven Sketches:
Abstract. This is a collection of introductory, expository notes on applied category theory, inspired by the 2018 Applied Category Theory Workshop, and in these notes we take a leisurely stroll through two themes (functorial semantics and compositionality), two constructions (monoidal categories and decorated cospans) and two examples (chemical reaction networks and natural language processing) within the field.
Check it out!
• Options
1.
Comment Source:Woohoo more fun reading! Thanks for the link!
• Options
2.
Looking forward to reading it! Still need to close many blind spots in this class, but this reading will further help :) Happy to see so much material approachable to layman persons! :)
Comment Source:Looking forward to reading it! Still need to close many blind spots in this class, but this reading will further help :) Happy to see so much material approachable to layman persons! :)
• Options
3.
In page 24 there is helpful insight on the hints of the end paragraph of previous lecture. Efforts like this aimed at a wider categorical outreach out of closed circles shine. Thanks to the author!
Comment Source:In page 24 there is helpful insight on the hints of the end paragraph of previous [lecture](https://forum.azimuthproject.org/discussion/2342/lecture-75-chapter-4-the-grand-synthesis#latest). Efforts like this aimed at a wider categorical outreach out of closed circles shine. Thanks to the author!
• Options
4.
edited September 2018 | {
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The links provided in the further reading section are also great especially this blog which was mentioned during John's lectures as well. Now with more understanding of the subject the blog seems to be extremely enlightening.
Also, I came across a paper written by John Baez and James Dolan called From Finite Sets to Feynman Diagrams. I found it to be very illuminating as well. I never thought too deeply about $$\mathbf{FinSet}$$ but it is a great place to start for newbies and I regret not doing so earlier.
Comment Source:The links provided in the further reading section are also great especially [this blog](http://www.math3ma.com) which was mentioned during John's lectures as well. Now with more understanding of the subject the blog seems to be extremely enlightening. Also, I came across a paper written by John Baez and James Dolan called [From Finite Sets to Feynman Diagrams](https://arxiv.org/pdf/math/0004133.pdf). I found it to be very illuminating as well. I never thought too deeply about \$$\mathbf{FinSet}\$$ but it is a great place to start for newbies and I regret not doing so earlier.
• Options
5.
Great!! This was very helpful in explaining -- in a mathematically entertaining way -- the pith of some nice topics in applied category theory.
Comment Source:Great!! This was very helpful in explaining -- in a mathematically entertaining way -- the pith of some nice topics in applied category theory.
• Options
6.
=D> !
Comment Source:=D> !
• Options
7.
To help me digest these ideas, I thought I'd write a summary of Tai-Danae Bradley's booklet. I've added my own reactions and I've tried link the ideas to what we've discussed in this course.
## Two Themes
Tai-Danae starts with two basic themes in applied category. (1) Functorial semantics and (2) compositionality.
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### Functorial Semantics
Modeling natural human language is a good example (and Tai-Danae discusses this example at length). Roughly speaking, language consists of words ("semantics") and ways to combine to words to create sentences/new words ("grammar"). This distinction can be captured by a functor between two categories!
$$F : C \to D$$ The category $$C$$ specifies the grammar (how words act, how they can be combined; see compositionality) while the functor assigns meaning to diagrams in $$C$$ (where the category $$D$$ represents meaning). So assigning individual words their meaning.
Another good example in John's Lectures and Seven Sketches in Compositionality is Chapter 3: Database Schemas. A database schema was a category presented by a graph. The schema specified objects (data) and some morphisms (relations; functions).
The database schema tells us the grammar of our database. It says what data is required, what actions can be performed, how different types of data are related, etc. But it doesn't say what the data actually is (or more abstractly, how we should interpret the objects and morphisms). We get an actual database by constructing a functor which sends our database schema to the category $$\mathbf{Set}$$ (assigning sets to objects; and functions to morphisms).
I think the most important idea here (especially for modeling/ theoretical science) is that we can separate how complex systems combine (this is the grammar) and how those complex systems function (this is the semantics). And we can interpret the same grammar with different semantics (which allows us to be somewhat agnostic about the fundamentals of a complex system). | {
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Electric circuits are a good example of this last point. We can model a electric circuit using a labeled graph, then construct a category where these labeled graphs are morphisms so we can compose them (this is done with cospans). Each edge in the graph is thought of as some basic component (like a resistor). These constructions don't tell us how voltages or currents distribute over the circuit (i.e., they don't tell us what a resistor really means), just how circuits combine. Choosing how components relate currents and voltages (Ohm's law for linear resistors) is then a functor which sends the labeled graphs to some other category that says how the circuit behaves.
But the grammar of electric circuits constrains how the circuits can behave, because whatever semantics we choose, the semantics has to preserve the functor axioms (intuitively, it has to distribute over our grammar). So if a resistor behaves like a linear relation between two vector spaces, then sticking resistors together still has to be a linear relation on vector spaces.
### Compositionality
This idea is relatively straight-forward. Essentially, complex things can be constructed out of simpler parts. But as pointed out in Chapter 1 of Seven Sketches, how you combine parts to form a whole matters! The whole is not necessary the sum of its parts (but it might be up to isomorphism!). A system's components might be behave a bit differently when studied separately than when glued together.
Biology has many, many examples of this phenomenon! Chemical reactions studied in vitro (in a beaker) aren't the same as in vivo (in a cell). The environment/context matters! The cell has many other reactions that influence the reaction under consideration. Making measurements of system components might not agree with measurements of the system as a whole. | {
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But this problem affects other fields! Quantum mechanics has this very issue with quantum interference (probabilities depend on the whole system; and can't be added up from measuring decoupled components; this is very roughly speaking as I am not really much of a physicist! ). Even the more hands-on case of electric circuits has this same problem. Adding components like resistors changes what you measure elsewhere in the network.
A goal in applied category theory is to identify the right algebraic structure for systematically building a system out of some component pieces. Using functorial semantics, algebraic structure in our "syntax" category is preserved. So how systems can be combined partially determines how they behave.
### The Yoneda Lemma (bonus theme)
Tai-Danae mentions the Yoneda Lemma is her discussion of natural language modeling. The Yoneda Lemma gets the heart of why category theory is useful for modeling! I won't give the formal statement, but I will also quote John Firth:
You shall know a word by the company it keeps.
Essentially this is what the Yoneda lemma says: "You shall know an object by the company it keeps." Or "You shall know an object by how it interacts with other objects." In other words, you can define/study an object by how it interacts with other objects. And in category theory, "how it interacts with other objects" means "the morphisms between this object and others".
To quote from the movie Forrest Gump:
Stupid is as stupid does
Similarly, "$$X$$ is as $$X$$ does" is a wonderful encapsulation of the Yoneda Lemma.
What does this mean for modeling? Sometimes you don't need to know what a thing is, just what it does!
## Two constructions
Tai-Danae discusses two constructions: monoidal categories and decorated cospans. I won't really summarize these constructions, but discuss how they could be used in modeling.
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### Monoidal Categories
A monoidal category gives us the syntax/grammar to talk about systems where we can talk about composition in series (via morphisms) and in parallel (via tensoring). We've have many, many examples in this course:
• Monoidal preorders: e.g. integers with order $$\le$$ (as morphisms) under addition (as tensoring)
• Resource monoidal preorder: stuff (like pie ingredients) and mixtures of stuff (tensoring) with processes/reactions turning stuff/mixtures into other stuff/mixtures (morphisms)
• Database schema: data types with combined data types (tensoring) and relationships/actions between types (morphisms)
• Co-design diagrams
along more traditional math examples (Tai-Danae gives some good examples).
Another way to think about the monoid part of a monoidal category is to notice that in several examples the monoidal product lets us combine or mix together basic stuff. For example, chemical reactions turn mixtures of certain substances to other substances. We can describe reactions as morphisms (morphisms are good for processes) while mixing stuff together is tensoring.
### Decorated Cospans
Let's imagine a pipe (or a wire). Pipes have no natural distinction between "input" and "output". Water can go in either direction. But morphisms are always directional! If we want to turn "direction agnostic" processes into a morphisms, we have to find a way to make distinction between input and output not really matter. Decorated cospans are the answer (well an answer).
A cospan is the following diagram: $$X \rightarrow N \leftarrow Y$$ where the object $$N$$ represents our pipe (potentially bidirectional process). What these objects are depends on the category (finite sets is a common choice since we will have to be able to compute colimits for any objects to get a category; and finite sets always have colimits). | {
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The cospan can be decorated with some extra structure (which might say things about how our pipe can be combined with other pipes). Decorating our cospan is the same idea as the functorial semantics. | {
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Comment Source:To help me digest these ideas, I thought I'd write a summary of Tai-Danae Bradley's booklet. I've added my own reactions and I've tried link the ideas to what we've discussed in this course. ## Two Themes Tai-Danae starts with two basic themes in applied category. (1) Functorial semantics and (2) compositionality. ### Functorial Semantics Modeling natural human language is a good example (and Tai-Danae discusses this example at length). Roughly speaking, language consists of words ("semantics") and ways to combine to words to create sentences/new words ("grammar"). This distinction can be captured by a functor between two categories! $F : C \to D$ The category \$$C\$$ specifies the grammar (how words act, how they can be combined; see compositionality) while the functor assigns meaning to diagrams in \$$C\$$ (where the category \$$D\$$ represents meaning). So assigning individual words their meaning. Another good example in John's Lectures and *Seven Sketches in Compositionality* is Chapter 3: Database Schemas. A database schema was a category presented by a graph. The schema specified objects (data) and some morphisms (relations; functions). The database schema tells us the grammar of our database. It says what data is required, what actions can be performed, how different types of data are related, etc. But it doesn't say what the data actually is (or more abstractly, how we should interpret the objects and morphisms). We get an actual database by constructing a functor which sends our database schema to the category \$$\mathbf{Set}\$$ (assigning sets to objects; and functions to morphisms). I think the most important idea here (especially for modeling/ theoretical science) is that we can separate how complex systems combine (this is the grammar) and how those complex systems function (this is the semantics). And we can interpret the same grammar with different semantics (which allows us to be somewhat agnostic about the fundamentals of a complex | {
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different semantics (which allows us to be somewhat agnostic about the fundamentals of a complex system). Electric circuits are a good example of this last point. We can model a electric circuit using a labeled graph, then construct a category where these labeled graphs are morphisms so we can compose them (this is done with cospans). Each edge in the graph is thought of as some basic component (like a resistor). These constructions don't tell us how voltages or currents distribute over the circuit (i.e., they don't tell us what a resistor really means), just how circuits combine. Choosing how components relate currents and voltages (Ohm's law for linear resistors) is then a functor which sends the labeled graphs to some other category that says how the circuit behaves. But the grammar of electric circuits constrains how the circuits can behave, because whatever semantics we choose, the semantics has to preserve the functor axioms (intuitively, it has to distribute over our grammar). So if a resistor behaves like a linear relation between two vector spaces, then sticking resistors together still has to be a linear relation on vector spaces. ### Compositionality This idea is relatively straight-forward. Essentially, complex things can be constructed out of simpler parts. But as pointed out in Chapter 1 of *Seven Sketches*, how you combine parts to form a whole matters! The whole is not necessary the sum of its parts (but it might be up to isomorphism!). A system's components might be behave a bit differently when studied separately than when glued together. Biology has many, many examples of this phenomenon! Chemical reactions studied *in vitro* (in a beaker) aren't the same as *in vivo* (in a cell). The environment/context matters! The cell has many other reactions that influence the reaction under consideration. Making measurements of system components might not agree with measurements of the system as a whole. But this problem affects other fields! Quantum | {
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not agree with measurements of the system as a whole. But this problem affects other fields! Quantum mechanics has this very issue with quantum interference (probabilities depend on the whole system; and can't be added up from measuring decoupled components; this is very roughly speaking as I am not really much of a physicist! ). Even the more hands-on case of electric circuits has this same problem. Adding components like resistors changes what you measure elsewhere in the network. A goal in applied category theory is to identify the right algebraic structure for systematically building a system out of some component pieces. Using functorial semantics, algebraic structure in our "syntax" category is preserved. So how systems can be combined partially determines how they behave. ### The Yoneda Lemma (bonus theme) Tai-Danae mentions the Yoneda Lemma is her discussion of natural language modeling. The Yoneda Lemma gets the heart of why category theory is useful for modeling! I won't give the formal statement, but I will also quote John Firth: > You shall know a word by the company it keeps. Essentially this is what the Yoneda lemma says: "You shall know an object by the company it keeps." Or "You shall know an object by how it interacts with other objects." In other words, you can define/study an object by how it interacts with other objects. And in category theory, "how it interacts with other objects" means "the morphisms between this object and others". To quote from the movie Forrest Gump: > Stupid is as stupid does Similarly, "\$$X\$$ is as \$$X\$$ does" is a wonderful encapsulation of the Yoneda Lemma. What does this mean for modeling? Sometimes you don't need to know what a thing is, just what it does! ## Two constructions Tai-Danae discusses two constructions: monoidal categories and decorated cospans. I won't really summarize these constructions, but discuss how they could be used in modeling. ### Monoidal Categories A monoidal category gives us the | {
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discuss how they could be used in modeling. ### Monoidal Categories A monoidal category gives us the syntax/grammar to talk about systems where we can talk about composition in series (via morphisms) and in parallel (via tensoring). We've have many, many examples in this course: + Monoidal preorders: e.g. integers with order \$$\le\$$ (as morphisms) under addition (as tensoring) + Resource monoidal preorder: stuff (like pie ingredients) and mixtures of stuff (tensoring) with processes/reactions turning stuff/mixtures into other stuff/mixtures (morphisms) + Database schema: data types with combined data types (tensoring) and relationships/actions between types (morphisms) + Co-design diagrams along more traditional math examples (Tai-Danae gives some good examples). Another way to think about the monoid part of a monoidal category is to notice that in several examples the monoidal product lets us combine or mix together basic stuff. For example, chemical reactions turn mixtures of certain substances to other substances. We can describe reactions as morphisms (morphisms are good for processes) while mixing stuff together is tensoring. ### Decorated Cospans Let's imagine a pipe (or a wire). Pipes have no natural distinction between "input" and "output". Water can go in either direction. But morphisms are always directional! If we want to turn "direction agnostic" processes into a morphisms, we have to find a way to make distinction between input and output not really matter. Decorated cospans are the answer (well an answer). A cospan is the following diagram: $X \rightarrow N \leftarrow Y$ where the object \$$N\$$ represents our pipe (potentially bidirectional process). What these objects are depends on the category (finite sets is a common choice since we will have to be able to compute colimits for any objects to get a category; and finite sets always have colimits). The cospan can be decorated with some extra structure (which might say things about how our pipe can | {
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The cospan can be decorated with some extra structure (which might say things about how our pipe can be combined with other pipes). Decorating our cospan is the same idea as the functorial semantics. | {
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# Prove that $R\subseteq X\times X$ is an equivalence relation and construct its equivalence class
Prove that relation $$R\subseteq X\times X$$, where $$X= \mathbb{R}\times\mathbb{R}$$, is an equivalence relation and construct its equivalence class. $$R$$ is defined as: $$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \Longleftrightarrow x_1^2+y_1^2 = x_2^2+y_2^2$$
The way I did prove that it is an equivalence relation was based on equality relationship between two or more quantities. To make it easier to read, we can also substitute $$a = x_1^2+y_1^2$$, $$b = x_2^2+y_2^2$$, $$c = x_3^2+y_3^2$$, at least I think we can.
The first property is reflexivity:
$$\langle x_1, y_1\rangle R \langle x_1, y_1\rangle \Longleftrightarrow x_1^2+y_1^2 = x_1^2+y_1^2$$, which is always true, as $$a=a$$.
Then we check for symmetry:
$$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \Longrightarrow \langle x_2, y_2\rangle R \langle x_1, y_1\rangle$$, so if $$x_1^2+y_1^2 = x_2^2+y_2^2$$, then $$x_2^2+y_2^2 = x_1^2+y_1^2$$. Using the substitution we can say that $$(a=b) \Rightarrow (b=a)$$, which is also true.
The last one is transitivity:
$$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \wedge \langle x_2, y_2\rangle R \langle x_3, y_3\rangle \Longrightarrow \langle x_1, y_1\rangle R \langle x_3, y_3\rangle$$, which means $$x_1^2+y_1^2 = x_2^2+y_2^2 \wedge x_2^2+y_2^2 = x_3^2+y_3^2 \Longrightarrow x_1^2+y_1^2 = x_3^2+y_3^2$$, using substitutions it is $$(a=b \wedge b=c) \Longrightarrow (a=c)$$.
Up to this moment I think that my line of thinking is quite correct, but I am not sure whether that proof is good enough and it is what I would like to know.
When it comes to equivalence class of $$R$$, I would say that the relation tells us about points on the common circle, but I have to write it up with the definition, which is:
Equivalence class of element $$a\in A$$ in regard to equivalence relation $$R\subseteq A\times A$$ is a set $$[a]_R = \{b\in A\: |\: a R b\}$$. | {
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$$[\langle a,b \rangle]_R = \{\langle a, b\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$$ I would like you to tell me whether my proof (also usage of substitutions) and construction of equivalence class is right, also how could I get grasp of it, as lectures I attend are not good enough.
• You first part(the relation part) looks good, the second part it should be $[\langle a,b \rangle]_R = \{\langle x, y\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$: this is the set of all elements of $A\times A$ that are in relation with $\langle a,b\rangle$(here $A$ is indeed $\Bbb R\times \Bbb R$) – Holo Dec 16 '18 at 21:14
I like your proof, but I'd write the equivalence classes as follows:
$$[(a,b)]_R=\{(x,y)\in \mathbb{R}^2|a^2+b^2=x^2+y^2\}$$
which translates to: "The equivalence class of $$(a,b)$$ is the set of all points $$(x,y)$$ having the same distance from 0 as $$(a,b)$$.
Your idea is quite good and can be better formalized.
Consider a set $$X$$ and a map $$f\colon X\to Z$$, $$Z$$ any set. Then you can define a relation $$\sim_f$$ on $$X$$ by decreeing that $$x\sim_f y\quad\text{if and only if}\quad f(x)=f(y)$$ Then $$\sim_f$$ is clearly an equivalence relation: the proof is easily based on the properties of equality, which is essentially what you did just with more complicated symbols.
For $$a\in X$$, its equivalence class is $$[a]_{\sim_f}=\{x\in X:f(x)=f(a)\}$$.
In your case, $$X=\mathbb{R}\times\mathbb{R}$$, $$Z=\mathbb{R}$$ and $$f(x,y)=x^2+y^2$$ (square of the distance of $$(x,y)$$ from the origin).
Thus the equivalence class of $$(a,b)\in X$$ is the set of all points in $$X=\mathbb{R}\times\mathbb{R}$$ that share the same distance from the origin as $$(a,b)$$.
• Is it more correct to say "and a map $f\colon X\to Z$, where $Z$ is any set"? – manooooh May 23 at 19:14
• @manooooh As correct. – egreg May 23 at 19:32 | {
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185 views
Given an array of 1,000,000 distinct integers you are going to perform 100,000 searches, one at a time, using the linear search algorithm. Each element you are searching for is in the array. What is the expected, total number of comparisons performed by the 100,000 linear searches?
Given Ans: 5x10^10 + 5x10^4
Expected number of searches for a single linear search of $n$ elements
$$= 1.\frac{1}{n} + 2.\frac{1}{n} + \dots n.\frac{1}{n}$$
as elements are distinct and the probability of any element being equal to the searched element is $\frac{1}{n}$ as element is guaranteed to be present
$= \frac{n+1}{2}$.
Here, $n = 1,000,000$. So, expected number of comparisons $= 1,000,001/2$
Expected no. of comparisons for 100,000 such searches $= 100,000 \times 1,000,001/2 \\=5 \times 10^{10} + 5 \times 10^4$
selected
good question and good explaination ,thanks :)
@Arjun Sir, after seeing the question,my approach was like:
For 1st search, the number of comparisons would be O(n).
For 2nd search, the number of comparisons will be O(n-1) since now, 1 element has been decreased from the array.
Similarly, the searches would continue in the form of O(n)+O(n-1)+O(n-2)+O(n-3)+........+1=O(n^2).
So, for 100,000 searches, Total comparisons= (n^2)*100,000
Given, n=1000,000, so, Total Comparisons= 100,000 *100,000 *100000=10^15. | {
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Given, n=1000,000, so, Total Comparisons= 100,000 *100,000 *100000=10^15.
Kindly guide me where am I going wrong since this is not the answer. :(
@Bongbirdie, the question does not mention that once an element is found in the array, it is removed from it. So even after x number of searches are performed, the array still has 1,000,000 entries. Even if you keep track of the indices at which your search succeeded, to skip those indices in subsequent searches, you have to make a comparison anyway. For example, suppose the first search succeeded at index 200 and you remember this fact for performing the second search. Now while performing the second search, if you decide that I will skip the index 200, then each index has to be compared to 200 to decide whether or not to skip it. So instead of reducing the number of comparisons, you end up increasing it.
For the above two reasons, the number of possible comparison counts does not decrease in consecutive iterations of the linear search.
Hope this helps :) | {
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# Tag Archives: calculus
## Many Roads Give Same Derivative
A recent post in the AP Calculus Community expressed some confusion about different ways to compute $\displaystyle \frac{dy}{dx}$ at (0,4) for the function $x=2ln(y-3)$. I share below the two approaches suggested in the original post, proffer two more, and a slightly more in-depth activity I’ve used in my calculus classes for years. I conclude with an alternative to derivatives of inverses.
### Two Approaches Initially Proposed
1 – Accept the function as posed and differentiate implicitly.
$\displaystyle \frac{d}{dx} \left( x = 2 ln(y-3) \right)$
$\displaystyle 1 = 2*\frac{1}{y-3} * \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{y-3}{2}$
Which gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ at (x,y)=(0,4).
2 – Solve for y and differentiate explicitly.
$\displaystyle x = 2ln(y-3) \longrightarrow y = 3 + e^{x/2}$
$\displaystyle \frac{dy}{dx} = e^{x/2} * \frac{1}{2}$
Evaluating this at (x,y)=(0,4) gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ .
### Two Alternative Approaches
3 – Substitute early.
The question never asked for an algebraic expression of $\frac{dy}{dx}$, only the numerical value of this slope. Because students tend to make more silly mistakes manipulating algebraic expressions than numeric ones, the additional algebra steps are unnecessary, and potentially error-prone. Admittedly, the manipulations are pretty straightforward here, in more algebraically complicated cases, early substitutions could significantly simplify work. Using approach #1 and substituting directly into the second line gives
$\displaystyle 1 = 2 * \frac{1}{y-3} * \frac{dy}{dx}$.
At (x,y)=(0,4), this is
$\displaystyle 1 = 2 * \frac{1}{4-3}*\frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{1}{2}$
The numeric manipulations on the right side are obviously easier than the earlier algebra.
4 – Solve for $\frac{dx}{dy}$ and reciprocate. | {
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4 – Solve for $\frac{dx}{dy}$ and reciprocate.
There’s nothing sacred about solving for $\frac{dy}{dx}$ directly. Why not compute the derivative of the inverse and reciprocate at the end? Differentiating first with respect to y eventually leads to the same solution.
$\displaystyle \frac{d}{dy} \left( x = 2 ln(y-3) \right)$
$\displaystyle \frac{dx}{dy} = 2 * \frac{1}{y-3}$
At (x,y)=(0,4), this is
$\displaystyle \frac{dx}{dy} = \frac{2}{4-3} = 2$, so
$\displaystyle \frac{dy}{dx} = \frac{1}{2}$.
### Equivalence = A fundamental mathematical concept
I sometimes wonder if teachers should place much more emphasis on equivalence. We spend so much time manipulating expressions in mathematics classes at all levels, changing mathematical objects (shapes, expressions, equations, etc.) into a different, but equivalent objects. Many times, these manipulations are completed under the guise of “simplification.” (Here is a brilliant Dan Teague post cautioning against taking this idea too far.)
But it is critical for students to recognize that proper application of manipulations creates equivalent expressions, even if when the resulting expressions don’t look the same. The reason we manipulate mathematical objects is to discover features about the object in one form that may not be immediately obvious in another.
For the function $x = 2 ln(y-3)$, the slope at (0,4) must be the same, no matter how that slope is calculated. If you get a different looking answer while using correct manipulations, the final answers must be equivalent.
### Another Example
A similar question appeared on the AP Calculus email list-server almost a decade ago right at the moment I was introducing implicit differentiation. A teacher had tried to find $\displaystyle \frac{dy}{dx}$ for
$\displaystyle x^2 = \frac{x+y}{x-y}$ | {
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$\displaystyle x^2 = \frac{x+y}{x-y}$
using implicit differentiation on the quotient, manipulating to a product before using implicit differentiation, and finally solving for y in terms of x to use an explicit derivative.
1 – Implicit on a quotient
Take the derivative as given:\$
$\displaystyle \frac{d}{dx} \left( x^2 = \frac{x+y}{x-y} \right)$
$\displaystyle 2x = \frac{(x-y) \left( 1 + \frac{dy}{dx} \right) - (x+y) \left( 1 - \frac{dy}{dx} \right) }{(x-y)^2}$
$\displaystyle 2x * (x-y)^2 = (x-y) + (x-y)*\frac{dy}{dx} - (x+y) + (x+y)*\frac{dy}{dx}$
$\displaystyle 2x * (x-y)^2 = -2y + 2x * \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{-2x * (x-y)^2 + 2y}{2x}$
2 – Implicit on a product
Multiplying the original equation by its denominator gives
$x^2 * (x - y) = x + y$ .
Differentiating with respect to x gives
$\displaystyle 2x * (x - y) + x^2 * \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$
$\displaystyle 2x * (x-y) + x^2 - 1 = x^2 * \frac{dy}{dx} + \frac{dy}{dx}$
$\displaystyle \frac{dy}{dx} = \frac{2x * (x-y) + x^2 - 1}{x^2 + 1}$
3 – Explicit
Solving the equation at the start of method 2 for y gives
$\displaystyle y = \frac{x^3 - x}{x^2 + 1}$.
Differentiating with respect to x gives
$\displaystyle \frac{dy}{dx} = \frac {\left( x^2+1 \right) \left( 3x^2 - 1\right) - \left( x^3 - x \right) (2x+0)}{\left( x^2 + 1 \right) ^2}$
Equivalence
Those 3 forms of the derivative look VERY DIFFERENT. Assuming no errors in the algebra, they MUST be equivalent because they are nothing more than the same derivative of different forms of the same function, and a function’s rate of change doesn’t vary just because you alter the look of its algebraic representation.
Substituting the y-as-a-function-of-x equation from method 3 into the first two derivative forms converts all three into functions of x. Lots of by-hand algebra or a quick check on a CAS establishes the suspected equivalence. Here’s my TI-Nspire CAS check. | {
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Here’s the form of this investigation I gave my students.
### Final Example
I’m not a big fan of memorizing anything without a VERY GOOD reason. My teachers telling me to do so never held much weight for me. I memorized as little as possible and used that information as long as I could until a scenario arose to convince me to memorize more. One thing I managed to avoid almost completely were the annoying derivative formulas for inverse trig functions.
For example, find the derivative of $y = arcsin(x)$ at $x = \frac{1}{2}$.
Since arc-trig functions annoy me, I always rewrite them. Taking sine of both sides and then differentiating with respect to x gives.
$sin(y) = x$
$\displaystyle cos(y) * \frac{dy}{dx} = 1$
I could rewrite this equation to give $\frac{dy}{dx} = \frac{1}{cos(y)}$, a perfectly reasonable form of the derivative, albeit as a less-common expression in terms of y. But I don’t even do that unnecessary algebra. From the original function, $x=\frac{1}{2} \longrightarrow y=\frac{\pi}{6}$, and I substitute that immediately after the differentiation step to give a much cleaner numeric route to my answer.
$\displaystyle cos \left( \frac{\pi}{6} \right) * \frac{dy}{dx} = 1$
$\displaystyle \frac{\sqrt{3}}{2} * \frac{dy}{dx} = 1$
$\displaystyle \frac{dy}{dx} = \frac{2}{\sqrt{3}}$
And this is the same result as plugging $x = \frac{1}{2}$ into the memorized version form of the derivative of arcsine. If you like memorizing, go ahead, but my mind remains more nimble and less cluttered.
One final equivalent approach would have been differentiating $sin(y) = x$ with respect to y and reciprocating at the end.
### CONCLUSION
There are MANY ways to compute derivatives. For any problem or scenario, use the one that makes sense or is computationally easiest for YOU. If your resulting algebra is correct, you know you have a correct answer, even if it looks different. Be strong!
## CAS and Normal Probability Distributions | {
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## CAS and Normal Probability Distributions
My presentation this past Saturday at the 2015 T^3 International Conference in Dallas, TX was on the underappreciated applicability of CAS to statistics. This post shares some of what I shared there from my first year teaching AP Statistics.
MOVING PAST OUTDATED PEDAGOGY
It’s been decades since we’ve required students to use tables of values to compute by hand trigonometric and radical values. It seems odd to me that we continue to do exactly that today for so many statistics classes, including the AP. While the College Board permits statistics-capable calculators, it still provides probability tables with every exam. That messaging is clear: it is still “acceptable” to teach statistics using outdated probability tables.
In this, my first year teaching AP Statistics, I decided it was time for my students and I to completely break from this lingering past. My statistics classes this year have been 100% software-enabled. Not one of my students has been required to use or even see any tables of probability values.
My classes also have been fortunate to have complete CAS availability on their laptops. My school’s math department deliberately adopted the TI-Nspire platform in part because that software looks and operates exactly the same on tablet, computer, and handheld platforms. We primarily use the computer-based version for learning because of the speed and visualization of the large “real estate” there. We are shifting to school-owned handhelds in our last month before the AP Exam to gain practice on the platform required on the AP.
The remainder of this post shares ways my students and I have learned to apply the TI-Nspire CAS to some statistical questions around normal distributions.
FINDING NORMAL AREAS AND PROBABILITIES | {
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FINDING NORMAL AREAS AND PROBABILITIES
Assume a manufacturer makes golf balls whose distances traveled under identical testing conditions are approximately normally distributed with a mean 295 yards with a standard deviation of 3 yards. What is the probability that one such randomly selected ball travels more than 300 yards?
Traditional statistics courses teach students to transform the 300 yards into a z-score to look up in a probability table. That approach obviously works, but with appropriate technology, I believe there will be far less need to use or even compute z-scores in much the same way that always converting logarithms to base-10 or base-to use logarithmic tables is anachronistic when using many modern scientific calculators.
TI calculators and other technologies allow computations of non-standard normal curves. Notice the Nspire CAS calculation below the curve uses both bounds of the area of interest along with the mean and standard deviation of the distribution to accomplish the computation in a single step.
So the probability of a randomly selected ball from the population described above going more than 300 yards is 4.779%.
GOING BACKWARDS
Now assume the manufacturing process can control the mean distance traveled. What mean should it use so that no more than 1% of the golf balls travel more than 300 yards?
Depending on the available normal probability tables, the traditional approach to this problem is again to work with z-scores. A modified CAS version of this is shown below.
Therefore, the manufacturer should produce a ball that travels a mean 293.021 yards under the given conditions.
The approach is legitimate, and I shared it with my students. Several of them ultimately chose a more efficient single line command: | {
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But remember that the invNorm() and normCdf() commands on the Nspire are themselves functions, and so their internal parameters are available to solve commands. A pure CAS, “forward solution” still incorporating only the normCdf() command to which my students were first introduced makes use of this to determine the missing center.
DIFFERENTIATING INSTRUCTION
While calculus techniques definitely are NOT part of the AP Statistics curriculum, I do have several students jointly enrolled in various calculus classes. Some of these astutely noted the similarity between the area-based arguments above and the area under a curve techniques they were learning in their calculus classes. Never being one to pass on a teaching moment, I pulled a few of these to the side to show them that the previous solutions also could have been derived via integration.
I can’t recall any instances of my students actually employing integrals to solve statistics problems this year, but just having the connection verified completely solidified the mathematics they were learning in my class.
CONFIDENCE INTERVALS
The mean lead level of 35 crows in a random sample from a region was 4.90 ppm and the standard deviation was 1.12 ppm. Construct a 95 percent confidence interval for the mean lead level of crows in the region.
Many students–mine included–have difficulty comprehending confidence intervals and resort to “black box” confidence interval tools available in most (all?) statistics-capable calculators, including the TI-Nspire.
As n is greater than 30, I can compute the requested z-interval by filling in just four entries in a pop-up window and pressing Enter.
Convenient, for sure, but this approach doesn’t help the confused students understand that the confidence interval is nothing more than the bounds of the middle 95% of the normal pdf described in the problem, a fact crystallized by the application of the tools the students have been using for weeks by that point in the course. | {
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Notice in the solve+normCdf() combination commands that the unknown this time was a bound and not the mean as was the case in the previous example.
EXTENDING THE RULE OF FOUR
I’ve used the “Rule of Four” in every math class I’ve taught for over two decades, explaining that every mathematical concept can be explained or expressed four different ways: Numerically, Algebraically, Graphically (including graphs and geometric figures), and Verbally. While not the contextual point of his quote, I often cite MIT’s Marvin Minsky here:
“You don’t understand anything until you learn it more than one way.”
Learning to translate between the four representations grants deeper understanding of concepts and often gives access to solutions in one form that may be difficult or impossible in other forms.
After my decades-long work with CAS, I now believe there is actually a 5th representation of mathematical ideas: Tools. Knowing how to translate a question into a form that your tool (in the case of CAS, the tool is computers) can manage or compute creates a different representation of the problem and requires deeper insights to manage the translation.
I knew some of my students this year had deeply embraced this “5th Way” when one showed me his alternative approach to the confidence interval question:
I found this solution particularly lovely for several reasons. | {
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I found this solution particularly lovely for several reasons.
• The student knew about lists and statistical commands and on a whim tried combining them in a novel way to produce the desired solution.
• He found the confidence interval directly using a normal distribution command rather than the arguably more convenient black box confidence interval tool. He also showed explicitly his understanding of the distribution of sample means by adjusting the given standard deviation for the sample size.
• Finally, while using a CAS sometimes involves getting answers in forms you didn’t expect, in this case, I think the CAS command and list output actually provide a cleaner, interval-looking result than the black box confidence interval command much more intuitively connected to the actual meaning of a confidence interval.
• While I haven’t tried it out, it seems to me that this approach also should work on non-CAS statistical calculators that can handle lists.
(a very minor disappointment, quickly overcome)
Returning to my multiple approaches, I tried using my student’s newfound approach using a normCdf() command.
Alas, my Nspire returned the very command I had entered, indicating that it didn’t understand the question I had posed. While a bit disappointed that this approach didn’t work, I was actually excited to have discovered a boundary in the current programming of the Nspire. Perhaps someday this approach will also work, but my students and I have many other directions we can exploit to find what we need. | {
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