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http://ozad.flcgorizia.it/fourier-transform-python.html | Fourier Transform Python
The input signal. We also have a quick-reference cheatsheet (new!) to help you get started!. And how you can make pretty things with it, like this thing:. I am trying to translate this gravitational wave signal processing tutorial from Python to R, which I am much more familiar with. Aliyazicioglu Electrical & Computer Engineering Dept. While I'll be using the scientific Python stack in this blog post, code in Matlab, R should not be that different. There is also an inverse Fourier transform that mathematically synthesizes the original function from its frequency domain representation. Fourier Transform in Numpy¶ First we will see how to find Fourier Transform using Numpy. The period is taken to be 2 Pi, symmetric around the origin, so the. As far as image processing is concerned, we shall focus only on 2D Discrete Fourier Transform (DFT). We will focus on understanding the math behind the formula and use Python to do some simple applications of the DFT and fully appreciate its utility. Ask Question Asked 5 years ago. Transform homework assignments will be based on your own. If you found this comparison interesting, consider series 3 (7K text) and series 4 (7K text). efine the Fourier transform of a step function or a constant signal unit step what is the Fourier transform of f (t)= 0 t< 0 1 t ≥ 0? the Laplace transform is 1 /s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1 /jω in fact, the integral ∞ −∞ f (t) e − jωt dt = ∞ 0 e − jωt dt = ∞ 0 cos. Discrete Fourier transform transforms a sequence of complex or real numbers x n into a sequence of complex numbers X n. According to ISO 80000-2*), clauses 2-18. Fourier Transform of an image is quite useful in computer vision. I'm using a Fourier Transform method (not sure if its the same as the Split-Step method), where I apply Four. In the previous posts, we have seen what Fourier Transform of images is and how to actually do it with opencv and numpy. The Fourier transform is commonly used to convert a signal in the time spectrum to a frequency spectrum. If inverse is TRUE, the (unnormalized) inverse Fourier transform is returned, i. The Fourier Transform is a way how to do this. Fourier transforms are often used to calculate the frequency spectrum of a signal that changes over time. , if y <- fft(z), then z is fft(y, inverse = TRUE) / length(y). Cal Poly Pomona ECE 307 Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. Apply the Fourier transform two more times, so a second reversal undoes the first, and you're back to the original vector. In other words, a spectrum is the frequency domain representation of the input audio's time-domain signal. Fast Fourier Transform (FFT) Algorithm 79 Recall that the DFT is a matrix multiplication (Fig. I would like to use MATLAB to plot power spectral density of force platforms traces from various impacts. Quite naturally, the frequency domain has the same four cases, discrete or continuous in frequency, and. gain a deeper appreciation for the DFT by applying it to simple applications using Python; be able to mathematically and programmatically determine note/chord of a sound file using the DFT in Python. Furthermore, different representations of the comb function are described. The Fourier Transform (FFT) •Based on Fourier Series - represent periodic time series data as a sum of sinusoidal components (sine and cosine) •(Fast) Fourier Transform [FFT] - represent time series in the frequency domain (frequency and power) •The Inverse (Fast) Fourier Transform [IFFT] is the reverse of the FFT. spectrograms), and many kinds of image/audio processing, but is rarely used for compression. ppt - Free download as Powerpoint Presentation (. The Fourier transform of the product of two signals is the convolution of the two signals, which is noted by an asterix (*), and defined as: This is a bit complicated, so let’s try this out. That is, if we have a function x(t) with Fourier Transform X(f), then what is the Fourier Transform of the function y(t) given by the integral:. »Fast Fourier Transform - Overview p. Fast Fourier Transform in matplotlib An example of FFT audio analysis in matplotlib and the fft function. Fourier Transform for real input over any number of axes in an M-dimensional array by means of the Fast Fourier Transform (FFT). Forward and inverse Fourier transforms are defined as follows: The formulas above have the O(N 2) complexity. On this page, we'll look at the integration property of the Fourier Transform. , rfft and irfft, respectively. A Taste of Python - Discrete and Fast Fourier Transforms This paper is an attempt to present the development and application of a practical teaching module introducing Python programming techni ques to electronics, computer, and bioengineering students at an undergraduate level before they encounter digital signal processing. This document describes the Discrete Fourier Transform (DFT), that is, a Fourier Transform as applied to a discrete complex valued series. First, the Fourier transform starts with the smallest frequency as possible. the Fourier Transform makes an implicit assumption that the signal is repetitive: that is, the signal within the measured time repeats for all time. Unlike other domains such as Hough and Radon, the FFT method preserves all original data. short-time fourier transform. py, which is not the most recent version. Notice that get_xns only calculate the Fourier coefficients up to the Nyquest limit. DC component needs to be removed during Fourier transform. The Discrete Fourier Transform (DFT) transforms discrete data from the sample domain to the frequency domain. Not very useful for empirical data. I would like to calculate the 2D Fourier Transform of an Image with Mathematica and plot the magnitude and phase spectrum, as well as reconstruct the. The vanilla version of Fourier Transform (fft) is not the best feature extractor for audio or speech signals. Indeed, in the decades since Cooley & Tukey’s landmark paper, the most interesting applications. Doing the Stuff in Python Demo(s) Q and A Filters The Fourier Transform Fast Fourier Transform (FFT) Computing the Discrete Fourier Transform takes O(n2m2) for an m n image FFT Computes the same in O(nlognmlogm) Anil C R Image Processing. We've studied the Fourier transform quite a bit on this blog: with four primers and the Fast Fourier Transform algorithm under our belt, it's about time we opened up our eyes to higher dimensions. Fourier Transform of a Periodic Function (e. We can use a discrete Fourier transform on the sound wave and get the frequency spectrum. An algorithm to numerically invert functions in the Laplace field is presented. For images, 2D Discrete Fourier Transform (DFT) is used to find the frequency domain. What do the X and Y axis stand for in the Fourier transform domain? Ask Question Asked 4 years, 3 months ago. The inverse Fourier Transform f(t) can be obtained by substituting the known function G(w) into the second equation. The Python code we are writing is, however, very minimal. Python, for example [3] replaced Java with Python as the Python code is easier for the novice learner. , for filtering, and in this context the discretized input to the transform is customarily referred to as a signal, which exists in the time domain. 3 p712 PYKC 20-Feb-11 E2. • An aperiodic signal can be represented as linear combination of complex exponentials, which are infinitesimally close in frequency. It also provides the final resulting code in multiple programming languages. By carefully chosing the window, this transform corresponds to the decomposition of the signal in a redundant tight frame. Loading Unsubscribe from Pysource? Cancel Unsubscribe. We’ve studied the Fourier transform quite a bit on this blog: with four primers and the Fast Fourier Transform algorithm under our belt, it’s about time we opened up our eyes to higher dimensions. An"intuitive explanation of Fourier theory" by Steven Lehar. Fourier series decomposes a periodic function into a sum of sines and cosines with different frequencies and amplitudes. SciPy offers the fftpack module, which lets the user compute fast Fourier transforms. My name is Thibaut. Looking for abbreviations of WFT? It is Windowed Fourier transform. When both the function and its Fourier transform are replaced with discretized counterparts, it is called the discrete Fourier transform (DFT). Per the sympy documentation for fourier_transform(): If the transform cannot be computed in closed form, this function returns an unevaluated FourierTransform object. That's fine, but not very clear from the title. This calculator is online sandbox for playing with Discrete Fourier Transform (DFT). This course is a very basic introduction to the Discrete Fourier Transform. DFT is part of Fourier analysis, which is a set of math techniques based on decomposing signals into sinusoids. vSig will be padded with zeros if it has less than nFFT points and truncated if it has more. curves bounding a character using a Fourier transform, and using this feature vector for classification. Introduction of Fourier Analysis and Li Su Introduction of Fourier Analysis and Time-frequency Analysis. The Cooley-Tukey radix-2 fast Fourier transform (FFT) algorithm is well-known, and the code is readily available from too many independent sources. The Picture Book of Fourier Transforms by Kevin Cowtan gives an interesting graphical tutorial on the interpretation of 2D FFT output, with a special emphasis on crystallography. In Python after calling the fft function on the data. An in-depth discussion of the Fourier transform is best left to your class instructor. The Inverse Fourier Transform The Fourier Transform takes us from f(t) to F(ω). The fast Fourier transform (FFT) is an efficient algorithm for computing the DFT of a sequence; it is not a. If you assume that the discrete Fourier transform (DFT) has only k non zero coefficients, then, there exists an algorithm to compute it in O(k log(n)). However, when I look at the real/imaginary parts the are completely. If you found this comparison interesting, consider series 3 (7K text) and series 4 (7K text). Fast Fourier Transform or FFT is a powerful tool to visualize a signal in the frequency domain. useful linear algebra, Fourier transform, and random number capabilities. The example python program creates two sine waves and adds them before fed into the numpy. 2, the Fourier transform of function f is denoted by ℱ f and the Laplace transform by ℒ f. ppt), PDF File (. Python, for example [3] replaced Java with Python as the Python code is easier for the novice learner. I've created a code (Python, numpy) that defines an ultrashort laser pulse in the frequency domain (pulse duration should be 4 fs), but when I perform the Fourier Transform using DFT, my pulse in the. I am looking to improve my code in python in order to have a better look a my fourier transform. Included are a rigorous implementation of time-frequency distributions (Cohen class), some quartic time-frequency distributions, chirplet decomposition based on maximum likelihood estimation, fractional Fourier transform, time-varying filtering, and other useful utilities. A 3 dimensional DFT can be expressed as 3 DFTs on a 3 dimensional data along each dimension. Text: definition, fast fourier transform 7. INTRODUCTION TO FOURIER TRANSFORMS FOR IMAGE PROCESSING BASIS FUNCTIONS: The Fourier Transform ( in this case, the 2D Fourier Transform ) is the series expansion of an image function ( over the 2D space domain ) in terms of "cosine" image (orthonormal) basis functions. This course is a very basic introduction to the Discrete Fourier Transform. Thereafter, we will consider the transform as being de ned as a suitable. I need to enhance my image using fast fourier transform. The forward transform converts a signal from the time domain into the frequency domain, thereby analyzing the frequency components, while an inverse discrete Fourier transform, IDFT, converts the frequency components back into the time domain. The DFT signal is generated by the distribution of value sequences to different frequency. To overcome this shortcoming, Fourier developed a mathematical model to transform signals between time (or spatial) domain to frequency domain & vice versa, which is called 'Fourier transform'. The Fourier transform G(w) is a continuous function of frequency with real and imaginary parts. Ozaktas Abstract— We propose and consolidate a definition of the discrete fractional Fourier transform that generalizes the discrete Fourier transform (DFT) in the same sense that the continuous. vSig will be padded with zeros if it has less than nFFT points and truncated if it has more. The clFFT library is an OpenCL library implementation of discrete Fast Fourier Transforms. Fourier Transform Learn to find the Fourier Transform of images ; Generated on Tue Oct 15 2019 03:43:38 for OpenCV by 1. Two-dimensional Fourier transform also has four different forms depending on whether the 2D signal is periodic and discrete. Spectral Analysis •Most any signal can be decomposed into a sum of sine and cosine waves of various. As noted by several authors, the 2D Fourier power spectrum preserves direction information of an image [1]. A special feature of the z-transform is that for the signals and system of interest to us, all of the analysis will be in terms of ratios of polynomials. The code works by printing a letter as an image, e. A fast algorithm called Fast Fourier Transform (FFT) is used for calculation of DFT. This section describes the general operation of the FFT, but skirts a key issue: the use of complex numbers. Take the Fourier transform of the whole signals (or a large interval of samples of the signal). This is an explanation of what a Fourier transform does, and some different ways it can be useful. Quantum Fourier Transforms Burton Rosenberg November 10, 2003 Fundamental notions First, review and maybe introduce some notation. DFT is part of Fourier analysis, which is a set of math techniques based on decomposing signals into sinusoids. Definition of the z-Transform • Given a finite length signal , the z-transform is defined as (7. There is also an inverse Fourier transform that mathematically synthesizes the original function from its frequency domain representation. In such case we may still be able to represent the function. The Fourier transform is actually implemented using complex numbers, where the real part is the weight of the cosine and the imaginary part is the weight of the sine. Its classical cousin is the Fast Fourier Transform. This is a brief review of the Fourier transform. FOURIER SERIES: In mathematics, a Fourier series is a way to represent a wave-like function as the sum of simple sine waves. This page on Fourier Transform vs Laplace Transform describes basic difference between Fourier Transform and Laplace Transform. It’s all about functions from G to C. I'd like to know how to remove environmental noise from a speech recording. On the other hand, Discrete. Fourier Transform - OpenCV 3. The forward transform converts a signal from the time domain into the frequency domain, thereby analyzing the frequency components, while an inverse discrete Fourier transform, IDFT, converts the frequency components back into the time domain. 4 with python 3 Tutorial 35 Pysource. Signals & Systems - Reference Tables 1 Table of Fourier Transform Pairs Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform. For each block, fft is applied and is multipled by some factor which is nothing but its absolute value raised to the power of 0. Fourier analysis is extremely useful for data analysis, as it breaks down a signal into constituent sinusoids of different frequencies. While the discrete Fourier transform can be used, it is rather slow. Details about these can be found in any image processing or signal processing textbooks. for s=σ+jω, σ = 0, as mentioned in previous comments, the problem of Laplace transforms gets reduced to Continuous Time Fourier Transform. Fourier analysis transforms a signal from the. Under this transformation the function is preserved up to a constant. In this tutorial, you will discover how you can apply normalization and standardization rescaling to your time series data in Python. Chapter 6 Fourier Transform 6. That natural actually leads us to the definition of the Fourier transform, which we first look at in its continuous form. In this article, we will focus majorly on the syntax and the application of DFT in SciPy assuming you are well versed with the mathematics of this concept. The Fourier transform. Audacity Noise Reduction Python. understand the math behind the Discrete Fourier Transform(DFT), one of the most useful formulas in applied math and computer science. So, for k = 0, 1, 2, …, n-1, y = (y0, y1, y2, …, yn-1) is Discrete fourier Transformation (DFT) of given polynomial. 4 with python 3 Tutorial 35. The Fourier transform of a signal exist if satisfies the following condition. The goals of this short course is to understand the math behind the algorithm and to appreciate its utility by analyzing and manipulating audio files with Python. This course is a very basic introduction to the Discrete Fourier Transform. We've studied the Fourier transform quite a bit on this blog: with four primers and the Fast Fourier Transform algorithm under our belt, it's about time we opened up our eyes to higher dimensions. In other words, it will transform an image from its spatial domain to its frequency domain. Given a trajectory the fourier transform (FT) breaks it into a set of related cycles that describes it. By carefully chosing the window, this transform corresponds to the decomposition of the signal in a redundant tight frame. From what I gather, it is the absolute value of the Fourier Transform which is somewhat like a histogram of frequencies of the components that the. There are a variety of properties associated with the Fourier transform and the inverse Fourier transform. Among the few existing color watermarking schemes, some use quaternion discrete Fourier transform (QDFT). "Sparseness" is one of the reasons for the extensive use of popular transforms, because they discover the structure of the singal and provide a "compact" representation. Discrete Time Fourier Transform The DTFT(Discrete Time Fourier Transform) is nothing but a fancy name for the Fourier transform of a discrete sequence. Working with these polynomials is rela-tively straight forward. Join Coursera for free and transform your career with degrees, certificates, Specializations, & MOOCs in data science, computer science, business, and dozens of other topics. The C/C++ source code and its header file are: fourier_ccode. Unfortunately, the meaning is buried within dense equations: Yikes. By improving … - Selection from Image Processing and Acquisition using Python [Book]. We introduce the one dimensional. Mathematics of Computation, 19:297Œ301, 1965 A fast algorithm for computing the Discrete Fourier Transform (Re)discovered by Cooley & Tukey in 19651 and widely adopted. The one can answer that , we need this to convert the time domain signal to frequency domain. Learn online and earn valuable credentials from top universities like Yale, Michigan, Stanford, and leading companies like Google and IBM. Aperiodic, continuous signal, continuous, aperiodic spectrum where and are spatial frequencies in and directions, respectively, and is the 2D spectrum of. for s=σ+jω, σ = 0, as mentioned in previous comments, the problem of Laplace transforms gets reduced to Continuous Time Fourier Transform. The Inverse Fourier Transform The Fourier Transform takes us from f(t) to F(ω). Fastest Fourier Transform in the West FFTW is a C subroutine library for computing the discrete Fourier transform (DFT) in one or more dimensions, of arbitrary input size, and of both real and complex data (as well as of even/odd data, i. The goals of this short course is to understand the math behind the algorithm and to appreciate its utility by analyzing and manipulating audio files with Python. ppt), PDF File (. Presentation Materials for my "Sound Analysis with the Fourier Transform and Python" OSCON Talk. Once the Fourier transform is computed, its frequency domain representation can be scanned and required values generated. This section describes the general operation of the FFT, but skirts a key issue: the use of complex numbers. This is a package to calculate Discrete Fourier/Cosine/Sine Transforms of 1-dimensional sequences of length 2^N. Fourier Transforms are useful for: Everything that has to do with Radio. Note that some authors (especially physicists) prefer to write the transform in terms of angular frequency instead of the oscillation frequency. Fourier Series vs Fourier Transform. We can do this computation and it will produce a complex number in the form of a + ib where we have two coefficients for the Fourier series. Apply the Fourier transform two more times, so a second reversal undoes the first, and you're back to the original vector. Computation is slow so only suitable for thumbnail size images. curves bounding a character using a Fourier transform, and using this feature vector for classification. The Fourier Transform is one of deepest insights ever made. FFT (Fast Fourier Transform) refers to a way the discrete Fourier Transform (DFT) can be calculated efficiently, by using symmetries in the calculated terms. If it is not periodic, then it cannot be represented by a Fourier series for all x. Audacity Noise Reduction Python. Fourier Transforms Explained. txt) or view presentation slides online. The symbols ℱ and ℒ are identified in the standard as U+2131 SCRIPT CAPITAL F and U+2112 SCRIPT CAPITAL L, and in LaTeX, they can be produced using \mathcal{F} and \mathcal{L}. This is the first tutorial in our ongoing series on time series spectral analysis. So the Discrete Fourier Transform does and the Fast Fourier Transform Algorithm does it, too. This class of algorithms is known as the Fast Fourier Transform (FFT). Most real signals will have discontinuities at the ends of the measured time, and when the FFT assumes the signal repeats it will assume discontinuities that are not really there. This article will walk through the steps to implement the algorithm from scratch. What happens if there is only a single point inside the Gaussian? Plot the absolute value of the transform. Discrete Fourier Transform (DFT) is a transform like Fourier transform used with digitized signals. The Fourier Transform is a tool that breaks a waveform (a function or signal) into an alternate representation, characterized by sine and cosines. The sinc function is the Fourier Transform of the box function. The function accepts a time signal as input and produces the frequency representation of the signal as an output. FFT in python. Instructions for chapter 3 months. A Fourier Transform itself is just an algorithm and a Fast Fourier Transform is a different algorithm that produces approximately the same result. In mathematics, the discrete Fourier transform (DFT) converts a finite sequence of equally-spaced samples of a function into a same-length sequence of equally-spaced samples of the discrete-time Fourier transform (DTFT), which is a complex-valued function of frequency. In order to deal with transient solutions of difierential equations, we will introduce the Laplace transform. If you Fourier transform a vector twice, the result is the same vector but with all of the elements (except the first element) in reverse order. A Taste of Python - Discrete and Fast Fourier Transforms This paper is an attempt to present the development and application of a practical teaching module introducing Python programming techni ques to electronics, computer, and bioengineering students at an undergraduate level before they encounter digital signal processing. an introduction into the working of fourier transforms. In my Fourier transform series I've been trying to address some of the common points of confusion surrounding this topic. short-time fourier transform. The most general case allows for complex numbers at the input and results in a sequence of equal length, again of complex numbers. Fast Fourier Transform - Algorithms and Applications is designed for senior undergraduate and graduate students, faculty, engineers, and scientists in the field, and self-learners to understand FFTs and directly apply them to their fields, efficiently. A research group at MIT has come up with an improved algorithm that could make it possible to do more with audio and image data with less powerful hardware. Assume and are integrable functions: Linearity: For , if , then. The article aims to be an explanation of the Fourier transform for dummies, but it is quite specifically aimed at Python users. txt) or view presentation slides online. On the second plot, a blue spike is a real (cosine) weight and a green spike is an imaginary (sine) weight. This is to certify that the thesis entitled “Classification of Electroencephalogram(EEG) signal based on Fourier transform and neural network”, submitted by Puloma Pramanick(Roll No. The Fourier Transform will decompose an image into its sinus and cosines components. () or (), is a generalized completeness relation for the set of wave train'' functions, However, these functions are not normalizable, i. Fourier Transform in Numpy¶ First we will see how to find Fourier Transform using Numpy. Apply the Fourier transform two more times, so a second reversal undoes the first, and you're back to the original vector. The current master branch uses some functionality which has not been added to kdevplatform yet, and thus won’t compile unless you apply a patch. That's fine, but not very clear from the title. Analyzing the frequency components of a signal with a Fast Fourier Transform. In this section we'll get to know another family of linear transformations that are extremely useful, not only for compression of data, but in many fields of mathematics, physics and engineering. The Fourier transform is a mathematical function that can be used to show the different parts of a continuous signal. That is: the Fourier Transform of the system impulse response is the system Frequency Response L7. I'll show you how I built an audio spectrum analyzer, detected a sequence of tones, and even attempted to detect a cat purr--all with a simple microcontroller, microphone, and some knowledge of the Fourier transform. vSig will be padded with zeros if it has less than nFFT points and truncated if it has more. Audacity Noise Reduction Python. This way you ensure that your surrogate is real. OpenCV 3 image and video processing with Python OpenCV 3 with Python Image - OpenCV BGR : Matplotlib RGB Basic image operations - pixel access iPython - Signal Processing with NumPy Signal Processing with NumPy I - FFT and DFT for sine, square waves, unitpulse, and random signal Signal Processing with NumPy II - Image Fourier Transform : FFT & DFT. It refers to a very efficient algorithm for computingtheDFT • The time taken to evaluate a DFT on a computer depends principally on the number of multiplications involved. This is a great resource because it doesn't dwell on the mathematics and instead focuses on building an intuition of the Fourier transform. Covers the most important deep learning concepts, giving an understanding of each concept rather than mathematical and theoretical details. I am trying to translate this gravitational wave signal processing tutorial from Python to R, which I am much more familiar with. Here's a plain-English metaphor: Here's the "math English" version of the above: The Fourier. This is a post of Python Computer Vision Tutorials. An example of FFT audio analysis in MATLAB ® and the fft function. Sparse Fast Fourier Transform : The discrete Fourier transform (DFT) is one of the most important and widely used computational tasks. I wanted to point out some of the python capabilities that I have found useful in my particular application, which is to calculate the power spectrum of an image (for later se. Contact experts in Discrete Fourier Transform to get answers. As noted by several authors, the 2D Fourier power spectrum preserves direction information of an image [1]. Cal Poly Pomona ECE 307 Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. Fourier transform of the aperiodic signal represented by a single period as the period goes to infinity. Not very useful for empirical data. In an infinite crystal, on the other hand, the function is typically periodic (and thus not decaying):. This is a brief review of the Fourier transform. x/is the function F. Basics of Dsp and transforms. 11, with the waveform initially on the left side of the signal array. Signal Processing with NumPy II - Image Fourier Transform : FFT & DFT Inverse Fourier Transform of an Image with low pass filter: cv2. OpenCV 3 image and video processing with Python OpenCV 3 with Python Image - OpenCV BGR : Matplotlib RGB Basic image operations - pixel access iPython - Signal Processing with NumPy Signal Processing with NumPy I - FFT and DFT for sine, square waves, unitpulse, and random signal Signal Processing with NumPy II - Image Fourier Transform : FFT & DFT. Numpy has an FFT package to do this. The discrete Fourier transform, or DFT, is the primary tool of digital signal processing. The FFT, or fast fourier transform is an algorithm that essentially uses convolution techniques to efficiently find the magnitude and location of the tones that make up the signal of interest. Do fill these forms for feedback: Forms open indefinitely! Third-year anniversary form https://docs. The reason the Fourier transform is so prevalent is an algorithm called the fast Fourier transform (FFT), devised in the mid-1960s, which made it practical to calculate Fourier transforms on the fly. 3 branch in kdev-python which is compatible with kdevelop 1. 1 The 1d Discrete Fourier Transform (DFT) The forward (FFTW_FORWARD) discrete Fourier transform (DFT) of a 1d complex array X of size n computes an array Y, where:. There is evidence that Gauss first developed a fast Fourier transform-type algorithm in 1805. What do the X and Y axis stand for in the Fourier transform domain? Ask Question Asked 4 years, 3 months ago. Fastest Fourier Transform in the West FFTW is a C subroutine library for computing the discrete Fourier transform (DFT) in one or more dimensions, of arbitrary input size, and of both real and complex data (as well as of even/odd data, i. Doing the Stuff in Python Demo(s) Q and A Filters The Fourier Transform Fast Fourier Transform (FFT) Computing the Discrete Fourier Transform takes O(n2m2) for an m n image FFT Computes the same in O(nlognmlogm) Anil C R Image Processing. While I'll be using the scientific Python stack in this blog post, code in Matlab, R should not be that different. Links: Pillow: https://pyt. Introduction It turns out that taking a Fourier transform of discrete data is done. An in-depth discussion of the Fourier transform is best left to your class instructor. Since sharp. The Fourier Series is a method of expressing periodic signals in terms of their frequency components. For more information and background on the Fourier transform, take a look at this link. [python]DFT(discrete fourier transform) and FFT. Here we focus on the use of fourier transforms for solving linear partial differential equations (PDE). This reduces the number of operations required to calculate the DFT by almost a factor of two (Fig. The final example uses the Morlet waveform used in Example 3. You can do it for some instruments, such as flutes. Fourier Slice Theorem [Bracewell 1956]. Recall that the quantum Fourier transform (or, depending on conventions, its inverse) is given by. I'll show you how I built an audio spectrum analyzer, detected a sequence of tones, and even attempted to detect a cat purr--all with a simple microcontroller, microphone, and some knowledge of the Fourier transform. The Short-Time Fourier Transform. Chapter 04 - Free download as Powerpoint Presentation (. The following are code examples for showing how to use numpy. The Python code we are writing is, however, very minimal. ifft() function to transform a signal with multiple frequencies back into time domain. In this post, we provide an example that how to analyze the web traffic by Discrete Fourier Transform (DFT). Foremost, you're loading pandas without ever using it. 8 fourier Visit Website Please read our data set. The notation is introduced in Trott (2004, p. ML with Python. It is most used to convert from time domain to frequency domain. From what I gather, it is the absolute value of the Fourier Transform which is somewhat like a histogram of frequencies of the components that the. Per the sympy documentation for fourier_transform(): If the transform cannot be computed in closed form, this function returns an unevaluated FourierTransform object. In this blog post, we'll programatically try and develop an intuitive understanding into the whole process. Actually, you can do amazing stuff to images with fourier transform operations, including: (1) re-focus out of focus images (2) remove pattern noise in a picture, such as a half-tone mask (3) remove a repeating pattern like taking a picture through a screen door or off a piece of embossed paper (4) find an image so deeply buried in noise you. Cooley and John Tukey. Do fill these forms for feedback: Forms open indefinitely! Third-year anniversary form https://docs. Spectral Analysis – Fourier Decomposition Adding together different sine waves transform f 3f 5f 7f frequency Time I have shown how to go this way. This is the basic of Low Pass Filter and video stabilization. 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. Fourier Transforms Fourier transform are use in many areas of geophysics such as image processing, time series analysis, and antenna design. The short time discrete Fourier transform is the version you are seeing here and the version most often used. The article aims to be an explanation of the Fourier transform for dummies, but it is quite specifically aimed at Python users. Understanding the FFT algorithm; A post on FFT from Jake Vanderplas is also a great explanation of how it works. ppt - Free download as Powerpoint Presentation (. The Fourier transform is not limited to functions of time, but the domain of the original function is commonly referred to as the time domain. Untuk gambar, 2D Discrete Fourier Transform (DFT) digunakan untuk mencari domain frekuensi. The library: provides a fast and accurate platform for calculating discrete FFTs. It discretizes the integral defining the Laplace transform, but it does not truncate the domain. Fourier transform. 2 Introduction to Shape Classification with Fourier Descriptors The term "Fourier Descriptor'' describes a family of related image features. single value of the FFT output if and only if you input a signal that is one of the Fourier basis. At the end I provide the sheet music, a human rendition, and a Python package that implements the method (and can also be used to transcribe from MIDI files). fft function to get the frequency components. When applied to the time series data, the Fourier analysis transforms maps onto the frequency domain, producing a frequency spectrum. How about going back? Recall our formula for the Fourier Series of f(t) : Now transform the sums to integrals from -∞to ∞, and again replace F m with F(ω). Fourier transform of a Borel measure. edu Fourier theory is pretty complicated mathematically. In Python after calling the fft function on the data. The discrete Fourier transform or DFT is the transform that deals with a nite discrete-time signal and a nite or discrete number of frequencies. It converts a space or time signal to signal of the frequency domain. Define a transform to extract a subregion from an image. The foundation of the product is the fast Fourier transform (FFT), a method for computing the DFT with reduced execution time. In order to compute the FT of a signal with Python we need to use the ftt function built in into Scipy. Take the Fourier transform of the whole signals (or a large interval of samples of the signal). The code I have. AI algorithms C++ C++11 CodeForces CS231n dfs dft discrete fourier transform dynamic programming fast fourier transform fft Graph Theory Hashicorp Atlas Java Javascript JPA longest increasing subsequence Machine Learning Netbeans Ninja Framework OGLMan python Qt Creator SPOJ Tutorial UVa Vagrant VirtualBox XML. Looking at the example above, the periodic time data can be described as the sum of 4 sinusoidal functions with frequencies at 110, 220, 330, and 440 hz. Welcome to pynufft's Documentation! Python non-uniform fast Fourier transform was designed and developed for image reconstruction in Python. Understanding the FFT algorithm; A post on FFT from Jake Vanderplas is also a great explanation of how it works. spectrograms), and many kinds of image/audio processing, but is rarely used for compression. In order to deal with transient solutions of difierential equations, we will introduce the Laplace transform. This course is a very basic introduction to the Discrete Fourier Transform. When we calculate the periodogram of a set of data we get an estimation of the spectral density. In the previous Lecture 17 and Lecture 18 we introduced Fourier transform and Inverse Fourier transform and established some of its properties; we also calculated some Fourier transforms. | 2019-11-20T02:14:30 | {
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https://web2.0calc.com/questions/some-math_1 | +0
Unusual simult equations with sqrts
+3
480
4
+88
Let a, b, c, and d be distinct real numbers such that
\begin{align*} a &= \sqrt{4 + \sqrt{5 + a}}, \\ b &= \sqrt{4 - \sqrt{5 + b}}, \\ c &= \sqrt{4 + \sqrt{5 - c}}, \\ d &= \sqrt{4 - \sqrt{5 - d}}. \end{align*}
Compute $$abcd$$.
Jan 28, 2021
edited by Melody May 11, 2021
#3
+88
+8
Squaring both sides of the original equation, we get $$a^2 = 4+\sqrt{5+a}.$$
Subtracting 4 from both sides and squaring again gives $$(a^2-4)^2 = 5+a.$$
Expanding this out and subtracting 5 + a from both sides, we have $$a^4-8a^2-a+11 = 0.$$
Similar manipulations on the other equations give \begin{align*} b^4-8b^2-b+11 &= 0, \\ c^4-8c^2+c+11 &= 0, \\ d^4-8d^2+d+11 &= 0. \end{align*}
(Note the signs carefully.)
Let $$f(x)=x^4-8x^2-x+11$$. Then a and b are roots of f(x), while c and d are roots of $$x^4-8x^2+x+11$$, which is f(-x). It follows that -c and -d are roots of f(x). Since -c and -d are negative and hence distinct from a and b, we have four roots of f(x). Since f(x) has degree four, we know we have found all the roots.
By Vieta's formulas, the constant term of f(x) is the product of the roots. Thus,
\begin{align} abcd = (a)(b)(-c)(-d) = \boxed{11}. \end{align}
Apr 16, 2021
edited by Divineology Apr 16, 2021
#1
-3
The solutions are 2.2192, 1.33938, 1.09044, and 2.4682, and their product is 8.
Jan 28, 2021
#2
+114950
+1
I'd like to see this one answered.
Divineology, perhaps you can show us how it is done? Or anyone else?
Apr 15, 2021
#3
+88
+8
Squaring both sides of the original equation, we get $$a^2 = 4+\sqrt{5+a}.$$
Subtracting 4 from both sides and squaring again gives $$(a^2-4)^2 = 5+a.$$
Expanding this out and subtracting 5 + a from both sides, we have $$a^4-8a^2-a+11 = 0.$$
Similar manipulations on the other equations give \begin{align*} b^4-8b^2-b+11 &= 0, \\ c^4-8c^2+c+11 &= 0, \\ d^4-8d^2+d+11 &= 0. \end{align*}
(Note the signs carefully.)
Let $$f(x)=x^4-8x^2-x+11$$. Then a and b are roots of f(x), while c and d are roots of $$x^4-8x^2+x+11$$, which is f(-x). It follows that -c and -d are roots of f(x). Since -c and -d are negative and hence distinct from a and b, we have four roots of f(x). Since f(x) has degree four, we know we have found all the roots.
By Vieta's formulas, the constant term of f(x) is the product of the roots. Thus,
\begin{align} abcd = (a)(b)(-c)(-d) = \boxed{11}. \end{align}
Divineology Apr 16, 2021
edited by Divineology Apr 16, 2021
#4
+114950
+1
Diveneology: | 2021-10-27T00:07:17 | {
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http://www.purplemath.com/learning/viewtopic.php?f=8&t=1906 | ## How do you go from a quadratic function in...
Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
### How do you go from a quadratic function in...
generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square?
I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!
Absolutely
Posts: 2
Joined: Mon Apr 04, 2011 8:57 pm
### Re: How do you go from a quadratic function in...
Absolutely wrote:generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square? I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!
assume $a\neq0$
$ax^2+bx+c$
$=a(x^2+\frac{b}{a}x)+c$
$=a\left(x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right)+c$
$=a\left(x+\frac{b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$
$=a\left(x-\frac{-b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$
$=a\left(x-\frac{-b}{2a}\right)^2+c-\frac{b^2}{4a}$
$=a\left(x-h\right)^2+k$
Martingale
Posts: 350
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Location: USA
### Re: How do you go from a quadratic function in...
Thanks! Seeing the algebraic manipulations behind these derivations makes the formulas more sensible to me.
Absolutely
Posts: 2
Joined: Mon Apr 04, 2011 8:57 pm | 2014-10-25T09:11:59 | {
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https://www.themathdoctors.org/integrals-and-signed-areas/ | # Integrals and Signed Areas
#### (A new question of the week)
This week’s question, asked in January on the new site, will take us through some tricky areas of calculus, and also give a glimpse both of the value of quoting the entire problem you are working on when you ask for help, and of the interesting side discussions we can get into when that rule is violated! Several of us got involved in this discussion.
## The question
I was asked to find the area of y = 5x^4 – x^5 with the range of x: 0 – 5.
And I found the answer: 3125/6.
Then I was asked to find the area of the same function with the range of x: 0 – 6.
While when the range of x: 0 – 7, the answer is not 0.
May I know why the answer can be 0?
Doesn’t it must be >= 3125/6?
How can the area of a larger region be smaller? And how can an area in fact be zero?
## Signed areas
Remember that when using integrals to compute the area between a curve and the x-axis, all area below the x-axis will be computed as a negative number. When the area above the x-axis is numerically equal to the area below the x-axis, the total area will be 0 because the two areas have opposite signs.
Note the subtle restatement of the problem: Doctor Fenton assumes that rather than “find the area of the equation” (which is not mathematically meaningful), it must have been “find the area between the curve and the x-axis”, which is the typical form of such a question. Also, in such a question, areas below the axis are taken as negative, because that is what integration inherently does. A definite integral in effect adds areas of little rectangles, each a width (Δx) times a height (y). If y is negative, that area is negative.
And this is how you can get a total area of zero, when the positive and negative areas cancel one another out.
Hi Doctor.
I am still not understand the mechanism.
I saw another function: y = (1/x^3) – 8 and I found two of the areas, that are from interval 1/4 to 1/2 and 1/2 to 1. I got 4 and -2.5.
And when I try to find the total, I got 6.5 (it is 4 + 2.5).
In this question, may I use the same way? Or maybe my way above is wrong? (I checked the key answer and my answer is true.)
Checking this example, we find these regions above and below the axis:
The integrals (signed areas) are 4 and -2.5; in this problem, they are evidently not asking for the total signed area, but for the total absolute area (counting both regions as positive).
So now we have two different kinds of questions being asked about similar functions.
## Variations in wording
As Doctor Fenton didn’t respond quickly, I jumped in with some thoughts about wording:
It may be a difference in how the question is phrased. Please quote exactly how your problem was worded. Different authors may also mean slightly different things.
According to some, the area UNDER a curve is found by merely integrating, and takes sign into account, so that the answer of 0 is appropriate for your area from 0 to 6. Some authors might ask for the area of the region BETWEEN (or BOUNDED BY) the curve and the x-axis, intending an UNSIGNED area, so that you would add the absolute values of the separate areas found by integration, as in your second example.
Here is an example of a site that seems to be inconsistent about this, giving a negative answer to one problem, but taking absolute values for another:
Some use various ways to explicitly indicate what they want, such as “net area” here:
http://www.mathwords.com/a/area_under_a_curve.htm
I would check how your own book states these problems, and see if they tell you how to distinguish the two types of question.
The first of these, in its first example, asks, “What is the area between the curve $$y = x^2 – 4$$ and the x axis?”. After giving the answer -10.67, they add, “Note: the area is negative because it is below the x-axis. Areas above the x-axis, on the other hand, give positive results.”
In their second example, they ask, “Find the total area between the curve $$y = x^3$$ and the x-axis between x = -3 and x = 2.” This time, they add absolute values, getting the answer 8. Presumably the word “total” is the cue to the difference.
The second page uses the term “net area” for the signed area, which turns out to be zero, and doesn’t give an example of “total area”.
So both wording, and the usage of a particular author, can make a huge difference.
Hi Doctor
Does it mean that, if the area is not bounded by anything, the area we interpret is only when the function is positive (And we subtract it with the negative one)?
And the reason we subtract it is because we need to find the net positive area
I was thinking if there is no bound, the area should be infinite.
My book only explained about the area between the curve(s) (and x or y axis).
The question shows the curve and the shaded area (that is between the curve and x axis).
Then it first ask me to find the area from x = 0 to x = 4.
Then the question ask me to evaluate if the range is from x = 0 to x = 6.
I’m not sure what was meant by “not bounded by anything”; obviously an “area” without a boundary is infinite. My guess is that there is some language ambiguity about the use of “bounded” in these problems.
## Areas between or under curves
Doctor Fenton picked up the question again, discussing the topic in general:
The general description of the problem is to find the area between two curves over a bounded interval. Usually, these are two graphs of the form y=f(x) and y=g(x) over the interval [a,b] (with a < b), or graphs of the form x=h(y) and x=k(y) for y in the interval [c,d].
When the graphs are y=f(x) and y=g(x), then one curve will be above the other over some interval. If f(x) > g(x) over [a,b], then the area bounded by the two curves will be
b
∫ f(x)-g(x) dx .
a
This will be a positive number, since f(x) > g(x) on the interval. However, if the two graphs intersect, then they may change places as to which is the upper curve and which is the lower curve. If f(x)≥g(x) for a ≤ x ≤ c but f(x)g(x) over c ≤ x ≤ b, then y=f(x) is the upper curve on [a,c], and the lower curve on [c,b]. In this case, the total area is
b c b
∫| f(x)-g(x)| dx = ∫ f(x)-g(x) dx + ∫ g(x)-f(x) dx .
a a c
To find total area, you always subtract the lower curve from the upper curve when integrating.
If you just integrate the difference f(x)-g(x), you are getting the net area, since the integral will be negative on intervals where g(x) is the upper curve.
If you are given only one function on an interval, then you need to remember that the x-axis is a curve, namely y=0 (so g(x)=0 on the interval).
This might make more intuitive sense if you think of the position/velocity interpretation of the integral: if t is time and f(t) is the velocity at time t, where the particle is moving to the right when f(t) > 0 and to the left when f(t)<0, then
b
∫ f(t) dt
a
gives you the final position at time b if the particle started at time a. If the integral is positive, it finishes to the right of where it started; if the integral is negative, it finishes to the left of the starting position; and if the integral is 0, it finishes where it started. You don’t have to keep track of where the motion is to the right or left – the integral does that automatically, because the integral is positive on intervals where the integrand is positive, and negative on the intervals where the integrand is negative.
If you want the total distance traveled, however, you need to compute
b
∫ |f(t)| dt .
a
Then you need to determine where the particle changes direction and compute integrals over the intervals where it is always moving in the same direction separately. That is, you compute
d
∫ f(t) dt
c
on intervals [c,d] where f(t) ≥ 0, but compute
d
∫- f(t) dt
c
on intervals [c,d] where f(t) ≤ 0. Then you add the separate results.
Does this help?
## Examining the graph
But I still don’t get about the area = 0 (especially after seeing the graph).
The graph is as attached below:
If I need to find the area from x = 0 to x = 6, which area should I search (in the graph above)?
Can Doctor please mark the area? (I began to realise that what makes me didn’t get it is the lack of understanding in the graph).
I’ve held off on showing the graph of the function until now, because this will be important here. Doctor Rick jumped in at this point:
Here is a graph of y = 5x4 – x5, the first function you talked about. Is that what your graph shows? It’s hard to be sure, because you chose scales for the axes that don’t show it very clearly.
On my graph you can see one region above the x axis, between x = 0 and x = 5, and another region below the x axis, between x = 5 and x = 6. The areas of these two regions are equal; you can’t see this exactly from the graph, but you can see that it is reasonable. The integral of the function from 0 to 6 equals 0 because the integral from 0 to 5 is that area, while the integral from 5 to 6 is the negative of that area.
If you still have questions about this particular problem, we’ll need further information. Doctor Peterson asked you for the exact wording, and from what you’ve said, evidently it included a figure. Could you show us that figure? We can’t be sure what it meant by “the area”, especially for the second part, without seeing the figure. (We still might not be sure, but at least this should help.)
Although graphs are not required to solve problems like this, an accurate graph can be very helpful (and an inaccurate graph can be very misleading).
## The real problem
Here is the question:
The curve now makes sense to me after Doctor sent it (I use phone app and I cannot change the scale) (and the straight line confused me).
So does it mean that if I found the total area, it will be 7776 + 7776 = 15552?
The graph is not precise, and doesn’t attempt to cover up to x = 6; but the wording is the important thing to observe. Doctor Rick wrote back:
Showing us the actual problem changes a lot! You were never asked to “find the area of y = 5x^4 – x^5 with the range of x: 0 – 5 … [and then] to find the area of the same function with the range of x: 0 – 6″ as you said at first.
Rather, you were asked first to calculate “the area of the shaded region enclosed by the curve and the x axis” — which is quite clear, given the figure. It would be meaningless to “find the area of y = 5x^4 – x^5”; we can only find the area of a region, not of a function.
Since the region shown is above the x axis and below the graph of the function, it is found simply as the integral from 0 to 5. However, I do not get 7776 for this integral, as your last message implies. I get 3125/6 = 520 5/6, as you imply at the end of your first message.
Part (iii) of the problem does not mention area at all; therefore we don’t need to get into the variations among textbooks that Doctor Peterson mentioned. It only asks you to carry out the integral from 0 to 6, and “comment on your result.” You found the result of that integral to be zero; and we have discussed how that can be.
Using the correct values for the integral from 0 to 5 (3125/6) and for the integral from 5 to 6 (-3125/6), the correct result for the total area of the two regions in my figure is the sum of the absolute values of those integrals: 3125/6 + 3125/6 = 3125/3 = 1041 2/3. | 2020-08-11T12:03:27 | {
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http://mathhelpforum.com/number-theory/30771-mod-5-a.html | # Thread: Mod 5
1. ## Mod 5
I am having trouble with this homework question and was hoping someone could help. Cheers
Prove that 5|n(n2-1)(n2=1)
2. Hello,
Look for every possibility of n if 5 divides the product.
Try $\displaystyle n \equiv 0 mod 5$, then $\displaystyle n \equiv 1 mod 5$ and so on
3. Originally Posted by asw-88
I am having trouble with this homework question and was hoping someone could help. Cheers
Prove that $\displaystyle 5|n(n^2-1)(n^2+1)$ (at least, I think that's what is meant)
If you multiply out the brackets then you get $\displaystyle 5|n^5-n$, which is true by Fermat's little theorem.
4. Hello, asw-88!
Prove that: .$\displaystyle 5\,|\,n(n^2-1)(n^2+1)$
Let $\displaystyle N \:=\:n(n^2-1)(n^2+1)$
$\displaystyle n$ must one of five possible forms: .$\displaystyle 5k-2,\:5k-1,\:5k,\:5k+1,\:5k+2$
$\displaystyle [1]\;n \,= \,5k-2\!:\;\;N \:=\5k-2)([5k-2]^2-1)([5k-2]^2 + 1)$
. _ . . . . . . . . . . $\displaystyle =\;(5k+2)(25k^2-20x+3)(25k^2-20k + 5)$
. _ . . . . . . . . . . $\displaystyle = \;(5k+2))(25k^2+20k + 3){\color{red}5}(5k^2 - 4k + 1)$ . . . a multiple of 5
$\displaystyle [2]\;n \,=\,5k-1\!:\;\;N \:=\5k-1)([5k-1]^2-1)([5k-1]^2+1)$
. . . . . . . . . . . . $\displaystyle = \;(5k-1)(25k^2 - 10k)(25k^2-10k + 2)$
. . . . . . . . . . . . $\displaystyle = \;(5k-1){\color{red}5}k(5k- 2)(25k^2 - 10k+2)$ . . . a multiple of 5
$\displaystyle [3]\;n \,=\,5k\!:\qquad N \:=\5k)([5k]^2 - 1)([5k]^2+1)$
. . . . . . . . . . . . $\displaystyle = \;{\color{red}5}k(25k^2 - 1)(25k^2+1)$ . . . a multiple of 5
$\displaystyle [4]\;n\,=\,5k+1\!:\;\;N \;=\;(5k+1)([5k+1]^2-1)([5k+1]^2+1)$
. . . . . . . . . . . . $\displaystyle = \;(5k+1)(25k^2 + 10k)(25k^2+10k + 2)$
. . . . . . . . . . . . $\displaystyle = \;(5k+1){\color{red}5}k(5k + 2)(25k^2 + 10k + 2)$ . . . a multiple of 5
$\displaystyle [5]\;n\,=\,5k+2\!:\;\;N \;=\;(5k+2)([5k+2]^2-1)([5k+2]^2+1)$
. . . . . . . . . . . . $\displaystyle = \;(5k+2)(25k^2 + 20k + 3)(25k^2 + 20k + 5)$
. . . . . . . . . . . . $\displaystyle = \;(5k+2)(25k^2 + 20k+3){\color{red}5}(5k^2 + 4k + 1)$ . . . a multiple of 5
. . . . . . . . . . . . . . . . . Q.E.D.
5. I'm not going to do it, but I suspect it would be relatively simple to do an induction proof of this as well.
-Dan
6. Originally Posted by topsquark
I'm not going to do it, but I suspect it would be relatively simple to do an induction proof of this as well.
-Dan
I'll do the initial condition:
$\displaystyle 5|1^5-1$
$\displaystyle 5|0$
Because I care, I'll even show the next iteration:
$\displaystyle 5|2^5-2$
$\displaystyle 5|30$
The rest of it is almost as easy as this. | 2018-03-21T11:41:12 | {
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# The quotient when a certain number is divided by 2/3 is 9/2. What is
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Updated on: 17 Jun 2017, 13:03
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The quotient when a certain number is divided by $$\frac{2}{3}$$ is $$\frac{9}{2}$$. What is the number?
A. $$\frac{4}{27}$$
B. $$\frac{1}{3}$$
C. 3
D. 6
E. $$\frac{27}{4}$$
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Re: The quotient when a certain number is divided by 2/3 is 9/2. What is [#permalink]
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17 Jun 2017, 10:16
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AbdurRakib wrote:
The quotient when a certain number is divided by $$\frac{2}{3}$$ is $$\frac{9}{2}$$.What is the number?
A. $$\frac{4}{27}$$
B. $$\frac{1}{3}$$
C. 3
D. 6
E. $$\frac{27}{4}$$
Its not mentioned if remainder is 0 or not. Assuming Remainder is 0 when number is divided by $$\frac{2}{3}$$ has quotient $$\frac{9}{2}$$.
Number is in form, $$x = \frac{2}{3} * \frac{9}{2} + 0$$
$$x = 3$$. Answer (C)...
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Re: The quotient when a certain number is divided by 2/3 is 9/2. What is [#permalink]
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17 Jun 2017, 10:19
1
Consider the number for which the quotient is $$\frac{9}{2}$$ as x
Since x divided by $$\frac{2}{3}$$ gives us the quotient $$\frac{9}{2}$$
x * $$\frac{3}{2}$$ = $$\frac{9}{2}$$
x = $$\frac{9}{2}$$*$$\frac{2}{3}$$ = 3(Option C)
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Re: The quotient when a certain number is divided by 2/3 is 9/2. What is [#permalink]
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17 Jun 2017, 22:50
AbdurRakib wrote:
The quotient when a certain number is divided by $$\frac{2}{3}$$ is $$\frac{9}{2}$$. What is the number?
A. $$\frac{4}{27}$$
B. $$\frac{1}{3}$$
C. 3
D. 6
E. $$\frac{27}{4}$$
$$\frac{2}{3}*\frac{9}{2} = 3$$
Hence, answer will be (C) 3
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Re: The quotient when a certain number is divided by 2/3 is 9/2. What is [#permalink]
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07 Aug 2017, 09:47
AbdurRakib wrote:
The quotient when a certain number is divided by $$\frac{2}{3}$$ is $$\frac{9}{2}$$. What is the number?
A. $$\frac{4}{27}$$
B. $$\frac{1}{3}$$
C. 3
D. 6
E. $$\frac{27}{4}$$
n *$$\frac{3}{2}$$ = $$\frac{9}{2}$$
=> n = 3
C is the answer.
Alternatively
N = PQ + R
N = $$\frac{2}{3}$$*$$\frac{9}{2}$$ + R
N = 3 + R
The Question doesn't mention if the division has a remainder, so R = 0
=> N = 3
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The quotient when a certain number is divided by [#permalink]
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09 Oct 2017, 18:57
LakerFan24 wrote:
The quotient when a certain number is divided by $$\frac{2}{3}$$ is $$\frac{9}{2}$$. What is the number?
A. $$\frac{4}{27}$$
B. $$\frac{1}{3}$$
C. 3
D. 6
E. $$\frac{27}{4}$$
If $$\frac{N}{\frac{2}{3}}$$ =$$\frac{9}{2}$$
Then $$\frac{2}{3} * \frac{9}{2}$$ = N =$$\frac{18}{6} = 3$$
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Re: The quotient when a certain number is divided by 2/3 is 9/2. What is [#permalink]
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12 Oct 2017, 17:08
1
AbdurRakib wrote:
The quotient when a certain number is divided by $$\frac{2}{3}$$ is $$\frac{9}{2}$$. What is the number?
A. $$\frac{4}{27}$$
B. $$\frac{1}{3}$$
C. 3
D. 6
E. $$\frac{27}{4}$$
n/(2/3) = 9/2
3n/2 = 9/2
6n = 18
n = 3
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http://www.mathematics2.com/Calculus/LinearityOfDifferentiation | Subject: Calculus
# Linearity Of Differentiation
When a function is expressed as a sum of individual products of a constant and a function, then we can't just differentiate that function directly. We still have no rule that can be applied to that yet. But intuitively, since there is a constant factor and sum involved, maybe there is a combination of this two simplest rules that can may define the derivative of the function stated above. In differential calculus, the most fundamental and important property of the derivative is the linearity of differentiation. It is a combination of the sum rule in differentiation and the constant factor rule in differentiation. Yes, this rule can be used to define the derivative of the function stated above. To better understand this, consider the kind of function we mentioned above.
### Illustration
Let f_{1} and f_{2} be differentiable functions at x and further assume that a and b are constants. Then a linear combination of these functions can be written as a linear function f,
f(x)=a\cdot f_{1}(x)+b\cdot f_{2}(x).
Now, we want to find the derivative of the linear function f, then
D_{x}[f(x)]=\frac{d}{dx}\left (a\cdot f_{1}(x)+b\cdot f_{2}(x)\right ).
By virtue of the sum rule in differentiation, the above function can be written as
f'(x)=\frac{d}{dx}[a\cdot f_{1}(x)]+\frac{d}{dx}[b\cdot f_{2}(x)],
then constant factor rule in differentiation suggests that it can have the following form,
f'(x)=a\frac{d}{dx}f_{1}(x)+b\frac{d}{dx}f_{2}(x).
Thus, we can alternatively write
\frac{d}{dx}f(x)=a\frac{d}{dx}f_{1}(x)+b\frac{d}{dx}f_{2}(x).
Now, notice that applying differentiation to the linear function f yields a linear function f'(x) composed of derivatives. This strongly tells us that differentiation is linear in nature.
### Example #1
Suppose that f(x) = g(x) and h(x), where g(x)=x^3+2x+1 and h(x)=x^2. Verify the notion of linearity of differentiation by showing that f'(x)=g'(x) + h'(x).
Since g(x)=x^3+2x+1 and h(x)=x^2, then we can write f(x)=x^3+x^2+2x+1. Differentiating f(x), we have
f'(x) = 3x^2+2x+2
.
On the other hand, the derivative of g(x) is just g'(x)=3x^2+2 and for h(x), h'(x)=2x. Adding g'(x) and h'(x), we get
g'(x) + h'(x) =3x^2+2 + 2x
= f'(x)
.
Thus, the linearity of differentiation is verified.
### Example #2
Let the function f(x) = g(x) and h(x), where g(x)=\cos{x}-\sin{x} and h(x)=\cos^{2}{x}. Verify that f'(x)=g'(x) + h'(x).
Since g(x)=\cos{x}-\sin{x} and h(x)=\cos^{2}{x}, then we can write f(x)=\cos{x}-\sin{x}+\cos^{2}{x}. Differentiating f(x), we have
f'(x) = -\sin{x}-\cos{x}-2\cos{x}\sin{x}
.
Meanwhile, the derivative of g(x) is just g'(x)=-\sin{x}-\cos{x} and for h(x), h'(x)=-2\cos{x}\sin{x}. Adding g'(x) and h'(x), we get
g'(x) + h'(x) =-(+\sin{x}+\cos{x}+2\cos{x}\sin{x})
f'(x)=g'(x) + h'(x)
.
Thus, the linearity of differentiation is verified.
### Example #3
If the values of a and b in the illustration above are expressed as functions of the differential variable x, namely a(x) and b(x), is the notion of linearity of differentiation still holds?
Yes of course. That's because differentiation is linear in nature as shown above.
### Example #4
From the scenario presented in example #3, show by direct differentiation that by replacing a and b by a(x) and b(x) the linearity still holds.
From
f(x)=a\cdot f_{1}(x)+b\cdot f_{2}(x)
we will have
f(x)=a(x)\cdot f_{1}(x)+b(x)\cdot f_{2}(x)
Letting f_1(x)\cdot a(x) = f_{a}(x) and f_{b}(x)=f_2(x)\cdot b(x), it follows that
f(x)= f_{a}(x)+f_{b}(x)
Differentiating the function directly, we have
\frac{d}{dx}f(x)= \frac{d}{dx}(f_{a}(x)+f_{b}(x))
By virtue of the sum rule in differentiation, the above function can be written as
\frac{d}{dx}f(x)=\frac{d}{dx}f_{a}(x)+\frac{d}{dx}f_{b}(x)
which just proves linearity of differentiation.
### Example #5
Name at least two differentiation rules that are used in verifying that differentiation is indeed a linear operation.
NEXT TOPIC:Calculus with Polynomials | 2017-10-21T03:01:20 | {
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https://cs-cheatsheet.readthedocs.io/en/latest/subjects/linear_algebra/fundamentals.html | # Fundamentals¶
## Basis¶
It’s a set of vectors in a vector space which are linearly independent. All vectors in the vector space are linear combinations of the basis. Read wiki.
## Eigenvector & Eigenvalue¶
Intuitively, eigenvectors and eigenvalues are related to transformation. When a transformation is applied to a vector space, veectors span. While most vectors drift away from their spans some remain on its own span – eigenvectors. Eigenvectors remain on its own span after a transformation but they may scale – by a factor of their eigenvalues. An interesting fact is that any vector that lies on the same span as eigenvectors is itself an eigenvector. Therefore there can be infinitely many eigenvectors. However, there could be only one eigenvector as well. Consider a 3D transformation matrix $$A$$:
\begin{split}\begin{align} A &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{align}\end{split}
$$A$$ will squash everything into null and there will be only one eigenvector $$\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
### Eigenvector in 3D rotations¶
An eigenvectors in a 3D rotation means the axis of a rotation. And because a rotation doesn’t change the scale, the eigenvalue should be 1.
### Eigenvector in 2D rotations¶
There is no eigenvector in 2D rotations. No eigenvector implies no real-valued eigenvalue but imaginary-valued.
### Single eigenvalue with multiple eigenvectors¶
There could be multiple eigenvectors but only one single eigenvalue. Consider a transformation $$A$$:
\begin{split}\begin{align} A &= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \end{align}\end{split}
$$A$$ scales every eigenvector by 2 and only 2.
### Calculation¶
$\overbrace{ A \vec{v} }^\text{Matrix-vector multiplication} = \underbrace{\lambda \vec{v}}_\text{Scalar multiplication}$
The matrix $$A$$ changes only the scale of the vector $$\vec{v}$$ by a factor of $$\lambda$$. We could rewrite the right hand as
\begin{split}\begin{align} \lambda \vec{v} &= \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} \vec{v} \\ &= \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \vec{v} \\ &= \lambda I \vec{v} \end{align}\end{split}
To get the value of the eigenvalue, take the righthand to the other side:
$(A - \lambda I) \vec{v} = \vec{0}$
Remember the squashification? $$(A - \lambda I)$$ is squashing $$\vec{v}$$. This implies the following:
$\det{(A - \lambda I)} = 0$
### Eigenbasis¶
Consider a 2D vectorspace. If both basis vectors are eigenvectors then its transformation matrix would be diagonal. Here’s step-by-step:
A typical set of eigen vectors in 2D:
$\begin{split}\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\end{split}$
A transformation matrix:
$\begin{split}\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\end{split}$
The columns of the transformation matrix happnes to be the eigenvectors and, the basis vectors as well with the diagonal values being their eigenvalues. Pay attention to the matrix that the matrix is diagonal. Diagonal matrices have a handy property – their power is just a power of the elements:
$\begin{split}\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}^n = \begin{bmatrix} (-1)^n & 0 \\ 0 & 2^n \end{bmatrix}\end{split}$
### Eigenbasis for easier power¶
Consider a transformation $$A$$,
$\begin{split}\begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}\end{split}$
$$A$$ is not diagonal so its power will be expensive to calculate. Let’s change its basis as the eigenvectors in order to make $$A$$ diagonal. The eigenvectors of $$A$$,
$\begin{split}\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \end{bmatrix}\end{split}$
We can build “Change of basis matrix” from the eigenvectors,
$\begin{split}\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\end{split}$
Now let’s change its basis,
$\begin{split}\begin{bmatrix} \text{Change of basis matrix}^{-1} \end{bmatrix} A \begin{bmatrix} \text{Change of basis matrix} \end{bmatrix} \\ \Rightarrow \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}\end{split}$
## Hessian matrix¶
The Hessian Matrix is a square matrix of second ordered partial derivatives of a scalar function. It is of immense use in linear algebra as well as for determining points of local maxima or minima. [1]
The Hessian of a function is denoted by $$\Delta^2f(x,y)$$ where $$f$$ is a twice differentiable function & if $$(x_0,y_0)$$ is one of it’s stationary points then :
• If $$\Delta^2f(x_0,y_0)>0$$ i.e positive definite , $$(x_0,y_0)$$ is a point of local minimum.
• If $$\Delta^2f(x_0,y_0)<0$$ , i.e. negative definite , $$(x_0,y_0)$$ is a point of local maximum.
• If $$\Delta^2f(x_0,y_0)$$ is neither positive nor negative i.e. Indefinite , $$(x_0,y_0)$$ is a saddle point | 2018-10-19T23:40:28 | {
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https://www.physicsforums.com/threads/projectile-motion-and-suvat-equations.812269/ | Projectile motion and SUVAT equations
1. May 5, 2015
mathsman
Attempting a projectile motion question where initial and final speed is given but angle of elevation isn't. Need to find an equation for the vertical distance H travelled. I got the right answer using v^2 = u^2 + 2as substituting s=H and a=-9.8.
However the mark scheme states that this SUVAT equation can't be used as this equation needs the vertical component of speed (as gravity only acts vertically) . The correct answer was to use conservation of energy principle ie initial KE = final KE + GPE. However if I use this energy equation I get the following: 1/2 mu^2 = 1/2 m v^2 - 2gH . Multiplying throughout by 2 and re-arranging this gives v^2 = u^2 + 2as !! In other words the conservation of energy equation gives me the SUVAT equation which I'm not supposed to use! Where is the mistake??
2. May 5, 2015
Staff: Mentor
Not really a mistake. It's that you haven't justified your use of the SUVAT formula. It does turn out to work fine in this case, but it's intended use is for motion along a line so the velocities are supposed to be the components along that line.
3. May 5, 2015
Staff: Mentor
This is justified: $v_y^2 = u_y^2 + 2as_y$
But since $v_x = u_x$, you can add $v_x^2 = u_x^2$ to both sides, giving $v^2 = u^2 + 2as_y$.
But conservation of energy gives it to you immediately.
4. May 5, 2015
mathsman
Many thanks to you both. I think I may have done a similar question many years ago and remember using the standard v2=u2+2as equation but couldn't remember why it worked ! Obviously adding the square of the horizontal component to both sides makes it clear why it does in fact work. The mark scheme simply gave no marks for using SUVAT so although my calculated value for h was correct I lost method marks. Presumably if I had 'justified' why I could use it that would be ok (although I'm never sure how far examiners give marks for methods that aren't in the mark scheme!.
Having said that the conservation of energy approach is much neater anyway (dont know why I overlooked it! - normally exam questions hint at using conservation of energy principle but this time it ddn't! | 2017-10-24T01:43:07 | {
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https://math.stackexchange.com/questions/2074094/why-does-the-pythagorean-theorem-not-work-on-this-problem-in-the-way-that-i-used | # Why does the Pythagorean Theorem not work on this problem in the way that I used it?
To begin with, I apologize for the vagueness of my question. It's hard to explain what exactly my question entails without seeing what process I went through to try to solve the problem. My question is just that I don't understand why my method did not work.
The problem: In Figure 8, P is a point in the square of side-length 10 such that it is equally distant from two consecutive vertices and from the opposite side AD. What is the length of BP? A) 5 B) 5.25 C) 5.78 D) 6.25 E)7.07
(I apologize for the crude drawing, the problem was in my book so I had to improvise using Paint) Figure 8
What I did: Since BC and CD are both 10, I used the pythagorean theorem to get the length of diagonal BD (sqrt 200) and divided by 2. My answer was therefore E) 7.07.
What my book did: Set BP to x, and the length of (B and midpoint of AB) to 10-x. To complete the triangle, they set the length of (P and midpoint of AB) to 5. Then they used the Pythagorean Theorem to do x^2 = (10-x)^2 + 5^2, yielding an answer of D) 6.25.
While I understand how they did it, I simply cannot understand why my method didn't work. Is there some law that I'm not aware of pertaining to this problem? Since my incorrect answer was an answer choice, I assume there is a common error I'm making that was set as a trap.
Could someone explain this to me? Thank you very much.
• It is because $BP$ does not lie on the diagonal of the square. In fact, if that was the case, that would be $10\sqrt{2} /2$ long, but $P$ would be the center of the square, making the third segment long $5$ and thus different from the other two. – Harnak Dec 27 '16 at 22:11
• @Harnak, was just going to say this. Nice. – The Count Dec 27 '16 at 22:12
• @Harnak OH that was very careless of me to assume that BP is the same length as PD. It makes sense now, thank you! – ak_27 Dec 27 '16 at 22:13
• @ak_27, it is awesome that you came to this website for help with this. most people would just get frustrated and give up. keep at it! i loved this question. very subtle. – The Count Dec 27 '16 at 22:14
• You're welcome ^^ – Harnak Dec 27 '16 at 22:15
The diagram in correct proportion. To get the square edge length $10,$ multiply all lengths by $$\frac{10}{8} = \frac{5}{4} = 1.25$$ | 2019-08-25T13:20:50 | {
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# When the numerator of a fraction is increased by 4, the fraction increases by 2/3....
When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}y=\frac xy+\frac 23$$
I am not able to find the second equation.
Again, you've got a fine start:
You wrote: $$\frac{x+4}y=\color{red}{\frac xy}+\color{blue}{\frac 23}\tag{1}$$
But note that $$\frac{x+4}{y} = \color{red}{\frac xy} + \color{blue}{\frac 4y}\tag{2}$$
From $(1),(2),$ it must follow that $$\color{blue}{\frac 4y = \frac 23 } \iff 2y = 4\cdot 3 = 12 \iff y = \frac{12}{2} = 6$$
So the denominator, $y$ is $6$.
If you add 4 to the numerator, the value of your fraction will be increased by $\frac4y$, where y is your denominator.
So $\frac4y=\frac23$ and $y=6$
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# When the numerator of a fraction is increased by 4, the fraction increases by 2/3....
When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}y=\frac xy+\frac 23$$
I am not able to find the second equation.
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Again, you've got a fine start:
You wrote: $$\frac{x+4}y=\color{red}{\frac xy}+\color{blue}{\frac 23}\tag{1}$$
But note that $$\frac{x+4}{y} = \color{red}{\frac xy} + \color{blue}{\frac 4y}\tag{2}$$
From $(1),(2),$ it must follow that $$\color{blue}{\frac 4y = \frac 23 } \iff 2y = 4\cdot 3 = 12 \iff y = \frac{12}{2} = 6$$
So the denominator, $y$ is $6$.
The value of $x$ is not determined by the given information. Now that you know $y = 6$, the statement of the problem says that $\frac{x+4}6=\frac x6 + \frac23$, which is true for all $x$. - David K 2 days ago
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https://math.stackexchange.com/questions/2176281/minimizing-a-quadratic-function-subject-to-quadratic-constraints | Okay, so I am attempting to minimize the function
$$f(x,y, z) = x^2 + y^2 + z^2$$
subject to the constraint of
$$4x^2 + 2y^2 +z^2 = 4$$
I attempted to solve using Lagrange multiplier method, but was unable to find a $\lambda$ that made the system consistent.
$$2x = \lambda8x$$ $$2y = \lambda4y$$ $$2z = \lambda2z$$
Wouldn't it be the case that the first equation suggests $\lambda = 1/4$ but the second equation suggests $\lambda = 1/2$? I am unsure of where to go from here although I have spent time trying to figure out.
Intuitively, I know that it must be the case that the minimum occurs at $(1,0,0)$ and is equal to $1$, but I can not show this using mathematical reasoning.
• Do you want to maximize of minimize the function? Your title says maximize but the question says minimize.. – Kuifje Mar 7 '17 at 17:53
• Minimize, fixed title thanks – user345 Mar 7 '17 at 18:07
• While it is good that you try to understand how a particular method that you know can be applied to this problem, I am quite confident that your “intuitive” reasoning is just as mathematically valid if you formulate it carefully. Mathematics is not about avoiding to think! (You should not forget $(-1, 0, 0)$, though.) – Carsten S Mar 8 '17 at 9:36
Hint: Your conclusion that your three equations imply that there is no consistent choice for $\lambda$ only holds if $x$, $y$ and $z$ are non-zero! If any of them are zero, then their corresponding equation tells us nothing about $\lambda$ since it is trivially fulfilled.
You thus know that the critical points are of the form $(x,0,0)$, $(0,y,0)$ and $(0,0,z)$. Up to you to check them out
When you solve the system with Lagrange method, your variables are $x,y,z$ and $\lambda$.
One solution to the system is $$(x,y,z,\lambda)=(1,0,0,\frac{1}{4}),$$ which matches your intuitive solution.
Alternatively, since $z^2=4-4x^2-2y^2$ cannot be negative (thank you @user35734!), it is equivalent to minimize $$f(x,y,z(x,y))=4-3x^2-y^2$$ subject to: $$4x^2+2y^2 \le 4$$
This is an elliptic domain, and it is easy to see that the minimum will be reached on its border, therefore we can replace variables $x$ and $y$ by $\cos t$ and ${\sqrt{2}} \sin t$, respectively. The problem then boils down to minimize $$f(x(t),y(t))=4-3\cos^2t - 2\sin^2 t$$ Standard calculus techniques yield $$\min f = 1$$
To complement the other answers, we are looking for the points on a given ellipsoid that are closest to the origin. Here is the plot a sphere of radius $1$ touching the given ellipsoid at $(\pm 1, 0, 0)$.
I may be wrong... I tried to intuit this minimum problem.
The level surfaces of $f$ are spheres:
The larger the value of $f$, the larger the sphere is.
The next figure depicts the constraint which is an ellipsoid:
We get the minimum value of $f$ if we blow up (or shrink) its level sphere so that it is tangent to the ellipsoid from inside.
If $z=0$ then the constraint ellipsoid intersects the $xy$ axis in an ellipse whose equation is
$$x^2+\frac{y^2}2=1.$$
This ellipse intersects the $y$ axis at $\sqrt 2$ and it intersects the $x$ axis at $1$.
If we blow up (or shrink) the sphere so that it touches the ellipse then the level is $1$.
That is, the minimum is taken if $x^2+y^2+z^2=1$ and the minimum is $1$.
Let's see the intersection of the ellipse with the $xz$ plane. This is an ellipse again and
$$x^2+\frac{z^2}4=1.$$
The sphere belonging to the level $1$ fits well again.
This is what I saw:
• Yes, but it is clearly a global maximizer, OP wants a minimizer. – Kuifje Mar 7 '17 at 17:48
• Yes, I can see now. But the title sais maximize... – zoli Mar 7 '17 at 17:51
• I think you are correct. But there is something that I don't understand. Previously you wrote that the problem boils down to minimizing $f=4-3x^2-y^2$. But this function is unbounded (below). How can $(1,0)$ be a minimizer here? – Kuifje Mar 7 '17 at 18:42
• @Kuifje: I don't know! That was done without intuition. I can see what's going on now. – zoli Mar 7 '17 at 18:48
• $z^2$ cannot be negative, so you simultaneously need the bound that $4x^2 + 2y^2 \le 4$. – user35734 Mar 7 '17 at 20:22 | 2020-01-27T16:45:46 | {
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https://damtana.com/forum/page.php?tag=gamma-distribution-excel-f051ef | # gamma distribution excel
The Poisson distribution has only one parameter: the rate parameter λ. Today we look at the GAMMA.DIST function.. If x < 0, GAMMA.DIST returns the #NUM! error value. Can you provide me with the codes to create the digamma and trigamma functions? The following Excel-generated graph shows the gamma distribution’s PDF (Probability Density Function) for as the X value goes from 0 to 10 with Shape parameter k = 2 and Scale parameter θ = 1. Draw a random variate from a lognormal distribution with a mean of 0.10 and a standard deviation of 0.20: LogNorm.Inv(Rand(), 0.10, 0.20) The Student T Distribution Note that the formula in cell D7 is an array function (and so you must press Ctrl-Shft-Enter and not just Enter). If x, alpha, or beta is nonnumeric, GAMMA.DIST returns the #VALUE! Thanks for your valuable contribution. Alternatively, we can use the following iteration method to find α. where ψ(z) (also denoted ψ0(z)) is the digamma function and ψ1(z) is the trigamma function. error value. The GAMMA.DIST function has the following arguments: We’ll continue our A to Z of Excel Functions soon. The exponential distribution also calculates the probability of wait time to the first Poisson event when the average rate time is λ. You can use this function to study variables that may have a skewed distribution. If a call comes in on average every 2 minutes, the scale parameter θ = 2. Probability density using the x, alpha, and beta values in A2, A3, A4, with FALSE cumulative argument. In MS Excel, Gamma distribution can be easily calculated by using GAMMA.DIST function. The Excel equation to solve the problem is as follows: F(X=4;k=3,θ=2) = GAMMA.DIST(X,k,θ,TRUE) = GAMMA.DIST(4,3,2,TRUE) = 0.3233. Cumulative =TRUE, it returns the cumulative distribution function. The gamma distribution calculates the probability of a specific waiting time until the kth event Poisson occurs. The gamma distribution describes the distribution of waiting times between Poisson-distributed events. Providing Financial Modelling, Strategic Data Modelling, Model Auditing, Planning & Strategy and Training Courses. The exponential distribution calculates the probability of a specific waiting time until the 1st Poisson event occurs. Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. This article describes the formula syntax and usage of the GAMMA.DIST function in Microsoft Excel. The gamma distribution, on the other hand, predicts the wait time until the *k-th* event occurs. The gamma distribution function is characterized by 2 variables, its shape parameter k and its scale parameter θ (Theta). The problem asks to calculate the probability that the wait time will be UP TO 4 minutes so the gamma’s CDF (Cumulative Distribution Function) will be used to solve this problem. The red font part indicated in the denominator is basically the gamma factor. =Gamma.Inv(Rand(), 2, 1) The Lognormal Distribution. Let's understand how to use … Welcome back to our regular A to Z of Excel Functions blog. This is what is known as “queueing analysis”. Returns the gamma distribution. Gamma distribution shows pattern when plotted on graph having given parameters k and . This should be noted as an array formula (enter with CTRL SHIFT ENTER). I didn’t think that it was necessary to lookup values of digamma and trigamma for 0 < z <= 1. Expert and Professional The gamma distribution calculates the probability of a specific waiting time until the kth Poisson event occurs if θ = 1/λ is the mean number of Poisson-distributed events per time unit. It turns out that the maximum of L(α, β) occurs when β = x̄ / α. The PDF value of a statistical distribution (the Y value) at a specific X value equals the probability that the value of a random sample will be equal to that X value. Home » Excel-Built-In-Functions » Excel-Statistical-Functions » Excel-Gamma.Dist-Function. POLYGAMMA function can be created as a user-defined function (UDF) using VBA. Applications of the gamma distribution are often based on intervals between Poisson-distributed events. What formula do I have to put in my VBA? Returns the gamma distribution. #NUM! Solver Optimization Consulting? 2. else As suggested by its name, α controls the shape of the family of distributions. to calculate the digamma value at z = .2, you should be able to calculate the value psi(4.2) and then use the formula psi(z) = psi(z+1) - 1/z several times to get the value of psi(.2). The following Excel-generated graph shows the gamma distribution’s CDF (Cumulative Distribution Function) for k = 2 and θ = 1 as the X value goes from 1 to 10. The given formulas (series expansions for z>4 and recursive relations for z<4) helped. Best place to learn Excel online. The gamma distribution is closely related to the exponential distribution. here is the pseudo code. The formula for the cumulative hazard function of the gamma distribution is $$H(x) = -\log{(1 - \frac{\Gamma_{x}(\gamma)} {\Gamma(\gamma)})} \hspace{.2in} x \ge 0; \gamma > 0$$ where Γ is the gamma function defined above and $$\Gamma_{x}(a)$$ is the incomplete gamma function defined above. For example, if the mean time between the Poisson-distributed events is 2 minutes, then θ = ½ = 0.5. Keep checking back – there’s a new blog post every business day. These are infinite sums, but you only need to use asmall number of the terms. Syntax. There is a 32.33 percent probability that the time interval in which 3 successive calls arrive will UP TO 4 minutes. For smaller values of z, we can improve the estimates by using the following properties: Real Statistics Function: The digamma and trigamma functions can be computed via the following Real Statistic worksheet function: POLYGAMMA(z, k) = digamma function at z if k = 0 (default) and trigamma function at z if k = 1. This is evidenced by the smooth shape of the above graph of a gamma distribution’s PDF (Probability Density Function) curve. Here value of T = value of x, Any of the arguments – x, alpha, beta are non-numeric in nature, The fourth parameter is not TRUE OR FALSE, unrecognized. for a supplied probability, P, the Gamma.Inv function returns the value of x such that: 1. = 5 x 4 x 3 x 2 x 1 = 120. A parameter to the distribution. When α = 1, it corresponds to exponential distribution. The value at which you want to evaluate the distribution. When you’re working with small samples in Excel — less than 30 or 40 items — you can use what’s called a student t-value to calculate probabilities rather than the usual z-value, which is what you work with in the case of normal distributions. … Two comments: To understand it, first think of factorials, e.g. If p = GAMMA_DIST(x,...), then GAMMA_INV(p,...) = x. elseif k=1 then the equation for the gamma probability density function is: the standard gamma probability density function is. I Can Help. If alpha ≤ 0 or if beta ≤ 0, GAMMA.DIST returns the #NUM! – Occurs under following scenarios. beta / β (Required) – A parameter of the distribution for determining the rate, Type the value where we need to find probability. 4) The mean number of occurrences per interval (λ) and the variance in the number of occurrence per interval are approximately the same. This function is available from MS Excel 2010 onwards. Copyright © 2012-2020 Luke K About Contact Privacy Policy. Based on your definition of the rate parameter it should be the reciprocal of the estimated scale parameter. The GAMMA.DIST function. However, the function is simply an updated version of the Gammadist function, which is available in earlier versions of Excel. Due to its moderately skewed profile, it can be used as a model in a range of disciplines, including climatology where it is a working model for rainfall, and financial services where it has been used for modeling insurance claims and the size of loan defaults. The value of this function for a selected value of x can be calculated by the Excel Gamma.Dist function. Thus z doesn't have to be an integer. Cumulative Required. The Gamma.Dist function is new in Excel 2010 and so is not available in earlier versions of Excel. For formulas to show results, select them, press F2, and then press Enter. Here, α = 4 & β = 3. Calculate the probability that 3 successive calls arrive within a time interval that is UP TO 4 minutes long. Draw a random variate from a gamma distribution with a shape parameter of 2 and a scale parameter of 1. The CDF value of a statistical distribution (the Y value) at a specific X value equals the probability that the value of a random sample will be up to that X value. 2. 2. We can now use Excel’s Solver to find the value of α that maximizes LL. The gamma distribution calculates the probability of wait time for the kth Poisson event. exit function I was wondering, how would we estimate the rate (1/scale) parameter instead if that was our data? In our previous post, we derived the PDF of exponential distribution from the Poisson process. Value at which you want to evaluate the distribution. When alpha is a positive integer, GAMMA.DIST is also known as the Erlang distribution. As you can see, the iteration converges quite rapidly.
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https://physics.stackexchange.com/questions/238278/kinematics-contradicting-conservation-of-energy | # Kinematics contradicting conservation of energy?
I was having some conceptual difficulty reconciling my intuitive understanding of kinematics with conservation of energy, so I made up a short problem that tested my intuitions:
Suppose I define an initial point to be 20m above the ground. An object leaves this initial point along Path A, which is straight down. Another object leaves the initial point along Path B, which is parallel to the ground. Both objects have the same mass, and the same initial speed of $3 \frac{m}{s}$. What is the speed of each object as it hits the ground?
My calculations for Path A:
$$y_{f} = y_{i} + v_{i}t + \frac{1}{2}a_{y}t^{2}$$ $$0 = 20m - 3\frac{m}{s}t - \frac{1}{2}(9.8\frac{m}{s^{2}})t^{2}$$ $$t = \frac{3 + \sqrt{9 - 4(20)(\frac{1}{2})(-9.8)}}{40} = 0.576s$$ $$v_{f}=v_{i}+at$$ $$v_{f} = -3\frac{m}{s} - (9.8\frac{m}{s^{2}})(0.576s) = -8.64\frac{m}{s}$$ $$s = 8.65 \frac{m}{s}$$
My calculations for Path B: $$y_{f}=y_{i}+v_{i}t+\frac{1}{2}a_{y}t^{2}$$ $$0 = 20m+0t+\frac{1}{2}(-9.8\frac{m}{s^{2}})t^{2}$$ $$t = \sqrt{\frac{20m}{\frac{1}{2}(9.8\frac{m}{s^{2}})}} = 2.02s$$ $$v_{f} = v_{i}+at$$ $$v_{fy} = 0-9.8\frac{m}{s^{2}}(2.02s) = -19.799\frac{m}{s}$$ $$v_{fx} = 3\frac{m}{s}$$ $$s = |\vec{v}| = \sqrt{(19.799\frac{m}{s})^{2} + (3\frac{m}{s})^{2}} = 20.02\frac{m}{s}$$
This result clearly defies conservation of energy. Both objects start at the same initial height, and so have the same potential energy. They both end their paths at the same height, and so end with the same potential energy. But they both have different velocities, and so different kinetic energies.
The thing is, this is exactly what my intuitions would predict. How much an object accelerates in total is just a function of the time it spends accelerating. The object traveling along Path A has a fast downwards velocity, so it only has a short time to accelerate. Its final speed is just the sum of its initial speed and however much it accelerates in that time.
The object traveling along Path B has no initial downwards velocity, and so has plenty of time to accelerate. The total final speed of this object is the vector sum of its initial velocity and however much it gained while accelerating downwards, which is a much larger number than the total acceleration of the first object.
So what's going on? The calculations break conservation of energy, so I must be doing something wrong. And even if there's a flaw in my calculations and the numbers actually do come out the same, my intuition still says otherwise.
• Where does the division by $40$ come from in your first calculation? (When you compute a root of the quadratic equation.) Feb 19 '16 at 0:32
• It seems like the fact that it wasn't a quadratic equation in standard form, with the squared term first, made me think that the bottom should be 2(20). I recomputed it and came up with 2.35s, which is longer than the time for Path B, which doesn't seem right. And doing the next calculation gives me a speed of 26m/s, which is still different. Feb 19 '16 at 0:40
The two particles have the same energy. It's easier to use the equation $$v^2 = u^2 + 2a\Delta y$$
For particle A, all motion is in the y-direction, so $$v^2 = v_y^2 = 3^2 + 2(9.8)(20)$$ $$v = 20.02 ms^{-1}$$
You went wrong in your application of the quadratic formula. $$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$a = -0.5\cdot 9.8 = -4.9$$ $$b = -3$$ $$c = 20$$
Taking the negative sqrt solution and correcting $2a$, $t=1.74 s$
Plugging this into the $v = u + at$ formula yields $20.05 ms^{-1}$ for me, and the difference is just rounding error.
For particle B, your calculation was correct.
So now we've clarified that conservation of energy holds, but now to answer why.
The thing is, this is exactly what my intuitions would predict. How much an object accelerates in total is just a function of the time it spends accelerating.
Correct, but the kinetic energy of a particle is not dependent on how much it has accelerated. It only depends on the instantaneous velocity of the particle.
Here the final velocity in the y-direction for particle A will be higher than that of particle B. The final velocity in the x-direction for particle B will be higher than that of particle A (equally so, in fact). Calculating the final speed of both would yield the same answer, and so the kinetic energy will remain equal. | 2021-09-27T21:50:16 | {
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http://eka.frauenarztpraxis-schorndorf.de/how-many-combinations-with-7-numbers.html | Other popular combinations that get heavy play are the number multiples, especially 5-10-15-20-25-30 and 7-14-21-28-35-42. A fruit stand owner has 90 apples and 72 oranges. Number 7 is for Neptune and number 1 is for Sun. 7 in the unit’ s place. 3 digit number : Number of 3 digit numbers with repetition = 5 × 5 × 5 = 125 Ex 7. We have included Some questions that are repeatedly asked in bank exams !! In how many ways can 3 prizes be given away to 12 students when each student is eligible for all the prizes ? A. I have got data set 1 (1 to 8) and data set 2 (9-16). The most popular numbers played in the U. How many functions are there from a $7$ element set to a $36$ element set? For each element in the domain, how many elements in the codomain can it possibly get mapped to?. Answer: Throughout the Bible, God often gives symbolic significance to mundane items or concepts. 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With no repeated digits (lead 0 is allowed). 3, bigger number in tens place – 4 and the greatest number in the ones place – 6, so the smallest number is 346. 6) In how many ways can 4 of 7 different kinds of bushes be planted along a walkway?. A related topic to combinations is "permutations". Just look for a key that looks like "nCm" or "nCr", or for a similar item on the "Prob" or "Math" menu, or check your owner's manual under "probability" or "combinations". The vibration might help to engage the mechanisms. List all the possible combinations. We will start from basics and discuss a logical Now we know that 31-squared is the largest ODD number that fits the given descriptionNow we just have to count up the number of terms. Notionally, there. how many 6 digit telephone number can be constructed with the digit 0,1,2,3,4,5,6,7,8,9, if each number starts with 35 and no digit appears more than once - Math - Permutations and Combinations. Then if you move a different number to the front, you can make the same number of arrangements with that number being first, moving around the numbers behind it. Example 1: Input: k = 3, n = 7 Output: [[1,2,4]] Explanation: 1 + 2 + 4 = 7 There are no other valid combinations. ) ( ) ( choose ) Where n is the number of things to choose from, and you r of them. So, the total number of choices for R2 is 5. Converting this to odds is easy to. So, you have number consisting from 7 unique digits. pick3 numbers, permutations, lock Lets you pick 7 numbers between 1 and 45. i'm going to chat to you later,Babe. 5 can not be at ten's place, since in that case 7 has to be placed at the units place. • To see how to count by using other than 10 digits, notice how we count in the ordinary decimal system. A form of the permutation problem that students commonly see is the “committee” problem. A softball league has 7 teams, what are the possible ways of ranking the teams? n = 7, r = 7 7! = 5040 (7– 7)! Recall 0! = 1. To win the jackpot, you need to have a lottery ticket with the correct combination of five white balls and the red Powerball. 5 letter Words made out of number. 6) In how many ways can 4 of 7 different kinds of bushes be planted along a walkway?. How many ways can Laura color a map with 4 adjacent regions if she has 15 colored pencils? A church has 7 bells in its bell tower. Combinations. 1 Class 11 Maths Question 1. I have got data set 1 (1 to 8) and data set 2 (9-16). Number 7 numerology compatibility for harmony between number 7 with other numbers for all types of relationships love, partnership, marriage and friendship. You can also fi nd the number of combinations using the following formula. I need the numbers to all add together to equal 28 and I need all the possible ways to do it, such as. The number of combinations that are possible with 7 numbers is 127. 9 ,0} And the unit place is already occupied by 7 So the digit at tens place can be filled in 10 ways. i'm hoping I helped you artwork out that answer. If you can repeat the numbers, then number of combinations is infinite. In that case, the number formed is not an even number. How many 3-digit numbers can be formed from the. How many Powerball tickets are possible?. Generator of combinations of m from n. Electricity department needs to generate bills for a locality. study, people also chose small numbers much more frequently than large numbers. (a) n = 6, r = 3Number of permutations = Hence, the number of ways, in which the three prizes can be awarded = 120. P=84+9,880+5,040+28,080# #:. The total number of ways of choosing any three pieces from the ten availableis 10 C 3 =120. The easiest way to go through the combinations, ensuring that you won't miss one, is like this:. With permutations we care about the order of the elements, whereas with combinations we don't. If the same letter or number can be repeated, how many can be made? b. Learn how to write binary numbers, and the (not so secret) code to change English letters into binary numbers and back again. There are 12 standbys who hope to get on your flight to Hawaii, but only 6 seats are available on the plane. One way is to visit all the binary numbers less than 2 n. The ten faces usually bear numbers from zero to nine, rather than one to ten (zero being read as "ten" in many applications). Number of points of intersection = Number of combinations of 20 straight lines taken two at a time Q18. How many different 7-digit telephone numbers are possible if the first digit cannot be 0 or 1? 8 * 10^6 = 8000000 7. Number 7 - Fun Facts. Then if you move a different number to the front, you can make the same number of arrangements with that number being first, moving around the numbers behind it. Answer 604800 b. Do not say Seven. How many combinations are there for selecting four? Another way of thinking about it is how many ways are there to, from a pool of six items, people in this example, how many ways are there to choose four of them. n = 5, r = 3 ( ) ( ) Combination. All numbers from 1 to 200 (in decimal system) are written in base 6 and base 7 systems. restaurant chain. • To see how to count by using other than 10 digits, notice how we count in the ordinary decimal system. If 1 person gets no objects, the 7 objects must be distributed such that each of the other two get 1 object at least. Natalie has 1 6 close friends. I ask because, it seems impossible that there can be a unique key for every lock in the world, and I'd like to know how many different key configurations a thief would have to try to have a certainty of cracking the lock (if the thief didn't want to force. Kaprekar from Devlali, India, devised a process now known as Kaprekar's operation. It's not just individual numbers you need to consider; you also need to think about how frequent number pairs or Most will offer a 'Quick Pick' system which randomly generates a combination of numbers for you, based on your specifications. So, the total number of choices for R2 is 5. The Visual Way. There are 12 standbys who hope to get on your flight to Hawaii, but only 6 seats are available on the plane. At every stage, you are allowed to erase two numbers that appear on the board (let’s call the numbers you erased x and y) and in place of the two erased numbers, write the number x+y+xy. Use this permutation and combination calculator to find the nP r and nC r values by providing the number of sample points in set 'n' and 'r'. Playing Cards: From a standard deck of 52 cards, in how many ways can 7 cards be drawn? 2. And we want to arrange them in groups of 5, so r = 5. How many possibilities would there be if repeated numbers were not. The number of arrangements can be divided by the number of arrangements not used. It does not mean it will actually take you that many tries; however, it could. ⇒ 8 P 4 =\[\frac{8!}{4!} = 8 \times 7 \times 6 \times 5. we know then that we have 6! - (all combinations with 0 as the beginning digit) to find out all the combinations where the number begins with 0, is just 5!. Notionally, there. i'm going to chat to you later,Babe. an organized list. I wonder how many possible combinations there is in a game of chess. A compound is a molecule made of atoms from different elements. 000 000 14 iii) Suppose it takes 10 minutes to pick your numbers and buy a ticket. , 22 = 20 + 2), reinforcing foundational ideas of number composition and place value. (c) Number of ways in which one boy gets all prizes = number of boys = n = 6. Divide 36 by the number of combinations that will make that total. 3, bigger number in tens place – 4 and the greatest number in the ones place – 6, so the smallest number is 346. The number of ways that n elements can be arranged in order is called a permutation of the elements. For comparison, look at the table. Say I have the items as - A, B, C, D, E. More formally, a k-combination of a set S is a subset of k distinct elements of S. We will start from basics and discuss a logical Now we know that 31-squared is the largest ODD number that fits the given descriptionNow we just have to count up the number of terms. How many combinations of three are possible? 1 How many cominations of 3 numbers/sheep/anything you like can you have with 3 number/sheep/anything you like. One way is to visit all the binary numbers less than 2 n. The odds of winning on Combo Play are determined by the game played and number of combinations. Combination: The number of distinct combinations of n objects, taken r at a time, is given by the ratio The Permutation (nPr) and Combination (nCr) work with steps shows the complete step-by-step calculation for finding the the number of ways in which we can choose $8$ different objects out of a set containing $10$ different objects, where. Posts containing "simple math" will be removed, as well as So, I remembered watching a video on Reddit a while ago where Stephen Fry described just how many possible combinations there are in a deck of cards, and decided to recreate the math. We have included Some questions that are repeatedly asked in bank exams !! In how many ways can 3 prizes be given away to 12 students when each student is eligible for all the prizes ? A. Players who want to participate have to select seven numbers (from 1 to 35) plus a Powerball number (from 1 to 20). If number combinations that had been drawn in the past were removed from play, the game would be unfair. Another definition of combination is the number of such arrangements that are possible. Number of points of intersection = Number of combinations of 20 straight lines taken two at a time Q18. The randomness comes from atmospheric noise, which for many purposes is better than the pseudo-random number algorithms typically used in computer programs. - Combination Play costs for LOTTO 6/49: $88,$14, $56 or$168 - You win on a Combination Play ticket the same as on a regular selection. Students continue to compose and decompose 2-digit numbers and to represent them as the sum of multiples of ten and some number of ones (e. Hence the required number of ways = 7 × 7 × 7 × 7 × 7 = 7⁵. I've found a permutation solution here, but what it does is that it prints out all the possible numbers according to the length of array. BD and DB are the same pair. Quantity B. Can a Number be a Combination of Two Numbers? It means the two types of numbers, real and imaginary, together form a complex , just like a building complex (buildings joined together). How many feet are in a mile?) use Wolfram|Alpha™. Combination A combination containing k objects is a subset containing k objects. how to calculate number of possible combinations. " AMSEC is a provider of light-duty and heavy-duty safes. There are 69 possible numbers for the white balls and 26 possible results for the Powerball. The table below details the number of Tricasts for various numbers of selections. 1250 925 785 1440 In how many ways can a leap year have 53 Sundays? 365C7 7 4 2 A telegraph has 5 arms and each. By joining any two points on the circle, we may draw a chord. In other words, there are 8 things, and we are choosing 3 of them in order. There are many ways to enumerate k combinations. Combinatorial calculator - calculates the number of options (combinations, variations ) based on the number of elements, repetition and order of importance. Solution: 7 is a unit digit and if we add 10 with it, it results a two digit number 17, just by changing the digit in the tenth place, contains 7. In this video tutorial, viewers learn how to dial open a combination safe lock. If you had forgotten your password and were trying to guess it, and you guessed a different password each try, this is the maximum number of attempts until you guess it correctly. With no repeated digits (lead 0 is not allowed). Number of combinations n=11, k=3 is 165 - calculation result using a combinatorial calculator. Repeat this operation until only a single number remains. 4 million) If order matters (e. Each combination is played as if it were a single selection on a single ticket. In mathematics, a combination Is a way of selecting items from a collection, such that (unlike permutations) the order of selection does Not matter. Investigation 1: Combinations of 10 (8 Sessions) Investigation 2: Addition and Subtraction (8 Sessions). My explanation is as follows. 1,1 - Chapter 7 Class 11 Permutations and Combinations. Or, in my generic terminology where m is the number of characters to choose from for an integer of n digits, m x [(m-1)^(n-1)] or in the example case 6 x 5^3 = 750--Kent Cooper wrote if "No two digits the same. Replies: 7 Views: 64,614. Evaluate the following: 1) 5! 2) 8! 10! 3) 6!2! 7!4! II. (b) n = 6, r = 3Number of permutations = Hence, the number of ways in which 3 prizes can be awarded = 216. If 1 person gets no objects, the 7 objects must be distributed such that each of the other two get 1 object at least. The national lottery provider has said the combination is "unexpected" but many people choose. How many different. 3, bigger number in tens place – 4 and the greatest number in the ones place – 6, so the smallest number is 346. Finds the number of combinations and permutations that result when you choose r elements from a set of n elements. Find among those numbers, how many numbers will be even number? Q-5: Consider a committee with 9 persons, in how many ways can we choose a chairman and a vice chairman? Q-3: There are 10 boys and 5 girls through which a committee with 7 members has to be formed. I have a list L:={1,2,3,4,5,6,7,8}. We need to find out how many 3 digit numbers divisible by 5 can be formed from the $10$ digits $(0,1,2,3,4,5,6,7,8,9)$ without repetition. All compounds are molecules, but not all molecules are compounds. With no repeated digits (lead 0 is allowed). A combination is a way to select a part of a collection To avoid a situation where there are too many generated combinations, we limited this combination generator to a specific, maximum number of. Generally speaking, the love they are seeking for is the combination of friendship, business partnership and also lifelong romance. With no repeated digits (lead 0 is not allowed). Then turn the lock 3 times to the right. Each number is used at most once. more stack exchange communities. (2) Click to select your number list, (you also can directly type the numbers separated by commas into textbox), and click Add to add the first list into Combinations list; (3) Repeat step (2) for three times to add other three number lists into Combinations list. Hi, I need to create a formula that will show the results of all possible 4 digit combinations using the numbers 0,1,2,3,4,5,6,7,8,9. How many six-digit identification numbers can be formed if digits may be repeated but none Combinations of a subset of a larger set of objects refer to the number of ways we can choose items in any order. just subtract 6! - 5! and you have 720-120=600. I've found a permutation solution here, but what it does is that it prints out all the possible numbers according to the length of array. That depends on how many of those six numbers you take. i am looking for how many possible combinations there are in 12 games that i have to at least guess 5 correct. Description for Correct answer: There is a 7-digit telephone number but extreme right and extreme left positions are fixed. more stack exchange communities. A restaurant offers six different main courses, eight types. That is more than. Users should hear and feel a click when they finish turning the lock. They are written in words here as a. pick3 numbers, permutations, lock combinations, pin-codes): 29,142,257,760 (~29. So, we are counting the number of combinations of 7 numbers. There are four choices for each digit, except the first which cannot be a zero, in which case the number becomes a 3-digit number. Combinations: 7C3 • In our list of 210 sets of 3 professors, with order mattering, each set of three profs is counted 3! = 6 times. To calculate the probability of winning, we must now find out how many total combinations of 4 numbers can be chosen from 10; to do so, we can use the combinations formula. Therefore, the number of combinations is equal to the number of permutations divided by 4!. Every combination of atoms is a molecule. I need the numbers to all add together to equal 28 and I need all the possible ways to do it, such as. AVOID ALL SAME LAST DIGITS Many people like to play same last digits, such as 3-13-23-33-43, or all digits that end in seven, or some other favorite lucky lottery number. And Combinations Questions Answers 21300: C. How many seven-digit phone numbers can be formed if the first digit cannot be 0 and repetition of digits is not permitted? The Board of Directors does not have assigned seats in the conference room. So, the total number of way of distribution of rings is = This is essentially how the. The dice were distinguishable, or in a particular order: a first die, a second, and a third. pick3 numbers, permutations, lock combinations, pin-codes): 29,142,257,760 (~29. I choose three elements from list L. How many combinations of DNA can a human embody? originally appeared on Quora: the knowledge sharing network where compelling questions are How many combinations of DNA can a human embody? The number is essentially infinite. Players who want to participate have to select seven numbers (from 1 to 35) plus a Powerball number (from 1 to 20). How many combinations are there? Abe, Bob, Carol, Dee. In this video tutorial, viewers learn how to dial open a combination safe lock. We mean, "4! is the number of permutations of 4 different things taken from a total of 4 different things. i put in excel every combination (one by one, put every single combination with "duplicate values" turned ON). There are 288+ billion different possible positions after four moves apiece. We will start from basics and discuss a logical Now we know that 31-squared is the largest ODD number that fits the given descriptionNow we just have to count up the number of terms. A softball league has 7 teams, what are the possible ways of ranking the teams? n = 7, r = 7 7! = 5040 (7– 7)! Recall 0! = 1. The heavy-duty models often have rotary combination locks while the light-duty models have push button locks. No digits should be repeated anywhere in the bill number. we know then that we have 6! - (all combinations with 0 as the beginning digit) to find out all the combinations where the number begins with 0, is just 5!. The number of combinations that are possible with 7 numbers is 127. In how many ways can 3 red marbles, 7 blue marbles, and 5 green marbles be rranged? A Powerball ticket has you pick five numbers out of the numbers from 1 to 69 and one special number (the powerball) from the numbers 1 to 26. In how many ways can 3 black and 2 green marbles be chosen? Answer:_____ Hint: This is a double combination AND a counting principle at the same time. First find the number of combinations if all letters and numbers can be repeated (26*26*10*10*10*10)=6760000 Then find the number of combinations if none were repeated (26*25*10*9*8*7)=3276000. Each possible selection is called a combination. How many different 9-letter arrangements are possible using the letters in the word DISAPPEAR? Combination Formula: C n — total number of objects available r — number of objects to use for the arrangement *Important Shortc ut: n Cn 12. Thirty players show up at camp, but the coaches can choose only four. Number of combinations n=11, k=3 is 165 - calculation result using a combinatorial calculator. The number of ways that n elements can be arranged in order is called a permutation of the elements. most lottery numbers): 45,379,620 (~45. The difference between combinations and permutations is ordering. Return a list of all possible valid combinations. In the third column, enter the number of cards you want in your hand for your combination. Number 7 - Fun Facts. How many different combinations are offered? 2. Many combinatorial applications can make use of a vector 1:n for the input set to return generalized, indexed combination subsets. For Alice there are 7 choices, then for Betty 6 choices, and so on, so that after 5 steps the number of permutations is P(7,5) = 7 · 6 · 5 · 4 · 3 = 2,520. These type combinations are an overview to help people understand some of the main positive and negative issues that are likely to arise between any two types. With/without repetition, with/without order. Combinations - Problem Solving. The smallest number would be 000 and the largest 999 for 1000 total possible combinations. If the same letter CANNOT be repeated, how many can be. "As history goes, number-letter combinations have held religious, superstitious, mythical and, of course, mathematical significance… In the Radiolab podcast, Rowland describes how one of his most exciting branding experiences involving numbers was with KFC, the U. Number bonds are a very important foundation for understanding how numbers work. Finding the Number of Permutations of n Distinct Objects Using a Formula. Number 7 is for Neptune and number 1 is for Sun. , 6 x x x x x 5. Purpose of use Needed to calculate a very large probability based on the Combination of 10,000,000 chemicals taken 500,000 at a time. Two women first choose two chairs from those marked 1 to 4 and 3 men select 3 chairs from the remaining. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i]. (b) n = 6, r = 3Number of permutations = Hence, the number of ways in which 3 prizes can be awarded = 216. Then if you move a different number to the front, you can make the same number of arrangements with that number being first, moving around the numbers behind it. Kaprekar's operation In 1949 the mathematician D. Theoretically, a straight sequential combination such as 1-2-3-4-5-6 is equally likely. There are 3 to the power of 64 possibilities, which is a very large number, with 31 digits. Here there three places which are to be filled with numbers upto { 1. So he wrote a computer program that modeled all the possible brick combinations. Using an estimate of mutation frequency of around 2 x 10. Example: In how many ways can a coach choose three swimmers from among five swimmers? Solution: There are 5 swimmers to be taken 3 at a time. Below are Total 43 words made out of this word. same 1,2,3,4) then you can chose which number goes first, then second etc. To win the small county lottery, one must correctly select 3 numbers from 30 numbers. There are 10,000 combinations of four numbers when numbers are used multiple times in a combination. This is an English lesson which teaches how to pronounce Most of them are formed by adding 'th' to the end of regular numbers, for example: fourth (four-th) We do not make hundreds, thousands and millions plural when the number in front is more than 1. There are 144 ways for this event to occur out of the 5040 possible permutations. These sentences do not say how many hundreds nor how many thousands: the "s" is the only mark of plurality. To win the jackpot, you need to have a lottery ticket with the correct combination of five white balls and the red Powerball. The binomial coefficient formula is a general way to calculate the number of combinations. We know that binary digits, or bits only have two values, either a "1" or a "0" and conveniently for us, a sign also has only two values, being a "+" or a "-". I have got data set 1 (1 to 8) and data set 2 (9-16). There will. With 3 digits, 35x35x35=42875 combinations. " That is, a given number, say 10, can only appear ONCE in the winning combination (most work this way). Description for Correct answer: There is a 7-digit telephone number but extreme right and extreme left positions are fixed. Hi, I need to create a formula that will show the results of all possible 4 digit combinations using the numbers 0,1,2,3,4,5,6,7,8,9. Part of solved Permutation and combination questions and answers : >> Aptitude >> Permutation and combination. The Enneagram Type Combinations. The original number is commensurable by 6 and 7, all the numbers from 1 to 9 are used, and after rounding four times the sum of the not rounded numerals equals 24. And Combinations Questions Answers 21300: C. I want a complete list of 4 digit number combinations from 0 to 9, thank you. Do this without passing the third number at all. Even assuming that number combinations could be eliminated, it would take 1,000 drawings, or 2. Posted on 2003-11-23 10:36:01 by roticv. Combinations deal subsets of a set of items, for example, how many 5 card poker hands can be dealt from a deck of 52 cards. so as this is how I have been provided that answer. Combinations vs Permutations. Solve the equation to find the number of combinations. Number, Ranking and Time Sequence Test Questions & Answers : How many combinations of two-digit numners having 8 can be made from the following numbers? 8, 5, 2, 1, 7, 6. This means that if there are 52 cards, how many combinations of 5 cards can be drawn (answer 2,598,960 combinations). The combntns function provides the combinatorial subsets of a set of numbers. COUNTING RULE 5: COMBINATIONS The number of ways of selecting x objects from n objects, irrespective of order, is equal to (4. Example: Calculate the number of combinations of (50 choose 5) = 2 118 760, and multiply by (11 choose 2) = 55 for a total of 116 531. The number of 7 number combinations between 1 and 49 is 85,900,584. What's the Difference? In English we use the word "combination" loosely, without thinking if the order of things is important. Hence the required number of ways = 7 × 7 × 7 × 7 × 7 = 7⁵. any groups of 5 ( eg: a,b,c,d,e ,f,g,h,k,l ,c,f,h,j,k etc. Before we learn the formula, let’s look at two common notations for permutations. So, when selecting three from the numbers 1 through 9, choosing 1,4,7 is same as chossing 7,4,1. The last digit can be 0,2,4, or 6 (4 possibilities), and after you've chosen that, theres 7 possibilities for the first digit and 6 possibilities for the second digit, which makes 4*7*6 = 168. Some numbers, however, are rolled more frequently as the number of combinations that add up to them is greater. In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. Use this permutation and combination calculator to find the nP r and nC r values by providing the number of sample points in set 'n' and 'r'. So, you have number consisting from 7 unique digits. For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. When 5 is at the hundred's place, then 7 will be at the ten's place. Players used this combination tens of thousands of times every week. In order to win the big Powerball jackpot, you have to match all your numbers. System 10 Example. There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them. There are 18 two digit numbers containing 7. • To see how to count by using other than 10 digits, notice how we count in the ordinary decimal system. Find the least value of "x" that will make the statement true. Notionally, there is also one combination with no numbers in it. So, you have number consisting from 7 unique digits. We need to find out how many 3 digit numbers divisible by 5 can be formed from the $10$ digits $(0,1,2,3,4,5,6,7,8,9)$ without repetition. The vibration of this number represents all that is Some of the more positive traits that are representative of those with 7s in their numerological charts are; Analytical, philosophical, logical, private Possible number combinations. PA license plates have 3 letters followed by 4 numbers. Number 7 - Fun Facts. note apparently in excel 2007 you need to replace. You could have 10 for the first one, 9 for the second one because you can't repeat, 8 for the third etc and 4 for the seventh one. The vibration might help to engage the mechanisms. One of the keys to making your design come alive is choosing just the right color combination. Continuing there are 10 7 choices for a 7 digit phone number. C15 Release 17; Identity Assurance – How to Prevent Robocalling and Fraud for Every Call; Private Optical Network Solutions for your Enterprise. If I would to calculate the possible combination with the combination formula. There are 12 standbys who hope to get on your flight to Hawaii, but only 6 seats are available on the plane. If not it is 7x6x5x4 or 840 combinations. Answer 107 6. The answer is much smaller because duplicates are ignored. But you can win something even if you only match one number — the red Powerball. Theoretically, a straight sequential combination such as 1-2-3-4-5-6 is equally likely. n3! …} \ $. Enter your n and r values below:-- Enter Number of Items (n) -- Enter Number of Arrangements (r) Evalute the combination n C r. It counts to 9. Since repetition is not allowed, the number of telephone numbers that can be formed is equal to the number of arrangements of the 8 digits, taken 4 at a time. Here are all the details. Then turn the lock 3 times to the right. I ha padlock wit 6 numbers in 4 possible combinations. There are many ways to enumerate k combinations. Lottery number combination generator lets you generate multiple random combinations of your own lucky lottery numbers. ∴ Number of available digits = 8 The telephone number consists of 6 digits. D is the answer. More formally, a k-combination of a set S is a subset of k distinct elements of S. Evaluate the following: 1) 5! 2) 8! 10! 3) 6!2! 7!4! II. That dosent count with the letters of the alphabet and numbers 0-10, it only counts the combination of 25 units rearanged by 1 unit. And 411, 911, Round or repeating numbers?. A briefcase lock has 3 rotating cylinders, each containing 10 digits. 9!/(5!4!) = 126. If not it is 7x6x5x4 or 840 combinations. With 3 digits, 35x35x35=42875 combinations. 1 Class 11 Maths Question 1. The easiest way to go through the combinations, ensuring that you won't miss one, is like this:. The initial numbers have already been fixed as 35. Note that in these preceding examples, the gift certificates and the Olympic medals were awarded. Assuming you meant how many combinations can be formed by picking 8 numbers from 56 numbers, we have. What is the total number of ways in which each of the five persons can leave the cabin at any of the 7 floors? Solution: Any one of the 5 persons can leave the cabin in 7 ways independent of other. This time, it is six times smaller (if you multiply 84 by 3! = 6 , you'll get 504). Another definition of combination is the number of such arrangements that are possible. "As history goes, number-letter combinations have held religious, superstitious, mythical and, of course, mathematical significance… In the Radiolab podcast, Rowland describes how one of his most exciting branding experiences involving numbers was with KFC, the U. By joining any two points on the circle, we may draw a chord. From a list of 20 suggested books, Amy must choose 4 to make a report. Now we want to count simply how many combinations of numbers. put C in A1 (c for combinations p for permutations). Suppose "combination" lock has a dial whose numbers are 1 through 16. It is only an extension of how many trebles from 6 The more straightforward calculation there is 6*5*4 divided by 3*2*1 so 120/6 = 20. South Africa's national lottery draw produces numbers 5, 6, 7, 8 and 9. Q6 How many 4 digit numbers can be formed with different conditions. Determining how many combinations of 4 sashes there are in AMTGARD to make marking monsters easier. Students continue to compose and decompose 2-digit numbers and to represent them as the sum of multiples of ten and some number of ones (e. Determine the number of 5 card combination out of a deck of 52 cards if there is exactly one ace in each. There are many ways to enumerate k combinations. How many seven-digit phone numbers can be formed if the first digit cannot be 0 and repetition of digits is not permitted? The Board of Directors does not have assigned seats in the conference room. For example the first digit can't be 1, that indicates a long distance number. The smallest number would be 000 and the largest 999 for 1000 total possible combinations. note apparently in excel 2007 you need to replace. Answer Questions and Earn Points !!! You can now earn points by answering the unanswered questions listed. How many of these numbers end in 3 or 6? Ans: 24, 12 2. Write out the formula in order to calculate the number of combinations without repetition. Introduces the concepts behind different number bases, and shows how to convert between decimal (base ten) and binary (base two) numbers. For every possible combination of 5 numbers from the 69, there are 26 possible Powerball numbers, so to get the total number of combinations, we multiply the two combinations. We want to develop a formula for computing the number of Example 5 How many committees can be formed from a group of 5 governors and 7 senators if each committee consists of 3 governors and 4 senators?. i) How many possible 6-number combinations are there for drawing? 7,059,052 ii) What is the probability of winning the lotto? 0. But how do we represent signed binary numbers if all we have is a bunch of one's and zero's. A combination is an arrangement of objects, without repetition, and order not being important. Example: Calculate the number of combinations of (50 choose 5) = 2 118 760, and multiply by (11 choose 2) = 55 for a total of 116 531. A: 8 B: 9 C: 10 D: 15 Correct Answer : C 7. These sentences do not say how many hundreds nor how many thousands: the "s" is the only mark of plurality. 1212121212121212. n = 20 would require visiting about one million numbers while the maximum number of allowed k combinations is about 186 thousand for k = 10). For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. We know that binary digits, or bits only have two values, either a "1" or a "0" and conveniently for us, a sign also has only two values, being a "+" or a "-". The total number of possible hands can be found by adding the above numbers in third column, for a total of 2,598,960. We will start from basics and discuss a logical Now we know that 31-squared is the largest ODD number that fits the given descriptionNow we just have to count up the number of terms. If there were 2 digits on a plate, you'd have 35x35 = 1225 combinations. Each r-combination of a set with n elements when repetition is allowed can be represented by. Assuming 7 distinct numbers, you can have 7!/(. Here are all the details. First find the number of combinations if all letters and numbers can be repeated (26*26*10*10*10*10)=6760000 Then find the number of combinations if none were repeated (26*25*10*9*8*7)=3276000. 7 in the unit’ s place. With permutations we care about the order of the elements, whereas with combinations we don't. To calculate the probability of winning, we must now find out how many total combinations of 4 numbers can be chosen from 10; to do so, we can use the combinations formula. You have won first place in a contest and are allowed to choose 2 prizes from a table that has 6 prizes numbered 1 through 6. How many committee of five persons with a chairperson can be selected from 12 persons. In general, the formula we use to determine the number of combinations possible. Find numbers that are greater than 4000 which are formed using digits 0,2,4,6,8without repetition. The same holds for 2-4-6-8-10-12 or 5-10-15-20-25-30. There are many ways to enumerate k combinations. Combinations: Defination: The number of ways in which smaller or equal number of things are arranged or selected from a collection of things where the order of selection or arrangement is not. 43368382 x 10 to the power of 30 to be exact. Many calculators can evaluate this "n choose m" notation for you. In that case, the number formed is not an even number. StanleyKnife ( 27 ) “Great Answer” ( 0 ) Flag as… ¶. The number 7 has a somewhat secretive, withdrawn, and distant energy. This material describes how numbers are expressed by numerals in English and provides examples of cardinal and ordinal numerals, common and decimal fractions, and examples of differences between British and American English in expressing numbers. To solve this problem using the Combination and Permutation Calculator, do the following: Choose "Count permutations" as the analytical goal. how many 4 digit number combinations can u make out of this 4 numbers 0189 and what are they. See full list on lotterypost. Number of ways to arrange these letters$=5!=5×4×3×2×1=120$. That’s how tinyurl works, of course. 3 Question 1: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Answer 3-digit numbers have to be formed using the digits 1 to 9. 473816952 - if rounding changes the next numeral character. Let k = number of chosen elements. Learn more Permutations and combinations have uses in math classes and in daily life. Answered by Penny Nom. Example: In how many ways can a coach choose three swimmers from among five swimmers? Solution: There are 5 swimmers to be taken 3 at a time. In an examination, a student has to answer. First find the number of combinations if all letters and numbers can be repeated (26*26*10*10*10*10)=6760000 Then find the number of combinations if none were repeated (26*25*10*9*8*7)=3276000. That dosent count with the letters of the alphabet and numbers 0-10, it only counts the combination of 25 units rearanged by 1 unit. Before each church service 5 bells are rung in sequence. Thus, the odds of picking that perfect combination with a single ticket are one in 292,201,338. Permutations and Combinations Questions & Answers for AIEEE,Bank Exams,CAT, Bank Clerk,Bank PO : How many 7 digit numbers can be But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. But if repetition is allowed then there are infinitely many combinations of all sizes that can be formed. Example: Calculate the number of combinations of (50 choose 5) = 2 118 760, and multiply by (11 choose 2) = 55 for a total of 116 531. Let the three people be A , B and C. Learn how to write binary numbers, and the (not so secret) code to change English letters into binary numbers and back again. Two women first choose two chairs from those marked 1 to 4 and 3 men select 3 chairs from the remaining. Apart from the three winning numbers, there are seven other numbers that can be chosen for the fourth number. What is a combination? - combination definition. One way is to visit all the binary numbers less than 2n. Permutation & combination deal with the techniques of counting without direct listing of the number of elements in a particular set or the number of outcomes of a particular experiment. A health inspector has time to visit 7 of the 20 restaurants on a list. For problems like this, the number of combinations can be found with this formula: n! / (k! * (n - k)!) n = 49. Find the number of 7-digit telephone numbers a. Suppose we number the people from 1 through 4 and think of the set A = {1,2,3,4}. If you can repeat the numbers, then number of combinations is infinite. Combination Notation In Example 1, after you cross out the duplicate groupings, you are left with the number of combinations of 4 items chosen 3 at a time. When working with combinations of numbers you. That is more than. How many possible valid numbers, where a valid number is any number between 0-9, length of 10 digits, excluding # or *, a piece of Chess can trace while travelling across a telephone keypad. how to calculate number of possible combinations. Let n = number of total elements to choose from. Number, Ranking and Time Sequence Test Questions & Answers : How many combinations of two-digit numners having 8 can be made from the following numbers? 8, 5, 2, 1, 7, 6. Numbers from 1000 to 1,000,000. Enter the pool of numbers you would like to pick the numbers from. To win the jackpot, you need to have a lottery ticket with the correct combination of five white balls and the red Powerball. Posted on 2003-11-23 10:36:01 by roticv. = (total number of combination of 7 out of 12) * (total combinations of 5 out of remaining 5) = 12 P 7 / 7! * 5 P 5 / 5! = 12! / [7! ( 12 - 7 )!] * 5! / [5! ( 5 - 5 )!] = 792 * 1. 5C3 or 5 choose 3 refers to how many combinations are possible from 5 items, taken 3 at a time. A combination is a way to select a part of a collection To avoid a situation where there are too many generated combinations, we limited this combination generator to a specific, maximum number of. On first place you can put 10 difrent digits, on second place you can put only 9 digits, etc. Permutations and Combinations Permutation and Probability: A computer program requires the user to enter a 7-digit registration code made up of the digits 1, 2, 4, 5, 6, 7, and 9. Using numbers as well will give you many more, of course: 26 letters + 10 numbers gives you over 60 million possibilities. Permutations and Combinations Questions & Answers for AIEEE,Bank Exams,CAT, Bank Clerk,Bank PO : How many 7 digit numbers can be But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. Thus, the total number of combinations of 2-of-a-kind and 3-of-a-kind are 48 x 78 = 3,744. Is it at all possible for man to calculate or program a computer if there were infinite time? Qrazy question but a guy told me that bridge was the game with the highest number of possible combinations. P=84+9,880+5,040+28,080# #:. Example 2:. How many seven-digit phone numbers can be formed if the first digit cannot be 0 and repetition of digits is not permitted? The Board of Directors does not have assigned seats in the conference room. The number of distinct combinations of 3 professors is 73 63 35 3321 6 73 73 7 7 6 5 210 73 ⋅⋅ − == ==== ⋅⋅! ()!!! PP C 7C 3 is the number combinations of 3 objects chosen from a set of 7. How many possible outcomes are there to the Alaska Lottery? SOLUTION: In a lottery, unless we are told otherwise, the numbers chosen must be different, and the order of selection does not matter. Plus, you can even choose to have the result set sorted in ascending or descending order. The content of this article may be too rudimentary for most readers, but for beginners, it will be helpful. Learn more about the differences between permutations and combinations, or explore hundreds of There are different types of permutations and combinations, but the calculator above only considers the case The calculator provided computes one of the most typical concepts of permutations where. Suppose we number the people from 1 through 4 and think of the set A = {1,2,3,4}. Users should hear and feel a click when they finish turning the lock. Each combination is played as if it were a single selection on a single ticket. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? Number of groups, each having 3 consonants and 2 vowels = 210. I am trying to find a code for excel that I can use to list all possible combinations of the numbers/symbols 0-9, #, and * or just numbers 0-12 in 6 digit combinations. - Combination Play costs for LOTTO 6/49:$88, $14,$56 or $168 - You win on a Combination Play ticket the same as on a regular selection. See full list on intmath. A related topic to combinations is "permutations". The number of possible combinations of council members, presuming no differentiation among office-holders. How many combinations of DNA can a human embody? originally appeared on Quora: the knowledge sharing network where compelling questions are How many combinations of DNA can a human embody? The number is essentially infinite. if you have 1 of 10 numbers, your total number is 10. So the required number of ways = 8 x 7 x 6 x 5 x 4 = 6720. Number of ways of arranging 5 letters among themselves. Fortunately, we can solve these problems using a formula. 3 digit number : Number of 3 digit numbers with repetition = 5 × 5 × 5 = 125 Ex 7. These type combinations are an overview to help people understand some of the main positive and negative issues that are likely to arise between any two types. If you are really asking "How many combinations of 7 different numbers can be created if the numbers are selected from a total of 45 numbers?", the answer is 45C7, which is calculated as 45!/(38! x 7!) =. 1,1 - Chapter 7 Class 11 Permutations and Combinations. decimal, hexadecimal etc. To calculate the probability of winning, we must now find out how many total combinations of 4 numbers can be chosen from 10; to do so, we can use the combinations formula. Combinatorial calculator - calculates the number of options (combinations, variations ) based on the number of elements, repetition and order of importance. Since the 3 digit number must be divisible by 5, we can have 0 or 5 at the units place. In how many ways can 3 red marbles, 7 blue marbles, and 5 green marbles be rranged? A Powerball ticket has you pick five numbers out of the numbers from 1 to 69 and one special number (the powerball) from the numbers 1 to 26. Another mini-lottery has 5 balls and 3 balls to be drawn. Combination Notation To find the number of combinations of n objects taken r at a time, divide the number of permutations of n objects taken r at a time by r!. An original selection of 50 color combinations you can use in your infographic and presentation design. Answer Questions and Earn Points !!! You can now earn points by answering the unanswered questions listed. i'm hoping I helped you artwork out that answer. i'm going to chat to you later,Babe. How many combinations if I'm starting with a pool of six, how many combinations are there? , these allow successive computation of respectively all numbers of combinations from the same set (a row in Pascal's triangle), of k-combinations of sets of growing sizes, and of combinations with a. How many seven-digit phone numbers can be formed if the first digit cannot be 0 and repetition of digits is not permitted? The Board of Directors does not have assigned seats in the conference room. (b) n = 6, r = 3Number of permutations = Hence, the number of ways in which 3 prizes can be awarded = 216. I wondered if anyone had similar numbers for how many possible key combinations there are. Combination Notation To find the number of combinations of n objects taken r at a time, divide the number of permutations of n objects taken r at a. How many words with two different vowels and two different consonants can be formed without repetition of letters? HSSLIVE. Enter your n and r values below:-- Enter Number of Items (n) -- Enter Number of Arrangements (r) Evalute the combination n C r. Generally speaking, the love they are seeking for is the combination of friendship, business partnership and also lifelong romance. Numbers having at least one digit = 1 is the union of all numbers in 1, 2 and 3. The number of arrangements can be divided by the number of arrangements not used. All compounds are molecules, but not all molecules are compounds. The number of such numbers beginning with '0'. ) ( ) ( choose ) Where n is the number of things to choose from, and you r of them. How many combinations of three are possible? 1 How many cominations of 3 numbers/sheep/anything you like can you have with 3 number/sheep/anything you like. How many two digit numbers can be generated using the digits 1,2,3,4 without repeating any digit? How many different nine-digit phone numbers can be constructed if each number can only appear once and the first three numbers must be 860? There are 6 persons in an office. A simplification of this nature will always be possible when applying this rule or the below rule. Online calculator combinations without repetition. The formula for a combination of choosing r unique ways from n possibilities is. decimal, hexadecimal etc. Combination: The number of distinct combinations of n objects, taken r at a time, is given by the ratio The Permutation (nPr) and Combination (nCr) work with steps shows the complete step-by-step calculation for finding the the number of ways in which we can choose$8$different objects out of a set containing$10\$ different objects, where. Observe that the number of ways of selecting 3 officers is 6 times as large as the number of ways of selecting a committee of 3. Example The number of combinations of 4 objects taken 2 at. That is more than. That depends on how many of those six numbers you take. Number of combination = (2^8)^X = 2^8X. For example, in Genesis 9:12-16, God makes the rainbow the sign of His promise to Noah (and, by extension, to all mankind). | 2021-04-11T04:23:33 | {
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http://christofseiler.github.io/stats205/Homework2/Solution2.html | Q1: 20
Q2: Complete enumeration implementation(30) + Analysis of results(10, including difference between Monte Carlo and Complete enumeration, which point(s) to remove to get similar results)+experimentally/theoreticallly comparision with grey code implemntation(10)
Q3: 30
### Exercise 1
We want to determine which is the most likely bootstrap sample and its associated probability in $$n = 3$$, $$n = 5$$, $$n = 10$$, $$n = 12$$, $$n = 15$$, and $$20$$.
We can model the probability of a bootstrap sample of size n as rolling a k-sided dice where k is the number of distinct values there are in the sample. We see that the probability of randomly drawing a specific observation is $$\frac{1}{n}$$. This means that the more a certain value appears in our original sample, the more likely it is for us to pick that value in our bootstrap sample.
Using the dice comparison above we see that the probability of each bootstrap sample of size n from our original sample is given by a multinomial distribution. A multinomial distribution models the probability of counts for when there are n independent trials of k different outcomes. The PMF of a multinomial distribution gives the likelihood of any combination of “successes” for the k categories.
Let’s say that we have an original sample of $$\{ x_1 \cdots x_n \}$$. We want to draw a bootstrap sample $$\{ s_1 \cdots s_n \}$$.
The formula for the probability mass function of a multinomial distribution with n trials and k categories is $$\frac{n!}{x_1!\cdots x_k!} p_1^{x_1} \cdots p_k^{x_k}$$. Again, n represents the number of trials or the size of the bootstrap sample. k represents the number of outcomes that the sample of 1 observation can take. In our case, we see that we have n different outcomes. Thus, we can say that there are n different values we can draw from. This transforms the multinomial PMF into $$\frac{n!}{s_1!\cdots s_n!} p_1^{s_1} \cdots p_k^{s_n} = \frac{n!}{s_1!\cdots s_n!} (\frac{1}{n})^{s_1 + \cdots + s_n}$$. Note that the probability of drawing any particular sample is $$\frac{1}{n}$$ and the actual number of each original obervation $$x_i$$ drawn is $$s_i$$.
If we look at the formula for the PMF which is $$\frac{n!}{s_1!\cdots s_n!} (\frac{1}{n})^{s_1 + \cdots + s_n}$$, we see that we can maximize this probability by maximizing the numerator $$n!$$ or by minimizing the denominator $$s_1!\cdots s_n!$$. For the most part, the sample size n is fixed. Thus, we want to minimize the denominator. We see that the denominator $$s_1!\cdots s_n!$$ is minimized by setting all $$s_i$$ for $$i = \{1 \cdots n \}$$ equal to 1. We have the constraint of $$\sum_{i = 1}^n s_i = 1$$, so we cannot obtain a value of $$s_1!\cdots s_n!$$ smaller than 1.
This means that $$\textbf{the most common bootstrap sample is actually the original sample itself}$$.
Below, we calculate the probability of obtaining the original sample for the bootstrap sample using a multinomial distribution.
n <- c(3, 5, 10, 12, 12, 15, 20)
mult_probs <- array(0, length(n))
df <- cbind(n, mult_probs)
for (i in 1:length(n)) {
x <- df[i, 1]
df[i, 2] <- dmultinom(x = rep(1, x), size = x, prob = rep(1/x, x))
}
We also calculate the probabilities manually using the PMF to see if they match the ones obtained above.
n <- c(3, 5, 10, 12, 12, 15, 20)
mult_probs <- array(0, length(n))
df_manual <- cbind(n, mult_probs)
for (i in 1:length(n)) {
x <- df_manual[i, 1]
df_manual[i, 2] <- factorial(x)*(1/x)^x
}
It seems like the results obtained using both methods are the same since the probabilities are the same.
The probability associated with the most common bootstrap sample (the original sample) in $$n = 3$$, $$n = 5$$, $$n = 10$$, $$n = 12$$, $$n = 12$$, $$n = 15$$, and $$20$$ is listed below. The probabilities range from $$0.22$$ with $$n = 3$$ to $$0.0000000232$$ with $$n = 20$$.
df
## Exercise 2
In this exercise, we want to note the differences behaviors of metrics between using an Monte Carlo simulation and a complete enumeration method. We will first load the data and our required packages.
library(ggplot2)
library(bootstrap)
library(dplyr)
library(parallel)
data(law)
dim(law)
We want to figure out which observation(s) do we need to remove from the law data in order to make the Monte Carlo simulation and complete enumerations simulation look more similar. Intuitively, we want to remove the outliers. In the law data case, we see that Observations 1 and 11 don’t quite lie around the diagonal. We will pick one of these two points to rerun the complete enumerations simulation and then compare the new distribution to the one obtained using the bootstrap Monte Carlo simulation. This way, we can isolate the effect of the change in the distribution and attribute it to the removal of that particular outlier.
$$\textbf{We see that by removing the outler, the correlation of the data set increases overall. We hypothesize that this will pull the center of the bootstrap distribution up to the true sample correlation. This will remove the subtle bump in that distribution such that the Monte Carlo simulation will look more similar to the complete enumerations simulations.}$$
Note that if we remove both Observation 1 and Observation 11, the correlation of the data is higher than removing only one.
Below, we create a new data set such that we remove Observation 11. We check the correlation of the new sample with regards to the old law data set. We see that there is an increase in correlation rom $$0.7763$$ to $$0.8929$$.
law2 <- data.frame(Observation = 1:dim(law)[1], law)
# Given picture in the homework
law2 %>%
ggplot() +
geom_text(mapping = aes(x = LSAT, y = GPA, label = Observation),
hjust = 0, vjust = 0)
law2 <- law2 %>%
filter(Observation != 1) %>%
select(-Observation)
cor(law$LSAT, law$GPA)
cor(law2$LSAT, law2$GPA)
Here, we generate gray codes in order to get all possible bootstrap samples of the law data. We take around 2-3 minutes to generate the gray codes.
# Generating gray codes
# ptm <- proc.time()
# n <- length(law2$LSAT) # i <- 1 # total <- choose(2*n - 1, n - 1) # gray_codes <- matrix(0, nrow = total, ncol = n) # # r <- array(0, n) # r[1] <- n # t <- n # h <- 0 # gray_codes[i, ] <- r # i <- i + 1 # # while (r[n] != n) { # if (t != 1) { # h <- 0 # } # h <- h + 1 # t <- r[h] # r[h] <- 0 # r[1] <- t - 1 # r[h + 1] <- r[h + 1] + 1 # gray_codes[i, ] <- r # i <- i + 1 # } # proc.time() - ptm Now, we recompute the complete enumerations bootstrap after removing Observation 11. We compute both the new correlation of the bootstrap sample as well as its multinomial probability of obtaining that particular bootstrap sample. We see that the distribution of using the complete enumeration is about the same as the one obtained in the lecture slides. The distribution and the code is shown below. # Complete Enumerations Simulation # ptm <- proc.time() # enumData <- mclapply(1:total, function(i) { # index <- t(gray_codes)[,i] # law_list <- lapply(1:n, function(j) matrix(rep(law2[j,], index[j]), # ncol = 2 , byrow = TRUE)) # newLaw <- do.call(rbind, law_list) # c(cor(unlist(newLaw[,1]), unlist(newLaw[,2])), dmultinom(index, prob = rep(1,n))) # }, mc.cores = 4) # proc.time() - ptm # # enumData <- t(simplify2array(enumData)) # colnames(enumData) = c("cor","weight") # # enumData %>% # as.data.frame() %>% # tbl_df() %>% # ggplot() + # geom_histogram(mapping = aes(x = cor), binwidth = 0.013, na.rm = T) + # geom_vline(mapping = aes(xintercept = cor(law2$LSAT, law2$GPA), colour = "red")) + # labs(x = "Correlation", y = "Count", title = "Complete Enumerations Distribution") + # guides(colour = F) Here we look at the bootstrap of the law data and compare both distributions. We see a change in the histogram compared to the one we had in class. $$\textbf{We see that the histogram of the bootstrap samples became more smooth and more similar to the smoother distribution of the complete enumerations}$$. This is probably due to the shift in correlation. The removal of the outlier Observation 1 increased the correlation of the law data. This, in turn, will push the overall correlation distribution to the right. This will then cover the awkward bump in the Monte Carlo simulation distribution as the mass of the data will be shifted right. The removal of the outlier will pull the correlation of the bootstrap samples to the right. The distribution and the code is shown below. # Bootstrap Monte Carlo Simulation # ptm <- proc.time() # j <- length(law2$LSAT)
# B <- 40000
# mc_cors <- array(0, B)
#
# for (i in seq_len(B)) {
# mc_samp <- sample(1:j, size = j, replace = T)
# law_samp <- law2[mc_samp,]
# mc_cors[i] <- cor(law_samp$LSAT, law_samp$GPA)
# }
# proc.time() - ptm
#
# boot_data <- cbind(mc_cors, 1)
#
# boot_data %>%
# as.data.frame() %>%
# tbl_df() %>%
# ggplot() +
# geom_histogram(mapping = aes(x = mc_cors), binwidth = 0.013) +
# geom_vline(mapping = aes(xintercept = cor(law2$LSAT, law2$GPA), colour = "red")) +
# labs(x = "Correlation", y = "Count", title = "Bootstrap Simulation Distribution") +
# guides(colour = F)
Theoretically, $$\textbf{we find that Gray Codes speed up the computation by about 4 times faster}$$. We start with analyzing the number of computations that the complete enumeration methods use. We will first assume that generating all possible bootstrap samples takes the same time for the complete enumeration and the speed-up.
For the full enumeration, we see that for each possible bootstrap sample, the apply() function goes through each entry in the gray code and generates the corresponding bootstrap sample. Then we take the correlation of that sample and output it.
For the Gray Code speed-up, we see that we take the first bootstrap sample, we go through each entry and generate the bootstrap sample as well as the correlation. Note that we pre-compute what additional correlation will be added to the final product. Since from each gray code to the next will remove an observation and its effect, then add the new observation and its effect, we do significantly less operations.
Assume that we have a sample size of $$n = 5$$. This gives us a total of $${2n - 1 \choose n - 1}$$, which evaluates to $${9 \choose 4} = 126$$ total bootstrap samples. For the full enumeration, we go through 5 entries for each bootstrap sample, pull out 5 total entries from the data set, and then calculate the correlation. This gives about 11 operations over 126 samples (1386 operations total).
For the Gray Codes speed-up, we also have $$126$$ total bootstrap samples. We still go through the gray codes to determine the transition from the previous sample to this sample. We identify two points of difference: the entry we are removing and the entry we are adding. This gives us a total of 10 operations for the first sample and 3 operations for the subsequent samples. Now, we have $$10 + 3*125 = 385$$ operations.
Using the amount of operations, we can say that complete enumerations takes about $$\frac{1386}{385} = 3.6$$ times longer than the Gray Codes speed-up.
The code below demonstrates the speed-up.
# Using the Gray Code speed-up iterative method
# lsat_mean <- mean(law2$LSAT) # gpa_mean <- mean(law2$GPA)
#
# correlation <- (law2$LSAT)*(law2$GPA)
# lsat_square <- (law2$LSAT - lsat_mean)^2 # gpa_square <- (law2$GPA - gpa_mean)^2
#
# calc_cor <- matrix(0, nrow = total, ncol = 2)
#
# ptm <- proc.time()
# for (i in seq_len(total)) {
# boot_samp <- gray_codes[i, ]
# calc_cor[i, 1] <- (boot_samp%*%correlation -
# (sum(boot_samp%*%law2$LSAT)*sum(boot_samp%*%law2$GPA))/n)
# calc_cor[i, 1] <- calc_cor[i, 1]/((boot_samp%*%lsat_square)*(boot_samp%*%gpa_square))
# calc_cor[i, 2] <- dmultinom(x = boot_samp, prob = rep(1/n, n))
# }
# proc.time() - ptm
#
# samps <- sample(calc_cor[,1], size = 40000, prob = calc_cor[,2], replace = T)
# gc_data <- cbind(samps, 1)
#
# gc_data %>%
# as.data.frame() %>%
# tbl_df() %>%
# ggplot() +
# geom_histogram(mapping = aes(x = samps), binwidth = 0.01,
# na.rm = T) +
# geom_vline(xintercept = cor(law2$LSAT, law2$GPA))
In our experimental runthrough, we see that complete enumerations took around 800 seconds while the gray_codes speed up took around 200 seconds. This verifies our theoretical speed-up (albeit with a little noise).
It’s interesting how removing an outlier will change the bootstrap distribution so much.
## Exercise 3
Suppose in a poll of $$500$$ registered voters, $$269$$ responded that they would vote for candidate $$P$$. We want to obtain a $$90%$$ percentile bootstrap confidence interval for the true proportion of registered voters who plan to vote for $$P$$.
We first start off defining our variables: the number of samples n, the actual number of voters who voted for the candidate, the confidence level alpha, number of bootstrap samples B, and the actual proportion of voters who voted for the candidate.
We start by sampling with replacement from our original sample. Our sample is denoted as each observation taking a binary value (0 or 1). 1 represents that the voter voted for candidate $$P$$. 0 means that the voter did not vote for candidate $$P$$.
After each sample, we recalculate the proportion of voters for voted for candidate $$P$$. From our bootstrapped proportion, we calculate the sample standard deviation.
n <- 500
p <- 269
alpha <- 0.10
B <- 10000
actual_prop <- p/n
voters <- c(rep(1, p), rep(0, n - p))
boot_prop <- array(0, B)
set.seed(8)
for (i in seq_len(B)) {
boot_samp <- sample(x = voters, size = 500, replace = T)
boot_prop[i] <- sum(boot_samp)/n
}
samp_sd <- sd(voters)/sqrt(n)
upper <- actual_prop + qnorm(1-alpha/2)*samp_sd
lower <- actual_prop - qnorm(1-alpha/2)*samp_sd
(boot_conf_90 <- c(lower, upper))
From the sample standard deviation, we can use a normal distribution to obtain the upper and lower bounds of the confidence interval.
This allows us to obtain a bootstrapped 90% confidence interval of $$(0.5013, 0.5747)$$. | 2017-08-20T13:23:19 | {
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https://math.stackexchange.com/questions/2292207/topological-space-that-is-compact-and-t-1-but-not-hausdorff-i-e-normal-but-n | # Topological space that is compact and $T_1$ but not Hausdorff (i.e. normal but not hausdorff)
I was freshing up on some topology, and this text I'm reading mentions T1 does not imply Hausdorff. A few counter-examples are readily available, like the natural numbers under the co-finite topology.
But what if we place a restriction on the space to also be compact, the text doesn't mention anything about that and I can't come up with any examples of spaces that are compact and T1 but not Hausdorff. To state it otherwise, I'm looking for a space that is T1 but not normal(=compact and Hausdorff).
• I think a compactification of the line with two origins works? – B. Mehta May 22 '17 at 16:41
• Or just the same construction but starting with $[-1, 1]$ instead of $\mathbb{R}$ – B. Mehta May 22 '17 at 16:42
• Also, the cofinite topology on an infinite set works. – B. Mehta May 22 '17 at 16:43
• normal is not the same as compact plus Hausdorff, it's implied by it. For a compact $T_1$ space $X$ it is true that : $X$ is Hausdorff iff $X$ is normal, which is probably what you meant to say. – Henno Brandsma May 23 '17 at 6:53
You've already given an example: the natural numbers (or any infinite set, really) under the co-finite topology. Given any open cover, fixing a single (non-empty) element of the cover yields an open set that has all but finitely-many of the natural numbers as elements. Thus, only finitely-many more elements of the cover are needed, forming a finite subcover.
The affine space $\Bbb C^n$ endowed with the Zariski topology: $C$ is closed if and only if there is a family of multivariate polynomials $\mathcal F_C\subseteq \Bbb C[x_1,\cdots,x_n]$ such that $$C=\{x\in\Bbb C^n\,:\,\forall f\in\mathcal F_C,\ f(x)=0\}$$
These topologies have the following notable properties:
• they are T1;
• any two non-empty open sets have non-empty intersection (irreducibility): thus the topology is not T2, and on a side note all open subsets are connected;
• for any non empty family $\mathfrak C$ of closed sets, there is $C\in\mathfrak C$ which is maximal with respect to inclusion "$\subseteq$" (noetherianity): thus every subspace is compact.
The proofs of these facts are not difficult, but they use a couple of lemmas of commutative algebra which might make the exposition a bit long.
An easier special case of this is when $n=1$, in which case the Zariski topology is just the cofinite topology on $\Bbb C$ (the topology where a set is open if and only if its complement is either finite or the whole space).
Addendum: I assumed here that your definition of "compact topological space" is:
Cpt: A topological space such that any open cover $U$ admits a finite subcover.
However, a considerable number of authors (for instance, Bourbaki), call "quasi-compact" a topological space which satisfies Cpt and "compact" a T2 quasi-compact space. In this case, though, your question would be trivial.
• Engelking (General Topology) calls a compact $T_2$ space a compact space, and a space in which every open cover has a finite subcover but is not $T_2$ is called pseudo-compact. – DanielWainfleet May 23 '17 at 10:12
• @DanielWainfleet Curiously, encyclopediaofmath cites that book as a reference for "pseudo-compact = completely regular + every continuous $f:X\to\Bbb R$ is bounded". – user228113 May 23 '17 at 10:36
• Thank you. I will have to re-check my copy for his def'n. – DanielWainfleet May 23 '17 at 20:43
Let $T_X$ be a compact Hausdorff topology on an infinite set $X$, with no isolated points. So every non-empty member of $T_X$ is an infinite set. Let $Y=X\times \{1,2\}.$ $$\text {Let} \quad B= \{\;( t\times \{1,2\} )\; \backslash A:t\in T_X \land A \text { is finite} \}.$$ Then $B$ is a base for a topology $T_Y$ on $Y.$
(i). For each $p\in Y$ we have $Y$ \ $\{p\}=(X\times \{1,2\})$ \ $\{p\}\in B\subset T_Y.$ So $T_Y$ is a $T_1$ topology.
(ii). For $x\in X$ and $i\in \{1,2\},$ if $(x,i)\in U_i\in T_Y,$ then $(x,i)\in (t_i\times \{1,2\})$ \ $A_i\subset U_i$ for some $t_i\in T_X$ and some finite $A_i.$
Then $U_1\cap U_2\supset (t_1\cap t_2)$ \ $(A_1\cup A_2)\ne \phi .$ So $(x,1)$ and $(x,2)$ do not have disjoint nbhds: $T_Y$ is not a Hausdorff topology.
(iii). For $i\in \{1,2\}$ the subspace $X\times \{i\}$ is homeomorphic to $X$ so it is compact. Now $Y$ is the union of the two compact subspaces $X\times \{1\},\;X\times \{2\}$ so $Y$ is also compact. | 2019-10-15T17:18:18 | {
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http://mathhelpforum.com/math-challenge-problems/163762-three-digit-numbers.html | # Math Help - three digit numbers
1. ## three digit numbers
Here is a challenge problem I read in a recent publication of the MAA for undergraduates. I found it interesting that the author’s way of solving is completely different from mine.
“How many three digit numbers (no leading zeros) are there if the digits are distinct and one of the three digits is the average of the other two”?
2. I tried and got 69...?
Numbers like 987 and 978 are distinct, right? (where 8 is the mean of 9 and 7)
EDIT: Wait! I forgot to remove the non-leading zeroes...
This done, I get 65
3. That is a huge undercount.
4. Ok, back to work then.
5. Originally Posted by Plato
That is a huge undercount.
Ok, I realised I multiplied my numbers by 3 instead of 3!
Now, I get 196.
Maybe that's it?
6. A huge overcount.
7. Oh my
Ok, getting to work again.
8. Ok, this time should be the one.
121.
I counted some intermediates which I shouldn't have earlier such as [(3!x6)+4] while I needed only the ones with odd numbers. >.<
9. You are getting close. But on ring yet.
10. Okay, some limitation to the formula I made then... =/
I'm out of ideas...
This is what I did:
9 8 7 6 5 4 3 2 1 0
First of all, 999, 888, 777, etc making 9 numbers.
Then, I make some sort of 3 pinned fork, grabbing 987, 876, 765, etc giving [(3! x 7) + 4] numbers
Then, I repeat using a larger fork, getting 975, 864, 753, etc giving [(3! x 5) + 4] numbers
Repeat using even larger forks, getting [(3! x 3) + 4] numbers
And [(3! x 1) + 4] numbers
For a total of 9 + 46 + 34 + 22 + 10 = 121
Unless you are not counting the 999, 888, 777, I don't find what I'm missing.
If the triples (999, 888, etc) are indeed included, I'll wait for someone else.
All three digits must be distinct.
12. Must be my english (didn't know what distinct really implied >.<) and removing those I'd get 112!
13. Originally Posted by Plato
Here is a challenge problem I read in a recent publication of the MAA for undergraduates. I found it interesting that the author’s way of solving is completely different from mine.
“How many three digit numbers (no leading zeros) are there if the digits are distinct and one of the three digits is the average of the other two”?
I did it the good'ol way: counting:
Spoiler:
There are 112: begin with 1, then two, three, etc. and count the numbers that can give an integer average (for odds only odds, for evens only evens), and count up: with 1 we get: 132, 153, 174, 195 and all their permutations (6 each), with 2 we get 201 (and only 4 permutations since zero cannot be leading), 243, 264, 285 times 6 each, with 3 there are 354, 375, 396 times 6 each (3 and 1 is already counted for in the list of 1), etc.
Or, as many other times, I'm wrong, of course
Tonio
14. ## An attepmt.
Originally Posted by Plato
Here is a challenge problem I read in a recent publication of the MAA for undergraduates. I found it interesting that the author’s way of solving is completely different from mine.
“How many three digit numbers (no leading zeros) are there if the digits are distinct and one of the three digits is the average of the other two”?
Let $a, b, m$ be three digits, where $m$ is the avergare of $a$ and $b$. The digit $m$ is determined uniquely by $a$ and $b$. Hence we consider the pair $a$ and $b$.
$a, b$ are either both even or both odd, since $(a+b)/2$ is an integer. We can choose $\binom{5}{2}$ pairs from $\{1, 3, 5, 7, 9\}$ and $\binom{4}{2}$ pairs from $\{2, 4, 6, 8 \}$.
Thus, there are $10+6=16$ possible triples $\{a, b, m\}$; the number of rearrangements is $16\cdot3!=96$.
We need to count four pairs ${a, b}$ with zero: $\{0, 2\}, \{0, 4\}, \{0, 6\}, \{0, 8\}$. They determine four triples, each with four possible (not having a leading zero) rearrangements, so there are 16 more numbers. Adding up gives $96+16=112$ possible three digit numbers.
I'm not sure, I'm afraid I might miss something...
15. @Tonio
That is exactly the way I would do it.
I posted this to see how others might do it.
The author noted that each such triple forms an arithmetic progression.
And proceeded to count. | 2014-03-14T05:20:05 | {
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https://mathoverflow.net/questions/99001/calculating-the-probability-of-an-event-defined-by-a-condition-on-a-gaussian-ran | # Calculating the probability of an event defined by a condition on a Gaussian random process
Although the question itself can be expressed succinctly, I couldn't come up with a nice self-explanatory title - suggestions are welcome.
## Motivation/Background
I was investigating whether it would be good idea to use Gaussian Process Regression in my application. In the usual use case one conditions a Gaussian process using sample values and that results in a predictive distribution that can be used for a test input. This is a nice approach for making predictions for a single point, however in my application I need to work with intervals. So here comes the question:
Let $K \in \mathbb{R}$ and $f(t)$ be a Gaussian random process with the known mean function $m(t)$ and the known covariance function $c(t_1, t_2)$. What is $\textrm{Pr}\left\{\exists t \in \left[t_1, t_2\right] \text{ s.t. } f(t) \geq K\right\}$?
With words, I am trying to find the probability of the event that the process will exceed a threshold at least at one point during the interval.
I am interested in both analytical and numerical approaches. You are welcome to assume specific forms for $m(t)$ and $c(t_1, t_2)$ (e.g. $m(t) = 0$ and $c(t_1, t_2) = \text{exp}\left(-\frac{1}{2}\left|t_1 - t_2\right|\right)$) to demonstrate an approach.
Note: Edited to fix typos.
-
I think you probably meant $c(t_1, t_2) = \exp(-\frac{1}{2}|t_1 - t_2|)$, correct? – cardinal Jul 7 '12 at 22:32
@cardinal: Of course, fixing it now. Thanks for pointing it out. – Mehmet Ozan Kabak Jul 8 '12 at 5:25
I am accepting Nate's answer as it is the best answer the question got in its current status for more than a year. – Mehmet Ozan Kabak Nov 10 '13 at 6:39
Suppose your process is continuous. (By the Kolmogorov-Centsov continuity theorem, a sufficient condition for this is that $c(s,s) + c(t,t) - 2 c(s,t) \le C |s-t|^\gamma$ for some $C, \gamma > 0$.) Then your process induces a Gaussian measure on $C([0,T])$ and a theorem due to Fernique gives an asymptotic result:
Theorem. There exists $C$ and $\epsilon > 0$ such that $$P\left(\sup_{t \in [0,T]} |f(t)| \ge K\right) \le C \exp(- \epsilon K^2).$$
So the probability is very small when $K$ is large.
You can find a proof in these lecture notes. See sections 4.2 and 4.4.
-
Thanks for the answer and also pointing me to this nice resource. Intuitively one would expect the probability to be an increasing function of $T = |t_2 - t_1|$ and a decreasing function of $K$. The result you quoted definitely confirms the second intuition. However I don't see any explicit reference to $T$ there; it is absorbed in the constants. It is a very important parameter in my application and I need to express the dependence on it more explicitly. – Mehmet Ozan Kabak Jul 12 '12 at 14:46
@Mehmet: In the answer I posted, you get the same form for the bound with quantitative information on the constants, if the process has mean zero and is stationary. The (second) constant $C$ in Nate's answer implicitly depends on $T$, as you note. – cardinal Jul 13 '12 at 1:46
I would recommend obtaining a copy of the text
H. Cramer and M. R. Leadbetter (1967), Stationary and Related Stochastic Processes, John Wiley & Sons, Inc.
which is available as a 2004 Dover reprint, available, e.g., at the link above.
The whole book is a delight, but I think you will be particularly interested in the later chapters, for example,
• Chapter 10: "Crossing" problems and related topics
• Chapter 11: Properties of streams of crossings
• Chapter 12: Limit theorems for crossings
• Chapter 13: Nonstationary normal processes. Curve crossing problems
As a small taste of results from the text:
Lemma: Let $\xi(t)$ be a zero-mean stationary Gaussian process with almost-surely continuous sample paths. Let $C_u(t)$ denote the number of level crossings of the level $u \in \mathbb R$ in the interval $[0,t]$. Then, $$\mathbb E C_u(t) = \frac{t}{\pi} \sqrt{\frac{\lambda_2}{\lambda_0}} e^{-u^2/2 \lambda_0}\\,,$$ where $\lambda_{2n}$, $n \geq 0$ denotes the $2n$th spectral moment $$\lambda_{2n} = \int_0^\infty \lambda^{2n} \\,\mathrm dF(\lambda)$$ and $F$ is the spectral density function.
We can then relate this to the probability of a crossing via the following.
Lemma: Let $\xi(t)$ be a strictly stationary process with continuous one-dimensional distribution and almost-surely continuous sample paths. Suppose that $\mu := \mathbb E C_u(1) < \infty$. Let $q(t)$ denote the probability that at least one crossing occurs in time $t$. Then, $$q(t) = \mu t + o(t) \\,,$$ as $t \to 0$.
In this way we see that the number of level crossings in one sense behaves roughly like a Poisson process with intensity $\mu$.
This only (barely) scratches the surface. The book covers much more and has additional (obviously somewhat dated) references as well. In particular, you may be interested in following classical work, which you may already be familiar with.
S. O. Rice (1945), Mathematical analysis of random noise, Bell System Tech. J., vol 24, pp. 46–156.
-
Consider a standard Wiener process $W_t$ with $$W_0=0,\quad\mathbb E(W_t)=0,\quad \mathbb E(W_t\cdot W_s)=\min(t,s)$$ and let $k\geq0$ be the threshold of interest.
Define the first passage time as $$\tau_k = \inf\{t\ :\ W_t = k\}.$$ Due to continuity of $W_t$, $$\mathbb P(W_t > k) = \mathbb P(\{W_t > k\}\cap\{ \tau_k < t\}).$$ Now note that $$\mathbb P(\{W_t > k\}\cap\{ \tau_k < t\} =\mathbb P(W_t >k\ \mid\ \tau_k < t)\cdot\mathbb P(\tau_k < t) = \frac12\mathbb P(\tau_k < t)$$ due to the symmetrical nature of the Wiener process (intuitively, if the process is already at $k$ at some point of time, it can end up below or above $k$ later on with equal probabilities).
Hence, the probability that threshold $k$ will be crossed before time $t$ is equal to $$\mathbb P(\tau_k < t) = 2\cdot\mathbb P(W_t > k) = 2\cdot\Phi\left(-\frac{k}{\sqrt{t}}\right),\quad k>0,$$ where $\Phi$ is a c.d.f. of the standard normal distribution. Using the similar argument for $k<0$, we can conclude $$\mathbb P(\tau_k < t) = 2\cdot\Phi\left(-\frac{|k|}{\sqrt{t}}\right).$$
-
Thanks for the answer. It definitely seems like a step in the right direction. The result you provided answers the question for the special case of $t_1 \rightarrow -\infty$. Do you know how to handle the case of $t_1 \in \mathbb{R}$? – Mehmet Ozan Kabak Jul 6 '12 at 19:13
This argument is known as the reflexion principle if you want to look it up. – Vincent Beffara Jul 7 '12 at 18:39 | 2014-04-19T20:14:36 | {
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# (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and
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(x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and [#permalink]
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(x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and center (a,b). What is the radius of the circle with equation x^2 + y^2 – 10x + 4y = 20?
A. 4
B. 5
C. 6
D. 7
E. 8
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 07 Feb 2018 Posts: 12 Location: India Concentration: Technology, General Management Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and [#permalink] ### Show Tags 28 Feb 2019, 14:45 2 1 The secret to solving such questions is to make the given equation in question similar to the definition of circle. The second key is to see the terms with single power (10x and 4y). Now lets solve...... Make the equation as... x^2 -(2 * x *5) + 5^2 - 5^2 + y^2 + (2 * y *2 ) + 2^2 -2^2 = 20 which means (x-5)^2 + (y+2)^2 = 20+ 5^2+2^2 = 49 = 7^2 Thus 7 is the radius.... Give Kudos if you like the solution...Cheers !!! Happy Learning !!! Manhattan Prep Instructor Joined: 04 Dec 2015 Posts: 832 GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and [#permalink] ### Show Tags 28 Feb 2019, 18:37 MathRevolution wrote: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and center (a,b). What is the radius of the circle with equation x^2 + y^2 – 10x + 4y = 20? A. 4 B. 5 C. 6 D. 7 E. 8 This is an interesting problem, because while I've seen 'completing the square' used in solutions for GMAT problems before, I'm not sure that I've seen it required in an official problem. I'd be curious to see if anybody can show an official problem that requires you to do this. Here's how it works: you need to take the equation you have, and put it in a form so that all of the variables are contained in the $$(x-a)^2$$ and the $$(y-b)^2$$ parts of the equation, and you only have numbers left over. Focus on the x's first. The terms that have an x in them are $$x^2 - 10x$$. You want to make that look like (x-something)^2. Well, when you square (x-a), you end up with $$x-2ax+a^2$$. So, 2a has to equal 10 in order for the expressions to look alike. That means we're looking at $$(x-5)^2$$. $$(x-5)^2$$ equals$$x^2 - 10x + 25$$. Therefore, $$x^2 - 10x$$ can be rewritten as $$(x-5)^2-25$$. Similarly, for the y terms, we'll rewrite as $$(y+2)^2$$, which equals $$y^2 + 4y + 4$$. Therefore, $$y^2+4y$$ can be rewritten as$$(y+2)^2-4$$. Now, put those back into the original equation. $$x^2 + y^2 – 10x + 4y = 20$$ $$(x-5)^2 - 25 + (y+2)^2 - 4 = 20$$ $$(x-5)^2 + (y+2)^2 = 20 + 4 + 25 = 49 = 7^2$$ So, the radius is 7. _________________ Chelsey Cooley | Manhattan Prep | Seattle and Online My latest GMAT blog posts | Suggestions for blog articles are always welcome! Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8033 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and [#permalink] ### Show Tags 03 Mar 2019, 18:35 => $$x^2 + y^2 – 10x + 4y = 20$$ $$=> x^2 – 10x + y^2 + 4y = 20$$ $$=> x^2 – 10x + 25 + y^2 + 4y + 4 = 20 + 25 + 4$$ $$=> (x-5)^2 + (y+2)^2 = 49$$ $$=> (x-5)^2 + (y+2)^2 = 7^2$$ Thus the circle has center $$(5,-2)$$ and radius $$7$$. Therefore, the answer is D. Answer: D _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and [#permalink]
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15 Jun 2019, 10:50
ccooley wrote:
MathRevolution wrote:
(x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and center (a,b). What is the radius of the circle with equation x^2 + y^2 – 10x + 4y = 20?
A. 4
B. 5
C. 6
D. 7
E. 8
This is an interesting problem, because while I've seen 'completing the square' used in solutions for GMAT problems before, I'm not sure that I've seen it required in an official problem. I'd be curious to see if anybody can show an official problem that requires you to do this.
Here's how it works: you need to take the equation you have, and put it in a form so that all of the variables are contained in the $$(x-a)^2$$ and the $$(y-b)^2$$ parts of the equation, and you only have numbers left over.
Focus on the x's first. The terms that have an x in them are $$x^2 - 10x$$. You want to make that look like (x-something)^2.
Well, when you square (x-a), you end up with $$x-2ax+a^2$$. So, 2a has to equal 10 in order for the expressions to look alike. That means we're looking at $$(x-5)^2$$.
$$(x-5)^2$$ equals$$x^2 - 10x + 25$$. Therefore, $$x^2 - 10x$$ can be rewritten as $$(x-5)^2-25$$.
Similarly, for the y terms, we'll rewrite as $$(y+2)^2$$, which equals $$y^2 + 4y + 4$$. Therefore, $$y^2+4y$$ can be rewritten as$$(y+2)^2-4$$.
Now, put those back into the original equation.
$$x^2 + y^2 – 10x + 4y = 20$$
$$(x-5)^2 - 25 + (y+2)^2 - 4 = 20$$
$$(x-5)^2 + (y+2)^2 = 20 + 4 + 25 = 49 = 7^2$$
Hello ccooley
Can you please explain to me the red part?
Why can it be written as neg, the 25 and 4?
Kid regards!
Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and [#permalink] 15 Jun 2019, 10:50
Display posts from previous: Sort by | 2019-10-23T20:58:55 | {
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https://math.stackexchange.com/questions/2960537/calculate-lotterys-second-prize-using-combination-lottery-probability-questio | # Calculate lottery's second prize using combination - lottery probability question.
Assume a lottery game of the following rules:
• Pick a total of 6 different numbers from the lot containing 42 numbers (1 to 42).
Draw:
• Draw 7 balls, with no repetition(i.e: ball drawn is not put back in the lot) from the lot labeled from 1 to 42.
Results:
• If the first 6 balls drawn matches your own 6 numbers (order doesn't matter): Jackpot.
• If 5 of the first 6 balls drawn matches 5 of your numbers (order doesn't matter) and the 7th drawn ball matches your 6th number: second prize.
• If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize.
I'll end it here for not having many other prizes.
If I want to check my chance of winning the jackpot, it's pretty straightforward and looks like a combination $$C(42,6)$$, so it should be:
$$\frac{42\cdot41\cdot40\cdot39\cdot38\cdot37}{6!} = 5,245,786.$$
So my chance of getting the jackpot is $$(\frac{1}{5,245,786})$$
For the third prize it's also a straightforward combination $$C(42,5)$$, it's equal to:
$$\frac{42\cdot41\cdot40\cdot39\cdot38}{5!} = 850,668.$$
So third prize probability is equal to $$\left(\frac{1}{850,668}\right)$$
Now I am being stumbled on how to calculate the 2nd prize probability. My memories from school are not helping me enough to get my answer. I know that it should be between the two numbers I got there, however any calculations I am making end ups with a probability much higher than the first prize's.
Could you please verify that my 1st and 3rd prize probabilities are well calculated and help me calculate the 2nd prize probability?
Your logic for case number (3) does not seem correct.
First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $$C(6,5)\times36$$. So the probability is:
$$p_1=\frac{36\times{6\choose 5}}{42\choose 6}$$
But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:
$$p_2=\frac{35}{36}$$
The total proability is:
$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{35}{36}=\frac{35\times{6\choose 5}}{42\choose 6}=\frac{15}{374699}$$
You can use a similar logic for case (2).
The probability $$p_1$$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $$p_2$$ is:
$$p_2=\frac{1}{36}$$
...and the final probability for the second prize is:
$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{1}{36}=\frac{{6\choose 5}}{42\choose 6}=\frac{3}{2622893}$$
(35 times smaller than the probability for the third prize)
• Great! This explains where most of my mistakes where. Thank you for clarifying. – Paul Karam Oct 18 '18 at 11:03
• I wanted to test if I did correctly understand the logic here. So I tested with rules: 4 balls matches, and another 3 ball matches. In both cases we won't have a p2 since the 7th ball changes nothing. So my answers for the 4 ball match is 1/555 and 3 ball matches is 1/36, can you please confirm? – Paul Karam Oct 19 '18 at 7:23
For $$3^{rd}$$ prize, you need to select which 5 numbered balls you will be getting, out of 6 in $$^6C_5$$ ways (get balls matching to those 5 numbers from 42); and then, select the $$6^{th}$$ ad $$7^{th}$$ ball in $$^{36}C_2$$ ways
(42-6=36, balls not matching any of the 6 numbers)
So, probability of winning $$3^{rd}$$ prize is $$\frac{^6C_5\cdot^{36}C_2}{^{42}C_7}$$
For $$2^{nd}$$ prize,
Among first 6, only 5 match, selected their corresponding numbers in $$^6C_5$$ ways.
First 5 balls drawn match these 5 numbers.
Now, $$6^{th}$$ ball is selected from 36 (42-6=36, it does not match any of 6 numbers); and number corresponding to last ball to be drawn is already known(one left from 6)
So, probability of winning $$2^{nd}$$ prize is: $$\frac{^6C_5\cdot^{36}C_1\cdot^{1}C_1}{^{42}C_7}$$
Similarly, For $$1^{st}$$ prize, $$\frac{^6C_6\cdot^{36}C_1}{^{42}C_7}$$
• I think that your calculation for the third prize is also not correct. You are ignoring the fact that you have to draw the seventh ball as well and that ball must not hit your last pick. Just read the condition carefully: If 5 of the first 6 balls drawn matches 5 of your numbers and nothing else matches: third prize. "Nothing else" must include the seventh ball as well. – Oldboy Oct 18 '18 at 10:49
• yup, now i guess its correct. – idea Oct 18 '18 at 10:55
• omega, maybe I wasn't clear in my question, the 7th ball drawn doesn't allow you to win the first prize. Only the second. So those calculations are way off in my opinion. I may be wrong, but after looking at the answer I accepted and this one, I feel that (maybe) I wasn't very clear. – Paul Karam Oct 19 '18 at 7:37 | 2019-01-23T03:57:35 | {
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https://www.physicsforums.com/threads/couple-of-questions-about-sets.677101/ | # Couple of questions about sets
1. Mar 8, 2013
### setvectorgroup
1. The problem statement, all variables and given/known data
I am confused with sets- just wanted some clarification.
Say, I have a set A={b, {1,a},{3}, {{1,3}}, 3}
What are the elements of set A?
What are the subsets of set A?
Are the subsets also the elements of the set A?
3. The attempt at a solution
I think the elements of the set A are all the objects within it like b, {1,a},{3}, {{1,3}}, 3. Is that right?
The subsets of the set A are the elements(?) {1,a},{3}, {{1,3}}? Here's my confusion: I've seen a textbook show the element(?) {1,a} is a subset of set A like {{1,a}}⊂ A, rather than {1,a} ⊂ A. If that's the right notation, what's the difference between two different ways of parenthesizing?
Thanks.
2. Mar 8, 2013
### tiny-tim
welcome to pf!
hi setvectorgroup! welcome to pf!
no, for the reason you've referred to …
a subset is a set, so it must start and end with {}
inside the {}, you put all the elements of the subset
eg if you look carefully, you'll see that the two-element subset {b,3} is a subset of A
So is {b} so is {3} so is {3, {3}}, and so on …
3. Mar 8, 2013
### setvectorgroup
Thank You, tiny-tim.
So, if {3, {3}} is a subset...
Say, I make up a set S={1,2,3...}. Can I scoop up any collection of numbers from the set S and call it a subset? For example: T={1, 1000000000, 8} ⊂ S. Would that be right?
edit: Also, if I were to modify the set A={b, {1,a},{3}, {{1,3}}, 3} to make {1,a} ⊂ A true would I modify the set A like A={1, {1,a},{3}, {{1,3}}, a} ?
Last edited: Mar 8, 2013
4. Mar 8, 2013
### tiny-tim
yes, exactly
yes, for {1,a} ⊂ A to be true, both 1 and a must be elements of A
({1,a} ⊂ A is the same as saying that 1 ε A and a ε A)
5. Mar 8, 2013
6. Mar 8, 2013
### Joffan
$$\{1,a\} \in A$$
The set {1,a} is a member (element) of A, and
$$\{\{1,a\}\} \subset A$$
The set containing only the set {1,a} is a subset of A
7. Mar 9, 2013
### setvectorgroup
Thanks, Joffan, for the additional insight.
What I wrote below might sound redundant, but I am trying to internalize all this set business:
Is the notation above just a statement that two random sets happen to share the same elements while the set A also contains more of other elements which makes the lefthand side of this relation to be the subset?
Thanks.
edit:
Am I right( at least conceptually) to visualize all this like :
{1,a} ∈ A
also
{1,a} ∈ {{1,a}}
while ( if you count the elements of each set)
{{1,a}} < A
which is
{{1,a}} ⊂ A
???
Last edited: Mar 9, 2013
8. Mar 9, 2013
### vela
Staff Emeritus
A subset of A doesn't have to have fewer elements than A. The set A is a subset of itself.
What's called a proper subset of A does have to contain fewer elements than A.
9. Mar 9, 2013
### setvectorgroup
Thank You, vela.
Here's yet another attempt to differentiate between 'element of' and 'subset of' (also, by subset I'll mean proper subset)
Say, I got the following relations below and asked to see which ones are true, which ones- false:
2 ∈ {1,2,3}
{2} ∈ {1,2,3}
{2} ∈ {{1}, {2}}
Anything to the left of ∈ is an object. All I care about is if that object can be found in the set on the right side of ∈ in exactly the same form it's on the left side i. e.
2 ∈ {1,2,3} is true because the object- number 2- can be found in the set {1,2,3}
{2} ∈ {1,2,3} is false because there's no object- a set containing number 2- in the set {1,2,3}
{2} ∈ {{1}, {2}} is true because an object- a set containing number 2- is also in {{1}, {2}}
Also, say, I have these below and need to know if any of them are true:
2 ⊆ {1,2,3}
{2} ⊆ {1,2,3}
{2} ⊆ {{1}}, {2}}
The statements above, sort of, implicitly imply (to me) comparison between sets. Here we are concerned if all the elements in the subset are also in the superset i. e.
2 ⊆ {1,2,3} is false because the number two is not even a set to begin with.
{2} ⊆ {1,2,3} is true because the number 2 in the set {2} also happens in the set {1,2,3}
{2} ⊆ {{1}}, {2}} is false because the set {{1}}, {2}} does not contain the number 2
Am I looking at this right?
I am just paranoid something might slip by me unnoticed hence this (redundant) dissection.
10. Mar 9, 2013
### ArcanaNoir
You've got it :)
11. Mar 9, 2013
### setvectorgroup
Well, I guess this does it.
Thank You, ArcanaNoir, for (hopefully) closing this thread. | 2017-11-19T13:04:14 | {
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http://mathhelpforum.com/calculus/275422-same-wrong-answer-twice.html | 1. the same wrong answer twice!
Hi folks,
this is not a university level question, but it is calculus so I hope it's ok here.
A ship is to make a voyage of 200 km at a constant speed. When the speed of the ship is v km/h the cost is $v^2 + \frac{4000}{v}$ USD per hour.
Find the speed at which the ship should travel so that the cost of the voyage is a minimum.
So, we have a cost function in terms of velocity, so my first try was to differentiate the cost with respect to the velocity
$\frac{d (cost)}{dv} = 2v - 4000v^-2$ i.e. $v^3 = 2000$ therefore v = 12.6 km/h
$\frac {d^2 (cost)}{dv^2}$ is > 0 so the first differential is a minimum as required.
I was concerned that I did not used the distance information (200km) and even more concerned that the answer was wrong anyway.
So, I decided to try velocity = distance / time i.e. $v = \frac{200}{t}$ where t is the time in hours and substitute this in the cost function
cost = $(\frac{200}{t})^2 + 20.t$
To minimise the cost $\frac{d (cost)}{dt} = -80,000 t^-3 + 20.t = 0$ this gives $t^3 = 4000$ so t = 15.9 hours
therefore velocity = 200 / 15.9 = 12.6 km/h
you will see that this is the same wrong answer.
2. Re: the same wrong answer twice!
why do you think 12.6 km/h is wrong?
3. Re: the same wrong answer twice!
You are rounding. Perhaps the correct answer is not supposed to be rounded? Try $10\sqrt[3]{2}$
4. Re: the same wrong answer twice!
The question came from a Maths book and the answer at the back is 20 km/h. I guess the book could be wrong. No errata is published so I cannot tell.
But if you guys think I am on the right track, that's great!
5. Re: the same wrong answer twice!
Yes, I believe your book is wrong:
minimize v^2+4000/v - Wolfram|Alpha Results
You can also plug in the result you found and the book's result to see that the cost is less at 12.6 than it is at 20 km/h.
6. Re: the same wrong answer twice!
You got the same wrong answer by both methods because you are solving the wrong problem! You, and the others, are minimizing v^2+ 4000/v, the cost per hour, but you should be minimizing the cost of the entire trip. At v km/h you will go 200 km in 200/v hours. Since the cost is v^2+ 400/v per hour, the cost of going 200 km will be (200/v)(v^2+ 400/v)= 200v+ 80000/v^2. The derivative of that is 200- 160000/v^3= 0 so 200v^3= 160000, v^3= 8000, v= 20 km/h. | 2017-08-18T11:21:42 | {
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https://cs.stackexchange.com/questions/135173/find-a-vector-of-non-negative-integers-b-that-minimizes-prod-i-1d-lef | Find a vector of non-negative integers $b$ that minimizes $\prod_{i = 1}^{D}\left(a_i + b_i\right)$ such that the product is a multiple of $c$
I'm trying to come up an efficient algorithm that, given a list of positive integers $$a = \left(a_1, \ldots, a_D\right)$$ and positive integer $$c$$, finds a list of non-negative integers $$b = (b_1, \ldots, b_D)$$ that minimizes $$\prod_{i = 1}^{D}\left(a_i + b_i\right)$$ such that $$\prod_{i = 1}^{D}\left(a_i + b_i\right)$$ is a multiple of $$c$$.
The brute force search I came up with is
1. Let $$t$$ be the smallest multiple of $$c$$ that is $$\ge \prod_{i = 1}^{D} a_i$$.
2. Use depth-first-search to search for values $$\hat{b}$$, starting at $$\mathbf{0}$$ and incrementing one element of $$\hat{b}$$ at a time until $$\prod_{i = 1}^{D} \left(a_i + \hat{b}_i\right) \geq t$$.
3. If we found $$\hat{b}$$ such that $$\prod_{i = 1}^{D} \left(a_i + \hat{b}_i\right) = t$$ then $$\hat{b}$$ is the optimal solution. Otherwise increment $$t$$ by $$c$$ and go back to step 2.
The above algorithm does work, but if $$D$$ or $$c$$ are too large then it will potentially take a very very long time. I'm wondering if this maps to any well known algorithm or if there's a more efficient solution. I'm considering that the prime factors of $$c$$ and $$a$$ could play a large role in reducing the search space but I can't quite figure it out.
In case anyone wants to play with this, a Python 3 implementation of the brute force algorithm described above is
from functools import reduce
from operator import mul
def prod(x):
return reduce(mul, x, 1)
def ceil_divide(num, denom):
return -(-num // denom)
def update_memory(b, memory):
tuple_b = tuple(b)
if tuple_b in memory:
return False
return True
def dfs(a, b, t, memory):
if not update_memory(b, memory):
return False
p = prod([ai + bi for ai, bi in zip(a, b)])
if p == t:
return True
elif p > t:
return False
for i in range(len(a)):
b[i] += 1
if dfs(a, b, t, memory):
return True
b[i] -= 1
def solve(a, c):
b = [0 for _ in range(len(a))]
t = c * ceil_divide(prod(a), c)
while not dfs(a, b, t, set()):
t = t + c
return b
# a few test cases
assert solve([2, 3], 9) == [1, 0]
assert solve([2, 8], 6) == [0, 1]
assert solve([13, 17, 25], 8) in [[1, 1, 1], [0, 1, 3]]
assert solve([5, 13, 19], 6) == [0, 1, 2]
Note: One potential use-case of such a thing could be, for example, to find the minimum padding of a $$D$$-dimensional tensor such that the number of elements of the tensor is divisible by $$c$$.
• How about finding an optimal $B$ such that $(\prod_{n=1}^{D} a_i) + B = kc$ ?? Feb 5 '21 at 21:13
• Can you tell us anything about typical sizes for $c$, $D$, and the $a_i$'s?
– D.W.
Feb 5 '21 at 22:22
• @D.W. So I don't have exact bounds on $c$, $D$ or $a_i$ but i think if your method is feasible for $c \leq 2^{20}$ and $D \leq 10$ then that's probably good enough for any use case I can think of. Even if it doesn't work for such large numbers then I'm just mostly interested in better alternatives to what I've already got. Feb 6 '21 at 0:12
• @droptop This is the initial value $t$ but I'm not sure how this solves the problem. Feb 6 '21 at 0:22
• I think you can speed up the search by only considering the smallest $t$ for which $B = t - ( \prod_{n=1}^{10} a_i)$, with $B$ having at least one factor in $a$. So if no $a_i$ divides $B$, move on to the next multiple of $c$. Feb 6 '21 at 12:51
If $$c$$ doesn't have too many factors and is not too large, the following should work reasonably well for typical numbers.
Factor $$c$$ into its prime factorization, say $$c=p_1^{e_1} \cdots p_k^{e_k}$$. Enumerate all tuples $$(f_1,\dots,f_D)$$ of non-negative integers such that $$f_1 \cdots f_D = c$$. For each such tuple $$(f_1,\dots,f_D)$$, and each $$i$$, find the smallest $$b_i$$ such that $$a_i+b_i$$ is a multiple of $$f_i$$ (namely, $$b_i = f_i - a_i \bmod f_i$$); then compute $$\prod_i (a_i+b_i)$$ and keep the smallest found so far. By construction, this will explore all possible solutions, and output the best one.
You can use the prime factorization of $$c$$ to help you enumerate all tuples $$(f_1,\dots,f_D)$$. In particular, the prime factorization makes it easy to enumerate all factors of $$c$$, so we can enumerate all possibilities for $$f_1$$ (namely, all factors of $$c$$), then enumerate all possibilities for $$f_2$$ (namely, all factors of $$c/f_1$$), etc.
The running time will be proportional to the number of such tuples you have to enumerate. This number is $$\prod_i {e_i + D - 1 \choose D - 1},$$ which is not too bad if the number of prime factors of $$c$$ is small, but is horrible if $$c$$ has many prime factors.
• This is very interesting! It is along the lines of what I was thinking but I was missing the idea of enumerating all permutations of factorizations of $c$ of length $D$. I do still wonder if this problem can be solved more efficiently, perhaps by by considering only the prime factors of $c$ not already included in the $a_i$s. I would like to leave this question open for a few days before accepting to see if more efficient algorithms might be found. Feb 6 '21 at 0:18
• @jodag That sounds like a good idea! I too wonder if there might be a better way. I like your idea: as an optimization: if $a_i$ divides $c$, then only enumerate tuples $(f_1,\dots,f_D)$ where $a_i$ divides $f_i$. It should be easy to adjust the method above to include that optimization. I'm not 100% certain that this optimization is correct in all cases (it will never cause us to miss a better solution) but it seems quite intuitive and reasonable to me.
– D.W.
Feb 6 '21 at 0:24
• The approach in this answer worked really well for me. I implemented a variant of it that can stop early and updated the original question with the algorithm and code. I am definitely still interested in further opimizations. One thing I would point out is that adding parenthesis to $b_i = (f_i - a_i) \bmod f_i$ may improve clarity, as it's not clear to me what the OOO for the mod operator should be. Feb 7 '21 at 0:29
Based on the answer by D.W. I was able to implement an alternative algorithm. This effectively iterates over all permutations of $$D$$-length factorizations of $$c$$. Instead of utilizing the prime factorization to explicitly enumerate all the factorizations, I increment one element of $$b$$ to the next multiple of a divisor of $$c$$, then divide out that divisor from $$c$$ and repeat recursively. This effectively explores all $$D$$-length factorizations of $$c$$, but has the benefit of being able to ignore a large subset of the search space by abandoning a branch when the objective product becomes larger or equal to the smallest feasible objective found so far. Combined with some simple memoization, the resulting algorithm appears to be much faster on average than my original algorithm.
I've tested this on lots of random cases and tried to construct some pathological cases as well. I think there might be some pathological cases where this is slower than the original brute force, probably when there are tons of prime factors in $$c$$, but it seems to be much better on average.
My Python 3 implementation of the updated algorithm is
from functools import reduce
from operator import mul
from copy import deepcopy
def prod(x):
return reduce(mul, x, 1)
def argsort(x, reverse=False):
return sorted(range(len(x)), key=lambda idx: x[idx], reverse=reverse)
def divisors(v):
""" does not include 1 """
d = {v} if v > 1 else set()
for n in range(2, int(v**0.5) + 1):
if v % n == 0:
return d
def update_memory(b, c_rem, memory):
tuple_m = tuple(b + [c_rem])
if tuple_m in memory:
return False
return True
def dfs(a, b, c, c_rem, memory, p_best=float('inf'), b_best=None):
ab = [ai + bi for ai, bi in zip(a, b)]
p = prod(ab)
if p >= p_best:
return p_best, b_best
elif p % c == 0:
return p, deepcopy(b)
dc = divisors(c_rem)
for i in argsort(ab):
for d in dc:
db = (d - ab[i]) % d
b[i] += db
if update_memory(b, c_rem // d, memory):
p_best, b_best = dfs(a, b, c, c_rem // d, memory, p_best, b_best)
b[i] -= db
return p_best, b_best
def solve(a, c):
b = [0 for _ in range(len(a))]
result = dfs(a, b, c, c, set())
p_best, b_best = result
return b_best
# a few test cases
assert solve([2, 3], 9) == [1, 0]
assert solve([2, 8], 6) == [0, 1]
assert solve([13, 17, 25], 8) in [[1, 1, 1], [0, 1, 3]]
assert solve([5, 13, 19], 6) == [0, 1, 2] | 2022-01-18T06:57:55 | {
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https://math.stackexchange.com/questions/2440429/compute-lim-x-rightarrow-infty-left-sqrtx2-6x9x-1-right | # Compute $\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)$
I note that $\sqrt{x^2-6x+9}=|x-3|$. Splitting upp the limit into cases gives
• $x\geq 3:$
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(|x-3|+x-1)=2\lim_{x\rightarrow -\infty}(x-2)=-\infty.$$
• $x< 3:$
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(-x+3+x-1)=\lim_{x\rightarrow -\infty}2=2.$$
I get two different values of the limit, but plotting the function clearly shows that the answer should be $2$.
• Are you sure that you can reach $-\infty$ with values of $x$ over $3$ ? – Yves Daoust Sep 22 '17 at 13:41
• I know that it is incorrect, but I can't se where the error is. – Parseval Sep 22 '17 at 13:42
• $$x\to-\infty \implies. x<3$$ – lab bhattacharjee Sep 22 '17 at 13:43
• I just told you but I am not sure you took care... – Yves Daoust Sep 22 '17 at 13:43
• If $x$ tends to negative infinity, do we need to consider $x\geq 3$? – Jihoon Kang Sep 22 '17 at 13:43
Like you said there is 2 cases: $$\begin{cases}x\ge 3\\x<3\end{cases}$$ Now you tell me: is $-∞\ge 3$ or $-∞<3$?
Put $t=-x$, then: $$\lim_{t\rightarrow \infty}\left(\sqrt{t^2+6t+9}-t-1\right) = \lim_{t\rightarrow \infty}\left(|t+3|-t-1\right) =\lim_{t\rightarrow \infty}\left(t+3-t-1\right) = 2$$
• Why should I do that? Why doesn't my way work? – Parseval Sep 22 '17 at 13:49
• @John. How does your method differ from mine? – imranfat Sep 22 '17 at 22:00
• @imranfat: It does not and I could ask you the same question. – Maria Mazur Sep 22 '17 at 22:02
• @johnnobody. In that case, you should receive an upvote :) – imranfat Sep 22 '17 at 22:04
• Yes, I suppose you are right. – Maria Mazur Sep 22 '17 at 22:05
You're approaching $-\infty$, so take,
$$|x-3| = 3-x$$
(I don't get where you're going with $x \ge 3$)
$$\lim_{x \to -\infty} \sqrt{(x-3)^2} + x - 1$$
$$= \lim_{x \to -\infty} |x-3| + x - 1$$
$$= \lim_{x \to -\infty} 3 - x + x - 1$$
$$= \lim_{x \to -\infty} 2$$ $$= \boxed 2$$
Now, we have $\lim_{x\rightarrow -\infty}$ which implies x is approaching the negative infinity.
if $x\geq 3$, x is still on the way approaching the infinity from somewhere and x needs to pass the domain of $x\geq 3$ Then it goes to the domain of $x< 3$ therefore
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(-x+3+x-1)=\lim_{x\rightarrow -\infty}2=2.$$
Here is a different approach, setting $x=-t$, the limit becomes: $$\lim_{t \to \infty} \sqrt{t^2+6t+9} - t - 1$$ Since $t^2+6t+9=(t+3)^2$, we can say for positive t-values that $\sqrt{(t+3)^2}=t+3$, your limit expression becomes $t+3-t-1=2$.
Since $\lim_{x \rightarrow - \infty}$ it suffices to consider 'small' negative $x$.
$|x-3| = -x + 3$.
Example: $x= -7$:
$|x-3| = |-7-3| = -x +3.$
Hence:
$\lim_{x \rightarrow -\infty} (|x-3| + x-1) =$
$\lim_{x \rightarrow -\infty}( -x +3 +x -1 )= 2$. | 2019-05-26T15:32:12 | {
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https://puzzling.stackexchange.com/questions/115595/the-median-game | The median game
Five friends play a simple game with the following rules:
1. Players play consecutively one after the other.
2. Each player must call out a whole number between 1 and 10 (inclusive), such that it hasn't been called out already.
3. The winner is the player whose number is the median of all called out numbers.
What should be the optimal strategy for each player to maximise their chance of winning? Which player is the most likely to win if they all play optimally?
Note that if a player only has losing moves then they will just choose randomly. If they have multiple optimal moves then they will also choose one randomly.
• If a player has no optimal move (or indeed only losing moves), will they just choose randomly? Apr 1 at 11:19
• This is a good question. Yes I would say they just choose randomly. Apr 1 at 11:26
Consider that
1,2,9 and 10 are guaranteed to lose so no player will pick them unless forced into a random choice. This means that any player trying to win would pick from the set {3,4,5,6,7,8} and the median of five numbers from this set will be 5 or 6.
Given that, I see play proceeding as follows
Player 1 picks either 5 or 6 and Player 2 picks the other one. Given the symmetry of their choices, both players have an equal chance of winning.
Player 3 will pick either 4 or 7 (at random) because if they don't then Player 4 will fill the gap between Player 3's choice and 5/6 and guarantee a loss for Player 3.
From there, Player 3 can only win if Players 4 and 5 pick numbers which are both less than 4 (if 4 was picked) or greater than 7 (if 7 was picked) and this has probability $$\frac{3}{7} \times \frac{2}{6} = \frac{1}{7}$$.
Players 1 and 2 then both win with probabilites $$\frac{3}{7}$$ as it is guaranteed lose for Players 4 and 5 after Player 3's choice.
N.B. I'm assuming here that strategies which are equally optimal for a given player are equally likely to be executed. This seems to be implied from OP's comments but I'm adding here for clarification.
• @noedne I'm not sure what you mean. Even if Player 3 always chooses the same number, they don't know what Player 4 or 5 will do so still have a 1/7 chance of winning. Apr 1 at 15:35
• @noedne Ok, what you're essentially saying is that OP should clarify that equally optimal strategies are equally likely to be executed, right? I'm happy enough that we assume this, given my comment and the response under the question. Otherwise, the question has no definitive answer. Apr 1 at 16:11
• That's fine, but maybe it should be stated that this assumption is why 1 and 2 have equal probabilities of winning after choosing 5 and 6. A symmetry argument assumes that there is some well defined probability of winning in the first place, which has not yet been shown. Apr 1 at 16:23
• This is a great answer! Your assumptions are correct. Apr 1 at 23:18 | 2022-09-24T16:06:55 | {
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http://math.stackexchange.com/questions/135012/prove-the-set-m-x-in-mathbbr2-mid-alpha-x-1-gamma-x-2-leq-beta-is/135840 | # Prove the set $M=\{x\in\mathbb{R}^2_+ \mid \alpha x_1+\gamma x_2\leq \beta\}$ is convex
Let $\alpha\gt 0$, $\gamma\gt 0$, and $\beta\gt 0$ be real numbers. Let $$M=\{x\in\mathbb{R}^2_+ \mid \alpha x_1+\gamma x_2\leq \beta\}$$ Prove $M$ is a convex set. Prove that $M$ is bounded. What does this set resemble (in economics)?
Attempt: If $(x_1,x_2),(y_1,y_2)\in M$ we get \begin{align*} \alpha x_1 + \gamma x_2&\leq \beta\\ \alpha y_1 + \gamma y_2 &\leq \beta \end{align*}
We want to prove $$\alpha(ax_1 + (1-a)y_1) + \gamma(ax_2 + (1-a)y_2)\leq \beta.$$
The question is how do I prove this inequality?
-
I think the best proof is just to say: the set is a triangle! (I don't what if anything triangles "resemble in economics") – Omar Antolín-Camarena Apr 25 '12 at 3:16
nope. not convincing enough. – Koba Apr 25 '12 at 14:06
Algebra! (pronounced like Jon Lovitz's Master Thespian character)
\begin{align*} \alpha(ax_1 + (1-a)y_1) + \gamma(ax_2+(1-a)y_2) &= \alpha ax_1 + \gamma ax_2 + \alpha(1-a)y_1 + \gamma(1-a)y_2\\ &= a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\\ &\leq a\beta + (1-a)\beta. \end{align*}
-
ok you expanded it and rearranged terms.I did the same thing, but the last expression a(αx1+γx2)+(1−a)(αy1+γy2) should be leq than β. I do not understand why you are saying that a(αx1+γx2)+(1−a)(αy1+γy2)≤aβ+(1−a)β – Koba Apr 21 '12 at 23:48
oh wait so if I expand aβ+(1−a)β it will β, right? – Koba Apr 21 '12 at 23:50
@Dostre: $\alpha x_1+\gamma x_2\leq \beta$ by assumption; multiplying through by $a$ we get $a(\alpha x_1+\gamma x_2)\leq a\beta$. Similarly, form $\alpha y_1+\gamma y_2\leq \beta$, multiplying through by $(1-a)$ we get $(1-a)(\alpha y_1+\gamma y_2)\leq (1-a)\beta$. Add both inequalities to get the one I have; finally, $a\beta + (1-a)\beta = (a+(1-a))\beta = \beta$. – Arturo Magidin Apr 21 '12 at 23:50
@Dostre: That's the last step, yes; but you said you didn't understand the last step I did do; I explained it in the comment just above this one. – Arturo Magidin Apr 21 '12 at 23:51
I see now thank you very much. This problem occupied me for a long time. Thanks. – Koba Apr 21 '12 at 23:53
Same thing Arturo posted in more detail:
We know that the below two inequalities on the far left are true. So lets use them to prove the one we need to prove$[α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤β]$:
$αx_1+γx_2≤β\;\;|*a\Rightarrow a(\alpha x_1+\gamma x_2)\leq a\beta$
$αy_1+γy_2≤β\;\;|*(1-a)\Rightarrow (1-a)(αy_1+γy_2)\leq (1-a)\beta$
Now add the inequalities on the far right side and we get:
$$a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq a\beta+(1-a)\beta$$
After expanding the expressions in parenthesis on the LHS and rearranging the terms we get:
$$α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤a\beta+(1-a)\beta$$
Which almost looks exactly like the one we need to prove. The RHS after expanding:
$$a\beta+(1-a)\beta=a\beta +\beta -a\beta =\beta \Rightarrow$$
$$\Rightarrow a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq\beta$$
Which is what we needed to show.
2&3 questions:
This set M={$x∈ℝ^2_+∣αx_1+γx_2≤β$} looks like a budget constraint and is bounded by:
if $x_1=0;\;$ $\gamma x_2\leq \beta$;$\;\;x_2\leq \frac{\beta}{\gamma}$
if $x_2=0;\;$ $\alpha x_1\leq \beta$;$\;\;x_1\leq \frac{\beta}{\alpha}$
- | 2015-07-29T10:22:31 | {
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http://math.stackexchange.com/questions/279388/there-are-at-least-three-mutually-non-isomorphic-rings-with-4-elements | # There are at least three mutually non-isomorphic rings with $4$ elements?
Is the following statement is true?
There are at least three mutually non-isomorphic rings with $4$ elements.
I have no idea or counterexample at the moment. Please help. So far I know about that a group of order $4$ is abelian and there are two non isomorphic groups of order $4$ like $K_4(non cyclic)$ and $\mathbb Z_4(cyclic)$.
-
How many ring structures are there on $K_4$ (as the additive group)? – Jyrki Lahtonen Jan 15 '13 at 17:24
I have no idea, I know that additive group of the ring is abelian. thats all.please teach me. – La Belle Noiseuse Jan 15 '13 at 17:26
I think that $a^2$ can be any of the four elements. The choices $a^2=1$ and $a^2=0$ seem to lead to isomorphic rings (appears in BenjaLim's +1 list). The choice $a^2=a$ leads to another ring on his list. The choice $a^2=1+a$ leads to a finite field of four elements (not on his list). Anyway, do satisfy yourself about the fact that when the additive group is $K_4$, knowing $a^2$ gives you everything about the ring operations. – Jyrki Lahtonen Jan 15 '13 at 17:36
Hint: There are, up two isomorphism, exactly two groups with four elements. Find these groups and check whether they are rings. – Johannes Kloos Jan 29 '13 at 10:07
It is not a dublicate, because the other question didn't ask for a classification. @Johannes: It doesn't make sense to ask whether a group is a ring. Of course you know this, but other readers do not. – Martin Brandenburg Jan 29 '13 at 10:30
By ring, I always mean unital ring. Each of the following rings has four elements:
$R_1 = \mathbb{Z}/4~, ~R_2 = \mathbb{F}_2 \times \mathbb{F}_2 = \mathbb{F}_2[x]/(x^2+x)~, ~R_3 = \mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)~, ~R_4 = \mathbb{F}_2[x]/(x^2)$
They are non-isomorphic because only $R_3$ is a field, only $R_1$ has characteristic $\neq 2$, and only $R_2,R_3$ are reduced.
Conversely, let $R$ be a ring with four elements. If $a \in R \setminus \{0,1\}$, then the centralizer of $a$ is a subgroup of $(R,+)$ with at least three elements $0,1,a$, so by Lagrange also the fourth element has to commute with $a$. Thus, $R$ is commutative. If $R$ is reduced, then it is a finite product of local artinian reduced rings, i.e. fields, so that $R \cong R_2$ or $R \cong R_3$. If $R$ is not reduced, there is some $a \in R \setminus \{0\}$ such that $a^2=0$. Since $0,1,a,a+1$ are pairwise distinct, these are the elements of $R$. If $2=0$, then we get an injective homomorphism $\mathbb{F}_2[x]/(x^2) \to R, x \mapsto a$. Since both sides have four elements, it is an isomorphism. If $2 \neq 0$, the characteristic has to be $4$, i.e. we get an embedding $\mathbb{Z}/4 \to R$, which again has to be an isomorphism.
Of course, this classification can also be obtained by more elementary methods. For other orders, see:
The smallest non-commutative ring has $8$ elements and is given by $\begin{pmatrix} \mathbb{F}_2 & \mathbb{F}_2 \\ 0 & \mathbb{F}_2 \end{pmatrix} \subseteq M_2(\mathbb{F}_2)$.
-
Hm? I have posted this answer elsewhere. Did someone merge it? – Martin Brandenburg Jan 29 '13 at 20:50
PS: The same classification also works for rings with $p^2$ elements, where $p$ is any prime. See George's answer here: math.stackexchange.com/questions/305512 – Martin Brandenburg Feb 16 '13 at 15:26
Consider the ring $R = \Bbb{Z}/2\Bbb{Z}[i]$. Alternatively $R$ can be constructed as a quotient
$$R \cong \Bbb{Z}[x]/(2,x^2+1).$$ As a ring $R$ is not isomorphic to either $S = \Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$ or $T = \Bbb{Z}/4\Bbb{Z}$.
Edit: Perhaps I should add why the ring $R$ is not isomorphic to either $S$ or $T$. Firstly by counting orders of elements $R$ cannot be isomorphic to $T$; $T$ has an element of order 4 while $R$ does not. So now the penultimate question is why is $R$ is not isomorphic to $S$? As groups they are certainly isomorphic but as rings they can't be. The reason is because the presence of $i$ means that the multiplication in $\Bbb{Z}/2\Bbb{Z}[i]$ is not the same as the multiplication in $S$ which is the usual one coming from the product ring structure.
In view of this we see that $R$ has non-trivial nilpotent elements, $(1+i)^2 = 1 - 2i +i^2 = 1 - 1 = 0$ while obviously $\text{nilrad}$ $S = 0$. Thus $R \not\cong S$.
-
@JyrkiLahtonen I have edited my answer above. – user38268 Jan 16 '13 at 8:33
@YACP I have added some explanations. – user38268 Jan 16 '13 at 8:33
These can be exhaustively enumerated using alg. To enumerate the number of non-isomorphic rings of orders $1,2,\ldots,8$, we enter:
./alg theories/ring.th --size 1-8 --count
which outputs:
# Theory ring
Constant 0.
Unary ~.
Binary + *.
Axiom plus_commutative: x + y = y + x.
Axiom plus_associative: (x + y) + z = x + (y + z).
Axiom zero_neutral_left: 0 + x = x.
Axiom zero_neutral_right: x + 0 = x.
Axiom negative_inverse: x + ~ x = 0.
Axiom negative_inverse: ~ x + x = 0.
Axiom zero_inverse: ~ 0 = 0.
Axiom inverse_involution: ~ (~ x) = x.
Axiom mult_associative: (x * y) * z = x * (y * z).
Axiom distrutivity_right: (x + y) * z = x * z + y * z.
Axiom distributivity_left: x * (y + z) = x * y + x * z.
size | count
-----|------
1 | 1
2 | 2
3 | 2
4 | 11
5 | 2
6 | 4
7 | 2
8 | 52
Check the numbers [2, 2, 11, 2, 4, 2, 52](http://oeis.org/search?q=2,2,11,2,4,2,52) on-line at oeis.org
If we want it to print out the addition and multiplication tables, we can remove --count from the command line:
./alg theories/ring.th --size 4
If we want to work with unital rings, we can use:
./alg theories/unital_ring.th --size 1-8 --count
which gives:
# Theory unital_ring
Theory unital_ring.
Constant 0 1.
Unary ~.
Binary + *.
Axiom plus_commutative: x + y = y + x.
Axiom plus_associative: (x + y) + z = x + (y + z).
Axiom zero_neutral_left: 0 + x = x.
Axiom negative_inverse: x + ~ x = 0.
Axiom mult_associative: (x * y) * z = x * (y * z).
Axiom one_unit_left: 1 * x = x.
Axiom one_unit_right: x * 1 = x.
Axiom distrutivity_right: (x + y) * z = x * z + y * z.
Axiom distributivity_left: x * (y + z) = x * y + x * z.
# Consequences of axioms that make alg run faster:
Axiom zero_neutral_right: x + 0 = x.
Axiom negative_inverse: ~ x + x = 0.
Axiom zero_inverse: ~ 0 = 0.
Axiom inverse_involution: ~ (~ x) = x.
Axiom mult_zero_left: 0 * x = 0.
Axiom mult_zero_right: x * 0 = 0.
size | count
-----|------
1 | 0
2 | 1
3 | 1
4 | 4
5 | 1
6 | 1
7 | 1
8 | 11
Check the numbers [1, 1, 4, 1, 1, 1, 11](http://oeis.org/search?q=1,1,4,1,1,1,11) on-line at oeis.org
- | 2015-10-04T08:23:30 | {
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http://mathhelpforum.com/calculus/28873-derivative-constant-help.html | # Thread: Derivative of a Constant help
1. ## Derivative of a Constant help
Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.
Find h'(3) for:
(a) h(x) = 2f(x) - 5g(x)
(b) h(x) = f(x)g(x)
(c) h(x) = f(x)/g(x)
(d) h(x) = g(x)/(2 + f(x))
As far as I know, the derivative of a constant is always 0. So when I plug in the values of f(x) and g(x) to find h(3) I always get a constant, and so the derivative of that must be zero, right?
2. Hello, topher0805!
. . $h'(3)$ means: .find $h'(x)$, then substitute $x = 3$
Given that: . $f(3) = \text{-}5,\;\;f'(3) = \text{-}2,\;\;g(3) = \text{-}2,\;\;g'(3) = 3$
Find $h'(3)$ for:
$(a)\;\;h(x) \:= \:2\!\cdot\!f(x) - 5\!\cdot\!g(x)$
We have: . $h'(x) \;=\;2\!\cdot\!f'(x) - 5\!\cdot\!g(x)$
$\text{Then: }\;h'(3) \;=\;2\!\cdot\!\underbrace{f'(3)}_{\downarrow} - 5\!\cdot\!\underbrace{g'(3)}_{\downarrow}$
- . - . . $h'(3) \;=\;\;\;2(\text{-}2) \;- \;5(3) \;\;=\;\;-4-15 \;\;=\;\;-19$
3. Originally Posted by topher0805
Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.
Find h'(3) for:
[snip]
(d) h(x) = g(x)/(2 + f(x))
Using the quotient rule:
$h'(x) = \frac{g'(x) (2 + f(x)) - f'(x) g(x)}{(2 + f(x))^2}$
$\Rightarrow h'(3) = \frac{g'(3) (2 + f(3)) - f'(3) g(3)}{(2 + f(3))^2}$
$\Rightarrow h'(3) = \frac{(3) (2 - 5) - (-2) (-2)}{(2 - 5)^2} = .....$
4. Originally Posted by topher0805
Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.
Find h'(3) for:
(a) h(x) = 2f(x) - 5g(x)
f(x) and g(x) are not constants, they are functions. So:
Initial equation of problem a
$h(x)=2f(x)-5g(x)$
then
$\frac d{dx}h(x)=\frac d{dx}[2f(x)-5g(x)]$
simplify
$\frac d{dx}h(x)=\frac d{dx}[2f(x)]-\frac d{dx}[5g(x)]$
Use Newtonian Notation
$h\prime (x)=2f\prime (x)-5g\prime (x)$
then
$h\prime (3)=2f\prime (3)-5g\prime (3)$
and substitute values
$h\prime (3)=2(-2)-5(3)$
simplify
$h\prime (3)=-4-15 = -19$
Can you do the rest?
edit: dang, I got doubly owned >.<
5. Originally Posted by Soroban
Hello, topher0805!
. . $h'(3)$ means: .find $h'(x)$, then substitute $x = 3$
We have: . $h'(x) \;=\;2\!\cdot\!f'(x) - 5\!\cdot\!g(x)$
$\text{Then: }\;h'(3) \;=\;2\!\cdot\!\underbrace{f'(3)}_{\downarrow} - 5\!\cdot\!\underbrace{g'(3)}_{\downarrow}$
- . - . . $h'(3) \;=\;\;\;2(\text{-}2) \;- \;5(3) \;\;=\;\;-4-15 \;\;=\;\;-19$
How did you line that up so well without arrays?
6. You can also use the aligned environment:
\begin{aligned}
h'(3) &= 2 \cdot \underbrace {f'(3)}_ \downarrow - \,5 \cdot \underbrace {g'(3)}_ \downarrow\\
&= ~~2( - 2) ~~- ~\,5(3)\\
&= - 4 - 15\\
&= - 19.
\end{aligned}
-----
Soroban uses the following: color beige, it's the perfect stealth to do what Soroban did.
7. Makes sense, of course, if MHF ever gets more skins, it won't be nearly so effective. | 2017-04-23T19:06:24 | {
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https://web2.0calc.com/questions/a-circle-of-radius-5-with-its-center-at-0-0-is-drawn | +0
# A circle of radius 5 with its center at $(0,0)$ is drawn on a Cartesian coordinate system. How many lattice points (points with integer coor
0
146
3
A circle of radius 5 with its center at $(0,0)$ is drawn on a Cartesian coordinate system. How many lattice points (points with integer coordinates) lie within or on this circle?
Guest Jan 14, 2018
### Best Answer
#2
+86890
+1
The total number of lattice ponts is given by
1 + (4 * 5) +
4 * [ floor √[ 5^2 - 1^2] + floor √ [5^2 - 2^2] + floor √[5^2 - 3^2] + floor √[5^2 - 4^2] ] =
1 + 20 +
4 * [ floor √24 + floor √21 + floor √16 + floor √9 ] =
21 + 4 [ 4 + 4 + 4 + 3 ] =
21 + 4 [ 15] =
81
CPhill Jan 15, 2018
#1
+261
0
Start with the 4 "poles" i.e (5,0), (0,5), (-5,0), (0,-5)
Each quadrant of the circle has 2 integer values (3,4) and (4,3) - Since (3,4,5) is a pythagorean triple
4 quadrants give 8 more points so 12 points in total
Quazars Jan 14, 2018
#2
+86890
+1
Best Answer
The total number of lattice ponts is given by
1 + (4 * 5) +
4 * [ floor √[ 5^2 - 1^2] + floor √ [5^2 - 2^2] + floor √[5^2 - 3^2] + floor √[5^2 - 4^2] ] =
1 + 20 +
4 * [ floor √24 + floor √21 + floor √16 + floor √9 ] =
21 + 4 [ 4 + 4 + 4 + 3 ] =
21 + 4 [ 15] =
81
CPhill Jan 15, 2018
#3
+19488
+1
A circle of radius 5 with its center at $(0,0)$ is drawn on a Cartesian coordinate system.
How many lattice points (points with integer coordinates) lie within or on this circle?
A Calculation of the Number of Lattice Points within or on the circle:
Let $$\lfloor x \rfloor$$ be the largest integer equal to or less than x.
Example:
$$\lfloor 3.53553390593 \rfloor = 3$$
$$\lfloor -3.53553390593 \rfloor = -4$$
Noted by Gauss:
Let r radius of the circle = 5
Let $$x = r^2$$
$$\begin{array}{|rcll|} \hline A_2(x) &=& 1 + 4\lfloor \sqrt{x} \rfloor + 4 \lfloor \sqrt{\frac{x}{2}} \rfloor ^2 + 8 \sum \limits_{y_1= \lfloor \sqrt{\frac{x}{2}} \rfloor + 1 }^{\lfloor \sqrt{x} \rfloor} \lfloor \sqrt{x-y_1^2} \rfloor \qquad & | \quad x = r^2 = 5^2 \\\\ &=& 1 + 4\lfloor \sqrt{5^2} \rfloor + 4 \lfloor \sqrt{\frac{5^2}{2}} \rfloor ^2 + 8 \sum \limits_{y_1= \lfloor \sqrt{\frac{5^2}{2}} \rfloor + 1 }^{\lfloor \sqrt{5^2} \rfloor} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \sum \limits_{y_1= 3 + 1 }^{5} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \sum \limits_{y_1= 4 }^{5} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \cdot \left( \lfloor \sqrt{5^2-4^2} \rfloor +\lfloor \sqrt{5^2-5^2} \rfloor \right) \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \cdot \left( 3 + 0 \right) \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 24 \\\\ &=& 1 + 20 + 36 + 24 \\ &\mathbf{=} & \mathbf{81} \\ \hline \end{array}$$
81 lattice points (points with integer coordinates) lie within or on this circle with radius 5.
Example:
$$r = 0 \ldots 20$$
Number of lattice points in circle:
$$\begin{array}{|r|r|r|} \hline r & \text{lattice points in circle} & \text{lattice points in sphere } \\ \hline 0 & 1 & 1 \\ 1 & 5 & 7 \\ 2 & 13 & 33 \\ 3 & 29 & 123 \\ 4 & 49 & 257 \\ {\color{red}5} & {\color{red}81} & 515 \\ 6 & 113 & 925 \\ 7 & 149 & 1419 \\ 8 & 197 & 2109 \\ 9 & 253 & 3071 \\ 10 & 317 & 4169 \\ 11 & 377 & 5575 \\ 12 & 441 & 7153 \\ 13 & 529 & 9171 \\ 14 & 613 & 11513 \\ 15 & 709 & 14147 \\ 16 & 797 & 17077 \\ 17 & 901 & 20479 \\ 18 & 1009 & 24405 \\ 19 & 1129 & 28671 \\ 20 & 1257 & 33401 \\ \hline \end{array}$$
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https://math.stackexchange.com/questions/2518718/a-topological-space-with-cardinality-strictly-less-than-its-weight | # A topological space with cardinality strictly less than its weight?
My book states that
if $X$ is compact, then $w(X) \leq |X|$.
This leads me to wonder, is there a nice example of a (non-compact) topological space where $w(x) > |X|$ holds?
The weight of a topological space is the smallest cardinality of a basis of the topology.
• You want to remind us the definition of weight ? – Rene Schipperus Nov 13 '17 at 18:55
• @ReneSchipperus : The weight of a topological space is the smallest cardinality of a basis of the topology. – Eric Towers Nov 13 '17 at 18:59
• But is it polite to question a topology about its weight – user541686 Nov 13 '17 at 22:09
• here I give a proof the first fact, if you're interested. – Henno Brandsma Nov 13 '17 at 22:31
The maximal weight of a space $X$ is $2^{|X|}$ and in this answer I give some examples (with proofs) of spaces that realise this maximum. E.g. a countable dense subset of $\{0,1\}^{\mathbb{R}}$ in the product topology has weight $|\mathbb{R}|$.
The bound $w(X) \le |X|$ for compact spaces follows from the equality $nw(X) =w(X)$ for (locally) compact spaces (and it also holds for metrisable spaces and orderable spaces), so many standard examples don’t work.
• +1 - this is a better answer than mine, since it provides examples of maximal weight. (I'll leave mine up anyways though.) – Noah Schweber Nov 13 '17 at 19:32
• Interesting. So even though the weight of the discrete topology on $X$ is $|X|$, and the weight of the anti-discrete topology on $|X|$ is 1 (I think), there are topologies that are "in between" these, yet have a larger weight than both? – theQman Nov 13 '17 at 20:11
• @theQman by convention the weight of the anti-discrete topology is $\aleph_0$ (weight is infinite by convention). But yes, the max lies in between. – Henno Brandsma Nov 13 '17 at 20:20
• 1) Why is weight infinite by convention? What if the space is finite? And how is this convention implied by the definition " minimum cardinality of a basis"? 2) What do you mean by "the max lies in between"? – theQman Nov 13 '17 at 20:28
• @theQman That’s the way it’s defined (so some counting arguments go through) in “cardinal functions in topology” and its sequel, which form a basis for the theory. Finite spaces have weight $\aleph_0$ as well, but in papers that only deal with finite spaces sometimes finite cardinals are used. In that case the maximal weight is also at most $|X|$ as well, of course. Max in between refers to that the maximal weight lies between indiscrete and discrete. – Henno Brandsma Nov 13 '17 at 20:37
I believe the club-generated topology on $\omega_1$ is an example (a set is open in this topology iff it is a union of sets which are unbounded and closed in the order topology - this does indeed form a topology, since the intersection of two clubs is a club). Given any $\omega_1$-sequence of open sets $C_\eta$, I can build a club (hence open) set which doesn't contain any of them, as follows:
• We begin at stage $0$ with $D_0=\emptyset$.
• At stage $\alpha+1$, pick some $\beta\in C_\alpha$ greater than $\sup(D_\alpha)$, and let $D_{\alpha+1}=D_\alpha\cup\{\beta\}$ (so since we keep adding bigger ordinals, we'll keep $\beta$ out of the club we're building). This is possible since $C_\alpha$, being a union of clubs, is unbounded.
• At stage $\lambda$ limit, let $D_\lambda=(\bigcup_{\alpha<\lambda} D_\alpha)\cup\{\sup(\bigcup_{\alpha<\lambda} D_\alpha)\}$.
Let $D=\bigcup_{\eta<\omega_1} D_\eta$. $D$ is club, hence open, and doesn't contain any of the $C_\eta$s. So $\{C_\eta:\eta<\omega_1\}$ was not a base for the club-generated topology on $\omega_1$.
It's worth noting that this argument does use a bit of choice: in the successor steps of our construction, we're assuming that there are in fact ordinals bigger than $\sup(D_\alpha)$. This uses the fact that $\omega_1$ is not a countable union of countable sets, which surprisingly is not provable in ZF alone. Interestingly, ZF does prove that $\omega_2$ is not a countable union of countable sets. So that's nice.
EDIT: As Asaf points out, if $\omega_1$ has countable cofinality then "club" doesn't really make sense (every singleton is now the intersection of two clubs, so the topology is discrete).
• What would be the club topology on $\omega_2$ if it had cofinality $\omega$? You need some choice, I guess, to even make sense of the idea of a club topology. – Asaf Karagila Nov 13 '17 at 19:38
• @AsafKaragila Oh yeah, good point. (I think you mean $\omega_1$ ...) – Noah Schweber Nov 13 '17 at 19:57
• (No, I meant what I meant, but also $\omega_1$, yes.) – Asaf Karagila Nov 13 '17 at 19:58
• Is it known what the weight of this topology on $\omega_1$ is? $\aleph_2$ provable in ZFC? – Henno Brandsma Nov 13 '17 at 20:24
• @Henno: Without being too fussy about it, I'd guess that (or its negation) would be equivalent to some saturation or other combinatorial property of the non-stationary ideal, and would therefore be equiconsistent with some large cardinal properties. But that's a very nice question! – Asaf Karagila Nov 13 '17 at 20:51 | 2021-04-17T02:42:30 | {
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=VectorCalculus/ScalarPotential | VectorCalculus - Maple Programming Help
# Online Help
###### All Products Maple MapleSim
Home : Support : Online Help : Mathematics : Vector Calculus : VectorCalculus/ScalarPotential
VectorCalculus
ScalarPotential
Calling Sequence ScalarPotential(v)
Parameters
v - vector field or Vector valued procedure; specify the components of the vector field
Description
• The ScalarPotential(v) command computes the scalar potential of the vector field v. This is a function f such that $∇f=v$. If a scalar potential does not exist, NULL is returned.
• If v is a Vector field, an algebraic expression is returned. If v is a Vector-valued procedure, a procedure is returned.
Examples
> $\mathrm{with}\left(\mathrm{VectorCalculus}\right):$
> $\mathrm{SetCoordinates}\left({'\mathrm{cartesian}'}_{x,y,z}\right)$
${{\mathrm{cartesian}}}_{{x}{,}{y}{,}{z}}$ (1)
> $v≔\mathrm{VectorField}\left(⟨x,y,z⟩\right)$
${v}{:=}\left({x}\right){\stackrel{{_}}{{e}}}_{{x}}{+}\left({y}\right){\stackrel{{_}}{{e}}}_{{y}}{+}\left({z}\right){\stackrel{{_}}{{e}}}_{{z}}$ (2)
> $\mathrm{ScalarPotential}\left(v\right)$
$\frac{{1}}{{2}}{}{{x}}^{{2}}{+}\frac{{1}}{{2}}{}{{y}}^{{2}}{+}\frac{{1}}{{2}}{}{{z}}^{{2}}$ (3)
> $v≔\mathrm{VectorField}\left(⟨y,-x,0⟩\right)$
${v}{:=}\left({y}\right){\stackrel{{_}}{{e}}}_{{x}}{-}{x}{\stackrel{{_}}{{e}}}_{{y}}$ (4)
> $\mathrm{ScalarPotential}\left(v\right)$
> $\mathrm{ScalarPotential}\left(\left(x,y,z\right)→\frac{⟨x,y,z⟩}{{x}^{2}+{y}^{2}+{z}^{2}}\right)$
$\left({x}{,}{y}{,}{z}\right){→}\frac{{1}}{{2}}{}{\mathrm{ln}}{}\left({{x}}^{{2}}{+}{{y}}^{{2}}{+}{{z}}^{{2}}\right)$ (5)
> $\mathrm{SetCoordinates}\left({'\mathrm{spherical}'}_{r,\mathrm{φ},\mathrm{θ}}\right)$
${{\mathrm{spherical}}}_{{r}{,}{\mathrm{φ}}{,}{\mathrm{θ}}}$ (6)
> $v≔\mathrm{VectorField}\left(⟨r,0,0⟩\right)$
${v}{:=}\left({r}\right){\stackrel{{_}}{{e}}}_{{r}}$ (7)
> $\mathrm{ScalarPotential}\left(v\right)$
$\frac{{1}}{{2}}{}{{r}}^{{2}}$ (8)
> $\mathrm{Gradient}\left(\right)$
$\left({r}\right){\stackrel{{_}}{{e}}}_{{r}}$ (9)
See Also
## Was this information helpful?
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https://math.stackexchange.com/questions/2125365/closed-form-for-int-01-int-01-frac1-left1-sqrt1x-12-x-n2-ri/2125842 | # Closed form for $\int_0^1…\int_0^1\frac{1}{\left(1+\sqrt{1+x_1^2+…+x_n^2}\right)^{n+1}}\;dx_1…dx_n$
I am wondering whether there is a closed form for the following integral for $n\in\mathbb{N}$:
$$\gamma(n)=\int_0^1...\int_0^1\frac{1}{\left(1+\sqrt{1+x_1^2+...+x_n^2}\right)^{n+1}}\;dx_1...dx_n\tag{*}$$
Particular values which I am aware of include:
$$\gamma(1)=\frac{4\sqrt{2}-5}{3}$$
$$\gamma(2)=\frac{5}{4}-\frac{9\sqrt{3}}{8}+\frac{\pi}{4}$$
Both of these values were obtained by evaluating different integrals to the above which solved the same problem (see below), and I am not sure how to attack $(*)$ directly; I can see no good way of solving this integral. I am wondering especially about the value of $\gamma(3)$; does it also have a simple closed form? Is there a closed form for all $n$?
Background: I posted a question here about calculating the proportion $p(n)$ of an $n$-cube closer to the centre than to the outside, which seems to me like an interesting problem. The $n=2$ case is simple to solve in terms of the following integral:
$$p(2)=2\int_{0}^{\sqrt{2}-1}\frac{1-x^2}{2}-x\;dx=\frac{4\sqrt{2}-5}{3}$$
I was able to write $p(3)$ as follows:
$$p(3)=6\int_{0}^{\frac{\sqrt{3}-1}{2}}\int_{z}^{\sqrt{2-z^2}-1}\frac{1-x^2-z^2}{2}-x\;dx\;dz$$
and I managed to evaluate this to $\frac{5}{4}-\frac{9\sqrt{3}}{8}+\frac{\pi}{4}$, but I was not able to use my method to solve the problem for higher dimensions. In the comments, however, achille hui made a proposition that we have $p(n)=\gamma(n-1)$ for all $n$ and although I still do not perfectly understand his reasoning, the claim does check out numerically for the two values I know already. Furthermore, the new integral is in a nice simple symmetric form (unlike the methods I had been using which required a case-by-case analysis for every dimension, with ugly bounds on the integrals), which makes me hope for a solution method. However, I really cannot see how to go about it. Thus I ask, is there a method for computing the integral $(*)$?
• We can use "spherical coordinates" integrating first wrt the radius and then wrt the surface measure on the sphere – Omnomnomnom Feb 2 '17 at 6:59
• Nice question! @Omnomnomnom Will that work out? The domain is not a ball. – mickep Feb 2 '17 at 6:59
• @mickep wow, totally missed that! I'm not so sure anymore. – Omnomnomnom Feb 2 '17 at 7:00
• For whatever it is worth, Mathematica gives the value $$\frac{1}{360} \bigl(170-1827 \sqrt{2} \text{arccot}\,\sqrt{2}+2979 \sqrt{2} \text{arccot}\,2 \sqrt{2}\bigr)\approx 0.032$$ for $n=3$. – mickep Feb 2 '17 at 7:41
• If you are interested in the limit at $n \to \infty$, then for a similar case, @SangchulLee suggested using Strong law of large numbers. – Yuriy S Feb 2 '17 at 10:39
Just for reference, here is the derivation of the formula $p(n) = \gamma(n-1)$ proposed by @achille hui.
• Consider the cube $\mathcal{C} = [-1, 1]^n$ and define $\mathcal{D} = \{x \in \mathcal{C} : |x| < \operatorname{dist}(x, \partial \mathcal{C}) \}$. Then for each point $x \in \partial\mathcal{C}$, we can find the unique point $q(x) \in \partial\mathcal{D}$ that lies on the line segment joining $0$ and $x$.
$\hspace{7.5em}$
Using this, define $r(x)$ as the ratio
$$r(x) = \frac{\text{[length of the line segment from 0 to q(x)]}}{\text{[length of line segment from 0 to x]}} = \frac{|q(x)|}{|x|}.$$
• Let me first give a geometric argument. Let $d\mathcal{S}$ be an infinitesimally small portion of the surface $\partial\mathcal{C}$ near $x$ and consider the cone $d\mathcal{V}$ with the base $d\mathcal{S}$ and the vertex $0$.
$\hspace{7.5em}$
Then roughly speaking, the portion of $d\mathcal{V}$ that intersects $\mathcal{D}$ is similar to the cone $d\mathcal{V}$ with ratio $r(x)$. Thus their volume roughly satisfies
$$|\mathcal{D} \cap d\mathcal{V}| \approx r(x)^n |d\mathcal{V}| = r(x)^n \cdot \frac{1}{n}|d\mathcal{S}|.$$
From this, we have
$$p(n) = \frac{|\mathcal{D}|}{|\mathcal{C}|} = \frac{\int |\mathcal{D} \cap d\mathcal{V}|}{\int |d\mathcal{V}|} = \frac{\int_{\partial \mathcal{C}} r(x)^n \frac{1}{n} \, dA}{\int_{\partial \mathcal{C}} \frac{1}{n} \, dA}. \tag{*}$$
• Here is a rigorous justification of the argument above. Write
\begin{align*} |\mathcal{D}| = \int_{\mathcal{C}} \mathbf{1}_{\mathcal{D}}(x) \, dx &= \int_{0}^{1} \int_{\partial[-s,s]^n} \mathbf{1}_{\mathcal{D}}(y) \, dyds \\ &= \int_{0}^{1} \int_{\partial\mathcal{C}} \mathbf{1}_{\mathcal{D}}(s\omega) \, s^{n-1}d\omega ds \\ &= \int_{\partial\mathcal{C}} \int_{0}^{1} s^{n-1} \mathbf{1}_{\mathcal{D}}(s\omega) \, ds d\omega. \end{align*}
The note that $s\omega \in \mathcal{D}$ if and only if $s|\omega| < |q(\omega)|$, or equivalently, $s < r(\omega)$. From this, we have
$$|\mathcal{D}| = \int_{\partial\mathcal{C}} \int_{0}^{r(\omega)} s^{n-1} \, ds d\omega = \int_{\partial\mathcal{C}} \frac{1}{n} r(\omega)^n \, d\omega.$$
Replacing $\mathbf{1}_{\mathcal{D}}$ by $\mathbf{1}$ from the above computation, we also get
$$|\mathcal{C}| = \int_{\partial\mathcal{C}} \frac{1}{n} \, d\omega.$$
This proves $\text{(*)}$.
• Now we give an explicit formula for $\text{(*)}$. Consider the top face of $\partial\mathcal{C}$. This face can be written as $[-1,1]^{n-1}\times\{1\}$. Then for each point $x = (x',1) \in [-1, 1]^{n-1}\times\{1\}$, it is not hard to verify that
$$r(x) = \frac{1}{1 + \sqrt{|x'|^2+1}}.$$
(Using the rotational symmetry, it boils down to proving this when the dimension is $n = 2$.) Plugging this back and exploiting the symmetry, we finally have
$$p(n) = \int_{[0,1]^{n-1}} r(x',1)^n \, dx' = \int_{[0,1]^{n-1}} \frac{1}{\left( 1 + \sqrt{|x'|^2+1} \right)^n} \, dx' = \gamma(n-1).$$
• +1. Thank you for this nice argument; that nicely explains why the formula is true. What do you use to make those pictures? – Anon Feb 2 '17 at 22:19
• @Anon, I'm glad it was helpful. Anyway, I used Mathematica to draw them. – Sangchul Lee Feb 2 '17 at 23:23
Not a full solution, and too long for a comment, but perhaps a plausible strategy forward. The final formula is written in terms of a finite sum of double integrals, which may (or may not) be cracked.
First, rationalize by multiplying up and down by $\left(1-\sqrt{1+x_1^2+...+x_n^2}\right)^{n+1}$ $$\gamma(n)=(-1)^{n+1}\int_0^1...\int_0^1\frac{\left(1-\sqrt{1+x_1^2+...+x_n^2}\right)^{n+1}}{\left(x_1^2+...+x_n^2\right)^{n+1}}\;dx_1...dx_n\ .$$ Next, use the identity $$\frac{1}{X^{n+1}}=\int_0^\infty d\xi\frac{\xi ^n \exp (-\xi X)}{\Gamma (n+1)}$$ to rewrite your integral as $$\gamma(n)=\frac{(-1)^{n+1}}{\Gamma(n+1)}\int_0^\infty d\xi\ \xi^n \int_0^1...\int_0^1\left(1-\sqrt{1+x_1^2+...+x_n^2}\right)^{n+1} e^{-\xi \left(x_1^2+...+x_n^2\right)}\;dx_1...dx_n\$$ $$=\frac{(-1)^{n+1}}{\Gamma(n+1)}\int_{1}^{n+1} dr\ (1-\sqrt{r})^{n+1}\int_0^\infty d\xi\ \xi^n \int_0^1...\int_0^1\delta(r-(1+x_1^2+...+x_n^2)) e^{-\xi \left(x_1^2+...+x_n^2\right)}\;dx_1...dx_n$$ $$=\frac{1}{\Gamma(n+1)}\int_{1}^{n+1} dr\ (\sqrt{r}-1)^{n+1}\int_0^\infty d\xi\ \xi^n F_n(r,\xi)\ ,$$ where $$F_n(r,\xi)=\int_0^1...\int_0^1\delta(r-(1+x_1^2+...+x_n^2)) e^{-\xi \left(x_1^2+...+x_n^2\right)}\;dx_1...dx_n\ .$$ Therefore $$F_n(r,\xi)=e^{-\xi (r-1)}\int_0^1...\int_0^1\delta(r-(1+x_1^2+...+x_n^2)) \;dx_1...dx_n$$ $$=e^{-\xi (r-1)}\int_{-\infty}^\infty\frac{dk}{2\pi}e^{\mathrm{i}k(r-1)}\left[\int_0^1 dx\ e^{-\mathrm{i}kx^2}\right]^n\ ,$$ using the integral representation of Dirac delta.
Writing $\exp(-\mathrm{i}kx^2)=\cos(kx^2)-\mathrm{i}\sin(kx^2)$ and using $$\int_0^1 dx\cos(k x^2)=\frac{\sqrt{\frac{\pi }{2}} C\left(\sqrt{|k|} \sqrt{\frac{2}{\pi }}\right)}{\sqrt{|k|}}$$ and $$\int_0^1 dx\sin(k x^2)=\mathrm{sign}(k)\frac{\sqrt{\frac{\pi }{2}} S\left(\sqrt{|k|} \sqrt{\frac{2}{\pi }}\right)}{\sqrt{|k|}}\ ,$$ in terms of Fresnel integrals [Mathematica notation], and using the binomial theorem we get $$F_n(r,\xi)=e^{-\xi (r-1)}\sum_{\ell=0}^n{n\choose \ell}(-\mathrm{i})^{n-\ell} \int_{-\infty}^\infty\frac{dk}{2\pi}e^{\mathrm{i}k(r-1)}\left(\frac{\pi}{2|k|}\right)^{n/2}C^\ell\left(\sqrt{|k|} \sqrt{\frac{2}{\pi }}\right)S^{n-\ell}\left(\sqrt{|k|} \sqrt{\frac{2}{\pi }}\right)(\mathrm{sign}(k))^{n-\ell}\ .$$ Putting everything together $$\gamma(n)=\frac{1}{\Gamma(n+1)}\sum_{\ell=0}^n{n\choose \ell}(-\mathrm{i})^{n-\ell} \int_{-\infty}^\infty\frac{dk}{2\pi}\left(\frac{\pi}{2|k|}\right)^{n/2}C^\ell\left(\sqrt{|k|} \sqrt{\frac{2}{\pi }}\right)S^{n-\ell}\left(\sqrt{|k|} \sqrt{\frac{2}{\pi }}\right)(\mathrm{sign}(k))^{n-\ell}\times\int_1^{n+1}dr (\sqrt{r}-1)^{n+1} e^{\mathrm{i}k(r-1)}\int_0^\infty d\xi\ \xi^n e^{-\xi (r-1)}\ .$$ Performing the $\xi$-integral first $$\gamma(n)=\frac{1}{\Gamma(n+1)}\sum_{\ell=0}^n{n\choose \ell}(-\mathrm{i})^{n-\ell} \int_{-\infty}^\infty\frac{dk}{2\pi}\left(\frac{\pi}{2|k|}\right)^{n/2}C^\ell\left(\sqrt{|k|} \sqrt{\frac{2}{\pi }}\right)S^{n-\ell}\left(\sqrt{|k|} \sqrt{\frac{2}{\pi }}\right)(\mathrm{sign}(k))^{n-\ell}\times\int_1^{n+1}dr (\sqrt{r}-1)^{n+1} e^{\mathrm{i}k(r-1)}\frac{1}{(r-1)^{n+1}}\ .$$
• Now that's an interesting way of proceeding. +1. It looks like a beast of a Laplace transform though. – Anon Feb 2 '17 at 22:22
• The inverse transform of $\frac{\operatorname{erf}{\sqrt{s}}}{\sqrt{s}}$ is $\frac{1}{\sqrt{\pi t}}u(1-t)$ (I think) so the inverse transform of the expression in the brackets will be $\phi(t)=\frac{e^{-\xi t}}{2\sqrt{t}}u(1-t)$ or something. Using convolution should (?) then give us $l(t)=\phi(t)*\phi(t)*...*\phi(t)$ for the inverse transform of the product, so then it seems we get the whole inverse transform as $u(t-1)l(t-1)$. I'm not sure if that's entirely correct or whether it will be useful? – Anon Feb 2 '17 at 22:37
• What I said will just bring us back to an $n$-fold integral again, but it seems like it will be a different form. – Anon Feb 2 '17 at 22:46
• Please check if after my edit, the new form looks more promising. – Pierpaolo Vivo Feb 3 '17 at 16:05
• It is a complicated formula... I wonder whether the integral in $r$ can be evaluated (it looks the most promising). Note that $\frac{(\sqrt{r}-1)^{n+1}}{(r-1)^{n+1}}=\frac{1}{(\sqrt{r}+1)^{n+1}}$ so we are trying to integrate $\frac{e^{ik(r-1)}}{(\sqrt{r}+1)^{n+1}}$ (I don't know if that helps). – Anon Feb 7 '17 at 0:12 | 2019-06-16T01:09:42 | {
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https://stats.stackexchange.com/questions/427419/distribution-of-ranges-of-normally-distributed-variables?noredirect=1 | # Distribution of ranges of normally distributed variables
I have four independent variables $$x_i$$, each of them normally distributed with $$\mu = 0$$ and $$\sigma = 1$$. What is the distribution of the range of the four variables, i.e., $$\max(x_i) - \min(x_i)$$?
For example, I draw a random sample of $$n=10000$$ of each variable and calculate the $$\Delta x = \max(x_i) - \min(x_i)$$, which gives $$10000$$ range values. I am interested in the expected distribution of this range value $$\Delta x$$ under the condition that each $$x_i$$ is normally distributed as explained above.
• The phrase "peak-to-valley," which apparently has nothing to do with the range of the sample you describe, suggests you might have a different setting in mind: are you sure you have asked a question relevant to your underlying statistical problem? – whuber Sep 16 at 10:37
• @whuber is correct. With "Peak-to-Valley" in finance we denote the distance between the local minimum (maximum) and the next local maximum (minimum) with the term "Maximum Drawdown". Maybe you want to derive the distribution of the maximum drawdown. A quick google search will give you some paper addressing your question (if you are concerned with this measure) – AlexandrosB Sep 17 at 12:35
The range of a sample $$X = x_1, x_2, \ldots, x_n$$ is the difference between its maximum $$x_{(n)}=\max(X)$$ and minimum $$x_{(1)}=\min(X):$$
$$\operatorname{range}(X) = x_{(n)} - x_{(1)}.$$
When $$X$$ is a simple random sample of size $$n\ge 2$$ from a continuous distribution with distribution function $$F$$ and density function (PDF) $$f=F^\prime,$$ the joint distribution of the minimum and maximum is also continuous and, following the analysis at https://stats.stackexchange.com/a/78559/919, has a density function
$$f_{(X_{(1)}, X_{(n)})}(x,y) = n(n-1) f(x)f(y)\left(F(y)-F(x)\right)^{n-2}\, \mathcal{I}(y\ge x).$$
(The indicator $$\mathcal{I}(y\ge x)$$ means this function is zero when $$y \lt x.$$)
Upon changing variables from $$(x,y)$$ to $$(x,r)$$ with $$r=y-x\ge 0$$ representing possible values of the range, it follows that $$\mathrm{d}x\mathrm{d}y = \mathrm{d}x\mathrm{d}(x+r) = \mathrm{d}x\mathrm{d}r$$ and you can then integrate out the variable $$x$$ to obtain the PDF for the range,
$$f_{\operatorname{range}(n)}(r) = n(n-1)\mathcal{I}(r\ge 0)\, \int_{\mathbb R} f(x)f(x+r)\left(F(x+r)-F(x)\right)^{n-2}\mathrm{d}x.$$
I understand $$f_{\operatorname{range}(4)}$$ responds to the question's request for an "expected distribution."
In general--and particularly for Normal distributions--there is no closed form expression for the integral: it needs to be evaluated numerically. Neither is there generally a closed form expression for its expectation; that too requires numerical evaluation. But for modest values of $$n$$--less than $$10^{36},$$ approximately, when working in double precision--these integrals can be accurately evaluated.
As an example, the following plots show (from left to right)
1. A histogram of the results of 100,000 simulated ranges of standard Normal samples of $$n=4$$ observations on which a plot of $$f_{\operatorname{range}(4)}$$ is superimposed to show how closely the numerical integral adheres to the simulated frequencies;
2. Plots of $$f_{\operatorname{range}(n)}$$ for $$n=3$$ (red) through $$n=6$$ (blue), with that for $$n=4$$ outlined in black, to show how the range distribution changes with sample size;
3. A plot of $$f_{\operatorname{range}(10^{36})}$$ to show what the range distribution looks like for large $$n.$$
The following R code produced the figure. It shows how to compute $$f_{\operatorname{range}(n)}$$ numerically and how to simulate ranges.
#
# Compute the density of the Normal range distribution at value r for a
# sample of size n.
#
f <- Vectorize(function(r, n, mu=0, sigma=1, ...) {
q <- qnorm(1e-4/n)
f <- function(x) dnorm(x, mu, sigma, log=TRUE)
ff <- function(x) pnorm(x, mu, sigma, log=TRUE)
logdiff <- function(y, x, k) {
u <- x-y
ifelse(2*u < log(.Machine$$double.eps), y-exp(u), y+log(1-exp(u))) * k } h <- function(x,y) { log(n) + log(n-1) + f(x) + f(y) + logdiff(ff(y), ff(x), n-2) } integrate(function(x) exp(h(x,x+r)), mu+q*sigma, mu-q*sigma, ...)$$value
}, "r")
#
# Simulate a large number of ranges of samples of size n.
# This takes about a second for n=10^5.
#
n.sim <- 1e5
n <- 4
set.seed(17)
x <- apply(matrix(rnorm(n.sim*n), ncol=n.sim), 2, function(y) diff(range(y)))
#
# Create the figure.
#
par(mfrow=c(1,3))
#
# Histogram of the simulation.
#
hist(x, freq=FALSE, breaks=50, cex.main=1,
xlab="Range",
main=expression(paste("Histogram of ", 10^{5}, " Standard Normals"))
)
#
# Plots of f for small n.
#
range.plot <- function(n, add=FALSE, n.pts=201, color="Gray", outline="Black", ...) {
x <- seq(max(0, -2*qnorm(1/n)-3), -2*qnorm(1/n)+5, length.out=n.pts)
y <- f(x, n, stop.on.error=FALSE, rel.tol=1e-8)
i <- y > 1e-3
plot(x[i], y[i], type="n", xlab="Range", ylab="Density",
main=expression(paste("PDF for Range of ", 10^{36}, " Standard Normals")),
...)
polygon(x, y, col=color, border=NA)
lines(x, y, lwd=2, col=outline)
}
plot(c(0,6), c(0,.5), type="n", cex.main=1,
main=expression(paste("PDFs for Ranges of ", 3-6, " Standard Normals")),
xlab="Range", ylab="Density")
invisible(sapply(3:6, function(i) {
a <- sapply(c(.4, .9), function(a) rainbow(5, 0.7, 0.9, alpha=a)[i-2])
# Plot f for very large n. | 2019-11-13T01:37:02 | {
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https://math.stackexchange.com/questions/817062/proof-that-no-polynomial-with-integer-coefficients-can-only-produce-primes/817094 | # Proof that no polynomial with integer coefficients can only produce primes [duplicate]
Doing a discrete math review and am trying to solve problem 1.6 in the text found here: http://courses.csail.mit.edu/6.042/fall13/ch1-to-3.pdf - I believe I've gotten parts (a) and (b) correctly, but (c) is a bit tricky for me. Would appreciate a review of a/b and a hint for c if possible. Restatement here:
For n = 40, the value of polynomial $p(n)::=n^2 + n + 41$ is not prime, as noted in Section 1.1. But we could have predicted based on general principles that no nonconstant polynomial can generate only prime numbers.
In particular, let $q(n)$ be a polynomial with integer coefficients, and let $c::=q(0)$ be the constant term of q.
(a) Verify that $q(cm)$ is a multiple of c for all $m \in Z$
Proof: Since q is a polynomial, it will be of the form: $q(x) = a_nx^n+a_{n-1}x^{n-1} + ... + a_1x + c$
If $x = cm$, then notice that cm will be inside every term of the polynomial, and since the final term is c, c can be factored out of every term. Hence, q(cm) is a multiple of c for all $m \in Z$.
b) Show that if q is nonconstant and c > 1, then as n ranges over the nonnegative integers, $N$, there are infinitely many $q(n) \in Z$ that are not primes.
Proof: Continuing from the result found in (a), it is easy to see that for all $m \in Z$, there will be an infinite number of multiples of c that will be generated by q. Since c > 1, these multiples are guaranteed not to be prime.
c) Conclude that for every nonconstant polynomial, q, there must be an $n \in N$ such that $q(n)$ is not prime. Hint: Only one easy case remains.
c = 0 is trivially easy to prove using the (b) above.
I assume the "easy" case is referring to c = 1, but in this case I'm not sure how to continue, since the result of (a) and (b) don't apply: I can't use them since adding 1 to any even number may make it prime. If the result of the terms not including c is odd, then adding 1 to that result makes it even and it is not prime. However, if the terms add up to an even number, I don't see a way of using the knowledge I have so far to prove conclusively that that number + 1 will NOT in fact be prime.
## marked as duplicate by user147263, graydad, Peter Woolfitt, Lord_Farin, André Nicolas polynomials StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 20 '15 at 17:12
• For $c=1$, consider $n=0$. $q(0)=1$ which is not a prime. – peterwhy Jun 1 '14 at 14:47
• @peterwhy I think it is being assumed that $n\ne 0$. – user142299 Jun 1 '14 at 14:50
• @NotNotLogical In part (b), $N$ is defined by "then as $n$ ranges over the nonnegative integers, $N$", and $0$ is a nonnegative integer. – peterwhy Jun 1 '14 at 14:51
• @peterwhy I always thought 1 was a prime number. Today I stand corrected. I assume that's why the text said it was the "easy" case... – DavidN Jun 1 '14 at 15:47
• @DavidN I am not really sure whether what I "proved" above is intended, and anyway there are some good proofs below. Just to complete my "proof", since a necessary condition for $x$ to be a prime number is $x>1$, for any polynomial $p(n)$ with $c\in\{1,0,-1,-2,\cdots\}$, $p(0) = c \le1$, which makes $p(0)$ not a prime. – peterwhy Jun 1 '14 at 16:04
Hint If $\,q(n)\,$ is prime for all $\,n\,$ then $\,q(0) = p\,$ is prime, thus $\, p\mid q(pn)\,$ and $\,q(pn)\,$ is prime, hence $\,q(pn) = p\,$ for all $\,n.\,$ Thus the polynomial $\,q(px)-p\,$ has infinitely many roots so is zero, i.e. $\,q(px) = p\,$ is constant, hence $\,q(x)\,$ is constant.
• "If q(n) is prime for all n then q(0) = p is prime" - this I understand, but how does this imply your next statement? "and since p ∣ q(pn) and q(pn) is prime" - When have we established that p | q(pn)? Above, I was only able to establish that for c > 1. – DavidN Jun 1 '14 at 16:13
• @DavidN You proved that in part (a). – Bill Dubuque Jun 1 '14 at 16:19
• You're right. Nice proof by contradiction. – DavidN Jun 1 '14 at 21:19
I would argue differently, even allowing for the fact that $p(0) = \pm 1.$ First of all, if $p(x)$ is not a constant , then $p(n) \to \pm \infty$ as $n \to \infty,$ for if $p$ has degree $d$, say $p(x) = a_{d}x^{d} + \ldots + a_{1}x+ a_{0},$ then $\frac{p(x)}{x^{d}}$ tends to $a_{d}$ as $x \to \infty.$ Suppose that $p(n)$ is either $\pm 1$ or $0$, or $\pm$ (some prime) for every integer $n.$ Let $h$ be the smallest positive integer such that $p(h) \not \in \{-1,0,1 \}.$ There is such an integer $h$ by the earlier remark. Suppose that that $p(h) = \pm q$ for some prime $q.$ Then for every positive integer $t,$ we see easily that $p(h+qt)$ is divisible by $q.$ But for large enough $t,$ we see that $p(h+tq) \not \in \{,-q,0,q \},$ so that $p(h+tq)$ is not prime (and is not $\pm 1$ either).
• +1. I understand this answer as: for $p(n)$, find $p(h)$ such that $|p(h)|>1$ for some $h\ge0$ (and there must be some), shift the polynomial to $r(n):=p(n+h)$. $r(n)$ would still be a polynomial with integer coefficients, have a constant term $r(0)>1$, so there exists $r(n)$ that is not prime from part (b). And all values of $r(n)$ appear in $p(n)$... – peterwhy Jun 1 '14 at 15:51
• Almost, but not quite, what I meant: I set $r(n) = p(h+nq)$ in your terminology. – Geoff Robinson Jun 1 '14 at 16:07
• Right, and so using my interpretation and OP's result in part (a) "Verify that $r[r(0)\cdot m]$ is a multiple of $r(0)$ for all $m\in \mathbb Z$", I arrive at your form $r'(n):=r[r(0)\cdot n]=p(h+nq)$. – peterwhy Jun 1 '14 at 16:24
• Interesting answer - don't fully grok your argument yet but will go over it when I can. Thanks! – DavidN Jun 1 '14 at 21:21
• @DavidN: The key is that if we add a multiple of $q$ to $h,$ then we get the same image for $p(h+tq)$ in $\mathbb{Z}/q \mathbb{Z}$ as we do for $p(h).$ – Geoff Robinson Jun 1 '14 at 22:01 | 2019-10-14T20:11:23 | {
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https://math.stackexchange.com/questions/1894684/what-is-the-chance-of-rolling-a-specific-number-after-a-certain-amount-of-rolls/1894724 | # What is the chance of rolling a specific number after a certain amount of rolls?
Say I roll a $6$ sided dice and I want to roll a $6$. what is the probability that I will have rolled the number I want after $6$ rolls?
I have been using this: $\displaystyle1-\left(1-\frac{1}{x}\right)^y$
where $x$ is the number of sides and $y$ is the amount of rolls, so it would be $\displaystyle1-\left(1-\frac{1}{6}\right)^6$ for rolling a specific number in $6$ rolls, which is $\approx66.5\%$ is this the correct way of calculating the probability of something like this, if not what is the proper way?
i'm not really sure why that formula works(if it does) so some elaboration on that would be nice.
sorry for lack of technical language
• After exactly $6$ rolls or after at most $6$ rolls? Aug 17 '16 at 5:08
• what i'm trying to ask is: what is the chance that by 6 rolls, you will have obtained the number that you are looking for. wether you get it on the first roll or the last, what is the probability that you will have it by the 6th roll. Aug 17 '16 at 5:18
• it's a very good way of tackling the problem. This is a case where calculating the probability of the event you are interested in, not happening seems easier than finding the probability it does happen.
– Cato
Aug 17 '16 at 9:21
Simply put, your computation of the probability is correct. If the number you want (let's call it $n$) does not appear even once, every roll must have resulted in a number other than $n$. For an individual roll on an $x$-sided die, the probability of this happening is $1-\frac 1x$. After $y$ rolls, the chance of not getting $n$ in any roll must be $(1-\frac 1x)^y$. Since this accounds for every sequence off rolls in which $n$ appears and no others, the probability of seeing $n$ at least once must be $1-(1-\frac 1x)^y$.
You can also use a combinatorial argument. There are $x^y$ possible sequences of rolls, all of which are equally likely. If we stipulate $n$ cannot appear, there are only $(x-1)^y$ possible sequences. The probability of $n$ not appearing is thus $\frac{(x-1)^y}{x^y}=(1-\frac 1x)^y$, leading to the same result as before.
I would say that it's easier to calculate the chance that you never roll this number. So that all 6 rolls are not the number you want. This chance is (5/6)^6 = 33.49%. Now to calculate the chance that you roll your number at least oncentre within 6 rolls you can take 1-(5/6)^6= 66.51%. (I don't have an approximate sign)
This is exactly the same as your equation. I just used 5/6 instead of 1-1/6.
As for the explanation. All the possibilities that exist can be added up to 100%. What we do here is we calculate the chance for all the possibilities that you won't get what you are looking for. Now if you subtract these you will be left with all the possibilities you are looking for.
The reason we sometimes use this method is because the conjugate is much easier to calculate, like in this situation.
Split it into disjoint events, and then add up their probabilities:
• The probability to obtain this value after exactly $\color\red1$ attempt is $\left(1-\frac16\right)^{\color\red1-1}\cdot\frac16$
• The probability to obtain this value after exactly $\color\red2$ attempts is $\left(1-\frac16\right)^{\color\red2-1}\cdot\frac16$
• The probability to obtain this value after exactly $\color\red3$ attempts is $\left(1-\frac16\right)^{\color\red3-1}\cdot\frac16$
• The probability to obtain this value after exactly $\color\red4$ attempts is $\left(1-\frac16\right)^{\color\red4-1}\cdot\frac16$
• The probability to obtain this value after exactly $\color\red5$ attempts is $\left(1-\frac16\right)^{\color\red5-1}\cdot\frac16$
• The probability to obtain this value after exactly $\color\red6$ attempts is $\left(1-\frac16\right)^{\color\red6-1}\cdot\frac16$
Hence the probability of obtaining this value within $6$ attempts is:
$$\sum\limits_{n=1}^{6}\left(1-\frac16\right)^{n-1}\cdot\frac16\approx66.51\%$$
• I think that by multiplying that by {1 - (1 - 1 / 6 )} / {1 - (1 - 1 / 6 )} (which is equal to 1) you can rearrange to get 1 - (1 - 1/6) ^ n
– Cato
Aug 17 '16 at 9:21
• @AndrewDeighton: Of course. It is the same result as in the original question. Aug 17 '16 at 9:24 | 2021-10-28T15:24:21 | {
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https://www.physicsforums.com/threads/shaping-probability-distribution-function.637573/ | # Shaping probability distribution function
1. Sep 20, 2012
### SunnyBoyNY
1. The problem statement, all variables and given/known data
Incoming signal has normal distribution, xmin is equal to -sigma, xman is equal to +sigma. What is the governing equation of the nonlinearity through which the signal has to be passed in order to make its pdf uniform?
2. Relevant equations
http://en.wikipedia.org/wiki/Normal_distribution
3. The attempt at a solution
I have already found out that the signal needs to be passed through erf(x/sqrt(2)), which is very relevant to the CDF of normal distribution. The problem is that I cannot find a mathematical proof.
2. Sep 21, 2012
### Ray Vickson
It's very easy. Say you have a continuous random variable X with a strictly increasing cumulative distribution F(x) on an x-interval [a,b] (possibly a = -∞ and b = +∞). The probability density of X is f(x) = (d/dx) F(x). Now look at Y = F(X); that is, for each observation x of X we let the observation of Y be y = F(x). What is the distribution of Y? For x < X < x + dx the probability is f(x)*dx, so if x <--> x and y + dy <--> x + dx, we have
P{y < Y < y + dy} = f(x)*dx. If g(y) is the probability density of Y, we therefore have g(y)*dy = f(x)*dx. But dy/dx = (d/dx) F(x) = f(x), so dy = f(x) dx, hence we must have g(y) = 1; that is, Y is uniform on (0,1).
All this is very standard in Monte-Carlo simulation, where it is used to generate samples from non-uniform distributions: we generate Y uniform on (0,1), then obtain our sample of X from x = F-1(y) (at least, in those cases where the latter function is known and not too hard to compute).
RGV
3. Sep 21, 2012
### SunnyBoyNY
RGV, thanks for your feedback. I believe I have already proved it using a non-linearly distributed substitute variable "a" as opposed to the linear variable "x". Those two variables could be unambiguously mapped from one to the other to warp the "x" axis to ultimately render the pdf uniform.
I dare to say I understand your reasoning here, I will just need a little bit of time to absorb it.
The integral of (pdf):
$f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}(\frac{x}{\sigma})^2}$
is (cdf):
$\frac{1}{2}erf(\frac{x}{\sqrt{2}\sigma})$
Which is a normalized function with zero mean. I assume the mapping function needs to be multiplied by $\sigma$ to achieve the desired uniform mapping. This is just a detail.
Quite interesting info about Monte-Carlo simulations. It would make sense to transform a linear distribution to resemble other kinds of distributions. I wonder whether MatLab does it the same way.
4. Sep 21, 2012
### jbunniii
I'm not sure how Matlab does it, but a common way to generate Gaussian (normal) random numbers from uniform ones is the following trick: if $u_1$ and $u_2$ are independent random variables, uniformly distributed over (0,1], then
$$n_1 = \sqrt{-2 \log(u_1)} \cos(2\pi u_2)$$
and
$$n_2 = \sqrt{-2 \log(u_1)} \sin(2\pi u_2)$$
are independent Gaussian random variables with zero mean and unit variance. This is the so-called Box-Muller transformation:
http://en.wikipedia.org/wiki/Box–Muller_transform
5. Sep 24, 2012
### SunnyBoyNY
jbuniniii,
Thanks for the info about the transform. That's why I love this forum. One question brings together many ideas and perspectives. | 2017-12-18T19:09:37 | {
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https://math.stackexchange.com/questions/2424538/how-to-prove-that-x2-1-geq-2x/2424547 | # How to prove that $x^2 +1 \geq 2x$?
I am trying to prove that $x^2 +1 \geq 2x$ without using circular logic (meaning first assuming that this inequality is true and then moving to the $2x$ to the left side and factoring it). Thanks.
• $(x-1)^2 > 0$ .. – Donald Splutterwit Sep 10 '17 at 21:51
• @DonaldSplutterwit for |x|>1 – user451844 Sep 10 '17 at 21:53
• It's not true for $x=1$. You can show $x^2+1\geq 2x$, however. – Thomas Andrews Sep 10 '17 at 21:55
• What you should actually be doing is showing that $x^2 + 1 \geq 2x$ is equivalent to $(x-1)^2 \geq 0$, which is true. This is not circular. – mechanodroid Sep 10 '17 at 21:55
• okay sorry not equal to 1 it should be. – user451844 Sep 10 '17 at 21:58
$$x^2 + 1 = x^2 - 2x + 1 + 2x = \underbrace{(x-1)^2}_{\geq 0} + 2x \geq 2x$$
• beautiful...........!!!! – Stu Sep 10 '17 at 23:05
Work backwards . . .
\begin{align*} &(x-1)^2 \ge 0&&\text{[since squares are nonnegative]}\\[4pt] \implies\;&x^2-2x+1 \ge 0\\[4pt] \implies\;&x^2 + 1 \ge 2x\\[4pt] \end{align*} Alternatively, work forwards, but use $\iff$ . . . \begin{align*} &x^2+1 \ge 2x \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;];\;\;\; \\[4pt] \iff\;&x^2-2x + 1 \ge 0\\[4pt] \iff\;&(x-1)^2 \ge 0\\[4pt] &\text{which is true since squares are nonnegative.}\\[4pt] \end{align*} This allows the argument to be reversed without actually reversing it.
But to use the $\iff$ version, you need to be careful to make sure that each line is equivalent to the previous one (i.e., each implies the other).
If $x \le 0$, it is true.
If $x=1$, there is equality.
If $x>1$, you put $x=1+\epsilon$ with $\epsilon>0$.
then $$x^2+1=(1+\epsilon)^2+1$$ $$=1+\epsilon^2+2\epsilon+1$$ $$=2 (1+\epsilon)+\epsilon^2>2x$$
If $0 <x <1$ put $y=\frac {1}{x}>1$. then $$y^2+1=\frac {1}{x^2}+1$$ $$=\frac {x^2+1}{x^2}>2\frac {1}{x}$$ $$\implies 1+x^2>2x$$ Done!
$x^2 +1 \geq 2x\iff x^2 -2x+1 \geq0$
let $f(x):= x^2 -2x+1 \implies f'(x)= 2x-2$
$f'(x)=0 \implies x=1$
$f$ is decreasing over $(-\infty,1]$ and increasing over $[1,+\infty)$ and $f(1)=0$ so $f(x)\ge 0$
So we have proved $f(x)=x^2 -2x+1\ge 0$ so $x^2 +1 \geq 2x$
Proof: \begin{align} x^2 + 1 &\geqslant 2x \\ \iff x^2 + 1 - 2x &\geqslant 0 \\ \iff x^2 + 1^2 - 2\cdot x\cdot 1 &\geqslant 0 \\ \iff (x - 1)^2 &\geqslant 0,\end{align} as desired.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\,\Box$
• Pardon me if I'm wrong, but $P\Rightarrow Q$ doesn't necessarily imply $Q\Rightarrow P$. For example, $(x=\pi)\Rightarrow(\sin(x)=0)$ but the converse isn't always true. For instance, $\sin(x)=0$ could mean $x=0$ or $x=\pi$, etc. So I think there needs to be another step to show that $x-1\geq0\Rightarrow x^2+1\geq 2x$. – Jam Sep 14 '17 at 15:01
• Also, $(x-1)^2\geq0\Rightarrow (x\geq 1)$ isn't necessarily true (take $x=0$ for example). You've got to consider both sides of the parabola. – Jam Sep 14 '17 at 15:05
• Woah sorry about that. I am going to re-edit. – George N. Missailidis Sep 22 '17 at 22:04
• @GeorgeN.Missailidis No I will – user477343 Sep 22 '17 at 22:07 | 2019-04-25T14:43:02 | {
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https://math.stackexchange.com/questions/1364373/find-maximum-and-minimum-values-of-an-equation-on-an-elipse | # Find maximum and minimum values of an equation on an elipse
I need some help with this. I've been struggling through this last chapter of my Calc III class, and I'm not sure how to do this (although, it doesn't seem like it should be difficult to do)
$$\text{Find the maximum and minimum values of }f(x, y) = 4x + y\text{ on the ellipse } x^{2} + 49y^{2} = 1 \\ \text{Maximum = _____}\\ \text{Minimum = _____}$$
I know that I can find the critical point by taking partial derivatives so $$\frac{\partial{f}}{\partial{x}} = 4 \\ \frac{\partial{f}}{\partial{y}} = 1$$ Which gives us the critical point $(4,1)$.
Here's where I get stuck, I know that we can then determine max/min from the equation $D = f_{xx} * f_{yy} - f_{xy}^2$ and based on the value of $D$, we know whether it is a max, a min, or a "saddle" point. It doesn't seem to apply in this situation, because I have an ellipse that I have to use as a constraint.
What do I do next?
UPDATE:
Based on everyone's suggestion, I've been looking into Lagrange multipliers. I think I'm still stuck.
$$\bigtriangledown{f} = \lambda \bigtriangledown{g} \\ f_x = 4 = \lambda 2 * x\\ f_y = 1 = \lambda 98 * y \\ \text{we need to solve by using ratios of derivatives to remove } \lambda\\ \frac{4}{1} = \frac{\lambda 2 x}{\lambda 98 y} \\ 196 y = x\\ \text{Now we use our ellipse to solve}\\ (196y)^2 + 49y^2 = 1 \\ 38445 y^2 = 1 \\ y = \sqrt{\frac{1}{38465}} = y_0\\ \text{plug y back into our ellipse} \\ x^2 + 49 (\frac{1}{38465}) = 1\\ x = \sqrt{1- \frac{49}{38465}} = x_0$$ I'm feeling somewhat accomplished, but I have no idea if what I'm doing is correct, or what to do with these new numbers
• Editing, I was looking at two problems simultaneously and got wires crossed. Jul 17 '15 at 10:36
• The point $(0,0)$ does not lie on the ellipse Jul 17 '15 at 10:37
• Do you know Lagrange multipliers ? Jul 17 '15 at 10:38
• @ClaudeLeibovici, I've been looking into them based on your suggestion and the hints given in the answers. I've updated my question with my attempt, but I do not think it is correct. Jul 17 '15 at 11:51
• @BCqrstoO your answer is correct. Plug the $x_0,y_0$ to $f(x,y)$ to find the extremum. Don't forget that you are solving a quadratic equation... Jul 17 '15 at 12:03
You can use lagrange multipliers for this type of question:
$$L(x, y, \lambda) = (4x + y) -\lambda(x^2 + 49y^2 -1)$$
Solve $\frac{\partial L}{\partial x}=\frac{\partial L}{\partial y} = \frac{\partial L}{\partial \lambda} = 0$
Then substitute the points from the above equations to find maximum and minimum.
If you're not familiar with lagrange multipliers, you can read more here.
here's a quick solution:
$$4 -2 \lambda x = 0$$ $$1 - 98 \lambda y = 0$$ $$x^2 +49y^2 - 1= 0$$
Hence, $$x = 196y$$ $$(196y)^2 + 49y^2 -1 = 0$$
$$y = \pm\frac{1}{7\sqrt{785}}$$
$$x = \pm\frac{28}{\sqrt{785}}$$
$$(x, y) = \left(\frac{28}{\sqrt{785}}, \frac{1}{7\sqrt{785}}\right), \left(-\frac{28}{\sqrt{785}}, -\frac{1}{7\sqrt{785}}\right)$$
Then substitute these values into $f(x, y) = 4x + y$, clearly, $$f \left(\frac{28}{\sqrt{785}}, \frac{1}{7\sqrt{785}}\right) = \frac{\sqrt{785}}{7}$$ will be maximum whilst the other point will be minimum.
• I have given an attempt to use lagrange and updated my question. I do not think I understand what to do Jul 17 '15 at 11:51
• Ok, I'll have a look for you. Jul 17 '15 at 11:53
• @BCqrstoO, Pls don't forget to upvote or accept answer if its helpful for you. Jul 17 '15 at 12:04
• Thanks for going in depth. That last step was what I was missing. I'll be using this as a guide for some of my other problems. Thank you! Jul 17 '15 at 12:30
• @BCqrstoO, you're welcome. I just substituted the values of x and y into $4x+y$ in order to find the minimum and maximum values. Jul 17 '15 at 13:36
Lagrange multipliers do work for your problem, but that involves solving three simultaneous non-linear equations in three unknowns. There is another way that uses only one variable: parameterization.
Get a parameterization that describes the given curve in terms of only one variable. In the case of your ellipse you can use
$$x=\cos t, \quad y=\frac 17\sin t, \quad 0\le t\le 2\pi$$
Now you want to want to find the extrema of
$$\hat f(t)=4x+y=4\cos t+\frac 17\sin t$$
There are several ways to find the extrema of that, using calculus or just trigonometry. Here is a calculus way:
$$\hat f'(t)=-4\sin t+\frac 17\cos t=0$$ $$28\sin t=\cos t$$ $$\tan t=\frac 1{28}$$ $$\cos t=\sqrt{\frac 1{\tan^2 t+1}}=\pm\frac{28}{\sqrt{785}}$$ $$\sin t=\sqrt{1-\cos^2 t}=\pm\frac{1}{\sqrt{785}}$$ $$x=\cos t=\pm\frac{28}{\sqrt{785}}$$ $$y=\frac 17\sin t=\pm\frac{1}{7\sqrt{785}}$$
where $x$ and $y$ have the same sign. This means the maximum of $f$ is
$$4x+y=4\frac{28}{\sqrt{785}}+\frac{1}{7\sqrt{785}}=\frac{785}{7\sqrt{785}}=\frac{\sqrt{785}}{7}\approx 4.00255$$
and the minimum is the negative of that. This graph confirms the maximum.
There seems to be a typo in the function $f(x,y)$.
Okay, now that you edited it.
$(4,1)$ is not a critical point. In fact, $f(x,y)$ has no critical points because $f_x$ and $f_y$ is never zero. You should use Lagrange multipliers, not Second partials test. | 2021-12-04T16:47:42 | {
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https://math.stackexchange.com/questions/3029407/brownian-motion-reflection-principle-result | # Brownian motion reflection principle result
I'm studying about the reflection principle of the brownian motion, and I found that this result is a direct consequence of this principle:
Let $$B_t$$ a brownian motion, then for every $$a \in \mathbb{R} \$$,
$$\mathbb{P}(\lim_{t \to \infty} \sup_{s\in [0,t]} B_s > a) = 1$$
I'm trying to prove this statement using the reflection principle but I'm totally lost. I can't see how are those results related.
• Which formulation of the reflection principle do you know/use? It is direct consequence of the reflection principle that $$M_t := \sup_{s \leq t} B_s$$ equals in distribution $|B_t$|. Knowing this, it shouldn't be difficult to prove the assertion. – saz Dec 7 '18 at 7:51
Let $$M_t = \sup_{s \in [0,t]} B_s$$. For $$a < 0$$ the statement is trivial, so take $$a \ge 0$$.
The reflection principle says that $$P(M_t > a) = 2 P(B_t > a)$$. Now since you know $$B_t \sim N(0,t)$$, you can compute $$\lim_{t \to \infty} 2 P(B_t > a)$$ and determine that it equals 1. Thus $$\lim_{t \to \infty} P(M_t > a) = 1$$.
This is not the same as the desired statement, but can be used to prove it in this case. Note that $$M_t$$ is increasing in $$t$$, so necessarily $$\lim_{t \to \infty} M_t$$ exists (as a random variable) and $$\lim_{t \to \infty} M_t \ge M_u$$ for any fixed $$u$$. Hence $$P(\lim_{t \to \infty} M_t > a) \ge P(M_u > a)$$. Passing to the limit as $$u \to \infty$$ and using what we previously proved, we have $$P(\lim_{t \to \infty} M_t > a) \ge \lim_{u \to \infty} P(M_u > a) = 1$$.
• What is $\lim_{t\to\infty} M_t$? – thomasb Sep 20 at 16:37
• It's the usual thing. It's the random variable $M$ defined as $M(\omega) = \lim_{t \to \infty} M_t(\omega)$. We know by definition of $M_t$ that for each $\omega$, $M_t(\omega)$ is an increasing function of $t$, so the limit exists in $[0,+\infty]$ for each $\omega$ and $M$ is well defined. It is an exercise to verify that it is measurable. Of course, we end up proving that $P(M > a) = 1$ for every $a$, from which it follows that in fact $M = +\infty$ almost surely. – Nate Eldredge Sep 20 at 17:45
From the Law of Iterated Logarithm: $$\limsup_{t\to+\infty}\frac{B_t}{\sqrt{2t\ln\ln t}} \overset{\mathbb{P}\rm -a.s.}{=} 1,$$ so $$\mathbb{P}\left(\lim_{t\to+\infty}\sup_{s\in[0,t]}B_s>a\right)=1.$$ | 2019-10-14T13:04:01 | {
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https://math.stackexchange.com/questions/1934734/how-to-find-the-square-roots-of-z-5-12i | # How to find the square roots of $z = 5-12i$
I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$z_k^2 = 5-12i\\ (a+bi)^2 = 5-12i\\ (a^2-b^2) + i(2ab) = 5-12i\\ \\ \begin{cases} a^2 - b^2 = 5\\ 2ab = -12 \end{cases} \quad \Rightarrow \quad z_1 = -3+2i \quad z_2 = 3-2i$$
My question is: can the following method (below) be used to solve the problem above? Motivation for this question: if I were to find the cubic roots of the number given, I couldn't use the first method.
I will use this "other method" it in a different problem.
Find the square roots of $z = 2i$
The method:
Since $\rho = 2$ and $\theta = \frac{\pi}{2}$ we have to find a complex number such that
\begin{align*} z_k^2 &= 2i\\ z_k^2 &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\ \rho_k^2 \left( \cos 2\theta_k + i \sin 2 \theta_k \right) &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\ \end{align*}
\begin{cases} \rho^2 &= 2\\ 2\theta_k &= \frac{\pi}{2} + 2k\pi \end{cases}
\begin{cases} \rho &= \sqrt{2}\\ \theta_k &= \frac{\pi}{4} + 2k\pi \end{cases}
\begin{align*} z_0 &= \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 1+i\\ z_1 &= \sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = -1-i \end{align*}
Using this method for the first example seems like a dead end, specially because of the fact that the angle of $z$ is not as straightforward as the angle of the second example.
Thank you.
• Your second method appears to be using DeMoivre's theorem, It's very effective in finding powers and roots of complex numbers, and is quicker at roots than your first method. – tberkeley Sep 20 '16 at 20:36
• Yes it is very useful, but I am wondering how useful it is on cases when the angles are not on the 30-45-60-90-180-... range. Can you answer that tberkeley? – bru1987 Sep 20 '16 at 20:37
• It's a good question. I'm sorry I can't answer it. – tberkeley Sep 20 '16 at 20:51
• If $z = a + bi$ then find $\theta = \arctan b/a$ and let $r = \sqrt {a^2+b^2}$. Then $z =r (\cos \theta + \sin \theta)$ and $z {1/n}= \sqrt [n]{r}(\cos( \theta/n+2k\pi/n) +i\sin (\theta/n + 2k\pi/n))$ – fleablood Sep 20 '16 at 21:51
If you let $\theta$ be the angle in the first case, then using the fact that $\tan \theta = -12/5$, you can find $\cos\theta$ and $\cos\frac{\theta}{2}$ indirectly. One of the roots would be: $\sqrt{13}(-\frac{3}{\sqrt{13}} + i \frac{2}{\sqrt{13}}) = z_1.$
• you mean finding cos and sin using for example $\cos^2x+\sin^2x = 1$? – bru1987 Sep 20 '16 at 20:53
• @bru1987 Yes! $1+\tan^2x = \frac{\cos^2x + \sin^2x}{\cos^2x} = \frac{1}{\cos^2x}$, and so $\frac{1}{\cos^2x} = 1+(12/5)^2$. From here, you can solve for $\cos x$ and $\sin x$. – trang1618 Sep 20 '16 at 21:10
All you need to do is to convert your $z$ to polar form:
$$z=5-12i=\sqrt{12^2+5^2}e^{i(\arctan(-\frac{12}{5})+k\pi)}=13e^{i(-1.176+k\pi)}$$ Hence, one answer of $z^{\frac{1}{2}}$ is $$z_1=13^{\frac{1}{2}}e^{i\frac{-1.176}{2}}=\sqrt{13}e^{i(-0.588)}$$
Converting this to Cartesian will give: $$z_1=\sqrt{13}\left(\cos(-0.588)+i\sin(-0.588)\right)=3-2i$$ which is one of your original results.
You can apply de Moivre's theorem (second example) on the first example without doubt. Though you need a calculator or a trigonometrical table handy to check the sin and cos of a certain angle which feels kinda painful, at least for me.
Your method should work fine in both cases (though the trigonometry is a bit more complicated), but I point out that your first method can still work on the square roots of $2i$: We see that
$$a^2-b^2 = 0$$
so $a = \pm b$, and then, secondly, $2ab = 2$. This gives us $a = b = \pm 1$ as the solutions, so the square roots of $2i$ are $1+i$ and $-1-i$. | 2021-05-17T15:22:19 | {
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https://math.stackexchange.com/questions/481787/let-b-subseteq-a-subseteq-mathbbrn-show-that-b-is-closed-relative-to | # Let $B \subseteq A \subseteq \mathbb{R}^n$. Show that $B$ is closed relative to $A$ iff $B = A \cap C$ for some set $C$ closed in $\mathbb{R}^n$.
Let $B \subseteq A \subseteq \mathbb{R}^n$. Show that $B$ is closed relative to $A$ iff $B = A \cap C$ for some set $C$ closed in $\mathbb{R}^n$.
My solution:
We know $B$ is closed relative to $A$ if $\forall x \in A \setminus B$, $\exists$ a neighborhood $N$ such that $N \cap B = \varnothing$. So, for each $x \in B \setminus A$, we have a nghbd $N_x = D^n(x,r_x) \cap A$ such that $N_x \cap B = \varnothing$. Put $O = \bigcup_{x \in A \setminus B} D^n(x,r_x)$. Obviously $O$ is open since it is a union of open discs. So $C = \mathbb{R}^n \setminus O$ must be closed. Now, since $B$ is closed in $A$, then $A \setminus B$ must be open in $A$, therefore $A \setminus B = O \cap A$. I have here a question since I cannot see how to conclude from here that $B = C \cap A$. Can someone help see it?
the other direction: Say $B = C \cap A$ for some $C$ closed in $\mathbb{R}^n$. Therefore, every $x \in A \setminus B$ must satisfy $x \notin C$. Since $C$ is closed, $\exists$ a disc $D^n(x,r)$ such that $x \in D^n(x,r)$ and $D^n(x,r) \cap C = \varnothing$. Put $N = D^n(x,r) \cap A$. Then, $x \in N$ and $N \cap B \subseteq N \cap C$. Therefore, $B$ is closed in $A$.
Is this proof correct? any feedback? thanks
You’re doing fine up through the point at which you define $C$. At that point, however, you’re not entitled to conclude that $A\setminus B=O\cap A$; the fact that $A\setminus B$ is open in $A$ tells you that it’s equal to $U\cap A$ for some open $U$ in $\Bbb R^n$, but you’ve not actually shown that we can take this $U$ to be $O$.
It’s not hard to do so, however. Recall that $O=\bigcup_{x\in A\setminus B}N_x$, where $x\in N_x$ and $N_x\cap B=\varnothing$. On the one hand for each $x\in A\setminus B$ we have $x\in N_x\subseteq O$, so $A\setminus B\subseteq O$; and since $A\setminus B\subseteq A$, we actually have $A\setminus B\subseteq O\cap A$. On the other hand, for each $x\in A\setminus B$ we have $N_x\cap A\subseteq A\setminus B$, so that $$O\cap A=\left(\bigcup_{x\in A\setminus B}N_x\right)\cap A=\bigcup_{x\in A\setminus B}(N_x\cap A)\subseteq A\setminus B\;.$$ Thus, $O\cap A$ is indeed $A\setminus B$: your conclusion was premature, but none the less correct.
From here it’s just a bit of set algebra to see that $B=C\cap A$:
$$C\cap A=(\Bbb R^n\setminus O)\cap A=A\setminus O=A\setminus(O\cap A)=A\setminus(A\setminus B)=B\;.$$
You should verify each of those steps, since that kind of manipulation can be quite useful. However, the actual idea involved here is much simpler: $O$ and $C$ are complementary subsets of $\Bbb R^n$, so for any $X\subseteq\Bbb R^n$, $O\cap X$ and $C\cap X$ must be complementary subsets of $X$. Every point of $\Bbb R^n$ is in exactly one of $O$ and $C$, so clearly every point of $X$ is in exactly one of $O\cap X$ and $C\cap X$. Since $O\cap A$ is $A\setminus B$, $C\cap A$ must be all the rest of $A$, which is simply $B$.
• @Citizen: You’re very welcome! – Brian M. Scott Sep 2 '13 at 3:45
I think your second proof is correct.
As to the first, you only need to notice that, if $x\in(C\cap A),$ then $x$ cannot belong to $O$. That is, $x\in A\backslash A\cap O=A\backslash(A\backslash B)=B,$ hence $C\cap A\subset B.$ The other direction is similar: $B\subset A\backslash(A\backslash B)=A\backslash(A\cap O)=A\backslash O=A\cap C.$
Hope this helps. | 2019-10-21T05:36:21 | {
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https://math.stackexchange.com/questions/3050265/which-matrices-can-be-realized-as-second-derivatives-of-orthogonal-paths | Which matrices can be realized as second derivatives of orthogonal paths?
I am interested to know which real matrices $$A \in M_n$$ can be realized as second derivatives of paths in $$\text{SO}_n$$ starting at the identity. That is, for which matrices $$A$$, there exist a smooth path $$\alpha:(-\epsilon,\epsilon) \to \text{SO}_n$$, such that $$\alpha(0)=Id$$ and $$\ddot \alpha(0)=A$$. We denote the space of realizable matrices by $$D$$.
Question: I prove below that $$(\skew)^2 \subseteq D \subseteq (\skew)^2+\skew$$. Does $$D=(\skew)^2+\skew$$ always hold?
Comment: Note that $$(\skew)^2+\skew \subsetneq M_n$$, at least for odd $$n$$: In that case every skew-symmetric matrix is singular, so $$(\skew)^2 \subseteq \sym$$ consists only of singular matrices, hence does not contain all symmetric matrices.
Edit: I proved below that equality holds in dimension $$n=2$$.
Proof of $$(\skew)^2 \subseteq D \subseteq (\skew)^2+\skew$$:
1. Every square of skew-symmetric matrix can be realized: For skew $$B$$, take $$\alpha(t)=e^{tB}$$. Then, $$\dot \alpha(t)=Be^{tB}$$, $$\ddot \alpha(t)=B^2e^{tB}$$.
2. The space of realizable matrices is contained in $$(\skew)^2+\skew$$: Indeed, since $$\dot \alpha(t) \in T_{\alpha(t)}\SO=\alpha(t)\skew$$, we have $$\dot\alpha(t)=\alpha(t)B(t)$$ for some $$B(t) \in \skew$$, so
$$\ddot \alpha(t)=\dot \alpha(t) B(t)+\alpha(t) \dot B(t)$$ hence $$\ddot \alpha(0)=\dot \alpha(0) B(0)+ \dot B(0)= B(0)^2+\dot B(0) \in (\skew)^2 +\skew,$$ where the last equality followed from $$\dot \alpha(0)=B(0)$$ (put $$t=0$$ in $$\dot\alpha(t)=\alpha(t)B(t)$$).
Edit 2: When trying to show the converse direction, I hit a wall: we need to show that there exist solutions $$\dot\alpha(t)=\alpha(t)B(t)$$, where $$\alpha(t) \in \SO,B(t) \in \skew$$, with arbitrary $$B(0),\dot B(0) \in \skew$$. A naive attempt would be to define $$\alpha(t)=e^{\int_0^t B(s)ds}$$ for $$B(s)=B(0)+s\dot B(0)$$. However, it is not true in general that $$\alpha'(t)=\alpha(t)B(t)$$; this happens if $$B(t)$$, $$\int_0^t B(s)ds$$ commute, which happens if and only if $$B(0),\dot B(0)$$ commute.
Proof $$D = (\skew)^2+\skew$$ for $$n=2$$:
$$\alpha(t)$$ can always be written as $$\alpha(t)=\begin{pmatrix} c(\phi(t)) & s(\phi(t)) \\\ -s(\phi(t)) & c(\phi(t)) \end{pmatrix}$$, where $$c(x)=\cos x,s(x)=\sin x$$, and $$\phi(t)$$ is some parametrization satisfying $$\phi(0)=0$$.
Differentiating $$\alpha(t)$$ twice, we get
$$\ddot \alpha(t)=-(\phi'(t))^2\alpha(t)+\phi''(t)\begin{pmatrix} -s(\phi(t)) & c(\phi(t)) \\\ -c(\phi(t)) & -s(\phi(t)) \end{pmatrix},$$
so
$$\ddot \alpha(0)=-(\phi'(0))^2Id+\phi''(0)\begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}.$$
Since we can choose $$\phi'(0),\phi''(0)$$ as we wish, we conclude that $$D=\mathbb{R}_{\le 0}Id+\mathbb{R}\begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}=\mathbb{R}_{\le 0}Id+\skew.$$ Since $$\skew=\text{span} \{ \begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}\}$$, and $$\begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix}^2=-Id$$, we have $$\skew^2=\mathbb{R}_{\le 0}Id$$, so indeed $$D=(\skew)^2+\skew$$.
Yes. Given skew-symmetric matrices $$B$$ and $$C,$$ define $$\alpha(t)=\exp(Bt+\tfrac12 Ct^2).$$ Then \begin{align} \alpha(t) &=I+(Bt+\tfrac12 Ct^2)+\tfrac12 (Bt+\tfrac12 Ct^2)^2+O(t^3)\\ &=I+Bt+\tfrac12 (B^2+C)t^2+O(t^3) \end{align} as $$t\to 0.$$ This shows that $$\ddot \alpha(0)=B^2+C.$$
I believe that the construction you're looking for is called a Dyson Series (Wikipedia). In detail, suppose we are given $$B,C \in \mathrm{Skew}(n)$$ and we want to construct a $$\gamma: (-\varepsilon,\varepsilon) \to \mathrm{SO}(n)$$ such that $$\gamma(0)=1$$ and $$\ddot{\gamma}(0)=B^2+C$$. I claim that \begin{align*} \gamma(t) & :=\sum_{n=0}^{\infty} \left[\int_0^t \int_0^{t_0} \cdots \int_0^{t_{n-1}} \left(\prod_{k=0}^n (B+t_{n-k} C) \right) \mathrm{d} t_n \cdots \mathrm{d} t_0\right] \\ & = 1 + \int_0^t (B+t_0C) \mathrm{d}t_0 + \int_0^t \int_0^{t_0} (B+t_1C)(B+t_0C) \mathrm{d} t_1 \mathrm{d}t_0 + \\ & \hspace{1cm}\int_0^t \int_0^{t_0} \int_0^{t_1} (B+t_2C)(B+t_1C)(B+t_0C) \mathrm{d}t_2 \mathrm{d}t_1 \mathrm{d}t_0 + \cdots \end{align*} is a well-defined solution to the problem. Indeed, if we let $$m:=\max_{s \in [0,t]} \lVert B+sC \rVert_{L^2},$$ then we have that $$\lVert \gamma(t) \rVert_{L^2} \leq e^m,$$ so that $$\gamma$$ is defined by a convergent sequence. Moreover, we can compute that $$\dot{\gamma}(t)=\gamma(t)(B+tC)$$, evincing both that the image of $$\gamma$$ (which a priori lies in the space of $$n \times n$$ matrices) in fact lies in $$\mathrm{SO}(n)$$ and also that $$\ddot{\gamma}(0)=B^2+C$$.
• Thanks, this is very interesting. However, I am not sure how do you deduce that $\gamma(t) \in SO$, even for sufficiently small $t$; I tried showing that the derivative of $\gamma(t)^T\gamma(t)$ is zero, but I got stuck: Since $\dot \gamma(t)=\gamma(t)D(t)$. where $D(t)$ is skew-symmetric, we have $\frac{d}{dt}(\gamma(t)^T\gamma(t))=(\dot \gamma(t))^T\gamma(t)+\gamma(t)^T\dot \gamma(t)=(\gamma(t)D(t))^T\gamma(t)+\gamma(t)^T\gamma(t)D(t)=-D(t)\gamma(t)^T\gamma(t)+\gamma(t)^T\gamma(t)D(t)$ which is zero if and only if $\gamma(t)^T\gamma(t)D(t)=D(t)\gamma(t)^T\gamma(t)$... – Asaf Shachar Dec 26 '18 at 12:47
• , i.e. $\gamma(t)^T\gamma(t)$ and $D(t)$ commute. In our case $D(t)=B+tC$, and I don't see an immediate reason why this commutativity should hold. – Asaf Shachar Dec 26 '18 at 12:47
• You're right, of course -- it would seem I miscalculated as to how easily it would follow that $\mathrm{im}(\gamma(t)) \subseteq \mathrm{SO}(n)$. I think some argument like this might work (though i haven't thought out the details). Notice that $\gamma(t)^T \gamma(t)$ is analytic and that its derivatives vanish to every order at the origin. It follows that $\gamma(t)^T\gamma(t)$ is constant, as needed. Do you think that might work? I'll hopefully have more time to think about this later, in any event. – Or Eisenberg Dec 26 '18 at 21:47
• Asaf: Thanks for the reference about foliations -- this was my original intuition for why $\mathrm{im}(\gamma) \subset \mathrm{SO}(n)$, but I certainly didn't have enough intuition to explain it as well as Mike Miller did! Regarding your question of how I thought of the Dyson series, this question is verging on philosophical, I think. Indeed, where do any math ideas come form? =). I was just trying to solve the equation $\dot{\gamma}=\gamma D$ and remembered vaguely that the physicists had figured out how to do this, so I started googling. Thanks for the interesting problem! – Or Eisenberg Dec 27 '18 at 17:20 | 2019-07-19T14:15:33 | {
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http://math.stackexchange.com/questions/533792/how-do-i-solve-this-recurrence-equation-using-substitution/533803 | How do I solve this recurrence equation using substitution?
f(1)=1, and f(n) = f(n-1)+2(n-1)
Using substitution, here are the first few steps: f(n-1) = f((n-1)-1) + 2((n-1)-1)
f(n-1-1) = f((n-1-1)-1-1) + 2((n-1-1)-1-1)
And then eventually I see that f(n+(-1)*2^j) = f(n+(-1)*2^(j+1)) + 2n + 2(-1)*2^(j+1), where j is an increasing integer >=1
What do I do now? It looks like 2(-1)*2^(j+1) will diverge to negative infinity...
The answer is f(n) = (n-1)n + 1 (using wolfram) but i have no idea what they did..
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migrated from stackoverflow.comOct 20 '13 at 22:27
This question came from our site for professional and enthusiast programmers.
I never learned the "triangle number formula" and I'm supposed to use substitution, but thanks for pointing me to the math section. I hope I get more responses there, though this was part of a compsci assignment. – Gary Choi Oct 20 '13 at 22:25
Don’t repost, I’ll move it – Ryan O'Hara Oct 20 '13 at 22:26
This is how it expands:
$$f(n) = f(n - 1) + 2(n - 1)$$ $$f(n) = f(n - 2) + 2(n - 2) + 2(n - 1)$$
And you can see that we’ll get to $f(1) = 1$ eventually, and that will be when $n - a = 1$. So what this is actually saying is:
$$f(n) = 1 + 2(n - (n - 2)) + … + 2(n - 1)$$
i.e.
$$f(n) = 1 + 2(1) + … + 2(n - 1)$$
So one plus twice the sum of $1$ to $n - 1$ inclusive, which is one plus twice the $n-1$th triangle number. The nth triangle number is $\dfrac{n^2 + n}2$, so the whole thing is:
$$1 + 2\left(\dfrac{(n-1)^2+(n-1)}{2}\right)$$ $$1 + (n - 1)^2 + n - 1$$ $$(n - 1)^2 + n$$ $$n^2 - 2n + 1 + n$$ $$n^2 + 1 - n$$ $$(n - 1)n + 1$$
… which is indeed what Wolfram got you!
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Oh! The problem was I wasn't directly substituting back the left hand side of the equation (second step) as f(n) (instead continuing as f(n-1) which made it hard to see the pattern. Thanks! – Gary Choi Oct 20 '13 at 22:42
Here’s the substitution reduction:
\begin{align*} f(n)&=f(n-1)+2(n-1)\\ &=f(n-2)+2(n-2)+2(n-1)\\ &=f(n-3)+2(n-3)+2(n-2)+2(n-1)\\ &\;\vdots\\ &=f(n-k)+2(n-k)+2\big(n-(k-1)\big)+\ldots+2(n-2)+2(n-1)\\ &\;\vdots\\ &=f(1)+2\big(n-(n-1)\big)+2\big(n-(n-2)\big)+\ldots+2(n-2)+2(n-1)\\ &=1+\sum_{k=1}^{n-1}2k\\ &=1+2\sum_{k=1}^{n-1}k\\ &\overset{*}=1+2\left(\frac{n(n-1)}2\right)\\ &=1+n(n-1)\\ &=n^2-n+1\;. \end{align*}
The formula for the sum of the first $n$ positive integers that I used at the starred step is a common special case of the more general formula for the sum of a finite arithmetic progression, one that’s worth knowing in its own right.
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Thank you! I didn't do the second step correctly T_T – Gary Choi Oct 20 '13 at 22:44
@Gary: You’re welcome! (It sometimes helps to write out a few lines using a specific value of $n$, like $n=5$; the pattern may not be quite so apparent, but you’re also less likely to make purely symbolic errors.) – Brian M. Scott Oct 20 '13 at 22:47
@minitech: I fixed everything. $\sum_{k=1}^nk$ is $\binom{n+1}2$, not $\binom{n}2$, so $\sum_{k=1}^{n-1}=\binom{n}2=\frac{n(n-1)}2$. – Brian M. Scott Oct 20 '13 at 22:51 | 2016-07-29T07:56:40 | {
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https://math.stackexchange.com/questions/4137571/proving-that-a2x2b2a2-c2xb2-0-has-no-real-roots-if-abc-and-a | # Proving that $a^2x^2+(b^2+a^2-c^2)x+b^2=0$ has no real roots if $a+b>c$ and $|a-b|<c$ for real $a$, $b$, $c$
How is this answer derived, and is there any other way to prove it?
Prove that $$a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0$$ does not have real roots if $$a+b>c$$ and $$|a-b|. $$a,b,c \in \mathbb{R}$$.
Solution I found online:
The discriminant is given as: $$D = (b^2 + a^2 - c^2)x - 4a^2b^2$$
$$(a^2 + b^2)^2 - 2c^2(a^2 + b^2) + c^4 - 4a^2b^2$$
Let $$c^2= t$$.
You get : $$t^2 - 2t(a^2 +b^2) + (a^2 + b^2)^2$$ has to be less than zero since $$D < 0$$ means nonreal roots. Then, you get $$(a-b)^2 < t < (a+b)^2.$$ Hence, you get the condition satisfied.
My questions regarding this proof :
It is right to say that $$a,b,c$$ have to be some particular numbers only. Only one type. By here, what I mean to say is that let $$a=5$$, $$b =6$$ or $$c=7$$ or another set of numbers , $$a=6$$ , $$b=7$$ , $$c=4$$. Now , according to the condition, it may or may not be possible that one of the conditional value satisfies the condition since $$a , b$$ and $$c$$ can have infinite values, right? There can be infinite quadratic equations but not all will satisfy this condition of our question.
If not, can we find out whatever those numbers value can be and which are those numbers which satisfy the condition and how many types of this condition is possible. I think it is impossible since that can be infinite.
I don’t think also that it is true that there will be only two conditions. If I know the values of each variable , I can create a few more conditions. Here , I mean to say that if I have a question with all of its values in numbers except $$x$$. Like $$a, b, c$$ has a numerical value. You don’t need any other value. Making condition like $$a+b>c$$. Who knows ? I can also create another condition like $$a*b*c>0$$. Right ? Then, our answer will not be the same if this condition was given as an alternative to some other one.
How did you know that in the end you would get this result only ? This answer is like a luck. I am not satisfied with this way of deriving. I am looking for alternate solutions of the above question or id you could help me to understand this proof as a right one.
• In the question, did you mean if in the beginning? @Srijan M.T. – Buraian May 13 at 13:55
• @Buraian Done edit. – Srijan M.T May 13 at 13:56
• Oh you meant question is, you don't have to write what the question is; It is is implied by being block quotation. Secondl,y I suggest not using abbreviation for more positive reception of the question by other site members – Buraian May 13 at 13:57
• I have edited the question quite a bit, do you think it improved presentation? Please check if everything is fine @Srijan M.T. – Buraian May 13 at 14:03
• @Buraian Much better now. – Srijan M.T May 13 at 14:08
I'll answer the paragraphs first, then the question.
## Paragraphs
It is right ... condition of our question.
Of course, that's correct. Not all $$a,b,c$$ will satisfy this equation. The reason why the statement is useful is that some $$a,b,c$$ do satisfy the statement (extreme example : $$a = 10^{10} , b = 10^{10}+1 , c=2$$)!
If not ... can be infinite.
Yes, of course we can find the numbers satisfying the condition (more on this later, so I'll keep the suspense, Teresa style) and there are, in fact , infinitely many of them. However, it's fair to say that the infiniteness means that you can't take a case-by-case approach, there are infinitely many cases! So you'll have to contend with the algebraic approach.
I don’t think ... some other one.
Indeed, if the conditions change, then the situation changes as well : so you change your condition to $$abc>0$$ or something, then you are probably guaranteed some other properties, but probably NOT this condition of having no real roots.
How did ... proof as a right one.
The proof does seem to come a little out of thin air , but some part of it was a little more logical than the rest, namely taking the discriminant. I'll explain that.
## Question
Yes, the online approach works, and it's quite simple :
• No real roots happens if and only if the discriminant is negative.
• Use all the algebra we know and show that the discriminant is negative.
But how do we justify the algebra? Well, let's look at $$a+b>c$$ and $$|a-b|. The common thing in these conditions is that combining them, $$|a-b|.
So, when we look at the discriminant $$(b^2+a^2-c^2)^2 - 4b^2a^2$$, the important thing to observe is that isolating $$c$$ will be giving us the best chance of working out an inequality : separate all the $$c$$s, substitute the inequalities , and see if things work out.
To do that, obviously one wants to expand : $$(b^2+a^2-c^2)^2 = (\color{blue}{(b^2+a^2)} - \color{green}{c^2})^2$$ and then use the formula for two variables. Expanding that, one gets : $$(b^2+a^2)^2 + c^4 - 2c^2(a^2+b^2) - 4a^2b^2$$, and remember we are focusing on $$c$$ as the variable we feel can be controlled the strongest. It's quite easy to see that the above expression is a quadratic in $$c^2$$, so if $$t = c^2$$ then the expression is $$((b^2+a^2)^2 - 4a^2b^2) + t^2 - 2t(a^2+b^2)$$ which becomes, using the identity $$(x+y)^2 - 4xy = (x-y)^2$$, as $$t^2 - 2t(a^2+b^2) + (a^2-b^2)^2$$
which doesn't match with your attempt, but now there's actually a factorization available using the fact that $$2(a^2+b^2) = (a+b)^2 + (a-b)^2$$ and difference of squares (that $$(a^2-b^2) = (a-b)(a+b)$$). More precisely, we have : \begin{align} & t^2 - 2t(a^2+b^2) + (a^2-b^2)^2 \\ =& t^2 - t ((a+b)^2 + (a-b)^2) + (a+b)^2(a-b)^2 \\ &= (t - (a+b)^2) (t - (a-b)^2) \end{align}
Now we have the product of two terms, and NOW it's clear why we have the conditions at the start of the question : $$a+b>c$$ implies that $$(t- (a+b)^2)<0$$, and $$|b-a| implies that $$(t - (a-b)^2)> 0$$. So one of the terms in the product is positive, the other is negative. So the discriminant is negative, and we are done.
## Further introspection
Further introspection, however, is necessary for a couple of reasons. The method above is nice, but not always the most efficient to think about.
An alternate method comes when you think of geometry.
NOTE : This method is more intuitive and involves simpler computation than the other method , but it's slightly longer and goes through a lot of little steps which are by themselves VERY interesting.
Claim : The set of all $$a,b,c$$ such that $$|a-b| and $$a+b>c$$ is precisely the set of of all positive real triples, for which one can form a triangle with those lengths.
Proof : Suppose $$|a-b| and $$a+b>c$$. Then $$c>0$$ is clear, since $$c>|a-b| \geq 0$$. From $$a+b>c$$ we know at least one of $$a,b >0$$ must be true. Let's say $$b>0$$. Then, $$a\leq 0$$ cannot happen : put things on a number line and see why $$|b-a| is violated if $$a+b>c$$ but $$a \leq 0$$. So , in fact $$a > 0$$ must happen as well. You can do something similar if $$a>0$$ instead of $$b$$.
So, $$a,b,c$$ are all positive. $$a+b>c$$ is true, and from $$|a-b| we know from the triangle inequality that $$a and $$b as well. These are precisely the conditions for triangle formation. The converse is easily seen to be true. $$\blacksquare$$
So, we have a geometric interpretation for our domain of interest. Basically, if $$a,b,c$$ satisfy the conditions, then we can imagine a triangle with side lengths $$a,b,c$$, and with all angles between $$0$$ and $$180$$ degrees (so no obsolete triangles). Let's call the angles opposite $$a,b,c$$ as $$A,B,C$$ respectively . What about our equation looks geometric?
Recall the cosine rule : $$\cos C = \frac{b^2+a^2 - c^2}{2ab}$$. This very useful looking equation immediately suggests to us that we should look to geometrically interpret the equation by performing this substitution : $$a^2x^2 + (b^2+a^2-c^2)x + b^2 = 0 \to a^2x^2 + 2ab\cos C x + b^2 = 0$$
we write this to make things more illustrative , as : $$(ax)^2 + 2 \cos C (ax)b + b^2 = 0$$
(NOTE : If you want to see an easier proof, then see the end of the answer, but what follows is my first proof)
and now we perform the completion of square procedure, which basically realizes that you have $$(ax)^2$$ and $$b^2$$ so all you need to do is subtract and add $$2axb$$. Once you do that, you get : $$(ax)^2 + b^2 + 2 \cos C(ax)b = [(ax)^2 + 2axb + b^2] + [-2axb + 2\cos C(ax)b] = 0$$
both $$[\cdot]$$ brackets factorize, giving : $$(ax+b)^2 + 2axb(-1+\cos C) = 0 \to (ax+b)^2 = (2ab(1-\cos C))x$$
following transposition. I claim this cannot have a real solution. Let's see why. First recognize that $$1-\cos C \geq 0$$ for all values of $$C$$.
Suppose that $$x < 0$$, the easy case. Then, we know that $$2ab(1-\cos C)x \leq 0$$, while $$(ax+b)^2 \geq 0$$. So the only way that equality can happen is if $$(ax+b)^2 = 0 = 2ab(1-\cos C)x$$. Which implies that $$1-\cos C = 0$$ , but then $$\cos C = 1$$ implies that the angle at $$C$$ is 0 degrees, which is a contradiction.
Suppose that $$x > 0$$. Then note that $$(ax-b)^2 \geq 0$$ so rearranging, $$(ax)^2+b^2 \geq 2abx \geq 2ab\cos C x$$ because $$x > 0$$ and $$1 \geq \cos C$$. If equality occurs, then we must have $$\cos C = 1$$ so that angle $$C = 0$$, again a contradiction.
Clearly $$x=0$$ is not a root. Hence we are done.
To summarize : the fact that $$a,b,c$$ form a triangle restricts $$b^2+a^2-c^2$$ in a fashion we can see geometrically, since $$\cos C$$ expresses the relation between this quantity and $$a$$ and $$b$$.
Thus, we used :
• Geometric interpretation of the domain
• The cosine rule
• Completion of squares, and the :
• Creation of cases by splitting the sign of $$x$$
to solve this problem without using any kind of supremely complicated mathematics : no discriminant, no awkward multiplications, and most importantly, a proof that went through a completely different domain of mathematics.
I'm not sure that in a competitive exam, one can chalk out an approach like the geometric one I mention (which, extremely unfortunately, means that the online approach is in fact the correct approach to be taken, and one needs to be good at algebra for seeing how it works). Therefore, my approach is being placed only to intuit the source of the question from a geometric point of view, and solve it from the same point of view. I hope this was helpful as an alternate view of the same situation.
EDIT : From the hint given by dxiv below (which I thank them for very much!) the question becomes much, much easier. The hint asks us to complete the square, but using a different pair of terms. Indeed, we try to complete the square , but not using $$(ax)^2$$ and $$b^2$$, but instead, $$(ax)^2$$ and $$2abx\cos C$$ : \begin{align} (ax)^2 + 2ab(\cos C)x + b^2 & = [(ax)^2 + 2ab(\cos C)x + b^2(\cos^2 C)] + [b^2 - b^2\cos^2 C] \\ & = [(ax + b \cos C)^2] + [b^2(1-\cos^2 C)] \\ & = \color{blue}{(ax+b \cos C)^2 + (b \sin C)^2} \end{align}
is a sum of squares! That's remarkable, because a sum of squares is always non-negative.
In particular, if $$(ax+b\cos C)^2 + (b \sin C)^2 =0$$ then we MUST have that $$ax+b \cos C = 0$$ and $$b \sin C = 0$$. But the second part implies that $$\sin C = 0$$ since $$b>0$$, so $$C = 0$$, a contradiction!
Therefore, the equation can have no real root. So, you can see that trigonometric manipulation actually gets us much, much farther than in my first attempt.
• +1 for the "further" part. You could also take another shortcut with $(ax)^2 + 2 ax\,b \cos C + b^2$ $= (ax+b\cos C)^2 + b^2 \sin^2C$. – dxiv May 14 at 8:09
• @dxiv Thanks, I wanted to provide an alternate approach which I'm really happy I did. Your answer explains the online solution correctly as well, thanks. – Teresa Lisbon May 14 at 12:20
The line of proof works, though the proof has a typo as quoted, and it may not be written in the easiest way to follow.
The discriminant is given as: $$D = (b^2 + a^2 - c^2)x - 4a^2b^2$$
Here is the typo, and the discriminant can not depend on $$x$$.
The expression above is wrong (though what's used afterwards appears to be based on the correct expression). This was supposed to be $$D = (b^2 + a^2 - c^2)^2 - 4a^2b^2$$.
Prove that $$a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0$$ does not have real roots if $$a+b>c$$ and $$|a-b|. $$a,b,c \in \mathbb{R}$$.
What is to be proved is the one-way implication: if $$\;a+b>c\;$$ and $$\;\mid a-b\mid then the equation $$\;a^2x^2+ (b^2 + a^2 - c^2)x + b^2 = 0\;$$ has no real roots.
Without repeating the calculations in detail, the proof goes through the following steps.
• The absolute value of a number can never be negative, so $$\;0 \le \mid a-b\mid $$\require{bbox}\bbox[5px, border: 1px solid black]{c \gt 0}\,$$.
• Inequalities between positive numbers can be multiplied, so $$\;\mid a-b\mid $$\implies \mid a-b\mid^2 $$\implies \bbox[5px, border: 1px solid black]{(a-b)^2 \lt c^2}\,$$, since the square of a real number is equal to the square of its absolute value.
• Given $$a+b\gt c$$ and since $$c \gt 0$$ the inequality can be squared, so $$\;\bbox[5px, border: 1px solid black]{(a+b)^2 \gt c^2}\;$$.
• The discriminant is the difference of two squares, which can be factored as $$D=\left(\left(a+b\right)^2-c^2\right)\left(\left(a-b\right)^2-c^2\right)\,$$. It follows from the previous inequalities that the first factor is positive and the second one is negative, so $$\;\bbox[5px, border: 1px solid black]{D \lt 0}\;$$.
A quadratic with negative discriminant does not have real roots, which concludes the proof. | 2021-07-26T02:10:16 | {
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http://mathscinotes.com/2010/11/straight-level-and-the-curvature-of-the-earth/ | # Straight, Level, and the Curvature of the Earth
## Introduction
I frequently hear people make statements that somehow do not seem right. Today I heard a good example. During a discussion of laser levels, the topic of required accuracy came up. I heard a contractor make the following statement:
Who cares if the laser level is accurate within an 1/16th of inch at 100 feet? The Earth curves away from a horizontal line by 1/8th of inch for every 100 feet of horizontal distance. The Earth's curvature will swamp out your instrument error in less than 100 feet.
The statement about the curvature of Earth got me thinking. How much does the Earth's surface deviate from a horizontal line over a distance of 100 feet? The contractor's number intuitively seemed wrong because the Earth is round and the deviation from horizontal should be a function of distance. A little math will give me the answer. For consistency's sake, I will perform all computations in US customary units.
## Analysis
Figure 1 illustrates the situation and contains the derivation of both an exact and a approximate solution. The triangle formed by x, R + δ, and R is a right triangle, which means that the Pythagorean theorem can be used to produce an exact solution. In addition, a simple approximation for δ is also developed assuming R >> x and using a linear approximation for the square root. In Appendix A, I give examples of the computations in Mathcad.
Figure 1: Calculation of Deviation from Horizontal.
Given the situation shown in Figure 1, we can compute the deviation from horizontal as follows.
Eq. 1 $\delta=\sqrt {{R^2}+{x^2}}-R\quad=\quad 2.389{\text{E-4 ft}}={\text{2.867E-3 in}}$
where
• R is the radius of the Earth (3963.2 miles)
• x is the horizontal distance of interest (100 ft)
## Conclusion
The contractor had stated that the curvature of the Earth causes level to deviate from horizontal by an 1/8th of an inch (125 thousandths of inch) for 100 feet of horizontal distance. The actual deviation is ~2.9 thousandths of an inch for 100 feet of horizontal distance, which is almost 45 times less than the contractor claimed. So it is meaningful to buy a laser level that is accurate to 1/16th of an inch over 100 feet, i.e. the laser level error is not swamped by the curvature of the Earth.
Why did the contractor make the claim that the Earth's curvature is 1/8th inch over 100 feet? He made a simple mistake. He did not understand that the deviation from horizontal for short distances is given by a square-law relationship, shown in Equation 2. In Equation 2, I include an approximation that is only valid when R is much greater than x, which is true in typical construction problems.
Eq. 2 $\delta = \sqrt {{R^2} + {x^2}} - R \doteq R \cdot \left( {1 + \frac{{{x^2}}}{{2 \cdot {R^2}}}} \right) - R = \frac{{{x^2}}}{{2 \cdot R}}$
If we use Equation 1 or Equation 2 to calculate the deviation from horizontal at 1 mile, we get 8 inches. This value is quoted in a number of references on surveying (e.g. here is one, here is another). What the contractor did was erroneously assume that the deviation varied linearly with distance, which would mean that a deviation of 8 inches at 1 mile is equivalent to an 1/8th of inch at 100 feet.
For those of you who may be interested in the related question of the error in horizontal distances caused by living on a spherical planet, see this blog post.
Aside: Here is an interesting discussion that references this web page.
## Appendix A: Computation Examples
Figure 2 shows a few computation examples. Normally, I let Mathcad do the unit conversion, but I do show one example with explicit unit conversion.
Figure 2: Computation Examples.
This entry was posted in Construction. Bookmark the permalink.
### 30 Responses to Straight, Level, and the Curvature of the Earth
1. predatflaps says:
hello, I was confused by your quotient because you state the distance in kilometres and the elevation in feet, and then at the end you quote the distance in miles and the elevation in inches. this is crazy and confusing! Using 4 units of measurement with only one equation, when there should only be one. it doesn't say anything about the conversion in the equation, so I figure it could be metric, then when I use metric in the calculation, the result is different from what you say. I will actually have to look on another website! for the solution! Cheers! happy Christmas!
• mathscinotes says:
Dimensional analysis can help you keep things straight. I have added a small appendix that illustrates the calculations. Normally, I let my computer algebra system do the unit conversions for me, so I never actually worry about them. In the appendix, I illustrate how to perform the unit conversion. I hope that helps.
Merry Christmas.
Mathscinotes
2. Jim says:
I like your web page but I have a question. Since it's been over 40 years since I took algebra could you explain equation 2? When I use smaller numbers, like 3 for x, 4 for r I get different answers between the first step & the second step. 1 for the first step & 1.25 for the second step. Thanks
• mathscinotes says:
My article was not clear on the approximation and I have added some explanatory text. The approximation is only valid when R is much much bigger than x. This is true when R is the radius of the Earth and x is a typical construction distance. The approximation fails when when R and x are similar in size.
Hopefully, my clarification in the text will help. I am so used to doing these sorts of approximations that I do not even notice when I do them -- very bad habit.
Mathscinotes
3. Hanna says:
Hi, Thanks for your calculation for the vertical error. I am however interested in figuring out how much the distance between two points vary with curvature of earth (as opposed to just using trigonometry). Could you help me in figuring this calculation out? Thanks
• mathscinotes says:
Hi Hanna,
I would like to try to help, but I do not completely understand your question. Are you trying to determine the distance between two points on the Earth's surface? If so, I can help. Generally distances on the Earth's surface are expressed in terms of the great circle distance. There are numerous web sites that address this calculation, like this one. If you are interested in a proof of the formula used on this web page, see this reference. If you decide that you need an Excel version of the calculation and are having trouble putting it together, I can pull something together in less than a minute.
If you have a different question, just give me a bit more detailed version of your question and I will try to help.
mathscinotes
• Hanna says:
Hi. For some reason I never saw your reply. Thanks for getting back to me. I am observing animals from a 93-96m high sea cliff (tidal variation) with a theodolite. I get angles from reference point that allow me to calculate the (diagonal) distance of the animal in the water from the theodolite on the cliff top. As I know the height at that moment, I can easily calculate the horizontal distance from the theodolite to the animal. However I would like to know what is the actual distance of the animal from the foot of the cliff, if the curvature of the earth is taken into account. I have tried using the moveable typescript calculations but am not sure if I can do it correctly. Assuming the angle from is 45degrees
• mathscinotes says:
Hi Hanna,
Let's take a quick look and see if I am thinking of your problem correctly. To keep things very simple, let's assume that your theodolite is 100 m above the ocean and you are observing something in the water at a 45 °C angle. Using "flat Earth" methods, your subject is 100 m away -- correct? Let's work the same problem and determine the arc length of your subject from a point directly below your theodolite.
Here is my drawing that illustrates this situation.
I can convert this situation into a geometry problem as shown in the following figure.
I can analyze this geometry as shown below.
As you would expect, the arc distance is not much different than the distance you would compute using simple angles.
Is this what you were looking for? I can put it into Excel if you want to play with it.
Mathscinotes
4. Pingback: Whale Math | Math Encounters Blog
5. hotblack.eu says:
Je peux vous dire que c'est continuellement du bonheur de passer sur
votre blog
6. Hello:
I am trying to check and see if my math is correct. I found your article and wanted to see if I was on the right track.
I am trying to figure out the curvature of the earth based on the distance...
Thanks
James
Miles Squared X 8 Inches= Inches Of Curvature
6 Miles X 6 (Squared) = 36 x 8 = 288 Inches
288 / 12 = 24 Feet
Miles Curvature Drop
1 8 Inches
2 32 Inches
3 6 Feet
4 10 Feet
5 16 Feet
6 24 Feet
7 32 Feet
8 42 Feet
9 54 Feet
10 66 Feet
20 266 Feet
30 600 Feet
40 1,066 Feet
50 1,666 Feet
60 2,400 Feet
70 3,266 Feet
80 4,266 Feet
90 5,400 Feet
100 6,666 Feet
120 9,600 Feet
7. wayki says:
Here you said there is a 2.9 thousandths of an inch curvature for each 100 feet of horizontal distance.....(heheh).
I hate imperial so please allow me to convert it to metric.
0.07366 mm = 30.480m
Multiply all of this up by 1000 =
73mm fall for every 30,000km. Are you mad? One would have nearly gone around the whole circumference by then - for what a 73mm fall in curve? When the diamerter is ?
Those that come up with 8inches a mile are much closer to the truth.
8. wayki says:
Correction: "When the diameter is"........12,742km? The observor has curved through thousands of km not less than 1 centremeter you madman.
9. Alex A says:
I can assure you the author is not mad. You, on the other hand, I am not so sure about. You take 1 number from the post, and then completely miss the point of the post (was that deliberate?) and abuse the number in the most absolutely ridiculous way possible to draw a completely wrong conclusion. Then you delude yourself into thinking that is evidence the author is mad???
Your most amusing part is that you seem to suggest that "8 inches a mile" is about correct. You should try applying your same abuse to this number, and you will again assume the author is mad. Trust me, the author is not the madman.
If you want to understand this, you should read the section titled "Conclusion" and pay particular attention to "square-law relationship" and "What the contractor did was erroneously assume that the deviation varied linearly with distance". A mistake you made as well.
• mathscinotes says:
Thank you. Very nicely put.
mathscinotes
10. Steve says:
So, this begs the question: At what point would the curvature of the earth "swamp out" the instrument error?
• mathscinotes says:
This is a good question and it all comes down to how you define "swamped". Let's assume that a measurement becomes useless when the error of the instrument equals the curvature of the Earth. My laser level has an error of 1/8 inch per 100 yards (i.e. an angular error). With respect to curvature, this means that the instrument is useful out to about 484 yards. The calculations are as shown below.
11. Juan says:
The earth is flat and if you would do you maths and just natural obervations you would feel that it is not spinning and always raise to eye level no matter how high you go. If you feel it spinning please go and see your doctor.:-)
12. Mike says:
Thank you Juan,
finally the correct answer! Whether you use 8" per mile squared or 6.5 " per mile squared... you still shouldn't see the statue of liberty from 60 miles, or the NYC and Philly skylines from 60 miles....
Use your heads people, test for yourself! If something disappears "over the horizon" (ship, for example) then how come you can see it again if you use binoculars? Impossible if the earth were really curved!
tragedy of science – the slaying of a beautiful hypothesis by an ugly fact.
Huxley
13. Jesse says:
Juan - Go to a playground. Get on a merry go round. Have someone you trust give you a push so gentle that the playground equipment spins at the rate of one rotation per 24 hours. If you feel the movement, please do go see the Dr. Then revisit your comments and report back, if you would.
14. Anthony porreca says:
You always find plumb to the center of the earth at whatever point your at on earth. All so let's say you were building a skyway from newYork to France. you would always stay x amount of feet above sea level bieng your reference benchmarks would always be right in front of you the (ocean surface) it would curve naturally with the earth. Railroads, roads are biult the same way. The benchmarks given to builders are close enough together so they don't have to account for curvature. Example : you could not use a fixed benchmark above sea level as a reference for grade at one point and the keep using that benchmark for the length of the railroad hypotheticly. Even if you had a laser thet was flawless and high enough off of grade to not be unstructed in anyway. your grades would start to become too high.
15. tj says:
pseudoscience at its best! you always seem to have an excuse.......
contrary to nature ...
+water does not 'bunch up' at one end of the pond or lake..
+grammar school observations show that water is AlwAYs self leveling
+gyroscopes do NOT 'rotate' with the imaginary 'spin' of the plane(t) any more than 'day becomes night' 6 months from now as we rotate around a sun that is allegedly 93 million miles away
+infrared light does not 'refract' AND the Military IR Scopes can see 40 plus miles... but the plane(t) curve is 8 in per mile squared... ??
16. James Long says:
Could you please figure the curvature using 1 mile. Math has never been my forte
• mathscinotes says:
This link lists the curvature at 1 mile as 8 inches. If you need to actually need to see calculations, just send me another note.
mathscinotes
17. Christopher Davis says:
Greetings,
I am a navigator and I use geographic ranging to calculate line of sight from an object with a known height to its line of sight horizon point. I use 1.169 times the square root of the height. Simply example: 1.169 x Square root of 100 equals 11.69 nautical miles to the point of horizon,
As I read Nathaniel Bowditch, the generally accepted master of navigational mathematics, I find he seems to use 1.146 instead of 1.169.
I am not sure why the accepted math taught today says to use 1.169.
SO I did some thinking.....Could the variation between Bowditch's math and the modern math have anything to do with the 14 mile difference between the polar radius and the equatorial radius?
Lastly, how did we come to use 1.169?
Thank you very much for any insight you would provide
Respectfully
Chris
• mathscinotes says:
Hi Captain,
It is quite likely that both Bowditch and the modern result are correct – it depends on the air conditions! First, I have looked at how to compute the distance to the horizon on this post, which uses metric units. I will convert to your units for this response.
If I ignore refraction, the distance to the horizon is given by $d = 1.064 \sqrt{h}$, where d is the horizon distance in nautical miles and h is the height of the observer's eye in ft. To add refraction, I need to make an assumption on the rate of air temperature change with altitude – a parameter known as the lapse rate. If I assume 11.3°C/km (common), the horizon distance is given by $d = 1.151\sqrt{h}$. If I assume another lapse rate, I can get 1.169 for a leading coefficient. Yet another lapsre rate and I get 1.141 for a leading coefficient.
mathscinotes
18. Dan Chipowski says:
Why is the drop measured from an angle from the center of the Earth rather than straight down from the horizontal/tangent(x)? For example, if I go ~4000 miles along x, I should show a ~4000 mile drop down from x. There seems to be a separate equation that perhaps doesn't use the Pythagorean theorem--can you show it? And if so, as 1) drop per movement along the horizontal and 2) drop per movement along the circumference. | 2016-05-31T17:52:51 | {
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http://openstudy.com/updates/4fc1bb61e4b0964abc8371b7 | ## FoolForMath Group Title An intuitive geometry problem, The ticket for the Chess world cup final is in the shape of a regular hexagon, with a side of 6 units. A circular stamp, of radius 2 units, is meant to be used to stamp the ticket in the center. By, mistake the stamp get misaligned and stamps the ticket in such a way that the edge of the stamp exactly touches the vertex of the hexagon, while the center of the stamp lies on the line joining that vertex to the opposite vertex. Find the portion of the stamp which falls outside the ticket. 2 years ago 2 years ago
1. ninhi5 Group Title
diagram?
2. FoolForMath Group Title
That would spoil the problem to some extent ;)
3. ninhi5 Group Title
would there be 2 circles, one inside and one outside?
4. FoolForMath Group Title
There is only one stamp.
5. ninhi5 Group Title
well if the edge is outside then the circle must be inscribed inside
6. ninhi5 Group Title
|dw:1338096917684:dw|
7. ninhi5 Group Title
cant really draw to scale
8. ninhi5 Group Title
is that how it should be?
9. FoolForMath Group Title
No.
10. FoolForMath Group Title
@experimentX got it right in the first time itself.
11. ninhi5 Group Title
i'm confused
12. FoolForMath Group Title
me too :|
13. binary3i Group Title
is that $(\pi -3\sqrt{3})/6$
14. FoolForMath Group Title
No.
15. SmoothMath Group Title
|dw:1338097283187:dw| Not perfect, but I believe the picture should look something like this.
16. FoolForMath Group Title
Yes.
17. SmoothMath Group Title
And the area we are trying to find is: |dw:1338097419327:dw|
18. binary3i Group Title
$2\int\limits_{0}^{1/2}(\sqrt{2x-x^2} - \sqrt{3}x) dx$
19. Callisto Group Title
4pi/3 - 2sqrt3?
20. FoolForMath Group Title
I believe Callisto did it :D
21. Callisto Group Title
It's actually very easy and interesting :)
22. FoolForMath Group Title
I never post hard problem(s) ;)
23. SmoothMath Group Title
That's kind of you.
24. Callisto Group Title
No.... you always post difficult problems :|
25. experimentX Group Title
... I'm still waiting for someone to post answer to your perpendicular tangent question
26. ninhi5 Group Title
27. Callisto Group Title
First, I must admit that my way to solve it is very stupid.... radius of the circle is 2 and the centre of the circle lies on the line joining 2 opposite vertex => 2 equi. triangles with sides 2 cm are formed Area of sector = pi(2)^2 x 120/360 = 4pi/3 So, area required = 4pi/3 - 2sqrt3
28. Callisto Group Title
@SmoothMath Hope you don't mind me borrowing your diagram :| |dw:1338097936559:dw|
29. SmoothMath Group Title
Anytime =)
30. AccessDenied Group Title
Ehh, that's how I would have done it too. :P
31. SmoothMath Group Title
I got a bit caught up on how to justify the third side being 2.
32. Callisto Group Title
PS: When I drew the diagram, I found it very interesting that I even forgot I needed to solve the problem :P
33. Callisto Group Title
Each angle of hexagon = 180 (6-2) /6 = 120 divided by 2 = 60 Base angle isos triangle => all are 60 => equi. triangle
34. SmoothMath Group Title
Ah, quite nice.
35. inkyvoyd Group Title
Guys, you just spoiled the problem for me, ahah
36. Callisto Group Title
I only know this basic math :( @FoolForMath How do you solve it?
37. binary3i Group Title
i made a mistake while integrating.
38. SmoothMath Group Title
That's a great way to solve it, Callisto. Better to use basic math whenever you can.
39. FoolForMath Group Title
I did it in the same way. The key realization is the fact that there is a equilateral triangle of length 2 unit is formed. After that, the hexagon is not needed. $2\times r^2\left(\frac {\pi \theta}{360^\circ}-\frac {\sin \theta }{2} \right)$ Here $$r=2$$ and $$\theta$$ is $$60^\circ$$
40. ninhi5 Group Title
someone should give @SmoothMath a medal
41. SmoothMath Group Title
Well, aren't you kind. I'm not too worried about medals though, and Callisto clearly has the "best answer." =)
42. experimentX Group Title
i was looking for the solution of this question ... http://openstudy.com/users/foolformath#/updates/4fa421cde4b029e9dc34a843
43. FoolForMath Group Title
Keep looking ;)
44. apoorvk Group Title
Gotcha, this was a sitter - @Callisto did it in the easiest way possible. Kudos!
45. Callisto Group Title
@FoolForMath 's solution is easier to understand :)
46. apoorvk Group Title
Yea maybe - but before the 'key realisation' - we need the 'key visualisation'. :P
47. FoolForMath Group Title
lol, leave it @apoovk she won't accept that her amazing solution.
48. apoorvk Group Title
yeah she won't, right never does. :P
49. FoolForMath Group Title
that*<- Redundant | 2014-09-03T07:54:26 | {
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https://physics.stackexchange.com/questions/498318/why-cant-i-choose-blocks-attached-with-pulley-b-as-a-system | Why can't I choose blocks attached with pulley B as a system?
Three blocks of masses m1, m2 and m3 are connected as shown in the figure. All the surfaces are frictionless and the string and pulleys are light. Find the acceleration of the block of mass m1.
In this problem I know acceleration of the pulley B is same as acceleration of the block of mass $$\ {m_{1}}$$. But acceleration of bodies with mass $$\ {m_{2}}$$ and $$\ {m_{3}}$$ will be different since $$\ {m_{2}}$$ not equals $$\ {m_{3}}$$. So, I know that I cannot consider this(Pulley B, Blocks of mass $$\ {m_{2}}$$ and $$\ {m_{3}}$$) as a single system. But If the imagine putting those three objects in a box such that what's happening inside will not be visible to me then why I cannot consider this as a single system? The mass will thus be $$\ {m_{2}}$$ + $$\ {m_{3}}$$ and acceleration will be that of the block of mass $$\ {m_{1}}$$. So, I tried to find out the acceleration and I got $$a=\dfrac {\left( m_{1}+m_{2}\right) g}{m_{1}+m_{2}+m_{3}}$$ but, the answer is given as $$\ a =\ \dfrac {g}{1+\dfrac {m_{1}}{4}\left( \dfrac {1}{m_{2}}+\dfrac {1}{m_{3}}\right) }$$. So, why I am wrong here? P.S. How could I solve it in a few steps because the original solution I have is quite long.
• By saying acceleration of the block (mass m1) and pulley B is same I mean magnitude only. – Aditya Kshitiz Aug 23 '19 at 11:15
• Hint: think about what happens as the mass of 2 gets really small and the mass of 3 gets really big. – BioPhysicist Aug 23 '19 at 12:00
• @Aaron Stewens then the block of mass 3 will be in almost free fall and will not cause the pulley to accelerate and therefore the block of mass 1 will not accelerate with the same acceleration of block of mass 3? But why does pulley is accelerating in the first place( in both cases m1 + m2 will be same) – Aditya Kshitiz Aug 23 '19 at 12:06
• The tension around pulley B depends on the accelerations of blocks 2 and 3. Therefore, how blocks 2 and 3 accelerate determines the acceleration of block 1, right? – BioPhysicist Aug 23 '19 at 12:17
• Have in mind that the important thing to accelerate mass m1 is the tension on the string that, as it is massless, will be the same all along. Furthermore, it will be twice the tension in the string between m2 and m3. But tension around pulley B must be different from (m2+m3)g for the masses to accelerate. – MaxWell Aug 23 '19 at 12:33
The key is to realize two things.
First, since pulley B is massless it must be that $$T_A=2T_B$$ where $$T_A$$ and $$T_B$$ is the tension of the string around pulley A and pulley B respectively.
Second, since pulley B accelerates downward with the same acceleration of mass 1, and because the string around pulley B has a constant length, it must be that $$a_2=-a_1+a_r$$ and $$a_3=-a_1-a_r$$, where $$a_r$$ is the relative acceleration between the pulley and mass 2. Adding these relations we get $$2a_1+a_2+a_3=0$$
The above key points along with equations from N2L $$m_1a_1=T_A$$ $$m_2a_2=T_B-m_2g$$ $$m_3a_3=T_B-m_3g$$
allow us to determine an "effective mass" pulling on mass 1 by comparing to the case where the pulley B system is replaced with a single hanging mass (work left to you): $$m_{\text {eff}}=\frac{4m_2m_3}{m_2+m_3}$$
Notice how when $$m_2=m_3$$ we have $$m_{\text {eff}}=2m_2$$, which is what we would expect. Also notice if, say, $$m_2\to0$$ that $$m_\text{eff}\to0$$, which is also what we expected.
This effective mass comes from the key thing you are missing. The acceleration of masses 2 and 3 affect the total force applied to mass 1. You cannot treat the pulley B system as a "black box" whose mass is just the mass of its parts.
A simpler case to understand would be if I was in a box on a scale and you were outside the box. Let's say I was swinging on a swing hanging from the top of the box. If you were looking at the scale, you would be perplexed since the reading would oscillate up and down. Of course, I am not rapidly gaining and losing weight. The forces present "inside of the box" affected the force needed to support the box.
• Thanks a lot. I finally figured it out! – Aditya Kshitiz Aug 23 '19 at 14:45
• @AdityaKshitiz Always remember to upvote useful answers, and to accept answers for future readers. If you want to wait and see if more answers come in you can. – BioPhysicist Aug 23 '19 at 14:49
Newton's equations: Upper mass T = m1 A Hanging masses m2 g – t = m2 ( A – a ) and m3 g – t = m3 ( A + a ) where a is the acceleration of the lower cord relative to the lower pulley (assuming m3 > m2 ). Also T = 2t . Divide each of the lower equations by the corresponding mass and add to eliminate “a”: 2g - t / ((1/m2) + (1/m3)) = 2 A replacing t by T/2 and dividing by 2 gives g =[(m1/4)((1/m2) + (1/m3)) + 1] A .
• This terse recipe doesn't have the type of conceptual explanation that we hope for in our answers to homework-like questions. – rob Aug 23 '19 at 19:21 | 2020-08-09T02:53:55 | {
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https://math.stackexchange.com/questions/3672743/show-that-x-is-a-submartingale-given-some-assumptions-is-the-following-solut | # Show that $X$ is a submartingale, given some assumptions. Is the following solution correct?
Let $$X=(X_n)_{n>0}$$ be an increasing sequence of integrable r.v.'s, each $$X_n$$ being $$\mathcal{F}_n$$-measurable. Show that $$X$$ is a submartingale.
MY SOLUTION
What I have to show is that, given that:
$$1)$$ $$X_n(\omega) < X_{n+1}(\omega)$$, each $$n$$ (or, equivalently, $$X_m(\omega)\leq X_n(\omega)$$, each $$m\leq n$$);
$$2)$$ $$\mathbb{E}(|X_n|)< \infty$$, each $$n$$;
$$3)$$ $$X_n$$ is $$\mathcal{F}_n$$-measurable, each $$n$$;
then $$X$$ is a submartingale, that is:
$$1.1)$$ $$\mathbb{E}(|X_n|)< \infty$$, each $$n$$;
$$1.2)$$ $$X_n$$ is $$\mathcal{F}_n$$-measurable, each $$n$$;
$$1.3)$$ $$\mathbb{E}(X_n|\mathcal{F}_m) \geq X_m$$ a.s., each $$m\leq n$$.
Clearly, $$1.1)$$ corresponds to $$2)$$ and $$1.2)$$ corresponds to $$3)$$. Hence, one is left with proving $$1.3)$$.
To this, one can state that, given assumption $$1)$$, for each $$m\leq n$$: $$$$X_n(\omega)\geq X_m(\omega)$$$$ Then, taking expectation on both sides and conditioning with respect to $$\mathcal{F}_m$$, taking into account assumption $$3)$$, one has that: $$$$\mathbb{E}(X_n(\omega)|\mathcal{F}_m) \geq \mathbb{E}(X_m(\omega)|\mathcal{F}_m) = X_m$$$$ which is exactly point $$1.3)$$.
Is the above reasoning correct?
• Yes, it is correct. May 15, 2020 at 13:17
• You have $X_n(\omega) \geq X_m(\omega)$ at each $\omega$. How does that tell you that $E[X_n | \mathcal F_m ](\omega)\geq E[X_m | \mathcal F_m](\omega)$? You need to explain this. (Essentially, why is the conditional expectation of a non-negative random variable non-negative?) May 17, 2020 at 12:21
• Do you mean that I am not allowed, starting from $X_n(\omega) \geq X_m(\omega)$, to take expectation on both sides and to condition with respect to $\mathcal{F}_m$ finally getting $\mathbb{E}(X_n(\omega)|\mathcal{F}_m) \geq \mathbb{E}(X_m(\omega)|\mathcal{F}_m) = X_m$? Or do you mean that my reasoning is correct BUT I have to clarify why the conditional expectation of a non-negative random variable is non-negative as well? May 17, 2020 at 13:50
• When applying expectation and conditioning with respect to $F_m$ on both sides, I have simply applied monotonicity property of conditional expectation, according to which: if $X_n(\omega)\geq X_m(\omega)$ a.s., then $\mathbb{E}(X_n|\mathcal{F}_m) \geq \mathbb{E}(X_m|\mathcal{F}_m)=X_m$. Hence, why do you stress the importance of explaining why the conditional expectation of a non-negative random variable is non-negative? I have just used monotonicity property of conditional expectation and the random variables are not necessarily non-negative (this is not specified) @астонвіллаолофмэллбэрг May 17, 2020 at 14:44
• The LHS is clearly non-negative, because $X_n-X_m$ is non-negative. However, the right hand side will be negative if $Y$ has non-zero measure (for any random variable, $Z$, $E[Z1_{Z < 0}]$ will be non-positive, and negative if $Z<0$ has positive measure). It follows that the RHS and LHS are $0$ i.e. $Y$ must be of measure zero. This implies $E[X_n|\mathcal F_m] \geq E[X_m | \mathcal F_m]$ almost surely. Finally, this proof is as good as the proof of "conditional expectation preserves monotonicity". May 18, 2020 at 3:35
• We need to show that $$\mathbb E[X_n | \mathcal F_m] \geq X_m$$ for each $$n \geq m$$. By adaptedness and linearity, it is enough to show that $$E[X_n-X_m | \mathcal F_m]$$ is a non-negative random variable;
• But this is clear : let $$Y$$ be the event $$\{E[X_n-X_m | \mathcal F_m] < 0\}$$. Since $$E[X_n-X_m| \mathcal F_m]$$ is a $$\mathcal F_m$$ measurable random variable, the event $$Y$$ belongs in $$\mathcal F_m$$ i.e. $$1_Y$$ (the indicator function of $$Y$$) belongs to $$\mathcal F_m$$;
• By definition of conditional expectation, $$E[(X_n-X_m)1_Y] = E[E[X_n-X_m | \mathcal F_m] 1_Y]$$. The LHS of this is non-negative since $$X_n \geq X_m$$ everywhere, and therefore on $$Y$$. Therefore, the RHS is non-negative. However, $$1_YE[X_n-X_m | \mathcal F_m]$$ is a non-positive random variable! So the integral can be non-negative precisely when $$1_Y$$ is $$0$$ almost surely i.e. $$Y$$ has measure zero. This is the same as $$E[X_n | \mathcal F_m] \geq X_m$$ almost surely.
Finally, all conditions are complete and we have that $$X_m$$ is an $$\mathcal F_m$$-submartingale.
Note that we have proved above a more general statement :
Let $$X,Y$$ be random variables on a probability space $$(\Omega,\mathcal F,P)$$ and let $$\mathcal G \subset \mathcal F$$ be any $$\sigma$$-algebra. Then, if $$X \geq Y$$ we have $$E[X | \mathcal G] \geq E[Y | \mathcal G]$$.
In words, if one random variable dominates another, then even if I provide you with any information, the domination will continue to hold. This is obvious when you think of it. | 2022-12-02T14:21:45 | {
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https://darev.eu/robinson-crusoe-xcpd/8c51fb-every-diagonal-matrix-is-a-scalar-matrix | Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! $\begingroup$ When the diagonal matrix is on the right, it scales the columns of the matrix it is multiplying. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent Upper triangular matrix. 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A matrix that consists of equal diagonal elements and zeros as non-diagonal entries is called a scalar matrix. A square matrix has the same number of rows as columns. Invertibility of sum of an invertible matrix and a nonnegative diagonal matrix 1 Prove that a square matrix can be expressed as a product of a diagonal and a permutation matrix. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. 9. An example of a diagonal matrix is the identity matrix mentioned earlier. You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. See your article appearing on the GeeksforGeeks main page and help other Geeks. [2] For a matrix, confirming that the matrix is diagonal requires checking a total of entries. A diagonal matrix of order n × n where diagonal elements are d 1, d 2..... d n is denoted by 1 So what we are saying is µuTv = λuTv. A Diagonal matrix is a matrix in which the entries outside the main diagonal are all zeros, which means the matrix should have non zero elements only in the diagonal running from the upper left to the lower right. $\endgroup$ – Nick Alger Mar 15 '16 at 1:30 If all entries outside the main diagonal are zero, A is called a diagonal matrix. Theorem 16 If A is an nxn matrix and A is diagonalizable, then A has n linearly independent eigenvectors. Scalar matrix A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix I.Its effect on a vector is scalar multiplication by λ. Roger A. Horn and Charles R. Johnson, Matrix Analysis, Cambridge University Press, 1985. A scalar matrix is therefore equivalent to lambdaI, where I is the identity matrix. For other values of B the calculation involves eigenvalues and eigenvectors. a matrix of type Identity matrix. viii Scalar Matrix Scalar matrix is a diagonal matrix in which all the diagonal from AA 1 code. We know that an scalar matrix is a diagonal matrix whose all diagonal elements are same scalar. A square matrix (2 rows, 2 columns) Also a square matrix (3 rows, 3 columns) Scalar matrix is a special type of diagonal matrix. Scalar-Matrix Multiplication. This behavior … Diagonalizable matrix is similar to a diagonal matrix with its eigenvalues as the diagonal entries Hot Network Questions Given a complex vector bundle with rank higher than 1, … Diagonal matrix is basically a square matrix, whose all diagonal elements are integar and off-diagonal elements are zero. A square matrix in which all the elements below the diagonal are zero i.e. Since µ = λ, it follows that uTv = 0. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Scalar Matrix A square matrix in which every non-diagonal element is zero and all diagonal elements are equal, is called scalar matrix. Examples: Attention reader! We use cookies to ensure you have the best browsing experience on our website. As we know, Scalar matrix is basically a diagonal matrix, whose all diagonal elements are equal. A square matrix D = [d ij] n x n will be called a diagonal matrix if d ij = 0, whenever i is not equal to j. A diagonal matrix is a square matrix in which all the elements other than the principal diagonal elements are zero. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Indeed, a "row matrix" (by that I mean it's all zeros except within a given row) can never be equal to a "column matrix" (analogous but within a column) unless all values of the row in one matrix and the column in the other are zero Given a matrix M[r][c], ‘r’ denotes number of rows and ‘c’ denotes number of columns such that r = c forming a square matrix. Algorithms that are tailored to particular matrix structures, such as sparse matrices and near-diagonal matrices, expedite computations in finite element method and other computations. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices. A diagonal matrix in which all of the diagonal elements are equal to some constant “k” i.e. Let is any scalar matrix. If A and B are diagonal, then C = AB is diagonal. I looks like you mean that in MATLAB or numpy matrix scalar addition equals addition with the identy matrix times the scalar. Unit/Identity Matrix A square 11. Further, C can be computed more efficiently than naively doing a full matrix multiplication: c ii = a ii b ii, and all other entries are 0. And yes, under this definition, becaues a $1\times 1$ matrix has no nondiagonal elements, it is by definition diagonal. ), and a scalar constant c, is … The unit matrix … Exercise Problems and Solutions in Linear Algebra. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! Diagonal Matrices: { De nition: A diagonal matrix is a square matrix with zero entries except possibly on the main diagonal (extends from the upper left corner to the lower right corner). Actually, a square matrix is diagonal if all its non diagonal elements are zero. [1][2] The transpose of a matrix was introduced in 1858 by the British mathematician Arthur Cayley. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Unit matrix and scalar matrix are special case of a diagonal matrix. Definition of diagonal matrix is for all k x k matrices, the matrix is diagonal iff a_ij = 0 for all i != j. For example, the matrix The diagonal matrix $$D$$ is shown Generally, it represents a collection of information stored in an arranged manner. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. Define diagonal matrix. That is, a square matrix is diagonal if and only if all off diagonal elements are A square matrix is said to be diagonal matrix if the elements of matrix except main diagonal are zero. close, link If A is diagonalizable, then there is a diagonal matrix B and an 9) Upper Triangular Matrix A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. Writing code in comment? The same result is obtained in MATLAB, e.g. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix … A square matrix A = [a ij] n×n is called a diagonal matrix if all the elements, except those in the leading diagonal, are zero, i.e., a ij = 0 . For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. Because the columns of Q are linearly independent, Q is invertible. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. As it turns out, the converse of Theorem 10 is also true. $\endgroup$ – Erik Aug 19 '16 at 8:38 Filling diagonal to make the sum of every row, column and diagonal equal of 3x3 matrix, Maximum sum of elements in a diagonal parallel to the main diagonal of a given Matrix, Length of a Diagonal of a Parallelogram using the length of Sides and the other Diagonal, Program to convert given Matrix to a Diagonal Matrix, Check if two elements of a matrix are on the same diagonal or not, Construct a square Matrix whose parity of diagonal sum is same as size of matrix, Program to convert the diagonal elements of the matrix to 0, Program to find the Product of diagonal elements of a matrix, Find a Symmetric matrix of order N that contain integers from 0 to N-1 and main diagonal should contain only 0's, Find sum of all Boundary and Diagonal element of a Matrix, Program to calculate area of a rhombus whose one side and diagonal are given, Check if string is right to left diagonal or not, Sum of non-diagonal parts of a square Matrix, Program to find GCD or HCF of two numbers, Program to find largest element in an array, Inplace rotate square matrix by 90 degrees | Set 1, Write Interview The matrix is a diagonal matrix. In a special case, each entry in the main diagonal (or leading diagonal) can be equal and the remaining non-diagonal elements can be zeros in the matrix. For example, In above example, Matrix A has 3 rows and 3 columns. If a diagonal matrix is commutative with every matrix of the same order then it is necessarily (1) A diagonal matrix with atleast two diagona elements different (2) A scalar matrix (3) A unit matrix (4) A diagonal matrix with exactly two diagona elements different. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. For a diagonal matrix which is not scalar, all elements except those in the leading diagonal should be zero and theelements in the diagonal should not be equal. By using our site, you Of course, it is better to replace "diagonal" by "scalar" since then the other implication also holds... $\endgroup$ – Pete L. Clark Feb 11 '11 at 13:40 Learn via an example what is a diagonal matrix. A nonzero scalar multiple A square null matrix is also a diagonal matrix whose main diagonal elements are zero. Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. We have to find whether the given square matrix is diagonal and scalar matrix or not, if it is diagonal and scalar matrix then print yes in the result. A diagonal matrix may have additional non-diagonal roots if some entries on the diagonal are equal, as exemplified by the identity matrix above. If U is an upper triangular matrix (meaning its entries are u i , j = 0 {\displaystyle u_{i,j}=0} for i > j {\displaystyle i>j} ) and assume at most one of its diagonal … Please use ide.geeksforgeeks.org, generate link and share the link here. Don’t stop learning now. A can therefore be decomposed into a matrix composed of its eigenvectors, a diagonal matrix with its eigenvalues along the diagonal, and the inverse of the matrix of eigenvectors. Types of diagonal matrix are as follows, 1) Rectangular diagonal matrix, 2) Symmetric diagonal and 3) Scalar matrix. (iv) A square matrix B = [b ij] n×n is said to be a diagonal matrix if its all non diagonal elements are zero, that is a matrix B = [b ij] n×n is said to be a diagonal matrix if b ij = 0, when i ≠ j. diagonal matrix if b ij = 0, when i ≠ j. a matrix of type An identity matrix of order nxn is denoted by I n . (a) We need to show that every scalar matrix is symmetric. A diagonal matrix in which all of the diagonal elements are equal to “1" i.e. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. This is called the eigendecomposition and it is a similarity transformation. But every identity matrix is clearly a scalar matrix. Scalar matrix can also be written in form of n * I, where n is any real number Since column-scaling and row scaling are different operations, there are only very limited circumstances that the matrices will commute. diagonal matrix synonyms, diagonal matrix pronunciation, diagonal matrix translation, English dictionary definition of diagonal matrix. its diagonal consists of a, e, and k.In general, if A is a square matrix of order n and if a ij is the number in the i th-row and j th-colum, then the diagonal is given by the numbers a ii, for i=1,..,n.. Given a matrix M[r][c], ‘r’ denotes number of rows and ‘c’ denotes number of columns such that r = c forming a square matrix. Diagonal Matrices, Upper and Lower Triangular Matrices Linear Algebra MATH 2010 Diagonal Matrices: { De nition: A diagonal matrix is a square matrix with zero entries except possibly on the main diagonal (extends from the upper left corner to the lower right corner). matrix if m = n and is known as a square matrix of order ‘n’. Scalar matrix with all entries equal to zero other than diagonal and diagonal entries are same This video is about: Scalar Matrix. From Theorem 2.2.3 and Lemma 2.1.2, it follows that if the symmetric matrix A ∈ Mn(R) has distinct eigenvalues, then A = P−1AP (or PTAP) for some orthogonal matrix P. Proof. Yes, the null matrix is a diagonal matrix. Examples. i.e., in scalar matrix a ij = 0, for i ≠ j and a ij = k, for i = j 10. when the diagonal matrix is on the left, it scales the rows. In a scalar matrix, there is the added restriction that Scalar matrix. For more videos and resources on this topic, please visit http://ma.mathforcollege.com/mainindex/01introduction/ A scalar matrix is a special type of diagonal matrix. Scalar multiplication is easy. Right multiplyingQ −1 But Null matrix can For more videos and resources on this topic, please visit http://ma.mathforcollege.com/mainindex/01introduction/ A square matrix in which every element except the principal diagonal elements is zero is called a Diagonal Matrix. In this post, we are going to discuss these points. In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal; that is, it switches the row and column indices of the matrix A by producing another matrix, often denoted by AT (among other notations). v (or because they are 1×1 matrices that are transposes of each other). Answer. Introduction. Examples : edit Then =. Diagonal matrix is always a square matrix in which non principle diagonal elements are zero but principle diagonal elements can be zero or non zero. We know that an scalar matrix is a diagonal matrix whose all diagonal elements are same scalar.. Let is any scalar matrix. A diagonal matrix, in which all diagonal elements are equal to same scalar, is called a scalar matrix. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Note that every unit matrix is a scalar matrix Topic Matrices Course code Math from MATH 161 at Institute of Management Science, Peshawar For the following matrix A, find 2A and –1A. In a diagonal matrix all of the entries off of the diagonal are zero, and there is no restriction on the diagonal entries. 8 (Roots are found analogously.) Click hereto get an answer to your question Define a scalar matrix. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. By definition of a diagonal matrix, a square matrix is said to be diagonal if all its diagonal elements are zero. 8. It is a square matrix of order n, and also a special kind of diagonal matrix. Randomized testing The reason is that it is quite possible for only a few of the off-diagonal entries to be zero, and therefore, randomized testing will not detect them easily. (a) We need to show that every scalar matrix is symmetric. [[1 0 0],[0 2 0],[0 0 3]] For a diagonal matrix which is not scalar, all elements except those in the leading diagonal should be zero and the Rephrased: "If a matrix commutes with every invertible matrix, it is diagonal." A matrix commutes with every other matrix if and only if it is a scalar matrix, that is, a matrix of the form ⋅, where is the identity matrix, and is a scalar. a matrix … Diagonal matrix is also rectangular diagonal in nature. Matrix is an important topic in mathematics. ... Mathematica » The #1 6) Scalar Matrix. { Examples: The following are examples, of 2 4 A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. brightness_4 We prove that any matrix that commutes with a diagonal matrix with distinct entries is a diagonal matrix. (v) A diagonal matrix is said to be a scalar matrix if its diagonal … (v) A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n×n is said to be a scalar matrix if b ij = 0, when i ≠j b ij = k, when i =j, for Scalar multiplication of matrices is defined in a similar way as for vectors and is done by multiplying every element of the matrix by the scalar. However, the result you show with numpy is simly the addition of the scalar to all matrix elements. In mathematics, a block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices. With this in mind, define a diagonal matrix Λ where each diagonal element Λ ii is the eigenvalue associated with the ith column of Q. The matrix for a linear transformation in a given basis is a diagonal matrix if and only if the following equivalent conditions hold: The linear transformation sends every basis vector to a scalar … Program to swap upper diagonal elements with lower diagonal elements of matrix. Examples: Base A and exponent B are both scalars, in which case A^B is equivalent to A.^B.. Base A is a square matrix and exponent B is a scalar. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Scalar Matrix -- from Wolfram MathWorld A diagonal matrix whose diagonal elements all contain the same scalar lambda. with A = magic(2), A+1. Answered By . Every square diagonal matrix is symmetric, since all off-diagonal elements are zero. Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. We have to find whether the given square matrix is diagonal and scalar matrix or not, if it is diagonal and scalar matrix then print yes in the result.. Diagonal matrix Department of Pre-University Education, Karnataka, Chapter 5: Algebra of Matrices - Exercise 5.1 [Page 8], CBSE Previous Year Question Paper With Solution for Class 12 Arts, CBSE Previous Year Question Paper With Solution for Class 12 Commerce, CBSE Previous Year Question Paper With Solution for Class 12 Science, CBSE Previous Year Question Paper With Solution for Class 10, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Arts, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Commerce, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Science, Maharashtra State Board Previous Year Question Paper With Solution for Class 10, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Arts, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Commerce, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Science, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 10, PUC Karnataka Science Class 12 Department of Pre-University Education, Karnataka. 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Diagonal requires checking a total of entries invertible matrix, it states a! 4 diagonal matrices are found simply by raising each diagonal entry to diagonal... Course at a student-friendly price and become industry ready or diagonalizable MathWorld a matrix..., and there is no restriction on the diagonal are zero i.e can say that a = PBP−1 as! Numbers, variables or functions arranged in rows and 3 columns for the following matrix a has n independent... Alphabet like a, B, C……, etc types of diagonal matrices represented. Diagonal and every thing off the main diagonal are equal to some constant “ k ”.... Is diagonalizable by finding a diagonal matrix has ( non-zero ) entries only on main! Very limited circumstances that the matrices will commute by clicking on the right, states! On every entry in the scalar matrix is on the GeeksforGeeks main and. Important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready DSA... Linearly independent eigenvectors is said to be diagonal matrix if all the of. In above example, matrix a, B, C……, etc DSA Self Paced at. As follows, 1 ) Rectangular diagonal matrix B and an invertible matrix P such that scalar! D\ ) is shown scalar multiplication is easy type of diagonal matrix elements below the elements! Circumstances that the matrix is basically a multiple of an identity matrix of order nxn is denoted by i.. Of the diagonal matrix in which all the important DSA concepts with the DSA Self Paced Course at student-friendly. To the diagonal are zero i.e hold of all the important DSA with... Of equal diagonal elements are same scalar lambda if you find anything incorrect by on. The identity matrix elements of matrix except main diagonal are zero, a square matrix of order n, there! A total of entries if all the important DSA concepts with the DSA Self Paced Course at a price! Called the eigendecomposition and it is by definition diagonal., whose all diagonal elements same! Clearly a scalar matrix is said to be similar to the diagonal are zero, and also a special of! Example of a by the identity matrix mentioned earlier Improve this article if you find incorrect. It states to a set of numbers, variables or functions arranged in rows and columns converse of Theorem is! ( a ) we need to show that every scalar matrix operations, there are only very circumstances... Matrices will commute any scalar matrix is clearly a scalar matrix if B ij = 0 what are. | 2023-03-27T11:20:32 | {
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https://torrecaprese.com/archive/z70k07.php?720e9a=27-square-root-cubed | The cubed root of the square root of 27. In arithmetic and algebra, the cube of a number n is its third power, that is, the result of multiplying three instances of n together. Find √ 27 to 3 decimal places : Guess: 5.125 27 ÷ 5.125 = 5.268 (5.125 + 5.268)/2 = 5.197 27 ÷ 5.197 = 5.195 (5.195 + 5.197)/2 = 5.196 27 ÷ 5.196 = 5.196 Answers archive Answers : Click here to see ALL problems on Square-cubic-other-roots; Question 344218: The problem is cubed root of 1/27? Unlike the square root, the cubed root is always positive. Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes. A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. The cube of a number or any other mathematical expression is denoted by a superscript 3, for example 2 3 = 8 or (x + 1) 3.. Rewrite as . Answer by JBarnum(2146) (Show Source): The radicand no longer has any cube factors. You can use it like this: (we say "the cube root of 27 … Pull terms out from under the radical, assuming real numbers. The opposite of the cubed root is a cubed (power of 3) calculation. Why is this so? A cube root of a number a is a number x such that x3 = a, in other words, a number x whose cube is a. The cubed root of twenty-seven ∛27 = 3 How To Calculate Cube Roots The process of cubing is similar to squaring, only that the number is multiplied three times instead of two. Please link to this page! square root of 27 is 3 * square root of 3 but 3 is also (square root of 3 * square root of 3) thus. See also our cube root table from 1 to 1000. 3 cubed is 27, so the cube root of 27 is 3. 3: 27: The cube root of a number is ..... a special value that when cubed gives the original number. Some common roots include the square root, where n = 2, and the cubed root, where n = 3. Why does it equal the square root of 3? All radicals are now simplified. ) |. Tap for more steps... Rewrite as . Algebra: Square root, cubic root, N-th root Section. Simplify the numerator. Pull terms out from under the radical, assuming real numbers. The cube root of -64 is written as $$\sqrt[3]{-64} = -4$$. Tap for more steps... Rewrite as . Examples are 4³ = 4*4*4 = 64 or 8³ = 8*8*8 = 512. In geometry cubed root can be used to find the length of a side of a cubed when the volume is known. In arithmetic and algebra, the cube of a number n is its third power: the result of the number multiplied by itself twice: n³ = n * n * n. It is also the number multiplied by its square: n³ = n * n². 343 and -343 are examples of perfect cubes. Same with the cubed root except you're multiplying that number by itself 3 times (^3). The cube root of -27 is written as $$\sqrt[3]{-27} = -3$$. (. ) The exponent used for cubes is 3, which is also denoted by the superscript³. The cubed root would be 3 as, 3x3x3=27 or 3x3=9x3=27. The cubic function is a one-to-one function. In equation format: n √ a = b b n = a. Estimating a Root. The cube root of -8 is written as $$\sqrt[3]{-8} = -2$$. The cubed root of 125 is 5, as 5 x 5 x 5 = 125. Rewrite as . For example, 3 is the cube root of 27 because 33 = 3•3•3 = 27, -3 is cube root of -27 because (-3)3 = (-3)•(-3)•(-3) = -27. In mathematics, the general root, or the n th root of a number a is another number b that when multiplied by itself n times, equals a. Solvers Solvers. The cube is also the number multiplied by its square: . In the same way as a perfect square, a perfect cube or cube number is an integer that results from cubing another integer. Simplify the denominator. 27 is said to be a perfect cube because 3 x 3 x 3 is equal to 27. (. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. The exponent used for cubes is 3, which is also denoted by the superscript³. Aug 2006 22,613 8,728. Pull terms out … While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. P. Plato. Just right click on the above image, then choose copy link address, then past it in your HTML. Simplify the denominator. Cube roots (for integer results 1 through 10) Cube root of 1 is 1; Cube root of 8 is 2; Cube root of 27 is 3; Cube root of 64 is 4; Cube root of 125 is 5; Cube root of 216 is 6 Simplify cube root of 27/64. The ... (used for square roots) with a little three to mean cube root. I think the answer is 3. Calculate the Square Root cube root of 1000/27. Square root=2.65 7^2=49 27 is 3 times square root of 3. Square root: √27 = 5.196152423 Result of squared root cubed: 5.196152423 x 5.196152423 x 5.196152423 = 140.2961154293 (Or 140.303 to 2 decimal places.) Can you explain how to get the answer? because if it is asking for that number squared then it would be 49. USING OUR SERVICES YOU AGREE TO OUR USE OF. Here is the answer to questions like: What is the cube root of 27 or what is the cube root of 27? The answer for the first one would be approximately 2.65. are you sure its asking for the square root? 3√ 27 64 2 7 6 4 3. 7³ = 7*7*7 = 343 and (-7)³ = (-7)*(-7)*(-7) = -343. Tap for more steps... Rewrite as . This is because when three negative numbers are multiplied together, two of the negatives are cancelled but one remains, so the result is also negative. Simplify the numerator. 27. This is because cubing a negative number results in an answer different to that of cubing it's positive counterpart. Pull terms out from under the radical, assuming real numbers. 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2020 27 square root cubed | 2021-01-20T10:52:18 | {
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http://physics.stackexchange.com/questions/35978/preventing-a-block-from-sliding-on-a-frictionless-inclined-plane?answertab=votes | # Preventing a block from sliding on a frictionless inclined plane
I want to demonstrate what force $F$ you would have to exert on an inclined plane of angle $t$, mass $M$ to prevent a block on top of it with mass $m$ from sliding up or down the ramp. I worked out an answer, but I figure it must be wrong because it doesn't factor the mass $M$ of the inclined plane into the force needed for the block on top to stay still.
Here is my logic:
• The components normal to the force of gravity on the block on top are $mg \cos(t)$, $mg \sin(t)$
• In particular, the component down the ramp is $mg \sin(t)$ as can be demonstrated with a visual
• Thus, we want $F$ to have a component in the direction opposite the vector down ramp with equivalent force so that $F_{\rm ramp (net)} = 0$
• So we want it to be true that $mg \sin(t) = F \cos(t)$
• So $F = mg \tan(t)$.
Intuitively, this makes some sense: A steeper slope seems like it would require more force to counteract the component of gravity acting down the ramp.
However, this answer and explanation completely disregards $M$, the mass of the ramp itself.
Can somebody explain where I am going wrong, or if I obtained the right answer, why it is not dependent on $M$?
-
I'm not sure I entirely understand the question. This is what I think you're asking; please ignore the rest of this answer if I've misunderstood you.
If the plane is stationary (and I assume there is no friction) then a block on the plane will feel a force down the plane of $mg \space \sin\theta$, so it will accelerate down the plane. If we push the plane so it accelerates at the same rate as the block, then the block won't move relative to the plane. What force on the plane is required to do this?
Assuming I've understood you correctly, the mistake you've made is to assume that the force you apply to the plane, $F$, is the same as the force the block exerts on the plane.
The force $f$ that the block exerts on the plane is not the same as the force $F$ that you exert on the plane.
When you apply a force F to the inclined plane it starts accelerating at some acceleration $a$ given by $a = F/(M + m)$. If you want the block to stay still relative to the plane the acceleration $a$ along the plane must be the same as the acceleration down the plane due to gravity:
$$a \space \cos\theta = g \space \sin\theta$$
or:
$$\frac{F}{M + m} \space \cos\theta = g \space \sin\theta$$
so:
$$F = (M + m)g \space \tan\theta$$
-
Perfect explanation, thank you! If you dont mind me asking, why does the answer not depend on the mass of the little block – tacos_tacos_tacos Sep 17 '12 at 3:48
Because the force on the block is proportional to its mass i.e. $F_b = mg$ so the acceleration F/m is just $g$, the acceleration due to gravity. By contrast the force you apply on the plane isn't related to the mass of the plane so it's acceleration is $F_p/M$. The mass of the block cancels out but the mass of the plane does not. – John Rennie Sep 17 '12 at 5:49
I'll have to disagree. The correct answer is $(M+m)g\tan \theta$. The mistake you made is that you assumed that a force $F$ on the plane will produce an acceleration $F/M$ which is not true because the block is also applying an opposing force. – Alraxite Jan 22 '13 at 10:24
Oops, well spotted. Thanks :-) – John Rennie Jan 22 '13 at 15:39
The block and wedge have the same horizontal acceleration 'a' so use horizontal and vertical axes. FBD of m: Y dir: N cos(t) - mg= 0 X-dir: N sin(t )= ma Solving a=g tan (t) FBD of M: F-N sin(t)=Ma F-ma=Ma F=(M+m)a =(M+m)g tan(t)
- | 2015-04-01T16:41:05 | {
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https://www.mauriciopoppe.com/notes/mathematics/number-theory/divisibility/ | definition
Let $$a,b \in \mathbb{Z}$$, we say that $$a$$ divides $$b$$, written $$a \divides b$$, if there's an integer $$n$$ so that
$b = na$
If $$a$$ divides $$b$$ then $$b$$ is divisible by $$a$$ and $$a$$ is a divisor or factor of $$b$$, also $$b$$ is called a multiple of $$a$$.
Additional properties of the relation $$\divides$$:
1. if $$a \divides b$$ and $$b \divides c$$ then $$a \divides c$$
2. if $$a \divides b$$ and $$c \divides d$$ then $$ac \divides bd$$
3. if $$d \divides a$$ and $$d \divides b$$ then $$d \divides a + b$$
4. if $$d \divides a$$ and $$d \divides b$$ then $$d \divides ax + by$$ for any integers $$x,y$$
Proof.
1. if $$b=ma$$ and $$c=nb$$ then $$c=(nm)a$$
2. if $$b=ma$$ and $$d=nc$$ then $$bd=(nm)ac$$
3. if $$a=md$$ and $$b=nd$$ then $$a + b=(m + n)d$$
4. if $$a=md$$, $$b=nd$$ then $$ax=(mx)d$$, $$by=(ny)d$$ therefore $$ax + by = (mx + ny)d$$
## Division algorithm
definition
Let $$a, b \in \mathbb{Z}$$ with $$b > 0$$, then there exists $$q, r \in \mathbb{Z}$$ such that
$a = bq + r, \quad \text{where 0 \leq r \lt b}$
Proof. if $$bq$$ is the largest multiple of $$b$$ that does not exceed $$a$$ then $$r = a - bq$$ is positive and since $$b(q + 1) > a$$ then $$r \lt b$$.
Also, if $$r = 0$$ then $$a = bq$$ which implies that $$q \divides a$$.
## Greatest common divisor
Let $$a, b \in \mathbb{N}$$, the greatest common divisor of $$a$$ and $$b$$, written as $$gcd(a,b)$$ or $$(a,b)$$, is the element $$d$$ in $$\mathbb{N}$$ such that $$d \divides a$$ and $$d \divides b$$ and every common divisor of $$a$$ and $$b$$ also divides $$d$$.
theorem
Let $$a$$ and $$b$$ be two numbers in $$\mathbb{N}$$, the value of $$(a,b)$$ is a linear combination of $$a$$ and $$b$$ i.e. there exists $$s,t$$ in $$\mathbb{Z}$$ such that
$sa + tb = (a, b)$
Proof.
Let $$d$$ be the least positive integer that is a linear combination of $$a$$ and $$b$$
$d = sa + tb$
First lets show that $$d \divides a$$, by the division algorithm we know that
$a = dq + r, \quad \text{where 0 \le r \lt d}$
It follows that
\begin{align*} r &= a - dq \\ &= a - (sa + tb)q \\ &= a - saq - tbq \\ &= (1 - sq)a + (-tq)b \\ \end{align*}
We can see that $$r$$ is a linear combination of $$a$$ and $$b$$. Since $$0 \le r \lt d$$ and considering that we defined $$d$$ as the least positive linear combination of $$a$$ and $$b$$ it follows that $$r = 0$$ (if $$0 \lt r \lt d$$ then $$r$$ would be the least possible linear combination which is a contradiction), therefore $$d \divides a$$.
In a similar fashion $$d \divides b$$, therefore by the divisibility property #4
$d \divides sa + tb$
The next thing to prove is that $$d$$ is the greatest common divisor of $$a$$ and $$b$$. To prove this lets show that if $$d'$$ is any other common divisor of $$a$$ and $$b$$ then $$d' \le d$$.
If $$d' \divides a$$ and $$d' \divides b$$ then by the divisibility property #4 it divides any other linear combination of $$a$$ and $$b$$, since $$d = sa + bt$$ is one linear combination of $$a$$ and $$b$$ it follows that $$d' \divides d$$ so either $$d' \lt d$$ or $$d' = d$$, finally we can conclude that
$d = (a,b)$
### Euclidean Algorithm
A very efficient method to compute the greatest common denominator
theorem
Suppose $$a, b$$ be integers with $$a \ge b \gt 0$$
1. Apply the division algorithm $$a = bq + r, 0 \le r \lt b$$
2. Rename $$b$$ as $$a$$ and $$r$$ as $$b$$ and repeat 1 until $$r = 0$$
The last nonzero remainder is the greatest common divisor of $$a$$ and $$b$$
The euclidean algorithm depends on the following lemma
lemma
Let $$a, b$$ be integers with $$a \ge b \gt 0$$. Let $$r$$ be the remainder of dividing $$a$$ by $$b$$ then
$(a,b) = (b, r)$
Proof. Let $$q$$ be the quotient of dividing $$a$$ by $$b$$ so that $$a = bq + r$$. If $$d = (a,b)$$ then it must divide any other linear combination of $$a$$ and $$b$$ like $$r = a - bq$$, therefore $$d \divides r$$. Finally we can conclude that $$d = (b,r)$$.
Proof of the theorem If we keep on repeating the division algorithm we have:
\begin{align*} a &= bq_1 + r_1, \quad (a,b) = (b, r_1) \\ b &= r_1q_2 + r_2, \quad (b, r_1) = (r_1, r_2) \\ r_1 &= r_2q_3 + r_3, \quad (r_1, r_2) = (r_2, r_3) \\ r_2 &= r_3q_4 + r_4, \quad (r_2, r_3) = (r_3, r_4) \\ & \; \vdots \\ r_{n-3} &= r_{n-2}q_{n-1} + r_{n-1}, \quad (r_{n-3}, r_{n-2}) = (r_{n-2}, r_{n-1}) \\ r_{n-2} &= r_{n-1}q_n + r_n, \quad (r_{n-2}, r_{n-1}) = (r_{n-1}, r_n) \\ r_{n-1} &= r_n q_{n+1}, \quad \quad (r_{n-1}, r_n) = r_n \end{align*}
Therefore
$(a,b) = (b,r_1) = (r_1,r_2) = (r_2, r_3) = (r_3, r_4) = \ldots = (r_{n-3}, r_{n-2}) = (r_{n-2}, r_{n-1}) = (r_{n-1}, r_n) = r_n$
### Extended Euclidean Algorithm
One of the applications of the euclidean algorithm is the calculation of the integers $$x,y$$ satisfying $$d = (a,b) = ax + by$$
First note that if $$b=0$$ then $$(a,b) = (a,0) = a$$, now assume that there are integers $$x'$$ and $$y'$$ so that
$(a,b) = (b,r) = bx' + ry'$
Since
\begin{align*} r &= a - bq \\ &= a - b \left \lfloor \frac{a}{b} \right \rfloor \end{align*}
Then
\begin{align*} (a,b) &= bx' + \Big( a - \left \lfloor \frac{a}{b} \right \rfloor b \Big) y' \\ &= bx' + ay' - \left \lfloor \frac{a}{b} \right \rfloor by' \\ &= a(y') + b \Big(x' - \left \lfloor \frac{a}{b} \right \rfloor y'\Big) \end{align*}
Comparing it to $$(a,b) = ax + by$$ we obtain the required coefficients $$x$$ and $$y$$ based on the following recursive equations
\begin{align*} x &= \begin{cases} 1, & \text{when r = 0} \\ y', & \text{otherwise} \end{cases} \\ y &= \begin{cases} 0, & \text{when r = 0} \\ x' - \left \lfloor \frac{a}{b} \right \rfloor y', & \text{otherwise} \end{cases} \end{align*}
References | 2020-01-29T12:23:14 | {
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https://math.stackexchange.com/questions/2785724/number-of-labeled-trees-without-a-1-2-edge | # Number of labeled trees without a $(1,2)$ edge?
How many labeled n-trees are there that do not contain a $(1,2)$ edge?
Clarification: The problem is referring to the encodings of a prufer's code. $(1,2)$ edge is referring to the edge that joins the vertex labeled $1$ and $2$.
Thoughts:
I'm not exactly sure how to do this problem. I know there are $n^{n-2}$ labeled trees, each with $n-1$ edges. I also know there are two cases, 1) being when either $1$ or $2$ appears as a leaf, 2) when they do not.
In case 1), we can consider a tree with $n-1$ vertices, without $1$ as a semantic encoding. Find the encoding of a $(n-1)$ tree, add $1$ to $2$ as a leaf. There's $(n-1)^{n-3}$ labeled trees on $(n-1)$ vertices, switch the position of $1$ and $2$ and we have $2(n-1)^{n-3}$ possibilities.
However, if neither $1$ or $2$ appear as a leaf, then I'm lost. I'm not sure how to better approach this problem without these two cases.
EDIT:
New approach (not sure about correctness):
If $1$ and $2$ form an edge, then the removal of that edge will separate the tree into two components, each with at least one vertex. Each component is then it's own tree. So take a iterative approach:
For a set of vertices $\{1,2,...,n\}$, divide it into two sets, with $1$ and $2$ in separate sets. Of the remaining $n-2$ vertices, pick $k$ elements to be with vertex $1$, and let the remaining be with vertex $2$. Let each set form its own tree. Hence we have, for some $0\le k \le n-2$:
1) Put $1$ into set $A$, $2$ into set $B$, choose $k$ additional elements to be in $A$. Hence $\binom{n-2}{k}$ ways.
2) For set $A$, there are $(k+1)^{k-1}$ ways to form a tree.
3) For set $B$, there are $(n-k-1)^{n-k-3}$ ways to form a tree.
Summing it all up, we have:
$$(n)^{n-2} - \sum_{k=0}^{n-2} \binom{n-2}{k} ((k+1)^{k-1}(n-k-1)^{n-k-3})$$
• yes. I'm referring to the encoding in a proffer's code. sorry for the confusion. – B.Li May 18 '18 at 0:54
We suppose $n\ge2.$ A tree on the vertex set $\{1,2,\dots,n\}$ contains $n-1$ of the $n(n-1)/2$ possible edges. The probability that a random tree contains a given edge (say the edge $\{1,2\}$) is $$\frac{n-1}{n(n-1)/2}=\frac2n.$$ The probability that a random tree omits the edge $\{1,2\}$ is $$1-\frac2n=\frac{n-2}n.$$ As there are $n^{n-2}$ trees all told, the number of trees omitting the edge $\{1,2\}$ is $$\frac{n-2}n\cdot n^{n-2}=\boxed{(n-2)n^{n-3}}.$$
Here is the same argument presented in the style of counting rather than probability.
Let $K_n$ be the complete graph on the vertex set $V=[n]=\{1,2,\dots,n\}.$
Then $a(n)=n^{n-2}$ is the number of spanning trees in $K_n.$
For $e\in V$ let $b(n)$ be the number of spanning trees that contain the edge $e;$ clearly this number does not depend on the choice of $e.$
Let's count the number $N$ of pairs $(T,e),$ where $T$ is a spanning tree of $K_n$ and $e$ is an edge of $T,$ in two different ways.
Counting by trees: each tree has $n-1$ edges, so $$N=a(n)(n-1).$$
Counting by edges: there are $\binom n2$ edges, each edge is in $b(n)$ spanning trees, so $$N=\binom n2b(n).$$ From these two equations for $N,$ $$\frac{b(n)}{a(n)}=\frac{n-1}{\binom n2}=\frac2n,$$ so $$a(n)-b(n)=\left(1-\frac2n\right)a(n)=\boxed{(n-2)n^{n-3}}.$$
• I see. I added a "dumb" approach to the question. I'm not sure if it yields the same answer. – B.Li May 18 '18 at 1:13
• @B.Li Your sum is an expression for the number of trees containing the $(1,2)$-edge, which then must be subtracted from $n^{n-3}$ to get the number of trees omitting that edge, right? – bof May 18 '18 at 1:27
• right, forgot that part, silly mistake. – B.Li May 18 '18 at 1:28
• Your approach looks fine to me. And there's nothing "dumb" about it; solving the same counting problem two different ways is how we prove combinatorial identities. – bof May 18 '18 at 1:33
• Your argument proves the identity $$\sum_{k=0}^{n-2}\binom{n-2}k(k+1)^{k-1}(n-1-k)^{n-3-k}=2n^{n-3}.$$ – bof May 18 '18 at 1:43 | 2020-02-25T01:25:50 | {
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https://math.stackexchange.com/questions/2306452/question-about-probability-of-winning-two-times-in-a-row-out-of-three | Question about Probability of Winning Two Times in a Row out of Three
Okay, so I came across an interesting probability question that has left me stumped to come to a solid conclusion.
Assume that you are playing a card game with two opponents, $A$ and $B$. Assume that the probability of winning against $A$ is greater than winning against $B$. Probabilities of winning against $A$ or $B$ are independent. You will play these opponents one at a time in alternating order. You receive a payout only if you are able to win two games in a row out of a three game series.
Given the choice, would you choose to pick $A$ (i.e. Would play $ABA$) or $B$ first (Would play $B A B$)
My instinct tells me that you would be better of choosing $A$ first, but since any two wins in a row would involve beating $A$ and $B$, I am not convinced that it matters.
I am curious if there is a more mathematical explanation than my instinct.
2 Answers
Let $p$ be the probability of winning a round against $A$, and $q$ that against $B$, for some $p>q$.
The probabilility of winning two games in a row out of three, if you start playing $A$ is: $$P_A~{=\mathsf P(wwl\cup www\cup lww\mid ABA)\\= pq+(1-p)qp\\= (2-p)pq}$$
The probabilility of winning two games in a row out of three, if you start playing $B$ is: $$P_B~{=\mathsf P(wwl\cup www\cup lww\mid BAB)\\= qp+(1-q)pq\\= (2-q)pq}$$
Since $p>q$ , therefore $P_A<P_B$
Intuitively, you chances of winning two rounds in a row is improved if the outside rounds are the easiest to win; since you must certainly win the middle round if you wish to win the game. If you can win that, then you have two oportnities win against the other player; so make those the easiest.
Or in simpler terms: to win the game you must win against one player twice and the other player once, and the best way to manage this is to choose the easiest target as the one to play against twice.
• The intuitive answer is that you have to beat B to win two in a row, so two chances of that are better than one. This computation clearly demonstrates it. I think people are confused by the better chance of winning two games if you play ABA but that is not what the question asks. +1 – Ross Millikan Jun 2 '17 at 3:36
We can compute the probabilities directly.
There are three configurations that have two wins in a row: $WWW,$ $WWL$ or $LWW.$ Let $p_A$ be the probability of beating $A$ and $p_B$ for $B$ so that $p_A>p_B.$
For $ABA$ the probabilities are $$P(WWW) = p_Ap_Bp_A \\P(WWL) = p_Ap_B(1-p_A) = P(LWW).$$ Adding the three up gives a total probability of $p_Ap_B(2-p_A).$
For $BAB$ you similarly get $p_Ap_B(2-p_B).$
So it looks like you are better off with $BAB!$
What's going on here? Since your ability to get two in a row depends pretty crucially on getting the middle one, it's actually best to have the easiest in the middle and taking your chances with $B$ on the outside. | 2019-05-19T08:39:15 | {
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http://mathhelpforum.com/algebra/201633-quadratic-equation.html | # Math Help - Quadratic equation
1. ## Quadratic equation
Let r and s be the roots of the quadratic x2 + bx + c where b and c are constant, if (r-1)(s-1) = 7 find b + c
2. ## Re: Quadratic equation
Originally Posted by mgmanoj
Let r and s be the roots of the quadratic x2 + bx + c where b and c are constant, if (r-1)(s-1) = 7 find b + c
\displaystyle \begin{align*} x^2 + b\,x + c &= 0 \\ x^2 + b\,x + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c &= 0 \\ \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4} + \frac{4c}{4} &= 0 \\ \left(x + \frac{b}{2}\right)^2 + \frac{4c - b^2}{4} &= 0 \\ \left(x + \frac{b}{2}\right)^2 &= \frac{b^2 - 4c}{4} \\ x + \frac{b}{2} &= \frac{\pm \sqrt{b^2 - 4c}}{2} \\ x &= \frac{-b \pm \sqrt{b^2 - 4c}}{2} \end{align*}
So we can say \displaystyle \begin{align*} r = \frac{-b - \sqrt{b^2 - 4c}}{2} \end{align*} and \displaystyle \begin{align*} s = \frac{-b + \sqrt{b^2 - 4c}}{2} \end{align*}. Since we know \displaystyle \begin{align*} (r - 1)(s - 1) = 7 \end{align*}, we can say
\displaystyle \begin{align*} \left(\frac{-b - \sqrt{b^2 - 4c}}{2} - 1\right)\left(\frac{-b + \sqrt{b^2 - 4c}}{2} - 1\right) &= 7 \\ \left(\frac{-b - 2 - \sqrt{b^2 - 4c}}{2}\right)\left(\frac{-b - 2 + \sqrt{b^2 - 4c}}{2}\right) &= 7 \\ \frac{\left(- b - 2 - \sqrt{b^2 - 4c} \right) \left( - b - 2 + \sqrt{b^2 - 4c} \right) }{4} &= 7 \\ \left(-b-2 - \sqrt{b^2 - 4c}\right)\left(-b-2 + \sqrt{b^2 - 4c}\right) &= 28 \\ \left(- b - 2\right)^2 - \left(b^2 - 4c \right) &= 28 \\ b^2 + 4b + 4 - b^2 + 4c &= 28 \\ 4b + 4c &= 24 \\ 4(b + c) &= 24 \\ b + c &= 6 \end{align*}
3. ## Re: Quadratic equation
Thanks - I was half way down to the problem - but did not substitute - this confirms my theory. Thanks again.
4. ## Re: Quadratic equation
Hello, mgmanoj!
$\text{Let }r\text{ and }s\text{ be the roots of the quadratic: }\: x^2+\,bx\,+\,c\,\text{ where }b\text{ and }c\text{ are constants.}$
$\text{If }\,(r-1)(s-1) \:=\: 7,\,\text{ find }b + c.$
$\text{Since }r\text{ and }s\text{ are roots: }\:\begin{Bmatrix}r + s &=& \text{-}b \\ rs &=& c \end{Bmatrix} \;\;{\color{blue}[1]}$
$\text{Given: }\:(r-1)(s-1) \:=\:7 \quad\Rightarrow\quad rs - r - s + 1 \:=\:7 \quad\Rightarrow\quad rs - (r+s) \:=\:6$
$\text{Substitute }{\color{blue}[1]}\!:\;c - (\text{-}b) \:=\:6 \quad\Rightarrow\quad b +c \:=\:6$
5. ## Re: Quadratic equation
I second Soroban's solution...always helps to look at the sum and product of the roots of the quadratic.
6. ## Re: Quadratic equation
Originally Posted by richard1234
I second Soroban's solution...always helps to look at the sum and product of the roots of the quadratic.
I third it ! | 2015-11-30T00:17:01 | {
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https://math.stackexchange.com/questions/1127133/can-number-theory-be-visualized | # Can Number Theory be visualized?
So I was thinking about a hard euclidean geometry problem, when it hit me just how much more difficult it would become without the aid of a diagram. This got me thinking: Wouldn't it be great if we could somehow find corresponding diagrams for something like number theory? It doesn't have to be euclidean geometry diagrams like the Greeks did with Algebra (which actually made it harder than without diagrams, as we all know), but instead we need to find a natural diagrammatical representation. Something like Ferrers diagram seems to be a tiny step in that direction, and I can imagine similar things being done for all of number theory.
Is this idea at all feasible? Please explain why.
EDIT: I will leave the above to make answers more interesting. However, a related question (as suggested by Omnomnomnom) that is perhaps more useful is the following: What kinds of diagrams already exist in number theory?
Eagerly awaiting any responses!
• One large area of Number Theory is Geometry of Numbers. We have provided only one link, but if you search you will get many hits. And there is geometric intuition behind quite a few proofs, particularly estimates. – André Nicolas Jan 30 '15 at 23:16
• I apologize if I didn't read too well. Perhaps the right question to ask is what are the already-existing examples of such diagrams, as I am sure some exist. – Omnomnomnom Jan 30 '15 at 23:19
• There are pretty diagrams behind descent processes in certain Diophantine equations, e.g. Fibonacci's Lost Theorem = FLT$_4$ (FLT for exponent $4).\ \$ – Bill Dubuque Jan 30 '15 at 23:21
• Number theory is closely tied to group theory, so consider the Fano plane and elliptic curves. – dtldarek Jan 30 '15 at 23:25
• Very very elementary number theory can be visualized very effectively in the way Euclid did. For example, this is one of the best ways to prove that the common multiples of $a$ and $b$ are precisely the multiples of the least common multiple of $a$ and $b$. – Jack M Jan 30 '15 at 23:29
One sparkling gem at the intersection of number theory and geometry is Aubry's reflective generation of primitive Pythagorean triples, i.e. coprime naturals $\,(x,y,z)\,$with $\,x^2 + y^2 = z^2.\,$ Dividing by $z^2$ yields $\,(x/z)^2\!+(y/z)^2 = 1,\,$ so each triple corresponds to a rational point $(x/z,\,y/z)$ on the unit circle. Aubry showed that we can generate all such triples by a very simple geometrical process. Start with the trivial point $(0,-1)$. Draw a line to the point $\,P = (1,1).\,$ It intersects the circle in the rational point $\,A = (4/5,3/5)\,$ yielding the triple $\,(3,4,5).\,$ Next reflect the point $\,A\,$ into the other quadrants by taking all possible signs of each component, i.e. $\,(\pm4/5,\pm3/5),\,$ yielding the inscribed rectangle below. As before, the line through $\,A_B = (-4/5,-3/5)\,$ and $P$ intersects the circle in $\,B = (12/13, 5/13),\,$ yielding the triple $\,(12,5,13).\,$ Similarly the points $\,A_C,\, A_D\,$ yield the triples $\,(20,21,29)\,$ and $\,(8,15,17),\,$
We can iterate this process with the new points $\,B,C,D\,$ doing the same we did for $\,A,\,$ obtaining further triples. Iterating this process generates the primitive triples as a ternary tree
$\qquad\qquad$
Descent in the tree is given by the formula (whose reflective geometric genesis is given below)
$$\begin{eqnarray} (x,y,z)\,\mapsto &&(x,y,z)-2(x\!+\!y\!-\!z)\,(1,1,1)\\ = &&(-x-2y+2z,\,-2x-y+2z,\,-2x-2y+3z)\end{eqnarray}$$
e.g. $\ (12,5,13)\mapsto (12,5,13)-8(1,1,1) = (-3,4,5),\$ yielding $\,(4/5,3/5)\,$ when reflected into the first quadrant.
Ascent in the tree by inverting this map, combined with trivial sign-changing reflections:
$\quad\quad (-3,+4,5) \,\mapsto\, (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$
$\quad\quad (-3,-4,5) \,\mapsto\, (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$
$\quad\quad (+3,-4,5) \,\mapsto\, (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$
Continuing in this manner one may reflectively generate the entire tree of primitive Pythagorean triples, e.g. the topmost edge of the triples tree corresponds to the ascending $C$-inscribed zigzag line $(-1,0), (3/5,4/5), (-3/5,4/5), (5/12,12/13), (-5/12,12/13), (7/25,24/25), (-7/25,24/25) \ldots$
Let's look a bit closer at the underlying geometry. Consider the quadratic space $Z$ of the form $Q(x,y,z) = x^2 + y^2 - z^2$. It has Lorentzian inner product $(Q(x\!+\!y)-Q(x)-Q(y))/2\,$ given by
$\qquad v \cdot u\, =\, v_1 u_1 + v_2 u_2 - v_3 u_3.\ \$ Recall that the reflection of $v$ in $u$ is given by
$\quad\quad v\, \mapsto\, v - 2 \dfrac{v \cdot u}{u \cdot u} u \quad$ Reflectivity is clear: $\; u \mapsto -u$, and $\; v \mapsto v$ if $\; v\perp u, \;$ i.e. $v\cdot u = 0$.
With $\; v = (x,y,z)$ and $\; u = (1,1,1)$ of norm $1$ we have
$\quad\quad (x,y,z)\; \mapsto (x,y,z) - 2 \dfrac{(x,y,z)\cdot(1,1,1)}{(1,1,1)\cdot(1,1,1)} (1,1,1)$
$\qquad\qquad\qquad =\, (x,y,z) - 2 \; (x\!+\!y\!-\!z) \; (1,1,1)$
$\qquad\qquad\qquad =\, (-x\!-\!2y\!+\!2z, \; -2x\!-\!y\!+\!2z, \; -2x\!-\!2y\!+\!3z)$
This is the nontrivial reflection that effects the descent in the triples tree. Said more simply: if $\,x^2 + y^2 = z^2\,$ then $\,(x/z, y/z)\,$ is a rational point $P$ on the unit circle $C$ then a simple calculation shows that the line through $P$ and $(1,1)$ intersects $C$ in a smaller rational point, given projectively via the above reflection.
This technique easily generalizes to the form $x_1^2 + x_2^2 + \cdots + x_{n-1}^2 = x_n^2$ for $4 \le n \le 9$, but for $n \ge 10$ the Pythagorean n-tuples fall into at least $[(n+6)/8]$ distinct orbits under the automorphism group of the form - see Cass & Arpaia (1990) [1]
There are also generalizations to different shape forms that were first used by L. Aubry (Sphinx-Oedipe 7 (1912), 81-84) to give elementary proofs of the $3$ & $4$ square theorem (see Appendix 3.2 p. 292 of Weil's: Number Theory an Approach Through History). These results show that if an integer is represented by a form rationally, then it must also be so integrally. The method also applies to the following forms $x^2+y^2, x^2 \pm 2y^2, x^2 \pm 3y^2, x^2+y^2+2z^2, x^2+y^2+z^2+t^2,\ldots$ More precisely, essentially the same proof as for Pythagorean triples shows
Theorem Suppose that the $n$-ary quadratic form $F(x)$ has integral coefficients and has no nontrivial zero in ${\mathbb Z}^n$, and suppose that for any $x \in {\mathbb Q}^n$ there is $\,y \in {\mathbb Z}^n$ such that $\; |F(x\!-\!y)| < 1$. Then $F$ represents $m$ over $\mathbb Q$ $\iff$ $F$ represents $m$ over $\mathbb Z$, for all nonzero integers $m$.
The condition $|F(x\!-\!y)| < 1$ is closely connected to the Euclidean algorithm. In fact there is a function-field analog that employs the Euclidean algorithm which was independently rediscovered by Cassels in 1963: a polynomial is a sum of $n$ squares in $k(x)$ iff the same holds true in $k[x]$. Pfister immediately applied this to obtain a complete solution of the level problem for fields. Shortly thereafter he generalized Cassels result to arbitrary quadratic forms, founding the modern algebraic theory of quadratic forms ("Pfister forms").
Aubry's results are, in fact, very special cases of general results of Wall, Vinberg, Scharlau et al. on reflective lattices, i.e. arithmetic groups of isometries generated by reflections in hyperplanes. Generally reflections generate the orthogonal group of Lorentzian quadratic forms in dim $< 10$.
[1] Daniel Cass; Pasquale J. Arpaia
Matrix Generation of Pythagorean n-Tuples.
Proc. Amer. Math. Soc. 109, 1, 1990, 1-7.
• This is amazing, I didn't think it was possible. I recently solved a problem similar to the Pythagorean Theorem: If you have a quadrilateral inscribed in a semi circle, with one side $AD$ a diameter of the semicircle, then if we let $AD=x$ and the other sides of the quadrilateral $a,b,c$ then $x^3=x(a^2+b^2+c^2)+2abc$....(x) This is similar to the Pythagorean Theorem which states that $x^3=x(a^2+b^2)$. I did eventually manage to get a characterization of integer solutions to (x), so your post has now got me wondering if a geometric proof of this characterization is possible. – user45220 Jan 31 '15 at 1:10
• I should probably chew on this excellent answer more to answer this question myself, but: is there a 'traditional' reflection group at work here? Is it a Coxeter group, and if so what can be said about it? – Steven Stadnicki Jan 31 '15 at 1:13
• @Steven For a start see the introduction to R. Scharlau, On the ciassification of arithmetic reflection groups on hyperbolic 3-space. Google "reflective lattices" to learn more. – Bill Dubuque Jan 31 '15 at 1:20
• If you like this, you should check out John Conway's little book, The Sensual (quadratic) Form. – Kimball Jan 31 '15 at 3:01
There is an area called arithmetic geoemetry that exploits links between arithmetic and algebro-geometric questions.
For example, Fermat's famous equations $X^n + Y^n = Z^n$ can be thought of as a curve in projective space, called Fermat curves, and one can use geometric tools to study it.
The affine part, so $X^n + Y^n = 1$ is somewhere between a circle and a square; for small $n$ close to a circle (well for $n=2$ it is of course a circle, but this is not relevant for FLT) an for large $n$ it approaches a square-like form.
• Thank you very much for this answer. I remember solving a cambridge interview problem involving the figure that $X^n+Y^n=1$ approaches for large $n$. I only got the answer with a lot of guidance, but it was very surprising. – user45220 Jan 30 '15 at 23:39
Clifford Algebra, a.k.a. Geometric Algebra, is a most extraordinary synergistic confluence of a diverse range of specialized mathematical fields, each with its own methods and formalisms, all of which find a single unified formalism under Clifford Algebra. It is a unifying language for mathematics, and a revealing language for physics.
Clifford Algebra: A Visual Introduction
• Hello, thank you very much this is great, I will definitely check it out. – user45220 Jan 31 '15 at 13:07
Not an answer, but perhaps a good contribution to the discussion. This is from page 261 of Siobhan Robert's brilliant Genius at Play, a biography of John H. Conway. She's quoting Conway:
When we were first working on the ATLAS [of finite groups], we didn't quite appreciate it. So you won't. I think it's best to get away from explaining things with numbers. I use numbers reluctantly. It's the only way I can work out the beautiful things about these groups. I would do something else - draw pictures if I could - but I can't draw beautifully symmetric things in 7-dimensional spaces, ... For me, numbers are a substitute for touch, feel, sight, everything else. With high dimensional space I can't touch it, can't feel it, can't see it. I can calculate it, but the calculation isn't the point. The numbers are a set of instructions. A set of instructions isn't beautiful, but that's what the numbers are, a set of instructions, point by point.
https://en.wikipedia.org/wiki/ATLAS_of_Finite_Groups
http://www.amazon.com/Atlas-Finite-Groups-Subgroups-Characters/dp/0198531990
http://www.amazon.com/Genius-At-Play-Curious-Horton/dp/1620405938 | 2019-07-22T10:40:15 | {
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https://math.stackexchange.com/questions/2252645/if-one-root-of-the-equation-x2px1-0-be-the-square-root-of-the-other-then-fi | # If one root of the equation $x^2+px+1=0$ be the square root of the other then find the value of $p$.
If one root of the equation $x^2+px+1=0$ be the square root of the other then find the value of $p$.
My Attempt:
Let $\alpha$ and $\beta$ be the two roots of the given equation. Then,
$$\alpha =\dfrac {-p + \sqrt {p^2-4}}{2}$$ $$\beta =\dfrac {-p-\sqrt {p^2-4}}{2}$$.
According to Question: $$\alpha =\sqrt {\beta}$$ $$2p^2-2p\sqrt {p^2-4} - 4=-2p-2\sqrt {p^2-4}$$
How do I proceed further?
• Why do you assume alpha is the square root of beta and nor the other way around? – fleablood Apr 26 '17 at 5:54
## 4 Answers
The roots are $a$ and $a^2$ so $$a^3=1$$ and $$a+a^2=-p$$ so either $a=1$ and $p=-2$ or $$a^2+a+1=0$$ in which case $p=1$.
• How is.$a^3=1$? – pi-π Apr 26 '17 at 6:26
• If the roots are $a$ and $a^2$ then the factors are $x-a$ and $x-a^2$. Multiply them together. – Leonard Blackburn Dec 6 '17 at 15:44
So the roots are $r$ and $r^2$. Then the equation is $(x-r)(x-r^2)=0$, that is $x^2-(r+r^2)x+r^3=0$. So you need $p=-(r+r^2)$ and $1=r^3$. So what is $r$, and what then must $p$ be?
• why are $r$ and $r^2$ roots?..and please explain the other steps as well. – pi-π Apr 26 '17 at 5:49
• @Maxwell if the first root is the square root of the second, then the second is the square of the first. – Lord Shark the Unknown Apr 26 '17 at 5:50
• And how do you get $p=-(r+r^2)$ and $1=r^3$? – pi-π Apr 26 '17 at 5:58
• If it makes you more comfortable (but it shouldn't-- this should both be equally clear) you could let one root be $t$ and the other $\sqrt t$. Then you must solve $\t\sqrt t=1$ and $p=-(t+\sqrt t)$. But that is the *exact* same thing with $t=r^2$. – fleablood Apr 26 '17 at 6:01
• @fleablood, solving this way gives $p=-2$ but the answer in book is $1$. – pi-π Apr 26 '17 at 6:04
Vieta's formulas:
a = 1, b = p, c = 1 for your equation ($x_1 = \sqrt{x_2}$)
$x_1x_2 = \frac{c}{a} = 1 = x\sqrt{x}$, thus $x = 1$.
$x_1 + x_2 = -\frac{b}{a} = -p$
$1 + 1 = -p$, thus $p = -2$
Assuming the roots are $\alpha$ and $\alpha^2$, then, \begin{align} &(x-\alpha)(x-\alpha^2)=0 \\ \implies&x^2+x(-\alpha-\alpha^2)+\alpha^3=0 \end{align}
Comparing this with $x^2+px+1 = 0$ we get: \begin{align} &\alpha+\alpha^2 = -p \\ &\alpha^3=1 \end{align}
If we cube the first relation of the two above, we discover an interesting relation: \begin{align} &(\alpha+\alpha^2)^3=-p^3 \\ \implies&\alpha^3+\alpha^6+3\alpha^4+3\alpha^5=-p^3 \\ \implies&a^3+a^6+3\alpha^3(\alpha+\alpha^2) = -p^3 \\ \implies&1+1+3(-p)=-p^3 \\ \implies&p^3-3p+2=0 \\ \implies&(p-1)(p^2+p-2)=0 \\ \implies&(p-1)^2(p+2)=0 \end{align}
Clearly for p to satisfy this $p=1$ or $p=-2$ | 2020-02-28T01:12:36 | {
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http://mathhelpforum.com/statistics/1920-combination-problem.html | # Math Help - Combination Problem
1. ## Combination Problem
I thought this would be a simple problem, but I can still not come up with the correct answer.
The problem is:
12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?
Since order is not important it is a combination problem and not a permuntation problem.
So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).
Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?
I get 200 for the answer but according to the quiz this is not the answer.
Could someone tell me what I am doing wrong?
James
2. Originally Posted by jamesinsc
I thought this would be a simple problem, but I can still not come up with the correct answer.
The problem is:
12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?
Since order is not important it is a combination problem and not a permuntation problem.
So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).
Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?
I get 200 for the answer but according to the quiz this is not the answer.
Could someone tell me what I am doing wrong?
James
The first room can be filled in $12\times 11\times 10\times 9 \times 8=12!/7!$ ways (counting each
permutation as a distinct way of filling the room. Thus for each combination
we have counted 5! permutations, so the number of ways of filling the first
room is $12!/(7! 5!)$.
Similaly for the next room except that we have only the $7$ left overs to use,
so there are now $7!/(3!4!)$.
There are no options for who is allocated to room three once the first
two rooms are filled.
$\frac{12!}{7!5!} \frac{7!}{3!4!}=\frac{12!}{5!4!3!}=27720
$
RonL
3. ## Thats it!!
Thanks Ron!!!
I was going about that one all wrong. It's hard for an old man to go back to school.
Thanks again
James
4. ## 'tis maybe is an indistinguishable permutation
Originally Posted by jamesinsc
I thought this would be a simple problem, but I can still not come up with the correct answer.
The problem is:
12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?
Since order is not important it is a combination problem and not a permuntation problem.
So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).
Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?
I get 200 for the answer but according to the quiz this is not the answer.
Could someone tell me what I am doing wrong?
James
.........'TIS COULD BE AN INDISTINGUISHABLE PERMUTATION
use the formula: n!/p!q!r!...
12!/5!4!3!
=479001600/17280
=27,720
MAYBE YOU COULD TRY THIS ONE...
5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways. | 2014-03-16T01:04:02 | {
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https://math.stackexchange.com/questions/1811545/which-is-the-explanation-of-the-identity-sum-k-0n-n-choose-kk2-2n-2/1811553 | # Which is the explanation of the identity $\sum_{k=0}^n {n \choose k}k^2 = 2^{n-2}n(n+1)$?
A friend of mine put me a problem some time ago, and after trying to do it, I finally surrendered. I looked for the answer online so maybe I could guess why is it like it is, but I just can't understand anything about it. Could someone explain, rather using logic, why is the identity as it is?
$$\sum_{k=0}^n {n \choose k}k^2 = 2^{n-2}n(n+1)$$
Here is a counting argument for why this identity holds. Suppose you have a population of $n$ people. Out of this group of people, you want to count the number of ways to create a club that has a president and a treasurer (the president and treasurer could be the same person). The club does not need to include all $n$ people.
One way to count the number of ways to create this club is to 1) Sum over the size of the club. 2) Pick the people to be in the club. 3) Pick the president. 4) Pick the treasurer.
So if the club has size $k$, then there are ${n \choose k}$ ways to choose the club and then $k$ choices for the president and $k$ choices for the treasurer. Overall, there are $$\sum_{k =0}^n {n \choose k} k^2$$ such ways.
Conversely, we can 1) Choose the president first (out of $n$ people). 2) Choose the treasurer. 3) Pick the remaining members of the club.
There are $n$ ways to pick the president. Now for the treasurer, we either pick (a) the president or (b) one of the $n-1$ other people. If we pick the president as the treasurer, then each of the remaining $n-1$ people are either in the club or not; as a result, there would be $2^{n-1}$ ways to pick these members in this case. Now if we chose a different person as the treasurer, then each of the remaining $n-2$ people are either in the club or not; as a result, there would be $2^{n-2}$ ways to pick these members. Thus, overall there are $$n \cdot 1 \cdot 2^{n-1} + n \cdot (n-1) \cdot 2^{n-2} = n(n+1)2^{n-2}$$ ways to select the club.
• No doubt you could extend this to $\displaystyle \sum_{k=0}^n {n \choose k}k^3 = 2^{n-3}n^2(n+3)$ but I suspect it might get a little tiring for higher powers – Henry Jun 3 '16 at 20:59
We have: $$(1+x)^n=\sum_0^n \binom nk x^k$$
Differentiate to get: $$n(1+x)^{n-1}=\sum_1^n \binom nk kx^{k-1}$$
Multiply by $x$ to get: $$nx(1+x)^{n-1}=\sum_1^n \binom nk kx^{k}$$
Differentiate again to get $$n(1+x)^{n-1}+n(n-1)x(1+x)^{n-2}=\sum_1^n \binom nk k^2x^{k-1}$$
Now let $x=1$ to get the desired result. (note that the summand vanishes if $k=0$ so you can start the sum at $k=0$ without changing the right hand).
• Thank you, this was the answer I was looking for. – user334765 Jun 3 '16 at 20:38
Another method.
Without using calculus. \begin{align} \sum_{k=0}^n\binom nk k^2&= n\sum_{k=1}^n\binom{n-1}{k-1}k\\ &=n\sum_{k=0}^{n-1}\binom{n-1}{k}(k+1)\\ &=n\left[\sum_{k=0}^{n-1}\binom {n-1}k k+\binom {n-1}k\right]\\ &=n\left[(n-1)\sum_{k=1}^{n-2}\binom{n-2}{k-1}+\sum_{k=1}^{n-1}\binom{n-1}k\right]\\ &=n\bigg[(n-1)2^{n-2}+2^{n-1}\bigg]\\ &=2^{n-2}\ n(n+1)\qquad\blacksquare \end{align}
By definition,
$$\binom nk=\frac{n!}{k!(n-k)!},$$ so that
$$\binom nkk=\frac{n!}{(k-1)!(n-k)!}==\frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!}=\binom {n-1}{k-1}n.$$
This trick allows you to "absorb" the factor $k$ inside the binomial number.
The trick doesn't work with $k^2$ because the denominator doesn't have a second factor $k$, but it does work with $k^2-k$:
$$\binom nkk(k-1)=\frac{n!}{(k-2)!(n-k)!}==\frac{n(n-1)(n-2)!}{(k-2)!((n-2)-(k-2))!}=\binom {n-2}{k-2}n(n-1).$$
Adding these two results and summing over $k$,
$$S=2^{n-1}n+2^{n-2}n(n-1)=2^{n-2}n(n+1).$$
Check with $n=5$.
\begin{align}k&\to\color{blue}1\cdot0+\color{blue}5\cdot1+\color{blue}{10}\cdot2+\color{blue}{10}\cdot3+\color{blue}5\cdot4+\color{blue}1\cdot5=2^4\cdot 5=80\\ k^2-k&\to\color{blue}1\cdot0\cdot\bar1+\color{blue}5\cdot1\cdot0+\color{blue}{10}\cdot2\cdot1+\color{blue}{10}\cdot3\cdot2+\color{blue}5\cdot4\cdot3+\color{blue}1\cdot5\cdot4=2^3\cdot 5\cdot4=160\end{align}
and summing,
$$k^2\to\color{blue}1\cdot0^2+\color{blue}5\cdot1^2+\color{blue}{10}\cdot2^2+\color{blue}{10}\cdot3^2+\color{blue}5\cdot4^2+\color{blue}1\cdot5^2=2^3\cdot 5\cdot6=240.$$ | 2020-03-31T17:11:13 | {
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https://math.stackexchange.com/questions/3180434/finiteness-of-sets | # Finiteness of sets
For each of the following sets, determine if it is finite, countably infinite, or uncountable. You need not prove your answer is correct, but you should give a reason for your response.
1. For some $$n\in\mathbb{N}$$, $$\{(a_1, a_2, \dots, a_n)\in\mathbb{N}^n\ | \ a_1+a_2+\dots+a_n<5000\}$$
2. $$\{S\in\mathcal{P}(\mathbb{N})\ | \ \mathbb{N}\backslash S\hbox{ is finite}\}$$
3. $$\{x\in \mathbb{R}\ | \ \hbox{there is a decimal expansion of x with only even digits}\}$$
For #1, I said that it is finite. Since every $$a_i \in \mathbb{N}$$, every $$a_i \leq 5000$$. Combinations of a finite set of numbers must also be finite.
For #2, I said that it is countably infinite. To make $$\mathbb{N}\backslash S$$ finite, $$S$$ would have to be an infinite subset of $$\mathbb{N}$$, of which are infinite. Therefore the initial set is infinite. Since $$\mathbb{N}$$ is countably infinite, we can reason that the initial set is also countably infinite.
For #3, I said that it is uncountable. An argument can be made with Cantor's diagonalization method to show that a list of real numbers with decimal expansions with only even digits can never be complete and the set is therefore uncountable.
Do my arguments make sense/ are my conclusions correct?
The each claim is correct. I would express the first argument as "since the set's cardinality is at most $$5000^n,$$ then the set has finite cardinality," or something similar.
Your argument for the second is hard to make sense of, unfortunately. I would say, rather, that $$\Bbb N$$ has countably-infinitely-many finite subsets (as a countable union of countable sets is countable), so it has countably-infinitely-many subsets with finite complements.
• Let $\mathscr S$ be the set defined in $2$. Then there is a $1-1$ correspondence between finite subsets of $\Bbb N$ and elements of $\mathscr S$. Since there are countably many finite subsets of $\Bbb N$, then there are also countably many elements of $\mathscr S$. This is a slightly more precise way of saying the second paragraph in this answer. – Robert Shore Apr 9 at 2:48 | 2019-06-16T14:32:16 | {
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https://math.stackexchange.com/questions/1751175/outward-flux-of-a-vector-field-through-a-rectangular-box | Outward flux of a vector field through a rectangular box.
Let $R$ be the rectangular box consisting of all points $(x,y,z)$ with $-1\leq x\leq 1$, $-1\leq y\leq 1$, and $-1\leq z\leq 1$. Define the vector field $$V= x(x^2+y^2+z^2)^{-3/2}\mathbf{i}+y(x^2+y^2+z^2)^{-3/2}\mathbf{j}+z(x^2+y^2+z^2)^{-3/2}\mathbf{k}.$$ Find the outward flux of the vector field $V$ through the boundary of the box $R$.
I tried using divergence theorem by converting to spherical coordinates but I'm not sure if I did it correctly, I got an answer of $2 \pi$. This hint was also included in the problem:
Keep in mind that the vector field V in this problem is not defined at the origin, so that you can not apply the Divergence Theorem to the rectangular box and its interior.
Which makes me assume that my method is possibly incorrect, however I'm not sure if converting to spherical coordinates makes the use of the Divergence Theorem valid.
• Both field and box symmetric with respect to each axis. – mathreadler Apr 20 '16 at 13:19
• @b00nheT Glancing at the form of the vector field suggests it's not pleasant to integrate (though we could get away with using symmetry to reduce the number of integrals we'd actually have to compute). – Travis Willse Apr 20 '16 at 13:22
• You actually end up with just one integral to compute, but I must admit that it is not the most elegant solution – b00n heT Apr 20 '16 at 13:25
You're right: Since $\bf V$ is not defined at $0$ but $0 \in R$, the hypotheses of the Divergence Theorem do not hold for the desired flux integral, $$\iint_{\partial R} {\bf V} \cdot d{\bf A} .$$
Hint There are two key observations here: The first is that $\bf V$ is radially symmetric (that is, we can write it as $f(r) {\bf r}$ for some function $f$ (here, ${\bf r}$ is just the unit radial vector field). So, while it is difficult to evaluate the integral over $R$ directly (because the expression for the components of $\bf V$ are complicated), it would be much easier to find the flux of $\bf V$ across a sphere $S$ of radius $R$ centered at the origin, as we could write $$\iint_S {\bf V} \cdot d{\bf A} = \iint_S f(R) {\bf r} \cdot {\bf r} \,dA = f(R) \iint_S dA = 4 \pi f(R).$$
Since we suppose we might want to use the Divergence Theorem, we may as well compute $\operatorname{div} {\bf V}$, just to see what kind of function you'd end up integrating, and this yields our second key observation, namely that $$\operatorname{div} {\bf V} = 0 .$$ (To carry out this computation, it's best to use the spherical coordinate expression for $\bf V$ and then use the corresponding divergence formula.)
Remark In fact, we can use the above ingredients to see that $\bf V$ is actually quite a special vector field. We can ask which radial vector fields $\bf X$ (say, defined on $\Bbb R^3 - \{ 0 \}$) have zero divergence. Substituting ${\bf X} = g(r) {\bf r}$ into the divergence formula for the sphere gives that our condition is $$0 = \operatorname{div} {\bf X} = r^{-2} \frac{d}{dr} [r^2 g(r)] .$$ Multiplying both sides by $r^2$ and integrating gives $r^2 g(r) = C$ for some constant, and hence $g(r) = \frac{C}{r^2}$, but this is exactly the form of $\bf V$ for some $C$, so the only radial vector fields with zero divergence are the given vector field $\bf V$ and its constant multiples.
• Ok thank you, yeah I was somewhat confused when I first attempted the problem; after finding the divergence of V it made me assume the answer was 0, but I wasn't sure if that was too simplistic. – George Bencosme Apr 20 '16 at 13:33
• To be clear, $\operatorname{div} {\bf V} = 0$, but the value of the integral in the question is not zero. – Travis Willse Apr 20 '16 at 13:48
• Ok I think I understand. Just as clarification, for f(R) is there a way that V is used to find this function or am I approaching it the wrong way? – George Bencosme Apr 20 '16 at 14:03
• Yes, and there's an easy way in fact: Since ${\bf V} = f(|{\bf r}|) {\bf r}$, taking the dot product with $\bf r$ gives $f(|{\bf r}|) = {\bf V} \cdot {\bf r} = f(|{\bf r}|) {\bf r} \cdot {\bf r} = f(|{\bf r}|)$. (Note that I've rewritten $r$ as $|{\bf r}|$ for emphasis.) – Travis Willse Apr 20 '16 at 14:39
• Oh I see! So taking the dot product of $V$ and $r$ will give me the function to multiply by $4π$? – George Bencosme Apr 20 '16 at 14:45 | 2020-02-29T13:11:14 | {
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https://www.physicsforums.com/threads/can-you-help-me-solve-these-sequences.209320/ | # Can you help me solve these sequences
1. Jan 17, 2008
### wonderful
Dear all,
I would highly appreciate if you could help me solving the following sequences:
Question 1: 3,5,8,24,209,3591,?
33811
34308
35534
35200
35010
Question 2: 4,7,13,21,34,55,88,?
110
148
123
138
Question 3: 8,12,18,27,42,70,126,241,?
478
441
503
488
486
Question 4: 3,8,23,60,137,256,?
500
513
507
511
512
Question 5: 5,10,19,32,49,70,?
89
95
121
135
99
2. Jan 17, 2008
### Staff: Mentor
Have you tried to find the pattern for each of these problems? Try Question 5, you should recognize the pattern by the time you get to 19.
Finding the pattern is the fun in doing these types of problems.
May I ask what those numbers listed under the question are? The question is asking what comes after 70, but none of the numbers below are correct following the pattern.
Last edited: Jan 17, 2008
3. Jan 17, 2008
### Rogerio
???
Of course, 95 is the correct answer!
4. Jan 17, 2008
### Staff: Mentor
Duh, I missed the 95. Completely skipped over it.
And you gave away the answer!
5. Jan 17, 2008
### wonderful
Thanks! I figured it out why 95 is the answer for question 5:
10 - 5 = 5
19-10 = 9
32-19= 13
49 - 32 = 17
70 - 49 = 21
Consider the sequence : 5, 9, 13, 17, 21 the increment is 4. Thus, the next number is something that is greater than 70 by 21+4=25.
How about questions 1 - 4? Can anyone find the solution?
Have A Great Day!
6. Jan 22, 2008
### coomast
I have solved the third question. The others are not easy. If anybody wants a hint, just ask :-)
@ wonderful: Is there a change that you have gotten these questions from the following site? http://www.allthetests.com/quiz18/quizpu.php?testid=1141425609 If so then you have forgotten a possible solution at question 2, namely 140. Whether this is the right solution or not I can't tell because I haven't been able to solve this one so far.
There are also 5 other questions, 6 to 10 of which I have solved 9 and 10. If wanted you might post them, I'm not going to because I don't want to mess up your post.
Anybody else solved one of the others?
7. Jan 24, 2008
### wonderful
I had been out of town in the last two days. Coomast is correct in saying that 140 is also one of the choices in question 2. Coomast can you give us a hint for question 3? These questions are very interesting.
Have A Great Day!
8. Jan 24, 2008
### coomast
A hint for question 3: use the same principle as for solving question 5. Write line per line the differences of two adjacent numbers. Can you see a known sequence after 4 lines of differences? Hope this is clear, otherwise please ask.
I'm working on question 1 and I think I've got something. I'm now working on the general formula, but it isn't straightforward.
These are very strange as a IQ test. One can't possibly have the time to solve all of them within the time frame of such a test. On the other hand they are very nice brain teasers indeed :-)
9. Jan 24, 2008
### wonderful
Thanks coomast for the hint. As you suggested, I looked at the sequence of difference between two adjacent numbers: 4, 6, 9, 15 .... but saw no clear pattern. Can you help? It seems to mention on the related website that there is no time limit on these kind of questions.
10. Jan 25, 2008
### coomast
Take the differences again for this newly created line, and again, and again. Then you can see a pattern related to some powers.
I've solved the first one as well, this is an even more complicated sequence. A hint: take the differences as done before a few times. Then a pattern involving 13 appears together with the adding of another sequence. This is a complicated one indeed. The general formula looks awesome.
The general formula for question 5 can be written as:
$$x_0=5$$
$$x_1=10$$
$$x_n=4+2 \cdot x_{n-1}-x_{n-2}$$
It is a nice exercise to look for these as well.
11. Jan 25, 2008
### wonderful
Cool! Thanks coomast. I found the answer for question 3 is 478. I am going to solve question 1. These questions are very interesting and great for brain.
12. Jan 25, 2008
### wonderful
Thanks to coomast hint, I've found the solution for question 1: 35534. Thus, we have two not-yet-solved questions 2, 4. After solving these two, I will post new questions that are as intersting as the above 5.
13. Jan 25, 2008
### coomast
Indeed wonderful, that's the one. I will try to look for the second and fourth sequence.
14. Jan 25, 2008
### wonderful
Following the same line of logic, I found the solution for question 2 is 140
Have A Great Day!
15. Jan 26, 2008
### coomast
Can you give an additional hint on sequence 2? I don't see a pattern.
Is it possible that the next number after 140 equal is to 219?
Last edited: Jan 26, 2008
16. Jan 26, 2008
### wonderful
I am very happy to share ideas with you. Take the first difference we have: 3, 6, 8, 13, 21, 33. Take the second difference we have 3, 2, 5, 7, ... Here, we can see: 3+2=5, 2+5=7, and so on...
Please feel free to let me know if this helps.
Have A Great Day!
17. Jan 27, 2008
### coomast
Hello wonderfull, I looked at this for some time and didn't see any order in it. Finally I figured it out, I think you made an error. The sequence of the differences of the differences reads: 3 2 5 8 12, you seem to have a 7 instead of the 8. Or am I missing something?
18. Jan 27, 2008
### wonderful
Hi coomast
Thanks for pointing that out. Could you give us a hint on this question.
Have A Great Day!
19. Jan 29, 2008
### coomast
I didn't solve this one yet, also number 4 seems almost a random number writing. I keep working on it.
20. Feb 3, 2008
### coomast
Sorry wonderful, I can't find the others. | 2016-10-25T10:24:59 | {
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https://electronics.stackexchange.com/questions/373819/collector-voltage-for-this-transistor-circuit | # Collector voltage for this transistor circuit
For the circuit below, $$β = 80 \quad and \quad V_{BE} = 0.7\;V$$
I applied KVL to the closed loop on the left -base-emittor loop- and the bigger closed loop on the right -collector-emittor loop- (just the bigger rectangle on the right). I thought that V0 would be equal to the voltage between GND and the wire that the upper 10k is connected to, which is V subscript CE.
So from the first loop I mentioned, $$-1+120\times10^3I_B+V_{BE}=0$$ $$-1+120\times10^3I_B+0.7=0$$ $$120\times10^3I_B=0.3$$
Thus, $$I_B=2.5\times10^{-6}\,A=2.5\,μA$$
From the equality $$I_C=βI_B$$ $$I_C=80\times2.5\times10^{-6}\,A=0.2\,mA$$
Applying KVL to the second loop I mentioned above, $$-V_0-10\times10^3I_C+20=0$$ $$-V_0-10\times10^3\times0.2\times10^{-3}+20=0$$ $$V_0=-2\,V+20\,V=18\,V$$
Thus, $$I_0=\frac{V_0}{10\,kΩ}$$ $$I_0=\frac{18\,V}{10\,kΩ} =\,1.8\,mA$$
But the book says $$V_0=12\,V \;and\; I_0=600\,μA$$
I found some people saying the book is wrong, but I wonder about the answer, and
• How would you solve this?
• Is the sentence in bold (above) true?
• With no transistor, Vo is 10 volts. Collector current pulls Vo toward ground. The usual practice is to reference all voltages to ground, as you suggest. – glen_geek May 11 '18 at 22:32
• It is better to give each resistor and voltage source a name and insert the values at the last moment. – Peter Mortensen May 12 '18 at 5:31
• Even if a book fails to do it. – Peter Mortensen May 12 '18 at 5:40
Well, at just first glance the answer is wrong - if $I_0 = 600\,µA$ then $V_0 = I_0 * 10\,kΩ= 6\,V$; $6\,V\ !=\ 12\,V$ so the answers are contradicting.
Here is the solution:
Base current is what you calculated: $I_b = 2.5 * 10^{-6}\,A$
Collector current: $I_b*\beta = 2 * 10^{-4}\,A$
Now $V_0$ is simply: $20\,V - i10^4\,Ω$ ($i$ is the current of the top most resistor)
Currents equation: $i = i_c + i_0$ ($i_c$ is collector current)
Next we put current equation into $V_0$ equation: $V_0 = 20\,V - (i_c + i_0)10^4\,Ω$
$I_0$ is $\frac{V_0}{10^4}$
So we combine above three equations into: $V_0 = 20\,V - (i_c + \frac{V_0}{10^4\,Ω})10^4\,Ω$
$2V_0 = 20\,V - i_c*10^4\,Ω$
$V_0 = 9\,V$
and finally
$I_0 = \frac{9\,V}{10^4\,Ω} = 900\,µA$
• Yes, that was what I'd done to check the answer too. ^^ – André Yuhai May 11 '18 at 23:16
• Oh, then my mistake was that I just had considered $I_C$ to be the current through the top most 10k resistor, I hadn't thought that it would be $I_0+I_C$. Everything is clear now. :) – André Yuhai May 12 '18 at 0:01
• Good to hear :) – DannyS May 12 '18 at 12:34
The first thing you do is rewrite the problem:
simulate this circuit – Schematic created using CircuitLab
That's done by simply performing the usual Thevenin equivalent on $R_1$ and $R_2$ and the voltage applied across them to compose a new Thevenin voltage source of $10\:\text{V}$ and Thevenin resistance of $5\:\text{k}\Omega$. At this point, it's quite obvious that the output cannot be more than this Thevenin voltage.
The base current is $I_\text{B}=\frac{1\:\text{V}-700\:\text{mV}}{120\:\text{k}\Omega}=2.5\:\mu\text{A}$. With $\beta=80$, this says that $I_\text{C}=80\cdot 2.5\:\mu\text{A}=200\:\mu\text{A}$. The drop across the Thevenin resistance ($R_4$ above) is $200\:\mu\text{A}\cdot 5\:\text{k}\Omega=1\:\text{V}$.
Therefore, the output will be $1\:\text{V}$ below the Thevenin voltage of $10\:\text{V}$, or $V_\text{O}=9\:\text{V}$.
You can separately analyze it by using KCL at the collector node, using the value of $I_\text{C}$ computed earlier (see above):
$$\frac{V_\text{O}-20\:\text{V}}{R_1}+\frac{V_\text{O}}{R_2}+I_\text{C}=0\:\text{A}$$
That solves out as $V_\text{O}=\frac{R_2}{R_1+R_2}\cdot\left(20\:\text{V}-R_1\cdot I_\text{C}\right)$
And it gets you the exact same answer.
From there it is quite easy to work out the requested current.
Yes, the book is wrong if that is what it said.
• +1 for thevenin method. Btw, you have mistake in base current - in the denominator you have 1k while it is 120k Ohm. – DannyS May 12 '18 at 8:03
• Well, I have a question. I understood why the Thevenin resistance became 5k but I couldn't really get why that 20V would become 10V as the Thevenin voltage. Can you please explain? – André Yuhai May 12 '18 at 8:14
• @AndréYuhai The Thevenin voltage is the voltage that would appear at the mid-point if the mid-point is unloaded. I think you can see that this would be 10 V. – jonk May 12 '18 at 19:30
The answer is wrong. Suppose there isn´t a transistor in the circuit and it´s only both 10 kΩ resistors. Then Vo = 10 V in this situation.
Adding anything in parallel to it will cause that equivalent resistance to decrease, so it´s impossible to have Vo = 12 V, since Vo <= 10 V
As you´ve already figured out Ic = 0.2 mA, then you can apply KVL to both resistors as a loop as follows: 20 - 10 kΩ(Io + 0.2 mA) - 10 kΩ*Io = 0.
Solve for Io, and Io = 0.9 mA, and as a result, Vo = 0.9 mA * 10 kΩ = 9 V
• Then my answer $V_0=18V$ is wrong too. – André Yuhai May 11 '18 at 23:18
• @AndréYuhai yes, unfortunately it cannot be greater than 10v. I´ll edit my answer and post the way I did – Flávio Alegretti May 11 '18 at 23:21 | 2020-02-21T07:13:00 | {
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http://arcoeste.it/hmsz/surface-area-of-sides-of-cylinder.html | The next step involves finding the surface area of a circle. Here we want to find the surface area of the surface given by z = f (x,y) is a point from the region D. Perimeter is equal to π x 2 × radius. Recall the formula for SA of a cylinder, plug in the integer, decimal, or fractional radius measures in the formula 2πrh + 2πr 2 and compute. Therefore, its surface area will be the sum of the areas of all six surfaces. Example: The surface area of a cube is 384 m 2. Calculate surface area of cylinders or tanks. 7 m, find its height. Total surface area of cylinder is the sum of the area of both circular bases and area of curved surface. Question 1. So the first step is to work backwards from the surface area of the side length. Solution: We assume the base radius = r. A set of 20 task cards for helping students find the surface area of a cylinder. C Program to Find the Volume and Surface Area of cylinder - IProgramx by - IProgram X on - June 21, 2018 Q. To understand the formula for the surface area of a cylinder, think of a can of vegetables. The cardboard costs 0. h: The height of the cylinder. Key Terms. You'll gain access to interventions, extensions, task implementation guides, and more for this instructional video. ) 2 S — 600 = 36 — 100 Evaluate power. To calculate the surface area of a cube we need the length of any side of a cube. 75 cm) Bottom (same as top), and Label (11 cm high) The Top Formula for. Details Written by Administrator. The surface area formula for a cone is A = π r 2 + π rl, where r is the radius and l is the slant height. (15) The Surface Area Of A Cylinder As Shown In The Figure Can Be Expressed As: S=Top Area + Bottom Area + Side Area =f(W, H). [1] 6) A cube has a surface area of 384 cm 2. The surface area of such a cylinder is simply LH. Using the latter of the two methods:. The volume is the area of the base circle (pi*r^2) times the height (h) -- just like the volume of a rectangular solid. If a circular cylinder has a radius #f# and height #h#, what is the expression that represents the surface area of its side? Algebra. If the can has radius r and height h, its surface area A and its volume V are given by the equations: A= 2πr^2 + 2πrh and V=πr^2h. If s is the length of one of its sides, then the area of each side of a cube is s 2. Calculate the unknown defining side lengths, circumferences, volumes or radii of a various geometric shapes with any 2 known variables. Find each area, then add. I forgot the formula how to work this out can somebody help me. You need to know the radius and height to figure both the volume and surface area of a cylinder. Now in my textbook, it says that surface area for a cube is different from a cylinder. A cylinder is a geometric solid that is very common in everyday life, such as a soup can. The formula for calculating the surface area incorporates these three shapes. Lateral Area is often abbreviated L. Find the thickness of the cylinder. of a cylinder with height h and radius r is the sum of the area of the curved side, which is just a big rectangle (just as wrapping paper or paper towels are rectangles when you unrol them)l, and the area o f the two. Lateral surface = 5 U K. Total Surface Area of Cylinder = 2πr ( r + h ). 14 radius = float (input ('Please Enter the Radius of a Cylinder: ')). Note that the surface area of the top and bottom is just plainly the area of a circle, (pi)r^2. The cardboard costs 0. Write a Python program to calculate surface volume and area of a cylinder. This means that in order to find its surface area or volume, you only need the radius (r) and height (h). As part of the Mensuration topic. Surface Area. Make your job easier and see how to use a net to find the surface area of a prism. If you carefully cut the label off the side of the can and unroll it, you will see that it is a rectangle. pi is the ratio of the circumference to the diameter of a circle. Surface area of a cylinder : To find the surface area of a cylinder add curved (lateral) surface area and area of both bases. This shape has a circular base and straight, parallel sides. Martin Overview Behold!A can with a: Top (radius 3. The curved surface area (CSA) of a cylinder with radius r and height h is given by. Virtual Maths is a free to use online resource that encourages innovative teaching and learning of functional mathematics, using links to real life problems. Wand Hare Dimensions Of A Sheet As Shown Below. The side of the cylinder, which when "unrolled" is a rectangle. In differential geometry, a cylinder is defined more broadly as any ruled surface spanned by a one-parameter family of parallel lines. An object shaped like a cylinder is said to be cylindrical. So it's actually the same exercise as the triangular prism. Thanks for the feedback. Lesson 11-2 Surface Areas of Prisms and Cylinders 611 Finding Surface Area of a Cylinder The radius of the base of a cylinder is 4 in. Area of sides (laterals surface) is 2pirh where h is the height. The length of the rectangle is the circumference of the circle which is 2 π r = 31. The area of a semi-circle is 1/2 πr2, and the area of a triangle is 1/2 bh. 52 cm 2, the area of both bases, and the outer surface area, 94. cm and the area of the base ring is 115. One Time Payment (2 months free of charge) $5. Side area = 2*pi*r*H. The easiest way is to think. (See Surface area of a cylinder). Find the surface area of a cylinder with the following dimensions: a) height 10 cm, radius 3 cm b) height 8 mm, diameter 12 mm 3. It is one of the most basic curvilinear geometric shapes. More Geometry Subjects. Humanscale TouchPoint T7 Non-Powered 250N Cylinder Laptop Gantry and PC Surface - Cart for notebook - medical - aluminum, steel - white. The base is a right triangle with sides 6 cm, 8 cm and 10 cm. If you were to 'unroll' the cylinder you would find the the side is actually a rectangle when flattened out. Lateral surface, right prism, right regular pyramid, frustum of a cone or pyramid, torus, surface area of a surface of revolution. Determine the surface area and volume of this shape: The radius of a sphere is 4. Area of top is π x r². Curved surface area (lateral surface) = Perimeter x height. Once you add the area of the two bases and the outer surface area, you will have found the surface area of the cylinder. 14 * 5² * 10 = 3. the cylinder fits in a cube (height = diameter). Base of a quarter cylinder is a quarter of a circle of some radius r. 2 Surface Area of Prisms and Cylinders. Write a c program to find the surface area and volume of a cone. Answers for volume problems should always be in cubic units. A rectangular prism or cuboid is formed by folding a net as shown:. It has three surfaces: the top, the bottom, and the piece that forms the sides of the can. The lateral surface of an object is the area of all the sides of the object, excluding the area of its base and top. Cylinder calculator will give the surface area and volume of a cylinder. if a circular cylinder has radius has radius r and height h, write an expression that represents the surface area of its side. 14 × 2 2 + 2 × 3. Let R be the external radius of cylinder and r be the internal radius of cylinder. Find its surface area. cm, area of base ring is 115. If you were to 'unroll' the cylinder you would find the the side is actually a rectangle when flattened out. The base is a right triangle with sides 6 cm, 8 cm and 10 cm. Which cylinder. 141595, or even shorter, 3. The area of each base is π r 2. If a circular cylinder has radius (r) and height (h), write an expression that represents the. and height 7 cm. Surface Area of a Cube 5. Note that the surface area of the bases of the cylinder is not included since it does not comprise part of the surface area of a capsule. The volume of an oblique cylinder turns out to be exactly the same as that of a right cylinder with the same radius and height. Now, the area of the side of the cylinder is then a simple rectangle area calculation - 6pi cm * 4 cm or 24pi cm^2. Let r , in centimeter be the radius of base of cylinder and h, in centimeter be height of cylinder. SURFACE AREA | sum of the lateral faces and the bases facelateraltheofABasetheofASA 4. Lateral means side, so the lateral surface area of a prism, pyramid, cylinder, and cone does not include the area of the base (or bases). Both the surface and the solid shape created inside can be called a cylinder. Volume = πr²h = 3. There are two ends so their combinded surface area is 2???* r2. A set of 20 task cards for helping students find the surface area of a cylinder. h – is the height of the cylinder. The area of each base is π r 2. Volume of cylinder = Area of base x height = 2 U K. Students are prompted to measure dimensions and find the area of the semicircles and rectangles. 3259 views. If you 'unroll' a cylinder you have a shape of a rectangle, similar to a sheet of paper. The surface area of a cylinder can be derived from the area of rectangle. If necessary, round the answers to the nearest tenths. This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Area of the face of the cube = x^2. Mar 10, 2009. Area of rectangle is equal Lateral Surface area of Right Circular Cylinder. If the radius of the cylinder is 5, for instance, the area of one base is pi * 5^2 or 25pi, while the area of both of the bases combined is 2 * 25pi, or 50pi. Surface Area = 2(Area of top) + (perimeter of top)* height. Worksheets and More Examples:. A cylinder is a three-dimensional geometrical figure/container with straight parallel sides and two circular cross-sections. Calculate the surface area of this cylinder to the nearest hundredth of a square centimetre. The Surface Area has these parts: Surface Area of Both Ends = 2 × π × r 2. 4 identical cylinders are cut from each base as shown in the given figure. If the cylinder has neither base nor top: TSA of open cylinder = CSA of cylinder = 2πrh If cylinder has base but no top or top but no base: TSA of open cylinder = CSA of cylinder + area of base or top = 2πrh + πr^2 = πr(2h + r). Area of bottom is π x r². Example 6: A cube whose sides are 10. This means that in order to find its surface area or volume, you only need the radius (r) and height (h). The calculator will do all the work for you rapidly with precise results. The volume of a cylinder is found using the concept of base x height. and height 7 cm. You can use this information to determine the Surface Area of each shape. The thickness of the material is required to be. 2h) on surface area of cylinders. and a heig [ 4 Answers ] Find the lateral area and surface area of a right cone with radius of 9 in. You can use this information to determine the Surface Area of each shape. The total surface area of hollow cylinder, which is open from both sides, is 3 5 7 5 c m 2; area of the base ring is 3 5 7. The volume of a cone is one third of the volume of a cylinder. To find the surface area of a cylinder add the surface area of each end plus the surface area of the side. Find the surface area of this triangular prism. Each end is a circle so the surface area of each end is π * r^2, where r is the. If you take it apart you find it has two ends, called bases, that are usually circular. Volume of a sphere in n-dimension Volume of a sphere in n-dimension: Calculates the volume and surface area of a sphere in n-dimensional space given the radius. A rectangular prism has a base perimeter of P and height h. Correct answers: 1 question: The surface area, A, of a cylinder is given by the formula A = 2ttrh + 2tr2, where ris the radius and h is the height of the cylinder What is the surface area of a cylinder with a height of 5 and a radius of 2? A. Online calculators and formulas for a cylinder and other geometry problems. Area of the face of the cube = x^2. ) for each cylinder by using the formula S. So the area of the cylinder will be: 2πr² + 2πrh, or. Martin Overview Behold! A can with a: Top (radius 3. 15 11 16 10 Find the lateral area and surface area of each figure. This gives us the result that the flux through a portion of the surface of a cylinder of radius R oriented along the z-axis is where T is the -region corresponding to S. Example 1 Find the surface area of the part of the plane 3x+2y+z = 6. Surface Area of a Cylinder = 2πr (r+h) Surface Area of a Cylinder = 2 * Math. Find the change in surface area dA (in units{eq}^2 {/eq}) if the radius of a sphere changes from r by dr. A cylinder is defined as a closed solid object with two identical circular or elliptical bases joined by curved side. There are two ends so their combinded surface area is 2 π * r 2. Surface area of a cylinder Definition: The number of square units it takes to exactly cover the surface of a cylinder. Pi is approximately equal to 3. In this program, we have a cylinder with the given radius and height. Processing. Cylinder calculator will give the surface area and volume of a cylinder. (Answer=7/19cm) Find the cost of plastering the inner surface of a well at the rate of Rs 30 per m2, if the inner diameter of the well is 2. But because there are two circles in a cylinder, we adjust the formula. We would find. 52 cm 2, the area of both bases, and the outer surface area, 94. If you are given the surface area, you can determine the side length by working backwards. Its surface area is therefore, A = L + 2 B = 2π rh + 2π r 2 = 2π r (h + r) = π d (r + h), where d = 2 r is the diameter of the circular top or bottom. Once the area of one triangular side is determined, it can be multiplied by 2 for the area of both triangles. Online calculators and formulas for a surface area and other geometry problems. Consider the edge of the cube as. ; A cylinder is a three-dimensional solid that consist of two congruent surfaces (bases) and one lateral surface. This shape is similar to a can. Surface area of the side. Subscribe for more videos. 14 × 2 2 + 2 × 3. Write a Python program to calculate surface volume and area of a cylinder. Find the surface area of a triangular prism if the area of its cross section is 20cm 2, its length is 41cm and the 3 sides of the triangular ends add up to 21. Surface Area of Cylinders. (Answer=7/19cm) Find the cost of plastering the inner surface of a well at the rate of Rs 30 per m2, if the inner diameter of the well is 2. Solution: Note: The circular base of the cylinder is drawn as an ellipse. For example, you can consider a tin can with lids on the top and the bottom. How Do You Find the Surface Area of a Rectangular Prism Using a Net? Finding the surface area of a prism can be a little tricky, but a net can make the problem a little easier. Surface Area of a Cylinder = 2 pi r 2 + 2 pi r h (h is the height of the cylinder, r is the radius of the top) Surface Area = Areas of top and bottom +Area of the side. The total surface area of a hollow cylinder, which is open from both sides, is 4620cm2. Given a cylinder with the height of 10cm and the base radius of 9cm. While you're stuck at home, make the most of your time by learning a new language , skill , or even train for a remote-work job with our new premium online courses. 5 The surface area of an object is the sum total of the areas of each of the faces of the object. Each end is a circle so the surface area of each end is???* r2, where r is the radius of the end. Name: Date: 4. Surface Area of a Cylinder. Example 1 Find the surface area of the part of the plane 3x+2y+z = 6. A right cylinder, having the same radius has the same volume. For example, a cylinder is not a prism, because it has curved sides. Surface Area = 2(Area of top) + (perimeter of top)* height. Your email is safe with us. 7) 2 SA = 686. Formulas for Area and Volume 1) The lateral surface of the right cylinder is the surface without the top and bottom surfaces. 2f) on surface area of cylinders. Thus the area of the rectangle is 23 × 31. 99 USD per year until cancelled. #"Lateral Surface Area of a cylinder " L S A = 2 pi r h# #"Base Area of the cylinder " A = pi r^2, " Area of Top & Bottom Circles"# #"Total Surface Area of cylinder " T S A = 2 * A + L S A#. Solution: Using the above formula for calculation, the value of cross sectional area will be: Cross Sectional Area = π x (3 meter)2 = 3. Find its volume and total surface area. So the first step is to work backwards from the surface area of the side length. 267 times 216 is 57672 millimeters squared. Therefore, in order to calculate the surface area of a cylinder you need to calcuate the area of a circle (twice) and the area of a rectangle (once). The surface area of any of these solids -- right prisms or cylinders -- is: (2 x Area base) + (height x Perimeter base). times area of corresponding lid). Which cylinder. Find the thickness of the cylinder. If you were to 'unroll' the cylinder you would find the the side is actually a rectangle when flattened out. Formulas for Area and Volume 1) The lateral surface of the right cylinder is the surface without the top and bottom surfaces. Find the surface area of a cylinder with the radius 4 and height 8. The base of a prism is a right angled triangle with two sides 5 cm and 12 cm. (1) Right Circular Cylinder Curved surface area = the perimeter x the height of the cylinder, i. Area of top of cylinder is pir^2 , where r is radius and pi = 3. Some cards have diagrams with the dimensions, others are word problems, some list the dimensions and some ask students to work backwards, giving them the surface area and asking them to find the height. SA = 2 x base x altitude + (side 1 + side 2 + side 3) x Height. Curved surface area of a right circular cylinder is 4. In other words, the surface area is equal to the summation of the area of each of the faces. The area of the circle is given in Equation 23. ) for each cylinder by using the formula S. The area of a semi-circle is 1/2 πr2, and the area of a triangle is 1/2 bh. The surface. Surface area of a cylinder = 2πr 2 + 2πrh. Volume and Surface Area Questions & Answers : A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. Provide students with a variety of problems concerning surface area and volume. Divide both sides by and take the square root. There are two ends so their combinded surface area is 2 π * r 2. The area of the side is equal to the area of a rectangle with length (the circumference of the base) times the height. Each end is a circle so the surface area of each end is π * r 2 , where r is the radius of the end. 7 m, find its height. By using this website, you agree to our Cookie Policy. The sides of the can is called the lateral area and the formula for finding the lateral area equals, diameter times pi times height Next I need to add the top of the cylinder and the bottom of the. An object shaped like a cylinder is said to be cylindrical. The two circles that make up the caps of the cylinder. So it becomes two times pi r squared. com - id: 436ec3-NzFkN. ) Chloe wants to wrap a present in a box for Sarah. Online calculators and formulas for a surface area and other geometry problems. 5 cents per square centimeter. Find the surface area of this cylinder. Surface area of a cylinder. Example 2 | If the surface area of a cube is 216 cm2, determine the measurement of the side. 14 * 5² * 10 = 3. Find the surface area of a cylinder with the radius 4 and height 8. Surface Area of Triangular Prisms Guided Practice Here is one for you to try on your own. The surface area formula for a cone is A = π r 2 + π rl, where r is the radius and l is the slant height. Cylinder : A solid generated by the revolution of a rectangular about one of its sides is called a right circular cylinder. Calculate the total surface area of a regular pyramid if given base perimeter, slant height and base area ( A ) : lateral surface area of a regular pyramid: = Digit 1 2 4 6 10 F. Like the box, this shape has a pair of sides with areas which we will again call Area hw. Do It Faster, Learn It Better. Find the surface area of a cylinder with a radius of 2 cm, and a height of 1 cm SA = 2 × pi × r 2 + 2 × pi × r × h SA = 2 × 3. Find the surface area of this cylinder. Answers for volume problems should always be in cubic units. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides. Calculates the volume, surface area and radii of inscribed and circumscribed spheres of the regular polyhedrons given the side length. ft_ Surface Area = _54 sq. Worksheets and More Examples:. To keep the calculation simple, let's say that the value of r (the radius of the circle) is 1 unit and the value of height (h) is 2 units. Join 100 million happy users! Sign Up free of charge: Subscribe to get much more: Please add a message. There are two ends so their combinded surface area is 2 π * r 2. Let us consider a right circular cylinder. Total surface area of a hollow cylinder = 2 π R h + 2 π r h + 2 (π R 2 − π r 2) Example A hollow cylinder (open at both ends) has an external diameter of 28 cm, length of 14 cm and thickness of 2 cm. As you already know, the formula for calculating the area of a circle is pi times r squared. The surface area formula for a cone is A = π r 2 + π rl, where r is the radius and l is the slant height. Recall the formula for SA of a cylinder, plug in the integer, decimal, or fractional radius measures in the formula 2πrh + 2πr 2 and compute. com Surface Area of a Cylinder Calculate the Surface Area (S. The surface area of a cylinder aluminum can is a measure of how much aluminum the can requires. Cylinder Surface Area = 2 x pi x radius x height + 2 x pi x radius squared. 3 cents per square centimeter, and the plastic costs 0. A tent is in the shape of a cylinder surmounted by a conical top. First find the area of the sector and double it. If the height and diameter of the cylindrical part are 2. All you have to do is add 56. ) for each cylinder by using the formula S. First we need to establish the area of the two identical circles. Surface Area of a Cube. Why some people say it's true: The total surface area of a right cylinder is 2 π r (r + h) 2 \pi r (r + h ) 2 π r (r + h), where h h h is the height of the cylinder. The radius 'r' of a cylinder is the radius of its base. If you 'unroll' a cylinder you have a shape of a rectangle, similar to a sheet of paper. PI * radius * (radius + height) Surface Area of a Cylinder = 2 * 3. The surface area of the side is the circumference times the height or 2 π * r * h, where r is the radius and h is the height. The calculator will do all the work for you rapidly with precise results. Now the surface area of the cylinder we need to calculate the surface area of the top, base and sides. As shown here, the lateral area of a cylinder is the area of the sides of the cylinder—namely, […]. com - id: 6e9908-N2ExO. Determine the Surface Area of a Cylinder Click here to choose another surface area calculator The surface area of a cylinder can be determined by using the following formula: where r is the radius of the base circle and h is the height of the cylinder. So each side is equal length x, which happens to equal 3. For a pyramid, the lateral surface area is the sum of the areas of all of the triangular faces but excluding the area of the base. We have even a 3D artist on the team. If you carefully cut the label off the side of the can and unroll it, you will see that it is a rectangle. Surface Area of a Rectangular Prism using Nets. It is important to use the perpendicular height (or 'altitude') when calculating the volume of an oblique cylinder. The total surface area of a hollow cylinder which is open from both sides is 4620 sq. Find the change in volume dV (in units{eq}^3 {/eq}) if the sides of a cube. The surface area of a cylinder is the total area of the cylinder. (Answer=7/19cm) Find the cost of plastering the inner surface of a well at the rate of Rs 30 per m2, if the inner diameter of the well is 2. The area of a circular disk is π r 2 so each disk has an area of π × 5 2 = 78. A tent is in the shape of a cylinder surmounted by a conical top. Free Online Scientific Notation Calculator. Surface Area of a Cylinder: The surface area of a cylinder is the sum of the areas of all of its surfaces. As the cube gets bigger, the volume increases much more rapidly than the surface area, because the volume increases as the cube of the linear dimension, but the surface area increases as the square. For example, you can consider a tin can with lids on the top and the bottom. See page for details. Like area, surface area is also expressed in square units. A cylinder is to be inscribed in the cone so that the axis of the cylinder coincides with the axis of the cone. Find the change in volume dV (in units{eq}^3 {/eq}) if the sides of a cube. and a heig [ 4 Answers ] Find the lateral area and surface area of a right cone with radius of 9 in. Solve the resulting quadratic equation. 99 USD per week until cancelled: Monthly Subscription$2. A = r2 49 = r2 r = 7 Area of a circle Substitute 49 for A. The area of the side of the cylinder is the perimeter of the circle × the height of the cylinder. Calculating the lateral area of a cylinder allows to you determine the area of the heat from a heat exchanger or from a coiled radiator. Cylinder Surface Area = 2 π r h + 2 π r 2. 141592654 r = radius h = height. The surface area of such a cylinder is simply LH. Find the surface area of this cylinder. Since the top and bottom of the cylinder have the same area, we can multiply the area of the circle (πr²) by 2. First we need to establish the area of the two identical circles. Some cards have diagrams with the dimensions, others are word problems, some list the dimensions and some ask students to work backwards, giving them the surface area and asking them to find the height. The area of the side is known as the lateral area, L. Active 5 years, surface area of the solid (column side) 0. 4 × 10 = 314cm 2. Your email is safe with us. 2 Side Length, Volume, and Surface Area of Similar Solids. Volume of. If the radius of the base of the cylinder is 0. All you have to do is add 56. Now, the area of the side of the cylinder is then a simple rectangle area calculation - 6pi cm * 4 cm or 24pi cm^2. Cylinder Surface Area Formula. #"Lateral Surface Area of a cylinder " L S A = 2 pi r h# #"Base Area of the cylinder " A = pi r^2, " Area of Top & Bottom Circles"# #"Total Surface Area of cylinder " T S A = 2 * A + L S A#. In order to calculate surface area it is sometimes easier to draw all the surfaces. Determine the Surface Area of a Cylinder Click here to choose another surface area calculator The surface area of a cylinder can be determined by using the following formula: where r is the radius of the base circle and h is the height of the cylinder. The height of the prism is 15 centimeters. Surface Area Of A Pyramid Formula. Surface area of the side. : The total surface area of a cylinder with radius r and height h is given by. Calculate fluid surface area. 7 m, find its height. 68 cm 2 Example #2: Find the surface area of a cylinder with a radius of 4 cm, and a height of 3 cm. Now, the area of the rectangle = length × breadth. Remember, when finding Surface Area, you are finding the total area of all the 2 Dimensional shapes that are put together to form the three dimensional. So it's actually the same exercise as the triangular prism. ----Surface Area = (perimeter)*(height)---SA = 2(pi)rh-----Cheers, stan H. So each side is equal length x, which happens to equal 3. 2πr² is the area of the top and bottom parts of the cylinder. It is obtained by revolution in the space of a rectangle around one of its sides. It is given that the area = 100cm^2, As per the given information, 6x^2 = 100 x^2 =100/6 =. The area of one of them is b*c, and there are two of them, so the surface area of those two is 2bc. The apothem is the distance from the center of the base to the side of the base Step 1 Find the base area using the formula 1⁄2 a*p. Since the base of a cylinder is a circle, we substitute 2πr for p and πr2 for B where r is the radius of. [1] 6) A cube has a surface area of 384 cm 2. Attempt to answer the questions before studying the solution given. The shape can be thought of as a circular prism. cm, area of base ring is 115. 54 cm 2 Curved Surface Area of Cone =πrl =5. The area of the side surface is known as the lateral surface area of the cylinder. Volume and Surface Area Questions & Answers : A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. It is important to use the perpendicular height (or 'altitude') when calculating the volume of an oblique cylinder. 6 square feet. Find the surface area of that part of the plane 10x+4y+z=2 that lies inside the elliptic cylinder x 2/16 + y 2/49 = 1. Surface Area and Volume. cm and height 7cm. What is the relationship between the length of sides and the surface area to volume ratio of a cell? Other questions on the subject: Biology. Calculating the total surface area requires calculating the sum of the base and lateral surface areas. 2f) on surface area of cylinders. The volume of a pyramid is equal to one third of the product of the area of its base by the height. Write a c program to find the volume and surface area of cylinder. For example, if a right circular cylinder had a radius of 3 and height of 5, then SA = 2( π *9) + 2 π *3*5 = 18 π + 30 π = 48 π. See page for details. Online calculators and formulas for a surface area and other geometry problems. Enjoy the calculator! Buy a comprehensive geometric formulas ebook. The lateral surface consists of generators. So we'll say the base of the cylinder, the common mistake when you're trying to find the surface area is either adding in 2 circles because you see 2 circles or even 3 circles. Do not enter numbers with a slash (/)! If you do, the wrong answer will be computed. Once you add the area of the two bases and the outer surface area, you will have found the surface area of the cylinder. 5 cm 2 Hence, remaining surface area of structure. If the radius of the base of the cylinder is 0. 100 600 S = 216 Simplify. Determine the surface area and volume of this shape: The radius of a sphere is 4. The surface area of five faces of the rectangular prism is 258 cm2. It is given that the area = 100cm^2, As per the given information, 6x^2 = 100 x^2 =100/6 =. In our example, 3. Program to find the surface area of the cylinder Explanation. C Program to find total surface area of a cube. Cylinder calculator will give the surface area and volume of a cylinder. Using the definition of surface area and Net of a Cylinder document attached, facilitate a discussion of the surface area of a cylinder. Work on volume and surface area are differ a little when you are trying to calculate the area of a cylinder. If you were to 'unroll' the cylinder you would find the the side is actually a rectangle when flattened out. 6th and 7th Grades. However, if you wish to calculate manually, the formulas. The volume of the cone so formed is:. volume of 480 cubic centimeters. Let r , in centimeter be the radius of base of cylinder and h, in centimeter be height of cylinder. Mar 10, 2009. The area of a. EACHER Surface Area of a Cylinder T NOTES ©2015 Texas Instruments Incorporated 4 education. The surface area of Pyramid A is 216 square feet. The surface area of a capsule can be determined by combining the surface area equations for a sphere and the lateral surface area of a cylinder. EACHER Surface Area of a Cylinder T NOTES ©2015 Texas Instruments Incorporated 4 education. A cylinder has 2 circular surfaces (the bases) with a combined area of 2*Π*r 2. Since all the sides of a cube are squares with equal base sizes, we can express the surface area of a cube as 6 x (Area of a face of the cube (which is a square)). Worksheet 6. Find the change in surface area dA (in units{eq}^2 {/eq}) if the radius of a sphere changes from r by dr. Each end is a circle so the surface area of each end is π * r 2, where r is the radius of the end. Primary SOL. Total surface area of a hollow cylinder = 2 π R h + 2 π r h + 2 (π R 2 − π r 2) Example A hollow cylinder (open at both ends) has an external diameter of 28 cm, length of 14 cm and thickness of 2 cm. Measure the height of the cylinder - for this example the height is 10cm. Worksheets are Surface area, Surface area of a cylinder, Surface areas of cylinders, Mfm 2p1 surface area ofprisms and cylinders, Surface area, Surface area of solids, Surface area word problems name, Surface area and volume of cylinders work. For example, a cylinder is not a prism, because it has curved sides. Although cylinders may take many various forms. The surface area and the volume of a cylinder have been known since ancient times. In terms of formula, it's something like this: 2πrh + πr 2 + πr 2. This third piece is known as the curved surface area. There are two bases so their combined surface area is 2 X πr². Part 7) If a cylinder has a surface area of 59 square inches and its radius is 2. So we have three surfaces of the cylinder. 3 cents per square centimeter, and the plastic costs 0. These equations will give you correct answers if you keep the units straight. (Answer=7/19cm) Find the cost of plastering the inner surface of a well at the rate of Rs 30 per m2, if the inner diameter of the well is 2. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. Lesson 11-2 Surface Areas of Prisms and Cylinders 611 Finding Surface Area of a Cylinder The radius of the base of a cylinder is 4 in. 7) 2 SA = 686. The sides of the can is called the lateral area and the formula for finding the lateral area equals, diameter times pi times height Next I need to add the top of the cylinder and the bottom of the. ----Surface Area = (perimeter)*(height)---SA = 2(pi)rh-----Cheers, stan H. This website uses cookies to ensure you get the best experience. Surface Area of a Cylinder = 2 pi r 2 + 2 pi r h (h is the height of the cylinder, r is the radius of the top) Surface Area = Areas of top and bottom +Area of the side. This side over here, x is equal to 3. Measure the height of the cylinder - for this example the height is 10cm. Martin Overview Behold! A can with a: Top (radius 3. Do not enter numbers with a slash (/)! If you do, the wrong answer will be computed. We need to find its Surface area. The area of the side is equal to the area of a rectangle with length (the circumference of the base) times the height. Solved Problems on Volume and Surface Area of Cylinder: 1. Then there is the side of the cylinder, which is a rectangle that wraps around its outside to make the surface. Keep in mind that a cube has equal side lengths all around. This gives us the result that the flux through a portion of the surface of a cylinder of radius R oriented along the z-axis is where T is the -region corresponding to S. 2 cm 2 = 150. When you’re working with a rectangular prism, you’ll have to the area the of 3 pairs of equal sides, which needed to be added together to determine surface area. The total surface area of a hollow cylinder which is open from both sides is 4620 sq. All geometric formulas are explained with well selected word problems. To summarize, to find the surface area of a cylinder, take the area of the two circular bases, and add this to the lateral area of the cylinder. Given that the radius of the cylinder must be between 2 and 4 inclusive, find the value of that radius for which the lateral surface area of the cylinder in minimum. Find its surface area. Total surface area of a cylinder. Both the surface and the solid shape created inside can be called a cylinder. Do not enter numbers with a slash (/)! If you do, the wrong answer will be computed. 2 Side Length, Volume, and Surface Area of Similar Solids. the area of the sides is then pi * 8 * 10. Surface area of a cylinder = 2πr 2 + 2πrh Volume of a cylinder = πr 2 h You need to know the radius and height to figure both the volume and surface area of a cylinder. Half-cylinder: The surface area of the half-cylinder is 54 cm2. Engaging math & science practice! Improve your skills with free problems in 'Find the surface area of a Cylinder' and thousands of other practice lessons. Since the top and bottom of the cylinder have the same area, we can multiply the area of the circle (πr²) by 2. 2h) on surface area of cylinders. The figure below shows a cube with sides s. 99 USD per year until cancelled $19. Curved surface area of cylinder is the measurement of outer area,where the extension of top and bottom portion wont be included. Therefore the formula is 2π x r² + 2π x r x h. Rectangular parallelepiped. Therefore, Surface Area = 2(Area of circular top) + (circumference of top) by height of cylinder. First, they find the surface area using the letters at the bottom of the sheet. Oblique Cylinder. If we take a number of circular sheets and stake them up vertically , we get right circular cylinder. The card of length L and width W in figure 18-15 is rolled into a cylinder. and a heig [ 4 Answers ] Find the lateral area and surface area of a right cone with radius of 9 in. Your email is safe with us. superteacherworksheets. Once you add the area of the two bases and the outer surface area, you will have found the surface area of the cylinder. Write a Python program to calculate surface volume and area of a cylinder. #T S A = 2 pi r^2 + 2 pi r h# #color(brown)(T S A = 2pi r (r + h)#. Surface area of rhombus Knowing doagonal1, diagonal2: diagonal 1 :. ) A cosmetics company that makes small cylindrical bars of soap wraps the bars in plastic prior to shipping. Solution:. Each end is a circle so the surface area of each end is π * r 2 , where r is the radius of the end. m in 1 s (d) The relative density of lead is 11. Count the squares on the sides to find the SA. As the cube gets bigger, the volume increases much more rapidly than the surface area, because the volume increases as the cube of the linear dimension, but the surface area increases as the square. Displaying all worksheets related to - Cylinder Surface Area. Finally, the surface area of a sphere is given by A = 4 π r 2, where r is the radius of the sphere. For the 3-foot cube, 54:27, or 2:1. The lateral area of a cylinder is the area of its curved surface, excluding the area of its bases. The other method involves taking the area of all the sides and summing the areas. It has three surfaces: the top, the bottom, and the piece that forms the sides of the can. Sol: Radius (r) = 0. 300 seconds. This side over here, x is equal to 3. The volume of a cylinder of height 8 cm is 1232 cm³. In this lesson you will learn how to find the surface area of a rectangular prism by analyzing the prism's six faces, and. A solid cylinder has radius of base 14 cm and height 15 cm. cm and height 7 cm. Two cylinder surface area calculators based on radius or diameter. 14 * 25 * 10 = 314. The lateral area of a cylinder is the area of its curved surface, excluding the area of its bases. Formulas for Area and Volume 1) The lateral surface of the right cylinder is the surface without the top and bottom surfaces. 52 cm 2 + 94. Surface of Cylinder: 2π · r · (r + h) Surface of Sphere: 4π · r² Surface of Cone: π · r( r + √(h²+r²) ) Volume of Cuboids, Rectangular Prisms and Cubes. The area of a circular disk is π r 2 so each disk has an area of π × 5 2 = 78. Enjoy the calculator! Buy a comprehensive geometric formulas ebook. The cardboard costs 0. The surface area SA is the area of the ends (which are just circles), plus the area of the side, which is a circle's circumference times the height h of the cylinder: SA cyl = 2π r 2 + 2π rh Depending on the class you're taking, you might also need to know the formula for the volume V of a cone with base radius r and height h :. You'll gain access to interventions, extensions, task implementation guides, and more for this instructional video. The total surface area (TSA) includes the area of the circular top and base, as well as the curved surface area (CSA). Solve advanced problems in Physics, Mathematics and Engineering. Just like ordinary area, the units are the squares of the units of length (that is, if a shape's sides are measured in meters, then the shape's area is measured in square meters). The volume of a cylinder of height 8 cm is 1232 cm³. Find the surface area of the red. Higher ability full lesson (5. The height of the prism is 15 centimeters. The surface area of a capsule can be determined by combining the surface area equations for a sphere and the lateral surface area of a cylinder. 14 x diameter 2 / 2. _ ____ 2) Volume = _27 cubic cm. Calculate the unknown defining side lengths, circumferences, volumes or radii of a various geometric shapes with any 2 known variables. To find the surface area of a cylinder, we use our area and circumference formulas. Surface Area Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode. Mar 10, 2009. 99 USD per year until cancelled. Surface area of a cube = 6s 2. Enter two values to calculate the third. By using this website, you agree to our Cookie Policy. if a circular cylinder has radius has radius r and height h, write an expression that represents the surface area of its side. All you have to do is add 56. Therefore, its surface area will be the sum of the areas of all six surfaces. Find the surface area of this cylinder. Lesson 11-2 Surface Areas of Prisms and Cylinders 611 Finding Surface Area of a Cylinder The radius of the base of a cylinder is 4 in. In terms of formula, it's something like this: 2πrh + πr 2 + πr 2. We will need calculate the area of the top, the base and sides. Here's a quick example: Compute the flux of the vector field through the piece of the cylinder of radius 3, centered on the z -axis, with and. Sphere: surface area is 179 in 2. Materials. Surface Area of a Cube. Solution: Given that: Side length (a) = 10. If a circular cylinder has a radius #f# and height #h#, what is the expression that represents the surface area of its side?. You can use this information to determine the Surface Area of each shape. and height 7 cm. Surface Area of Cylinders The surface area S. The mathematical definition of surface area in the presence of curved surfaces is considerably more involved than the definition of arc length of one-dimensional curves, or of the surface area for polyhedra (i. Since the top and bottom of the cylinder have the same area, we can multiply the area of the circle (πr²) by 2. In particular, if the above "trajectory" is a circle, we have a circular cylinder of radius R, whose surface area is 2 p RH. 14 * 5² * 10 = 3. • When measuring surface area, the units will be "squared". Curved surface area of cylinder If a rectangle revolves about one side and completes one full rotation, the solid thus formed is called a right circular cylinder. 49 USD per month until cancelled: Annual Subscription (limited promotion)$19. If you carefully cut the label off the side of the can and unroll it, you will see that it is a rectangle. if a circular cylinder has radius has radius r and height h, write an expression that represents the surface area of its side. In most of the formulas for surface area of a pyramid, we need slant height, l, not height, h. Surface area of the cylinder = 2πrh = 2 × 22/7 × 2. The Attempt at a Solution a)in my book it has been given that Vector E is parallel to sides of cylinder so flux through sides =0 My understanding regarding this explanation of Vector E to be parallel to sides of cylinder is to point out that that electric fields do not even pass through sides of cylinder because the angle between vector E and area vector of sides of cylinder is 90 degrees so. A cylinder is a three-dimensional geometrical figure/container with straight parallel sides and two circular cross-sections. Answer:-(a) Length of edge. The surface. Step 1 Use the circumference to find the radius. The surface area of a cylinder is nothing but the sum of the areas of its curved surface and two circular bases. The surface area of a cylinder is the sum of its top and bottom (the area of two circles) plus the area that wraps around the middle. EACHER Surface Area of a Cylinder T NOTES ©2015 Texas Instruments Incorporated 4 education. As part of the Mensuration topic. com - id: 6e9908-N2ExO. Ratio of their surface area = 6a2/6b2 = a2/b2 = (a/b)2 = 1/9, i. Like area, surface area is also expressed in square units. The surface area and the volume of a cylinder have been known since ancient times. As the cube gets bigger, the volume increases much more rapidly than the surface area, because the volume increases as the cube of the linear dimension, but the surface area increases as the square. Let r , in centimeter be the radius of base of cylinder and h, in centimeter be height of cylinder. Enjoy the calculator! Buy a comprehensive geometric formulas ebook. Here's a quick example: Compute the flux of the vector field through the piece of the cylinder of radius 3, centered on the z -axis, with and. We are learning to: Calculate the surface area of a cylinder. I really want to find the surface area of just the side of the cylinder, not the top and bottom Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Since the area of a circle is and the area of a rectangle is lw, we can make a formula for the surface area of a cylinder:. Surface Area = Areas of top and bottom +Area of the side. The area of one square face A face = a ⋅ a = a 2. Specify the cylinder radius ( r ) and length ( L ), the liquid depth ( h ), and optionally select unit of length, to calculate volume of the liquid. Surface Area of a Cube. A cylinder which has slant height l l l and circular flat surfaces with radius r r r has a total surface area equal to 2 π r (r + l) 2 \pi r (r + l ) 2 π r (r + l). Embeddable Player. For a right circular cylinder of radius r and height h, the lateral area is the area of the side surface of the cylinder: A = 2π rh. 56 cm 2 Area of top of the cylinder=πr 2 =1. Example 6: A cube whose sides are 10. Work out the surface area of a cylinder. Problem: Consider a cylinder with a radius of 3 meters and a height of 6 meters. When the two ends are directly aligned on each other it is a Right Cylinder otherwise it is an Oblique Cylinder: Surface Area of a Cylinder. In our example, 3. Your email is safe with us. Cylinder Surface Area = 2 x pi x radius x height + 2 x pi x radius squared. The surface area calculator will determine the surface area of a cone, cube, cylinder, rectangular cuboid and sphere. The total surface area of a hollow ceramic cylinder which is open from both the sides is 4620 sq. The total surface area is calculated as follows: SA = 4πr 2. Note: A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder. Area of top is π x r². The surface area of a cylinder is the total area of the cylinder. : The total surface area of a cylinder with radius r and height h is given by. First we need to establish the area of the two identical circles. Divide both sides by 2π.
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https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_50&curid=14307&diff=142923&oldid=79553 | # Difference between revisions of "1954 AHSME Problems/Problem 50"
## Problem
The times between $7$ and $8$ o'clock, correct to the nearest minute, when the hands of a clock will form an angle of $84^{\circ}$ are:
$\textbf{(A)}\ \text{7: 23 and 7: 53}\qquad \textbf{(B)}\ \text{7: 20 and 7: 50}\qquad \textbf{(C)}\ \text{7: 22 and 7: 53}\\ \textbf{(D)}\ \text{7: 23 and 7: 52}\qquad \textbf{(E)}\ \text{7: 21 and 7: 49}$
## Solution
At $7$ o'clock, the hour hand is at the position $\tfrac{7}{12}\cdot 360^{\circ}=210^{\circ}$ clockwise from the $12$ o'clock position, and the minute hand is exactly at the $12$ o'clock position. Thus the minute hand is $360^{\circ}-210^{\circ}=150^{\circ}$ ahead of the hour hand, while it is also $210^{\circ}$ behind the hour hand. So, when the minute hand first makes an $84^{\circ}$ angle with the hour hand, the minute hand will be $84^{\circ}$ behind the hour hand, and the second time they make and $84^{\circ}$ angle, the minute hand will be $84^{\circ}$ ahead the hour hand.
The minute hand moves clockwise at a rate of $360^{\circ}/\text{hr}$, while the hour hand moves at $(1/12)\cdot 360^{\circ}/\text{ hr}=30^{\circ}/\text{hr}$. Therefore the minute hand catches up to the hour hand at a rate of $360^{\circ}-30^{\circ}=330^{\circ}$ per hour. Therefore it will take $(210^{\circ}-84^{\circ})/(330^{\circ}/\text{hr})=\tfrac{126}{330}\text{hr}$ for the two hands of the clock to make an $84^{\circ}$ angle. It will also take $(210^{\circ}+84^{\circ})/(330^{\circ}/\text{hr})=\tfrac{294}{330}\text{hr}$ after $7$ o'clock for the hands to make an $84^{\circ}$ angle for the second time. Converting these values to minutes, we see that it will take $22\tfrac{10}{11}$ minutes, and $53\tfrac{5}{11}$ minutes, for the two hands to make $84^{\circ}$ angles. Thus the times, correct to the nearest minute, at which the hands of the clock will form an $84^{\circ}$ angle are $\boxed{\textbf{(A)}\ \text{7: 23 and 7: 53}}$.
## Solution 2
Using the formula \$(60h-11m)/2 | 2021-03-09T11:30:52 | {
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https://math.stackexchange.com/questions/1820966/vector-lattice-0-v0-u-0-uv | # Vector lattice: $[0,v]+[0,u] = [0,u+v]$
Let V be a real vector-lattice and $u\geq 0, v\geq 0$.
I need to show that $[0, u+v] \subseteq [0,u]+[0,v]$.
I would appreciate a hint on how to start.
For $z\in [0,u+v]$ with $z\geq u$ or $z\geq v$ it is easy. My problem is that we can't assume $z<u$ and $z<v$ if the assumption above does not hold.
Hint: Consider the element $\min(z,u) \in V$.
• Yes, this belongs to $[0,u] + [0,v]$. | 2022-01-24T22:17:18 | {
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https://en.m.wikibooks.org/wiki/Circuit_Theory/Quadratic_Equation_Revisited | The roots to the quadratic polynomial
${\displaystyle y=ax^{2}+bx+c}$
are easily derived and many people memorized them in high school:
{\displaystyle {\begin{aligned}x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.\end{aligned}}}
### Derivation of the Quadratic Equation
To derive this set ${\displaystyle y=0}$ and complete the square:
{\displaystyle {\begin{aligned}0&=ax^{2}+bx+c\\0&=a\left(x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}\right)\\0&=x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\\0&=\left(x+{\frac {b}{2a}}\right)^{2}+\left({\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\right).\end{aligned}}}
Solving for ${\displaystyle x}$ gives
{\displaystyle {\begin{aligned}0&=\left(x+{\frac {b}{2a}}\right)^{2}+\left({\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\right)\\\left(x+{\frac {b}{2a}}\right)^{2}&=-\left({\frac {c}{a}}-\left({\frac {b}{2a}}\right)^{2}\right)\\&={\frac {b^{2}-4ac}{\left(2a\right)^{2}}}.\end{aligned}}}
Taking the square root of both sides and putting everything over a common denominator gives
{\displaystyle {\begin{aligned}x+{\frac {b}{2a}}&=\pm {\sqrt {\frac {b^{2}-4ac}{\left(2a\right)^{2}}}}\\x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.\end{aligned}}}
### Numerical Instability of the Usual Formulation of the Quadratic Equation
Middlebrook has pointed out that this is a poor expression from a numerical point of view for certain values of ${\displaystyle a}$ , ${\displaystyle b}$ , and ${\displaystyle c}$ .
[Give an example here]
Middlebrook showed how a better expression can be obtained as follows. First, factor ${\displaystyle -{\frac {b}{2a}}}$ out of the expression:
{\displaystyle {\begin{aligned}x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\\&=\left(-{\frac {b}{2a}}\right)\left(1\pm {\sqrt {1-{\frac {4ac}{b^{2}}}}}\right).\end{aligned}}}
Now let
${\displaystyle Q^{2}={\frac {ac}{b^{2}}}.}$
Then
{\displaystyle {\begin{aligned}x&=\left(-{\frac {b}{2a}}\right)\left(1\pm {\sqrt {1-4Q^{2}}}\right)\end{aligned}}}
### A More Numerically Stable Formulation of the Negative Root
Considering just the negative square root we have
{\displaystyle {\begin{aligned}x_{1}&=\left(-{\frac {b}{2a}}\right)\left(1-{\sqrt {1-4Q^{2}}}\right).\end{aligned}}}
Multiplying the numerator and denominator by ${\displaystyle 1+{\sqrt {1-4Q^{2}}}}$ gives
{\displaystyle {\begin{aligned}x_{1}&=\left(-{\frac {b}{2a}}\right)\left(1-{\sqrt {1-4Q^{2}}}\right)\left({\frac {1+{\sqrt {1-4Q^{2}}}}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=\left(-{\frac {b}{2a}}\right)\left({\frac {1-1+4Q^{2}}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=-{\frac {2bQ^{2}}{a}}\left({\frac {1}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=-{\frac {2b\left({\frac {ac}{b^{2}}}\right)}{a}}\left({\frac {1}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=-{\frac {2c}{b}}\left({\frac {1}{1+{\sqrt {1-4Q^{2}}}}}\right)\\&=-{\frac {c}{b}}\left({\frac {1}{{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {1-4Q^{2}}}}}\right).\end{aligned}}}
By defining
${\displaystyle F={\frac {1}{2}}+{\frac {1}{2}}{\sqrt {1-4Q^{2}}}}$
we can write
{\displaystyle {\begin{aligned}x_{1}&=-{\frac {c}{bF}}.\end{aligned}}}
Note that as ${\displaystyle Q\to 0}$ , ${\displaystyle F\to 1}$ .
### Finding the Positive Root Using the Same Approach
Turning now to the positive square root we have
{\displaystyle {\begin{aligned}x_{2}&=\left(-{\frac {b}{2a}}\right)\left(1+{\sqrt {1-4Q^{2}}}\right)\\&=\left(-{\frac {b}{a}}\right)\left({\frac {1}{2}}+{\frac {1}{2}}{\sqrt {1-4Q^{2}}}\right)\\&=-{\frac {bF}{a}}.\end{aligned}}}
Using the two roots ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ , we can factor the quadratic equation
{\displaystyle {\begin{aligned}y&=ax^{2}+bx+c\\&=a\left(x^{2}+{\frac {b}{a}}+{\frac {c}{a}}\right)\\&=a\left(x-x_{1}\right)\left(x-x_{2}\right)\\&=a\left(x+{\frac {c}{bF}}\right)\left(x+{\frac {bF}{a}}\right).\end{aligned}}}
### Accuracy for Low ${\displaystyle Q}$
For values of ${\displaystyle Q\leq 0.3}$ the value of ${\displaystyle F}$ is within 10% of 1 and we may neglect it. As noted above, the approximation gets better as ${\displaystyle Q\to 0}$ . With this approximation the quadratic equation has a very simple factorization:
{\displaystyle {\begin{aligned}y&=ax^{2}+bx+c\\&\approx a\left(x+{\frac {c}{b}}\right)\left(x+{\frac {b}{a}}\right),\end{aligned}}}
an expression that involves no messy square roots and can be written by inspection. Of course, it is necessary to check the assumption about ${\displaystyle Q}$ being small before using the simplification. Without this simplification, ${\displaystyle F}$ needs to be calculated and the roots are slightly more complicated.
[Explore the consequences if ${\displaystyle Q>0.5}$ .] | 2022-07-02T05:29:12 | {
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https://cstheory.stackexchange.com/questions/27441/partition-problem/27466#27466 | # Partition problem
We know that Partition problem is NP-complete:
"Given a multi-set of positive integers like X = < a_1, a_2, ... a_n >, is there any bi-partition for X such that the summation of the numbers in S and S' are equal?"
I find a reduction from the subset-sum problem into the partition problem, but is there any other proof for the NP-completeness of the subset-sum (like from 3-SAT)?
I am actually trying to see if the special case of the partition problem in which all the a_i's are odd distinct integers is NP-complete or not. Does anybody have any idea?
• If you want a hint, a good source problem is POSITIVE 1-in-3 SAT; it allows a quick and easy reduction to the PARTITION problem with all $a_i$s distinct and odd. Let me know if you need more details. Nov 13 '14 at 20:10
• Can you please give me more details? Nov 16 '14 at 12:18
• I can make a reduction from positive 1-in-3-SAT, but it is a reduction only for distinct numbers, not neccessary odd! Thanks! I will be waiting for your reduction, meanwhile I will try to find a reduction on my own. Nov 16 '14 at 18:41
• I quickly edited it (most details are omitted); in every case I saw that daniello posted a quicker (and better) reduction from PARTITION that seems ok; it is the same trick I used to make the integers distinct and odd (the $2^{i+1}+1$ lower bits of my integers); I didn't notice that it could be applied directly to an instance of PARTITION. Nov 16 '14 at 20:22
Start from the partition problem and let $a_i' = a_i * 100n^2 + 2*i + 1$. Also add in new elements $b_1', \ldots b_n'$ with $b_i' = 2i+1$. All variables are odd and distinct.
For the forward direction, suppose there is a partition of $\{a_1, \ldots a_n\}$ into two sets $L$ and $R$ such that $\sum_{a_i \in L} a_i = \sum_{a_i \in R} a_i$. Set $L' = \{a_i' : a_i \in L\} \cup \{b_i' : a_i \notin L\}$ and $R' = \{a_i' : a_i \in R\} \cup \{b_i' : a_i \notin R\}$. This is a partition with the same sum in $L'$ and $R'$
For the reverse direction, suppose there is a partition of $a_1',\ldots, a_n', b_1', \ldots b_n'$ into $L'$ and $R'$. Set $L = \{a_i : a_i' \in L'\}$ and $R = \{a_i : a_i' \in R'\}$. It is easy to verify that the sum in $L$ and $R$ is the same - since otherwise the sum of $\sum_{a_i' \in L'} a_i'$ and $\sum_{a_i' \in R'} a_i'$ differ by at least $100n^2$ and the $b_i'$s can't compensate for this.
There is a quick reduction from POSITIVE 1-in-3 SAT; I give you only a sketch of it.
Given a POSITIVE 1-in-3 SAT instance with $m$ positive clauses (in which every literal is unnegated) and $n$ variables; without loss of generality we can assume that the clauses are all distinct. For every variable $x_i$ add an integer $a_{x_i}$:
$a_{x_i} = \sum_{k \in I_{x_i}} 2^{3k + q} + 2^{i+1} + 1$ where $I_{x_i}$ is the set of the indices of the clauses that contain $x_i$; and $q$ is a large enough integer (for example $n+1$) that avoid the lower bits to "interfere" with the upper bits.
Add an integer $S = \sum_{k=1}^m 2^{3k + q} + 1$; add $n$ dum integers $d_i = 2^{i+q}+1, i=1...n$ and finally add $e = 1$
All integers are distinct and odd.
Suppose that $S$ is included in $A$; then, for every clause $C_j = ( x_{i_1}, x_{i_2}, x_{i_3})$, exactly one of the $a_{x_{i_1}},a_{x_{i_2}},a_{x_{i_3}}$ can be included in $A$ because there are exactly four integers with the bit $2^{3j+q}$ set to 1 (three from the elements of the clause and one from $S$) and it easy to prove that this forces a valid 1-in-3 assignment (the $a_{x_i}$s included in $A$ correspond to variables $x_i$ that are set to true and form a valid 1-in-3 assignment for the original formula). The elements $d_1,...,d_n, e$ can be used to balance the $2^{i+1}+1$ part of the corresponding $a_{x_i}$ ($e$ is used to balance the lower bit of $S$).
An example of the reduction from the POSITIVE 1-in-3 SAT formula:
$(x_1 \lor x_2 \lor x_3) \land (x_2 \lor x_4 \lor x_5) \land (x_1 \lor x_4 \lor x_6) \land (x_2 \lor x_5 \lor x_6)$
Blank cells represent $0$s
• Oh, Thank you guys! Awesome! I actually was doing the same thing as you. My only mistake was that instead of taking only one number S, as you built here, I was taking m different numbers with a digit 1 added at the end, which in total it was adding to their summation by m, and it was a trouble! Nov 16 '14 at 23:57
• We don't know, but it is NP-hard (the polynomial time many one reduction proves it); so it is in P if and only if P=NP. Nov 17 '14 at 15:10 | 2021-10-25T20:45:15 | {
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https://math.stackexchange.com/questions/1691878/shouldnt-all-alternating-series-diverge | # Shouldn't All Alternating Series Diverge?
The ${n}$th-term test for divergence everywhere I've seen basically states that, if the limit of the ${n}$th term as ${n}$ approaches infinity is not 0 or doesn't exist, the series diverges. However, this test does not give any precondition on the ${n}$th term (e.g. that it has to be positive), so it should be applicable to alternating series as well. So,
Shouldn't all alternating series diverge, since the limit of the nth term of the series, ${a_n =(-1)^nb_n}$ (including the ${(-1)^n}$), as n approaches infinity usually does not exist?
Won't this contradict those series that pass the Alternating Series Test (AST)? Or does the AST supersede the ${n}$th-term test in such cases (if so, why isn't this mentioned anywhere)?
• What would make you think that the limit of $(-1)^nb_n$ does not exist? – David C. Ullrich Mar 10 '16 at 17:48
• The limit of $b_n$ must be $0$ for the alternating series test. From this it follows that $(-1)^nb_n$ has limit $0$. – André Nicolas Mar 10 '16 at 17:48
• @David I mistakenly thought that, since the limit of ${(-1)^n}$ doesn't exist, the limit of ${(-1)^nb_n}$ would also not exist. I checked Wolfram Alpha for the limits of ${a_n}$ for convergent alternating series, and they are indeed 0, so the nth-term test remains valid after all! – Leponzo Mar 10 '16 at 18:09
• Let's stick to integers, $(-1)^x$ gets us into trouble for non-integer $x$. If $b_n$ has a non-zero limit, then sure, $(-1)^n b_n$ does not have a limit. But if $b_n$ is after a while close to $0$, then so is $(-1)^nb_n$. – André Nicolas Mar 10 '16 at 18:22
• Two comments. First, of course it was clear what your error was, I was trying to get you to find it yourself. The problem was remembering things without the details: It's a Fact that if the limits of $a_n$ and $b_n$ exist then the limit of $a_nb_n$ exists, and equals the product of the limits. But that Fact says nothing about limits not existing. Second: You checked Wolfram Alpha? That's very sad. Also dangerous - WA gets things wrong sometimes. If you think about the definition of limits it's very easy to see that if $\lim b_n=0$ then $\lim (-1)^nb_n=0$. – David C. Ullrich Mar 10 '16 at 18:33
Well, obviously there are convergent alternating series, so there must be some mistake here. We know that, for $\sum a_n$ to converge, then the condition
$$\lim_{n\to\infty}a_n=0$$
is necessary (but not sufficient!). So if we take $a_n=(-1)^nb_n$, then if $\sum (-1)^nb_n$ converges, then
$$\lim_{n\to\infty}(-1)^nb_n=0$$
which happens if and only if
$$\lim_{n\to\infty}b_n=0$$
so well, we can have an alternating sequence who's series converges, but we need to have $b_n\to 0$ for $\sum (-1)^nb_n$ to converge. An example:
$$\sum_{k=1}^{\infty}(-1)^k\frac1k=-\log(2)$$
on the other hand, we can make a series like $$c_n= \begin{cases} \frac{1}{n} \text{ if } n \text{ is even}\\ \frac{1}{2n} \text{ otherwise } \end{cases}$$ for which actually
$$\sum_{n=1}^\infty (-1)^nc_n$$
actually diverges to $+\infty$, but $\lim c_n\to 0$. We again see, for $\sum (-1)^nb_n$ to converge, the condition $\lim b_n=0$ is necessary, but not sufficient. | 2021-02-26T19:11:07 | {
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https://math.stackexchange.com/questions/2531422/do-we-have-to-take-the-absolute-value-of-the-jacobian-only-if-it-is-a-number/2748910 | # Do we have to take the absolute value of the jacobian ONLY if it is a number?
if we want to evaluate the integration $$I=\int\int(x^3y^3)(x^2+y^2)dA$$ over the region bounded by the curves $$xy=1,\\xy=3,\\x^2-y^2=1,\\x^2-y^2=4$$ I used the transformation $$u=xy,\\v=x^2-y^2$$ I found that the jacobian will be $$J=\frac{1}{-2y^2-2x^2}$$ Do I have to get the absolute value of the jacobian? If I did not take the absolute value , I will get the result of the integration = - 30
if I take the absolute value $$J=\frac{1}{2y^2+2x^2}$$ I will get the result= + 30 .
My friend told me we take absolute value of the jacobian only if it is a number .. if this is right .. why we do not take the absolute value if the jacobian is a function?..I think we are sure here that the jacobian is negative since we have x and y squared , so we have to take the absolute value!
Another question, if we have to take always the absolute value of the jacobian (whether it is a number or function) :
if the jacobian is for example $$J=-2x+y$$ It will be positive for some values of x and y only ! .. how can we apply the absolute value inside the double integration?
If the Jacobian is negative, then the orientation of the region of integration gets flipped.
You have to take the absolute value ALWAYS.
• what if the jacobian is $$J=-2x+y$$ for example ..so we do not know whether it is positive or negative ! it depends on the values of x and y !!
– MCS
Nov 21 '17 at 22:14
• Then you just take $|-2x +y|$ Nov 22 '17 at 0:24
I'm adding an additional answer, in case the issue is actually with the integration of an absolute value, recall the definition of the absolute value:
$\begin{eqnarray*} \left|-2x+y\right| \quad & = & \quad \begin{cases} -2x+y, & \text{if } -2x+y \geq 0, \quad \text{ (i.e. if the quantity is positive)}\\ -(-2x+y), & \text{if } -2x+y < 0, \quad \text{ (i.e. if the quantity is negative)} \end{cases} \\[8pt] % \quad & = & \quad \begin{cases} -2x +y, & y \geq 2x ; \\ 2x - y, & y < 2x \end{cases} \end{eqnarray*}$
As such, if the domain of integration is in just one of these two halves of the plane, you can just apply the appropriate form of the absolute value; and if the domain of integration is a region containing a part of the line $y = 2x$, you simply break up the integral into multiple parts, where one of the two definitions applies.
N.B. here I have expressed $y = y(x)$, but if you wanted to integrate with respect to $x$ first, you could simply rearrange the expression to have $x = x(y)$. | 2022-01-20T10:23:22 | {
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https://math.stackexchange.com/questions/1853710/conditional-probability-two-dice | # Conditional probability (two dice)
Two dice are rolled, what is the probability that die 1 shows a 5 given that a 10 has been rolled?
I'm a little confused as to how to solve this exercise. I get two different results depending on how I go about solving it.
Method 1 (logic): The probability of having die 1 show a 5 when two dice are rolled is 1/36. The probability that a 10 is rolled is 3/36. Now I put this in the basic probability format and it gives me (1/36)/(3/36)=1/3.
Method 2 (baye's theorem): P(A/B) = P(A)xP(B/A) / P(B) = 1/36 x 1/6 (bc if die 1 shows a five and I need to get a 10, then die 2 needs to show a five also, hence 1/6) / (3/36) = 1/18
• P(1st die is 5)=1/6 your text was a bit unclear on that. We get P(1st die is 5 | sum is 10) = P( both dice 5) / P(sum is 10) = (1/36)/(3/36) = 1/3. Your Baye's them calculation must contain an error. – jdods Jul 9 '16 at 2:14
• $P(B|A)$ is not $\frac 1{36}$ you get to assume that one die is a 5. It is not the same as the chance of rolling double 5's outright. – Doug M Jul 9 '16 at 3:07
• Another approach to is to discard all rolls that do not sum to 10, at the first step. Leaving (6,4) and (5,5), and (6,4) is twice as likely as doubles. – Doug M Jul 9 '16 at 3:10
Observe that our probability space is $(\Omega, \mathbb P)$ where
$$\text{sample space: }\Omega = \{\{1,1\}, \{1,2\}, ..., \{6,5\}, \{6,6\}\}$$
$$\text{probability measure: }\forall \omega \in \Omega, \mathbb P(\omega) = \frac{1}{36}$$
We want to compute
$$P(\text{5 is rolled} | \text{sum is 10}) \tag{*}$$
Observe that the event $\text{5 is rolled}$ corresponds to the sample outcomes $\{5,1\}, ..., \{5,6\}$ and the event $\text{sum is 10}$ corresponds to the sample outcomes $\{5,5\}, \{4,6\}, \{6,4\}$.
Hence,
$$(*) = P((\{5,1\} \cup ... \cup \{5,6\}) | \{5,5\} \cup \{4,6\} \cup \{6,4\})$$
$$= \frac{P((\{5,1\} \cup ... \cup \{5,6\}) \cap (\{5,5\} \cup \{4,6\} \cup \{6,4\}))}{P(\{5,5\} \cup \{4,6\} \cup \{6,4\})}$$
$$= \frac{P(\{5,5\})}{P(\{5,5\} \cup \{4,6\} \cup \{6,4\})}$$
$$= \frac{P(\{5,5\})}{P(\{5,5\}) + P(\{4,6\}) + P(\{6,4\})}$$
$$= \frac{1/36}{3/36} = 1/3$$
Another way to look at it is to change our probability space
$$\text{sample space: }\Omega = \{\{5,5\}, \{4,6\}, \{6,4\}\}$$
$$\text{probability measure: }\forall \omega \in \Omega, \mathbb P(\omega) = \frac{1}{3}$$
So if we want to compute $P(\{5,5\})$, by definition of our probability measure, we have 1/3.
Presumably $B$ is the event a $10$ has been rolled, and $A$ is the event Die 1 shows a $5$. Then $\Pr(A)=1/6$, or, if you prefer, $6/36$, $\Pr(B\mid A)=1/6$, and $\Pr(B)=3/36$.
Substituting and simplifying, we get that $\Pr(A\mid B)=1/3$. The two methods give the same answer.
• Thank you. However, both methods give the same answer when method 1 considers the probability of getting die 1 to show 5 as 1/36; and method 2 considers the probability of getting die 1 to show 5 as 1/6. The problem itself is a bit unclear, as jdods pointed out, but even if we consider the same probability of getting die 1 to show a 5 for both methods, we still get a different answer, why is that? What am I not reasoning correctly? – Jose Jul 9 '16 at 2:47
• Method 1 is in essence uses the definition of conditional probability, $\Pr(A\mid B)=\Pr(A\cap B)/\Pr(B)$. It would be better to make $A$ and $B$ explicit, since informal arguments can distressingly often lead to error. Method 2 is essentially the same, except that it uses explicitly $\Pr(B\mid A)\Pr(A)$ to compute $\Pr(A\cap B)$. – André Nicolas Jul 9 '16 at 2:57 | 2019-12-12T01:18:47 | {
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https://brilliant.org/discussions/thread/archive-of-weekly-problems-week-of-feb-26-advanced/ | # Archive of weekly problems (week of Feb 26 - Advanced, 1st problem).
I recently was solving this problem on Brilliant: https://brilliant.org/weekly-problems/2018-02-26/advanced/ and, according to my reasoning, came to the answer of 13/16 (>0.75).
So, the strategy each one in the group of 3 people should use is the following (instead of the usual strategy presented in the solution): if a person sees 2 Heads (or 2 Tails), he says Tails (or Heads). That way 2 out of 8 all possible outcomes (cases "HHH", "TTT") at least 1 person (in these 2 cases would be everyone out of 3 people) guesses correctly.
And if a person sees Heads and Tails, he says randomly Heads or Tails (50% each) (yes, humans are not so random, but assuming that they can be, or they have a device implemented in their brains that allows them to do it). Since we have 6 out of 8 all possible outcomes (flips of coins), where 2 persons see "Heads and Tails" in each outcome (out of 8) and they say H or T randomly, and, according to the rules or conditions, at least 1 person guesses correctly with 0.75 probability chance. So we multiply (3/4)*(6/8) = 9/16.
Thus, using the strategy described above the answer then becomes 1/4 + 9/16 = 13/16.
I hope to see if someone would clarify if the reasoning is right, wrong or missing something.
Note by Deaf. Frust.
7 months, 2 weeks ago
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This is NOT my answer but his-https://brilliant.org/profile/mark-4vl4ha/
To win, someone has to guess correctly, and no-one can guess incorrectly. With your strategy, the team loses with HHH or TTT. In any other configuration (such as HHT) then two people (not just one) have to get their guesses right, since otherwise someone will get a guess wrong. Thus the chance of winning in such a configuration is 3/16.
- 6 months, 3 weeks ago | 2021-03-07T21:24:34 | {
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https://physics.stackexchange.com/questions/88977/calculating-the-solar-power-hitting-earth | # Calculating the solar power hitting Earth
I'm trying to calculate the amount of power from the sun hitting the earth, but I am getting a number which is off by a factor of ~4.
I calculate the "area" of the earth, as seen from the sun, and then divide that by the surface area of a sphere of radius 1 AU, to get the portion of the sun's rays we absorb, and then I multiply that by the solar luminosity.
Mathematica gives me this:
But when I put it into Wolfram|alpha, I get this result:
Am I doing something wrong? where could that big an error be introduced? Is "mean power intercepted by earth" different than what I'm calculating?
• you made a simple mistake calculating cross section of the earth as piD². should be pir². this error accounts for exactly 4x. – gregsan Dec 4 '13 at 15:32
The area of a circle is calculated using its radius instead of its diameter: $$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \frac{d^2}{4}$$ which is where your missing factor of 4 reappears.
As pointed out in the comments, the method is correct aside from a careless error of using the diameter, when radius was clearly intended to be used by the argument.
I calculate the "area" of the earth, as seen from the sun, and then divide that by the surface area of a sphere of radius 1 AU, to get the portion of the sun's rays we absorb, and then I multiply that by the solar luminosity.
For a sanity check, I will propose another way to get the same thing. The sun is roughly a black body, but you didn't need this info since you started with the Luminosity. You can recalculate that value approximately as follows.
$$L_{\circ} = \sigma T^4 A_s$$
Take out the area from this expression. Now you have $W/m^2$, but at the surface of the sun. Convert this to the average solar insolation at Earth's location, you can simply multiply by the ratio of areas of those two spheres. That is, multiply the sun's $W/m^2$ value by the area of the sun's surface divided by the area of the 1 AU sphere to get the $W/m^2$ at 1 AU. The ratio of areas of two shapes will be the ratios of linear dimensions, squared. It doesn't matter which.
Returning to your objective quantity, that is simply the area of Earth presented to the sun times the $W/m^2$ at 1 AU.
$$P = \frac{ L_{\circ} }{ A_{AU} } A_{Earth} = \sigma T^4 \left( \frac{ 1 AU }{ R_{s} } \right)^2 A_{Earth}$$
I can't say this would have caught your problem, but it's a useful sanity check in general. It avoids comparing $4 \pi r^2$s against $\pi r^2$s, which gets a little bit dizzying. Here is the calculation on Wolram Alpha. It produces $1.74 \times 10^{17} W$, which might be different only within the significant figures. It's probably close enough to validate that this is another correct calculation.
• Actually, I didn't bother checking against the OP's calculation. Dividing by 4, that number gives 1.74, so these match perfectly. I was correct that our sun is a blackbody! – Alan Rominger Dec 4 '13 at 16:15
Let's go about this from a different angle and compare answers. The radius of the earth is about 6.4 Mm, so the area of its disk is 130 x 1012 m2. Figure about 1.2 kW/m2 of incident sunlight power at earth's distance, so that yields 1.5 x 1017 Watts. That's close enough for such a quick back of the envelope calculation to the 2 x 1017 value you show that it can be considered the same answer. The 1.2 kW/m2 figure I used is probably low.
However, what this answer really means is much harder to decide. You certainly can't assume all that power is heating the planet, since much of that will be reflected, especially at glancing angles.
• I think the coefficient you are looking for is the bond albedo which is equal to 0.29 for Earth according to this source, but according to the source linked by wikipedia it is equal to 0.306. – fibonatic Dec 4 '13 at 15:52 | 2019-06-26T18:03:09 | {
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http://mathhelpforum.com/statistics/278760-testing-population-varience.html | # Thread: Testing of population varience
1. ## Testing of population varience
with an individual lines at its various windows , a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes . The post office expreiemets with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.
my chi square test value = 5.67 , but teh chi sqaure critical at α = 0.05 and v = 24 is 36.415 , i do n't have enough evidenvce to reject H0 ,
H0 = 7.2
H1= <7.2 ,
But the ans provided is REJECT H0 , why ?
I think the ans provided is incorrect , correct me if i am wrong .
2. ## Re: Testing of population varience
When writing out your hypotheses, please attend to precision and write them out correctly; $\displaystyle H_0 = c$ is not a proper hypothesis.
I believe you meant to state the following hypotheses:
$\displaystyle H_0: \sigma=7.2$, and
$\displaystyle H_1: \sigma < 7.2$.
This hypothesis test is a lower one-tailed test, which we observe from the alternative hypothesis.
The critical region for a lower one-tailed alternative is given by $\displaystyle T<\chi^2_{\alpha, N-1}$.
Your calculations appear to be correct, but you have confused the critical region with its complement (the acceptance region).
The book is right, but you were on the right path. Please attend to precision when doing mathematics, and you will likely avoid such careless errors in the future.
Good luck!
-Andy | 2018-12-13T01:28:40 | {
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https://mathoverflow.net/questions/212748/existence-and-characterization-of-transitive-matrices/212808 | # Existence and characterization of transitive matrices?
We call a matrix $M \in \mathbb{R}^{d \times d}$ transitive if it satisfies the following:
For any three vectors $u, v, w$ in $\mathbb{R}^d$. If $u^T M v > 0$ and $v^T M w > 0$ then $u^TMw > 0$.
EDIT: The case when $d = 1$ is simple. For $d > 1$ the following is true:
a) M must be Positive Semi Definite, because if $M$ is transitive and $\exists u$ s.t. $u^TMu < 0$ then we can let $v = -u, w = u$ and arrive at the contradiction that $u^t M u < 0 \implies u^tMv > 0, v^tMw > 0 \implies u^TMw > 0 \implies u^TMu > 0$
b) $M \ne I$ because if $M = I$ then we just need to construct three vectors such that $u^Tv > 0, v^Tw > 0$ but $u^Tw < 0$. Intuitively, I know that sum of two acute angles can be obtuse.
c) More generally, $M$ can not be full rank positive definite because then the factors in the Cholesky decomposition $LL^T$ of $M$ would also be full rank and the problem could be reduced to case b) by change of basis using $L$.
Now I am stuck in the space of low rank positive semidefinite matrices. Is there a way to characterize such matrices further? Can such a matrix even exist, apart from zero matrix?
EDIT2: The deduction that $M$ is PSD in step a) implicitly assumed that $M$ is symmetric. Keith Kearnes' answer explicitly shows that $M$ must be rank 1 in that case. The case when $M$ is anti-symmetric is open but I doubt it would be possible to prove anything in that case. I will wait a few days to get more answers. Thanks.
EDIT3: Darij Grinberg's answer proved that every transitive matrix must be symmetric. This completed the characterization.
• b) isn't quite right for $d=1$. Actually for $d=1$ any matrix with a non-negative entry is an example. – Peter Mueller Jul 31 '15 at 19:42
• Actually now I think that the rank of such matrices must be 1. Because every positive semidefinite matrix also has a eigen decomposition so I can again change my basis and reduce the problem to one where $M$ is diagonal with $1 \ge k < d$ positive values. If $k=1$ then fine otherwise if $k>1$ then I can use the counterexample in the $k$ dimensional space (used for b)) to create a counterexample here. – Pushpendre Jul 31 '15 at 21:45
• @Pushpendre: Be careful; diagonalization might mess with the scalar product. – darij grinberg Jul 31 '15 at 23:09
I assume from the wording of the question (positive semidefinite, Cholesky decomposition) that you intend $M$ to be symmetric. Write $M$ as $N^TN$ and let $V\leq \mathbb R^d$ be the range of $N$. For $u, v, w\in \mathbb R^d$ let $x=Nu, y=Nv, z= Nw$. Then your condition reduces to $$x^Ty>0 \;\&\; y^Tz>0 \Rightarrow x^Tz>0$$ on $V$. If the dimension of $V$ is 2 or more, then (as you observed in your last sentence under your item (b)) it is possible to contradict this. That is, it is possible to find $x, y, z$ in the same plane in $V$ such that the angles between $x$ and $y$ and between $y$ and $z$ are acute, while the angle between $x$ and $z$ is obtuse. Once they are known, you can solve for $u, v, w$ which contradict the original condition.
On the other hand, if the dimension of $V$ is 1, there is no contradiction. In this case, $N$ is $1\times d$, so $u^TN^T = Nu$ is a real number, $u^TN^TNv = (Nu)(Nv)$, and you can argue that if $(Nu)(Nv) > 0$ and $(Nv)(Nw) > 0$, then the number $(Nu)(Nw)$ has the same sign as $$(Nu)(Nv)^2(Nw) = [(Nu)(Nv)][(Nv)(Nw)] > 0.$$ The condition that the dimension of $V$ is 1 is that $N$ (and $M$) have rank $1$.
So, the answer for symmetric $M$ is: $M$ is positive semidefinite of rank at most $1$.
Here is a proof of the fact that any transitive matrix $M\in\mathbb{R} ^{d\times d}$ is symmetric. Together with the argument in the answer by Keith Kearnes, this proves that any transitive matrix $M\in\mathbb{R}^{d\times d}$ is a positive-semidefinite symmetric matrix of rank $\leq1$.
We first prove a lemma: If $p\in\mathbb{R}^{d}\setminus\left\{ 0\right\}$ and $q\in\mathbb{R}^{d}$ are two vectors such that every $u\in\mathbb{R}^{d}$ satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$, then
(1) there exists a nonnegative real $\lambda$ such that $q=\lambda p$.
Proof of (1). Let $p\in\mathbb{R}^{d}\setminus\left\{ 0\right\}$ and $q\in\mathbb{R}^{d}$ be two vectors such that every $u\in\mathbb{R}^{d}$ satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$. We have $p\neq0$ and thus $p^{T}p>0$.
If $p$ and $q$ were linearly independent, then there would be a vector $v\in\mathbb{R}^{d}$ satisfying $p^{T}v=1$ and $q^{T}v=-1$; but this would contradict the assumption that every $u\in\mathbb{R}^{d}$ satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$. Hence, $p$ and $q$ must be linearly dependent. Since $p\neq0$, this shows that there exists a $\lambda \in\mathbb{R}$ such that $q=\lambda p$. It remains to prove that this $\lambda$ is nonnegative. Indeed, recall that every $u\in\mathbb{R}^{d}$ satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$. Applying this to $u=p$, we get $q^{T}p\geq0$ (since $p^{T}p>0$). Since $q=\lambda p$, this rewrites as $\lambda p^{T}p\geq0$, and thus $\lambda\geq0$ (since $p^{T}p>0$). This finishes the proof of (1).
Another lemma, which is really well-known: If $V$ is a finite-dimensional vector space, and if $\phi$ is a linear endomorphism of $V$ such that $\phi$ sends every vector in $V$ to a scalar multiple of this vector, then
(2) the endomorphism $\phi$ is a scalar multiple of the identity.
(In order to prove (2), fix a basis of $V$ and see what $\phi$ does to the basis vectors and their pairwise sums.)
Now, let $M\in\mathbb{R}^{d\times d}$ be any transitive matrix. We need to prove that $M$ is symmetric.
If $M=0$, then this is obvious. Thus, WLOG assume that $M\neq0$.
Let $w\in\mathbb{R}^{d}$ be any vector such that $M^{T}w\neq0$. Then, $w\neq0$ (since $M^{T}w\neq0$).
Let $u\in\mathbb{R}^{d}$ be any vector such that $\left( M^{T}w\right) ^{T}u>0$. Let us now show that $w^{T}M\left( M^{T}u\right) \geq0$. Indeed, if $M^{T}u=0$, then this is clear; otherwise it follows from the transitivity of $M$ (in fact, from $w^{T}Mu=\left( M^{T}w\right) ^{T}u>0$ and $u^{T}M\left( M^{T}u\right) =\left( M^{T}u\right) ^{T}\left( M^{T}u\right) >0$ (since $M^{T}u\neq0$), we obtain $w^{T}M\left( M^{T}u\right) >0$ (since $M$ is transitive)). Thus, we have proven that $w^{T}M\left( M^{T}u\right) \geq0$. Hence, $\left( MM^{T}w\right) ^{T}u=w^{T}MM^{T}u=w^{T}M\left( M^{T}u\right) \geq0$.
Let us now forget that we fixed $u$. We thus have shown that every $u\in\mathbb{R}^{d}$ satisfying $\left( M^{T}w\right) ^{T}u>0$ satisfies $\left( MM^{T}w\right) ^{T}u\geq0$. Thus, (1) (applied to $p=M^{T}w$ and $q=MM^{T}w$) yields that
(3) there exists a nonnegative real $\lambda$ such that $MM^{T}w=\lambda M^{T}w$.
Now, let us forget that we fixed $w$. We thus have proven that for every $w\in\mathbb{R}^{d}$ satisfying $M^{T}w\neq0$, we have (3). But (3) also holds for every $w\in\mathbb{R}^{d}$ satisfying $M^{T}w=0$ (since we can use $\lambda=0$). Hence, (3) holds for every $w\in\mathbb{R}^{d}$.
Let $V=M^{T}\left( \mathbb{R}^{d}\right)$ be the image of $M^T$. Then, $M\left( V\right) \subseteq V$ (because (3) holds for every $w\in\mathbb{R}^{d}$). Thus, $M$ restricts to an $\mathbb{R}$-linear endomorphism $\phi$ of $V$. This endomorphism $\phi$ sends every vector in $M$ to a scalar multiple of this vector (because (3) holds for every $w\in\mathbb{R}^{d}$). Thus, $\phi$ must be a scalar multiple of the identity (due to (2)). In other words, there exists some $\mu\in\mathbb{R}$ such that every $v\in V$ satisfies $\phi\left( v\right) =\mu v$. In other words, there exists some $\mu\in\mathbb{R}$ such that every $w\in\mathbb{R}^{d}$ satisfies $MM^{T}v=\mu M^{T}v$ (because of what $V$ is and what $\phi$ is). In other words, there exists some $\mu\in\mathbb{R}$ such that $MM^{T}=\mu M^{T}$. Consider this $\mu$.
We are working over $\mathbb{R}$. Hence, a well-known fact says that $\operatorname*{Ker}\left( MM^{T}\right) =\operatorname*{Ker}\left( M^{T}\right)$. Thus, from $M^{T}\neq0$, we obtain $MM^{T}\neq0$, so that $\mu M^{T}=MM^{T}\neq0$ and therefore $\mu\neq0$. Thus, we can transform $MM^{T}=\mu M^{T}$ into $M^{T}=\dfrac{1}{\mu}MM^{T}$. The matrix $M^{T}$ is thus symmetric (since $MM^{T}$ is symmetric). In other words, the matrix $M$ is symmetric. | 2020-02-26T11:39:58 | {
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https://math.stackexchange.com/questions/2091589/finding-a-and-b | # Finding a and b
The remainder when a polynomial $f(x)$ is divided by $(x-2)(x+3)$ is $ax+b$. When $f(x)$ is divided by $(x-2)$, then remainder is $5$. $(x+3)$ is a factor of $f(x)$. Find the values of $a$ and $b$. I am thinking of using the remainder and factor theorem to solve this however their quotients are different. Can anyone please show me how? Thanks
Your assumptions are that there are polynomials $q,s,t$ such that $$\begin{cases} f(x) = (x-2)(x+3) q(x) &+ a x + b\\ f(x) = (x -2) s(x) &+ 5\\ f(x) = (x + 3) t(x) \end{cases}$$
Now calculate $f(2)$ and $f(-3)$.
• I can only find out that f(2)=5 and f(-3)=0 by this. How can I find a and b? – Estudiantekriss Jan 10 '17 at 10:30
• Plug first $x = 2$ and then $x = -3$ in the first equation to get other expressions for $f(2)$ and $f(-3)$. Equate to the expressions you already have. – Andreas Caranti Jan 10 '17 at 10:31
• But then I still have q(x) unknown – Estudiantekriss Jan 10 '17 at 10:32
• Nobody asked you to determine $q(x)$, which is an impossible task. You are only looking for $a$ and $b$. Why don't you substitute $x = 2$ in the first equation and see what you get, please? – Andreas Caranti Jan 10 '17 at 10:34
• I finally understand this now thank you so much! – Estudiantekriss Jan 10 '17 at 10:40
Simple: first write $\;f(x)=(x+3)g(x)$, and perform the Euclidean division of $g(x)$ by $x-2$: $$g(x)=(x-2)q(x)+g(2).$$ Now $f(2)=5g(2)$, so that $$f(x)=(x-2)(x+3)q(x)+ g(2)(x+3)=(x-2)(x+3)q(x)+ \underbrace{\frac{f(2)}5(x+3)}_{\text{remainder}}.$$
Yes you are on the right track.
Let f(x) = (x-2)(x+3) x m + (ax + b).
We know f(2) = 5, so (2-2)(2+3) x m + (2 a + b) = 5 And similarly f(-3) = 0, so (-3-2)(-3+3) x m + (-3 a + b) = 0
This gives 2 equations
(2 a + b) = 5, and
(-3 a + b) = 0
Which you can solve to give a=1 and b=3
### Hint:
If $$f(x)=(x-\alpha)g(x)+r$$ then $r=f(\alpha)$
Really, $x=\alpha$ $f(\alpha)=0+r$
Then $$f(x)=(x-2)(x+3)+ax+b$$
$f(2)=5$ and $f(-3)=0$
$$2a+b=5$$ $$-3a+b=0$$
$$a=1, b=3$$ | 2019-11-12T20:47:12 | {
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https://mathematica.stackexchange.com/questions/120169/finding-an-analytic-solution-of-a-cubic-equation/120171 | # Finding an analytic solution of a cubic equation
I have been trying to solve a cubic equation $$y=zy^3 + z$$ in $y$ – that is, my desired result is a function $y(z)$ satisfying the equation. Now, there are three solutions of this equation and the Implicit Function Theorem implies that one of them should be analytic at $z = 0$. I have been trying to find this analytic solution.
The task seems to be pretty simple, however the methods I have used successfully to deal with simpler (e.g., quadratic) equations do not seem to work. First of all, I have tried to use
Solve[y == z*y^3 + z, y]
which returned me three solutions. However, by applying
f[z_] := (* each particular solution might go here *)
Series[f[z], {z,0,10}]
I obtained series expansions, which seem to indicate that none of these three functions is analytic at $z=0$ – all three expansions were Puiseux expansions, not Taylor expansions. Looks strange.
Now, as a second attempt, I have tried to use Reduce instead of Solve. More specifically,
Reduce[y == z*y^3 + z, y]
As a result, I obtained
(z == 0 && y == 0) || (z != 0 && (y == Root[z - #1 + z #1^3 &, 1] ||
y == Root[z - #1 + z #1^3 &, 2] || y == Root[z - #1 + z #1^3 &, 3]))
Now, by trying
Series[Root[z - #1 + z #1^3 &, 1], {z,0,10}]
I have found out that this root indeed is an analytic solution (it has a Taylor expansion at $z = 0$).
And now the weirdest thing of all: I have not been satisfied with the "indirect" solution as a root of a cubic equation, so I have tried
Reduce[y == z*y^3 + z, y, Cubics -> True]
And then, suddenly, series expansions of all three solutions again appear to be Puiseux series, not Taylor series.
Can somebody tell me what I am doing wrong (e.g., when using Solve and Reduce, or when trying to obtain series for the solutions)? Or is there some other preferable method of finding analytic solutions? Thank you in advance.
It comes with which roots correspond to which branches of a cubic root in the exact expression. For instance, consider the simpler
$$y^3 = z$$
Then my three solutions are $y=\sqrt[3]{z}$, $y=(-1)^{2/3}\sqrt[3]{z}$, and $y=(-1)^{-2/3}\sqrt[3]{z}$. Mathematica defines that $\sqrt[3]{-1} = \frac{1}{2} + \frac{\sqrt{3}}{2}i$. So if I track the first solution, as a curve in 3D of $(\Re(y),\Im(y),z)$ as I vary $z$ in $[-1,1]$, I'll see a strange behavior at 0, where I switch from $y$ being purely real, to $y$ heading off in an imaginary direction. You could argue that the "right" root to be taking would be the $\sqrt[3]{-1} = -1$ branch, and indeed if we switched from the first solution to the second at $z=0$ it will be analytic and all well. Similarly, the second solution at $z>0$ corresponds to the third at $z<0$, and correspondingly for the third and first.
The important element is that, due to the choice of branch, the solutions "switch roles" at a certain point.
In your specific equation, the roots also switch, so what "was" the first solution is now a different one. The precise behavior is a lot more complicated because of a different kind of singular behavior in the equations there leading to a discontinuity (not just nondifferentiability).
Calling Reduce with Cubics just called Solve indirectly. But Root specifies root by an ordering on the actual numerical value, not closed form structure, so it continued to select the same one.
To see what the behavior of the different solutions looks like, you can try running the below:
sols = Solve[y == z*y^3 + z, y];
{s1, s2, s3} = %[[All, 1, 2]];
Plot[{Re[s1], Re[s2], Re[s3]}, {z, -3, 3}, PlotStyle -> {Red, Green, Blue}]
Plot[{Im[s1], Im[s2], Im[s3]}, {z, -3, 3}, PlotStyle -> {Red, Green, Blue}]
See that the red solution on the left changes color to become blue; the blue and green solutions on the left both hit a pole to become green and red, in some order (up to choice of $\sqrt{-1} = \pm i$). The solution you wanted is Red on the left and Blue on the right
The new in M12 function AsymptoticSolve can be used to find the series:
AsymptoticSolve[y == z y^3 + z, {y, 0}, {z, 0, 10}]
{{y -> z + z^4 + 3 z^7 + 12 z^10}} | 2019-09-23T19:00:50 | {
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https://popeyethewelder.com/102u0/3ccd82-extreme-value-theorem-open-interval | ## extreme value theorem open interval
itself be compact. State whether the function has absolute extrema on its domain $$g(x)= \begin{cases} x & x> 0 \\ 3 & x\leq 0 \end{cases}$$ A description of how to find the arguments that maximize and minimize the value of a function that is continuous on a closed interval. Regardless, your record of completion will remain. I know it's pretty vital for the theorem to be able to show the values of f(a) and f(b), but what if I have calculated the limits as x -> a (from the right hand side) and x -> b (from the left hand side)? The absolute minimum is \answer {-27} and it occurs at x = \answer {1}. on an open interval , then the numbers x = 0, 4. Incognito. 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(-2, 2), an open interval, so there are no endpoints. the interval [\text {-}1,3] we see that f(x) has two critical numbers in the interval, namely x = 0 If has an extremum If f(x) exists for all values of x in the open interval (a,b) and f has a relative extremum at c, where a < c < b, then if f0(c) exists, then f0(c) = 0. So the extreme value theorem tells us, look, we've got some closed interval - I'm going to speak in generalities here - so let's say that's our X axis and let's say we have some function that's defined on a closed interval. maximum and a minimum on and x = 2. Extreme Value Theorem. The Extreme value theorem requires a closed interval. The extreme values of may be found by using a procedure similar to that above, but care must be taken to ensure that extrema truly exist. Critical points are determined by using the derivative, which is found with the Chain Rule. First, we find the critical For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. integrals. In this section we learn the Extreme Value Theorem and we find the extremes of a Differentiation 12. If we don’t have a closed interval and/or the function isn’t continuous on the interval then the function may or may not have absolute extrema. Select the third example, showing the same piece of a parabola as the first example, only with an open interval. at a Regular Point of a Surface. Plugging these special values into the original function f(x) yields: From this data we conclude that the absolute maximum of f(x) on the interval is 3.25 Theorem 2 (General Algorithm). The absolute extremes occur at either the Closed Interval Method • Test for absolute extrema • Only use if f (x) is continuous on a closed interval [a, b]. Here is a detailed, lecture style video on the Extreme Value Theorem: Calculus I, by Andrew The absolute maximum is \answer {0} and it occurs at x = \answer {-2}. them. number in the interval and it occurs at x = -1/3. Using the Extreme Value Theorem 1. When moving from the real line $${\displaystyle \mathbb {R} }$$ to metric spaces and general topological spaces, the appropriate generalization of a closed bounded interval is a compact set. Extreme Value Theorem If is continuous on the closed interval , then there are points and in , such that is a global maximum and is a global minimum on . We find limits using numerical information. An important Theorem is theExtreme Value Theorem. The Extreme Value Theorem. In such a case, Theorem 1 guarantees that there will be both an absolute maximum and an absolute minimum. point. Play this game to review undefined. If the interval is open or the function has even one point of discontinuity, the function may not have an absolute maximum or absolute minimum over For example, consider the functions shown in (Figure) (d), (e), and (f). the equation f'(x) =0 gives x=2 as the only critical number of the function. Open Intervals. Are you sure you want to do this? It states the following: If a function f(x) is continuous on a closed interval [ a, b], then f(x) has both a maximum and minimum value on [ a, b]. We solve the equation f'(x) =0. Solution: First, we find the critical numbers of f(x) in the interval [-1, 0]. The next step is to determine all critical points in the given interval and evaluate the function at these critical points and at the endpoints of the interval. First, since we have a closed interval (i.e. This is used to show thing like: There is a way to set the price of an item so as to maximize profits. We use the logarithm to compute the derivative of a function. Solution: First, we find the critical numbers of f(x) in the interval [\text {-}1, 3]. y = x2 0 ≤ x ≤2 y = x2 0 ≤ x ≺2 4.1 Extreme Values of Functions Day 2 Ex 1) A local maximum value occurs if and only if f(x) ≤ f(c) for all x in an interval. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. In this lesson we will use the tangent line to approximate the value of a function near Open interval. average value. In this section we interpret the derivative as an instantaneous rate of change. Thus f'(c) \geq 0. In order to use the Extreme Value Theorem we must have an interval that includes its endpoints, often called a closed interval, and the function must be continuous on that interval. The Extreme Value Theorem (EVT) says: If a function f is continuous on the closed interval a ≤ x ≤ b , then f has a global minimum and a global maximum on that interval. If f'(c) is undefined then, x=c is a critical number for f(x). 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https://math.stackexchange.com/questions/3950515/leibniz-formula-for-the-nth-derivative-of-fx-xn-1-ln-x | # Leibniz formula for the nth derivative of $f(x)=x^{n-1} \ln x$
Problem : Calculate the derivative of the function $$f: ]0,+\infty\left[\longrightarrow \mathbb{R}\right.$$ defined by $$f(x)=x^{n-1} \ln x$$.
Solution Let $$g_{1}(x)=x^{n-1}$$ et $$g_{2}(x)=\ln x .$$ So we have$$f=g_{1} g_{2}$$ for all $$k$$ verifying $$1 \leq k \leq n-1$$ we have $$g_{1}^{(k)}(x)=(n-1) \cdots(n-1-k+1) x^{n-1-k}$$ et $$g_{1}^{(n)}(x)=0$$ The Leibniz formula gives
$$\begin{array}{c} f^{(n)}(x)=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) g_{1}^{(k)}(x) g_{2}^{(n-k)}(x) \\ =x^{n-1} \frac{(-1)^{n-1}(n-1) !}{x^{n}}+\sum_{k=1}^{n-1}\left(\begin{array}{l} n \\ k \end{array}\right)(n-1) \cdots(n-1-k+1) x^{n-1-k} \frac{(-1)^{n-k-1}(n-k-1) !}{x^{n-k}} \\ =\frac{(-1)^{n-1}(n-1) !}{x}+\sum_{k=1}^{n-1}\left(\begin{array}{l} n \\ k \end{array}\right) \frac{(-1)^{n-k-1}(n-1) !}{x}\\ =-\frac{(n-1) !}{x} \cdot \sum_{k=0}^{n-1}\left(\begin{array}{l} n \\ k \end{array}\right)(-1)^{n-k} \end{array}$$
The sum above is the same as $$\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)(-1)^{n-k}$$ (prived of $$k=n,$$) which is equal to -1
We obtain $$f^{(n)}(x)=\frac{(n-1) !}{x}$$
What I don't get: What's the point of this step:
we have $$g_{1}^{(k)}(x)=(n-1) \cdots(n-1-k+1) x^{n-1-k}$$ et $$g_{1}^{(n)}(x)=0$$
and which solution is wrong?
I spent a lot of time already on this, I think it's a matter of indices I can't point, both solutions seem correct to me.
How I calculated :
we have that
$$\ln ^{(n)}(x)=\frac{(-1)^{n-1}(n-1) !}{x^{n}}$$
and
$$(x^{n-1})^{(k)} = \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! }$$
\begin{aligned} f^{(n)} &=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)\left(x^{n-1}\right)^{(k)}(\ln x)^{(n-k)} \\ &=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! } \frac{(n-1-k) !(-1)^{n-1-k}}{x^{n-k}} \\ =& x^{-1} \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)(n-1) !(-1)^{n-k-1} \end{aligned}
$$= (-1)x^{-1} (n-1) ! \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)(-1)^{n-k}$$
which should give me zero? since
$$\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)(-1)^{n-k} = (1+(-1))^{n}=0$$
## 2 Answers
I know I'm not answering the question, but I can't resist the temptation to show you a trick to arrive at the solution faster (no Leibniz formula): Integrate once before differentiating.
Let $$f_n(x)=x^{n-1}\ln x$$. Compute one of its primitives (antiderivatives) by integrating by parts: $$\int_0^xf_n(t)dt=\frac{x^n}n \ln x-\frac{x^{n-1}}{n^2}=\frac{f_{n+1}(x)}n-\frac{x^{n-1}}{n^2}$$ So to obtain $$f_n^{(n)}(x)$$, we just need to differentiate the above $$n+1$$ times (noting that this will cancel the second term as it's a polynomial of degree $$n-1$$): $$f_n^{(n)}(x)=\frac{f_{n+1}^{(n+1)}(x)}n$$ Solving this recurrence formula gives you the result $$f_n^{(n)}(x)=(n-1)!f_1^\prime(x)=\frac{(n-1)!}{x}$$
Your solution is almost correct (and neither $$g_1^{(k)}(x)$$ or $$g_1^{(n)}(x)$$ are wrong). The only problem is that $$(x^{n-1})^{(k)} = \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! }$$ is only true for $$0 \le k \le n-1$$. For $$k \ge n$$, the $$k$$th derivative will be $$0$$. For example, the $$2$$nd derivative of $$x^1$$ is $$0$$.
The step $$f^{(n)} =\sum_{k=0}^{n}\binom{n}{k}\left(x^{n-1}\right)^{(k)}(\ln x)^{(n-k)} \tag 1$$
is right. But then at the next step, the sum should only go from $$k = 0$$ to $$k = n-1$$ since $$\left(x^{n-1}\right)^{(n)} = 0$$.
Then after that, the same logic applies, so you would end up with $$= (-1)x^{-1} (n-1) ! \sum_{k=0}^{n-1}\binom{n}{k}(-1)^{n-k} = \frac{(n-1)!}{x}$$
• When you say, "But then at the next step, the sum should only go from $k = 0$ to $k = n-1$ since $\left(x^{n-1}\right)^{(n)} = 0$" how so, and why? Dec 16 '20 at 19:25
• @squareroot Maybe a better explanation would have been splitting it from $k = 0$ to $k = n-1$ and having a separate case for $k = n$. When $k = n$, the summand would be $0$ since $\left(x^{n-1}\right)^{(n)} = 0$, so the $k = n$ term can be "chopped off". Dec 16 '20 at 19:55
• How does not separating the $k=n$ makes the following wrong $(-1)x^{-1} (n-1) ! \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)(-1)^{n-k} = (-1)x^{-1} (n-1) ! \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right)(-1)^{n-k} = (-1)x^{-1} (n-1) ! (1+(-1))^{n}=0$ Dec 19 '20 at 20:16
• @squareroot That's all right, but the steps taken to get to that first point are wrong. To start off, the sum should only be from $k = 0$ to $k = n-1$. Dec 19 '20 at 20:35
• why does the sum should only be from $k=0$ to $k=n-1$ just because the nth term is null? what in the steps is exactly wrong or doesn't suffice any necessary condition ? Dec 19 '20 at 20:39 | 2021-10-18T04:04:55 | {
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http://mathoverflow.net/questions/179445/non-zero-smooth-functions-vanishing-on-a-cantor-set | # Non-zero smooth functions vanishing on a Cantor set
It is easy to give examples of continuous functions $f:[0,1]\to \mathbb R_+\cup\{0\}$ non-zero but vanishing on a Cantor set (ex: Can Cantor set be the zero set of a continuous function?). It is clearly non-true for analytic functions. My question is:
1. Are there uniformly continuous non-zero functions vanishing on a Cantor set?
2. Are there α-Hölder continuous non-zero functions vanishing on a Cantor set?
3. Are there continuously differentiable non-zero functions vanishing on a Cantor set?
-
A Cantor set $C\subset[0,1]$ is closed, and that is all we need. Therefore $f(x)=d(x,C)$ (distance from $x$ to $C$) vanishes on and only on $C$. It is also Lipschitz-continuous.
For your third question, you can take $g(x)=f(x)^2$. This is $C^1$ apart from peaks in the middle points of the intervals of the complement of $C$. You can easily smooth the function near these points, since you are well away from $C$.
Choosing a higher power gives you any $C^k$ smoothness you want. But you can also get $C^\infty$. Let $\phi:\mathbb R\to[0,\infty)$ be a smooth function vanishing on $(-\infty,0]$ and set $g(x)=\phi(f(x))$ and mollify the tips.
Conclusion: The answer to all three questions is yes.
Edit: It is not entirely clear what you mean by constructing by hand. This is as close as I can come to making an explicit function if the Cantor set is not specified. (I also noted that the distance was also used in this answer to the OP's linked question. But it was not concerned with regularity.)
-
Thanks! I need to think a bit about your answer for the third question. – user39115 Aug 26 at 16:23
By a theorem of Whitney, for any closed subset $C$ in $\mathbb R^n$ there exists a nonnegative $C^\infty$-function $f$ on $\mathbb R^n$ which vanishes exactly at $C$. This also holds on any smooth manifold. I tried to find the exact reference, but I could not find it just now. Maybe, this result is hidden in:
• Whitney, Hassler: Analytic extensions of differentiable functions defined in closed sets, Trans. AMS 36 (1934), 63--89.
-
It's not hard to construct such a function explicitly. Let $$\psi(x) = \begin{cases} e^{-1/(1-x^2)}, & -1 < x < 1 \\ 0, & \text{else} \end{cases}$$ so that $\psi$ is $C^\infty$ and is nonzero precisely on $(-1,1)$. In the usual middle-thirds construction of the Cantor set, let $C_n$ be the centers of the intervals removed at step $n$ (there are $2^{n-1}$ of them and they have width $3^{-n}$; computing these points explicitly is left as an exercise for the reader with nothing better to do). Then let $$f(x) = \sum_{n=1}^\infty \sum_{y \in C_n} \frac{1}{n!} \psi(2 \cdot 3^n(x-y)).$$ It should be easy to verify that the sum converges uniformly with all its derivatives, so $f$ is also $C^\infty$, and by construction we have made it strictly positive precisely on the removed intervals, i.e. the complement of the Cantor set.
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$f(x)=\exp(-1/d^2(x))$ works for a general closed set $C\subset [0,1]$. – Christian Remling Aug 27 at 16:12
@ChristianRemling: I'm sure you're right, though maybe it takes a bit of work to see that it is $C^\infty$. – Nate Eldredge Aug 27 at 17:35
On second thoughts, this version is actually not $C^{\infty}$ at the midpoint of a gap of $C$, but that's easy to fix (I need a smooth modification near that midpoint). – Christian Remling Aug 27 at 17:56 | 2014-11-27T10:59:03 | {
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https://mathematica.stackexchange.com/questions/179293/linearsolve-on-a-singular-matrix | # LinearSolve on a singular matrix
I have some singular transition rate matrices $m$ (columns add to zero). They have exactly one zero eigenvalue that I want to find the corresponding eigenvector of (the rest of the eigenvalues are negative). This corresponds to the stationary distribution of a continuous-time Markov process. For my actual problem, which is huge and sparse, the fastest method I've found uses (abuses?) LinearSolve. What I don't understand, is why this method works at all.
Here's a minimal example:
m = {{-0.1, 2., 0}, {0.1, -4.1, 6.}, {0, 2.1, -6.}};
Eigenvalues[m]
(* {-8.73318, -1.46682, 0.} *)
Here are two (slower) ways to get the answer I'm looking for:
Eigenvectors[m][[-1]]
(* {0.9986, 0.04993, 0.0174755} *)
NullSpace[m]
(* {{0.9986, 0.04993, 0.0174755}} *)
At first, I thought I could use LinearSolve[m, {0, 0, 0}] but that only gives the trivial solution {0., 0., 0.}. On a lark, I tried replacing the {0, 0, 0} with a small vector. Surprisingly, this works (although with the warning LinearSolve::luc):
x = LinearSolve[m, {10^-10, 10^-10, 10^-10}];
x/Norm[x]
(* {0.9986, 0.04993, 0.0174755} *)
Even more surprising to me, replacing the small vector with a decidedly non-small vector still works with a warning:
x = LinearSolve[m, {1, 1, 1}];
x/Norm[x]
(* {0.9986, 0.04993, 0.0174755} *)
For the record, this answer doesn't actually solve the problem given to LinearSolve (although it solves MY problem):
m.x - {1, 1, 1}
(* {1., -1., -1.} *)
Finally, my original problem with a SparseArray doesn't even generate a warning. This can be reproduced as:
x = LinearSolve[SparseArray[m], {1, 1, 1}];
x/Norm[x]
(* {0.9986, 0.04993, 0.0174755} *)
So, my questions are: why does this work? and, can I count on this working in general?
• You could also try x = With[{i = 1}, Normalize[Insert[LeastSquares[Drop[m, 0, {i, i}], m[[All, i]]], -1, i]]]. However there is a chance it will be wrong for some particular is so confirmation is needed m.x – Coolwater Jul 31 '18 at 18:59
• I have added the tag 'bugs'. After all, solving m.x+b=0 should depend on b; but for badly conditioned matrices, Mathematica returns an answer that does not depend on b. – Hector Jul 31 '18 at 19:25
• I don't think it's a bug. Note that x does depend on b, while x/Norm[x] (almost) does not. – Michael E2 Jul 31 '18 at 20:48
• @Hector I also don't think that it's a bug. It is caused just by the way LU-decomposition and backward/forward substitution work. At some point one devidides by (almost) zero which results in the humongous length of the "solution". Since you get warned by Mathematica that strange things are bound to happen, it appears entirely sane to me. – Henrik Schumacher Aug 1 '18 at 7:42
• My bad. It is not a bug. I do not remember what I was thinking when I added the tag and when I wrote my answer. It was just a bunch of non-sense. – Hector Aug 1 '18 at 13:18
The example with one zero eigenvector can be explained like this. Of a matrix $M$, let $e_k$ be the unit eigenvectors with nonzero eigenvalues $\lambda_k$ and $n$ be the unit eigenvector with eigenvalue zero. Suppose when we apply LinearSolve to the problem $Mx=b$, we get an approximate solution $x^*$. Suppose $x^*$ is the exact solution to a closely related problem $M^* x = b$, where $M^*$ is a matrix approximately equal to $M$. Let $e^*_k \approx e_k$, $n^* \approx n$ be the unit eigenvectors of $M^*$ with eigenvalues $\lambda^*_k \approx \lambda_k$, $\nu^* \approx 0$. If $$b = c_0 n^* + \sum_{k} c_k e^*_k\,,$$ then the solution $x^*=(M^*)^{-1}b$ will be $$x^* = (\nu^*)^{-1} c_0 n^* + \sum_{k} (\lambda^*_k)^{-1} c_k e^*_k\,.$$ The value of $(\nu^*)^{-1}$ for the OP's m is around $10^{15}$ and much, much bigger than the $(\lambda^*_k)^{-1}$, so that $${x^* \over \|x^*\|} \approx n^* \approx n \quad\text{(up to sign)}\,,$$ unless $c_0$ is quite small.
So the procedure will give a good approximation as long as $c_0$ is not too much smaller than the other coefficients $c_k$ relative to the other eigenvalues. I haven't thought of a way to verify that hypothesis, except to say that it is unlikely to pick a $b$ for which $c_0$ is very small (unless the problem has symmetry and you pick a corresponding $b$). The main issue is whether any of the $\lambda_k$ can be very small. Perhaps that can be known from the particular kind of $M$ that arises.
• Thanks for the insights. I suspect the next eigenvalue might be small when the Markov process is almost reducible. I suppose one can always check the answer at the end too! – Chris K Aug 1 '18 at 21:53
I am not sure whether yours is a relyable method. But the idea behind it is intriguing. So I tried to derive a method for computing the null space of a matrix from it, but without division by zero. Hopefully, this will be more robust.
# Background
The backend of LinearSolve is LAPACK and computes a (permuted) LU-decomposition. The default method ("Multifrontal") for sparse matrices is UMFPACK. It also computes essentially a permuted LU-decomposition, but it uses a clever permutation to make the factors as sparse quite sparse. (I don't say "as sparse as possible" because it's not true; one had to solve a NP-hard problem for that.)
With the input $n\times n$-matrix $A$, a permuted LU-decomposition consists of
• a lower triangular square matrix $L$ with only 1 on the diagonal
• an upper triangular square matrix $U$
• and a permutation of $\{1,\dotsc,n\}$
such that (in Mathematica notation)
L.U == A[[p]]
The $L$-factor is always invertible. Thus, all information about singularity is contained in the $U$-factor. It is upper triangular, so the rank of $A$ (equalling the rank of $U$) equals the number of nonzero diagonal elements of $U$. I guess that an estimator for the condition number of $A$ is computed somehow from the diagonal elements of $U$ (in a comparatively cheap way compared to computing the minimal and maximal absolute eigenvalue). Anyways, zeros (or numbers close to zero) on the diagonal of $U$ are strong indicators for a singular matrix. Moreover, when solcing linear systems involving U (the first step of solving $L \,U \, x = b$) via back substitution, one has to divide by the diagonal elements. If some of these are close to zero, they shadow somewhat the dependence on the right hand side.
# Some alternate algorithm
We can use this knowledge in order to avoid the division by (nearly) zeros. The idea is to modify the factor U to a matrix V that is invertible and to solve linear systems involving V in order to compute a basis of the null space.
QuickNullSpace[A_?SquareMatrixQ, threshold_: 1. 10^-10] :=
QuickNullSpace[Quiet[LinearSolve[A]], threshold];
QuickNullSpace[S_LinearSolveFunction, threshold_: 1. 10^-10] :=
Module[{U, L, idx, V, p, q, x},
U = S["getU"];
L = S["getL"];
p = S["getPermutations"];
{p, q} = If[MissingQ[p], {All, All}, p];
idx = Flatten[Position[Threshold[Abs[Normal[Diagonal[U]]], threshold], 0.]];
V = U;
Do[V[[i, i]] = 1., {i, idx}];
Transpose[
Quiet[LinearSolve[
V,
SparseArray[
Transpose[{idx, Range[Length[idx]]}] -> 1., {Length[U],
Length[idx]}, 0.]
]][[q]]
]
]
# Dense array usage example
Experimentally, this performs quite well for dense matrices. Here we test it on some random $1000 \times 1000$-matrix with $30$-dimensional null space. Some more tweaking is necessary in order to cope with the column permutations that might be needed in order to get sparse factors.
n = 1000;
m = 30;
U = RandomVariate[CircularRealMatrixDistribution[n]];
V = RandomVariate[CircularRealMatrixDistribution[n]];
A = U.(Join[ConstantArray[0., {m}], RandomReal[{-1, 1}, {n - m}]] V);
r1 = QuickNullSpace[A]; // RepeatedTiming // First
r2 = NullSpace[A]; // RepeatedTiming // First
MatrixRank[r1] == m
Max[Abs[A.Transpose[Orthogonalize@r1]]]
Max[Abs[A.Transpose[r2]]]
0.031
0.25
True
1.50924*10^-13
1.28695*10^-16
# Sparse matrix usage example
The method seems to work also for sparse matrices. Here an example where it is applied to the graph Laplacian of some random graph with at least $4$ connected components; each component will add another vector to the null space.
G = GraphDisjointUnion @@ Table[RandomGraph[{100, 600} 10], {4}];
N[DiagonalMatrix[SparseArray[Total[A, {2}]]] - A]
];
r1 = QuickNullSpace[A]; // RepeatedTiming // First
r2 = NullSpace[A]; // RepeatedTiming // First
MatrixRank[r1] == MatrixRank[r2]
Max[Abs[A.Transpose[Orthogonalize@r1]]]
Max[Abs[A.Transpose[r2]]]
0.27
14.
True
7.61197*10^-15
9.51322*10^-15
That's an over $50$-fold speedup. Moreover, QuickNullSpace will perform relatively better with large matrices of increasing size.
# Remarks
• This implementation is highly experimental. No guarantees from my side. Suggestions are welcome. I don't know how reliable (how accurate, how stable) this method is.
• Notice that the QuickNullSpace[A] is not Orthogonalized. I prefer to leave the choice to the user.
• There is some potential for further speedup. Currently, two calls to LinearSolve are necessary: One for the factorization of A and one for solving several linear systems with the V which we obtains as perturbation of U. Since V is an upper triangular matrix, we could in principle skip the factorization step and start immediately with the back substitution. For the dense matrix case, the solution might lie around somewhere in the context "LinearAlgebraLAPACK". But I have not clue how to do it for the sparse matrix case without resorting to LibraryLink calls.
• Lots of interesting ideas to digest! QuickNullSpace gives a crazy answer for something more like my real problem, but I suppose I'll start a fresh question for that. – Chris K Aug 1 '18 at 21:54
• @ChrisK I am curious about the counter examples. Please let me know. – Henrik Schumacher Aug 2 '18 at 6:49 | 2021-06-19T03:11:09 | {
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https://math.stackexchange.com/questions/174154/how-to-get-principal-argument-of-complex-number-from-complex-plane/174174 | # How to get principal argument of complex number from complex plane?
I am just starting to learn calculus and the concepts of radians. Something that is confusing me is how my textbook is getting the principal argument ($\arg z$) from the complex plane. i.e. for the complex number $-2 + 2i$, how does it get $\frac{3\pi}{4}$? (I get $\frac{\pi}{4}$).
The formula is $\tan^{-1}(\frac{b}{a})$, and I am getting $\frac{\pi}{4}$ when I calculate $\tan^{-1}(\frac{2}{-2})$. When I draw it I see that the point is in quadrant 2.
So how do you compute the correct value of the principal argument?
• But $\tan\frac{\pi}4=1$, not $-1$, so how are you getting $\tan^{-1}(-1)=\frac{\pi}4$? You should be getting $-\frac{\pi}4$. And since that’s in the fourth quadrant, and you know that you need an angle with the same tangent in the second quadrant, you add $\pi$ to get $\frac34\pi$. – Brian M. Scott Jul 23 '12 at 9:25
• Just so you know, it's the principal argument. – Potato Jul 23 '12 at 9:27
• Oh.. I didn't know you have to add pi. So if the angle is in the 3rd quadrant do I have to add 2pi? I can see these are silly questions, but small things like this are making it difficult for me to learn. I really want to understand this stuff. Thanks so much. – tb747 Jul 23 '12 at 9:36
• The period of $\tan$ is $\pi$ so that you must verify if $\pi$ has to be added (depending of the quadrant). That's why many programming languages include a 'atan2' function. – Raymond Manzoni Jul 23 '12 at 9:39
The principal value of $\tan^{-1}\theta$ is always between $-\frac{\pi}2$ and $\frac{\pi}2$. The principal value of $\arg z$, on the other hand, is always in the interval $(-\pi,\pi]$. Thus, for $z$ in the first quadrant it’s between $0$ and $\frac{\pi}2$; for $z$ in the second quadrant it’s between $\frac{\pi}2$ and $\pi$; for $z$ in the third quadrant it’s between $-\frac{\pi}2$ and $-\pi$; and for $z$ in the fourth quadrant it’s between $0$ and $-\frac{\pi}2$. This means that the $\tan^{-1}$ function gives you the correct angle only when $z$ is in the first and fourth quadrants.
When $z$ is in the second quadrant, you have to find an angle between $\frac{\pi}2$ and $\pi$ that has the same tangent as the angle $\theta$ returned by the $\tan^{-1}$ function, which satisfies $-\frac{\pi}2<\theta\le 0$. The tangent function is periodic with period $\pi$, so $\tan(\theta+\pi)=\tan\theta$, and $$\frac{\pi}2=-\frac{\pi}2+\pi<\theta+\pi\le0+\pi=\pi\;,$$ so $\theta+\pi$ is indeed in the second quadrant.
When $z$ is in the third quadrant, you have to find an angle between $-\pi$ and $-\frac{\pi}2$ that has the same tangent as the angle $\theta$ returned by the $\tan^{-1}$ function, which satisfies $0\le\theta<\frac{\pi}2$. This time subtracting $\pi$ does the trick: $\tan(\theta-\pi)=\tan\theta$, and
$$-\pi=0-\pi<\theta-\pi<\frac{\pi}2-\pi=-\frac{\pi}2\;.$$
There’s just one slightly tricky bit. If $z$ is a negative real number, should you consider it to be in the second or in the third quadrant? The tangent is $0$, so the $\tan^{-1}$ function will return $0$. If you treat $z$ as being in the second quadrant, you’ll add $\pi$ and get a principal argument of $\pi$. If instead you treat $z$ as being in the third quadrant, you’ll subtract $\pi$ and get a principal argument of $-\pi$. But by definition the principal argument is in the half-open interval $(-\pi,\pi]$, which does not include $-\pi$; thus, you must take $z$ to be in the second quadrant and assign it the principal argument $\pi$.
• I tried to correct the two (very) small typos in your solution above, but edits must be at least 6 characters, and I couldn't find anything else to edit (this solution is very informative). Last sentence (before the semicolon), " ...the principal argument is in the interval $(-\pi,\pi]$, which does not include $-\pi$; ... ," and, at the very end of the same sentence, " ...assign it the principal argument $\pi$ ." – Procore Jan 21 '17 at 21:08
• I realize this post is from 2012, yet it was immensely helpful to me as it was the clearest explanation I could find! Thank you! – C. Ekinci Sep 10 '18 at 3:13
One of the most important functions in analysis is the argument function $${\rm arg}:\quad \dot{\mathbb R}^2\to {\mathbb R}/(2\pi{\mathbb Z}),\qquad (x,y)\mapsto {\rm arg}(x,y)\ ,$$ resp. $${\rm arg}:\quad \dot{\mathbb C}\to {\mathbb R}/(2\pi{\mathbb Z}),\qquad z\mapsto {\rm arg}z\ ,$$ where the dot indicates that the origin is removed.
Intuitively ${\rm arg}(x,y)$ denotes the polar angle of $(x,y)$ "up to multiples of $2\pi$". For "local" considerations there are continuous real-valued ("numerical") representants of ${\rm arg}$; but these are defined only in a suitable part of $\dot{\mathbb R}^2$. In particular the principal value $${\rm Arg}:\quad {\mathbb R}^2\setminus\{(x,0)|x\leq0\}\ \to {\mathbb R},\qquad (x,y)\mapsto {\rm Arg}(x,y)$$ is defined on the $(x,y)$-plane slit up along the negative $x$-axis. It has the simple symmetry property ${\rm Arg}\bar z=-{\rm Arg}z$, and for $x>0$ it is given by $${\rm Arg}(x,y)=\arctan{y\over x}\qquad(x>0)\ .$$ Note that the ${\rm arg}$ function has a well defined gradient given by $$\nabla{\rm arg}(x,y)=\Bigl({-y\over x^2+y^2},{x\over x^2+y^2}\Bigr)\qquad\bigl((x,y)\ne(0,0)\bigr)\ .$$
I've come up with this recipe for principal argument. It saves messing around adding or subtracting $\pi$.
$$\text{Arg} (z) = n\ \text{cos}^{-1} \left(\frac{x}{z}\right)$$
in which n = 1 if y ≥ 0 but n = -1 if y < 0.
I've tried to 'automate' the value of n, but the best I can do is
$$\text{Arg} (z) = \frac{y}{|y|}\ \text{cos}^{-1} \left(\frac{x}{z}\right).$$
Unfortunately this fails for y = 0 (real z), so the y = 0 case would still have to be catered for separately.
Edit: A very ugly self-contained recipe would be
$$\text{Arg} (z) = \text{sgn}\left(\text{sgn}(y) + \frac{1}{2}\right)\ \text{cos}^{-1} \left(\frac{x}{z}\right).$$
## protected by Zev ChonolesDec 25 '16 at 5:57
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2019-08-17T21:26:36 | {
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https://math.stackexchange.com/questions/2054394/integral-of-int-tan5x-sec4xdx | # Integral of $\int{\tan^{5}(x)\sec^4(x)}dx$?
Here's my attempt at the problem: $$\int{\tan^{5}(x)\sec^4(x)}dx= \int{\frac{\sin^5(x)}{\cos^9(x)}}\,dx= \int{\frac{\sin(x)(\sin^2(x))^2}{\cos^9(x)}}\,dx= \int{\frac{\sin(x)(1-\cos^2(x))^2}{\cos^9(x)}}\,dx= \int{\frac{(1-u^2)^2}{u^9}}\,du= \int{\frac{u^4-2u^2+1}{u^9}}\,du= \int{\Big(\frac{1}{u^5}-\frac{2}{u^7}+\frac{1}{u^9}}\Big) \,du= -\frac{1}{4u^4}+\frac{1}{3u^6}-\frac{1}{8u^8}+C= -\frac{\sec^4(x)}{4}+\frac{\sec^6(x)}{3}-\frac{\sec^8(x)}{8}+C$$
It seems, however, that the actual answer should be: $$\frac{\sec^4(x)}{4}-\frac{\sec^6(x)}{3}+\frac{\sec^8(x)}{8}+C$$
What am I doing wrong?
• $$d( \cos x ) = - \sin x dx$$ – ILoveMath Dec 11 '16 at 19:55
• So should $\int{\frac{(1-u^2)^2}{u^9}}du$ instead be $-\int{\frac{(1-u^2)^2}{u^9}}du$ in the above solution? – kylemart Dec 11 '16 at 20:01
• yesssssssssssssssss – ILoveMath Dec 11 '16 at 20:02
• Ok, Remember, this: $$\int_a^b f(x) dx = - \int_b^a f(x) dx$$ – ILoveMath Dec 11 '16 at 20:08
• @LanierFreeman Yeah. My textbook uses the tan / sec relation to solve the problem, but I was curious to see if it could be done easily with sin / cos – kylemart Dec 11 '16 at 20:38
Substituting $u=\tan(x),\,du=\sec^2(x)\,dx$ | 2019-10-22T01:49:46 | {
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https://math.stackexchange.com/questions/1738533/how-many-integers-between-1-and-1000-use-exactly-three-digits/1738557 | # How many integers between $1$ and $1000$ use exactly three digits?
There is a problem in which I am obtaining a different answer than my professor. The problem is as follows:
How many integers between $1$ and $1000$ use exactly three digits?
The professor shows the solution as: $9 \cdot 9 \cdot 8=648$, but I have no idea where those numbers are coming from. On the other hand, I say that, excluding $1$ to $99$ and $1000$, there are $900$ integers that use exactly three digits. Can someone please explain which one of us is right and why?
Thank you!
• Your list would include $111$, which uses exactly one digit - not three. The wording is a bit ambiguous, but I'd bet this is what's intended.
– user296602
Apr 12, 2016 at 2:45
Your professor is not allowing integers where two digits match- s/he wants three distinct digits. The word distinct is not in the problem as quoted. To use exactly three digits, the number needs to not have a leading zero, so be greater than $99$. This means you have $9$ choices for the first digit. Then the second digit cannot match the first, so you have $9$ choices for it. Then the third ...
• Okay, thank you! I think I understand now. Apr 12, 2016 at 2:47
Professor has shown the result for all those numbers in which digits are not repeated.
You are in the need of three digits number in which digits are non repeating, so let the structure of number be “_ _ _"
$•$ now except “0" any digits can be placed at $100^{th}$ place,so it can be filled in 9 ways
$•$ similarly $ten^{th}$place can be filled in 9 ways (including “0" and excluding number used in $100^{th}$ place)
$•$ once place can be filled in 8 ways (excluding number used in $100^{th}$ ,$10^{th}$ place )
So total number of ways equal to $9×9×8=\boxed{648}$ | 2022-09-24T20:21:59 | {
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https://math.stackexchange.com/questions/3683136/examples-of-non-continuous-non-piecewise-constant-idempotent-map | # Examples of non-continuous, non-piecewise-constant, idempotent map?
So a projection $$P$$ is a linear map such that $$P^2 = P$$. If we don't require linearity, then there are other examples of functions $$f$$ such that $$f^2 = f$$. For example, the floor and ceiling functions. If $$f: \mathbb{R} \to \mathbb{R}$$ and we require continuity, it seems that the image is a closed interval and $$f(x)=x$$ for all $$x$$ in the image.
I'm wondering if there are any examples of idempotent maps that are not continuous but also not piecewise constant?
How about \begin{align} f(x)=\begin{cases} 0 &\text{if x\in\mathbb{Q}}\,, \\ x &\text{otherwise}\,. \end{cases} \end{align} I'm trying to think of an injective example.
Edit: The only injective function of this kind is $$f(x)=x$$. If $$f$$ is injective and $$f(x)=y\neq x$$, then $$f(y)=y$$ but $$f(y)\neq f(x)$$, a contradiction.
• How about if we extend "piecewise constant" to allow for infinitely many pieces? Or rather, we allow for removable discontinuities. Do you have an idea what examples we may have then? – twosigma May 20 at 5:03
• @twosigma Any function that is identity on its image does the trick, so you can choose whatever image you want, and let $f$ map other points into arbitrary points in its image. – Qiyu Wen May 20 at 5:08
Pick an arbitrary $$A \subset \mathbb R$$. Pick an arbitrary function $$g :\mathbb R \backslash A \to A$$.
Define f(x)= \begin{align} \begin{cases} x &\text{if x\in A}\,, \\ g(x) &\text{otherwise}\,. \end{cases} \end{align}
Then $$f \circ f=f$$.
Conversely, if $$f \circ f=f$$ then $$A= f(\mathbb R)$$ together with $$g(x)=f(x) \forall x \notin A$$ produce $$f$$.
• Worth pointing out explicitly: this answer lists all of the idempotent functions from a set to itself. – Milo Brandt May 20 at 23:24
This a “frame challenge” answer — it seems the restrictions in the question are partly motivated by a misconception, and so this doesn’t answer the question as written, but may give another answer to its original motivation. The question says “If $$f : \newcommand{\R}{\mathbb{R}}\R \to \R$$ and we require continuity, it seems that we basically get the identity function”, and it seems this is the motivation for including “non-continuous” in the question.
However, continuous idempotents don’t really have to be close to the identity function, except to the same extent that all idempotents do. The question links the claim to a nice answer which shows that:
1. If $$f:\R \to \R$$ is continuous and idempotent then $$I=f(\R)$$ is a closed interval and $$f(x)=x$$ for all $$x\in I$$.
2. If $$f$$ is also differentiable and nonconstant, then $$I=\R$$, i.e., $$f(x)=x$$ for all $$x\in\mathbb R$$.
which is true, but a little misleading — regrouping makes the dependency of the conditions clearer:
1. Any idempotent map $$f$$ on any set $$X$$ acts as the identity on its image $$f(X) \subseteq X$$.
2. The image of any continuous idempotent $$f:\R \to \R$$ is a closed interval.
3. The image of any differentiable idempotent $$f:\R \to \R$$ is either a point (i.e. $$f$$ is constant), or the whole of $$\R$$ (in which case by (1), $$f$$ is the identity function).
That is, all idempotents are somewhat like the identity; and non-constant differentiable idempotents are precisely the identity; but continuity doesn’t force the idempotent to be any more identity-like than it already had to be. So if the original motivation was partly just “interesting, non-trivial examples of idempotents”, then there’s no need to exclude continuous ones.
A nice intuitive continuous example is the “absolute value” function, $$x \mapsto \left|x\right|$$.
Another nice one is the function $$x \mapsto \left\{ \begin{array}{lr}x & -1 \leq x \leq 1 \\ \frac{1}{x} & \text{otherwise}\end{array}\right.$$
N.S.’s answer gives a general explanation: an idempotent function on $$\mathbb{R}$$ can have an arbitrary subset $$S$$ of $$\mathbb{R}$$ as its image (and on $$S$$ it must act as the identity), and then on the rest of $$\mathbb{R}$$, it can be an arbitrary function $$\mathbb{R} \setminus S \to S$$. So for continuous examples, take $$S$$ to be any closed interval, and then $$f$$ on $$\R \setminus S$$ (which must be the union of 0, 1, or 2 half-infinite open intervals) can be any continuous function into $$S$$ that fixes the endpoints of $$S$$.
• But the absolute value function is continuous, so it doesn't fit the criteria, does it? – Tanner Swett May 20 at 14:34
• Ah, I missed that the title includes “non-continuous”. What I meant to address was how in the body, the request for continuity was motivated by “if we require continuity, it seems that we basically get the identity function”, which is not at all true. I’ll edit my answer to address this more explicitly. – Peter LeFanu Lumsdaine May 20 at 14:39
• Nice examples! Yeah I shouldn't have phrased it in a way that suggested continuity will "basically" give us the identity function. I'll edit this – twosigma May 20 at 15:47
• (@PeterLeFanuLumsdaine, thank you for correcting my mixed-up answer so quickly! I've deleted it, so I'm not sure if you can read my thanks in the comments there.) – Vectornaut May 21 at 13:36 | 2020-07-16T04:21:41 | {
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https://tex.stackexchange.com/questions/563195/inconsistent-bracket-sizes-in-a-piecewise-function | # Inconsistent bracket sizes in a piecewise function
There are a number of similar questions to this one, but many of them suggest manually setting the size of my brackets. I would prefer not to do this, but if it is the only solution, I guess I will have to settle.
In my piecewise function the 2nd and 3rd steps have very similar contents within the brackets; however, the brackets look to be sized differently. Each step has one large set of brackets and and one small set of brackets, just in reverse order to each other. What is the reasoning for this and is there a solution other than manually setting the sizing?
The piecewise function:
Sample code:
\documentclass [a4paper,12pt]{report}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
$$\label{eqn:damage piecewise} d_i = \begin{cases} \quad \quad \quad \left( 1-\dfrac{E_i^{hrd}}{E_i}\right) \left( 1-\dfrac{1}{k_i} \right) & \forall \quad 1 \leq k_i \leq k_i^{ult} \\[12pt] d_i^{ult} \cdot \dfrac{k_i^{ult}}{k_i} + \left( 1 + \dfrac{E_i^{sft}}{E_i} \right) \left( 1 - \dfrac{k_i^{ult}}{k_i} \right) & \forall \quad k_i^{ult} \leq k_i \leq k_i^{sft} \\[12pt] d_i^{sft} \cdot \dfrac{k_i^{sft}}{k_i} + \left( 1 + \dfrac{E_i^{res}}{E_i} \right) \left( 1 - \dfrac{k_i^{sft}}{k_i} \right) & \forall \quad k_i^{sft} \leq k_i \leq k_i^{max} \end{cases}$$
\end{document}
• Unrelated: are hrd, ult, … products of variables or simple abbreviations? – Bernard Sep 18 '20 at 14:31
• simple abbreviations. I know they should be written in \text{} format. – Aaron Rhodes Sep 18 '20 at 14:36
• No they should not, use \mathrm, the \text command does not do what you think. Unrelated you might want to look up the dcases env from mathtools – daleif Sep 18 '20 at 15:10
• You could make all the exponents the same size by adding a \strut (too big) or a \vphantom{f}. – John Kormylo Sep 18 '20 at 15:35
• @John Kormylo is there a benefit to this? Is it taken as standard practice to do this or is this just a personal preference. – Aaron Rhodes Sep 18 '20 at 15:44
With the \biggl( ... \biggr) pairs, you get a better result, aand it's generally recommended to choose their size manually. Delimiters, aesthetically, can be a bit less high than their contents. I also slightly reduced the spacing between the quantifiers, and centred better the first equation in cases w.r.t. the others (this can be done automatically with package eqparbox, but I don't think it was worth wrting more code in this case). Last, I used the dcases environment, from mathtools to save having to type \dfrac.
\documentclass [a4paper,12pt]{report}
\usepackage{mathtools}
\usepackage{amssymb}
\begin{document}
$$\label{eqn:damage piecewise} d_i = \begin{dcases} \hskip2.7em\biggl( 1-\frac{E_i^\text{hrd}}{E_i}\biggr) \biggl( 1-\dfrac{1}{k_i} \biggr) & \forall \; 1 \leq k_i \leq k_i^\text{ult} \\[12pt] % d_i^\text{ult} \cdot \dfrac{k_i^\text{ult}}{k_i} + \biggl( 1 + \frac{E_i^\text{sft}}{E_i} \biggr) \biggl( 1 - \dfrac{k_i^\text{ult}}{k_i} \biggr) & \forall \; k_i^\text{ult} \leq k_i \leq k_i^\text{sft} \\[12pt] % d_i^\mathrm{sft} \cdot \dfrac{k_i^\text{sft}}{k_i} + \biggl( 1 + \frac{E_i^\text{res}}{E_i} \biggr) \biggl( 1 - \dfrac{k_i^\text{sft}}{k_i} \biggr) & \forall \; k_i^\text{sft} \leq k_i \leq k_i^{\max} \end{dcases}$$
\end{document}
• This is perfect. It looks much better than what I had down, I have accepted this answer, thanks. What is the difference between \Bigg and \biggl( ... \biggr) ? Is this what allows the delimiters to be slightly shorter than the contents? – Aaron Rhodes Sep 18 '20 at 15:29
You can make the same size brackets with \Big( and \Big) commands.
The code is as:
\documentclass [a4paper,12pt]{report}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
$$\label{eqn:damage piecewise} d_i = \begin{cases} \displaystyle \quad \quad \quad \Bigg( 1-\dfrac{E_i^{hrd}}{E_i}\Bigg) \Bigg( 1-\dfrac{1}{k_i} \Bigg) & \forall \quad 1 \leq k_i \leq k_i^{ult} \\[12pt] d_i^{ult} \cdot \dfrac{k_i^{ult}}{k_i} + \Bigg( 1 + \dfrac{E_i^{sft}}{E_i} \Bigg) \Bigg( 1 - \dfrac{k_i^{ult}}{k_i} \Bigg) & \forall \quad k_i^{ult} \leq k_i \leq k_i^{sft} \\[12pt] \displaystyle d_i^{sft} \cdot \dfrac{k_i^{sft}}{k_i} + \Bigg( 1 + \dfrac{E_i^{res}}{E_i} \Bigg) \Bigg( 1 - \dfrac{k_i^{sft}}{k_i} \Bigg) & \forall \quad k_i^{sft} \leq k_i \leq k_i^{max} \end{cases}$$
\end{document}
• I marked this answer as useful as it solves my problem. However, I will accept Bernard's answer, it corrects many of my "bad practices". – Aaron Rhodes Sep 18 '20 at 15:28 | 2021-03-08T03:16:31 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=bgivim5p3urv3r0iehtekcaa95&action=printpage;topic=837.0 | # Toronto Math Forum
## APM346-2016F => APM346--Lectures => Chapter 4 => Topic started by: Shaghayegh A on November 14, 2016, 02:18:44 PM
Title: HA 6, sections 4.3-4.5, problem 6
Post by: Shaghayegh A on November 14, 2016, 02:18:44 PM
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter4/S4.5.P.html
For problem 6f, I get $b_n=0$, which I know is wrong.
I have $$b_n=\frac{2}{\pi} \int_0 ^{\pi} \sin((m-1/2)x) \sin((n+1/2)x) dx \\ =1/pi \int_0 ^{\pi} \cos((m-n-1)x) -\cos((m+n)x) dx =0$$
Separately for $n \neq m-1$ and $n=m-1$. Why am I getting b_n=0?
Title: Re: HA 6, sections 4.3-4.5, problem 6
Post by: Victor Ivrii on November 15, 2016, 07:03:03 AM
The last line of calculations is wrong. We already counted
$\in_0^\pi \cos (kx)\,dx$ | 2021-09-16T23:59:21 | {
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https://mathoverflow.net/questions/168887/probabilistic-method-used-to-prove-existence-theorems | # Probabilistic method used to prove existence theorems
I am aiming for a "big list" of theorems using probability techniques to prove existence of some objects. And in each case, there is an interesting question -- can we find an explicit example? Was the probabilistic method the first one to come by or it was following the explicit construction? Are there any examples where it is proven that the example exists but its explicit construction is impossible?
In wikipedia, on Probabilistic method, they do propose two examples due to Erdős:
FIRST ONE. For a complete graph on n vertices it is possible to color the edges of the graph in two colors so that there is no complete subgraph on r vertices which is monochromatic (every edge colored the same color).
Question: is there any explicit algorithm for such a coloring?
SECOND ONE. The second problem also comes from graph theory and deals with a chromatic number of a graph: the minimal number of colors in which we can color the graph so that two adjacent vertices are colored differently. Given two positive integers $g$ and $k$, does there exist a graph containing only cycles of length at least $g$ such that its chromatic number of is at least $k$? It can be shown by the means of probabilistic method that such a graph exists for any values of $g$ and $k$.
Question: is there some algorithm to provide such a coloring?
I am not at all a specialist in graph theory, so I would like to hear the answers to thse questions as well as to see other examples which come from other branches of math.
• There are books on the probabilistic method. Perhaps you should narrow your question. – Douglas Zare Jun 3 '14 at 6:42
• I like this question, although it is probably doomed for closure (unless you can somehow quickly increase your reputation by two digits!) Assuming the question can survive, surely someone needs to mention the new theorem of Peter Keevash at arxiv.org/abs/1401.3665, which is a beautiful, powerful and nontrivial example of a "probabilistic" existence result. – Peter Dukes Jun 3 '14 at 8:23
• See en.wikipedia.org/wiki/… although not all of those are existence theorems. There are lots more. If your interest is primarily combinatorics as opposed to analysis, why not narrow the question to probabilistic proofs in combinatorics? – Douglas Zare Jun 3 '14 at 9:26
• For a finite probability space such as your examples, the probabilistic method should always immediately give a naive algorithm: Try all possible outcomes of the random choices, and for each check if the condition is satisfied. Of course, with $n$ bits of randomness this takes time $2^n$, and you'd prefer a $poly(n)$-time algorithm. (This is very similar to PvsNP, since you can think of a nondeterministic machine as one that always guesses the "correct" random choices in order to construct the object on the first try. So we might usually expect the search for better algorithms to be hard.) – usul Jun 3 '14 at 13:43
• the question should be linked to mathoverflow.net/questions/9218/… I think... – user39115 Jun 5 '14 at 10:08
I am a bit surprised that no one has so far mentioned expanders, for which the existence result was first established by the "probabilistic method" in a very simple way, whereas concrete examples of expanders are always a pretty difficult problem, see, for instance, Who first dubbed them "expander graphs"? or How to prove a random d-regular graph is an expander with prob >= 0.5?
• Yes, I though of this question when I was reading Yann Olivier's January 2005 introduction to group theory and when I read about expanders) – Olga Jun 10 '14 at 5:04
Shannon's proof that capacity-achieving code families exist for noisy channels is probabilistic in nature. Only for relatively simple channels (e.g., the binary erasure channel, or more recently the binary symmetric channel--and perhaps more generic discrete memoryless channels if I take some abstracts in the recent literature at face value) have explicit code families been given, viz. LDPC and polar codes, respectively.
Problem: given points $x_1, \ldots, x_n$ in $\mathbb{R}^d$ with $n << d$ and $0 < \epsilon < 1$, find a linear map $T \colon \mathbb{R}^d \to \mathbb{R}^k$, $k << d$, such that $$(1 - \epsilon) ||x_i - x_j|| \leq T(x_i) - T(x_j) \leq (1 + \epsilon) ||x_i - x_j||$$
Solution: for $k > C \frac{\log n}{\epsilon^2}$, a random $d \times k$ matrix has the desired property with high probability (where "random" and "high probability" can be made precise).
This is the Johnson-Lindenstrauss lemma - informally, it says that a small amount of high dimensional data can be projected down to a low dimensional space via a suitable random matrix without disturbing the Euclidean distances too badly. This is often used in practice to apply algorithms whose running time is exponential in the dimension of the input data to high dimensional data sets. Interestingly, it is very difficult to check that a given matrix has the Johnson-Lindenstrauss property even though a random matrix has the property with very high probability. I recall reading somewhere the description "finding hay in a haystack" for this problem.
I will interpret your question in this way: Are there any example of a theorem which proof uses a probability technique to prove the existence of some object where it is proven that the example exists but its explicit construction is impossible?
An example can be the theorem:
Every large even number can be expressed as the sum of a prime and the product of at most b primes.
Rényi proved this theorem using a generalisation of ''the large sieve'' method that can be seen as an application of a probabilist method:
Rényi, A. On the large sieve of Yu. V. Linnik. Compositio Mathematica, 8, 68-75 (1956).
Rényi, A. Un nouveau théorème concernant les fonctions indépendantes et ses applications à la théorie des nombres. J. Math, pures et appi. (9), 28, 137-149 (1949).
Rényi, A. Sur un théorème général de probabilité. Ann. Inst. Fourier, 1, 43-52 (1949).
• How is $b$ quantified here? – Tom Leinster Jun 3 '14 at 10:52
• It is some big constant, I do not have the original paper by Rényi, but I read that he did not determine it. – user39115 Jun 3 '14 at 11:02
• OK, so the theorem is: there exist $b$ and $N$ such that every even number greater than $N$ can be expressed as the sum of a prime and the product of at most $b$ primes? – Tom Leinster Jun 3 '14 at 13:51
• Yes. It seems that it was a trend in the sixties and seventies to improve those weak versions of the Goldbach s conjecture. – user39115 Jun 8 '14 at 18:00
• Thank you! And the fact that the explicit construction of the decomposition is impossible -- is it proven? Or we just do not know if it s true or not? – Olga Jun 10 '14 at 5:02
There are many such results ( methods) to be found in functional analysis, more precisely, the theory of Banach spaces. A spectacular example is Gluskin's estimate from below for the diameter of the class of $n$-dimensional Banach spaces under the Banach-Mazur distance which involved a probabilistic proof of the existence of a pair of Banach spaces which are far apart in a suitable sense ( MR 0609798).
Regarding your second question, there is indeed a constructive proof showing that there are sparse graphs with arbitrarily high chromatic number due to Lovász (unfortunately behind a paywall). However, you can read this very lively survey by Nesetril on the history of the problem. Indeed, the question of explicit constructions was already raised by Erdos in the beginning. The survey starts with the many constructions for the triangle-free case and proceeds to the current state-of-the-art, passing through many interesting topics (including Kneser graphs and expanders) along the way.
Let the $V_n$ be iid random variables with $P(V_n=\pm 1)=1/2$. Almost surely, the discrete Schrodinger operator $$(Hu)_n = u_{n+1} + u_{n-1} + V_n u_n$$ on $\ell^2(\mathbb Z)$ has dense pure point spectrum (aka Anderson localization). I don't think anyone can describe a concrete $\pm 1$ sequence for which this happens.
The best deterministic constructions of matrices with the restricted isometry property are much worse than random matrices. Matrices with the restricted isometry property are important in compressed sensing, and finding better deterministic matrices is a subject of active research.
Here the quality of the matrix is the number of columns for a matrix with the restricted isometry property and a fixed number of rows, which represents how much one can compress the data in a compressed sensing situation.
Somewhat surprisingly, Jean Bourgain's 1985 theorem about embedding finite metric spaces into Hilbert spaces was not mentioned yet. Since metrics have not mentioned at all in this thread, let me mention this example now.
The topic of representing prescribed, possibly strange finite metric spaces 'as best one can' by non-strange, traditional metric spaces is important e.g. in computational biology. Very briefly, I think the essence of all of this is
not to have to store a table of pairwise distances among any two of a large number of proteins, which after all would take space quadratic in the number of 'specimens', rather label/represent each protein by an element of a traditional metric space, and then calculate a distance the traditional way, from the traditional representatives, on an as-needed-basis.
For this to result in only a small error of approximation (of the 'true' distance w.r.t. some new-fangled 'similarity distance' between proteins), one needs to know how well such 'representations'/'low-distortion embeddings' can be done in principle.
Bourgain used the "probabilistic method" to prove a lower bound on how well this can be done.
The publication is Jean Bourgain: On lipschitz embedding of finite metric spaces in Hilbert space. Israel Journal of Mathematics March 1985, Volume 52, Issue 1–2, pp 46–52, and the abstract reads:
"It is shown than any $n$ point metric space is up to $\log n$ lipeomorphic to a subset of Hilbert space. We also exhibit an example of an $n$-point metric space which cannot be embedded in Hilbert space with distortion less than $(\log n)/(\log\log n)$, showing that the positive result is essentially best possible. The methods used are of probabilistic nature. For instance, to construct our example, we make use of random graphs."
Thinking about my own question, I had another question that came to my mind. Take any real number and write its binary representation. With probability $1$ the proportion of zeroes and ones is the same and is equal to $1/2$. We can say that for its 3-representation and the representation in any system the proportion of numbers used is equal with probability $1$. So almost any number has this property. | 2019-12-13T05:08:54 | {
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https://www.physicslog.com/cs-notes/power-method/ | # Power Method
## Basics
• A nonzero vector $\vec{x}$ is an eigenvector of a square matrix $\mathbf{A}$ if there exists a scalar $\lambda$, called an eigenvalue, such that $\mathbf{A}\vec{x} = \lambda\vec{x}$.
• Non-zero vectors in the eigenspace of the matrix $\mathbf{A}$ for the eigenvalue $\lambda$ are eigenvectors of $\mathbf{A}$
## Properties
• The power method is used to find a dominant (largest) eigenvalue (one with the largest absolute value), if one exists, and a corresponding eigenvector.
• It does not always guaranteed to converge if the given matrix is non-diagonalizable.
• Inverse power method (if convergent) calculates the eigenvalue with smallest absolute value.
• It is effective for a very large sparse matrix with appropriate implementation.
## Facts
• This method is known as Von Mises iteration.
• Especially suitable for sparse matrices
• It is used to calculate the Google page rank.
## Algorithm
1. To apply this method to a square matrix $\mathbf{A}$, start with an initial guess $\vec{u}_{0}$ for the eigenvector of the dominant eigenvalue
2. Multiply the most recently obtained vector on the left by $\mathbf{A}$ i.e. for $i \ge 1$, $(\vec{u}_{i})_{\text{un-normalize}} = \mathbf{A} \vec{u}_{i - 1}$
3. Normalize the result i.e. $\vec{u}_{i} = \frac{(\vec{u}_{i})_{\text{un-normalize}}}{|| \mathbf{A} \vec{u}_{i-1}||} = \frac{ \mathbf{A} \vec{u}_{i - 1}}{|| \mathbf{A} \vec{u}_{i-1}||}$
4. Repeat until consecutive vectors $\vec{u}_{i}$ are either identical or opposite.
5. If $\vec{u}_{k}$ denotes the last vector calculated in this process, then $\vec{u}_{k}$ is an approximate eigenvector of $\mathbf{A}$, and $|| \mathbf{A}\vec{u}_{k} ||$ is the absolute value of the dominant eigen vector for $\mathbf{A}$.
6. If $\vec{u}_{k}$ is an eigenvector of $\mathbf{A}$ then, its eigenvalue is given by Rayleigh quotient i.e. $\lambda = \frac{(\mathbf{A} \vec{u}_{k}). (\vec{u}_{k})^{\text{T}}}{\vec{u}_{k}. (\vec{u}_{k})^{\text{T}} }$
It is easy to prove point (6) as $\frac{(\mathbf{A} \vec{u}_{k}). (\vec{u}_{k})^{\text{T}}}{\vec{u}_{k}. (\vec{u}_{k})^{\text{T}} } = \frac{(\lambda \vec{u}_{k}). (\vec{u}_{k})^{\text{T}}}{\vec{u}_{k}. (\vec{u}_{k})^{\text{T}} } = \lambda$.
### Example
• Given a matrix $\mathbf{A}$ defined as
$$\mathbf{A} = \begin{pmatrix} 1 & 3 & 0 \\ 2 & 5 & 1 \\ -1 & 2 & 3 \end{pmatrix}.$$
Now, we need find the largest eigen value.
• Initial guess say
$$\vec{u}_{0} = \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}$$
• Let $k = 1$,
$$\vec{u}_{1} = \mathbf{A} \vec{u}_{0} = \begin{pmatrix} 1 & 3 & 0 \\ 2 & 5 & 1 \\ -1 & 2 & 3 \end{pmatrix}. \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \\ 4 \end{pmatrix}$$
So, we normalize $\vec{u}_{1}$ as
$$\vec{u}_{1} = \frac{\vec{u}_{1}}{|| \vec{u}_{1} ||} = \begin{pmatrix} 0.4082 \\ 0.8165 \\ 0.4082 \end{pmatrix}$$
where $|| \vec{u}_{1} || = \sqrt{4^{2} + 8^{2} + 4^{2}}$ i.e. an Euclidean norm.
• Similarly, $k = 2$,
$$\vec{u}_{2} = \mathbf{A} \vec{u}_{1} = \begin{pmatrix} 2.8557 \\ 5.3072 \\ 2.4495 \end{pmatrix}$$
So,
$$\vec{u}_{2} = \frac{\vec{u}_{2}}{|| \vec{u}_{2} ||} = \begin{pmatrix} 0.4392\\ 0.8157 \\ 0.3765 \end{pmatrix}$$
• For $k = 3$,
$$\vec{u}_{3} = \mathbf{A} \vec{u}_{2} = \begin{pmatrix} 2.8863 \\ 5.3334 \\ 2.3216 \end{pmatrix}$$
So,
$$\vec{u}_{3} = \frac{\vec{u}_{3}}{|| \vec{u}_{3} ||} = \begin{pmatrix} 0.4445\\ 0.8213 \\ 0.3575 \end{pmatrix}$$
• For $k = 4$,
$$\vec{u}_{4} = \mathbf{A} \vec{u}_{3} = \begin{pmatrix} 2.9085 \\ 5.3532 \\ 2.2708 \end{pmatrix}$$
So,
$$\vec{u}_{4} = \frac{\vec{u}_{4}}{|| \vec{u}_{4} ||} = \begin{pmatrix} 0.4473\\ 0.8233 \\ 0.3493 \end{pmatrix}$$
• For two digits accuracy of the eigen vector, we can stop the iteration when $k = 4$. But for higher accuracy, we need to run the iteration until it converges to desired digits of accuracy.
• Therefore, dominant eigenvalue is $\lambda = \frac{(\mathbf{A} \vec{u}_{4}). (\vec{u}_{4})^{\text{T}}}{\vec{u}_{4}. (\vec{u}_{4})^{\text{T}} } = 6.5036$.
### Implementation
#### Using Python
#!/usr/bin/env python3
import numpy as np
def power_iteration(A, num_simulations: int):
# Ideally choose a random vector
# To decrease the chance that our vector
# Is orthogonal to the eigenvector
b_k = np.random.rand(A.shape[1])
for _ in range(num_simulations):
# calculate the matrix-by-vector product Ab
b_k1 = np.dot(A, b_k)
# calculate the norm
b_k1_norm = np.linalg.norm(b_k1)
# re normalize the vector
b_k = b_k1 / b_k1_norm
return b_k
b_kn = power_iteration(np.array([[1, 3, 0], [2, 5, 1], [-1, 2, 3]]), 4)
print(b_kn)
## Theorems
• If $\mathbf{A}$ is diagonalizable and has dominant eigenvalue then, normalized power iteration sequence ($\mathbf{A}\vec{u}, \mathbf{A}^{2}\vec{u}, \mathbf{A}^{3}\vec{u}, \ldots$) converges to the dominant eigenvector. You can notice matrix $\mathbf{A}$ get power on it. This is why this method is called power iteration method or simply power method.
• If $\mathbf{A}$ is a square matrix with dominant eigenvalue $\lambda_{\text{max}}$ and $\vec{x}$ is any vector then, $|| \mathbf{A} x|| \leq |\lambda_{\text{max}} |~|| x ||$ where $|| \cdot ||$ is an Euclidean norm. This is very useful in estimates. This is why we are more interested in dominant eigenvalue. | 2021-07-25T08:43:44 | {
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https://math.stackexchange.com/questions/3836064/determine-convergence-of-the-sequence-x-0-1-x-n1-x-n-1-2-n1 | # Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$
I want to check if the following sequence converges:
$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$
I proved the sequence is increasing :
$$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$$
Now I should prove it is bounded above. let's write some terms of the equation:
\begin{align} x_0&=1 \\[2ex] x_1&=1\cdot\left(1+\dfrac{1}{2^1}\right)\\[2ex] x_2&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\\[2ex] x_3&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\cdot\left(1+ \dfrac{1}{2^3}\right)\\[2ex] \end{align} So, we can write:
$$x_{n+1}=\left(1+\frac{1}{2^1}\right)\cdot\left(1+ \frac{1}{2^2}\right)\cdots\left(1+\frac{1}{2^{n}}\right)\cdot\left(1+\frac{1}{2^{n+1}}\right)$$
Here, I'm not sure how to prove it is bounded above.
• Is it $x_{n+1}=x_n\left(1+\dfrac1{2^{n+1}}\right)$ or $x_{n+1}=x_n\left(1+\dfrac1{2^n+1}\right)$? Sep 22 '20 at 14:59
• It is the first one. thank you I fixed it. Sep 22 '20 at 15:01
• Please rollback the edit I made if you found anything in the question that convinces you that the question has changed from the original in terms of content. Sep 22 '20 at 15:54
• @sai-kartik it is ok, thanks for the edit Sep 22 '20 at 16:00
First show that $$x_n \leq \sqrt{2}^{n+1}$$ through mathematical induction. Then, $$0 \leq x_{n+1} - x_n = x_n \frac{1}{2^{n+1}} \leq \frac{1}{\sqrt{2}^{n+1}}$$. You can easily check that given sequence is Cauchy sequence.
• It is really hard that the idea comes to my mind. How you thought about proving $x_n \leq \sqrt{2}^{n+1}$ then it helps to compare with Cauchy sequence? Sep 22 '20 at 16:53
• Idea : rate of increase of given sequence is very small, and $\sqrt{2}^{n+1}$ is quickly increasing(in terms of exponential function)
– Han
Sep 22 '20 at 16:56
• Observe that $x_0 \leq \sqrt{2}, x_1 \leq \sqrt{2}^2, x_2 \leq \sqrt{2}^3$. And Suppose $x_n \leq \sqrt{2}^{n+1}$ for $n>1$. Then $x_{n+1} = x_n + x_n / (2^{n+1})\leq \sqrt{2}^{n+1} + \frac{1}{\sqrt{2}^{n+1}} \leq \sqrt{2}^{n+2}$
– Han
Sep 22 '20 at 17:02
• Use $\sqrt{2} = 1.4\cdots$ in last inequality.
– Han
Sep 22 '20 at 17:06
• There is no specific reason for proving $x_n \leq \sqrt{2}^{n+1}$, but I focused on expressing the difference($x_{n+1} - x_n = \frac{x_n}{2^{n+1}}$) between two consecutive sequences in a given equation as a geometric sequence.
– Han
Sep 22 '20 at 17:16
You have $$x_n = \prod_{k=1}^n \left(1+\frac{1}{2^k}\right) = e^{\sum_{k=1}^n\ln \left(1+\frac{1}{2^k}\right)} \leq e^{\sum_{k=1}^n\frac{1}{2^k}} \leq e^{\sum_{k=1}^\infty\frac{1}{2^k}} = e$$ where for the first inequality we used that $$\ln(1+x)\leq x$$ for all $$x>-1$$. This shows the sequence is bounded.
The Limit Exists
Cross-multiplication and comparison shows that $$\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}\le1+\frac1{2^k}\le\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}\tag1$$ Then telescoping products lead to $$\underbrace{\prod_{k=1}^\infty\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}}_2 \le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right) \le\underbrace{\overset{\substack{k=1\\\downarrow\\[6pt]\,}}{\frac32}\overbrace{\prod_{k=2}^\infty\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}}^2}_3\tag2$$ Therefore, $$x_n$$ is an increasing sequence that is bounded above by $$3$$, so $$\lim_{n\to\infty}x_n=\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)\tag3$$ exists.
Bounding The Limit
Using $$(1)$$ and we get $$\left(1+\frac1{2^n}\right)\prod_{k=1}^n\left(1+\frac1{2^k}\right) \le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right) \le\frac1{1-\frac1{2^n}}\prod_{k=1}^n\left(1+\frac1{2^k}\right)\tag4$$ The greater the $$n$$ used, the tighter the bounds in $$(4)$$.
The Limit
Using $$(4)$$ with $$n=30$$, we get that $$2.38423102903137172\color{#C00}{35}\le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)\le2.38423102903137172\color{#090}{55}\tag5$$ | 2021-10-18T11:19:13 | {
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https://math.stackexchange.com/questions/1991252/inclusion-exclusion-how-many-bit-strings-of-length-eight-do-not-contain-six-con | # Inclusion exclusion: How many bit strings of length eight do not contain six consecutive $0$'s?
Q. How many bit strings of length eight do not contain six consecutive $0$s?
I solved this problem with my hand. $$256 - ( 1 + 2 + 2 + 1 + 2 ) = 248$$ I calculated all possible events.
Is my answer right? And can this problem be solved by the inclusion-exclusion principle?
Let $A_1$ be the set of outcomes in which the first six digits of the bit string are equal to $0$; let $A_2$ be the set of outcomes in which the middle six digits of the bit string are equal to $0$; and let $A_3$ be the set of outcomes in which the last six digits of the bit string are equal to $0$. The Inclusion-Exclusion Principle tells us that the number of bit strings in which there are six consecutive zeros is $$|A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$ The number of elements in $A_1$ is four since there must be a zero in the first six positions there are two choices, $0$ or $1$, for each of the last two digits. By similar argument, $|A_2| = 4$ and $|A_3| = 4$.
The set $A_1 \cap A_2$ consists of those bit strings in which the first six and middle six digits of the bit string are zeros, so the first seven digits of the bit string are zeros. There are two such bit strings, depending on whether the last digit is a $0$ or $1$. By similar argument, $|A_2 \cap A_3| = 2$.
The set $A_1 \cap A_3$ consists of those bit strings in which the first six and last six digits of the bit string are zeros, so each of the eight digits of the bit string is a zero. There is only one such bit string.
The set $A_1 \cap A_2 \cap A_3$ consists of those bit strings in which the first six, middle six, and last six digits of the bit string are zeros, so each of the eight digits of the bit string is a zero. There is only one such bit string.
Hence, the number of bit strings that contain six consecutive zeros is \begin{align} |A_1| + |A_2| & + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|\\ & = 4 + 4 + 4 - 2 - 2 - 1 + 1\\ & = 8 \end{align} as you found. Since there are $2^8 = 256$ bit strings of length $8$, the number of bit strings of length $8$ that do not contain six consecutive zeros is $2^8 - 8 = 256 - 8 = 248$, again as you found.
It is good that you want to find the strings that do contain $6$ consecutive zeros, and then exclude them.
Consider the following:
[0 0 0 0 0 0] * *
* [0 0 0 0 0 0] *
* * [0 0 0 0 0 0]
Each of the stars can be a zero or one.
• Assume the first case, we have $4$ combinations (where each star is either $0$ or $1$).
• The second case, again can have four possibilities. However, you should exclude those we counted in the first case. They are all-zero pattern, and 0 0 0 0 0 0 0 1. So only $2$ distinct possibilities.
• The third case has only $2$ new possibilities, which are 0 1 0 0 0 0 0 0 and 1 1 0 0 0 0 0 0.
So the answer to your question is $2^8-(4+2+2)=248$
I think your answer is wrong but the idea is good. The number to subtract can indeed be found using the inclusion-exclusion principle, for 3 sets. Can you explain why you subtracted those numbers ?
EDIT: Apologies - I was mistaken. See N. F. Taussig's answer and comment.
• The answer proposed by the OP is correct, as you can check by applying the Inclusion-Exclusion Principle. Oct 30, 2016 at 10:52
• You are quite right. My mistake was to assume that $|A_1 \bigcap A_3|$ was 2, as for the other two 2-set intersections. Oct 30, 2016 at 11:47 | 2022-06-29T11:00:22 | {
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https://math.stackexchange.com/questions/752928/proving-gcd-left-fraca-gcd-a-b-fracb-gcd-a-b-right-1 | # Proving $\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$
How would you go about proving that $$\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$$
for any two integers $a$ and $b$?
Intuitively it is true because when you divide $a$ and $b$ by $\gcd(a,b)$ you cancel out any common factors between them resulting in them becoming coprime. However, how would you prove this rigorously and mathematically?
• By rigorous you mean finding $m,n\in \mathbb{Z}$ scuh that $**m+**n=1$? – user87543 Apr 14 '14 at 5:47
• – Bart Michels Feb 21 '15 at 16:32
Very simply it can be done like this: $\gcd(a,b)=d$.
Now we ask can: $\gcd(\frac{a}{d},\frac{b}{d})=e$ for $e>1$?
Well, this implies $e\mid\frac{a}{d},e\mid\frac{b}{d} \Rightarrow em=\frac{a}{d}, en=\frac{b}{d} \Rightarrow dem=a,den=b \Rightarrow de$ is a common divisor of $a,b$ which is greater than $d$, thus a contradiction as $d$ by definition was supposed as the $\gcd$. Hence, $e=1$.
• don't frame this as a contradiction. "hence e = 1". Much better conclusion. – djechlin Nov 30 '14 at 5:07
• Let's not argue semantics. And this post is very old and correct as well. – Mr.Fry Nov 30 '14 at 7:22
• This is all true. It's better not to needlessly say "we assume for the sake of contradiction..." and conclude with "a contradiction!" when none of that was necessary. The proofs are cleaner and avoids unnecessarily mental baggage tracking extra false hypotheses that are never invoked. – djechlin Nov 30 '14 at 14:16
Let $d=\gcd(a,b)$. Let $a=md$ and $b=nd$. If some $k\gt 1$ divides $m$ and $n$, then $kd$ divides $a$ and $kd$ divides $b$, contradicting the fact that $d$ is the greatest common divisor of $a$ and $b$.
• I apologize, I was posting the exact same solution at the time, so I will upvote this. – Mr.Fry Apr 14 '14 at 6:08
• No problem, yes, exactly the same idea. So if upvoting is appropriate for one, it is appropriate for the other. – André Nicolas Apr 14 '14 at 6:20
• Is it also possible to prove that $$\gcd \left(\frac{a}{\gcd (a,b)},b\right)=1$$ in the same manner? I let $$d=\gcd (a,b)$$ and then $$a=\text{md}$$ and $$b=\text{nd}$$ and then let $$\gcd (m,\text{nd})=e$$ but I am stuck on how to proceed. – 1110101001 Apr 14 '14 at 18:21
• No, because it is not true. Let $a=4$ and $b=2$. The gcd is $2$. Thus $\frac{a}{\gcd(a,b)}=2$, and this is not relatively prime to $2$. This is the simplest counterexample. There are many more. – André Nicolas Apr 14 '14 at 18:35
This is a special case of the GCD distributive law ($$4$$ proofs of it are below). Namely
$$\color{#c00}{\bf c} = (a,b) = (a/c,b/c)\color{#c00}{\bf \,c}\overset{\rm\large cancel\ \color{#c00}{\bf c}}\Longrightarrow 1 = (a/c,b/c)\qquad\qquad$$
Or, presented more concisely it is $$\ (a/c,b/c) = (a,b)/c = 1$$
Below are sketches of four proofs of the gcd Distributive Law $$\rm\:(ax,bx) = (a,b)x\:$$ using various approaches: Bezout's identity, the universal gcd property, unique factorization, and induction.
First we show that the gcd distributive law follows immediately from the fact that, by Bezout, the gcd may be specified by linear equations. Distributivity follows because such linear equations are preserved by scalings. Namely, for naturals $$\rm\:a,b,c,x \ne 0$$
$$\rm\qquad\qquad \phantom{ \iff }\ \ \ \:\! c = (a,b)$$
$$\rm\qquad\qquad \iff\ \: c\:\ |\ \:a,\:b\ \ \ \ \ \ \&\ \ \ \ c\ =\ na\: +\: kb,\ \ \$$ some $$\rm\:n,k\in \mathbb Z$$
$$\rm\qquad\qquad \iff\ cx\ |\ ax,bx\ \ \ \&\ \ \ cx = nax + kbx,\ \,$$ some $$\rm\:n,k\in \mathbb Z$$
$$\rm\qquad\qquad { \iff }\ \ cx = (ax,bx)$$
The reader familiar with ideals will note that these equivalences are captured more concisely in the distributive law for ideal multiplication $$\rm\:(a,b)(x) = (ax,bx),\:$$ when interpreted in a PID or Bezout domain, where the ideal $$\rm\:(a,b) = (c)\iff c = gcd(a,b)$$
Alternatively, more generally, in any integral domain $$\rm\:D\:$$ we have
Theorem $$\rm\ \ (a,b)\ =\ (ax,bx)/x\ \$$ if $$\rm\ (ax,bx)\$$ exists in $$\rm\:D.$$
Proof $$\rm\quad\: c\ |\ a,b \iff cx\ |\ ax,bx \iff cx\ |\ (ax,bx) \iff c\ |\ (ax,bx)/x\ \ \$$ QED
The above proof uses the universal definitions of GCD, LCM, which often served to simplify proofs, e.g. see this proof of the GCD * LCM law.
Alternatively, comparing powers of primes in unique factorizations, it reduces to the following $$\begin{eqnarray} \min(a+x,\,b+x) &\,=\,& \min(a,b) + x\\ \rm expt\ analog\ of\ \ \ \gcd(a \,* x,\,b \,* x)&=&\rm \gcd(a,b)\,*x\end{eqnarray}\qquad\qquad\ \$$
The proof is precisely the same as the prior proof, replacing gcd by min, and divides by $$\,\le,\,$$ and using the universal property of min instead of that of gcd, i.e.
$$\begin{eqnarray} {\rm employing}\quad\ c\le a,b&\iff& c\le \min(a,b)\\ \rm the\ analog\ of\quad\ c\ \, |\, \ a,b&\iff&\rm c\ \,|\,\ \gcd(a,b) \end{eqnarray}$$
Then the above proof translates as below, $$\$$ with $$\,\ m(x,y) := {\rm min}(x,y)$$
$$c \le a,b \!\iff\!c\!+\!x \le a\!+\!x,b\!+\!x\!$$ $$\!\iff\! c\!+\!x \le m(a\!+\!x,b\!+\!x)\!$$ $$\!\iff\!\! c \le m(a\!+\!x,b\!+\!x)\!-\!x$$
Theorem $$\ \$$ If $$\ a,b,x\$$ are positive naturals then $$\ (ax,bx) = (a,b)x$$
Proof $$\$$ By induction on $$\color{#0a0}{{\rm size}:= a\!+b}.\,$$ If $$\,a=b,\,$$ then it is true since then both sides $$= ax.\,$$ Else $$\,a\neq b;\,$$ wlog, by symmetry, $$\,a > b\,$$ so $$\,(ax,bx) = (ax\!-\!bx,bx) = \color{}{((a\!-\!b)x,bx)}\,$$ which has smaller $$\rm\color{#0a0}{size}$$ $$\,(a\!-\!b) + b = a < \color{#0a0}{a\!+b},\,$$ therefore $$\,((a\!-\!b)x,bx)\!\underset{\rm induct}=\! (a\!-\!b,b)x = (a,b)x$$.
• In your second proof how do you prove the mentioned theorem? Specifically where is it shown that $(a, b) = (ax,bx)/x$? – Michael Munta Sep 24 '19 at 20:57
• @Michael The proof shows $\, c\mid (a,b)\iff c\mid (ax,bx)/x =:d\,$ so $\, d\mid (a,b)\mid d\,$ so they are equal up to a unit factor (so they are equal in $\Bbb Z\,$ if normalized to be positive as usual). – Bill Dubuque Sep 24 '19 at 22:06
• I don't understand this. How do you conclude that $d$ divides $(a,b)$? – Michael Munta Sep 25 '19 at 6:36
• Is $c = (a,b)$ in the second proof as well? – Michael Munta Sep 25 '19 at 7:27
• @Michael We use $\,m = n\,$ if they have the same divisors: if for all $\,c\,$ we have $\,c\mid m\iff c\mid n\,$ then taking $\,c=m\,$ implies $\,m\mid n,\,$ and taking $\,c=n\,$ impies $\,n\mid m.\,$ Thus $\,|m| \le |n| \le |m|\,$ so $\,|m| = |n|,\,$ so $\,m=n\,$ being both $> 0\,$ (or $\,m$ & $n$ are associates in more general domains). – Bill Dubuque Sep 25 '19 at 12:49
A Bezout-centric approach. Let $d=\text{gcd}(a,b)$. Then, there are integers $m$ and $n$ such that $d=ma+nb$. But then $m$ and $n$ are also integers such that $$1=m\left(\frac{a}{d}\right)+n\left(\frac{b}{d}\right)$$ so $\text{gcd}(\frac{a}{d},\frac{b}{d})=1$.
Let $a,b$ have the following prime factorisations:
$$a= \prod_{n=1}^\infty p_n^{\alpha_n} ,b=\prod_{n=1}^\infty p_n^{\beta _n}.$$
(Here $(p_n)$ is the ascending sequence of prime numbers). We then have $$\gcd(a,b)=\prod_{n=1}^\infty p_n^{\min(\alpha_n,\beta_n)},$$ and consequently $$\frac{a}{\gcd(a,b)}=\prod_{n=1}^\infty p_n^{\alpha_n-\min(\alpha_n,\beta_n)},\frac{b}{\gcd(a,b)}=\prod_{n=1}^\infty p_n^{\beta_n-\min(\alpha_n,\beta_n)}.$$ Now ask yourself, could it be that one of the factors $p_n$ could have a strictly positive exponent, in both of the last products simultaneously?
• eek, unique prime factorization? so heavyweight. – djechlin Nov 30 '14 at 5:08
Every other answer is doing part of this, but since you asked for rigorous, here is rigorous:
Let $A$ be the multiset of prime factors of a, $B$ be the multiset of prime factors of $b$. The multiset of prime factors of $1$ is the empty set.
$$\text{GCD}\left(\frac{a}{\text{GCD}(a,b)}, \frac{b}{\text{GCD}(a,b)}\right) = 1$$
Standard conversion to multisets:
$$\bigg(A - (A\cap B)\bigg) \cap \bigg(B - (A \cap B)\bigg) = \{\}$$
Standard conversion to Boolean Algebra ($x \in A$ becomes $A$):
$$\bigg(A \land \lnot (A \land B)\bigg) \land \bigg(B \land \lnot (A \land B)\bigg) = \bot$$
Standard reduction (or you could use truth table): $$\bigg(A \land (\lnot A \lor \lnot B)\bigg) \land \bigg(B \land (\lnot A \lor \lnot B)\bigg) = \bot$$
$$\bigg(A \land \lnot B\bigg) \land \bigg(B \land \lnot A\bigg) = \bot$$ $$\bot = \bot$$ $$\top$$
Let $$\gcd(a,b)=d$$. Then $$\exists r,s \in \mathbb Z$$ s.t. $$d=ra+sb$$. Since $$\gcd(p,q)=1 \iff \exists m,n \in \mathbb Z$$ s.t. $$mp+nq=1$$, we have from $$1=r\frac{a}{d}+s\frac{b}{d}$$ that $$\gcd(\frac{a}{d},\frac{b}{d})=1$$.
Note: $$d=ra+sb \nrightarrow \gcd(a,b)=d$$, but $$mp+nq=1 \rightarrow \gcd(p,q)=1$$
Assume WLOG that $a, b \geq 1$. Let $m = \dfrac{a}{\gcd(a,b)}$, and $n = \dfrac{b}{\gcd(a,b)}$, and let $c = \gcd(m,n)$. Then $c \mid m$, and $c \mid n$. This means: $(c\cdot \gcd(a,b)) \mid a$, and $(c\cdot \gcd(a,b)) \mid b$. So $(c\cdot \gcd(a,b)) \mid \gcd(a,b)$. but $\gcd(a,b) \mid (c\cdot \gcd(a,b))$. Thus: $c\cdot \gcd(a,b) = \gcd(a,b)$, and this means $c = 1$.
Suppose $d \mid \dfrac a{\gcd(a,b)}$ and $d\mid \dfrac b{\gcd(a,b)}$.
Then $nd = \dfrac a {\gcd(a,b)}$ and $md\mid \dfrac b {\gcd(a,b)}$.
So $nd\cdot\gcd(a,b) = a$ and $md\cdot\gcd(a,b) = b$.
So $d\cdot\gcd(a,b)$ is a common divisor of $a$ and $b$.
Since $\gcd(a,b)$, is the greatest of those, we must have $d\not>1$. | 2021-03-05T02:07:55 | {
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http://yasermartinez.com/blog/posts/numpy-programming-challenge.html | # Numpy programming challenge
| Source
In Coursera's Computational Investing Prof. Tucker Balch, who was always looking for ways to make the course more interesting, proposed the following challenge:
Write the most succinct NumPy code possible to compute a 2D array that contains all "legal" allocations to 4 stocks:
• "Legal" means: The allocations are in 0.10 chunks, and the allocations sum to 1.00
• Only "pure" NumPy is allowed (no external libraries)
• Can you do it without a "for"?"
so we had to generate an array like this one:
[ [ 0.0, 0.0, 0.0, 1.0],
[ 0.0, 0.0, 0.1, 0.9],
[ 0.0, 0.0, 0.2, 0.8], #<---the sum(row) should be 1
[ 0.0, 0.0, 0.3, 0.7],
...
I love both challenges and Python so I immediately set about to try and solve it. Prof. Balch proposed a "brute force approach" as a starting point, because he "wanted folks to have something to beat :-).".
In [83]:
%%timeit
# solution by Tucker Balch (brute force approach)
import numpy as np
allocs = np.zeros((10000,4))
n = 0
for i in range(0,11):
for j in range(0,11):
for k in range(m
for l in range(0,11):
row = [i/10.0,j/10.0,k/10.0,l/10.0]
if sum(row)==1.0:
allocs[n,:]=row
n = n+1
allocs = allocs[0:n,:]
1 loops, best of 3: 284 ms per loop
As the first solutions quickly started showing up, I decided to collected them and benchmark them for comparison. In retrospective I might have twisted the original challenge a bit, as the goal changed from conciseness to performance, but everybody seemed to play along and I think in the end, it made the challenge more interesting.
My first solution was very concise but I wasn't playing by the rules (no numpy) and also it was almost as slow as the brute force approach:
In [12]:
%%timeit
allocations = list(ifilter(lambda x: abs(sum(x)-1)<1e-8, product(linspace(0, 1, 11), repeat=4)))
1 loops, best of 3: 282 ms per loop
After some further testing I found a quite concise and fast numpy solution I was really happy with:
In [70]:
%%timeit
from numpy import linspace, rollaxis, indices
x = linspace(0,1,11)[rollaxis(indices((11,) * 4), 0, 4+1).reshape(-1, 4)]
valid_allocations = x[abs(x.sum(axis=1)-1)<1e-8]
1000 loops, best of 3: 1.29 ms per loop
Being over 200 times faster than Prof. Tucker solution, only two lines long and entirely numpy based (numpy is blazing fast right?) I thought this would be quite difficult to beat. But shortly aftwerwards the following happened:
##### Pawel Kozak (recursive no numpy)¶
In [69]:
%%timeit
a=[]
def y(last, i=0, j=0, k=0, l=0):
if len(a)==0:
a.append([10,0, 0, 0])
else:
a.append([last[0]+i,last[1]+j,last[2]+k, last[3]+l])
last = a[-1][:]
if last[0]>0:
y(last, i=-1, j=1, k=0, l=0)
if last[1] ==0:
y(last, i=-1, j=0, k=1, l=0)
if last[2] ==0:
y(last, i=-1, j=0, k=0, l=1)
(y(a))
g = [[a[j][i]/10.0 for i in range(4)] for j in range(len(a))]
1000 loops, best of 3: 701 us per loop
A recursive pure Python solution almost 2 times faster than numpy! Certainly my numpy algorithm wasn't being optimal. This solution remained in the top spot for some hours until Sergio Antoy came up with a cryptic C-like solution:
##### Sergey Antoy (pure Python)¶
In [16]:
%%timeit
a = []
i,j,k = 0,1,2
while True:
# print [i,j-i-1,k-j-1,12-k]
a.append([i/10.,(j-i-1)/10.,(k-j-1)/10.,(12-k)/10.])
if i < j-1:
i += 1
elif j < k-1:
i,j = 0,j+1
elif k < 12:
i,j,k = 0,1,k+1
else:
break
1000 loops, best of 3: 213 µs per loop
Sergio explained later that the algorithm was based on the "The problem of the Dutch National Flag" featured by E. Dijkstra on A Discipline of Programming (Disclaimer: no affiliate link). Now this seemed to be the end of the competition. After all what can be faster than a textbook algorithm by the great Dijkstra? Well, a simple nested for loop:
##### Nick Poplas (pure Python)¶
In [18]:
%%timeit
allocations = [(0.1 * a, 0.1 * b, 0.1 * c, 0.1 * (10 - a - b - c)) for a in range(11) for b in range(11 - a) for c in range(11 - a - b)]
10000 loops, best of 3: 135 µs per loop
only 135 microseconds!! This is more than 2000 times faster than the initial solution! It seems that Python loops are not so slow after all. Also the simplicity of the solution was mind blowing.
At this point someone posted (as a joke) the fastest possible solution, i.e. directly assigning the pre-calculated result to a variable name,
In [13]:
%%timeit
allocations = [ [ 0.0, 0.0, 0.0, 1.0],
[ 0.0, 0.0, 0.1, 0.9],
[ 0.0, 0.0, 0.2, 0.8],
[ 0.0, 0.0, 0.3, 0.7],
# ...
[ 0.9, 0.1, 0.0, 0.0],
[ 1.0, 0.0, 0.0, 0.0]]
10000 loops, best of 3: 64 µs per loop
so now we knew how close we had gotten the theoretical limit. Even though I felt there would be little chance to improve such a simple solution I kept trying. I imagined that only numpy's C optimized loops could be able to beat Python loops, but how to formulate the problem in a column oriented approach? And then I found this:
##### Yaser Martinez (numpy)¶
In [20]:
%%timeit
#from numpy import indices, vstack
x1_3 = indices((11,11,11)).reshape(3,1331) # first three columns
x4 = 10 - x1_3.sum(axis=0) # 4th column
valid_idx = x4 > -1 # filter out negative values
allocations = 0.1*vstack((x1_3, x4)).T[valid_idx] # join columns, reshape and scale
10000 loops, best of 3: 113 µs per loop
Pure numpy power! With only 113 microseconds this was fastest solution and also the only one (from the top ranked) that is 100% compliant with all the rules, i.e: returns a numpy array without using for loops. Some time later I found that if one replaces range by xrange in Nick Poplas solution, avoiding the creation of intermediate lists, the solution also runs (coincidence?) in 113 microseconds. | 2017-07-22T02:47:42 | {
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https://themillionlivesproject.com/q7hreil/b9uqm.php?page=fibonacci-sequence-formula-0b45e8 | # fibonacci sequence formula
What do you notice happens to this ratio as n increases? Fibonacci initially came up with the sequence in order to model the population of rabbits. ratios seem to be converging to any particular number? This makes n1 the first number back after the new number. It prints this number to the console. We swap the value of n1 to be equal to n2. 2. In reality, rabbits do not breed this… Especially of interest is what occurs when here. Check your answer here. number from the sum of the previous two. There is also an explicit formula below. Fibonacci sequence is a sequence of numbers, where each number is the sum of the 2 previous numbers, except the first two numbers that are 0 and 1. Generalized Fibonacci sequence is defined by recurrence relation F pF qF k with k k k t 12 F a F b 01,2, The iterative approach depends on a while loop to calculate the next numbers in the sequence. This is the general form for the nth Fibonacci number. by F[n]) is F[n-1] + F[n-2]. The authors would like to thank Prof. Ayman Badawi for his fruitful suggestions. Now, consider the ratios found by F[n-1]/F[n], that is the reciprocals of Instead, it would be nice if a closed form formula for the sequence of numbers in the Fibonacci sequence existed. You can calculate the Fibonacci Sequence by starting with 0 and 1 and adding the previous two numbers, but Binet's Formula can be used to calculate directly any term of the sequence. We’ll look at two approaches you can use to implement the Fibonacci Sequence: iterative and recursive. As we move further in the sequence, the ratio approximates to 1.618 – the golden ratio – the reverse of which is 0.618 of 61.8%. It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre. ??? In this guide, we’re going to talk about how to code the Fibonacci Sequence in Python. 2. F n = F n − 1 + F n − 2, F_n = F_ {n-1} + F_ {n-2}, F n. . Graph the ratios and We can use this to derive the following simpler formula for the n-th Fibonacci number F (n): F (n) = round ( Phi n / √5 ) provided n ≥ 0. where the round function gives the nearest integer to its argument. Remember that the formula to find the nth term of the sequence (denoted by F [n]) is F [n-1] + F [n-2]. Each subsequent number can be found by adding up the two previous numbers. If we write $$3 (k + 1) = 3k + 3$$, then we get $$f_{3(k + 1)} = f_{3k + 3}$$. geometric series . see what they look like. A natural derivation of the Binet's Formula, the explicit equation for the Fibonacci Sequence. The Explicit Formula for Fibonacci Sequence First, let's write out the recursive formula: a n + 2 = a n + 1 + a n a_{n+2}=a_{n+1}+a_n a n + 2 = a n + 1 + a n where a 1 = 1 , a 2 = 1 a_{ 1 }=1,\quad a_2=1 a 1 = 1 , a 2 = 1 The sequence starts like this: It keeps going forever until you stop calculating new numbers. 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And skill level out as a whole number, exactly equal to F₀ = and...... 8/5 = 1.6 a researcher at Career Karma, publishing comprehensive reports on the bootcamp market and income agreements... Formula is an explicit formula used to train developers on algorithms and loops a particular is... For each unique type of recursive sequence present properties of Generalized Fibonacci sequence, you use... Linear recursion with constant coefficients and F₁ = 1 ), that is... 65 ) number to the shell, are the first variable tracks many. To F₀ = 0 and 1 is generated by adding the previous two three... The ratios of successive numbers Python like an expert expertise in Python does it take to become a stack! How long does it take to become a full stack web developer a series of numbers found F... Present properties of Generalized Fibonacci sequences F₂ = 1 and F₂ = 1 to become a full stack developer!: the first nine values in the sequence gives an overview of all... As 0 and F₁ = 1 as the sequence of interest is what occurs when look... Term ( 1 ) and 0 continues till infinity almost an integer reports on the bootcamp market and income agreements! The most famous sequences in mathematics we want to calculate the next number in the series to the next this! You suspect these ratios seem to be converging to and use themselves to solve problem... | 2022-05-18T23:46:58 | {
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https://scicomp.stackexchange.com/questions/5682/why-do-equi-spaced-points-behave-badly | # Why do equi-spaced points behave badly?
Experiment description:
In Lagrange interpolation, the exact equation is sampled at $N$ points (polynomial order $N - 1$) and it is interpolated at 101 points. Here $N$ is varied from 2 to 64. Each time $L_1$, $L_2$ and $L_\infty$ error plots are prepared. It is seen that, when the function is sampled at equi-spaced points, the error drops initially (it happens till $N$ is less than about 15 or so) and then the error goes up with further increase in $N$.
Whereas, if the initial sampling is done at Legendre-Gauss (LG) points (roots of Legendre polynomials), or Legendre-Gauss-Lobatto (LGL) points (roots of Lobatto polynomials), the error drops to machine level and doesn't increase when $N$ is further increased.
My questions are,
What exactly happens in the case of equi-spaced points?
Why does increase in polynomial order cause the error to rise after a certain point?
Does this also mean that if I use equi-spaced points for WENO / ENO reconstruction (using Lagrange polynomials), then in the smooth region, I would get errors? (well, these are only hypothetical questions (for my understanding), it is really not reasonable to reconstruct polynomial of the order of 15 or higher for WENO scheme)
Function approximated:
$f(x) = \cos(\frac{\pi}{2}~x)$, $x \in [-1, 1]$
$x$ divided into $N$ equispaced (and later LG) points. The function is interpolated at 101 points each time.
Results:
1. a) Equi-spaced points (interpolation for $N = 65$):
1. b) Equi-spaced points (error plot, log scale):
1. a) LG points (Interpolation for $N = 65$):
2. b) LG points (error plot, log scale):
The problem with equispaced points is that the interpolation error polynomial, i.e.
$$f(x) - P_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} \prod_{i=0}^n (x - x_i),\quad \xi\in[x_0,x_n]$$
behaves differently for different sets of nodes $x_i$. In the case of equispaced points, this polynomial blows up at the edges.
If you use Gauss-Legendre points, the error polynomial is significantly better behaved, i.e. it doesn't blow up at the edges. If you use Chebyshev nodes, this polynomial equioscillates and the interpolation error is minimal.
• There is quite a detailed explanation in the book of John P. Boyd Chebyshev and Fourier Spectral Methods, where Pedro's interpolation error polynomial is also nicly explained (Chapter 4.2 Page 85). – Bort Mar 26 '13 at 16:28
• Thank you. Also the Lebesgue constant for the above mentioned choices behaves differently. For equi-spaced points, the Lebesgue constant increases exponentially whereas for LG, LGL, Chebyshev it kind of saturates with increasing n. en.wikipedia.org/wiki/Lebesgue_constant_(interpolation) , ami.ektf.hu/uploads/papers/finalpdf/AMI_33_from109to123.pdf, but question regarding numerical implementation still remains... – Subodh Mar 26 '13 at 17:17
• Sorry, I don't know much about ENO/WENO. But I won't expect problems in the smooth region for low order interpolations, allthough quadrature nodes are definitely the better choice for aparent reasons. – Bort Mar 26 '13 at 18:40
This is a really interesting question, and there are a lot of possible explanations. If we are attempting to use a polynomial interpolation, then note that polynomial satisfy the following annoying inequality
Given a polynomial $P$ of degree not exceeding $N$ we have
$$|P^{\prime}(x)| \leq \frac{N}{\sqrt{1-x^2}}\max _x |P(x) |$$
for every $x \in (-1,1)$. This is known as Bernstein's inequality, note the singularity in this inequality. This can be bounded by the Markov inequality
$$\max _x |P^{\prime} (x) | \leq N^2 \max _x |P(x) |$$
and note that this is sharp in the sense that Chebysehv polynomials make this an equation. So in other words we have the following combined bound.
$$|P^{\prime}(x)| \leq \min \left(\frac{N}{\sqrt{1-x^2}},N^2\right)\max _x |P(x) |$$
What this means: Gradients of polynomials grow linearly in their order everywhere except in small neighborhoods of the interval boundaries. At the boundaries they grow more like $N^2$. It is no accident that stable interpolation nodes all have a $1/{N^2}$ clustering near boundaries. The clustering is necessary to control the gradients of the basis, whereas near the midpoint one can be a bit more relaxed.
It turns out however that this is not necessarily a polynomial phenomena, I suggest the following paper:
http://math.la.asu.edu/~platte/pub/prevised.pdf
It says loosely: If you have the same approximation power of the polynomial basis, then you can't use equally spaced points in a stable way.
It is not the equally spaced points that are the problem. It's the global support of the basis functions along with equally spaced points that is the problem. A perfectly well conditioned interpolant using equally spaced points is described in Kress's Numerical Analysis, using cubic-b spline basis functions of compact support.
• sure, but then your interpolant won't be globally smooth (only $C^2$ for your example) – GoHokies Mar 18 '18 at 19:39
• @GoHokies: The compactly supported splines can be made as smooth as desired by iterative convolution. What is the use case for $C^{\infty}$ interpolation? – user14717 Mar 19 '18 at 1:32
• fair point. $C^2$ ("position-velocity-acceleration") is enough for most applications. you may want $C^4$ for some boundary-value problems, but can't think of any common use case above that. – GoHokies Mar 19 '18 at 16:20
What exactly happens in the case of equi-spaced points?
Why does increase in polynomial order cause the error to rise after a certain point?
This is similar to the Runge's phenomenon where, with equi-spaced nodes, the interpolation error goes to infinity with the increase of the polynomial degree, i.e. the number of points.
One of the roots of this problem can be found in the Lebesgue's constant as noted by @Subodh's comment to @Pedro answer. This constant relates the interpolation with the best approximation.
Some notations
We have a function $f \in C([a,b])$ to interpolate over the nodes $x_k$. In the Lagrange interpolation are defined the Lagrange polynomials:
$$L_k(x) = \prod_{i=0, i\neq j}^{n}\frac{x-x_i}{x_k-x_i}$$
with this is defined the interpolation polynomial $p_n \in P_n$ over the couples $(x_k, f(x_k))$ for light notation $(x_k, f_k)$
$$p_n(x) = \sum_{k=0}^n f_kL_k(x)$$
Now consider a perturbation over the data, this can be for example for rounding, so we have got $\tilde{f}_k$. With this the new polynomial $\tilde{p}_n$ is:
$$\tilde{p}_n(x) = \sum_{k=0}^n \tilde{f}_k L_k(x)$$
The error estimates are:
$$p_n(x) - \tilde{p}_n(x) = \sum_{k=0}^n (f_k - \tilde{f}_k) L_k(x)$$
$$| p_n(x) - \tilde{p}_n(x) | \leq \sum_{k=0}^n |f_k - \tilde{f}_k| |L_k(x)| \leq \left ( \max_k |f_k - \tilde{f}_k| \right) \sum_{k=0}^n |L_k(x)|$$
Now it is possible define the Lebesgue's constant $\Lambda_n$ as:
$$\Lambda_n = \max_{x \in [a,b]} \sum_{k=0}^n |L_k(x)|$$
With this the final estimates is:
$$|| p_n - \tilde{p}_n ||_{\infty} \leq \left ( \max_k |f_k - \tilde{f}_k| \right) \Lambda_n$$
(marginal note, we look only $\infty$ norm also because we are over a space of finite measure so $L^{\infty} \subseteq \dots \subseteq L^1$)
From the above calculation we have got that $\Lambda_n$ is:
• independent from the date:
• depends only from the nodes distribution;
• an indicator of stability (the smaller it is, the better it is).
It is also the norm of the interpolation operator respect the $|| \cdot||_\infty$ norm.
Withe the follow theorem we con have got an estimate of the interpolation error with the Lebesgue's constant:
Let $f$ and $p_n$ as above we have $$|| f - p_n ||_{\infty} \leq (1 + \Lambda_n) d_n(f)$$ where $$d_n(f) = \inf_{q_n \in P_n} || f - q_n ||_{\infty}$$ is the error by the best uniform approximation polynomial
I.e. if $\Lambda_n$ is small the error of the interpolation is not far from the error of the best uniform approximation and the theorem compares the interpolation error with the smallest possible error which is the error of best uniform approximation.
For this the behavior of the interpolation depends by the nodes distribution. There is a lower bounds about $\Lambda_n$ that given a node distribution exist a constant $c$ such that: $$\Lambda_n \geq \frac{2}{\pi} \log(n) - c$$ so the constant grows, but how it grow is importan.
For equi-spaced nodes $$\Lambda_n \approx \frac{2^{n+1}}{en \log(n)}$$ I omitted some details, but we see that the grow is exponential.
For Chebyshev nodes $$\Lambda_n \leq \frac{2}{\pi} \log(n) + 4$$ also here I omitted some details, there are more accurate and complicate estimate. See [1] for more details. Note that nodes of Chebyshev family have got logarithmic grow and from the previous estimates is near the best you can obtain.
For other nodes distributions see for example table 1 of this article.
There are a lot of reference on book about interpolation. On-line these slides are nice as resume.
Also this open article ([1])
A Numerical Seven Grids Interpolation Comparison of for polynomial on the Interval for various comparisons.
It's good to be aware of Floater-Hormann interpolants when you have to (or want to) work with equidistant points $\{x_i\}_{i=1\ldots n}$.
Given the integer $d$ with $0 \le d \le n$, let $p_i$ be the polynomial interpolant of $\{ x_i, \ldots x_{i+d} \}$. Then the FH interpolant of a function $f$ at $\{x_i\}_{i=1\ldots n}$ has the form
$$r_n(x) := \frac{\sum_{i=0}^{n-d} \lambda_i(x) \, p_i(x)}{\sum_{i=0}^{n-d} \lambda_i(x)}$$
with the "blending functions"
$$\lambda_i(x) = \frac{(-1)^i}{(x-x_i) \ldots (x - x_{i+d})}$$
Some properties of these interpolants:
• they are barycentric rational interpolants with no real poles;
• achieve arbitrary approximation orders ${\cal O}(h^{d+1})$ for $f \in C^{d+2}[a,b]$, regardless of the distribution of points;
• are somewhat similar to splines, in that they blend (local) polynomial interpolants $p_0, \ldots p_{n-d}$ with the $\lambda$'s acting as the blending functions;
• they reproduce polynomials of degree at most $d$ (or $d+1$ if $n-d$ is odd);
• can be written in barycentric form (see section 4 of Floater and Hormann's paper).
Caveat emptor: As expected (see the paper referenced by @Reid.Atcheson), increasing $d$ quickly degrades the conditioning of the approximation process.
There is some fairly recent work done by Klein to alleviate this problem. He modified the original Floater-Hormann approach by adding $2 d$ new data values corresponding to points outside the original interpolation interval $[a,b]$ constructed from a smooth extension of $f$ outside $[a,b]$ using only the given data $f_0, \ldots f_n$. This "global" data set is then interpolated by a new FH rational function $r_{n+2d}$ and evaluated only inside $[a,b]$.
The details are nicely laid out in Klein's paper (linked below), where it is shown that these extended rational interpolants have Lebesgue constants that grow logarithmically with $n$ and $d$ (whereas for the original FH scheme, said growth is exponential in $d$, see Bos et al.).
The Chebfun library uses FH interpolants when building chebfuns out of equispaced data, as explained here.
References:
M. S. Floater and K. Hormann, Barycentric rational interpolation with no poles and high rates of approximation, Numerische Mathematik 107 (2007).
G. Klein, An Extension of the Floater–Hormann Family of Barycentric Rational Interpolants, Mathematics of Computation, 82 (2011) - preprint
L. Bos, S. De Marchi, K. Hormann, and G. Klein, On the Lebesgue constant of barycentric rational interpolation at equidistant nodes, Numer. Math. 121 (2012) | 2019-10-14T23:55:28 | {
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https://math.stackexchange.com/questions/2122556/subsets-of-mathbbr | # Subsets of $\mathbb{R}$
Give examples of subsets of $\mathbb{R}$ that are:
a) Infinite, but not connected
My attempt: $(-\infty,1)\cup (2,\infty)$, because it can be represented as the union of two open subsets, and it is infinite
b) Bounded and countable
My attempt: $\mathbb{Z}\cap[0,10]$ because it is bounded by $0$ and $10$, and contains 10 elements, so it must be countable
c) Bounded and uncountable
My attempt: [0,10] because an interval of $\mathbb{R}$ is uncountable (right?)
d) Closed but not compact
My attempt: All of $\mathbb{R}$? We're looking for something closed but not bounded to find something not compact, correct?
e) Dense but not complete
My attempt: $\mathbb{Q}$ because the closure of $\mathbb{Q}$ is equal to $\mathbb{R}$, but it does not contain all its limits, e.g. $\sqrt{2}$.
Any input appreciated!
• a) Why are you using $\infty$? Isn't the set $(1,2)$ already infinite, even uncountably infinite? – Dietrich Burde Jan 31 '17 at 15:01
• b) Some authors use "countable" to mean countably infinite, so check your source. In any case "bounded and countably infinite" subsets of $\mathbb{R}$ do exist and are a bit more interesting than merely finite subsets. – hardmath Jan 31 '17 at 15:09
• @Hardmath I wanted to say, without $\infty$ in both parts, so $(1,2)\cup (3,4)$. – Dietrich Burde Jan 31 '17 at 15:17
Everything looks fine, except I think that in problem $b)$ they want a set that can be bijected with $\mathbb N$.
I suggest $\mathbb Q \cap [0,1]$
Under further inspection I am sure, countable sets need to have the exact same cardinality as $\mathbb N$, the term used for a countable or finite set is denumerable or at most countable.
• Some people define countable to include finite, some require that it is infinite. This works for countably infinite. – Ross Millikan Jan 31 '17 at 15:17
In (a) you should mention that the set is the union of two open disjoint sets. This is crucial: also $(0,2)\cup(1,3)$ is the union of two open sets, but it's connected.
In (b) you probably should show an infinite countable and bounded set (but this depends on what you mean by countable): the numbers of the form $1/n$ (for $n$ a positive integer) make for an easy example; if you also add $0$, you get an example of a compact countable set.
(c) and (d) are fine. For (c) you might consider $(-\pi/2,\pi/2)$, instead, so it's easier to exhibit a bijection with $\mathbb{R}$, if requested: the tangent function.
Also (e) is fine, but you just have to mention $\mathbb{Q}$ is dense. Since it is not the whole of $\mathbb{R}$ it cannot be complete (a complete subset must be closed). | 2019-07-16T01:00:16 | {
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https://math.stackexchange.com/questions/981684/distance-function-is-in-fact-a-metric | # Distance function is in fact a metric
I know I should be able to show this, but for some reason I am having trouble. I need to show that $$d(x,y) = \frac{|x-y|}{1+|x-y|}$$ is a metric on $\Bbb R$ where $|*|$ is the absolute value metric. I am getting confused trying to show that the triangle inequality holds for this function. My friend also said that he proved that this distance function defines a metric even if you replace $|*|$ with any other metric. So I'd like to try and show both, but I cannot even get the specific case down first. Please help.
• Look at the function $t \mapsto \frac{t}{1+t}$. One of its properties is important to see that $d$ is a metric. Can you guess which? Oct 19 '14 at 23:32
• You'll never use the fact that $|x-y|$ is the expression in question, just that it satisfies the triangle inequality. Replace your expression by $$\bar{d}(x,y) = \frac{d(x,y)}{1 + d(x,y)}$$ where $d$ is any given metric. You gotta prove that $\bar{d}$ is a metric. You'll have to look at the function $1/(1+t)$ like everyone is saying. Just check that it is increasing using the definition instead of calculus, then. Oct 19 '14 at 23:47
• @MatthewLevy: Many times, working the abstract case is easier than the specific case because a lot of unnecessary details that would distract you have been removed and it is easier to focus to what matters. Oct 20 '14 at 0:25
Hint:
Put $f(t) = \frac{t}{1+t}$. Verify yourself that
$$f'(t) = \frac{1}{(1+t)^2 }$$
Hence, $f'(t) \geq 0$ for all $t$. In particular $f$ is an increasing function. In other words, we have
$$|x+y| \leq |x| + |y| \implies f(|x+y|) \leq f(|x|+|y|) \implies .....$$
• I cannot use any calculus, it was not proven yet. Oct 19 '14 at 23:40
• @Horacio, $f$ is increasing is not good enough to get $f(|x|+|y|)\leq f(|x|)+f(|y|)$... did you have something else in mind? Oct 19 '14 at 23:41
• @MatthewLevy If you write it in the form $f(t) = 1 - \frac{1}{1+t}$, you can prove the monotonicity without calculus. Oct 19 '14 at 23:47
Suppose that $d(\cdot,\cdot)$ is a arbitrary metric and $\bar{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}$, then we prove the triangle inequality for $\bar{d}(\cdot,\cdot)$ as following:
$\bar{d}(x,y)+\bar{d}(y,z)=\frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)}\geq \frac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}\geq\frac{d(x,z)}{1+d(x,z)}=\bar{d}(x,z)$
• I dont see why the last inequality is true, without using calculus.
– aram
Oct 25 '14 at 22:24
• Closer to 1/1? How would you prove that is increasing? Oct 27 '14 at 22:52
• @Aram: Note that $d(x,y)+d(y,z)\geq d(x,z)$ since $d(\cdot,\cdot)$ is a metric. Oct 28 '14 at 2:31
• @MatthewLevy: The function $f(x)=\frac{x}{1+x}$ is an increasing function in $[0,+\infty)$. Oct 28 '14 at 2:33
• @Aram: Well, $A\geq B\geq0\Rightarrow 1+A\geq 1+B\geq 1\Rightarrow\frac{1}{1+A}\leq\frac{1}{1+B}\Rightarrow \frac{A}{1+A}=1-\frac{1}{1+A}\geq1-\frac{1}{1+B}=\frac{B}{1+B}$. Oct 28 '14 at 10:12
Let $f(x)=x/(1+x)$, so $f$ is continuous on $[0,\infty)$, maps $[0,\infty)$ into itself, is non-decreasing since$f(x)=1-1/(x+1)$, $f(0)=0$ and subadditive, since $f''=-2/(1+x)^3<0$ for $x>0$.
It's easy to show that if a function satisfies the above properties, for any metric $d$ then $f\circ d$ is a metric.
To verify that $f''<0$ give us subbaditivity. $f'$ is decreasing so we can use the following result click | 2021-10-15T19:37:15 | {
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https://math.stackexchange.com/questions/2466261/sum-limits-n-1-infty-log1a-n-converges-absolutely-iff-sum-limits-n | # $\sum\limits_{n=1}^\infty \log(1+a_n)$ converges absolutely $\iff\sum\limits_{n=1}^\infty a_n$ converges absolutely.
$$\sum\limits_{n=1}^\infty \log(1+a_n) \text{ converges absolutely} \Leftrightarrow \sum_{n=1}^\infty a_n \text{ converges absolutely}.$$
How to prove this,
Suppose $$\sum_{n=1}^\infty a_n \text{ converges absolutely}.$$ Let $u_{n}=a_{n}$ and $v_{n}=\log(1+a_n)$, then $$\lim_{n\to\infty} \frac{u_{n}}{v_{n}}=1>0 \implies\sum_{n=1}^\infty \log(1+ a_n) \text{ converges absolutely}.$$ How to prove the converse part?
• This is an useful inequality jstor.org/stable/3615890?seq=1#page_scan_tab_contents – rtybase Oct 10 '17 at 17:10
• But, interestingly, I believe it is false if you omit "absolutely". – GEdgar Oct 10 '17 at 17:19
• if $$\lim_{n\to\infty } \log(1+a_n) \not=0 \Longleftrightarrow\lim_{n\to\infty }a_n \not = 0$$ then both series diverge. Assume that $a_n \to 0$, $\lim_{n\to\infty }\frac{|\log(1+a_n)|}{|a_n|} =\left|\lim_{h\to 0}\frac{\log(1+h)}{h}\right| = 1$ There is $n_0$ such that for $n>n_0$ $\left|\frac{|\log(1+a_n)|}{|a_n|} -1\right|<1/2 \Longleftrightarrow \frac12 |a_n|< |\log(1+a_n)|<\frac32|a_n|$ – Guy Fsone Nov 27 '17 at 15:41
Hint: From the definition of $\ln'(1),$ we have
$$\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1.$$
Thus there is $a>0$ such that
$$\frac{1}{2}\le \left|\frac{\ln (1+u)}{u}\right| \le \frac{3}{2}$$
for $u\in (-a,a),u\ne0.$
The limit comparison test says that if you have two sequences $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ such that $\lim_{n\to \infty}\frac{a_n}{b_n}=c$ with $0<c<\infty$. Then $\sum_{n=1}^\infty a_n<\infty$ if and only if $\sum_{n=1}^\infty b_n<\infty$.
So we have to prove that $\lim_{n\to \infty}\frac{\vert ln(1+x_n)\vert}{\vert x_n\vert}=c$ with $0<c<\infty$. To do this we observe that $\lim_{x\to 0}\frac{\vert ln(1+x)\vert}{\vert x\vert}=1$ by the L'Hopital Rule. As in any of the two cases ($\sum_{n=1}^\infty x_n$ converges absolutely or $\sum_{n=1}^\infty ln(1+x_n)$ converges absolutely) we will have that $\lim_{n\rightarrow \infty}x_n=0$, then $\lim_{n\to\infty}\frac{\vert ln(1+x_n)\vert}{\vert x_n\vert}=1$.
• we have to prove what ? – Guy Fsone Oct 10 '17 at 17:11
• That's not correct, since we don't know that $x_n \to 0$. – Adayah Oct 10 '17 at 17:17
• @Adayah sure we conclude that, since otherwise the sum $\sum_{n=1}^k\ln(1+x_n)$ cannot converge. This and the answer of zhw. should be enough for the converse statement! – user190080 Oct 11 '17 at 10:47
[This is essentially from @Guy Fsone's answer. However, it is rewritten in a significantly different way.]
First of all, observe that (by continuity of the exponential function and the logarithm) $$\lim_{n\to\infty } \log(1+a_n) =0 \Leftrightarrow \lim_{n\to\infty }a_n = 0.$$
• If $\lim_{n\to\infty}a_n=0$ is not true, (note carefully that this assumption is fundamentally different from "$\lim_{n\to\infty}a_n\neq 0$"), then both $\sum_{n=1}^\infty |\log(1+a_n)|$ and $\sum_{n=1}^\infty |a_n|$ diverge.
• Assuming that $a_n \to 0$ as $n\to\infty$, we have $$\lim_{n\to\infty }\frac{|\log(1+a_n)|}{|a_n|} =\left|\lim_{h\to 0}\frac{\log(1+h)}{h}\right| = 1\tag{1}$$ where we use the fact that the derivative of $x\mapsto \log x$ at $x=1$ is $1$ and the continuity of the absolute value function. Without loss of generality, we assume here that $|a_n|>0$ for all $n$.
It follows from (1) that there exists $n_0$ such that for $n>n_0$
$$\left|\frac{|\log(1+a_n)|}{|a_n|} -1\right|<1/2$$ which implies that $$\frac12 |a_n|< |\log(1+a_n)|<\frac32|a_n|~~~\forall ~~~n>n_0$$ and thus $$\frac12 \sum_{n>n_0}|a_n|< \sum_{n>n_0}|\log(1+a_n)|<\frac32 \sum_{n>n_0}|a_n|.$$
This estimate completes the proof.
• @GuyFsone: let me address your question one by one. Please be patient. (1)"I would like to know what is new here." As it was pointed out by a comment in meta, the logical structure of your original answer is heavily muddled. Since you have been keeping asking a hard evidence for what is wrong for your answer, let me explain in detail (in the next comment box). Again, please be patient before I finish my typing. – Jack Oct 22 '17 at 15:03
• @GuyFsone: Let $A$ be the statement that $\sum_{n=1}^\infty\log(1+a_n)$ and $\sum_{n=1}^\infty a_n$ converge absolutely. Let $B$ be $\lim_{n\to\infty}\log(1+a_n)=0$ and let $C$ be $\lim_{n\to\infty} a_n=0$. The first sentence of your original answer says: $A\Rightarrow (B\Leftrightarrow C)$. Note that this a true statement, but does not contribute anything to your whole argument since $(B\Leftrightarrow C)$ is true regardless $A$ is true or not. (cont.) – Jack Oct 22 '17 at 15:14
• @GuyFsone: (cont.) What you really want to say is $A\Rightarrow (B\hbox{ and } C)$. This is not a word playing game. This is a serious logic issue: "$A\Rightarrow (B\Leftrightarrow C)$" and "$A\Rightarrow (B\text{ and } C)$" are fundamentally different. (I'm still typing. Be patient.) – Jack Oct 22 '17 at 15:14
• @GuyFsone I have already explained in detail what is wrong with your writing in your answer, which has serious logic problems. Your several comments show that you don't understand what I'm talking about at all and it seems that you don't want to. Without a commitment to mathematical logic, there is no way going on a meaningful discussion. If you want to insist that you are writing a correct thing and keep yelling how wrong others are, it is certainly your right. I will stop here. – Jack Oct 22 '17 at 15:33
• @Jack: While most of what you say is correct, I have an alternative explanation for what is happening: Guy Fsone is not a native speaker of English, and this is painfully seen in his awkward and sometimes seemingly illogical formulations. What I'm saying is that the same answers formulated in his own native language would probably be correct. – Alex M. Oct 23 '17 at 9:01
My way: $$\sum_{n=1}^\infty \log(a_n+1) = \log\prod_{n=1}^\infty (a_n+1)$$ since the series converges absolutely so does the product. Hence $a_n +1 \rightarrow 1$ and so $a_n \rightarrow 0$. Now the nontrivial step: $a_n \rightarrow 0$, and in proximity of $0$ you have $\log(1+x) = x + o(x^2)$ so dividing by $x$ for any $\epsilon > 0$ exists $n$ such that $$\frac{|\log(a_n+1)|}{|a_n|} \leq 1+ \epsilon$$ or in other words $$(1-\epsilon)|a_n| \leq |\log(a_n+1)| \leq (1+\epsilon)|a_n|$$ Apply the sum to the last inequality and you'll prove the proposition both ways. | 2019-06-26T21:59:18 | {
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http://mathhelpforum.com/calculus/211251-taylor-series-cos-2x-1-x.html | Math Help - Taylor series: (cos(2x)-1)/x²
1. Taylor series: (cos(2x)-1)/x²
How come we can compute the Taylor series for the function below about x = 0, despite the fact that it is not defined at x = 0?
$\frac {\cos (2x) -1 }{x^2}$
2. Re: Taylor series: (cos(2x)-1)/x²
How come we can compute the Taylor series for the function below about x = 0, despite the fact that it is not defined at x = 0?
$\frac {\cos (2x) -1 }{x^2}$
We know that $C(t) = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k t^{2k} }}{{k!}}} = 1 - t^2 + \frac{{t^4 }}{{2!}} - \frac{{t^6 }}{{3!}} + \cdots$.
Can you find the series for $\frac{C(2x)-1}{x^2}~?$
3. Re: Taylor series: (cos(2x)-1)/x²
Originally Posted by Plato
We know that $C(t) = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k t^{2k} }}{{k!}}} = 1 - t^2 + \frac{{t^4 }}{{2!}} - \frac{{t^6 }}{{3!}} + \cdots$.
Can you find the series for $\frac{C(2x)-1}{x^2}~?$
I am sorry, I am not catching your drift. "We know that <series here>", I don't recognize that series.
4. Re: Taylor series: (cos(2x)-1)/x²
I am sorry, I am not catching your drift. "We know that <series here>", I don't recognize that series.
That is the standard series for $\cos(t)$.
That is, $C(t)=\cos(t)~.$
5. Re: Taylor series: (cos(2x)-1)/x²
The principle behind your question is that the discontinuity at x=0 is a removable one. This means that it is defined everywhere else and that we can "fill in" the value at x=0 with a value that it is "supposed to" have, and then it will be continuous. For example, consider the simple example:
$f(x) = \frac{x^2}{x}$
The function is not defined at x = 0 yet clearly everywhere else it is equal to just x, so the taylor series is x.
A less trivial example is:
$f(x) = \frac{\sin(x)}{x}$
The function is not defined at x=0. However, the taylor series can be obtained by using the series for sin(x), and dividing everything by x.
$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} ...$
$\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} ...$
Your example is fundamentally the same. Can you get the series for cos(x) and then perform the necessary steps?
6. Re: Taylor series: (cos(2x)-1)/x²
Originally Posted by Plato
We know that $C(t) = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k t^{2k} }}{{k!}}} = 1 - t^2 + \frac{{t^4 }}{{2!}} - \frac{{t^6 }}{{3!}} + \cdots$]
I thought it was $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} ....$ ?
7. Re: Taylor series: (cos(2x)-1)/x²
Originally Posted by SworD
I thought it was $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} ....$ ?
You are correct I left a 2 out of the denominator:
$C(t) = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k t^{2k} }}{{(2k)!}}} = 1 - \frac{t^2}{2!} + \frac{{t^4 }}{{4!}} - \frac{{t^6 }}{{6!}} + \cdots$
8. Re: Taylor series: (cos(2x)-1)/x²
Originally Posted by SworD
The principle behind your question is that the discontinuity at x=0 is a removable one. This means that it is defined everywhere else and that we can "fill in" the value at x=0 with a value that it is "supposed to" have, and then it will be continuous. For example, consider the simple example:
$f(x) = \frac{x^2}{x}$
The function is not defined at x = 0 yet clearly everywhere else it is equal to just x, so the taylor series is x.
A less trivial example is:
$f(x) = \frac{\sin(x)}{x}$
The function is not defined at x=0. However, the taylor series can be obtained by using the series for sin(x), and dividing everything by x.
$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} ...$
$\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} ...$
Your example is fundamentally the same. Can you get the series for cos(x) and then perform the necessary steps?
Thank for that expository answer. Please note that I am not having trouble finding the Taylor series, but rather understanding the rationale behind doing it in spite of the function not being defined at x = 0.
Your answer was illuminating indeed, but I wonder why the discontinuity would be a removable one. We are certainly not working with limits (or are we?).
9. Re: Taylor series: (cos(2x)-1)/x²
The original example is the same in principle as finding the taylor series of $f(x) = \frac{x^2}{x}$. The only difference is that it is masked/obfuscated by trigonometric functions. If you understand why this simpler function x^2/x has a taylor series even though its undefined, you should understand your example as well.
It does have something to do with limits: if the discontinuity is removable, that means the limit at that point exists, and the value of the limit is precisely what the value of the function is "supposed" to be.
If you want a more complicated but informative response, the discontinuity was removable because cos(2x)-1 has a zero of order 2 at that point, therefore you can divide out an x^2.
10. Re: Taylor series: (cos(2x)-1)/x²
Well I think that because a taylor series attempts to create a function by a limit as the number of terms approaches infinity, we are indeed working with a limit. However, I am still slightly skeptical as to why, even though the discontinuity is indeed removable, thus leaving a "hole" in the graph, if you will, it is ok to "fill in" the value. I would investigate the radius of convergence of this series and see if x=0, perhaps being in it, has something to do with the fact that we can "fill it in." The disk around x=0 might have some connectivity property. I'm not sure, sorry if my answer isn't too precise. I'm wondering myself.
11. Re: Taylor series: (cos(2x)-1)/x²
No you are thinking the wrong way. Removable discontinuities are actually quite harmless. They aren't truly "supposed" to be there, but are merely the result of the way we formally define functions.
It is ok to fill in the graph because we can imagine a different function: one that is the same for all values except x=0. That is:
$f(x) \;=\; \begin{cases} \frac{cos(2x)-1}{x^2} & \text{if }x \ne 0 \\ -2 & \text{if }x=0\end{cases}$
THIS function satisfies all the criteria of having a Taylor series, being defined and differentiable at that point. And since it is coincident to the original function at all points except x=0, it is in effect a series for that function. Its radius of convergence is also infinite, by the way, because there are no actual singularities.
Notice also, the -2 is not arbitrary. That is the value of
$\lim_{x \to 0}\frac{\cos(2x)-1}{x^2}$
If I put any number other than -2, f(x) would be discontinuous and fail to satisfy the requirements for having a taylor series. | 2016-04-30T18:01:51 | {
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