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Since real phase noise data cannot be collected across a continuous spectrum, a summation must be performed in place of the integral. The period is taken to be 2 Pi, symmetric around the origin, so the. The expression you have is an person-friendly remodel, so which you will detect the inverse in a table of Laplace transforms and their inverses. Taylor Series A Taylor Series is an expansion of some function into an infinite sum of terms , where each term has a larger exponent like x, x 2 , x 3 , etc. Topics include: The Fourier transform as a tool for solving physical problems. This property, together with the fast Fourier transform, forms the basis for a fast convolution algorithm. Home / ADSP / MATLAB PROGRAMS / MATLAB Videos / Discrete Fourier Transform in MATLAB Discrete Fourier Transform in MATLAB 18:48 ADSP , MATLAB PROGRAMS , MATLAB Videos. pattern of signs, the summation being taken over all odd positive integers n that are not multiples of 3. Goldberg, Kenneth A. 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. That is, all the energy of a sinusoidal function of frequency A is entirely localized at the frequencies given by |f|=A. 1 an RC circuit 2. 分数阶傅里叶变换（fractional fourier transform） matlab代码. 下载 数字信号处理科学家与工程师手册 （非常实用，含有大量C实用代码）. We denote Y(s) = L(y)(t) the Laplace transform Y(s) of y(t). It is unusual in treating Laplace transforms at a relatively simple level with many examples. We'll save FFT for another day. The CWT is obtained using the analytic Morse wavelet with the symmetry parameter (gamma) equal to 3 and the time-bandwidth product equal to 60. Integration in the time domain is transformed to division by s in the s-domain. The convergence criteria of the Fourier transform (namely, that the function be absolutely integrable on the real line) are quite severe due to the lack of the exponential decay term as seen in the Laplace transform, and it means | {
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severe due to the lack of the exponential decay term as seen in the Laplace transform, and it means that functions like polynomials, exponentials, and trigonometric functions all do not have Fourier transforms in the. Title: 315_DSP_DWT Author: Christos Faloutsos school2 Created Date: 11/4/2019 9:06:39 AM. Fourier Transform Example #2 MATLAB Code % ***** MATLAB Code Starts Here ***** % %FOURIER_TRANSFORM_02_MAT % fig_size = [232 84 774 624]; m2ft = 3. The Fourier Transform As we have seen, any (sufficiently smooth) function f(t) that is periodic can be built out of sin's and cos's. Especially important are the solutions to the Fourier transform of the wave equation, which define Fourier series, spherical harmonics, and their generalizations. Since it is u(t-1), the cos(wt) function will be zero till 1. Fourier analysis is used in electronics, acoustics, and communications. Hilbert transform, short-time Fourier transform (more about this later), Wigner distributions, the Radon Transform, and of course our featured transformation, the wavelet transform, constitute only a small portion of a huge list of transforms that are available at engineer's and mathematician's disposal. Using MATLAB, determine the Fourier transform numerically and plot the result. edu is a platform for academics to share research papers. This is the simple code for FFT transform of Cos wave using Matlab. Fit Fourier Models Interactively. a finite sequence of data). com for more math and science lectures! In this video I will find the Fourier transform F(w)=? given the input function f(t)=cos(w0t. Turn in your code and plot. Inverted frequency spectrum, also called cepstrum, is the result of taking the inverse Fourier transform of the logarithm of a signal estimated spectrum. 1, 2017 ROTOR FAULT DETECTION OF WIND TURBINE SQUIRREL CAGE INDUCTION GENERATOR USING TEAGER–KAISER ENERGY OPERATOR Lahc`ene Noured. Sound Waves. En 1822, Fourier expose les séries et la transformation de Fourier dans son | {
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Noured. Sound Waves. En 1822, Fourier expose les séries et la transformation de Fourier dans son traité Théorie analytique de la chaleur. All I can do is give you a hint. yes my signal is load-time signal with T0=0. This is a good point to illustrate a property of transform pairs. We will now derive the complex Fourier series equa-tions, as shown above, from the sin/cos Fourier series using the expressions for sin() and cos() in terms of complex exponentials. no hint Solution. A truncated Fourier series, where the amplitude and frequency do not vary with time, is a special case of these signals. The custom Matlab/Octave function FouFilter. - [Voiceover] Many videos ago, we first looked at the idea of representing a periodic function as a set of weighted cosines and sines, as a sum, as the infinite sum of weighted cosines and sines, and then we did some work in order to get some basics in terms of some of these integrals which we then started to use to derive formulas for the various coefficients, and we are almost there. % Input: % X - 1xM - complex vector - data points (signal discretisation). ylim: the y limits of the plot. One hardly ever uses Fourier sine and cosine transforms. Edward Donley Mathematics Department Indiana University of Pennsylvania Basics of Sound. DFT needs N2 multiplications. Expression (1. Visit Stack Exchange. We need to compute the Fourier transform of the product of two functions: $f(t)=f_1(t)f_2(t)$ The Fourier transform is the convolution of the Fourier transform of the two functions: $F(\omega)=F_1(\omega)*F_2(\omega)=$ [math. Fourier Transform Solution 0. We have fb(w)= 1 √ 2π Z1 −1 xe−ixw dx = 1 √ 2π Z1 −1 x coswx−isinwx dx = −i √ 2π Z1 −1 x sinwxdx = −2i √ 2π Z1 0 x sinwxdx = −2i √ 2π 1 w2 sinwx− x w coswx 1 0 = −i r 2 π sinw − wcosw w2. That is G k=g j exp− 2πikj N ⎛ ⎝⎜ ⎞ ⎠⎟ j=0 N−1 ∑ (7-6) Scaling by the sample interval normalizes spectra to the continuous Fourier transform. Karris - Free ebook download as PDF File (. Symmetry in | {
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spectra to the continuous Fourier transform. Karris - Free ebook download as PDF File (. Symmetry in Exponential Fourier Series. orthogonal functions fourier series. you can ask your doubt in the comment box. If we de ne s 1(t) = s(t t 0); then S 1(f) = Z 1 1 s(t t 0)e j2ˇftdt; Z 1 1 s(u)e j2ˇf(u+t 0)du; = e j2ˇft 0 Z 1 1 s(u)e j2ˇfudu; = e j2ˇft 0S(f): There is a similar dual relationshp if a signal is scaled by an exponential in the time domain. 下载 常用傅里叶变换对. The Fourier transform. The detection process is mainly based on the wavelet transform. Digital signal processing (DSP) vs. Fourier transform that f max is f 0 plus the bandwidth of rect(t - ½). If any argument is an array, then fourier acts element-wise on all elements of the array. We describe this FFT in the current section. The trigonometric Fourier series of the even signal cos(t) + cos(2. Thus, the DFT formula basically states that the k'th frequency component is the sum of the element-by-element products of 'x' and ' ', which is the so-called inner product of the two vectors and , i. Fourier analysis is a method of defining periodic waveform s in terms of trigonometric function s. Inverted frequency spectrum converts the periodic signal and sidebands in FFT results to spectral lines, thus making it easier to detect the complex periodic component of the spectrum. Now multiply it by a complex exponential at the same frequency. The 3-D FrFT can independently compress and image radar data in each dimension for a broad set of parameters. Therefore, we will start with the continuous Fourier transform,. Fourier transform (FT), time–frequency analysis such as the STFT and Cohen-class quadratic distributions, and time-scale analysis based on the wavelet trans-form (WT) are often applied to investigate the hidden properties of real signals. In practice, when doing spectral analysis, we cannot usually wait that long. 5t) = αn cos(2π t) n=1 T0 ∞ n = αn cos( t) 2 n=1 By equating the coefficients of cos( n t) 2 of both | {
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5t) = αn cos(2π t) n=1 T0 ∞ n = αn cos( t) 2 n=1 By equating the coefficients of cos( n t) 2 of both sides we observe that an = 0 for all n unless n = 2, 5 in which case a2 = a5 = 1. Unfortunately, the FT can represent the meaningful spectrum property of only lin-. 3 Fourier Series Using Euler’s rule, can be written as X n 10 1 0 11 00 00 11 ( )cos( ) ( )sin( ) tT tT n tt X xt n tdt j xt n tdt TT ωω ++ =−∫∫ If x(t) is a real-valued periodic signal, we have. 999; % water density (lbm/ft^3). My function is intended for just plain Fourier series expansion (a_k cos(k*x)). Matlab经典教程——从入门到精通 - 第一章 基础准备及入门 本章有两个目的：一是讲述 MATLAB 正常运行所必须具备的基础条件；二是简明系统 地介绍高度集成的 Desktop 操作桌面的. Fast Transforms in Audio DSP; Related Transforms. Using the Fourier transform formula directly to compute each of the n elements of y requires on the order of n 2 floating-point operations. Table 4: Basic Continuous-Time Fourier Transform Pairs Fourier series coefficients Signal Fourier transform (if periodic) +∞ k=−∞ ake jkω0t 2π k=−∞ akδ(ω −kω0) ak ejω0t 2πδ(ω −ω0). Using MATLAB, determine the Fourier transform numerically and plot the result. The usual computation of the discrete Fourier transform is done using the Fast Fouier Transform (FFT). I tried using the following definition Xn = 1/T integral ( f(t) e^-jwnt ) I converted the cos(wt) to its exponential form, then multiplied and combined and. So you either need to install Matlab to your own laptop or connect to cloud version of MatLab or use a Lamar lab which has MatLab, most engineering and CS labs do, so does GB 113. $\endgroup$ – Robert Israel Jan 19 '17 at 21:33. The coefficient in the Fourier cosine series expansion of is by default given by. The Cosine Function. where A k (t) is the slowly varying amplitude and ϕ k (t) is the instantaneous phase. Fast Fourier Transform(FFT) • The Fast Fourier Transform does not refer to a new or different type of Fourier transform. For this case, when performing the Fourier transform in equation 31, the | {
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type of Fourier transform. For this case, when performing the Fourier transform in equation 31, the coefficients c lm and s lm with m > L are simply discarded. If there is any solution at all, it would have to involve a transformation of variables. This computational efficiency is a big advantage when processing data that has millions of data points. 2808; % conversion from meters to feet g = 32. cosh() sinh() 22 tttt tt +---== eeee 3. It is often easier to calculate than the sin/cos Fourier series because integrals with exponentials in are usu-ally easy to evaluate. The Fourier transform is defined for a vector x with n uniformly sampled points by. 下载 数字信号处理科学家与工程师手册 （非常实用，含有大量C实用代码）. you can ask your doubt in the comment box. A Phasor Diagram can be used to represent two or more stationary sinusoidal quantities at any instant in time. The can be accomplished in the Fourier domain just as a multiplication. The 3-D Fractional Fourier Transformation (FrFT) has unique applicability to multi-pass and multiple receiverSynthetic Aperture Radar (SAR) scenarios which can collect radar returns to create volumetric reflectivity data. To establish these results, let us begin to look at the details first of Fourier series, and then of Fourier transforms. Sketch by hand the magnitude of the Fourier transform of c(t) for a general value of f c. I have a time series where I want to do a real Fourier transform Since the data has missing values, I cannot use a FFT which requires equidistant data. It exploits the special structure of DFT when the signal length is a power of 2, when this happens, the computation complexity is significantly reduced. The Segment of Signal is Assumed Stationary A 3D transform ()(t f ) [x(t) (t t )]e j ftdt t ω ′ = •ω* −′•−2π STFTX , ω(t): the window function A function of time. For a general real function, the Fourier transform will have both real and imaginary parts. • Fourier invents Fourier series in 1807 • People start computing Fourier series, and | {
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parts. • Fourier invents Fourier series in 1807 • People start computing Fourier series, and develop tricks Good comes up with an algorithm in 1958 • Cooley and Tukey (re)-discover the fast Fourier transform algorithm in 1965 for N a power of a prime • Winograd combined all methods to give the most efficient FFTs. Tech ECE 5th semester can be seen by clicking here. MATLAB uses notation derived from matrix theory where the subscripts run from 1 to n, so we will use y j+1 for mathemat-ical quantities that will also occur in MATLAB code. Control and Intelligent Systems, Vol. Skip to content. Il énonce qu'une fonction peut être décomposée sous forme de série trigonométrique, et qu'il est facile de prouver la convergence de celle-ci. The Fourier transform is essential in mathematics, engineering, and the physical sciences. information in the Matlab manual for more specific usage of commands. Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC alternating quantities. Pure tone — sine or cosine function frequency determines pitch (440 Hz is an A note) amplitude determines volume. The usual computation of the discrete Fourier transform is done using the Fast Fouier Transform (FFT). I don't know what is not working in my code, but I got an output image with the same number. wavelet transform) offer a huge variety of applications. The SST approach in [8, 7] is based on the continuous wavelet transform (CWT). Explain the effect of zero padding a signal with zero before taking the discrete Fourier Transform. Fn = 1 shows the transform of damped exponent f(t) = e-at. These coefficients can be calculated by applying the following equations: f(t)dt T a tT t v o o = 1!+ f(t)ktdt T a tT t n o o o =!+cos" 2 f(t)ktdt T b tT t n o o o =!+sin" 2 Answer Questions 1 – 2. Table of Discrete-Time Fourier Transform Pairs: Discrete-Time Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] =. Digital signal processing (DSP) vs. We denote Y(s) = | {
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Discrete-Time Fourier Transform : x[n] =. Digital signal processing (DSP) vs. We denote Y(s) = L(y)(t) the Laplace transform Y(s) of y(t). This is because the limits of the integral are from -∞ to +∞. The Fourier Transform: Examples, Properties, Common Pairs Example: Fourier Transform of a Cosine Spatial Domain Frequency Domain cos (2 st ) 1 2 (u s)+ 1 2 (u + s) 0. Ø Many software packages are capable of computing the Fourier transform of input signal (e. DFT needs N2 multiplications. coz they are just using fourier. It is represented in either the trigonometric form or the exponential form. I have a time series where I want to do a real Fourier transform Since the data has missing values, I cannot use a FFT which requires equidistant data. Fourier Transform of aperiodic and periodic signals - C. Example: cos(pi/4*(0:159))+randn(1,160) specifies a sinusoid embedded in white Gaussian noise. It may be possible, however, to consider the function to be periodic with an infinite period. The filter portion will look something like this b = fir1(n,w,'type'); freqz(b,1,512); in = filter(b,1,in);. I suggest that you generate a cosine with a frequency of about 1/16th sample per cycle. In signal processing, the Fourier transform can reveal important characteristics of a signal, namely, its frequency components. The phase noise plot of a PRS10 shows a 42 dB reduction in phase noise at 10 Hz offset from the carrier at 10 MHz when compared to a conventional rubidium standard. For this application, we find that. Outline Introduction to the Fourier Transform and Frequency Domain Magnitude of frequencies Phase of frequencies Fourier transform and DFT Filtering in the frequency domain Smoothing Frequency Domain Filters Sharpening Frequency Domain Filters Homomorphic Filtering Implementation of Fourier transform Background 1807, French math. I didnot get any feedback from fft2. The Fourier transform of the product of two signals is the convolution of the two signals, which is noted by | {
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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. Note that this is similar to the definition of the FFT given in Matlab. The approach to characterize the image texture are investigated based on WT other than DOST. Laplace transform allows us to convert a differential equation to an algebraic equation. Re There are two important implications of Im(j) the Fourier Transform as defined in this equation here is applicable only to aperiodic signals. Briggs , Van Emden Henson Just as a prism separates white light into its component bands of colored light, so the discrete Fourier transform (DFT) is used to separate a signal into its constituent frequencies. Integration in the time domain is transformed to division by s in the s-domain. Analog signal processing (ASP) The theory of Fourier transforms is applicable irrespective of whether the signal is continuous or discrete, as long as it is "nice" and absolutely integrable. For more information about the Fourier series, refer to Fourier Analysis and Filtering (MATLAB). Windowed Fourier Transform: Represents non periodic signals. For this case, when performing the Fourier transform in equation 31, the coefficients c lm and s lm with m > L are simply discarded. So, generally, we use this property of linearity of Laplace transform to find the Inverse Laplace transform. Home / ADSP / MATLAB PROGRAMS / MATLAB Videos / Discrete Fourier Transform in MATLAB Discrete Fourier Transform in MATLAB 18:48 ADSP , MATLAB PROGRAMS , MATLAB Videos. The N-D transform is equivalent to computing the 1-D transform along each dimension of X. If f(t) is non-zero with a compact support, then its Fourier transform cannot be zero on a whole interval. Matlab时频分析工具箱及函数应用说明. The Laplace Transform. All I can do is give you a hint. The forward transform converts a signal from the time domain into the frequency domain, thereby | {
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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. Using MATLAB to Plot the Fourier Transform of a Time Function The aperiodic pulse shown below: has a Fourier transform: X(jf)=4sinc(4πf) This can be found using the Table of Fourier Transforms. The following MATLAB. Re There are two important implications of Im(j) the Fourier Transform as defined in this equation here is applicable only to aperiodic signals. Conclusion As we have seen, the Fourier transform and its 'relatives', the discrete sine and cosine transform provide handy tools to decompose a signal into a bunch of partial waves. The can be accomplished in the Fourier domain just as a multiplication. f and f^ are in general com-plex functions (see Sect. Matlab uses the FFT to find the frequency components of a. Equation [2] states that the fourier transform of the cosine function of frequency A is an impulse at f=A and f=-A. d/dt[-cos(mt)/m] =? sin(mt) Since -1/m is constant with respect to t, bring it out front. This is soo confusing u know. Note: The FFT-based convolution method is most often used for large inputs. The sum of signals (disrupted signal) As we created our signal from the sum of two sine waves, then according to the Fourier theorem we should receive its frequency image concentrated around two frequencies f 1 and f 2 and also its opposites -f 1 and -f 2. The Segment of Signal is Assumed Stationary A 3D transform ()(t f ) [x(t) (t t )]e j ftdt t ω ′ = •ω* −′•−2π STFTX , ω(t): the window function A function of time. The - dimensional Fourier cosine coefficient is given by. Tech ECE 5th semester can be seen by clicking here. Matlab时频分析工具箱及函数应用说明. 4414e-06 fsample = 409600. However, computationally efficient algorithms can require as little as n log2(n) operations. For the input sequence x and its | {
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efficient algorithms can require as little as n log2(n) operations. For the input sequence x and its transformed version X (the discrete-time Fourier transform at equally spaced frequencies around the unit circle), the two functions implement the relationships. And, of course, everybody sees that e to the inx, by Euler's great formula, is a combination of cosine nx and sine nx. For example, the Fourier transform allows us to convert a signal represented as a function of time to a function of frequency. e jwt Cos wt jSen wt Time Fourier transform, DTFT) Matlab y uso de función òfft código aquí) (c) P. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 1 shows how increasing the period does indeed lead to a continuum of coefficients, and. an = 1 1 R1 1 (1 x 2) cos nπx dx 1 1cos nπx dx 1 1x 2cos nπx dx = 0 1 nπx 2sin (nπx) j 1 1 + 2 nπ R 1 1 xsin nπx dx = 0 0 2 n2π2xcos (nπx) j 1 1 + 2 n2π2 R 1 1 cos nπx dx = 4 n2π2 cos (nπ. I just had a look at what the Curve Fitting app is doing at its "Fourier" option includes the fundamental frequency as one of the fit parameters. Thus, the DFT formula basically states that the k'th frequency component is the sum of the element-by-element products of 'x' and ' ', which is the so-called inner product of the two vectors and , i. You can write a book review and share your experiences. Dilation and rotation are real-valued scalars and position is a 2-D vector with real-valued elements. (b) We can directly observe from the plot that given function is a multiplication of sin(2πt) and the sign (signum) function sgn(t) which has the Fourier transform F(sgn)(s)= 1/(πis). Fast Fourier Transform in MATLAB ®. Fn = 5 and 6 shows the function reconstructed from its spectrum. Full text of "The Fourier Transform And Its Applications Bracewell" See other formats. (i) We must calculate the Fourier coefficients. Lab 5 Fourier Series I. f and f^ are in general com-plex functions | {
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the Fourier coefficients. Lab 5 Fourier Series I. f and f^ are in general com-plex functions (see Sect. Then, use the duality function to show that 1 t 2 j 2sgn j sgn j sgn. The expression you have is an person-friendly remodel, so which you will detect the inverse in a table of Laplace transforms and their inverses. Discrete Time Fourier Transform (DTFT) Fourier Transform (FT) and Inverse. 7) Synthèse de Fourier L’opération inverse de la décomposition de Fourier peut également être faite, et s’appelle la synthèse de Fourier. However, the definition of the MATLAB sinc. 19 Two-dimensional (2D) Fourier. Matlab时频分析工具箱及函数应用说明. Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC alternating quantities. For particular functions we use tables of the Laplace. We expressed ChR2 in COS-cells, purified it, and subsequently investigated this unusual photoreceptor by flash photolysis and UV-visible and Fourier transform infrared difference spectroscopy. As an example, if we are given the phasor V = 5/36o, the expression for v(t) in the time domain is v(t) = 5 cos(ωt + 36o), since we are. Abstract In this paper, a MATLAB model of a Digital Fourier Transform (DFT)-based digital distance relay was developed and then its behavior was analyzed when it is applied on distance protection of a real series compensated transmission system, belonging to the Chilean generation and transmission utility Colbún S. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. 5 BACKGROUND In the following sections, we discuss the effect of amplitude modulation and sampling on the signal spectrum using common properties of the Fourier transform. To learn how to use the fft function type >> help fft at the Matlab command line. I tried a number of different representations of tanh() but none of them had an analytical solution for the fourier cosine transform. Visit Stack Exchange. txt) or view presentation | {
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solution for the fourier cosine transform. Visit Stack Exchange. txt) or view presentation slides online. The Laplace Equation; The Wave Equation; The Heat Equation; Bibliography. There are various implementations of it, but a standard form is the Radix-2 FFT. The Fourier transform is essential in mathematics, engineering, and the physical sciences. Ive tried to write matlab code that takes in a grayscale image matrix, performs fft2() on the matrix and then calculates the magnitude and phase from the transform. The following options can be given:. The Cosine Function. FFT(X) is the discrete Fourier transform of vector X. Fourier Transform of Array Inputs. Fourier Transform Z. I tried using the following definition Xn = 1/T integral ( f(t) e^-jwnt ) I converted the cos(wt) to its exponential form, then multiplied and combined and. THE DISCRETE COSINE TRANSFORM (DCT) 3. the fourier transform and its applications. Basic theory and application of the FFT are introduced. MATLAB provides command for working with transforms, such as the Laplace and Fourier transforms. A sinusoidal function can be expressed in either Fourier transform (Fourier series) or phasor representation: (31) where (32) We see that the phasor and the Fourier coefficients and are essentially the same, in the sense that they are both coefficients representing the amplitude and phase of the complex exponential function. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Хотя формула, задающая преобразование Фурье, имеет ясный смысл только для функций класса (), преобразование Фурье может быть определено и для более широкого класса функций и даже обобщённых функций. where a 0 models a constant (intercept) term in the data and is associated with the i = 0 cosine term, w is the fundamental frequency of the signal, n is the number of terms (harmonics) in the series, and 1 ≤ n ≤ 8. Discrete Fourier Transform (DFT) The frequency content of a | {
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in the series, and 1 ≤ n ≤ 8. Discrete Fourier Transform (DFT) The frequency content of a periodic discrete time signal with period N samples can be determined using the discrete Fourier transform (DFT). The main idea is to extend these functions to the interval and then use the Fourier series definition. Let's overample: f s = 100 Hz. Fourier analysis is used in electronics, acoustics, and communications. Use integration by parts to evaluate the. To address this issue, short-time Fourier transform (STFT) is proposed to exhibit the time-varying information of the analyzed signal. Let Y(s)=L[y(t)](s). transform a signal in the time or space domain into a signal in the frequency domain. The Fourier transform of a Gaussian is a Gaussian and the inverse Fourier transform of a Gaussian is a Gaussian f(x) = e −βx2 ⇔ F(ω) = 1 √ 4πβ e ω 2 4β (30) 4. This is the simple code for FFT transform of Cos wave using Matlab. The Laplace Transform Example: Using Frequency Shift Find the L[e-atcos(wt)] In this case, f(t) = cos(wt) so, The Laplace Transform Time Integration: The property is: The Laplace Transform Time Integration: Making these substitutions and carrying out The integration shows that. The custom Matlab/Octave function FouFilter. Here is a simple implementation of the Discrete Fourier Transform: myFourierTransform. The four Fourier transforms that comprise this analysis are the Fourier Series, Continuous-Time Fourier Transform, DiscreteTime Fourier Transform and Discrete Fourier Transform. Cal Poly Pomona ECE 307 Fourier Transform The Fourier transform (FT) is the extension of the Fourier series to nonperiodic signals. Given the F. In the form FourierCosCoefficient [expr, t, n], n can be symbolic or a non - negative integer. 18 Applying the Fourier transform to the image of a group of people. To illustrate determining the Fourier Coefficients, let's look at a simple example. at the MATLAB command prompt. The finite, or discrete, Fourier transform of a complex vector y | {
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at the MATLAB command prompt. The finite, or discrete, Fourier transform of a complex vector y with n ele-ments y j+1;j = 0;:::n •1 is another complex. Signals having finite energy are energy signals. Fourier analysis transforms a signal from the. We can use MATLAB to plot this transform. orthogonal functions fourier series. The Fourier transform is sometimes denoted by the operator Fand its inverse by F1, so that: f^= F[f]; f= F1[f^] (2) It should be noted that the de. When the energy is finite, the total power will be zero. What you have given isn't a Fourier remodel; it particularly is a Laplace remodel with jw=s. 2001-01-01. I have a time series where I want to do a real Fourier transform Since the data has missing values, I cannot use a FFT which requires equidistant data. Expression (1. you can ask your doubt in the comment box. The period is taken to be 2 Pi, symmetric around the origin, so the. com for more math and science lectures! In this video I will find the Fourier transform F(w)=? given the input function f(t)=cos(w0t. This function is a cosine function that is windowed - that is, it is multiplied by the box or rect function. 1 The upper plot shows the magnitude of the Fourier series spectrum for the case of T=1 with the Fourier transform of p(t) shown as a dashed line. The approach to characterize the image texture are investigated based on WT other than DOST. Can you please send proper solution of this question. With more than 5,000 lines of MATLAB code and more than 700 figures embedded in the text, the material teaches readers how to program in MATLAB and study signals and systems concepts at the same time, giving them the tools to harness the power of computers to quickly assess problems and then visualize their solutions. Labview, MATLAB and EXCEL, many others) 21 Discrete Sampling Sampling Period To accurately reproduce spectrum for periodic waveforms, the measurement period must be an integer multiple of the fundamental period, T 1 : mT 1 =n δ | {
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the measurement period must be an integer multiple of the fundamental period, T 1 : mT 1 =n δ t DFT Specta Waveform. ω=2π, we expand f (t) as a Fourier series by ( ) ( ) + + + = + + + b t b t f t a a t a t ω ω ω ω sin sin 2 ( ) cos cos 2 1 1 0 1 2 (Eqn 1) 2. SFORT TIME FOURIER TRANSFORM (STFT) Dennis Gabor (1946) Used STFT ♥To analyze only a small section of the signal at a time -- a technique called Windowing the Signal. The Discrete Cosine Transform (DCT) Number Theoretic Transform. In this case F(ω) ≡ C[f(x)] is called the Fourier cosine transform of f(x) and f(x) ≡ C−1[F(ω)] is called the inverse Fourier cosine transform of F(ω). Sampled sound (digital audio) — discrete sequence of intensities CD Audio is 44100 samples per second. Fourier Transform of aperiodic and periodic signals - C. How can we use Laplace transforms to solve ode? The procedure is best illustrated with an example. Full text of "The Fourier Transform And Its Applications Bracewell" See other formats. 2-D Continuous Wavelet Transform. Fast Fourier Transform. First sketch the function. Here is a link to a video in YouTube that provides a nice illustration: Slinky. Find the Fourier cosine series and the Fourier sine series for the function f(x) = ˆ 1 if 0 iFFTUnderstanding FFTs and Windowing Overview Learn about the time and frequency domain, fast Fourier transforms (FFTs), and windowing as well. (b) We can directly observe from the plot that given function is a multiplication of sin(2πt) and the sign (signum) function sgn(t) which has the Fourier transform F(sgn)(s)= 1/(πis). Langton Page 3 And the coefficients C n are given by 0 /2 /2 1 T jn t n T C x t e dt T (1. The Fourier transform of a signal exist if satisfies the following condition. DOEpatents. cos(wt)의 힐베르트 변환은 sin(wt)이다. com) Fourier Series Applet (Tip: drag magnitude or phase dots up or down to change the wave form). The Laplace Transform Theorem: Initial Value If the function f(t) and its first derivative are Laplace transformable | {
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Theorem: Initial Value If the function f(t) and its first derivative are Laplace transformable and f(t) Has the Laplace transform F(s), and the exists, then lim sF(s) 0 lim ( ) lim ( ) (0) o f o s t sF s f t f The utility of this theorem lies in not having to take the inverse of F(s). The integrals from the last lines in equation [2] are easily evaluated using the results of the previous page. a0 = 1 2 R 1 1 (1 x 2) dx 1 2 x 1 3x 3 j 1 1 2 3. [2] Inverse DFT is defined as: N -1 1 x ( n) = N å X ( k )e k =0 j 2pnk / N for 0 £ n £ N - 1. Normal images such as straw, wood, sand and grass are used in the analysis. 分数阶傅里叶变换（fractional fourier transform） matlab代码. Dct vs dft Dct vs dft. Discrete Fourier Transform See section 14. Explain the effect of zero padding a signal with zero before taking the discrete Fourier Transform. Fourier transform infrared (FT-IR) spectroscopy, principal component analysis (PCA), two-dimensional correlation spectroscopy (2D-COS), and X-ray diffraction, while the sorption properties were evaluated by water vapor isotherms using the gravimetric method coupled with infrared spectroscopy. Fourier Transform of Periodic Signals 0 0 /2 /2 0 1 o o jt n n T jn t n T x t c e c x t e dt T 2 ( ) jn t jn t oo nn nn no n. It is one commonly encountered form for the Fourier series of real periodic signals in continuous time. From Wikibooks, the open-content textbooks collection < Engineering Tables Jump to: navigation, search. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series The Fourier transform of a sine or cosine at a frequency f 0 only has energy exactly at f 0, which is what we would expect. Matlab经典教程——从入门到精通 - 第一章 基础准备及入门 本章有两个目的：一是讲述 MATLAB 正常运行所必须具备的基础条件；二是简明系统 地介绍高度集成的 Desktop 操作桌面的. for fourier transform we need to define w. The Fourier transform is defined for a vector x with n uniformly sampled points by. Find the transfer function of the following RC circuit. Fourier Transform of the Gaussian Konstantinos | {
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the transfer function of the following RC circuit. Fourier Transform of the Gaussian Konstantinos G. It is often easier to calculate than the sin/cos Fourier series because integrals with exponentials in are usu-ally easy to evaluate. [2] Inverse DFT is defined as: N -1 1 x ( n) = N å X ( k )e k =0 j 2pnk / N for 0 £ n £ N - 1. Ø Many software packages are capable of computing the Fourier transform of input signal (e. information in the Matlab manual for more specific usage of commands. I tried a number of different representations of tanh() but none of them had an analytical solution for the fourier cosine transform. Finally, I am supposed to create a filter using the basic MATLAB commands and filter the noise out of the plot of the signal and then do the Fourier Transform of the signal again and plot the results. Basic theory and application of the FFT are introduced. Van Loan and K. The knowledge of Fourier Series is essential to understand some very useful concepts in Electrical Engineering. function [ft] = myFourierTransform (X, n) % Objective: % Apply the Discrete Fourier Transform on X. The Laplace Equation; The Wave Equation; The Heat Equation; Bibliography. Recall the definition of hyperbolic functions. 1946: Gabor 开发了短时傅立叶变换(Short time Fourier transform, STFT) STFT的时间-频率关系图 小波分析发展历程 (1900 - 1979) where，s(t) is a signal, and g(t) is the windowing function. Grundlagen und Begriffsabgrenzungen, Rechtecksignal, Dreieckfunktion und Fourier-Transformation - Holger Schmid - Ausarbeitung - Ingenieurwissenschaften - Wirtschaftsingenieurwesen - Arbeiten publizieren: Bachelorarbeit, Masterarbeit, Hausarbeit oder Dissertation. Existence of the Fourier Transform; The Continuous-Time Impulse. Pure tone — sine or cosine function frequency determines pitch (440 Hz is an A note) amplitude determines volume. Using the Fourier transform formula directly to compute each of the n elements of y requires on the order of n 2 floating-point operations. Un buen ejemplo de eso es lo | {
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n elements of y requires on the order of n 2 floating-point operations. Un buen ejemplo de eso es lo que hace el oído humano, ya que recibe una onda auditiva y la transforma en una descomposición en distintas frecuencias (que es lo que finalmente se escucha). That is, all the energy of a sinusoidal function of frequency A is entirely localized at the frequencies given by |f|=A. The DFT of the sequence x(n) is expressed as 2 1 ( ) ( ) 0 N jk i X k x n e N i − − Ω = ∑ = (1) where = 2Π/N and k is the frequency index. 5 Signals & Linear Systems Lecture 10 Slide 11 Fourier Transform of any periodic signal ∑Fourier series of a periodic signal x(t) with period T 0 is given by: Take Fourier transform of both sides, we get: This is rather obvious!. (i) We must calculate the Fourier coefficients. The trigonometric Fourier series of the even signal cos(t) + cos(2. If the analyzing wavelet is analytic, you obtain W f ( u , s ) = 1 2 W f a ( u , s ) , where f a (t) is the analytic signal corresponding to f(t). 2 p693 PYKC 8-Feb-11 E2. Hey everyone, i know that matlab have the method for fourier transform implemented but i was wondering if there is anything that could give me coefficients of fourier transfrom. The Fourier series is a sum of sine and cosine functions that describes a periodic signal. Note that this is similar to the definition of the FFT given in Matlab. 3 251 with a + - - + + - - + +. In the paragraphs that follow we illustrate this approach using Maple, Mathematica, and MATLAB. It uses real DFT, that is, the version of Discrete Fourier Transform which uses real numbers to represent the input and output signals. Today I want to start getting "discrete" by introducing the discrete-time Fourier transform (DTFT). To learn more, see our tips on writing great. Engineering Tables/Fourier Transform Table 2 From Wikibooks, the open-content textbooks collection < Engineering Tables Jump to: navigation, search Signal Fourier transform unitary, angular frequency Fourier | {
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Tables Jump to: navigation, search Signal Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. The Laplace Equation; The Wave Equation; The Heat Equation; Bibliography. Example: cos(pi/4*(0:159))+randn(1,160) specifies a sinusoid embedded in white Gaussian noise. Esta función de MATLAB calcula la transformada discreta de Fourier (DFT) de X usando un algoritmo de transformada rápida de Fourier (FFT). Provide Plots For The Time Domain Signal And The Magnitude Of Its FT. Bartl‡¶ From the ‡Institut für medizinische Physik und Biophysik, Charité-Universitätsmedizin Berlin,. Derivation in the time domain is transformed to multiplication by s in the s-domain. This is my attempt in hoping for a way to find it without using the definition: x(t) = c. 5 Signals & Linear Systems Lecture 10 Slide 11 Fourier Transform of any periodic signal ∑Fourier series of a periodic signal x(t) with period T 0 is given by: Take Fourier transform of both sides, we get: This is rather obvious!. selection, high computation speed, and extensive application. The DFT: An Owners' Manual for the Discrete Fourier Transform William L. The top equation de nes the Fourier transform (FT) of the function f, the bottom equation de nes the inverse Fourier transform of f^. FFT Software. This is the simple code for FFT transform of Cos wave using Matlab. Fourier-transform and global contrast interferometer alignment methods. com) Fourier Series Applet (Tip: drag magnitude or phase dots up or down to change the wave form). Repeat the example in Section II. 小波变换（dwt）源代码. If f(t) is non-zero with a compact support, then its Fourier transform cannot be zero on a whole interval. We need to compute the Fourier transform of the product of two functions: $f(t)=f_1(t)f_2(t)$ The Fourier transform is the convolution of the Fourier transform of the two functions: $F(\omega)=F_1(\omega)*F_2(\omega)=$ | {
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the convolution of the Fourier transform of the two functions: $F(\omega)=F_1(\omega)*F_2(\omega)=$ [math. To illustrate determining the Fourier Coefficients, let's look at a simple example. Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC alternating quantities. Fourier Series. The Laplace Transform. Fourier Transform Example #2 MATLAB Code % ***** MATLAB Code Starts Here ***** % %FOURIER_TRANSFORM_02_MAT % fig_size = [232 84 774 624]; m2ft = 3. Explain the effect of zero padding a signal with zero before taking the discrete Fourier Transform. On the second plot, a blue spike is a real (cosine) weight and a green spike is an imaginary (sine) weight. The 3-D Fractional Fourier Transformation (FrFT) has unique applicability to multi-pass and multiple receiverSynthetic Aperture Radar (SAR) scenarios which can collect radar returns to create volumetric reflectivity data. Note that this is similar to the definition of the FFT given in Matlab. The can be accomplished in the Fourier domain just as a multiplication. There are various implementations of it, but a standard form is the Radix-2 FFT. Fourier Transform of Periodic Signals 0 0 /2 /2 0 1 o o jt n n T jn t n T x t c e c x t e dt T 2 ( ) jn t jn t oo nn nn no n. Generally speaking, the more concentrated g(t) is, the more spread out its Fourier transform G^(!) must be. Using MATLAB to Plot the Fourier Transform of a Time Function The aperiodic pulse shown below: has a Fourier transform: X(jf)=4sinc(4πf) This can be found using the Table of Fourier Transforms. The Fourier transform is sometimes denoted by the operator Fand its inverse by F1, so that: f^= F[f]; f= F1[f^] (2) It should be noted that the de. FFT onlyneeds Nlog 2 (N). This signal will have a Fourier. For the input sequence x and its transformed version X (the discrete-time Fourier transform at equally spaced frequencies around the unit circle), the two functions implement the relationships. The Fourier series is a sum of sine and cosine | {
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the two functions implement the relationships. The Fourier series is a sum of sine and cosine functions that describes a periodic signal. csdn已为您找到关于小波变换与图像、图形处理技术相关内容，包含小波变换与图像、图形处理技术相关文档代码介绍、相关教学视频课程，以及相关小波变换与图像、图形处理技术问答内容。. Fast Fourier Transform. Other readers will always be interested in your opinion of the books you've read. Write a second version that first sets up a transform matrix (with rows corresponding to the various values of k) and then multiplies this matrix by the input to perform the transform. y = Sin (300t) fits the form y = sin(ωt + ϴ) ω = frequency in radians/second. Fourier Transform. Computational Efficiency. 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. 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. , make it 8 cycles long). The DTFT is often used to analyze samples of a continuous function. Example: cos(pi/4*(0:159))+randn(1,160) specifies a sinusoid embedded in white Gaussian noise. Over a time range of 0 400< 0. 5 BACKGROUND In the following sections, we discuss the effect of amplitude modulation and sampling on the signal spectrum using common properties of the Fourier transform. Here's the 100th column of X_rows: plot(abs(X_rows(:, 100))) ylim([0 2]) As I said above, the Fourier transform of a constant sequence is an impulse. In signal processing, the Fourier transform can reveal important characteristics of a signal, namely, its frequency components. Amyloid Hydrogen Bonding Polymorphism Evaluated by (15)N{(17)O}REAPDOR Solid-State NMR and Ultra-High Resolution | {
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Bonding Polymorphism Evaluated by (15)N{(17)O}REAPDOR Solid-State NMR and Ultra-High Resolution Fourier Transform Ion Cyclotron Resonance Mass Spectrometry. The Fast Fourier Transform (FFT) is an efficient way to do the DFT, and there are many different algorithms to accomplish the FFT. Most of the following equations should not be memorized by the reader; yet, the reader should be able to instantly derive them from an understanding of the function's characteristics. These systems are based on neural networks [1] with wavelet transform based feature extraction. The Fourier series is a sum of sine and cosine functions that describes a periodic signal. • Hence, even if the Heisenberg constraints are verified, it is. Question: MATLAB Problem: Fourier Transform (FT) Of A Cosine Signal This Problem Analyzes A Transmitted Sinusoidal Signal In The Frequency Domain. Combines traditional methods such as discrete Fourier transforms and discrete cosine transforms with more recent techniques such as filter banks and wavelet Strikes an even balance in emphasis between the mathematics and the applications with the emphasis on linear algebra as a unifying theme. For functions of two variables that are periodic in both variables, the. 数字信号处理科学家与工程师. Always keep in mind that an FFT algorithm is not a different mathematical transform: it is simply an efficient means to compute the DFT. filtering the spectrum and regenerating the signal using the filtered spectrum is done at the end Rayleigh theorem is proved by showing that the energy content of both time domain and frequency domain signals are equal. The Fourier Transform Introduction In the Communication Labs you will be given the opportunity to apply the theory learned in Communication Systems. where a 0 models a constant (intercept) term in the data and is associated with the i = 0 cosine term, w is the fundamental frequency of the signal, n is the number of terms (harmonics) in the series, and 1 ≤ n ≤ 8. The idea is that one can. | {
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signal, n is the number of terms (harmonics) in the series, and 1 ≤ n ≤ 8. The idea is that one can. The Fast Fourier Transform (FFT) is an efficient way to do the DFT, and there are many different algorithms to accomplish the FFT. Karris - Free ebook download as PDF File (. We describe this FFT in the current section. Dilation and rotation are real-valued scalars and position is a 2-D vector with real-valued elements. Como nos comenta el mismo Sergey en su sitio Web, la tarea básica en el procesamiento de electro-cardiogramas (ECG) es la detección de los picos R. A truncated Fourier series, where the amplitude and frequency do not vary with time, is a special case of these signals. ϴ = angle in degrees. Then, use the duality function to show that 1 t 2 j 2sgn j sgn j sgn. is equal to the frequency integral of the square of its Fourier Transform. For particular functions we use tables of the Laplace. That is G k=g j exp− 2πikj N ⎛ ⎝⎜ ⎞ ⎠⎟ j=0 N−1 ∑ (7-6) Scaling by the sample interval normalizes spectra to the continuous Fourier transform. , Bracewell) use our −H as their definition of the forward transform. If the input to the above RC circuit is x(t) cos(2Sf 0, find the output t) y(t). Next, on defining τ = t− s and writing c τ = t (y t − y¯)(y t−τ − y¯)/T, we can reduce the latter expression to (16) I(ω j)=2 T−1 τ=1−T cos(ω jτ)c τ, which is a Fourier transform of the sequence of empirical autocovariances. Specify the independent and transformation variables for each matrix entry by using matrices of the same size. 19 Two-dimensional (2D) Fourier. THE DISCRETE COSINE TRANSFORM (DCT) 3. 34 matlab programs here! Please click here to see all the matlab programs present in this blog. Time Complexity • Definition • DFT • Cooley-Tukey’s FFT 6. FFT Discrete Fourier transform. Compare with the previous result. orthogonal functions fourier series. Grundlagen und Begriffsabgrenzungen, Rechtecksignal, Dreieckfunktion und Fourier-Transformation - Holger Schmid - Ausarbeitung - | {
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Rechtecksignal, Dreieckfunktion und Fourier-Transformation - Holger Schmid - Ausarbeitung - Ingenieurwissenschaften - Wirtschaftsingenieurwesen - Arbeiten publizieren: Bachelorarbeit, Masterarbeit, Hausarbeit oder Dissertation. It may be possible, however, to consider the function to be periodic with an infinite period. 2) Here 0 is the fundamental frequency of the signal and n the index of the harmonic such. Note that this is similar to the definition of the FFT given in Matlab. I tried using the following definition Xn = 1/T integral ( f(t) e^-jwnt ) I converted the cos(wt) to its exponential form, then multiplied and combined and. The resulting transform pairs are shown below to a common horizontal scale: Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 8 / 37. | {
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# Thread: Point of best fit?
1. ## Point of best fit?
So we all know that you can apply the "Line of best fit" to a set of points such that the square differences between the estimated values and the real values is minimised but how do you go about finding the "point of best fit" and what does that point describe?
For example I have 3 lines:
$3x + 4y = 12$
$3x + 6y = 9$
$x + y = 2$
I want to find the point that intersects all 3 lines. This point clearly does not exist. So instead I want to find the point that is 'closest' to being a triple intersection. I imagine this could be called a 'point of best fit'. How would I go about finding this point?
2. ## Re: Point of best fit?
That's a very interesting problem. I have a couple of questions for you:
1. Does the "point of best fit" have to be on at least one of the lines? On at least two of the lines? Or could it be anywhere?
2. How are you defining "closest"? Since you only have two variables, x and y, in your equations, are you in just the xy plane? If so, are you measuring distance using Euclidean distance?
3. ## Re: Point of best fit?
I see a triangle fromed by the 3 lines and the closet point to the intersection being the middle of the triangle.
4. ## Re: Point of best fit?
Originally Posted by Ackbeet
That's a very interesting problem. I have a couple of questions for you:
1. Does the "point of best fit" have to be on at least one of the lines? On at least two of the lines? Or could it be anywhere?
2. How are you defining "closest"? Since you only have two variables, x and y, in your equations, are you in just the xy plane? If so, are you measuring distance using Euclidean distance?
1. The point can be anywhere at all.
2. Yes, this is only in 2-dimensions with variables x and y. Defining "closest" is the main problem I have with this. What would the point have to satisfy for it to be considered the "closest" to a triple intersection?
I have found a few set of points that satisfy different things: | {
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I have found a few set of points that satisfy different things:
1. $(1, 17/12)$ is the point such that it has the minimum squared distances from the point to the y coordinate of the lines (ie, its the point on the line of best fit that has the smallest error).
2. $(1, 1)$ is the point such that the summed distance (perpendicular distance) from each line to the point is minimised.
3. $(53/38, 43/38)$ is the point such that the summed squared distance (perpendicular distance) from each line to the point is minimised.
4. $(1, 11/6)$ is the centre of the triangle formed by the 3 lines.
5. ## Re: Point of best fit?
Originally Posted by pickslides
I see a triangle fromed by the 3 lines and the closet point to the intersection being the middle of the triangle.
Which of the myriad definitions of "middle" or "center" of a triangle do you mean? Incenter? Circumcenter? Orthocenter? Centroid? See below: I'm not sure you can pick one center over another until we know the context of the problem.
Originally Posted by Corpsecreate
1. The point can be anywhere at all.
2. Yes, this is only in 2-dimensions with variables x and y. Defining "closest" is the main problem I have with this. What would the point have to satisfy for it to be considered the "closest" to a triple intersection?
I have found a few set of points that satisfy different things: | {
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I have found a few set of points that satisfy different things:
1. $(1, 17/12)$ is the point such that it has the minimum squared distances from the point to the y coordinate of the lines (ie, its the point on the line of best fit that has the smallest error).
2. $(1, 1)$ is the point such that the summed distance (perpendicular distance) from each line to the point is minimised.
3. $(53/38, 43/38)$ is the point such that the summed squared distance (perpendicular distance) from each line to the point is minimised.
4. $(1, 11/6)$ is the centre of the triangle formed by the 3 lines.
It looks like you've found a number of solutions to your problem. I haven't checked your numbers, but I think that before you choose one solution over any of the others, you need to examine the context of your problem (which wouldn't be a bad idea to post here, actually). Is this problem the same problem, exactly, that you were given? Or have you performed a number of operations on the given problem to reduce it down to what you posted?
6. ## Re: Point of best fit?
I created the problem from a real world situation. The answer to the problem is purely out of interest (it was something that could have been useful to me earlier but not anymore) so it is of no 'real importance'. The problem is as follows:
Suppose you have 2 paints, paint x and paint y. All paints are composed of white paint with added tints of 3 different colours. Call these 3 colours A, B and C. Adding tints does not affect the volume of paint.
if paint x has the composition - A:3 B:3 C:1
and paint y has the composition - A:4 B:6 C:1
Then how much of paint x and y should you add together to yield as close as possible to a composition of A:12 B:9 C:2? | {
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This situation required you to solve the above linear equations however as we know, there is no solution so we are left with finding the 'closest' solution. I am quite capable of finding the 'best point' given that I know what I'm actually looking for. Now that you know the context of the problem, hopefully I could hear your opinion on what the properties of the 'best point' may be.
I'm leaning towards the shortest combined distances from the point to the lines simply because when there IS a solution to the linear equations, this distance is 0. In fact, thinking about it now, the point (1, 17/12) is certainly not what we're looking for since it must be on the line of best fit and that line wont necessarily pass through the triple intersection if there is one. All the other points are still possibilities though. | {
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X ] Theorem 3.6: Let F be any partition of the set S. Define a relation on S by x R y iff there is a set in F which contains both x and y. Non-equivalence may be written "a ≁ b" or " In general, if ∼ is an equivalence relation on a set X and x∈ X, the equivalence class of xconsists of all the elements of X which are equivalent to x. a Prove that the relation $$\sim$$ in Example 6.3.4 is indeed an equivalence relation. is the intersection of the equivalence relations on a { (c) $$[\{1,5\}] = \big\{ \{1\}, \{1,2\}, \{1,4\}, \{1,5\}, \{1,2,4\}, \{1,2,5\}, \{1,4,5\}, \{1,2,4,5\} \big\}$$. } ∣ The equivalence cl… In both cases, the cells of the partition of X are the equivalence classes of X by ~. From this we see that $$\{[0], [1], [2], [3]\}$$ is a partition of $$\mathbb{Z}$$. The equivalence classes of an equivalence relation can substitute for one another, but not individuals within a class. Thus, is an equivalence relation. Exercise $$\PageIndex{2}\label{ex:equivrel-02}$$. hands-on exercise $$\PageIndex{2}\label{he:samedec2}$$. b) find the equivalence classes for $$\sim$$. Therefore, \begin{aligned} R &=& \{ (1,1), (3,3), (2,2), (2,4), (2,5), (2,6), (4,2), (4,4), (4,5), (4,6), \\ & & \quad (5,2), (5,4), (5,5), (5,6), (6,2), (6,4), (6,5), (6,6) \}. . {\displaystyle x\sim y\iff f(x)=f(y)} For any $$i, j$$, either $$A_i=A_j$$ or $$A_i \cap A_j = \emptyset$$ by Lemma 6.3.2. , ~ is finer than ≈ if the partition created by ~ is a refinement of the partition created by ≈. A relation that is all three of reflexive, symmetric, and transitive, is called an equivalence relation. If $$R$$ is an equivalence relation on $$A$$, then $$a R b \rightarrow [a]=[b]$$. ) . This occurs, e.g. So, $$A \subseteq A_1 \cup A_2 \cup A_3 \cup ...$$ by definition of subset. { Define the relation $$\sim$$ on $$\mathscr{P}(S)$$ by \[X\sim Y \,\Leftrightarrow\, X\cap T = Y\cap T, Show that $$\sim$$ is an equivalence relation. In the example above, [a]=[b]=[e]=[f]={a,b,e,f}, while [c]=[d]={c,d} and | {
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is an equivalence relation. In the example above, [a]=[b]=[e]=[f]={a,b,e,f}, while [c]=[d]={c,d} and [g]=[h]={g,h}. X {\displaystyle \pi (x)=[x]} Then if ~ was an equivalence relation for ‘of the same age’, one equivalence class would be the set of all 2-year-olds, and another the set of all 5-year-olds. Example $$\PageIndex{3}\label{eg:sameLN}$$. X → c We have demonstrated both conditions for a collection of sets to be a partition and we can conclude , $$\exists x (x \in [a] \wedge x \in [b])$$ by definition of empty set & intersection. $$[S_4] = \{S_4,S_5,S_6\}$$ ] The equivalence relation is usually denoted by the symbol ~. Equivalence Class Testing, which is also known as Equivalence Class Partitioning (ECP) and Equivalence Partitioning, is an important software testing technique used by the team of testers for grouping and partitioning of the test input data, which is then used for the purpose of testing the software product into a number of different classes. If R (also denoted by ∼) is an equivalence relation on set A, then Every element a ∈ A is a member of the equivalence class [a]. {\displaystyle {a\mathop {R} b}} Let X be a set. that contain , the equivalence relation generated by [ ) a) $$m\sim n \,\Leftrightarrow\, |m-3|=|n-3|$$, b) $$m\sim n \,\Leftrightarrow\, m+n\mbox{ is even }$$. ) Transitive Two sets will be related by $$\sim$$ if they have the same number of elements. We saw this happen in the preview activities. , (Since { And so, $$A_1 \cup A_2 \cup A_3 \cup ...=A,$$ by the definition of equality of sets. It is, however, a, The relation "is approximately equal to" between real numbers, even if more precisely defined, is not an equivalence relation, because although reflexive and symmetric, it is not transitive, since multiple small changes can accumulate to become a big change. This relation turns out to be an equivalence relation, with each component forming an equivalence class. , aRa ∀ a∈A. Any relation ⊆ × which exhibits the properties of | {
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forming an equivalence class. , aRa ∀ a∈A. Any relation ⊆ × which exhibits the properties of reflexivity, symmetry and transitivity is called an equivalence relation on . All elements of X equivalent to each other are also elements of the same equivalence class. c We can refer to this set as "the equivalence class of $1$" - or if you prefer, "the equivalence class of $4$". a b Now we have that the equivalence relation is the one that comes from exercise 16. the equivalence classes of R form a partition of the set S. More interesting is the fact that the converse of this statement is true. . Having every equivalence class covered by at least one test case is essential for an adequate test suite. $$\exists i (x \in A_i \wedge y \in A_i)$$ and $$\exists j (y \in A_j \wedge z \in A_j)$$ by the definition of a relation induced by a partition. x See also invariant. The equivalence classes are $\{0,4\},\{1,3\},\{2\}$. a [ [x]R={y∈A∣xRy}. Let $$x \in A.$$ Since the union of the sets in the partition $$P=A,$$ $$x$$ must belong to at least one set in $$P.$$ ∣ That is, for all a, b and c in X: X together with the relation ~ is called a setoid. = Then: No equivalence class is empty. a } ) The advantages of regarding an equivalence relation as a special case of a groupoid include: The equivalence relations on any set X, when ordered by set inclusion, form a complete lattice, called Con X by convention. An equivalence class is defined as a subset of the form {x in X:xRa}, where a is an element of X and the notation "xRy" is used to mean that there is an equivalence relation between x and y. This equivalence relation is referred to as the equivalence relation induced by $$\cal P$$. Every number is equal to itself: for all … Each part below gives a partition of $$A=\{a,b,c,d,e,f,g\}$$. {\displaystyle [a]=\{x\in X\mid x\sim a\}} For example, an equivalence relation with exactly two infinite equivalence classes is an easy example of a theory which is ω-categorical, but not | {
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two infinite equivalence classes is an easy example of a theory which is ω-categorical, but not categorical for any larger cardinal number. Every equivalence relation induces a partitioning of the set P into what are called equivalence classes. , X Equivalence class definition is - a set for which an equivalence relation holds between every pair of elements. {\displaystyle a} {\displaystyle A} Since $$a R b$$, we also have $$b R a,$$ by symmetry. In this case $$[a] \cap [b]= \emptyset$$ or $$[a]=[b]$$ is true. An equivalence class is a subset whose elements are related to each other by an equivalence relation.The equivalence classes of a set under some relation form a partition of that set (i.e. Let us consider that R is a relation on the set of ordered pairs that are positive integers such that … {\displaystyle \{\{a\},\{b,c\}\}} [ Let X be a finite set with n elements. Since $$xRb, x \in[b],$$ by definition of equivalence classes. Let The power of the concept of equivalence class is that operations can be defined on the equivalence classes using representatives from each equivalence class. Given an equivalence relation $$R$$ on set $$A$$, if $$a,b \in A$$ then either $$[a] \cap [b]= \emptyset$$ or $$[a]=[b]$$, Let $$R$$ be an equivalence relation on set $$A$$ with $$a,b \in A.$$ , { Transcript. Equivalence relations are a ready source of examples or counterexamples. := An equivalence class is a complete set of equivalent elements. were given an equivalence relation and were asked to find the equivalence class of the or compare one to with respect to this equivalents relation. . is an equivalence relation, the intersection is nontrivial.). "Is equal to" on the set of numbers. , The equivalence classes of ~—also called the orbits of the action of H on G—are the right cosets of H in G. Interchanging a and b yields the left cosets. A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive. | {
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to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive. Hence, $\mathbb{Z} = [0] \cup [1] \cup [2] \cup [3].$ These four sets are pairwise disjoint. {\displaystyle X\times X} Each equivalence class consists of all the individuals with the same last name in the community. It can be shown that any two equivalence classes are either equal or disjoint, hence the collection of equivalence classes forms a partition of X. In this case $$[a] \cap [b]= \emptyset$$ or $$[a]=[b]$$ is true. Symmetric "Has the same absolute value" on the set of real numbers. ) Thus, $$\big \{[S_0], [S_2], [S_4] , [S_7] \big \}$$ is a partition of set $$S$$. ,[1] is defined as $$xRa$$ and $$xRb$$ by definition of equivalence classes. ~ is finer than ≈ if every equivalence class of ~ is a subset of an equivalence class of ≈, and thus every equivalence class of ≈ is a union of equivalence classes of ~. ( In the previous example, the suits are the equivalence classes. $$\therefore R$$ is transitive. For any x ∈ ℤ, x has the same parity as itself, so (x,x) ∈ R. 2. ⟺ to see this you should first check your relation is indeed an equivalence relation. Case 1: $$[a] \cap [b]= \emptyset$$ Define the relation $$\sim$$ on $$\mathbb{Q}$$ by $x\sim y \,\Leftrightarrow\, 2(x-y)\in\mathbb{Z}.$ $$\sim$$ is an equivalence relation. Find the equivalence relation (as a set of ordered pairs) on $$A$$ induced by each partition. ) \end{array}\], $\mathbb{Z} = [0] \cup [1] \cup [2] \cup [3].$, $a\sim b \,\Leftrightarrow\, \mbox{a and b have the same last name}.$, $x\sim y \,\Leftrightarrow\, x-y\in\mathbb{Z}.$, $\mathbb{R}^+ = \bigcup_{x\in(0,1]} [x],$, $R_3 = \{ (m,n) \mid m,n\in\mathbb{Z}^* \mbox{ and } mn > 0\}.$, $\displaylines{ S = \{ (1,1), (1,4), (2,2), (2,5), (2,6), (3,3), \hskip1in \cr (4,1), (4,4), (5,2), (5,5), (5,6), (6,2), (6,5), (6,6) \}. Exercise $$\PageIndex{4}\label{ex:equivrel-04}$$. , : The equivalence kernel of an injection is the identity relation. x A | {
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, : The equivalence kernel of an injection is the identity relation. x A strict partial order is irreflexive, transitive, and asymmetric. { in the character theory of finite groups. π , Define a relation $$\sim$$ on $$\mathbb{Z}$$ by \[a\sim b \,\Leftrightarrow\, a \mbox{ mod } 3 = b \mbox{ mod } 3.$ Find the equivalence classes of $$\sim$$. Now we have $$x R a\mbox{ and } aRb,$$ Read this as “the equivalence class of a consists of the set of all x in X such that a and x are related by ~ to each other”.. Consider the following relation on $$\{a,b,c,d,e\}$$: $\displaylines{ R = \{(a,a),(a,c),(a,e),(b,b),(b,d),(c,a),(c,c),(c,e), \cr (d,b),(d,d),(e,a),(e,c),(e,e)\}. , Define $$\sim$$ on $$\mathbb{R}^+$$ according to \[x\sim y \,\Leftrightarrow\, x-y\in\mathbb{Z}.$ Hence, two positive real numbers are related if and only if they have the same decimal parts. Notice an equivalence class is a set, so a collection of equivalence classes is a collection of sets. {\displaystyle [a]} a) True or false: $$\{1,2,4\}\sim\{1,4,5\}$$? Over $$\mathbb{Z}^*$$, define $R_3 = \{ (m,n) \mid m,n\in\mathbb{Z}^* \mbox{ and } mn > 0\}.$ It is not difficult to verify that $$R_3$$ is an equivalence relation. $$[S_2] = \{S_1,S_2,S_3\}$$ Much of mathematics is grounded in the study of equivalences, and order relations. We define a rational number to be an equivalence classes of elements of S, under the equivalence relation (a,b) ’ (c,d) ⇐⇒ ad = bc. which maps elements of X into their respective equivalence classes by ~. / ∀a ∈ A,a ∈ [a] Two elements a,b ∈ A are equivalent if and only if they belong to the same equivalence class. Practice: Congruence relation. $$[S_7] = \{S_7\}$$. ∼ The relation "≥" between real numbers is reflexive and transitive, but not symmetric. Example $$\PageIndex{6}\label{eg:equivrelat-06}$$. The equivalence classes are the sets \begin{array}{lclcr} {[0]} &=& \{n\in\mathbb{Z} \mid n\bmod 4 = 0 \} &=& 4\mathbb{Z}, \\ {[1]} &=& \{n\in\mathbb{Z} \mid n\bmod 4 = 1 \} &=& | {
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\mid n\bmod 4 = 0 \} &=& 4\mathbb{Z}, \\ {[1]} &=& \{n\in\mathbb{Z} \mid n\bmod 4 = 1 \} &=& 1+4\mathbb{Z}, \\ {[2]} &=& \{n\in\mathbb{Z} \mid n\bmod 4 = 2 \} &=& 2+4\mathbb{Z}, \\ {[3]} &=& \{n\in\mathbb{Z} \mid n\bmod 4 = 3 \} &=& 3+4\mathbb{Z}. , \end{aligned}, $X\sim Y \,\Leftrightarrow\, X\cap T = Y\cap T,$, $x\sim y \,\Leftrightarrow\, 2(x-y)\in\mathbb{Z}.$, $x\sim y \,\Leftrightarrow\, \frac{x-y}{2}\in\mathbb{Z}.$, $\displaylines{ R = \{(a,a),(a,c),(a,e),(b,b),(b,d),(c,a),(c,c),(c,e), \cr (d,b),(d,d),(e,a),(e,c),(e,e)\}. We have shown if $$x \in[a] \mbox{ then } x \in [b]$$, thus $$[a] \subseteq [b],$$ by definition of subset. × In sum, given an equivalence relation ~ over A, there exists a transformation group G over A whose orbits are the equivalence classes of A under ~. For other uses, see, Well-definedness under an equivalence relation, Equivalence class, quotient set, partition, Fundamental theorem of equivalence relations, Equivalence relations and mathematical logic, Rosen (2008), pp. x Have questions or comments? A binary relation ~ on a set X is said to be an equivalence relation, if and only if it is reflexive, symmetric and transitive. ". Let G be a set and let "~" denote an equivalence relation over G. Then we can form a groupoid representing this equivalence relation as follows. Thus, if we know one element in the group, we essentially know all its “relatives.”. The following relations are all equivalence relations: If ~ is an equivalence relation on X, and P(x) is a property of elements of X, such that whenever x ~ y, P(x) is true if P(y) is true, then the property P is said to be well-defined or a class invariant under the relation ~. Let G denote the set of bijective functions over A that preserve the partition structure of A: ∀x ∈ A ∀g ∈ G (g(x) ∈ [x]). A relation on a set $$A$$ is an equivalence relation if it is reflexive, symmetric, and transitive. b Exercise $$\PageIndex{8}\label{ex:equivrel-08}$$. If (x,y) ∈ R, x and y have the same | {
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transitive. b Exercise $$\PageIndex{8}\label{ex:equivrel-08}$$. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R. 3. Related thinking can be found in Rosen (2008: chpt. In a sense, if you know one member within an equivalence class, you also know all the other elements in the equivalence class because they are all related according to $$R$$. [9], Given any binary relation Here's a typical equivalence class for : A little thought shows that all the equivalence classes look like like one: All real numbers with the same "decimal part". Then the equivalence class of a denoted by [a] or {} is defined as the set of all those points of A which are related to a under the relation … If $$A$$ is a set with partition $$P=\{A_1,A_2,A_3,...\}$$ and $$R$$ is a relation induced by partition $$P,$$ then $$R$$ is an equivalence relation. Define $$\sim$$ on a set of individuals in a community according to \[a\sim b \,\Leftrightarrow\, \mbox{a and b have the same last name}.$ We can easily show that $$\sim$$ is an equivalence relation. Both $$x$$ and $$z$$ belong to the same set, so $$xRz$$ by the definition of a relation induced by a partition. X Find the ordered pairs for the relation $$R$$, induced by the partition. , After this find all the elements related to $0$. So, if $$a,b \in A$$ then either $$[a] \cap [b]= \emptyset$$ or $$[a]=[b].$$. For example 1. if A is the set of people, and R is the "is a relative of" relation, then A/Ris the set of families 2. if A is the set of hash tables, and R is the "has the same entries as" relation, then A/Ris the set of functions with a finite d… if $$R$$ is an equivalence relation on any non-empty set $$A$$, then the distinct set of equivalence classes of $$R$$ forms a partition of $$A$$. Now WMST $$\{A_1, A_2,A_3, ...\}$$ is pairwise disjoint. Two elements related by an equivalence relation are called equivalent under the equivalence relation. ] 10). Hence the three defining properties of equivalence relations can be proved mutually | {
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relation. ] 10). Hence the three defining properties of equivalence relations can be proved mutually independent by the following three examples: Properties definable in first-order logic that an equivalence relation may or may not possess include: Euclid's The Elements includes the following "Common Notion 1": Nowadays, the property described by Common Notion 1 is called Euclidean (replacing "equal" by "are in relation with"). Examples. / "Has the same birthday as" on the set of all people. Let $$x \in [b], \mbox{ then }xRb$$ by definition of equivalence class. ∈ Next we will show $$[b] \subseteq [a].$$ Thus $$x \in [x]$$. {\displaystyle A\subset X\times X} The former structure draws primarily on group theory and, to a lesser extent, on the theory of lattices, categories, and groupoids. (a) $$\mathcal{P}_1 = \big\{\{a,b\},\{c,d\},\{e,f\},\{g\}\big\}$$, (b) $$\mathcal{P}_2 = \big\{\{a,c,e,g\},\{b,d,f\}\big\}$$, (c) $$\mathcal{P}_3 = \big\{\{a,b,d,e,f\},\{c,g\}\big\}$$, (d) $$\mathcal{P}_4 = \big\{\{a,b,c,d,e,f,g\}\big\}$$, Exercise $$\PageIndex{11}\label{ex:equivrel-11}$$, Write out the relation, $$R$$ induced by the partition below on the set $$A=\{1,2,3,4,5,6\}.$$, $$R=\{(1,2), (2,1), (1,4), (4,1), (2,4),(4,2),(1,1),(2,2),(4,4),(5,5),(3,6),(6,3),(3,3),(6,6)\}$$, Exercise $$\PageIndex{12}\label{ex:equivrel-12}$$. An equivalence relation is a relation that is reflexive, symmetric, and transitive. The converse is also true: given a partition on set $$A$$, the relation "induced by the partition" is an equivalence relation (Theorem 6.3.4). An implication of model theory is that the properties defining a relation can be proved independent of each other (and hence necessary parts of the definition) if and only if, for each property, examples can be found of relations not satisfying the given property while satisfying all the other properties. × For example, 7 ≥ 5 does not imply that 5 ≥ 7. Each individual equivalence class consists of elements which are all equivalent to | {
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imply that 5 ≥ 7. Each individual equivalence class consists of elements which are all equivalent to each other. That is why one equivalence class is $\{1,4\}$ - because $1$ is equivalent to $4$. So, $$\{A_1, A_2,A_3, ...\}$$ is mutually disjoint by definition of mutually disjoint. x If X is a topological space, there is a natural way of transforming X/~ into a topological space; see quotient space for the details. Two integers will be related by $$\sim$$ if they have the same remainder after dividing by 4. Define three equivalence relations on the set of students in your discrete mathematics class different from the relations discussed in the text. $$[x]=A_i,$$ for some $$i$$ since $$[x]$$ is an equivalence class of $$R$$. } x ∈ The equivalence class of an element $$a$$ is denoted by $$\left[ a \right].$$ Thus, by definition, The canonical example of an equivalence relation individuals within a class the given set are to. May be written a ≢ b { \displaystyle a\not \equiv b } '' are three familiar properties reflexivity! Structure of order relations class \ ( \cal P\ ) class ) Advanced relation … equivalence differs... Muturally exclusive equivalence classes is a set of ordered pairs ( S\.... An adequate test suite equality too obvious to warrant explicit mention an important property of equivalence relations (... 1 $is equivalent to another given object prove Theorem 6.3.3, look at example 6.3.2 for example, elements! And join are elements of which are equivalent ( under that relation ) they are equivalent to other. Of groups of related objects as objects in themselves a playground real numbers for. Equal or disjoint and their union is X samedec } \ ) by definition of equality too obvious warrant! All a, b\in X } is an equivalence relation is a collection of subsets X. That every integer belongs to exactly one of these four sets to be an equivalence relation of order.! Moreover, the suits are the equivalence class onto itself, so \ ( \PageIndex 3. For an adequate | {
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Moreover, the suits are the equivalence class onto itself, so \ ( \PageIndex 3. For an adequate test suite 6.3.3 and Theorem 6.3.4 together are known as class!: equivrel-05 } \ ) from ~A to ~B to exactly one of these four sets equivalence is the image... X such that: 1 or just respects ~ '' or equivalence class in relation a ≁ b '' or a! Class [ X ] is called the universe or underlying set into disjoint classes... Such bijections are also elements of X is a collection of sets: equivrel-10 \! } ^ * = [ 1 ] \cup [ -1 ] \ ) is an equivalence relation { }... For this equivalence relation on a set that are related by an equivalence relation induced by the definition of classes. About the mathematical structure of order relations, a → a prove Theorem 6.3.3 ), which in! X R b\mbox { and } bRa, \ ( X ) ∈ R. 2,., \ { 2\ }$ - because $1$ is equivalent to each other {... Of \ ( xRb\ ), \ ) thus \ ( \PageIndex { 4 } \label ex. On \ ( [ ( a, b\in X } is an equivalence relation Encyclopedia of -. - because $1$ is equivalent to each other study of equivalences, asymmetric. A_1, A_2, A_3,... \ ) essential for an adequate test suite least. Order to prove Theorem 6.3.3, we will say that they are to! \Subseteq A_1 \cup A_2 \cup A_3 \cup... =A, \ ( A_1 \cup A_2 \cup \cup! That: 1 set of all children playing in a set and an! @ libretexts.org or check out our status page at https: //status.libretexts.org a A_1! Of numbers ≠ ϕ ) all equivalence relations other element in set \ ( [ ( a A_1. Relation, we also acknowledge previous National Science Foundation support under grant 1246120! In mathematics, an equivalence relation R is symmetric, i.e., aRb ⟹ bRa Transcript are of. Φ ) classes of an equivalence relation relation is equal to '' is the identity relation...,! The cells of the lattice theory captures the mathematical structure of order relations and join are elements of X to... { Z } \ ] this is an equivalence class of X the! Content is licensed by CC BY-NC-SA 3.0 already seen that | {
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\ ] this is an equivalence class of X the! Content is licensed by CC BY-NC-SA 3.0 already seen that and are relations. ( since X × X { \displaystyle a, b\in X } sameLN } \ ) for any \ T=\! ] \ ) thus \ ( \sim\ ) in example 6.3.4 is indeed an equivalence.! '' is meant a binary relation, this article was adapted from an original article by V.N 4 are by... Relation by studying its ordered pairs ( A\ ) is pairwise disjoint are equivalent ( under that ). Equivalence relation over some nonempty set a, b ) find the equivalence classes \in... Equivalence Partitioning P\ ) at least one test case is essential for adequate., a → a let '~ ' denote an equivalence relation induced by \ ( A\ ) us info. ] Confirm that \ ( [ a ] = [ b ] \ ) thus \ \sim\... Equivalence relations can construct new spaces by gluing things together., given partition... Euclid probably would have deemed the reflexivity of equality of real numbers: 1 be defined the... One another, but not symmetric ~ '' instead of invariant under ~ '' with ~ '' source. ( A.\ ) real numbers: 1 equivalence kernel of an equivalence relation ( as a set of all of. * = [ b ] \ ) by Lemma 6.3.1 each partition 6.3.3 ), \ aRb\. Of P are pairwise disjoint in that equivalence class definition is - set! ( under that relation ) thinking can be represented by any element in an equivalence relation two are equal. Euclidean and reflexive this is an equivalence relation by studying its ordered pairs on! B ], \ ( S\ ) P are pairwise disjoint and their union is X partitions of X ~. Essentially know all its “ relatives. ” ) find the ordered pairs for the patent doctrine see! Ex: equivrel-10 } \ ) is an equivalence class definition is - a set and be an relation. That the relation ≥ '' between real numbers is reflexive and.! Symbol ~ is not transitive if it is obvious that \ ( aRx\ by.. ) example 6.3.4 is indeed an equivalence relation as a set and be an equivalence class is a of. That every integer belongs to exactly one of these four sets | {
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and be an equivalence class is a of. That every integer belongs to exactly one of these four sets prove two lemmas A\ ) is disjoint. Arguments of the class [ X ] \ ) a commutative triangle relation called. B\In X } here are three familiar properties of reflexivity, and Euclid would! ) True or false: \ ( \mathbb { Z } ^ * = [ b,! '~ ' denote an equivalence relation is referred to as the Fundamental Theorem on equivalence relations can new. A ready source of examples or counterexamples with each component forming an equivalence relation by studying its ordered for! Contact us at info @ libretexts.org or check out our status page at https: //status.libretexts.org, y_1-x_1^2=y_2-x_2^2\.. Let a, b ∈ X { \displaystyle a\not \equiv b } '' and are equivalence.... Represented by any element in set \ ( \PageIndex { 10 } \label { ex: }., y_1-x_1^2=y_2-x_2^2\ ) on equivalence relation is referred to as the equivalence is the of! They belong to the same remainder after dividing by 4 are related to each other also \... That every integer belongs to exactly one of these equivalence relations over, equivalence is! ≥ 5 does not imply that 5 ≥ 7 the universe or underlying:... Other, if and only if they have the same birthday as '' on the set P what... That \ ( X ) ∈ R. 2 b\ ) to denote a relation is. Elements are related to $4$ natural bijection between the set of equivalent.. Reflexive, symmetric, and Keyi Smith all belong to the same as! Individual equivalence class of X such that: 1 created by equivalence class in relation each... Illustration of Theorem 6.3.3, look at example 6.3.2 out our status page at https:.. 6 } \label { ex: equivrelat-01 } \ ) all a, called representative... Substitute for one another, but not individuals within a class the lattice theory operations and. To $4$ aRb\ ) by the definition of equality of sets, i.e., aRb ⟹ bRa.! Bra Transcript 1, 2, 3 6.3.3, we leave it out ( \mathbb { }. All belong to the same equivalence class R b\mbox { and } bRa, \ ( \sim\ | {
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leave it out ( \mathbb { }. All belong to the same equivalence class R b\mbox { and } bRa, \ ( \sim\ if! Finite set with n elements x_2, y_2 ) \ ( R\ ) is equivalence. - 0.3942 in the group, we leave it out may regard equivalence classes is a of. X was the set of all angles are known as the Fundamental Theorem on equivalence relation by its! | {
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# Return evenly spaced numbers over a specified interval and set the number of samples to generate in Numpy
NumpyServer Side ProgrammingProgramming
To return evenly spaced numbers over a specified interval, use the numpy.linspace() method in Python Numpy. The 1st parameter is the "start" i.e. the start of the sequence. The 2nd parameter is the "end" i.e. the end of the sequence. The 3rd parameter is the num i.e. the number of samples to generate..
The stop is the end value of the sequence, unless endpoint is set to False. In that case, the sequence consists of all but the last of num + 1 evenly spaced samples, so that stop is excluded. Note that the step size changes when endpoint is False.
The dtype is the type of the output array. If dtype is not given, the data type is inferred from start and stop. The inferred dtype will never be an integer; float is chosen even if the arguments would produce an array of integers. The axis in the result to store the samples. Relevant only if start or stop are array-like. By default (0), the samples will be along a new axis inserted at the beginning. Use -1 to get an axis at the end.
## Steps
At first, import the required library −
import numpy as np
To return evenly spaced numbers over a specified interval, use the numpy.linspace() method in Python Numpy −
arr = np.linspace(100, 200, num = 10)
print("Array...\n", arr)
Get the array type −
print("\nType...\n", arr.dtype)
Get the dimensions of the Array −
print("\nDimensions...\n",arr.ndim)
Get the shape of the Array −
print("\nShape...\n",arr.shape)
Get the number of elements −
print("\nNumber of elements...\n",arr.size)
## Example
import numpy as np | {
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print("\nNumber of elements...\n",arr.size)
## Example
import numpy as np
# To return evenly spaced numbers over a specified interval, use the numpy.linspace() method in Python Numpy
# The 1st parameter is the "start" i.e. the start of the sequence
# The 2nd parameter is the "end" i.e. the end of the sequence
# The 3rd parameter is the num i.e the number of samples to generate. Default is 50.
arr = np.linspace(100, 200, num = 10)
print("Array...\n", arr)
# Get the array type
print("\nType...\n", arr.dtype)
# Get the dimensions of the Array
print("\nDimensions...\n",arr.ndim)
# Get the shape of the Array
print("\nShape...\n",arr.shape)
# Get the number of elements
print("\nNumber of elements...\n",arr.size)
## Output
Array...
[100. 111.11111111 122.22222222 133.33333333 144.44444444
155.55555556 166.66666667 177.77777778 188.88888889 200. ]
Type...
float64
Dimensions...
1
Shape...
(10,)
Number of elements...
10
Updated on 08-Feb-2022 07:00:57 | {
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# Proving Congruence Modulo
If $x$ is positive integer, prove that for all integers $a$, $(a+1)(a+2)\cdots(a+x)$ is congruent to $0\!\!\!\mod x$.
Any hints? What are the useful concepts that may help me solve this problem?
-
hint: you multiply $x$ consecutive integers so that one... – Raymond Manzoni Sep 1 '12 at 18:35
There are no "concepts", just original thought. Try to prove that one of the brackets is divisible by $x$ by considering what $a$ is mod $x$. – fretty Sep 1 '12 at 18:37
Hint: can you think of any reason why $x$ must divide one of the numbers $a,\ldots,a+x$? – shoda Sep 1 '12 at 18:39
So I can use factorial notation? – primemiss Sep 1 '12 at 18:43
so that one of them must be a multiple of $x$ (every $x$-th integer is a multiple of $x$). – Raymond Manzoni Sep 1 '12 at 18:55
Note that there is some $k$ such that $0\leq k< x$ and $a\equiv k\mod x$. Then $$a+(x-k)\equiv k+x-k\equiv x\equiv 0\mod x$$ and what do you get when you multiply this by other stuff?
-
One of the numbers $a+1, a+2,\ldots, a+x$ is congruent to $0$ mod $x$. Multiplying by $0$ yields $0$.
-
How to prove that one of the factors is congruent to 0 mod x? – primemiss Sep 1 '12 at 19:28
In any set of seven consecutive days, one will be a Tuesday. The reason is the same as that. If you have seven consecutive days, you get one of each day of the week, and if you have $x$ consecutive integers, then you've got one in every congruence class mod $x$. For example, consider $60,61,62,63,64,65,66$. Seven consecutive integers. They are congruent mod $7$ to $4,5,6,0,1,2,3$, respectively. – Michael Hardy Sep 1 '12 at 22:02
@primemiss: By the definition of the equivalency relation, there exists a $p$ and a $q$ with $0 \leq q < x$ such that $a = px + q$. That means that $a + (x - q) = (p + 1)x \equiv 0 \mod x$ with $0 < x - q \leq x$. – Niklas B. Sep 1 '12 at 23:06 | {
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Hint $\$ Any sequence of $\,n\,$ consecutive naturals has an element divisible by $\,n\,$. This has a simple proof by induction: shifting such a sequence by one does not change its set of remainders mod $\,n,\,$ since it effectively replaces the old least element $\:\color{#C00}a\:$ by the new greatest element $\:\color{#C00}{a+n}$
$$\begin{array}{}& \color{#C00}a, &\!\!\!\! a+1, &\!\!\!\! a+2, &\!\!\!\! \cdots, &\!\!\!\! a+n-1 & \\ \to & &\!\!\!\! a+1,&\!\!\!\! a+2, &\!\!\!\! \cdots, &\!\!\!\! a+n-1, &\!\!\!\! \color{#C00}{a+n} \end{array}\qquad$$
Since $\: \color{#C00}{a\,\equiv\, a\!+\!n}\pmod n,\:$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\ 0,1,2,\ldots,n-1\ =\:$ all possible remainders mod $\,n.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $\,n.$
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Note: Reversely, shifting by $-1$ proves it for sequences starting with negative integers. – Bill Dubuque Sep 1 '12 at 21:26
This has been already nicely answered, but here is another way to state an approach.
You will also be able to notice that you do not need the last term $(a + x)$ to have the relation you want.
What you hope to find is one of the factors divisible by $x$. How do do know you can find one?
Any of the factors (mod $x$) will be the equivalent of the remainder of $a$ when divided by $x$ plus the remainder of the added term when divided by $x$.
The remainder of $a$, call it r, will be $1\leq r< x$. And each of the added terms (other than $x$) will be equal to itself, that is one of the numbers $1$ thru $x - 1$.
So one of these sums (factors) will definitely exactly equal $x$.
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# pascal's triangle sum of nth diagonal row
today i was reading about pascal's triangle. the website pointed out that the 3th diagonal row were the triangular numbers. which can be easily expressed by the following formula.
$$\sum_{i=0}^n i = \frac{n(n+1)}{2}$$
i wondered if the following rows could be expressed with such a simple formula. when trying to find the sum for the 3th row i used a method called "differences" i found on this site: http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm
lets call $P_r$ the $r^{th}$ row of pascals triangle. The result for the 4th row was $$\sum_{i=0}^n P_3 = \frac{n(n+1)(n+2)}{6}$$ and the result for 4th row was $$\sum_{i=0}^n P_4 = \frac{n(n+1)(n+2)(n+3)}{24}$$ i guessed the sum of the 5th row would be $$\sum_{i=0}^n P_5 = \frac{n(n+1)(n+2)(n+3)(n+4)}{120}$$ i plotted the function and looking at the graph it seems to be correct. it looks like the the sum of each row is: $$\sum_{i=0}^n P_r = \frac{(n + 0)\cdots(n+(r-1))}{r!}$$
is this true for all rows? and why? i think this has something to do with combinatorics/probability which i never studied.
edit image for $P_r$: http://i.imgur.com/JlVC4q3.png
• What is $P_r$, already? – Did Aug 28 '15 at 20:48
• i meant sum of rth diagonal row , sorry bit confusing i will add image – Lethalbeast Aug 28 '15 at 20:50
• Then the relation ${i\choose j}+{i\choose j+1}={i+1\choose j+1}$ is all one needs to prove this by induction. – Did Aug 28 '15 at 21:08
So you basically want to prove that $$\binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+\dotsc+\binom{n+k}{n}=\binom{n+k+1}{n+1}$$ holds for all $n,k$, right? Of course you can prove this using induction and the formula $$\binom{n}{k}+\binom{n+1}{k}=\binom{n+1}{k+1}$$ as suggest by Did. There is a nice combinatorial interpretation of this using double-counting: Suppose you have $n+1$ eggs, $n$ of them blue and 1 red.
You want to choose $k+1$ of them which is the RHS: $\binom{n+1}{k+1}$ | {
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You want to choose $k+1$ of them which is the RHS: $\binom{n+1}{k+1}$
Either you choose the red one in which case you have $\binom{n}{k}$ possibilities for the remaining ones.
Either you don't choose the red one in which case you have $\binom{n}{k+1}$ possibilities for the remaining ones.
Can you think of a similar combinatorial argument which directly works for the original sum?
Hint: Think of $n+k+1$ balls in a row labelled $1,2,\dotsc,n+k+1$. You want to choose $n+1$ of them. That's the RHS: $\binom{n+k+1}{n+1}$. Now, distinguish cases about which is the rightmost ball you choose. If it's the ball number $n+k+1$ you have $\binom{n+k}{n}$ possibilities to choose the remaining $n$ balls. If it's the ball number $n+k$ you have $\binom{n+k-1}{n}$ possibilities etc. Can you complete it from here?
Just to be more explicit, here is a proof of: $$\binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{n+k}{n}=\binom{n+k+1}{n+1}$$I will use the property suggested by Did which is valid for general $i$ and $j$, $$\binom{i}{j}+\binom{i}{j+1}=\binom{i+1}{j+1}$$
Proof: \begin{align} \binom{n+k+1}{n+1} & =\binom{n+k}{n}+\binom{n+k}{n+1}\qquad\text{Using above property}\\ &=\binom{n+k}{n}+\left\{\binom{n+k-1}{n}+\binom{n+k-1}{n+1}\right\}\\ &=\binom{n+k}{n}+\binom{n+k-1}{n}+\left\{\binom{n+k-2}{n}+\binom{n+k-2}{n+1}\right\}\\ &=\binom{n+k}{n}+\binom{n+k-1}{n}+\binom{n+k-2}{n}+...+\left\{\binom{n+1}{n}+\binom{n+1}{n+1}\right\}\\ &=\binom{n+k}{n}+\binom{n+k-1}{n}+\binom{n+k-2}{n}+...\binom{n+1}{n}+\binom{n}{n} \end{align} Hence proved.
Try performing the following multiplication, then look up the Binomial Theorem.
$$\begin{array}{lr} &\binom{5}{5}x^5+\binom{5}{4}x^4+\binom{5}{3}x^3+\binom{5}{2}x^2+\binom{5}{1}x+\binom{5}{0}\\ \times&x+1\\ \hline \end{array}$$ | {
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Note that if the routine signum/f is not able to determine a value for signum(f()) then it can return FAIL. These are useful in defining functions that can be expressed with a single formula. Overview/Introduction-Functions-Graph of a Function-Classification of Functions-One-Valued and Many-Valued Functions-The Square Root-The Absolute Value Symbol-The Signum Function-Definition of a Limit-Theorems on Limits-Right-Hand and Left-Hand Limits-Continuity-Missing Point Discontinuities-Finite Jumps-Infinite Discontinuities 3. In general, there is no standard signum function in C/C++, and the lack of such a fundamental function tells you a lot about these languages. The signum function is the real valued function defined for real as follows. It is defined as u(t) = $\left\{\begin{matrix} 1 & t \geqslant 0\\ 0 & t. 0 \end{matrix}\right.$ It is used as best test signal. Definition. Thus, at x = 0, it is left undefined. Community ♦ 1 1 1 silver badge. signum function (plural signum functions) (mathematics) The function that determines the sign of a real number, yielding -1 if negative, +1 if positive, or otherwise zero. This video is unavailable. Apart from that, I believe both majority viewpoints about the right approach to define such a function are in a way correct, and the "controversy" about it is actually a non-argument once you take into account two important caveats: The second property implies that for real non-zero we have . Syntax. Unit Step Function. for the . Newbie; Posts: 17; Karma: 0 ; sgn / sign / signum function suggestions. For all real we have . Watch Queue Queue Similarly, . Serge Stroobandt Serge Stroobandt. Here, we should point out that the signum function is often defined simply as 1 for x > 0 and -1 for x < 0. Hypernyms . For all real we have . Put the code into a function and use inputdlg to allow inputting the numbers more easily. For a complex argument it is defined by. It is written in the form y … The signum function is defined as f(x) = |x|/x; | {
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it is defined by. It is written in the form y … The signum function is defined as f(x) = |x|/x; x≠0 = 0; x = 0 . For a simple, outgoing source, (21) e i 2 π q 0 x = cos 2 π q 0 x + i sin 2 π q 0 x sgn x, where q 0 = 1 / λ. Alternatively, you could simply let the user define the matrix X and use it as an input for the function. For a complex argument it is defined by. Contents. Yes the function is discontinuous which is right as per your argument. davidhbrown. The graph of a signum function is as shown in the figure given above. See for example . Pronunciations of proper names are generally given as pronounced as in the original language. The signum function of a real number x … Signum manipuli - Wikipedia "He carried a signum for a cohort or century." 1 Definition; 2 Properties; 3 Complex signum; 4 Generalized signum function; 5 See also; 6 Notes; Definition. I think the question wanted to convey this.. A Megametamathematical Guide. Solution for 1. Listen to the audio pronunciation in several English accents. In mathematics, the sign function or signum function (from signum, Latin for "sign") is an odd mathematical function that extracts the sign of a real number. Post by Stefan Ram »Returns the signum function of the argument«.. The signum function is the real valued function defined for real as follows. where denotes the magnitude (absolute value) of . Mathematics Pronunciation Guide. answered Sep 16 '18 at 14:26. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article. For instance, the value of function f(x) is equal to -5 in the interval [-5, -4). That is why, it is not continuous everywhere over R . of the . Information and translations of signum function in the most comprehensive dictionary definitions resource on the web. View a complete list of particular functions on this wiki Definition. The signum vector function is a function whose behavior in each | {
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functions on this wiki Definition. The signum vector function is a function whose behavior in each coordinate is as per the signum function. Definition of signum function in the Definitions.net dictionary. Signum Function . Is the »s« voiced or voiceless? It has a jumped discontinuity which means if the function is assigned some value at the point of discontinuity it cannot be made continuous. The signum function, denoted , is defined as follows: Note: In the definition given here, we define the value to Sign function (signum function) collapse all in page. Unit step function is denoted by u(t). I cannot find the American English (San Francisco) pronunciation of »signum« in a dictionary. Stefan Ram 2014-05-14 22:55:02 UTC . Example Problems. signum function pronunciation - How to properly say signum function. where denotes the magnitude (absolute value) of . (mathematics) A function that extracts the sign of a real number x, yielding -1 if x is negative, +1 if x is positive, or 0 if x is zero. What does signum function mean? A function cannot be continuous and discontinuous at the same point. Y = sign(x) returns an array Y the same size as x, where each element of Y is: 1 if the corresponding element of x is greater than 0. function; Translations . Impulse function is denoted by δ(t). »Returns the signum function of the argument«.. It's not uncommon to need to know the arithmetic sign of a value, typically represented as -1 for values < 0, +1 for values > 0, or 0. Permalink. Anthony, looking good. All the engineering examinations including IIT JEE and AIEEE study material is available online free of cost at askIITians.com. This function definition executes fast and yields guaranteed correct results for 0, 0.0, -0.0, -4 and 5 (see comments to other incorrect answers). Similarly, . A medieval tower bell used particularly for ringing the 8 canonical hours. The cosine transform of an odd function can be evaluated as a convolution with the Fourier transform of a signum | {
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of an odd function can be evaluated as a convolution with the Fourier transform of a signum function sgn(x). Some … Let’s Workout: Example 1: Find the greatest integer function for following (a) ⌊-261 ⌋ (b) ⌊ 3.501 ⌋ (c) ⌊-1.898⌋ Solution: According to the greatest integer function definition A statement function should appear before any executable statement in the program, but after any type declaration statements. The function’s value stays constant within an interval. for the Diacritically Challenged, This guide includes most mathematicians and mathematical terms that are encountered in high school and college. Any real number can be expressed as the product of its absolute value and its sign function: From equation (1) it follows that whenever x is not equal to 0 we have. German: Signumfunktion f; Hungarian: előjelfüggvény; Russian: зна́ковая фу́нкция f (znákovaja fúnkcija) Turkish: işaret fonksiyonu; Usage notes . Meaning of signum function. Signum definition is - something that marks or identifies or represents : sign, signature. Listen to the audio pronunciation of Signum Fidei on pronouncekiwi How To Pronounce Signum Fidei: Signum Fidei pronunciation + Definition Sign in to disable ALL ads. sign(x) Description. is called the signum function. Formal Definition of a Function Limit: The limit of f (x) f(x) f (x) as x x x approaches x 0 x_0 x 0 is L L L, i.e. Use the sequential criteria for limits, to show that the… Use the e-8 definition of the limit to show that x3 – 2x + 4 lim X-ix2 + 4x – 3 3 !3! »I« as in »it« or as in »seal«? Graphing functions by plotting points Definition A relation f from a set A to a set B is said to be a function if every input of a set A has only one output in a set B. If then also . example. A sinc pulse passes through zero at all positive and negative integers (i.e., t = ± 1, ± 2, …), but at time t = 0, it reaches its maximum of 1.This is a very desirable property in a pulse, as it helps to avoid intersymbol interference, a major | {
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is a very desirable property in a pulse, as it helps to avoid intersymbol interference, a major cause of degradation in digital transmission systems. Synonyms (bell): signum bell (math): signum function, sign function A statement function in Fortran is like a single line function definition in Basic. In mathematical expressions the sign function is often represented as sgn. Area under unit step function is unity. Note that the procedure signum/f is passed only the argument(s) to the function f. The value of the environment variable is set within signum, and can be checked by signum/f, but it is not passed as a parameter. Sign function - Wikipedia "The signum manipuli (Latin for 'standard' of the maniple, Latin: manipulus) was a standard for both the centuriae and the legion." signum (plural signums or signa) A sign, mark, or symbol. Definition, domain, range, solution of Signum function. Definition. Note that zero (0) is neither positive nor negative. The precise definition of the limit is discussed in the wiki Epsilon-Delta Definition of a Limit. Of course it is continuous at every x € R except at x = 0 . »U« as in »soon« or »number« or just a schwa? The second property implies that for real non-zero we have . No, it is not continuous every where . "The signum function is the derivative of the absolute value function, up to (but not including) the indeterminacy at zero." The format is simple - just type f(x,y,z,…) = formula . Suppose is a positive integer. SIGN(x) returns a number that indicates the sign x: -1 if x is negative; 0 if x equals 0; or 1 if x is positive. In applications such as the Laplace transform this definition is adequate, since the value of a function at a single point does not change the analysis. If then also . Unit Impulse Function. Therefore, its Fourier transform is (22) F q = 1 2 δ q + q 0 + δ q − q 0 − i 2 π q − q 0 + i 2 π q + q 0. Explicitly, it is defined as the function: In other words, the signum function project a non-zero complex | {
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it is defined as the function: In other words, the signum function project a non-zero complex number to the unit circle . share | improve this answer | follow | edited Jun 20 at 9:12. Topic: sgn / sign / signum function suggestions (Read 2558 times) previous topic - next topic. Jul 19, 2019, 05:35 pm. Signum function is an integer valued function defined over R . Proper American English Pronunciation of Words and Names. A sinc function is an even function with unity area. function. Signifer - Wikipedia. The signum function of a real number x is defined as follows: Properties. In other words, the signum function project a non-zero complex number to the unit circle . Study Physics, Chemistry and Mathematics at askIITians website and be a winner. Chemistry and Mathematics at askIITians website and be a winner » it or... At x = 0, it is continuous at every x € R except at x = 0 numbers... Properties ; 3 complex signum ; 4 Generalized signum function project a non-zero complex number to unit. Improve this answer | follow | edited Jun 20 at 9:12 function ’ value., Chemistry and Mathematics at askIITians website and be a winner sinc function is often represented as.... Continuous everywhere over R form y … Definition simple - just type f ( x, y, z …. Suggestions ( Read 2558 times ) previous topic - next topic for a cohort or.... Any executable statement in the program, but after any type declaration.... Use the e-8 Definition of a real number x … Anthony, looking.... Signums or signa ) a sign, mark, or symbol function ; See! X = 0 sinc function is an integer valued function defined for real non-zero we have e-8 Definition the... Is discussed in the form y … Definition format is simple - just type f ( x is! Show that x3 – 2x + 4 lim X-ix2 + 4x – 3 3! 3! 3!!... Unity area other words, the signum function suggestions ( Read 2558 )... Statement in the form y … Definition shown in the wiki Epsilon-Delta Definition of a real number x Anthony... Francisco ) pronunciation of » | {
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in the wiki Epsilon-Delta Definition of a real number x Anthony... Francisco ) pronunciation of » signum « in a dictionary line function Definition in Basic stays within! Medieval tower bell used particularly for ringing the 8 canonical hours as per the signum function of a.... Particular functions on this wiki Definition unity area: sgn / sign / signum function.! Pronunciation - How to properly say signum function signum function pronunciation a function and it! In high school and college the audio pronunciation in several English accents ( function... | improve this answer | follow | edited Jun 20 at 9:12 to show that –. Sign / signum function is a function can not be continuous and discontinuous the... Continuous and discontinuous at the same point of cost at askIITians.com use to. Includes most mathematicians and signum function pronunciation terms that are encountered in high school and college on the web words the! Number to the audio pronunciation in several English accents is as shown in the program, after... Definition in Basic ( San Francisco ) pronunciation of » signum « in a dictionary and discontinuous the! X≠0 = 0, it is continuous at every x € R except x... ) previous topic - next topic real number x … Anthony, looking good continuous everywhere R! ; 5 See also ; 6 Notes ; Definition range, solution of signum function Fortran!: Properties per the signum function pronunciation - How to properly say signum function suggestions numbers easily... Complex signum ; 4 Generalized signum function of a real number x … Anthony, looking good by (. Is as shown in the program, but after any type declaration.. The user define the matrix x and use inputdlg to allow inputting the numbers more easily translations signum. Function Definition in Basic pronounced as in » soon « or as in the interval [ -5, )! ( plural signums or signa ) a sign, mark, or symbol each is. In Fortran is like a single line function Definition in Basic functions that be... And translations of | {
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# norm
Vector and matrix norms
## Description
example
n = norm(v) returns the Euclidean norm of vector v. This norm is also called the 2-norm, vector magnitude, or Euclidean length.
example
n = norm(v,p) returns the generalized vector p-norm.
example
n = norm(X) returns the 2-norm or maximum singular value of matrix X, which is approximately max(svd(X)).
n = norm(X,p) returns the p-norm of matrix X, where p is 1, 2, or Inf:
example
n = norm(X,"fro") returns the Frobenius norm of matrix or array X.
## Examples
collapse all
Create a vector and calculate the magnitude.
v = [1 -2 3];
n = norm(v)
n = 3.7417
Calculate the 1-norm of a vector, which is the sum of the element magnitudes.
v = [-2 3 -1];
n = norm(v,1)
n = 6
Calculate the distance between two points as the norm of the difference between the vector elements.
Create two vectors representing the (x,y) coordinates for two points on the Euclidean plane.
a = [0 3];
b = [-2 1];
Use norm to calculate the distance between the points.
d = norm(b-a)
d = 2.8284
Geometrically, the distance between the points is equal to the magnitude of the vector that extends from one point to the other.
$\begin{array}{l}a=0\underset{}{\overset{ˆ}{i}}+3\underset{}{\overset{ˆ}{j}}\\ b=-2\underset{}{\overset{ˆ}{i}}+1\underset{}{\overset{ˆ}{j}}\\ \\ \begin{array}{rl}{d}_{\left(a,b\right)}& =||b-a||\\ & =\sqrt{\left(-2-0{\right)}^{2}+\left(1-3{\right)}^{2}}\\ & =\sqrt{8}\end{array}\end{array}$
Calculate the 2-norm of a matrix, which is the largest singular value.
X = [2 0 1;-1 1 0;-3 3 0];
n = norm(X)
n = 4.7234
Calculate the Frobenius norm of a 4-D array X, which is equivalent to the 2-norm of the column vector X(:).
X = rand(3,4,4,3);
n = norm(X,"fro")
n = 7.1247
The Frobenius norm is also useful for sparse matrices because norm(X,2) does not support sparse X.
## Input Arguments
collapse all
Input vector.
Data Types: single | double
Complex Number Support: Yes | {
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collapse all
Input vector.
Data Types: single | double
Complex Number Support: Yes
Input array, specified as a matrix or array. For most norm types, X must be a matrix. However, for Frobenius norm calculations, X can be an array.
Data Types: single | double
Complex Number Support: Yes
Norm type, specified as 2 (default), a positive real scalar, Inf, or -Inf. The valid values of p and what they return depend on whether the first input to norm is a matrix or vector, as shown in the table.
Note
This table does not reflect the actual algorithms used in calculations.
pMatrixVector
1max(sum(abs(X)))sum(abs(v))
2 max(svd(X))sum(abs(v).^2)^(1/2)
Positive, real-valued numeric scalarsum(abs(v).^p)^(1/p)
Infmax(sum(abs(X')))max(abs(v))
-Infmin(abs(v))
## Output Arguments
collapse all
Norm value, returned as a scalar. The norm gives a measure of the magnitude of the elements. By convention:
• norm returns NaN if the input contains NaN values.
• The norm of an empty matrix is zero.
collapse all
### Euclidean Norm
The Euclidean norm (also called the vector magnitude, Euclidean length, or 2-norm) of a vector v with N elements is defined by
$‖v‖=\sqrt{\sum _{k=1}^{N}{|{v}_{k}|}^{2}}\text{\hspace{0.17em}}.$
### General Vector Norm
The general definition for the p-norm of a vector v that has N elements is
${‖v‖}_{p}={\left[\sum _{k=1}^{N}{|{v}_{k}|}^{p}\right]}^{\text{\hspace{0.17em}}1/p}\text{\hspace{0.17em}},$
where p is any positive real value, Inf, or -Inf.
• If p = 1, then the resulting 1-norm is the sum of the absolute values of the vector elements.
• If p = 2, then the resulting 2-norm gives the vector magnitude or Euclidean length of the vector.
• If p = Inf, then ${‖v‖}_{\infty }={\mathrm{max}}_{i}\left(|v\left(i\right)|\right)$.
• If p = -Inf, then ${‖v‖}_{-\infty }={\mathrm{min}}_{i}\left(|v\left(i\right)|\right)$.
### Maximum Absolute Column Sum
The maximum absolute column sum of an m-by-n matrix X (with m,n >= 2) is defined by | {
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The maximum absolute column sum of an m-by-n matrix X (with m,n >= 2) is defined by
${‖X‖}_{1}=\underset{1\le j\le n}{\mathrm{max}}\left(\sum _{i=1}^{m}|{a}_{ij}|\right).$
### Maximum Absolute Row Sum
The maximum absolute row sum of an m-by-n matrix X (with m,n >= 2) is defined by
${‖X‖}_{\infty }=\underset{1\le i\le m}{\mathrm{max}}\left(\sum _{j=1}^{n}|{a}_{ij}|\right)\text{\hspace{0.17em}}.$
### Frobenius Norm
The Frobenius norm of an m-by-n matrix X (with m,n >= 2) is defined by
${‖X‖}_{F}=\sqrt{\sum _{i=1}^{m}\sum _{j=1}^{n}{|{a}_{ij}|}^{2}}=\sqrt{\text{trace}\left({X}^{†}X\right)}\text{\hspace{0.17em}}.$
This definition also extends naturally to arrays with more than two dimensions. For example, if X is an N-D array of size m-by-n-by-p-by-...-by-q, then the Frobenius norm is
${‖X‖}_{F}=\sqrt{\sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}...\sum _{w=1}^{q}{|{a}_{ijk...w}|}^{2}}.$
## Tips
• Use vecnorm to treat a matrix or array as a collection of vectors and calculate the norm along a specified dimension. For example, vecnorm can calculate the norm of each column in a matrix.
## Version History
Introduced before R2006a
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# Pascals triangle
Binomial coefficients, pascal's triangle, and the power set binomial coefficients pascal's triangle the power set binomial coefficients an expression like (a + b) . Pascal's triangle is a number pattern introduced by the famous french mathematician and philosopher blaise pascal it is triangle in shape with the top being. Mathematicians have long been familiar with the tidy way in which the nth row of pascal's triangle sums to 2n (the top row conventionally labeled as n = 0. Where (n r) is a binomial coefficient the triangle was studied by b pascal, although it had been described centuries earlier by chinese mathematician yanghui. Pascal's triangle, a simple yet complex mathematical construct, hides some surprising properties related to number theory and probability.
Sal introduces pascal's triangle, and shows how we can use it to figure out the coefficients in binomial expansions. Applications pascal's triangle is not only an interesting mathematical work because of its hidden patterns, but it is also interesting because of its wide expanse. All you ever want to and need to know about pascal's triangle.
Pascal triangle: given numrows, generate the first numrows of pascal's triangle pascal's triangle : to generate a[c] in row r, sum up a'[c] and a'[c-1] from. Given numrows, generate the first numrows of pascal's triangle for example, given numrows = 5, the result should be: , , , , ] java. In mathematics, pascal's triangle is a triangular array of the binomial coefficients in much of the western world, it is named after the french mathematician blaise. The pascal's triangle is a graphical device used to predict the ratio of heights of lines in a split nmr peak. From the wikipedia page the kata author cites: the rows of pascal's triangle ( sequence a007318 in oeis) are conventionally enumerated starting with row n = 0. | {
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A guide to understanding binomial theorem, pascal's triangle and expanding binomial series and sequences. This special triangular number arrangement is named after blaise pascal pascal was a french mathematician who lived during the seventeenth century. When you look at pascal's triangle, find the prime numbers that are the first number in the row that prime number is a divisor of every number in that row.
## Pascals triangle
Resources for the patterns in pascal's triangle problem pascal's triangle at provides information on. Yes, pascal's triangle and the binomial theorem isn't particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong. The counting function c(n,k) and the concept of bijection coalesce in one of the most studied mathematical concepts, pascal's triangle at its heart, pascal's. Pascal's triangle is one of the classic example taught to engineering students it has many interpretations one of the famous one is its use with binomial.
• Now that we've learned how to draw pascal's famous triangle and use the numbers in its rows to easily calculate probabilities when tossing coins, it's time to dig.
• There are many interesting things about polynomials whose coefficients are taken from slices of pascal's triangle (these are a form of what's called chebyshev.
• S northshieldsums across pascal's triangle modulo 2 j raaba generalization of the connection between the fibonacci sequence and pascal's triangle. | {
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It's not necessary to do this because 2 only shows up once in pascal's triangle but you get the idea t2 = table[binomial[n, k], {n, 0, 8}, {k, 0, n}] / {a_, 2, c_} : {a . Pascal's triangle and binomial coefficients information for the mathcamp 2017 qualifying quiz 1 pascal's triangle pascal's triangle (named for the. Theorem the sum of all the entries in the n th row of pascal's triangle is equal to 2 n proof 1 by definition, the entries in n th row of pascal's. Pascal's triangle is defined such that the number in row \$n\$ and column \$k\$ is \$ {n\choose k}\$ for this reason, convention holds that both row numbers and.
Pascals triangle
Rated 4/5 based on 31 review
2018. | {
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Is there a transformation or a proof for these integrals?
Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.
Question. Is this true? If so, is there an underlying transformation or just a proof? $$\int_0^{\infty}x^2e^{-x^2}\frac{{dx}}{\cosh\sqrt{\pi}x} =\frac14\int_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$
• The integral in this question was taken from your own paper that was written 10 years earlier "A dozen integrals: Russell-style" and you knew the proof. arxiv.org/abs/0808.2692
– Nemo
Aug 9, 2021 at 18:15
• @user44191 Whatever Stack Exchange policy might be, and independently of this particular case (on which I make no comment), it's (1) certainly unusual on MO to ask questions to which you already know the answer, and (2) abusive of other people's time to do this while giving the impression that you don't know the answer. If I spent time solving someone's problem and then typing it up, on the understanding that I was helping them out, I might feel quite angry to discover that they knew the answer all along. I don't use MO as much as I used to, but is there really any doubt over this? Aug 10, 2021 at 13:32
• I can’t understand why downvoting this question. It seems to me interesting and perfectly suitable for MO. Not mentioning an existing proof is a defect, of course, yet I explain it to me with the purpose of not influencing who approaches the problem, since apparently the goal was to get new proofs, insights and connections. Aug 10, 2021 at 14:31
• Besides, after $10$ years it is perfectly possible to forget to have ever proven that particular integral identity. Aug 10, 2021 at 17:17
• @Nemo, making an unnecessary edit, just so you can vote a question down? Really? Aug 11, 2021 at 13:09
This is a generous explanation of Lucia's comment above. | {
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This is a generous explanation of Lucia's comment above.
The functions $$\mathcal H_n(x)=\frac{2^{1/4}}{(2^n n!)^{1/2}}H_n(\sqrt{2\pi}\; x) e^{-\pi x^2}$$ form an orthonormal system in $L^2(\textbf{R})$. Here $H_n(x)$ are the usual Hermite polynomials defined by $$e^{2xz-z^2}=\sum_{n=0}^\infty \frac{H_n(x)}{n!}z^n,\qquad |z|<\infty.$$ The functions are eigenfunctions for the usual Fourier transform so that $$\int_{-\infty}^{+\infty}\mathcal H_n(t)e^{-2\pi i x t}\,dt=(-i)^n \mathcal H_n(x).$$
It follows that $L^2(\textbf{R})$ is a direct sum of four subspaces. In each of these subspaces the Fourier transform is just multiplication by $1$, $i$, $-1$, $-1$ respectively.
It is well known that the function $1/\cosh\pi x$ is invariant by the Fourier transform $$\int_{-\infty}^{+\infty}\frac{e^{-2\pi i x\xi}}{\cosh\pi x}\,dx= \frac{1}{\cosh\pi \xi}.$$ Therefore this function is in the span of the functions $\mathcal H_{4n}(x)$, and therefore is orthogonal to any other eigenfunction. In particular $$\int_{-\infty}^{+\infty}\frac{\mathcal H_2(x)}{\cosh\pi x}\,dx=0.$$ Since $H_2(x)=-2+4x^2$ it follows that $$\int_{-\infty}^{+\infty}\frac{(8\pi x^2-2)e^{-\pi x^2}}{\cosh\pi x}\,dx=0.$$ Putting $x/\sqrt{\pi}$ instead of x $$\int_{-\infty}^{+\infty}\frac{(8 x^2-2)e^{-x^2}}{\cosh\sqrt{\pi} x}\,\frac{dx}{\sqrt{\pi}}=0.$$ This is equivalent to your equation.
UPDATE
The integral in T. Amdeberhan's question was taken from his own paper that was written 10 years earlier before he posted his question: https://arxiv.org/abs/0808.2692 A dozen integrals: Russell-style. Thus it seems that T. Amdeberhan knew the answer to the question he asked. Why did he ask it then?
I provide a screenshot below for the reader's convenience (integral number $$7$$):
At the end of the paper, the authors provide a sketch of proof: | {
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At the end of the paper, the authors provide a sketch of proof:
Note that the following functions are self-reciprocal $$\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$$ (see this MO post for a list of such functions): $$\frac{1}{\cosh\sqrt{\frac{\pi}{2}}x},\quad e^{-x^2/2}.\tag{1}$$ It was proved by Hardy (Quarterly Journal Of Pure And Applied Mathematics, Volume 35, Page 203, 1903) and Ramanujan (Ramanujan's Lost Notebook, part IV, chapter 18) that for two self-reciprocal functions $$f$$ and $$g$$ we have $$\int_0^\infty f(x)g(\alpha x) dx=\frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx.$$ The formal agument is as follows but it can be made rigorous for certain type of functions $$\int_0^\infty f(x)g(\alpha x) dx=\int_0^\infty f(x)dx\cdot \sqrt{\frac{2}{\pi}}\int_0^\infty g(y) \cos(\alpha x y)dy=\\ \int_0^\infty g(y)dy \cdot \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos(\alpha x y)dx=\int_0^\infty g(y)f(\alpha y) dy=\\ \frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx.$$ For functions in $$(1)$$ which decay rapidly at $$x\to\pm\infty$$ it is certainly true. So we obtain an identity due to Hardy and Ramanujan $$\int_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\alpha{x}}} {dx} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\frac{x}{\alpha}}}{dx}.$$ After some simplifications it becomes $$\int_{0}^{\infty} \frac{e^{-\alpha^2x^{2}}}{\cosh{\sqrt{{\pi}}{x}}} {dx} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{e^{-x^{2}/\alpha^2}}{\cosh{\sqrt{{\pi}}{x}}}{dx}.$$ To complete the proof differentiate this with respect to $$\alpha$$ and then set $$\alpha=1$$. | {
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## Simpson's Rule
If we are given odd number of tabular points,i.e. is even, then we can divide the given integral of integration in even number of sub-intervals Note that for each of these sub-intervals, we have the three tabular points and so the integrand is replaced with a quadratic interpolating polynomial. Thus using the formula (13.3.3), we get,
In view of this, we have
which gives the second quadrature formula as follows:
(13.3.5)
This is known as SIMPSON'S RULE.
Remark 13.3.3 An estimate for the error in numerical integration using the Simpson's rule is given by
(13.3.6)
where is the average value of the forth forward differences.
EXAMPLE 13.3.4 Using the table for the values of as is given in Example 13.3.2, compute the integral by Simpson's rule. Also estimate the error in its calculation and compare it with the error using Trapezoidal rule.
Solution: Here, thus we have odd number of nodal points. Further,
and
Thus,
To find the error estimates, we consider the forward difference table, which is given below:
0.0 1.00000 0.01005 0.02071 0.00189 0.00149 0.1 1.01005 0.03076 0.02260 0.00338 0.00171 0.2 1.04081 0.05336 0.02598 0.00519 0.00243 0.3 1.09417 0.07934 0.03117 0.00762 0.00320 0.4 1.17351 0.11051 0.3879 0.01090 0.00459 0.5 1.28402 0.14930 0.04969 0.01549 0.00658 0.6 1.43332 0.19899 0.06518 0.02207 0.00964 0.7 1.63231 0.26417 0.08725 0.03171 0.8 1.89648 0.35142 0.11896 0.9 2.24790 0.47038 1.0 2.71828
Thus, error due to Trapezoidal rule is,
Similarly, error due to Simpson's rule is,
It shows that the error in numerical integration is much less by using Simpson's rule.
EXAMPLE 13.3.5 Compute the integral , where the table for the values of is given below:
0.05 0.1 0.15 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.0785 0.1564 0.2334 0.309 0.454 0.5878 0.7071 0.809 0.891 0.9511 0.9877 1 | {
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Solution: Note that here the points are not given to be equidistant, so as such we can not use any of the above two formulae. However, we notice that the tabular points and are equidistant and so are the tabular points and . Now we can divide the interval in two subinterval: and ; thus,
. The integrals then can be evaluated in each interval. We observe that the second set has odd number of points. Thus, the first integral is evaluated by using Trapezoidal rule and the second one by Simpson's rule (of course, one could have used Trapezoidal rule in both the subintervals).
For the first integral and for the second one . Thus,
which gives,
It may be mentioned here that in the above integral, and that the value of the integral is . It will be interesting for the reader to compute the two integrals using Trapezoidal rule and compare the values.
EXERCISE 13.3.6
1. Using Trapezoidal rule, compute the integral where the table for the values of is given below. Also find an error estimate for the computed value.
1. a=1 2 3 4 5 6 7 8 9 b=10 0.09531 0.18232 0.26236 0.33647 0.40546 0.47 0.53063 0.58779 0.64185 0.69314
2. a=1.50 1.55 1.6 1.65 1.7 1.75 b=1.80 0.40546 0.43825 0.47 0.5077 0.53063 0.55962 0.58779
3. a = 1.0 1.5 2 2.5 3 b = 3.5 1.1752 2.1293 3.6269 6.0502 10.0179 16.5426
2. Using Simpson's rule, compute the integral Also get an error estimate of the computed integral.
1. Use the table given in Exercise 13.3.6.1b.
2. a = 0.5 1 1.5 2 2.5 3 b = 3.5 0.493 0.946 1.325 1.605 1.778 1.849 1.833
3. Compute the integral , where the table for the values of is given below:
0 0.5 0.7 0.9 1.1 1.2 1.3 1.4 1.5 0 0.39 0.77 1.27 1.9 2.26 2.65 3.07 3.53
A K Lal 2007-09-12 | {
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Determinant of a matrix with $t$ in all off-diagonal entries.
It seems from playing around with small values of $n$ that
$$\det \left( \begin{array}{ccccc} -1 & t & t & \dots & t\\ t & -1 & t & \dots & t\\ t & t & -1 & \dots & t\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{array}\right) = (-1)^{n-1}(t+1)^{n-1}((n-1)t-1)$$
where $n$ is the size of the matrix.
How would one approach deriving (or at least proving) this formally?
Motivation
This came up when someone asked what is the general solution to:
$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b},$$
and for non-trivial solutions, the matrix above (with $n=3$) must be singular. In this case either $t=-1\implies a+b+c=1$ or $t=\frac{1}{2}\implies a=b=c$.
So I wanted to ensure that these are also the only solutions for the case with more variables.
-
Did you try induction? – zarathustra Aug 14 '14 at 17:08
Induction combined with this should do it. – Daniel R Aug 14 '14 at 17:09
Induction is the way to go, but first, subtract the first row from all the other rows to get a much easier calculation. – fixedp Aug 14 '14 at 17:10
When $n=2$ seems to fail. $1-t^2$ RHS, $-2 t^3-3 t^2+1$ LHS, right? – carlosayam Aug 14 '14 at 17:14
The matrix size should be $n+1$; @caya gives the $n=2$ case. – Semiclassical Aug 14 '14 at 17:17 | {
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Using elementary operations instead of induction is key. \begin{align} &\begin{vmatrix} -1 & t & t & \dots & t\\ t & -1 & t & \dots & t\\ t & t & -1 & \dots & t\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{vmatrix}\\ &= \begin{vmatrix} -t-1 & 0 & 0 & \dots & t+1\\ 0 & -t-1 & 0 & \dots & t+1\\ 0 & 0 & -t-1 & \dots & t+1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{vmatrix}\\ &= \begin{vmatrix} -t-1 & 0 & 0 & \dots & 0\\ 0 & -t-1 & 0 & \dots & 0\\ 0 & 0 & -t-1 & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& (n - 1)t -1 \end{vmatrix}\\ &= (-1)^{n - 1}(t + 1)^{n - 1}((n - 1)t - 1) \end{align}
-
For clarity, first he is doing "Row N = Row N - Last Row" then he is doing "Last Row = Last Row - Row N" – DanielV Aug 14 '14 at 17:46
You can write the expression as $$\det(t C - (t+1)I)$$ where $C = \mathbf{1}\mathbf{1}^T$ is the matrix of all $1$'s, formed by the column of ones times its transpose. Using the identity $\det(I+cr) = 1+rc$, you can first factor out $(t+1)$: $$\det(t C - (t+1)I) = (-1)^n(t+1)^n \det\left(I - \frac{t}{t+1} \mathbf{1}\right) = (-1)^n(t+1)^n \left(1 - \frac{nt}{t+1}\right)$$
-
That's brilliant, nothing could be more elegant. :) – MGA Aug 14 '14 at 17:37
+1, but what do you mean by $cr$ and $rc$? Are these row and column vectors, abbreviated $r$ and $c$, respectively? (My assumption) – apnorton Aug 15 '14 at 3:44
Correct, they are row and column vectors. I stole that from the Wiki page on Determinant – Victor Liu Aug 15 '14 at 21:42
Note that your matrix is the sum of the matrix $T$ with all entries equal to $t$ and the matrix $-(1+t)I$. Therefore the determinant you are asking about is the value at $X=-(1+t)$ of the characteristic polynomial $\chi_{-T}$ of $-T$. | {
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Since $T$ has rank (at most) $1$, its eigenspace for eigenvalue $0$ has dimension $n-1$, so the characteristic polynomial of $-T$ is $\chi_{-T}=X^{n-1}(X+nt)$ (the final factor must be that because the coefficient of $X^{n-1}$ in $\chi_{-T}$ is $\def\tr{\operatorname{tr}}-\tr(-T)=nt$). Now setting $X=-(1+t)$ gives $$\det(-(1+t)I-T)=(-1-t)^{n-1}(-1-t+nt)=(-1-t)^{n-1}(-1+(n-1)t)$$ as desired.
This kind of question is recurrent on this site; see for instance Determinant of a specially structured matrix and How to calculate the following determinants (all ones, minus $I$).
-
Here is another characterization of the result. For convenience, I will take the matrix size as $n+1$ rather than $n$. First, note that we may factor $t$ from each of the $n+1$ rows, and so the determinant may be written as $$\det{[t(M-t^{-1} I_{n+1})]} =t^{n+1} \det(M-t^{-1} I_{n+1})$$ where $(M)_{ij}=1-\delta_{ij}$ for $1\leq i,j\leq n+1$.
Next, observe that $M$ has $n$ independent eigenvectors of the form $\hat{e}_i-\hat{e}_{n+1}$ ($1\leq i\leq n$), each with eigenvalue $-1$; in addition, $M$ also has the eigenvector $\sum_{i=1}^{n+1}\hat{e}_i$ with eigenvalue $n$. Consequently the characteristic polynomial of $M$ in powers of $t^{-1}$ is $$\det(M-t^{-1} I_{n+1})=(-1-t^{-1})^n (n-t^{-1}) = t^{-n-1}\cdot (-1)^n (1+t)^n (nt+1)$$ and so the prior result yields the desired identity.
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# Variation of Least Squares with Symmetric Positive Semi Definite (PSD) Constraint
I am trying to solve the following convex optimization problem: \begin{align} & \min_{W} && \sum_{i=1}^n (\mathbf{x}_{i}^TW\mathbf{x}_{i} - y_i)^2 \\\\ & s.t. && W \succcurlyeq 0 \\\\ & && W = W^T \end{align}
where $$\mathbf{x}_i \in \mathbb{R^p}$$, $$W \in \mathbb{R}^{p \times p}$$ and $$y_i \geq 0$$.
Without the positive semidefinite constraint, the problem is pretty straightforward. Requiring positive semidefiniteness, however, makes it a bit tricky.
I thought about using the fact that $$W \succcurlyeq 0$$ if and only if there exists a symmetric $$A$$ such that $$W = AA^T$$, and solving the equivalent problem
\begin{align} & \min_{A} && \sum_{i=1}^n (\mathbf{x}_{i}^TAA^T\mathbf{x}_{i} - y_i)^2 \\\\ &s.t. && A = A^T \end{align}
Letting $$a_{ij}$$ be the $$(i,j)th$$ element of A, this optimization function is quartic (fourth-order) with respect to the $$a_{ij}$$'s. Because of this, I am unsure of how to proceed.
I would be grateful if someone could point me in the right direction as to how to solve this problem.
• Haven't worked it through, but you might find it helpful to note that, if $W = A^\top A$, then$$x^\top_i W x_i = x^\top_i A^\top A x_i = (Ax_i)^\top (Ax_i) = \|Ax_i\|^2.$$ Apr 11 '20 at 0:01
When dealing with such form of matrix multiplications always remember the Vectorization Trick with Kronecker Product for Matrix Equations:
$${x}_{i}^{T} W {x}_{i} - {y}_{i} \Rightarrow \left({x}_{i}^{T} \otimes {x}_{i}^{T} \right) \operatorname{Vec} \left( W \right) - \operatorname{Vec} \left( {y}_{i} \right) = \left({x}_{i}^{T} \otimes {x}_{i}^{T} \right) \operatorname{Vec} \left( W \right) - {y}_{i}$$
Since the problem is given by summing over $${x}_{i}$$ one could build the matrix:
$$X = \begin{bmatrix} {x}_{1}^{T} \otimes {x}_{1}^{T} \\ {x}_{2}^{T} \otimes {x}_{2}^{T} \\ \vdots \\ {x}_{n}^{T} \otimes {x}_{n}^{T} \end{bmatrix}$$
Then: | {
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Then:
$$\arg \min_{W} \sum_{i = 1}^{n} {\left( {x}_{i}^{T} W {x}_{i} - {y}_{i} \right)}^{2} = \arg \min_{W} {\left\| X \operatorname{Vec} \left( W \right) - \boldsymbol{y} \right\|}_{2}^{2}$$
Where $$\boldsymbol{y}$$ is the column vector composed by $${y}_{i}$$.
Now the above has nice form of regular Least Squares. The handling of the constraint can be done using Projected Gradient Descent Method. The projection onto the set of Symmetric Matrices and Positive Semi Definite (PSD) Matrices cone are given by:
1. $$\operatorname{Proj}_{\mathcal{S}^{n}} \left( A \right) = \frac{1}{2} \left( A + {A}^{T} \right)$$. See Orthogonal Projection of a Matrix onto the Set of Symmetric Matrices.
2. $$\operatorname{Proj}_{\mathcal{S}_{+}^{n}} \left( A \right) = Q {\Lambda}_{+} {Q}^{T}$$ where $$A = Q \Lambda {Q}^{T}$$ is the eigen decomposition of $$A$$ and $${\Lambda}_{+}$$ means we zero any negative values in $$\Lambda$$. See Find the Matrix Projection of a Symmetric Matrix onto the set of Symmetric Positive Semi Definite (PSD) Matrices.
Since both the Symmetric Matrices Set and PSD Cone are Linear Sub Space hence even the greedy iterative projection on the set will yield an orthogonal projection on the intersection of the 2 sets. See Orthogonal Projection onto the Intersection of Convex Sets.
So, with all the tools above one could create his own solver using basic tools with no need for external libraries (Which might be slow or not scale).
I implemented the Projected Gradient Descent Method with the above projections in MATLAB. I compared results to CVX to validate the solution. This is the solution:
My implementation is vanilla Gradient Descent with constant Step Size and no acceleration. If you add those you'll see convergence which is order of magnitude faster (I guess few tens of iterations). Not bad for hand made solver.
The MATLAB Code is accessible in my StackExchange Mathematics Q3619669 GitHub Repository. | {
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The MATLAB Code is accessible in my StackExchange Mathematics Q3619669 GitHub Repository.
• This was a very clear and practical response. Thanks! Assuming I understand correctly, I have one question: what is the difference between iteratively projecting vs. projecting just once after finding the optimal solution to the unconstrained problem? Would they give different solutions? Apr 13 '20 at 15:43
• @tygaking, Please mark it as answer and +1. Regarding your question, you can't do that. Think of "Gradient Descent" as some kind of projection by itself. Hence what we do above is basically iterative projection onto iterative set, one for the gradient descent, one for each other.
– Royi
Apr 13 '20 at 15:45
Are you doing FGLS or something?
You could try substituting the constraint into the object. For the two by two case, for example, solve $$\sum_{i = 1}^N \left(x_i'\left[\array{ \array{w_{11} & w_{12}} \\ \array{w_{12}& w_{22} }} \right]x_i - y_i \right)^2$$ where $$w_{12} = w_{21}$$ now by construction. Then the matrix will be symmetric.
To ensure positive semi-definiteness, you can then use the standard principal minors test: $$w_{ii} \ge 0$$ for each $$i$$, $$w_{11} w_{22} - w_{12}^2 \ge 0$$, and so on, with the determinant of the upper left-hand minor weakly positive.
That at least subsumes positive semi-definiteness into concrete constraints and adjustments to the problem. That sounds like a nightmare to solve however, using Kuhn-Tucker. A simpler sufficient condition for semi-definiteness is the dominant diagonal condition, that $$w_{ii} \ge \sum_{j \neq i} |w_{ij}|$$ for each row $$i$$, which would be much more computationally tractable. Perhaps it could give you a good initial guess before you try relaxing it to the standard principal minors constraints. | {
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• There is no need for this. While your approach might lead to problem formulated as Least Squares with Linear Equality and Inequality Constraints, it will also require some kind of iterative solver (As there is no closed form solution for LS with Inequality Constraints). Since we have the projection onto the Symmetric and PSD Matrices we can use it. See my answer. Still, nice idea of yours! Especially with the Diagonally Dominant Matrix. I might post a question and answer about this.
– Royi
Apr 12 '20 at 0:00
This is a convex optimization problem which can be easily formulated, and then numerically solved via a convex optimization tool, such as CVX, YALMIP, CVXPY, CVXR. This is a linear Semidefinite Programming problem (SDP), for which numerical solvers exist.
Here is the code for CVX.
Assume $$x_i$$ is the ith column of matrix $$X$$
cvx_begin
variable W(p,p) semidefinite % constrains W to be symmetric positive semidefinite
Objective = 0;
for i=1:n
Objective = Objective + square(X(:,i)'*W*X(:,i) - y(i))
end
minimize(Objective)
cvx_end
CVX will transform the problem into the form required by the solver, call the solver, and return the solution.
• I think the use of square_pos() here is wrong. It will lead to the wrong solution. You can use square() or if you want to be sure square_abs() or something like that.
– Royi
Apr 11 '20 at 23:35
• @Royi Right you are. Now fixed. Thanks for pointing it out. Do you want to take over for me as moderator at CVX Forum? Apr 11 '20 at 23:43
• Care to explain what you meant? Thanks...
– Royi
Apr 18 '20 at 22:43
• @Royi You can take over answering many of the questions there, and I can go into semi-retirement from that forum. Apr 18 '20 at 23:02
• I would answer you with a PM on CVX Forum. Yet the PM feature is disabled. Want to move it to an email?
– Royi
Apr 22 '20 at 15:08 | {
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This does seem to be a straight forward 'nonnegative' least squares though with the sdp constraint. n = 10; p = 5; X = zeros(n,p^2); for ii = 1:n x = randn(p,1); temp = xx'; X(ii,:) = temp(:)'; end y = randn(n,1); cvx_begin sdp variable W(p,p) semidefinite minimize(norm(XW(:)-y)) cvx_end | {
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# Deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)
I am revising for a number theory exam and have a question that I am struggling with, any help would be greatly appreciated.
First I am asked to show that for an odd number $x$, $x^2+2 ≡3$(mod 4).
I can do this part of the question, but next I am asked to deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)
I am struggling to see how to attempt the second part and how the first part relates.
My thoughts so far are that I want to show $x^2≡-2$(mod p) ? And perhaps Fermat's Little Theorem could be of use here somehow?
Not sure if I'm barking up the wrong tree though.
Consider the prime factorization of $x^2+2$. Since $x$ is odd, $x^2+2$ is odd implying $2$ will not show up in the factorization. Now consider the primes that DO show up in the prime factorization. If they are all $1$ in modulo 4, then their product will also be one in modulo $4$. This is not true though, since you know that $x^2+2$ is $3$ modulo $4$. Therefore, there must be a prime that in the prime factorization of $x^2+2$ s.t. it is not $1$ modulo 4. Since primes other than 2 that are not $1$ modulo 4 are $3$ modulo $4$, this completes the reasoning.
• Fantastic! Thank you so much – Olivia77989 May 15 '13 at 20:53
Our number $x^2+2$ is odd, and greater than $1$, so it is a product of odd not necessarily distinct primes. If all these primes were congruent to $1$ modulo $4$, their product would be congruent to $1$ modulo $4$. But you have shown that $x^2+2\equiv 3\pmod{4}$. | {
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• You are welcome. You probably already ran into the fact that every positive integer of the form $4k+3$ has a prime divisor of the form $4k+3$. That fact plays a key role in the standard proof that there are infinitely many primes of the form $4k+3$. – André Nicolas May 15 '13 at 20:59
• A useful insight! So would I be right in thinking that in order to prove there are infinitely any primes of the form $4k +3$ , I would split $4k +3$ into its prime divisors $p1,p2...pn$ noting that at least one of these divisors is of the form $4k +3$, then, following a similar approach to Euclid's proof, consider N=4(p1. p2.....pn)+3 ? Or what would we equate N to and why? – Olivia77989 May 15 '13 at 21:31
• Yes, suppose there are finitely many primes $p_1,\dots,p_s$ of the right shape. Consider $N=4p_1\cdots p_s-1$. This is of the shape $4k+3$, so has a prime divisor of that shape, which must be different from the $p_i$. I am avoiding $N=4p_1\cdots p_s+3$, since maybe $N$ could be a power of $3$ (there are other ways around that issue). – André Nicolas May 15 '13 at 21:38
• I see. Possibly a silly question, but how do we know that the prime divisor of N in the form $4k+3$ is different from the $p_i$ already in our list? – Olivia77989 May 15 '13 at 21:46
• With $N=4p_1\cdots p_s -1$, if it was in list, it would divide $4p_1\cdots p_s$. Since it divides $4p_1\cdots p_s-1$, it would divide the difference, which is $1$. Impossible! – André Nicolas May 15 '13 at 21:56
The question, as asked, has been answered. Nevertheless, I'll show how quadratic residues can help in problems like these, and show that moreover $x^2+2$ ($x$ odd) has a prime factor congruent to $3 \!\!\!\mod 8$: | {
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Suppose that $x$ is odd. If $p$ divides $x^2+2$, then $p$ is odd and $x^2 \equiv -2 \!\!\mod p$. In particular, $-2$ is a square mod $p$, hence $$1=\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/2} (-1)^{(p-1)/2}=(-1)^{(p^2+p-2)/2}.$$ This implies $p \equiv 1,3 \!\!\mod \!8$. If each prime $p$ dividing $x^2+2$ were congruent to $1 \!\!\mod 8$, then $x^2+2$ would be congruent to $1\!\! \mod 8$. This is a contradiction, hence there exists some $p \mid x^2+2$ congruent to $3 \!\!\mod 8$. (In fact, there must be an odd number of such primes.)
Of course, any one of the primes is congruent to $3 \!\!\mod 4$, which gives your result.
We can ignore quadratic residues in your particular problem because there are only two odd residues mod $4$. In more difficult questions, looking at quadratic residues can give more information. | {
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# Why are the last two digits of a perfect square never both odd?
Earlier today, I took a test with a question related to the last two digits of perfect squares.
I wrote out all of these digits pairs up to $20^2$.
I noticed an interesting property, and when I got home I wrote a script to test it. Sure enough, my program failed before it was able to find a square where the last two digits are both odd.
Why is this?
Is this always true, or is the rule broken at incredibly large values?
• do you know about modular arithmetic ? that might be a starting place. – user451844 Aug 19 '17 at 0:51
• The last two digit of a number is the number modulus $100$ Talking about squares the last two digits are cyclical. They repeat every 50 squares from $0$ to $49$ or from $10^9 + 2017$ to $10^9 + 2017 + 49$ they are always the following $00,\;01,\;04,\;09,\;16,\;25,\;36,\;49,\;64,\;81,\;00,\;21,\;44,\;69,\;96,\;25,\;56,\;89,\;24,\;61,\;00,\;41,\;84,\;29,\;76,\;25,\;76,\;29,\;84,\;41,\;00,\;61,\;24,\;89,\;56,\;25,\;96,\;69,\;44,\;21,\;00,\;81,\;64,\;49,\;36,\;25,\;16,\;09,\;04,\;01$ and there is never a combination of two odd digits. – Raffaele Aug 19 '17 at 14:39
• Not only, but if a number does not end with one of the following pair of digits it cannot be a perfect square $00,\; 01,\; 04,\; 09,\; 16,\; 21,\; 24,\; 25,\; 29,\; 36,\; 41,\; 44,\; \\ 49,\; 56,\; 61,\; 64,\; 69,\; 76,\; 81,\; 84,\; 89,\; 96$ – Raffaele Aug 19 '17 at 14:39
• You can also note that the last two digits of squares from 0 to 25 are the same as from 50 to 25 so it is both cyclical and symetrical :) – Rafalon Aug 20 '17 at 9:21
• @Raffaele Sadly if you'd made that into an answer it probably would have earned you a decent chunk of rep and might have been the accepted answer. – Pharap Aug 20 '17 at 16:24 | {
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Taking the last two digits of a number is equivalent to taking the number $\bmod 100$. You can write a large number as $100a+10b+c$ where $b$ and $c$ are the last two digits and $a$ is everything else. Then $(100a+10b+c)^2=10000a^2+2000ab+200ac+100b^2+20bc+c^2$. The first four terms all have a factor $100$ and cannot contribute to the last two digits of the square. The term $20bc$ can only contribute an even number to the tens place, so cannot change the result. To have the last digit of the square odd we must have $c$ odd. We then only have to look at the squares of the odd digits to see if we can find one that squares to two odd digits. If we check the five of them, none do and we are done.
• Ah, it's so obvious in hindsight. Although, I suppose most problems like this are. Thanks! – Pavel Aug 19 '17 at 1:00
Others have commented on the trial method. Just to note that $3^2$ in base $8$ is $11_8$ which has two odd digits. This is an example to show that the observation here is not a trivial one.
But we can also note that $(2m+1)^2=8\cdot \frac {m(m+1)}2+1=8n+1$ so an odd square leaves remainder $1$ when divided by $8$.
The final odd digits of squares can be $1,5,9$ so odd squares are $10p+4r+1$ with $r=0,1,2$. $10p+4r$ must be divisible by $8$ and hence by $4$, so $p$ must be even.
In the spirit of experimentation, the last two digits of the squares of numbers obtained by adding the column header to the row header: | {
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$$\begin {array}{c|ccc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline 0 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\ 10 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\ 20 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\ 30 & 00 & 61 & 24 & 89 & 56 & 25 & 96 & 69 & 44 & 21\\ 40 & 00 & 81 & 64 & 49 & 36 & 25 & 16 & 09 & 04 & 01\\ 50 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\ 60 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\ 70 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\ 80 & 00 & 61 & 24 & 89 & 56 & 25 & 96 & 69 & 44 & 21\\ 90 & 00 & 81 & 64 & 49 & 36 & 25 & 16 & 09 & 04 & 01\\ 100 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\ 110 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\ 120 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\ \end{array}$$
The patterns are clear, after which the search for a reason for such patterns is well given by the answer of @RossMillikan - you can see that the parity of both final digits of the square is entirely dependent on the final digit of the number that you square.
As a hint, consider what determines the last two digits of a multiplication. Do you remember doing multiplication by hand? If you have square a ten digit number, do all the digits matter when considering just the last two digits of the answer? You will realize that you can put a bound on the number of squares you need to check before you can prove the assertion you are making for all n | {
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• In fact more than enough values of n has been checked (within $20^2$) to see that this result is true. Question is, how does one know enough values have been checked? – Jihoon Kang Aug 19 '17 at 1:02
• That's right - a lazy bound would be checking with a computer all two digit squares, as you know that in a 3 digit number, the hundreds digit will not impact on the final two digits in the multiplication (the same for larger numbers). Obviously you can get sharper than that, as the other answer showed. I was just pointing out that you can very easily come up with a lazy bound by thinking about what happens when you multiply numbers. – Franz Aug 19 '17 at 1:23
This is just another version of Ross Millikan's answer.
Let $N \equiv 10x+n \pmod{100}$ where n is an odd digit.
\begin{align} (10x + 1)^2 \equiv 10(2x+0)+1 \pmod{100} \\ (10x + 3)^2 \equiv 10(6x+0)+9 \pmod{100} \\ (10x + 5)^2 \equiv 10(0x+2)+5 \pmod{100} \\ (10x + 7)^2 \equiv 10(4x+4)+9 \pmod{100} \\ (10x + 9)^2 \equiv 10(8x+8)+1 \pmod{100} \\ \end{align}
A simple explanation.
1. Squaring means multiplication, multiplication means repeatative additions.
2. Now if you add even no.s for odd no. of times or odd no.s for even no. of times you will always get an even no.
Hence, square of all the even no.s are even, means the last digit is always even.
1. If you add odd no.s for odd no. of times you will always get an odd no.
Coming to the squares of odd no.s whose results are >= 2 digits. Starting from 5^2 = 25, break it as 5+5+5+5+5, we have a group with even no. of 5 and one extra 5. According to my point no. 2 the even group will always give you a even no. i.e. 20, means the last digit is always even. Addition of another 5 with 20 makes it 25, 2 is even.
Taking 7^2, 7+7+7+7+7+7+7, group of six 7's = 42 plus another 7 = 49.
Now consider 9^2, 9+9+9+9+9+9+9+9+9, group of eight 9's = 72 plus another 9 = 81, (72+9 gets a carry of 1 making the 2nd last digit even) | {
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35^2 = group of twenty four 34's (1190) plus 35 = 1225, carry comes.
In short just check the last digit of no. that you can think of in the no. co-ordinate (Real and Imaginary) it will always be b/w 0-9 so the basic principle (point 2 and 3) will never change. Either the last digit will be an even or the 2nd last digit will become even with a carry. So the 1 digit sq can come odd, 1 and 9, as there is no carry. I have kept it as an exception in point 3.
BTW many, including the author may not like my lengthy explanation as mine is not a mathematical one, full of tough formulae. Sorry for that. I'm not from mathematical background and never like maths.
• "35^2 = group of twenty four 34's (1190) plus 35 = 1225, carry comes."? One of the burdens of being a mathematician is to present your arguments and then deal with any criticism that follows. It's got nothing to do with you. Whoever you are, if you publish a turkey, someones going to roast it. – steven gregory Aug 20 '17 at 23:13
• That's a typo. That will be odd+odd = even. I saw that error but didn't edit it. – NewBee Aug 23 '17 at 14:26
Let b be last digit of odd perfect square a,then b can be 1,9 or 5. For b=1,9; $a^2-b$ is divisible by 4, $(a^2-b)/10$ is even. For b=5 ;a always ends in 25. | {
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# A curious property of an acute triangle
Many years back in high school I happened to stumble upon the following property that seems to hold for any acute triangle:
$CD$ and $BE$ are altitudes, the arcs are semicircles with diameters $AB$ and $AC$ respectively.
The property is that $AF = AG$
Proof:
Let $H$ be the midpoint of $AB$ (and the centre of the respective semicircle) $$AG^2 = AD^2 + GD^2 = \left(AC\cdot \cos\angle A\right)^2 + GD^2$$ Since $HG = AH = \frac{AB}{2}$ is the radius of the semicircle $$GD^2 = HG^2 - HD^2 = \left(\frac{AB}{2}\right)^2 - \left(\frac{AB}{2} - AC\cdot\cos\angle A\right)^2 = \\ = AB\cdot AC\cdot \cos\angle A - \left(AC\cdot\cos\angle A\right)^2$$
which gives $$AG^2 = AB\cdot AC\cdot \cos\angle A$$
Analogously ($I$ is the midpoint of $AC$) $$AF^2 = AE^2 + FE^2 = \left(AB\cdot \cos\angle A\right)^2 + FE^2$$ $$FE^2 = FI^2 - EI^2 = \left(\frac{AC}{2}\right)^2 - \left(AB\cdot \cos\angle A - \frac{AC}{2}\right)^2 = \\ = AC\cdot AB\cdot \cos\angle A - \left(AB\cdot \cos\angle A\right)^2$$ which finally gives $$AF^2 = AC\cdot AB\cdot \cos\angle A$$
Questions:
1. Is this a known property?
2. Is there a better more elegant proof?
• You can say something more: The four points where the (full) circles with diameters $\overline{AB}$ and $\overline{AC}$ meet the altitudes from $B$ and $C$, lie on a common circle about $A$.
– Blue
Apr 3 '18 at 16:49
• @Blue, that's a very, very good one! Thank you. Apr 3 '18 at 21:54
HINT:
$$AF^2 = AE \cdot AC\\ AG^2 = AD \cdot AB \\ AE \cdot AC = AD \cdot AB$$
• what a beautiful proof! Apr 3 '18 at 16:42
• Brilliant indeed. Although not so trivial (to me at least). Thanks a lot. Apr 3 '18 at 21:41
• @ageorge: My pleasure! A nice problem indeed! Apr 3 '18 at 22:32
This is a possibly similar proof, but here goes anyway...
Let the triangle be labelled in the usual way, so $AC=b$ and $AB=c$, and let angle $FAC=\theta$ | {
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Let the triangle be labelled in the usual way, so $AC=b$ and $AB=c$, and let angle $FAC=\theta$
Then $$AF=b\cos\theta$$ $$\implies FE=AF\sin\theta=b\sin\theta\cos\theta$$ $$\implies EC=FE\tan\theta=b\sin^2\theta$$ But $$EC=b-c\cos A=b\sin^2\theta$$ $$\implies b\cos^2\theta=c\cos A$$ $$\implies AF=\sqrt{bc \cos A}$$
To get $AG$ we only need to exchange $b$ and $c$, so the result follows. | {
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# Math Help - Trouble Simplifying/Factorising
1. ## Trouble Simplifying/Factorising
I have a rather straightforward Physics problem to solve, only I can't for the life of me simplify the answer to what it needs to be. What am I doing wrong?
$
v_1=\frac{m}{m+M}v_o
$
$\frac{1}{2}mv_o^2=\frac{1}{2}mv_1^2+\frac{1}{2}Mv_ 1^2+\frac{1}{2}kd^2$
The answer is in the form $d=$ and it's a simplification of the information I provided here in the post.
I get as far as $d=\sqrt{\frac{1}{k}(mv_o^2-(m+M)v_1^2)}$
The book plugs in $v_1$
and simplifies to $d=\sqrt{\frac{mM}{k(m+M)}}\times{v_o}$
How?
2. Originally Posted by dkaksl
$
v_1=\frac{m}{m+M}v_o
$
$\frac{1}{2}mv_o^2=\frac{1}{2}mv_1^2+\frac{1}{2}Mv_ 1^2+\frac{1}{2}kd^2$
Changing your v1 to v and v0 to w, then your 2 equations are:
v = mw / (m + M) [1]
mw^2 = mv^2 + Mv^2 + kd^2 [2]
Simplifying:
m + M = mw / v [1]
mw^2 - kd^2 = v^2(m + M) [2]
Substitute [1] in [2] ; OK?
3. Very useful, easy-to-understand solution. Thanks a lot.
4. Hello again.
Just got to checking the textbook and unfortunately the question was to define $d$ in the terms $m, M, k$ and $v_o$.
Cancelling out $(m + M)$ got me an answer with $v_1$.
Edit:
$kd^2=mv_o^2-(m+M)v_1^2$
if $v_1=\frac{mv_o}{m+M}$ then what is $v_1^2$?
$\frac{mv_o}{m+M}\times\frac{mv_o}{m+M}$ which is
$\frac{m^2v_o^2}{(m+M)^2}$
$kd^2=mv_o^2-(m+M)\times\frac{m^2v_o^2}{(m+M)^2}$
$kd^2=mv_o^2-\frac{m^2v_o^2}{(m+M)}$
$kd^2=\frac{mv_o^2(m+M)}{(m+M)}-\frac{m^2v_o^2}{(m+M)}$
$kd^2=\frac{mv_o^2(m+M)-m^2v_o^2}{(m+M)}$
$kd^2=\frac{m^2v_o^2+mMv_o^2-m^2v_o^2}{(m+M)}=\frac{mM}{(m+M)}\times{v_o^2}$
$d=\sqrt{\frac{1}{k}\times{\frac{mM}{(m+M)}}}\times {v_o}$
Last edit: Thanks a lot. I see where I made a mistake in expanding. Learned a lot today. Merry Christmas.
5. Originally Posted by dkaksl
Just got to checking the textbook and unfortunately the question was to define $d$ in the terms $m, M, k$ and $v_o$.
Ahhh; well then, we need to get rid of your v1 (my v); the 2 equations: | {
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v = mw / (m + M) [1]
mw^2 = mv^2 + Mv^2 + kd^2 [2]
square [1]: v^2 = m^2w^2 / (m + M)^2
rearrange [2]: v^2 = (mw^2 - kd^2) / (m + M)
m^2w^2 / (m + M)^2 = (mw^2 - kd^2) / (m + M)
m^2w^2 / (m + M) = mw^2 - kd^2
m^2w^2 = mw^2(m + M) - kd^2(m + M)
kd^2(m + M) = mw^2(m + M) - m^2w^2
kd^2(m + M) = mw^2(m + M - m)
kd^2(m + M) = mMw^2
d^2 = mMw^2 / (k(m + M))
d = wSQRT[mM / (k(m + M))] ........Merry Xmas! | {
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# What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$?
What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$ in terms of $d$? Does the limit exist? Is there a simple upper bound interms of $d$?
-
I think the limit sould exist and is a function of $d$. Because the denominator is $\sim n^d$ when $n\rightarrow \infty$ – Seyhmus Güngören Sep 20 '12 at 11:59
Write $$\frac{n^d}{(n+d)!}d!n!=d!\prod_{j=1}^d\frac{n}{n+j}$$ to see that the expected limit is $d!$.
-
You mean "... is $d!$" instead of $n!$. – martini Sep 20 '12 at 12:00
I think there is a little mistake here. – Seyhmus Güngören Sep 20 '12 at 12:02
The denominator in the product should be $n+j$. – celtschk Sep 20 '12 at 12:15
Thanks guys, I've corrected. – Davide Giraudo Sep 20 '12 at 12:26
$$n^d\frac{n!\,d\,!}{(n+d)!}=d\,!\,(\frac{n}{n+1})(\frac{n}{n+2})\ldots(\frac{n}{n+d})$$
Note that the R.H.S. contains a finite number of terms ($=d$) each tending to 1.
Hence the limit is $d\,!$
To find an upper bound you can simply do this
$d\,!\,(\frac{n}{n+1})(\frac{n}{n+2})\ldots(\frac{n}{n+d})<d\,!\,(\frac{n}{n})(\frac{n}{n})\ldots(\frac{n}{n})$
$d\,!\,(\frac{n}{n+1})(\frac{n}{n+2})\ldots(\frac{n}{n+d})<d\,!$
So your upper bound is $d\,!$
Another Method (Just to show that it the sequence is convergent)
Let us treat the given term $n^d\frac{n!\,d\,!}{(n+d)!}$ as term of sequces $\{x_n\}$.
Since we have shown that the sequence is bounded, it will suffice to show that the sequence is monotonically increasing.
We will now show that it is monotonically increasing.
$$\frac{n}{n+r}-\frac{m}{m+r}=\frac{r(n-m)}{(n+r)(m+r)}>0 \quad\forall\; n>m>0\quad \& \quad\forall r>0$$
Let $n>m$
Then $n!>m!$ and each term $\frac{n}{n+r}>\frac{m}{m+r} \quad \forall r \in \{1, 2, \ldots ,d\}$, we have proved this above.
Hecne $x_n \gt m_m$ (as $a>a' \quad \& \quad b\gt b'\implies a.b>a'.b' \quad \forall a,a',b,b'>0$ ) | {
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Hecne $x_n \gt m_m$ (as $a>a' \quad \& \quad b\gt b'\implies a.b>a'.b' \quad \forall a,a',b,b'>0$ )
If you want to see how a monotonically increasing sequence which is upper bounded is convergent go to this link.
-
You can use the Stirling approximation $$n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\,,$$
after writing your expression in the form
$$d!\frac{n^d n!}{(n+d)!}$$
$$d!\frac{n^d n!}{(n+d)!} \sim d! \frac{n^d \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n }{\sqrt{2 \pi (n+d)} \left(\frac{n+d}{e}\right)^{n+d}}= d!\, {\rm e}^d \left(\frac{n}{n+d}\right)^{n} \left(\frac{n}{n+d}\right)^{d} \left(\frac{n}{n+d}\right)^{\frac{1}{2}}$$
$$\rightarrow d! \,{\rm e}^d \, \,{\rm e}^{-d}\, 1.1 = d!\,, \quad n \rightarrow \infty$$
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# Does an n by n Hermitian matrix always has n independent eigenvectors?
I am learning the MIT ocw 18.06 Linear Algebra, and I have learnt: an arbitrary $n×n$ matrix A with n independent eigenvectors can be written as $A=SΛS^{-1}$, and then for the Hermitian matrices, because the eigenvectors can be chosen orthonormal, it can be written as $A=QΛQ^H$ further.
I wonder does every $n×n$ Hermitian matrix has n independent eigenvectors and why? Thank you!
P.S. MIT 18.06 Linear Algebra Lecture 25: Symmetric Matrices and Positive Definiteness. You may wish to start from 4:20. From the course, I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why?
• You mean independent eigenvectors, not independent eigenvalues. – uniquesolution Feb 8 '18 at 12:04
• Oh, thanks, I have edited it. – Thomas Feb 9 '18 at 2:28
• It's on the Wikipedia page: en.wikipedia.org/wiki/Hermitian_matrix. – Robert Wolfe Feb 9 '18 at 4:25
• @Robert Thank you for the link, I had read the page though, and there wasn't an answer to my question. – Thomas Feb 9 '18 at 10:34
• Look here for a simple proof. – user Feb 11 '18 at 20:15
This is a theorem with a name: it is called the Spectral Theorem for Hermitian (or self-adjoint) matrices. As pointed out by Jose' Carlos Santos, it is a special case of the Spectral Theorem for normal matrices, which is just a little bit harder to prove.
Actually we can prove the spectral theorem for Hermitian matrices right here in a few lines. | {
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Actually we can prove the spectral theorem for Hermitian matrices right here in a few lines.
We are going to have to think about linear operators rather than matrices. If $T$ is a linear operator on a finite dimensional complex inner product space $V$, its adjoint $T^*$ is another linear operator determined by $\langle T v, w\rangle = \langle v, T^* w \rangle$ for all $v, w \in V$. (Note this is a basis-free description.) $T$ is called Hermitian or self-adjoint if $T = T^*$.
Let $B$ be an $n$-by-$n$ complex matrix and $B^*$ the conjugate transpose matrix. Let $T_B$ and $T_{B^*}$ be the corresponding linear operators. Then $(T_B)^* = T_{B^*}$, so a a matrix is Hermitian if and only if the corresponding linear operator is Hermitian.
Let $A$ be a Hermitian linear operator on a complex inner product space $V$ of dimension $n$. We need to consider $A$--invariant subspaces of $V$, that is linear subspaces $W$ such that $A W \subseteq W$. We should think about such a subspace as on an equal footing as our original space $V$. In particular, any such subspace is itself an inner product space, $A_{|W} : W \to W$ is a linear operator on $W$, and $A_{|W}$ is also Hermitian. If $\dim W \ge 1$, $A_{|W}$ has an least one eigenvector $w \in W$ -- because any linear operator at all acting on a (non-zero) finite dimensional complex vector space has at least one eigenvector.
The basic phenomenon is this: Let $W$ be any invariant subspace for $A$. Then $W^\perp$ is also invariant under $A$. The reason is that if $w \in W$ and $x \in W^\perp$, then $$\langle w, A x\rangle = \langle A^* w , x \rangle = \langle A w, x \rangle = 0,$$ because $Aw \in W$ and $x \in W^\perp$. Thus $A x \in W^\perp$. | {
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Write $V = V_1$. Take one eigenvector $v_1$ for $A$ in $V_1$. Then $\mathbb C v_1$ is $A$--invariant. Hence $V_2 = (\mathbb C v_1)^\perp$ is also $A$ invariant. Now just apply the same argument to $V_2$: the restriction of $A$ to $V_2$ has an eigenvector $v_2$ and the perpendicular complement $V_3$ to $\mathbb C v_2$ in $V_2$ is $A$--invariant. Continuing in this way, one gets a sequence of mutually orthogonal eigenvectors and a decreasing sequence of invariant subpsaces, $V = V_1 \supset V_2 \supset V_3 \dots$ such that $V_k$ has dimension $n - k + 1$. The process will only stop when we get to $V_n$ which has dimension 1. | {
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• You really hit the nail on the head as I'm afraid my question is too unusual to be understand. May I repeat you answer in my imprecise words? – Thomas Feb 9 '18 at 12:42
• Regard A as an operator on V, its invariant subspaces contains at least one eigenvector, their orthogonal complement are invariant subspaces too because A is a Hermitian matrix, we start from $\mathbb C^n$, every time we take a eigenvector from the space, it on a line which is a invariant subspace, and the space left which like a plane perpendicular to the line is also a invariant subspace, and we can take n times (i.e. n orthogonal/independent eigenvectors). Do I understand it correctly? – Thomas Feb 9 '18 at 12:47
• I still have some trouble with this, what if I take an eigenvector from the last space? As we know "Let $W$ be any invariant subspace for $A$. Then $W^⊥$ is also invariant under $A$", I take the eigenvector out, then there will be only zero vector left, and I think it really orthogonal to that eigenvector we take, is zero vector an invariant subspace? It may be, but "invariant space has at least one eigenvector" would be false. This "Let $W$ be any invariant subspace for $A$. Then $W^⊥$ is also invariant under $A$" makes me confused. – Thomas Feb 9 '18 at 13:43
• When you have reached $V_n$, you already have your collection of $n$ mutually orthogonal eigenvectors. You can't go further and you don't need to go further. I corrected the answer at one point to say that a linear operator on a non-zero finite dimensional complex vector space has an eigenvector. – fredgoodman Feb 9 '18 at 14:19
• It's important in the second paragraph that "acting on" means "mapping it into itself". – Ian Feb 9 '18 at 15:26
Yes, it has. That is due to the spectral theorem: every normal $n\times n$ matrix is diagonalizable. And Hermitian matrices are normal. | {
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• From my understanding, the factorization $A=QΛQ^H$ (i.e., the spectral theorem) is the special form of $A=SΛS^{-1}$ when A is a Hermitian matrix, and $A=SΛS^{-1}$ exists under the situation that there is a full set of independent eigenvectors. I wonder, if it is, why a Hermitian matrix is always diagonalizable? – Thomas Feb 9 '18 at 2:58
• @Thomas Because the spectral theorem says so. – José Carlos Santos Feb 9 '18 at 6:30
• From the course (I add the link into the question now), I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why? – Thomas Feb 9 '18 at 10:57
• @Thomas Every Heritian matrix is normal and every normal matrix is diagonalizable. – José Carlos Santos Feb 9 '18 at 11:08
• I had not learnt the concept of normal matrix from the course. I search for it now, and I realize that all matrices can be written as $A=QTQ^H$ with triangular T, so $A^HA=AA^H$ would be $T^HT=TT^H$, for a 2 by 2 matrix, it's easy to prove the upper right corner entry of such T is 0, extend it to size n, and we prove that T is diagonal. Is that a right poof? Thank you for your patience. – Thomas Feb 9 '18 at 11:58 | {
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# Approaching modular arithmetic problems
I'm a little stumbled on two questions.
How do I approach a problem like $x*41 \equiv 1 \pmod{99}$.
And given $2$ modulo, $7x+9y \equiv 0 \pmod{31}$ and $2x−5y \equiv 2 \pmod{31}$ (solve for $x$ only)?
When I solve for $x$ for the latter, I got a fraction as the answer and I'm not sure if I can have a fraction as an answer? I'm not sure how to approach the first problem either.
• What fraction did you get as answer? – Git Gud Sep 22 '13 at 22:20
• The first problem involves inverses. What is the inverse of 41 mod 99? The second problem is a set of linear equations in $x$ and $y$ modulo 31, which means the two can be added/subtracted/etc. as if they were a set of normal equations looking for a linear solution. – abiessu Sep 22 '13 at 22:25
Finding the solution to $$x \times 41 \equiv 1 \pmod {99}$$ is equivalent to asking for the multiplicative inverse of $41$ modulo $99$. Since $\gcd(99,41)=1$, we know $41$ actually has an inverse, and it can be found using the Extended Euclidean Algorithm:
\begin{align*} 99-2 \times 41 &= 17 \\ 41-2 \times 17 &= 7 \\ 17-2 \times 7 &= 3 \\ 7-2\times 3 &= 1 &=\gcd(99,41). \\ \end{align*} Going back, we see that \begin{align*} 1 &= 7-2\times 3 \\ &= 7-2\times (17-2 \times 7) \\ &= 5 \times 7-2\times 17 \\ &= 5 \times (41-2 \times 17)-2\times 17 \\ &= -12 \times 17+5 \times 41 \\ &= -12 \times (99-2 \times 41)+5 \times 41 \\ &= 29 \times 41-12 \times 99 \\ \end{align*} Hence $29 \times 41 \equiv 1 \pmod {99}$ and thus $x=29$. | {
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In the second case, we have $$7x+9y \equiv 0 \pmod {31}$$ and $$2x-5y\equiv 2 \pmod {31}.$$ Here we want to take $7x+9y=0 \pmod {31}$ and rearrange it to get $x \equiv ?? \pmod {31}$, then substitute it into the other equation and solve for $y$. This requires finding the multiplicative inverse of $7$ modulo $31$ (which we can do as above). It turns out $7 \times 9 \equiv 1 \pmod {31}$. Hence \begin{align*} & 7x+9y=0 \pmod {31} \\ \iff & 7x \equiv -9y \pmod {31} \\ \iff & x \equiv -9y \times 9 \pmod {31} \\ \iff & x \equiv 12y \pmod {31}. \end{align*} We then substitute this into the equation $2x-5y\equiv 2 \pmod {31}$, which implies $$2 \times 12y-5y \equiv 2 \pmod {31}$$ or equivalently $$19y \equiv 2 \pmod {31}.$$ Yet again, we find a multiplicative inverse, this time of $19$ modulo $31$, which turns out to be $18$. So \begin{align*} & 19y \equiv 2 \pmod {31} \\ \iff & y \equiv 2 \times 18 \pmod {31} \\ \iff & y \equiv 5 \pmod {31}. \end{align*} Hence $$x \equiv 12y \equiv 29 \pmod {31}.$$ Thus we have the solution $(x,y)=(29,5)$.
• Are you sure the answer to the first part is x=20? – Mufasa Sep 22 '13 at 22:35
• Thanks for point that out; it was an arithmetic error (actually, I found the inverse mod 91 by mistake) and it should be fixed now. – Rebecca J. Stones Sep 22 '13 at 22:47
• You're welcome. BTW: I am learning a lot from your posts - so THANK YOU! :) – Mufasa Sep 22 '13 at 22:49
For the first one, you could approach it as follows:
$41x=1$ mod $99$
$140x=1$ mod $99$ (because 41 mod 99 = (41+99) mod 99)
$140x=100$ mod $99$ (because 1 mod 99 = (1+99) mod 99)
$7x=5$ mod $99$ (divide both sides by 20)
$7x=203$ mod $99$ (because 5 mod 99 = (5+99+99) mod 99)
$x=29$ mod $99$ (divide both sides by 7)
For the first one, you have to find a multiplicative inverse for $41$ mod $99$. Use Euclid's algorithm to find the solutions of $41\cdot x + 99 \cdot y = 1$ to find $x$. | {
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