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There is also something called the Gambler's Fallacy, which is the mistaken belief that the probability on the next roll changes because a particular outcome is "due." In the example above, the probability of rolling at least one 6 in the next roll of the three dice (after three rolls with no 6's) is still 42.13%. A (non-mathematician) gambler might think that the dice are "due," that in order to get the long-term average back up to 42%, the probability of the next roll getting at least one 6 must be higher than 42%. This is wrong, and hence it's called "the Gambler's Fallacy."
The important thing here is that things will “average out” in the long run, so that you get at least one 6 in 42.1% of the rolls, but not because at any one time there is a greater chance, to make up the average — it happens only because there is a very long time to do so. It is not because past events have any effect on future events, pulling them into line with the “averages”.
For more on the Law of Large Numbers, see
The Law of Large Numbers
Digging deeper
In 2008, we had the following detailed question about that from Konstantinos:
Do Prior Outcomes Affect Probabilities of Future Ones?What I want to know is if in matters of luck, such as games of dice, or lotteries, or flipping coins, the future outcomes have any relativity with past results. What I mean in each case is: If I flip a coin and get tails, don't I have bigger or smaller possibilities to get heads on the next roll? I mean I know that the possibility is always 1/2 but since I have already thrown the coin 5 times and rolled 5 tails there aren't possibilities that in the next throws there will be heads? The same question goes for dice rolls, and for lotto numbers. If a number has come more times than others, isn't it possible that for a limited amount of coming times, numbers that haven't come yet, will start showing more? | {
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What I find most confusing is that the relation between probabilities and past possibilities, and the outcome in real life. How can a coin come tails 5 or six times in a row when the possibility is always 1/2? Are probabilities only theoretical?
I tried noting down results of different trials of luck (dice, past lotteries, coins) but in the end they don't seem to make true to any theory I have heard. I would like to see the magic of numbers in real life and on this subject and how it works as to prove a theory. Do I have to toss the coin 10,000 times? And if I do will I see heads coming in a row after 10 subsequent tails?
I responded to this one:
What you are suggesting is called the Gambler's Fallacy--the WRONG idea that future results of a random process are affected by past results, as if probability deliberately made things balance out. The law of large numbers says that it WILL balance out eventually; but that does not happen by changing the probabilities in the short term. The long-term balance just swamps the short-term imbalance.
If you think about it, what could cause a coin to start landing heads up more often after a string of tails? There is no possible physical cause for this; the coin has no memory of what it, or any other coin, previously did. And probability theory does not make things happen without a physical cause; it just describes the results.
If I tossed a coin and it landed tails 5 times, I would just recognize that that is a perfectly possible (and not TOO unlikely) thing to happen. If I got tails 100 times, I would NOT expect the next to be heads; I would inspect the coin to make sure it actually has a heads side! An unlikely string of outcomes not only does not mean that the opposite outcome is more likely now; it makes it LESS likely, because it suggests statistically that the basic probability may not be what I was originally assuming. | {
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In response to the question about probabilities being merely theoretical, I explained what probability is and isn’t:
Probabilities are theoretical, but have experimental results, IN THE LONG RUN. The law of large numbers says that, if you repeat an experiment ENOUGH TIMES, the proportion of times an event occurs will PROBABLY be CLOSE to the predicted theoretical probability. Probability can't tell you what will happen the next time, but it does predict what is likely to happen on the average over the next, say, million times. If you started out with ten tails in a row, you will not necessarily get ten heads in a row at any point, or even more heads than tails in the next few tosses; you will just get enough heads in the next million to keep it balanced.
Note all the capitalized words, emphasizing the vagueness of this statement. A formal statement of this theorem will define those more clearly, but they will then all become statements of probability (my “probably”) and limits (my “enough” and “close”). See, for example, the formal statements of the Law of Large Numbers in the Wikipedia page (my emphases in the last paragraph):
The weak law of large numbers (also called Khinchin’s law) states that the sample average converges in probability towards the expected value $$\displaystyle \overline {X}_{n}\overset{P}{\rightarrow} \mu \ {\textrm {when}}\ n\to \infty .$$.
That is, for any positive number ε, $$\displaystyle \lim _{n\to \infty }\Pr \!\left(\,|{\overline {X}}_{n}-\mu |>\varepsilon \,\right)=0$$.
Interpreting this result, the weak law states that for any nonzero margin specified, no matter how small, with a sufficiently large sample there will be a very high probability that the average of the observations will be close to the expected value; that is, within the margin.
What it does not say is that over, say, 100 trials (or 10,000 trials), we can be sure that the average will be exactly what we expect.
I continued: | {
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I continued:
In particular, if you were to throw five coins at a time (to make it easier than throwing the same coin five times in a row), and do that 1000 times, you would expect that you would get 5 tails about 1/32 of those times (since the probability of all five being tails is 1/2^5). That's about 31 times! So it's not at all unreasonable to expect that it will occur once in a while.
On the other hand, the probability of getting 100 tails in a row is 1/2^100, or 1/1,267,650,600,228,229,401,496,703,205,376, which makes it very unlikely that it has ever happened in the history of the world, though it could!
Again, probability can't tell you what WILL happen, specifically; it is all about unpredictable events. But if you tossed a coin 1000 sets of 10 times, on the average one of those is likely to yield 10 tails. (The probability is 1/2^10 = 1/1024.) The probability of some ten in a row out of 10,000 tosses is a little bigger, but that's harder to calculate.
Regression to the mean
In an unarchived question from 2014, Adam brought in another idea:
If I flip a coin 10 times, the most likely outcome is that I will flip a total of 5 heads and a total of 5 tails. If each round of coin flipping (one round being 10 flips, in my example) is independent of previous rounds, then the probability of flipping a total of 5 heads and a total of 5 tails never changes. However, the concept of "regression to the mean" implies that "rare" events are likely to be followed by less rare events and vice versa. So, if I flip 10 heads in a row in round 1 (a rare event), the odds of flipping a total of 5 heads and 5 tails in round 2 are greater than if I'd flipped 5 heads and 5 tails in round 1. What is the correct way to see this, do the odds change or not? | {
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Coin flipping rounds are independent of one another, which implies that the probabilities never change. Regression to the mean states that rare events are likely to be followed by less rare events, implying that the probability of even random events does change.
I replied:
The odds don't change. You are NOT more likely to flip 5 heads if you previously flipped 10.
Regression to the mean only says that the most likely event on ANY toss will be about 5 heads, so if you got 10 on one toss, you are more likely to get something closer to 5 on the next toss, simply because something closer to 5 is ALWAYS more likely than 10. If you got 5 on the first toss, you are as likely to get 5 again as ever; but any deviation that does occur will be away from the mean, because there is no direction to go except away!
It is an entirely wrong interpretation of the phenomenon to think that, having tossed 10 heads, you are more likely to toss 5 heads than under other circumstances. It's just that the event that is always more likely will be less extreme than that first toss.
http://en.wikipedia.org/wiki/Regression_toward_the_mean
Past Events and ProbabilityBatting Averages | {
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complex fourier series of square wave. I'm trying to plot the fourier series following fourier series; f(t)=$$\sum_{k=0}^k \frac{(1)(\sin(2k+1)pi*t)}{2k+1}$$ equation. 2 Fourier Series Learning outcomes In this section we will learn how Fourier series (real and complex) can be used to represent functions and sum series. Now, we will write a Matlab code. A Fourier series (which rely on complex Fourier coefficients defined by integrals) uses the complex exponential e inx. The functions sin(nx) and cos(nx) form a sort of periodic table: they are the atoms that all other waves are built out of. When an and bn are given by ( 2 ), the trigonometric series ( 1 ) is called the Fourier series …. Fourier Series of waveforms. 16 Example: Find the complete Fourier series of the square wave function sqr(x). Example # 01: Calculate fourier series of the function given below: $$f\left( x \right) = L – x on – L \le x \le L$$ Solution: As,. s (1) and (2), is a special case of a more gen-eral concept: the Fourier series for a periodic function. Many component approximation to square wave. Then for all t g t+ p a =f a t+ p a =f(at+p)=f(at)=g(t). This is the fundamental frequency or f 1, that corresponds to the “pitch” of the sound Each of the higher-frequency simple waves …. Read Book Fourier Series Examples And Solutions Square Wave Fourier Series Examples And Solutions Square Wave When somebody should go to the books stores, search start by shop, shelf by shelf, it is in Complex fourier Series - Example Fourier Transform (Solved Problem 1) Fourier Analysis: Fourier …. 57 questions with answers in FOURIER SERIES. • stem – Draws discrete plots (as opposed to plot, which draws continuous plots). On the other hand, an imaginary number takes the general form. The derivation of this real Fourier series from (5. Read through the lab and pay special attention to the introductory parts about Fourier series …. Since this function is even, the coefficients Then. Sketch 3 cycles of the function | {
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series …. Since this function is even, the coefficients Then. Sketch 3 cycles of the function represented by. The main aim of the Fourier series is one period to many frequencies. Transcribed image text: (a) Find the complex Fourier series of the periodic square wave shown in Problem 5. Properties of Fourier Series John T. 1 A Historical Perspective By 1807, Fourier had completed a work that series …. Thus, the Fourier Series is an in nite superposition of imaginary exponentials with frequency terms that in-crease as n increases. A key difference, no matter how you want to formulate or describe it, goes something like this: 1) Start with a square wave with infinitely sharp transitions. PDF Odd 3: Complex Fourier Series. The first term of the Fourier Series will be a sinusoid with same phase and frequency as the square wave. University of California, San Diego J. 9 Even and Odd Functions The astute reader will have noticed that the Fourier series constructed in Secs. Complex Exponentials ejn t n t j n t cos 0 sin 0. A full FFT produces a complex number, so yes, phase is included. 1 Periodic Functions and Orthogonality Relations The differential equation y′′ + 2y =F cos!t models a mass-spring system with natural frequency with a pure cosine forcing function of frequency !. 4 first nonzero coefficients are used only, so the Square Wave approximation will be not sensational. I am trying to calculate in MATLAB the fourier series …. For more information about the Fourier series, refer to Fourier. Let's go back to our non-periodic driving force example, the impulse force, and apply the Fourier …. (You can figure out the last step and the casework for even and odd by drawing a little. Chapter 65, The complex or exponential form of a Fourier. Consider the periodic square wave x(t) shown in figure a. sum of sine waves each with different amplitudes waves. This is the idea of a Fourier series. The Fourier basis is convenient for us in that this series already separates these | {
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a Fourier series. The Fourier basis is convenient for us in that this series already separates these components. The next question is, given a complex periodic wave, how can we extract its component waves? Two Methods are Required. The Fourier Series is a shorthand mathematical description of a waveform. Here we see that adding two different sine waves make a new wave: When we add lots of them (using the sigma function Σ as a handy notation) we can get things like: 20 sine waves: sin (x)+sin (3x)/3+sin (5x)/5 + + sin (39x)/39: Fourier Series …. %Fourier series of rectangular wave clc; close all; clear all; j=1; T=4; %Time period of square wave tau=1; %2tau= On time of the square wave …. not just odd like the square wave. If the following condition (equation [5]) is true, then the resultant function g (t) will be entirely real: In equation [5], the. The series for f(x) defined in the interval (c, c+2π)and satisfying. N is the number of number of terms used to approximate the square wave. We will call it the real form of the Fourier series. Convert the real Fourier se- ries of the square wave f(t) to a complex series. The Complex Fourier Series. derived a real representation (in terms of cosines and sines) for from the complex exponential form of the Fourier series: (1) For example, in the lecture #13 notes, we derived the following Fourier coefficients for a triangle wave (sym-metric about the vertical axis), (2) and converted the complex exponential series…. The coefficients for Fourier series expansions of a few common functions are given in Beyer (1987, pp. So, responding to your comment, a 1 kHz square wave doest not include a component at 999 Hz, but only odd harmonics of 1 kHz. The Fourier transform and Fourier…. 8 Illustration of the superposition of terms in the. The Fourier transform of this image exhibits an "infinite" series of harmonics or higher order terms, although these do not actually go out to infinity due to the finite resolution of the original | {
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although these do not actually go out to infinity due to the finite resolution of the original image. So, responding to your comment, a 1 kHz square wave doest not include a. Since sines and cosines (and in turn, imaginary exponentials) form an orthogonal set1, this se-ries converges for any moderately well-behaved function f(x). Often in solid state physics we need to work with functions that are periodic. The square wave is a special case of a pulse wave which allows arbitrary durations at minimum and maximum amplitudes. In other words, Fourier series can be used to express a function in terms of the frequencies (harmonics) it is composed of. A few examples are square waves, saw-tooth waves, and triangular pulses. Example: Compute the Fourier series of f(t), where f(t) is the square wave …. In general, given a repeating waveform , we can evaluate its Fourier series coefficients by directly evaluating the Fourier transform: but doing this directly for sawtooth and parabolic waves …. They are helpful in their ability to imitate many different types of waves: x-ray, heat, light, and sound. The complex Fourier expansion coefficients are cn= 4 π 1 n nodd 0 neven 0 2 4 6 8 10 0. In case of a string which is struck so that say at x=a only the string has a …. Fourier Series using LabVIEW Fourier Series using LabVIEW Student-developed LabVIEW VI The student is then asked to approximate a square wave in both the time and frequency domain using a summed set of 5 sine waves The frequency and amplitude from the LabVIEW interface provide the coefficients of the Fourier Series …. ABSTRACT Fourier analysis and a computer were used to generate the Fourier series and coefficients for the transmission distribution of a square wave …. Complex Fourier Series Animation of the square wave The 4 upper rotating vectors correspond to the 4 lower formula components. From the real to the complex Fourier series Proposition The complex Fourier coe cients of f(x) = a0 2 + X1 n=1 an cosnx + bn | {
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Fourier series Proposition The complex Fourier coe cients of f(x) = a0 2 + X1 n=1 an cosnx + bn sinnx are cn = an ibn 2; c n = an + ibn 2: M. Fourier Series for a … Odd 3: Complex Fourier Series - Imperial College London It can be done by using a process called Fourier analysis. Draw a square wave of amplitude 1 and period 1 second whose trigonometric Fourier Series Representation consists of only cosine terms and has no DC component. m m! Again, we really need two such plots, one for the cosine series and another for the sine series. 3 1) Compute the Fourier series that corresponds to a square wave 1 - /4 /4 (0) 1. Square WaveFourier Series Examples And Solutions Square Wave Thank you enormously much for downloading fourier series examples and solutions square wave. Such a Fourier expansion provides an interpetation of the wave …. For comparison, let us find another Fourier series, namely the one for the periodic extension of g(x) = x, 0 x 1, sometimes designated x …. c) Write the first three nonzero terms in the Fourier expansion of f ( t). Jean Baptiste Joseph Fourier (21 March 1768 - 16 May 1830) Fourier series. He initialized Fourier series, Fourier transforms and their applications to problems of heat transfer and vibrations. 1) Compute the Fourier series that corresponds to a square wave. A Fourier series is nothing but the expansion of a periodic function f(x) with the terms of an infinite sum of sins and cosine values. The exact combination of harmonics will vary depending on the way the string is set in motion; e. Check the time t of the 10 rotations, t=24sec->T=2. Again, of course, you’re not going to get a perfect square wave with a finite number of Fourier terms in your series (in essence, it’s then not a …. 13 Applications of the Fourier transform 13. of Electrical Engineering, Southern Methodist University. The synthesis technique is also called additive synthesis. This chapter includes complex differential forms, geometric inequalities from one and | {
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synthesis. This chapter includes complex differential forms, geometric inequalities from one and several complex …. For instance, the square wave …. The ideas are classical and of transcendent beauty. An Introduction to Fourier Analysis Fourier Series, Partial Differential Equations and Fourier Transforms. As mentioned in the previous section, perhaps the most important set of orthonormal functions is the set of sines and cosines. s(t) = AN ODD SQUAREWAVE with DC. PDF Characterization of Signals Frequency Domain. Finding the Fourier series coefficients for the square wave sq T (t) is very simple. A Fourier series uses the relationship of orthogonality between the sine and cosine functions. , the output will always have a negative portion and positive portion. 1 The Real Form Fourier Series as follows: x(t) = a0 2 + X∞ n=1 an cosnω0t+bn sinnω0t (1) This is called a trigonometric series. In practice, the complex exponential Fourier series (5. (one period is T which is equal to 2PI) Looking at the figure it is clear that area bounded by the Square wave …. Plot one-sided, double-sided and normalized spectrum. continuation of part 1a - Introduction to Complex Fourier Series. The coefficients become small quickly for the triangle wave, but not for the square wave or the sawtooth. This means a square wave in the time domain, its Fourier …. Concept: The complex Exponential Fourier Series representation of a periodic signal x(t) with fundamental period To is given by, \$$x\\left(. In class, we mentioned how complex exponentials version of the Fourier Series can be used to represent a 2Tperiodic function f(t) as follows: can be used to represent a Fourier Series: f(t) = X1 n=1 c ne inˇ T t (1) The crux of Fourier Series analysis is to nd the c n values specifying how much the nth harmonic contributes to the function f(t). The synthesis of a complex wave …. 10 Trigonometric Fourier series of another square wave. Note that the Fourier coefficients are complex numbers, even | {
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Fourier series of another square wave. Note that the Fourier coefficients are complex numbers, even though the series in Equation [1], evaluated with the coefficients in Equation [4], result in a real function. Similarly, if G(x) is an odd function with Fourier coe cients a nfor n 0 and b n for n 1, then a n= 0 for all n 0, and a n= 2 L Z L 0 G(x)sin nˇx L dxfor all n 0(16) In particular, the fourier series of an even function only has cosine terms and the fourier series of an odd function only has sine terms. It is very easy to see that an vanishes if f is an odd function, while bn vanishes if f is even. Index Terms—Fourier series, periodic function, recursive. Based on what I read at this link: Any periodic . It is natural for complex numbers and negative frequencies to go hand-in-hand. Even Square Wave (Exponential Series) Consider, again, the pulse function. 3: Fourier Cosine and Sine Series, day 1 Trigonometric Fourier Series (Example 2) Complex fourier Series - Example Fourier Transform (Solved Problem 1) Fourier Analysis: Fourier Transform Exam Question ExampleFourier Series: Complex Version! Part 1 Fourier …. (a) Determine the complex exponential Fourier series of x (t). Young (translator), There are excellent discussions of Fourier series …. Since the time domain signal is periodic, the sine and cosine wave …. Write a computer program to calculate the exponential Fourier series of the half-wave rectified sinusoidal current of Fig. This function is a square wave; a plot shows the value 1 from x=p to x = 0 followed by the . 3 Square wave Analysis (breaking it up into sine waves). Summation of just five odd harmonics gives a fairly decent representation in Figure 15. Complex fourier Series - Example Fourier Transform (Solved Problem 1) Fourier Analysis: Fourier Transform Exam Question ExampleFourier Series: Complex …. One of the principles of Fourier analysis is that any imaginable waveform can be constructed out of a carefully chosen set of sinewave | {
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is that any imaginable waveform can be constructed out of a carefully chosen set of sinewave components, assembled in a particular way (the frequency -> time task). (Image will be uploaded soon) Laurent Series Yield Fourier Series (Fourier Theorem). The complex Fourier series Recall the Fourier series expansion of a square wave, triangle wave, and sawtooth wave that we looked at before. Schrödinger's Equation Up: Wave Mechanics Previous: Electron Diffraction Representation of Waves via Complex Numbers In mathematics, the symbol is conventionally used to represent the square …. So here's the final wave, listening to the final waveform. You can work out the 2D Fourier transform in the same way as you did earlier with the sinusoidal gratings. In fact, in many cases, the complex Fourier series is easier to obtain rather than the trigonometrical Fourier series In summary, the relationship between the complex and trigonometrical Fourier series …. To motivate this, return to the Fourier series…. Recall from Chapter 1 that a digital sound is simply a sequence of num- Fourier analysis with complex exponentials which will often result in complex square wave are now 0 and 1,contraryto−1 and 1 before. 5) The Laurent expansion on the unit circle has the same form as the complex Fourier series, which shows the equivalence between the two expansions. Fourier Series | Brilliant Math & Science Wiki. Also, as with Fourier Sine series…. Theorem 2 lim N→∞ sup 0≤x≤1 f −S N(f) = 0 holds for any continuous function g. Compute the Exponential Fourier Series for the square wave shown below assuming that. Louis, MO April 24, 2012 The Fourier series is a tool for solving partial differential equations. (We assume the reader is already at least somewhat familiar with these. Step 1: Obtain the Fourier series of F(t). magnitude of the square of the Fourier transform: SFEt {()}2 This is our measure of the frequency content of a light wave. Plot the function over a few periods, as well as a few | {
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of the frequency content of a light wave. Plot the function over a few periods, as well as a few truncations of the Fourier series. It is not a mathematical proof, and several terms are used loosely (particularly those in quotes). We can be confident we have the correct answer. %Fourier series of rectangular wave clc; close all; clear all; j=1; T=4; %Time period of square wave tau=1; %2tau= On time of the square wave w0=2*pi/T;. 3–81 Spectrum of a Sum of Cosine Signals spectrum cosines 3–82 3. I understand the bounds that he chooses and. If a square waveform of period T is defined by \left\{ \begin{array}{l l} f(t)= 1 \text{ if } t= T/2 \end{array} \right. Series coefficients c n (d) Fig. Discrete Time Fourier Transforms The discrete-time Fourier transform or the Fourier transform of a discrete–time sequence x[n] is a representation of the sequence in terms of the complex exponential sequence. There's also the infamous Square wave …. Since this wave is periodic, its harmonic content can be found using Fourier series as follows: The Fourier coefficients are,. Answer The function is discontinuous at t = 0, and we expect the series to converge to a value half-way between the upper and lower values; zero in this case. Let f(x) = {1 if -pPDF Fourier analysis for vectors. 2 Complex conjugate sqrt(x) Square root log(x) Natural logarithm Suppose we want to enter a vector x consisting of points Sine Wave 0. Example 1 Find the Fourier sine coefficients bk of the square wave SW(x). Fourier Analysis: Fourier Series with Complex Exponentials n The Complex Fourier series can be written as: where: n Complex cn n *Complex conjugate * n Note: if x(t) is real, c-n = cn 32 Fourier Analysis: Fourier Series Line Spectra n n Line Spectra refers the plotting of discrete coefficients corresponding to their frequencies For a periodic. Start by forming a time vector running from 0 to . See below Once rectified, it is even , so you only need the cosine series. Importantly there is nothing | {
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below Once rectified, it is even , so you only need the cosine series. Importantly there is nothing special about the square …. Where To Download Fourier Series Examples And Solutions Square Wave Fourier Series Examples And Solutions Square Wave As recognized, adventure as well as experience practically lesson, amusement, as skillfully as arrangement can be gotten by just checking out a books fourier series examples and solutions square wave …. The 8-term Fourier series approximations of the square wave and the sawtooth wave: Mathematica code: f[t_] := SawtoothWave[t] T = 1; Fourier series decomposition of a square wave using Phasor addition : It retains it’s form over several complex …. 3 Case 2: Some periodic functions (e. The Fourier series represents periodic, continuous-time signals as a weighted sum of continuous-time sinusoids. Not surprisingly, the even extension of the function into the left half plane produces a Fourier series that consists of only cos (even) terms. By the double angle formula, cos(2t) = 1 2sin2 t, so 1 + sin2 t= 3 2 1 2 cos(2t): The right hand side is a Fourier series; it happens to have only nitely many terms. A Fourier series ( / ˈfʊrieɪ, - iər /) is a sum that represents a periodic function as a sum of sine and cosine waves. The integral splits into two parts, one for each piece of , and we find that the Fourier coefficients are. Fourier Series in Mathematica Craig Beasley Department of Electrical and Systems Engineering Washington University in St. Jean Baptiste Joseph Fourier, a French mathematician and a physicist; was born in Auxerre, France. 005 (b) The Fourier series on a larger interval Figure 2. The Fourier series forthe discrete‐time periodic wave shown below: 1 Sequence x (in time domain) 0. On-Line Fourier Series Calculator is an interactive app to calculate Fourier Series coefficients (Up to 10000 elements) for user-defined piecewise …. The initial phase of the n-th oscillation θ θ n. (b) Consider the signal x(t) = | {
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piecewise …. The initial phase of the n-th oscillation θ θ n. (b) Consider the signal x(t) = sin(2πf0t), Find the complex Fourier series of x(t) and plot its frequency spectra. First, let x(t) be the zero-mean square wave. Wave Series Fourier Series Grapher; Square Wave: sin(x) + sin(3x)/3 + sin(5x)/5 + sin((2n−1)*x)/(2n−1) Sawtooth: sin(x) + sin(2x)/2 + sin(3x)/3 …. 1 Infinite Sequences, Infinite Series and Improper In-tegrals 1. This can be accomplished by extending the definition of the function in question to the interval [−L, 0] so that the extended function is either even (if one wants a cosine series) or odd (if one wants a sine series). As far as I know, Sage does not have a built-in method to find a “least-squares solution” to a system of linear equations. -L ≤ x ≤ L is given by: The above Fourier series formulas help in solving different types of problems easily. The periodic function shown in Fig. 23) all coefficients an vanish, the series only contains sines. HOWELL Department of Mathematical Science University of Alabama in Huntsville Principles of Fourier …. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Convert the real Fourier se-ries of the square wave f(t) to a complex series. Rad225/Bioe225 Ultrasound Fourier Series (review) Fall 2019. This section explains three Fourier series: sines, cosines, and exponentials e ikx. 5 The complex form of the Fourier series. 3 Square Wave–High Frequencies One application of Fourier series, the analysis of a “square” wave (Fig. Fourier series of the elementary waveforms. PDF The Exponential Form Fourier Series. It builds upon the Fourier Series. For the periodic bipolar, 50% duty-cycle square wave, the θ -averaging of this waveform over one θ -cycle is: QED. This subtle property is due to the symmetry of waveforms (except for the sawtooth, which is not symmetric). 01:6; >> fexact=4*(t<=3)-2*(t>=3); >> plot(t,fexact) and results in the | {
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which is not symmetric). 01:6; >> fexact=4*(t<=3)-2*(t>=3); >> plot(t,fexact) and results in the plot: Plotting the Truncated Fourier. Joseph Fourier showed that any periodic wave can be represented by a sum of simple sine waves. 5, an =0, n ≥1, , 1 [1 cos( )] ≥ − = n n n bn π π. com analysis are described in the chapter on musical tones. Fourier series are used to approximate complex functions in many different parts of science and math. Demonstrates Taylor series expansion of complex exponentials. Recall the Fourier series, in which a function f[t] is written as a sum of sine and cosine terms: One interpretation of the above Fourier transform is that F[Z] is the frequency spectrum of a sine wave signal f[t] which is varying in time; thus Z is the angular frequency. The free space loss for electromagnetic waves spreading from a point source is The Friis’ loss formula for antenna-to-antenna loss is given by Radio wave propagation in the atmosphere: (1) space-wave propagation (e. This series of sine waves always contains a wave called the "FUNDAMENTAL", that has the same FREQUENCY (repetition rate) as the COMPLEX WAVE being created. Fourier transform spectroscopy (cont. Complex Fourier Series • Complex Fourier Analysis Example • Time Shifting • Even/Odd Symmetry • Antiperiodic ⇒ Odd Harmonics Only • Symmetry Examples • Summary E1. The fourier series of a sine wave is 100% fundamental, 0% any harmonics. If X is a multidimensional array, then fft …. Even Square Wave (Exploiting Symmetry). (here: symmetric, zero at both ends) Series –Taylor and Fourier Seismological applications The Delta function Delta function – generating series The delta function Seismological applications Fourier Integrals The basis for the spectral analysis (described in the continuous world) is the transform pair: Complex fourier spectrum The complex …. The conversion of complex Fourier series into standard trigonometric Fourier series is based on Euler's formulas: sinθ = 1 2j ejθ − 1 2je − jθ = | {
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standard trigonometric Fourier series is based on Euler's formulas: sinθ = 1 2j ejθ − 1 2je − jθ = ℑejθ = Imejθ, cosθ = 1 2ejθ − 1 2e − jθ = ℜejθ = Reejθ. Try another waveform, including one of the complex ones (Triangle, Sawtooth, or Square). A complex exponential einx= cosnx+isinnxhas a small-est period of 2π/n. This lesson shows you how to compute the Fourier series …. PDF General Inner Product & Fourier Series. m % Description: m-file to plot complex (exponential) Fourier Series % representation of a square wave…. Recall that the Taylor series expansion is given by f(x) = ¥ å n=0 cn(x a)n, where the expansion coefficients are. 1 This pages contains exercises to practice computing the Fourier series of a CT signal. Since the coefficients of the Exponential Fourier Series of complex numbers we. Complex Fourier series Complex representation of Fourier series of a function f(t) with period T and corresponding angular frequency != 2ˇ=T: f(t) = X1 n=1 c ne in!t;where c n = 8 <: (a n ib n)=2 ;n>0 a 0=2 n= 0; (a jnj+ib jnj)=2 n<0 Note that the summation goes from 1 to 1. DC Value of a Square Wave The Fourier series coefficient for k = 0 has a special interpretation as the average value of the signal x(t). In short, the square wave’s coefficients decay more slowly with increasing frequency. Fourier series coefficients for a symmetric periodic square wave. Find the exponential Fourier series for the square wave of Figure 11. 10 Fourier Series and Transforms (2014-5543) Complex Fourier Series…. 2 Approximating the Square Wave Function using Fourier Sine Series. For the real series, we know that d = an = 0 and. Complex Fourier Series In an earlier module , we showed that a square wave could be expressed as a superposition of pulses. 1) translates into that the inverse of a complex (DFT on a square wave…. Example: Compute the Fourier series of f(t), where f(t) is the square wave with period 2π. The frequency of each wave in the sum, or harmonic, is an integer multiple of the | {
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with period 2π. The frequency of each wave in the sum, or harmonic, is an integer multiple of the periodic function's fundamental frequency. 11 Fourier Series of a Square Wave Co is a DC average. Fourier Series Square Wave Example The Fourier series of a square wave with period is University of California, San Diego J. Fourier series of a square wave of period T = 1. Lec1: Fourier Series Associated Prof Dr. It will definitely squander the time. Contribute to Abdul-Rahman-Ibrahim/Fourier-Series-Expansion-Complex-Coefficients development by creating an account on GitHub. Fourier analysis and predict with precision their behavior by just looking at the final expression, also called the design equation. Determine the Fourier series expansion for full wave rectified sine wave i. where the Fourier coefficients and are given by. Example: Find the complete Fourier series of the square wave function sqr(x). Keywords: Fourier analysis, Fourier coefficients, complex exponentials, discrete sine series, discrete cosine series, full and half wave rectifier, diode valve, square wave, triangular wave, and the saw-tooth wave. The complex algebra provides an elegant and compact re. The Fourier Transform is a method to single out smaller waves in a complex wave. If we let the numbers in the Fourier series get very large, we get a phenomenon, called This is called the complex Fourier series…. Fourier Series coefficients for a square wave. The main advantage of an FFT is speed, which it …. The series does not seem very useful, but we are saved by the fact that it converges rather rapidly. the Fourier representation of the square wave is given in Fig. Then mathematically, a T-periodic waveform v satisfies — a periodic waveform with period T (2) for all t. Discrete Fourier Transform (DFT)¶ From the previous section, we learned how we can easily characterize a wave with period/frequency, amplitude, phase. Show that the Fourier Series expansion of y (x) is given by y(t)= 4! sin"t+ sin3"t 3 | {
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phase. Show that the Fourier Series expansion of y (x) is given by y(t)= 4! sin"t+ sin3"t 3 + sin5"t 5 + # % & '(where ω = 2π/T. In this particular SPICE simulation, I’ve summed the 1st, 3rd, 5th, 7th, and 9th harmonic voltage sources in series …. Fourier Series and Coefficients Fourier series may be used to represent periodic functions as a linear combination of sine and cosine functions. 1: The cubic polynomial f(x)=−1 3 x 3 + 1 2 x 2 − 3 16 x+1on the interval [0,1], together with its Fourier series …. Example 2 Given a signal y(t) = cos(2t), find its Fourier Series coefficients. Let’s assume we have a square wave with following characteristics: P eriod = 2ms P eak−to −P eak V alue = 2 V Average V alue = 0 V P e r i o d = 2 m s P e a k − t o − P e a k V a l u e = 2 V A v e r a g e V a l u e = 0 V. The key here is that the Fourier basis is an orthogonal basis on a given interval. That is why we have programmed our free online Fourier series calculator to determine the results instantly and precisely. 6 Example: Fourier Series for Square Wave. Find the Fourier Complex Fourier Series . In the processing of audio signals (although it can be used for radio waves, light waves, seismic waves, and even images), Fourier analysis can isolate individual components of a continuous complex waveform, and concentrate. The function is reconstructed by the following summations over the fourier coefficients. In general square integrability will not guarantee convergence of the Fourier series to the original function. Fourier series are useful in a wide range of fields including acoustics, with representation of musical sounds as sums of waves of various frequencies (Nearing, 2020) or quantum mechanics, for the quantum wave function of a particle in a box. Step 3: Finally, substituting all the coefficients in Fourier …. Check back soon! Problem 5 Show that the complex Fourier series …. Any function can be written as the sum of an even and an odd function [ ( )]/2 A complex | {
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series …. Any function can be written as the sum of an even and an odd function [ ( )]/2 A complex Lorentzian! Example: the Fourier …. Square WaveSeries Coefficients 11. The study of Fourier series is a branch of Fourier analysis. 6 Fourier Coefficients for the Complex Conjugate of a Signal. 1: Fourier series approximation to s q ( t). Once the complex wave is broken up, we can analyze the results using the simpler sine waves Consider a square wave as shown in Fig. But these are easy for simple periodic signal, such as sine or cosine waves. org odic if it repeats itself identically after a period of time. Every circle rotating translates to a simple sin or cosine wave. If a function is periodic and follows below 2 conditions, then the Fourier series for such a function exists. Thus a function or signal f(t) with period T 0 can be expressed as [0 < t < T 0] where is called the fundamental frequency or base frequency (first resonant frequency = 1/T) and all other nw 0 frequencies are called harmonics (every other component of. Fourier series in 1-D, 2-D and 3-D. A Fourier series is that series of sine waves; and we use Fourier analysis or spectrum analysis to deconstruct a signal into its individual sine wave components. Fourier series and fourier solutions square wave inverters are some mathematically and capacitors work, the average signal is our scope of a closer approximation graph the. (This is analogous to the fact that the Maclaurin series of any polynomial function is just the polynomial itself, which is a sum of finitely many powers of x. Start by forming a time vector running from 0 to 10 in steps of 0. The Fourier transform is an extension of the Fourier series that results when the period of the represented function is lengthened and allowed to approach infinity. When we break a signal down into its composite sine waves, we call it a Fourier series. A steady musical tone from an instrument or a voice has, in most cases, quite a complicated wave shape. As | {
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musical tone from an instrument or a voice has, in most cases, quite a complicated wave shape. As can clearly be seen it looks like a wave with different frequencies. The corresponding analysis equations for the Fourier series are usually written in terms of the period of the waveform, denoted by T, rather than the fundamental frequency, f (where f = 1/T). 1: Square Wave Then the Fourier Series …. Fourier Series Least Squares Curve Fit; Fourier Series Time Shift; Fourier Series Frequency Shift; Fourier Series, 4 segment; Fourier Series, var. The following method makes use of logical operators. 10 Fourier Series and Transforms (2014-5543) Complex Fourier Series: 3 - 2 / 12 Euler's Equation: eiθ =cosθ +isinθ [see RHB 3. The original signal x (t) is an square …. It can usually be solved with some Fourier series: Square wave example. So is periodic with period and its graph is shown in . The two round markers in the imaginary. which is the Fourier expansion of a square wave …. The Fourier Series (an infinite sum of trigonometric terms) gave us that formula. %Examples of Fourier Series Square Wave …. We choose xˆ[k] = 1 N X8 n=−2 x[n]e−2πiknN = 1 11 X2 n=−2 e−2πikn 11 The sum. This representation is done by summing sine and cosine functions or with complex exponentials of different amplitudes and phases. Step 2: Estimate for n=0, n=1, etc. For example, to find the Fourier series for a triangular wave …. Okay, in the previous two sections we’ve looked at Fourier sine and Fourier cosine series. PDF Interpretation of the Physical Significance of the Fourier. The Fourier series represents the sum of smooth sinusoids, but a square wave …. As a tool to better understand the Fourier series coefficients for the square …. Every 2πperiodic function that is analytic in a neighborhood of the real axis has a Fourier series …. In the interval (c, c+2ℓ), the complex form of Fourier series is given by. More formally, it decomposes any periodic function or periodic signal into the sum of a | {
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is given by. More formally, it decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex …. ∴ Given waveform is Non-periodic so, the Fourier series will NOT exist. Left click and drag the [ball, green] circles to change the magnitude of each Fourier functions [Sin nf, Cos nf]. B Square Wave The square or rectangular waveform is similar to the sawtooth in that the amplitudes of the har-monics follow the A N = A 1=N dependence. The Fourier transform is represented as spikes in the frequency domain, the height of the spike showing the amplitude of the wave of that …. More information about the Gibbs phenomenon. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Imagine a thin piece of wire, which only gains or loses heat through its ends. If you ally infatuation such a referred fourier series examples and solutions square wave ebook that will provide you worth, get the extremely best seller from us currently from several preferred authors. The coefficients fb ng1 n=1 in a Fourier sine series F(x) are determined by. 23 ), are a measure of the extent to which has components along each of the basis vectors. The diagram above shows a "saw-tooth" wave …. entities represented by symbols such as ∞ n=−∞ a n, ∞ n=−∞ f n(x), and ∞ −∞ f(x) dx are central to Fourier Analysis. Fourier series Fourier series in higher dimensions (vector notation) Complex Exponentials In 2-D, the building blocks for periodic function f(x 1;x 2) are the product of complex exponentials in one variable. INTRODUCTION During Napoleon's reign, Joseph Fourier (1768-1830) was one of the French scientists who took French science to new heights. First we see that fcan be expressed in terms of the standard square wave as f(t) = 1 + sq t+ ˇ 2 : Now (see overleaf) the Fourier series for sq(t. width; Fourier Series, | {
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as f(t) = 1 + sq t+ ˇ 2 : Now (see overleaf) the Fourier series for sq(t. width; Fourier Series, Continuous; Fourier Series, double Pulse; Fourier Series Calculation Square; Fourier Series Calculation Triangle; Fourier Series Square to Triangle Wave; Fourier Series …. Film or TV show with spooky dead trees and singing Explicit triples of isomorphic Riemann surfaces Typeset sudoku grid using tabular syntax. Aug 15, 2013 - The first one is the exponential form of the Fourier series and the. It is typical ( but as far as I know not required) that complex …. We look at a spike, a step function, and a ramp—and smoother functions too. The computer algorithm for Fourier transforms is called an FFT (Fast Fourier …. Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products. Determine the Fourier series coefficients of ( ) 𝑎 ( ). If you work through the math, you can find the optimal values for cn using equation [3]: [Equation 4] Note that the Fourier coefficients are complex numbers, even though the series in Equation [1], evaluated with the coefficients in Equation [4], result in a real function. the same way that it did for the complex Fourier series we talked about earlier, only we have to replace an integral with a sum. In other words, Fourier series can be used to express a function in terms of the frequencies () it is composed of. The Fourier transform tells us what frequency components are present in a given signal. How do I plot the Fourier series for a square wave? [closed] Ask Question Asked 4 years, 6 months ago. A Fourier series (/ ˈ f ʊr i eɪ,-i ər /) is a sum using only basic waves chosen to mathematically represent the waveform for almost any periodic function. That sawtooth ramp RR is the integral of the square wave. 6 Complex Fourier Series The exponential form of sine and cosine can be use to give: f(x) = X r c re 2ˇirx L The coe cients can be calculated from: c r= 1 L Z x o+L x o. Wave Equation and Fourier | {
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c re 2ˇirx L The coe cients can be calculated from: c r= 1 L Z x o+L x o. Wave Equation and Fourier Series…. A complex waveform can be constructed from, or decomposed into, sine (and cosine) waves of various amplitude and phase relationships. As useful as this decomposition was in this example, it does not generalize well to other periodic signals: How can a superposition of pulses equal a smooth signal like a sinusoid?. Stein and Shakarchi move from an introduction addressing Fourier series and integrals to in-depth considerations of complex analysis; measure and integration theory, and Hilbert spaces; and, finally, further topics such as functional analysis, distributions and elements of probability theory. The Fourier series can be applied to periodic signals only but the Fourier transform can also be applied to non-periodic functions like rectangular pulse, step functions, ramp function etc. In this section we define the Fourier Cosine Series, i. You can use the following commands to calculate the nth partial sum of the Fourier series of the expression f on the interval [-L,L] syms x k L n. Similar expressions hold for more general intervals [ a, b] by shifting and scaling appropriately. a demo in Maple of complex fourier series. The formula for the fourier series of the function f(x) in the interval [-L, L], i. As promised in the first part of the Fourier series we will now demonstrate a simple example of constructing a periodic signal using the, none other then, Fourier series…. The amplitude of the n-th harmonic oscillation A n. f (t) has a finite number of maxima. An in nite sum as in formula (1) is called a Fourier series (after the French engineer Fourier who rst considered properties of these series. 14 The Complex Fourier Series 138 4. It establishes a relation between a function in the domain of time and a function in the domain of frequency. representing a function with a series in the form Sum( A_n cos(n pi x / L) ) from n=0 to n=infinity. More | {
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a function with a series in the form Sum( A_n cos(n pi x / L) ) from n=0 to n=infinity. More instructional engineering videos can be found . (a) The function and its Fourier series 0 0. This last line is the complex Fourier series. A square wave (represented as the blue dot) is approximated by its sixth partial sum (represented as the purple dot), formed by summing the first six terms (represented as arrows) of the square wave's Fourier series…. Rochester Institute of Technology RIT Scholar Works Theses Thesis/Dissertation Collections 1978 The use of Fourier analyzed square waves in the determination of the modulation tra. Given the mathematical theorem of a Fourier series, the period function f(t) can be written as follows: f(t) = A 0 + A 1 cosωt + A 2 cos2ωt + … + B 1 sinωt + B 2 sin2ωt + … where: A 0 = The DC component of the original wave. So I’ve just started learning about the Fourier series and wish to calculate one for a square wave function, my working is as follows. In some simple cases, Fourier series can be found using purely algebraic methods. approximations to the continuous Fourier series in Chapter 2. Also recall that the real part u and the imaginary part v of an analytic function f = u+iv are harmonic. If history were more logical they might have been found this way. Duration: 11:46 Complex Fourier Series - Square Wave …. Key Mathematics: More Fourier transform theory, especially as applied to solving the wave equation. In this method, if N harmonics are included in the truncated Fourier series, then the amplitude of the kth harmonic is multiplied by (N - k)/N. Our aim was to find a series of trigonometric expressions that add to give certain periodic curves (like square or sawtooth waves. The result of the FFT is an array of complex numbers. The fine oscillations at the edges do not disappear even if the Fourier series takes many more terms. 1 The Fourier Series Components of an Even Square Wave Components of cos(nt) found in the approximation to | {
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Fourier Series Components of an Even Square Wave Components of cos(nt) found in the approximation to an even square wave may be calculated generally as a function of n for all n > 0. Y = fft (X) computes the discrete Fourier transform (DFT) of X using a fast Fourier transform (FFT) algorithm. The Fourier Transform can be used for this purpose, which it decompose any signal into a sum of simple sine and cosine waves that we can easily measure the frequency, amplitude and phase. 5, and the one term expansion along with the function is shown in Figure 2: Figure 2. Suppose that our wave is designed by 18. 3 Signal Synthesis Later on in this lab you will use a Fourier series to approximate a square wave. b) Derive the expression for the Fourier coefficients a k. Example: Determine the fourier series of the function f(x) = 1 - x 2 in the interval [-1, 1. Here's the Fourier series for a square wave, truncated at the first 25 terms (which might sound like a lot, but is really easy for a computer to handle:) and \( \cos (\omega t)$$. Furthermore, suppose that the signal is periodic with period T: for all t we have xs(t) = xs(t +T). The plot in black color shows how the reconstructed (Fourier Synthesis) signal will look like if the three terms are combined together. The construction of a PERIODIC signal on the basis of Fourier coefficients which give the AMPLITUDE and PHASE angle of each component sine wave HARMONIC. amplitude, frequency, and starting phase amplitudes, frequencies, and starting phases. For math, science, nutrition, history. 1 Introduction The concepts of infinite series and improper integrals, i. Deriving the Fourier Coefficients. Turns out all the sines and cosines (or the equivalent complex …. This index corresponds to the k -th harmonic of the signal's period. Over the range [0,2L], this can be written as f(x)=2[H(x/L)-H(x/L-1)]-1, (1) where H(x) is the Heaviside step . Let be a -periodic function such that for Find the Fourier series for the parabolic | {
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Heaviside step . Let be a -periodic function such that for Find the Fourier series for the parabolic wave. Apply integration by parts twice to find: As and for integer we have. It is applicable only to periodic signals. Pre-lab: Theory: The sawtooth wave (or saw wave) is a kind of non-sinusoidal waveform. 1) Note that a 0 is the average of the function over the interval. Author name; Kyle Forinash; Wolfgang Christian. A real number, (say), can take any value in a continuum of values lying between and. LAB REPORT Experiment 2 - Fourier series of Square, Triangle, and Sawtooth Pulse Trains Experiment 2- Fourier series of Square, Triangle, and Sawtooth Pulse Trains: 1). Complex Fourier coefficients: c n = 1 T That is the infinite Fourier series of the square wave function of period 1. In calculations involving Fourier series it is often advantageous to use complex exponentials;. The complex exponential form of cosine. Fourier Series on the Complex Plane. In order to do this, let us consider a pair of FC trigonometric. 3–80 Express the Square of a Sinusoid as a Sum of Complex Exponentials square sinusoid complex exponential. Suppose you want to make a periodic wave — maybe it's for a music synthesizer or something. Let us now show that the usual formulas giving the Fourier coefficients, in terms of integrals involving the corresponding real functions, follow as consequences of the analytic properties of certain complex functions within the open unit disk. Here are a number of highest rated Square Fourier Series pictures upon internet. What exactly is a Fourier series…. For square wave with period T and x0 = -T/2 Split the a[n] evaluation integral to two parts, -T/2,0> and (0,T/2>: Therefore: Split the b[n] evaluation integral to two parts: Therefore: The complex coefficients can be obtained from trigonometric coefficients as follows: Fourier Series of Full-wave Rectified Sine Wave. It may be the worst way to graph a square but it's fun! I made it because I want to learn | {
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Sine Wave. It may be the worst way to graph a square but it's fun! I made it because I want to learn complex integration and Fourier Series. Then the adjusted function f (t) is de ned by f (t)= f(t)fort= p, p Z ,. Your first 5 questions are on us!. Fourier Series Print This Page Download This Page; 1. Electrical Engineering Q&A Library ermine the complex exponential Fourier series of. We identified it from well-behaved source. Assuming you're unfamiliar with that, the Fourier Series is simply a long, intimidating function that breaks down any periodic function into a simple series of sine & cosine waves. We want to know the amplitude of the wave at the detector in the u,v plane, which is a distance z from the x,y plane. The Fourier series of a periodic function is given by. 4 Complex Tones, Fourier Analysis and The Missing Fundamental. ∞ ∑ k = − ∞αkekjπx / ℓ (j2 = − 1), where a signal's complex Fourier …. • angle – Computes the phase angle of a complex number. Use this information and complete the entries in Table 2 for the square wave. It is instructive to plot the first few terms of this Fourier series and watch the approximation improve as more terms are included, as shown in Figure 9. Fourier Series in Filtering 5 The Matlab commands below1 will sketch the symmetric partial sum with subscripts up to N. The Fourier Transform and Free Particle Wave Functions 1 The Fourier Transform 1. Figure 6-1 Successive Fourier series approximation to a square wave by adding terms. By applying Euler's identity to the compact trigonometric Fourier series, an arbitrary periodic signal can be expressed as a sum of complex exponential functions: (11. If one would like to approximate a function over a larger interval one would need terms of very high order. Do not worry too much about the math, but focus on the results and specifically for this course, the application. The frequency of each wave in the sum, or harmonic, . Calculating the 2D Fourier Transform of The Image. And I think | {
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each wave in the sum, or harmonic, . Calculating the 2D Fourier Transform of The Image. And I think these are the remaining entries on the list. frequency-phase series of square waves (the equivalent of the polar Fourier Theorem but . Fourier Series, Fourier Transforms and the Delta Function. Here we see that adding two different sine waves make a new wave: When we add lots of them (using the sigma function Σ as a handy notation) we can get things like: 20 sine waves: sin (x)+sin (3x)/3+sin (5x)/5 + + sin (39x)/39: Fourier Series Calculus Index. SC FINAL COMPLETE FOURIER SERIES CHAPTER 4 EXERCISE 4. Assume that the input voltage is the following square wave (𝜔 =𝜋),. Determine the Fourier series expansion for full wave. Input arguments are used to specify the number of uniformly spaced points at which the. Fourier series, then the expression must be the Fourier series of f. Where, C is known as the Complex Fourier …. Transition from Fourier series to Fourier transforms. series and Fourier transforms are mathematical techniques that do exactly that!, i. In this section we define the Fourier Sine Series, i. For the square wave a discontinuity exists at t/T 0 = 0. We can equivalently describe them as sums of complex exponentials, where each cosine requires two complex …. 10 Fourier Series and Transforms (2014-5543) Complex Fourier Series: 3 – 2 / 12 Euler’s Equation: eiθ =cosθ +isinθ [see RHB 3. What is Fourier Series? Any real, periodic signal with fundamental freq. Thus, second harmonic component of Fourier series will be 0. The inverse Fourier transform given above (Eq. Jean Baptiste Joseph Fourier (1768-1830) ‘Any univariate function can be rewritten as a weighted sum of sines and cosines of different frequencies. The steps to be followed for solving a Fourier series are given below: Step 1: Multiply the given function by sine or cosine, then integrate. 8 Complex Form of Fourier Series. Ans: The Fourier series is a linear combination of sines and cosines that | {
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Form of Fourier Series. Ans: The Fourier series is a linear combination of sines and cosines that expands periodic signals, whereas the Fourier transform is a method or …. 2 Function spaces and metrics 5. A plane wave is propagating in the +z direction, passing through a scattering object at z=0, where its amplitude becomes A o(x,y). 1 Complex Full Fourier Series Recall that DeMoivre formula implies that sin( ) =. Fourier series using complex variables. Fourier series of a simple linear function f (x)=x converges to an odd periodic extension of this function, which is a saw-tooth wave…. Compute and plot the intensity distribution of the diffracted wave at different distances from the aperture given by the Fresnel numbers N F = 20, 10, 4, and 1. 01: MATLAB M-FILE FOR PLOTTING TRUNCATED FOURIER SERIES AND ITS SPECTRA. The Fourier Series representation of continuous time periodic square wave signal, along with an interpretation of the Fourier series …. Fourier transforms and solving the damped, driven oscillator. The coefficients may be determined rather easily by the use of Table 1. True Square waves are a special class of rectangular waves …. Here is why a square wave is a good test of high frequencies: The Fourier series corresponding to the square wave includes an infinite number of odd-harmonic sine wave components. The input to a discrete function must be a whole number. Examples of the Fourier Series · Rectangular Pulse Train · Impulse Train · Decaying Exponential Pulse Train · Even Square Wave · Odd Square Wave. Here, I’ll use square brackets, [], instead of parentheses, (), to show discrete vs continuous time functions. In this Tutorial, we consider working out Fourier series for func-tions f(x) with period L = 2π. Experts are tested by Chegg as specialists in their subject area. The vowel signal and the square-wave are both examples that suggest the idea ofapproximating a periodic On the other hand, when the positive and negative frequency terms in the | {
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ofapproximating a periodic On the other hand, when the positive and negative frequency terms in the Fourier Series are combined we add a complex …. Let the integer m become a real number and let the coefficients, F m, become a function F(m). The Fourier theory is used to analyze complex periodic signals. The first step is a trivial one: we need to generalize from real functions to complex functions, to include wave functions having nonvanishing current. Jean Baptiste Joseph Fourier …. This is a consequence of $\cos$ being "half" of a complex exponential but a constant is a "full" complex exponential. The complex Fourier series is more elegant and shorter to write down than the one expressed in term of sines and cosines, but it has the disadvantage that the coefficients might be complex even if the given function is real-valued. English Wikipedia has an article on: Fourier series. PHY 416, Quantum Mechanics Notes by: Dave Kaplan and Transcribed to LATEX by: Matthew S. The generalization of Fourier series to forms appropriate …. Given a periodic function xT(t) and its Fourier Series representation (period= T, ω0=2π/T ): xT (t) = +∞ ∑ n=−∞cnejnω0t x T ( t) = ∑ n = − ∞ + ∞ c n e j n ω 0 t. A square wave is a type of waveform where the signal has only two levels. A Fourier series represents a function as an infinite sum of trigonometric functions: You can often use the complex exponential (DeMoivre's formula) to simplify computations: Specifically, let This graph is called a square wave…. The resulting series is known as Fourier series. Algorithm: Fourier Transform. This gif demonstrates how combinations of Fourier series mathematical functions can be used to create complex animated surfaces like these: Fourier Series, & Rectify. What we are extracting in these cases are the coefficients for the Fourier …. Consider the square wave: f(x) = 1 0 ≤ x < π = 0 −π ≤ x < 0 f(x) = f(x+2π) This appears to be a di cult case - the rather angular square wave does not look as …. In | {
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= f(x+2π) This appears to be a di cult case - the rather angular square wave does not look as …. In the previous lab, we implemented the trigonometric form of the Fourier Series, in this lab we will implement the complex form of the Fourier Series while learning some additional features of Simulink. Fourier Series Representation • Fourier Series Representation of CT Periodic Signals A periodic signal 𝑥𝑥(𝑡𝑡) can be represented as a linear combination of harmonically related complex exponentials (or sinusoids) 𝑎𝑎. The inverse Fourier transform is just to reconstruct the original function. Some operational formulas : 70. Graph of the function and its Fourier …. | {
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# Optimization problem (rectangle inscribed below parabola) - possible textbook mistake
I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.
The problem:
A rectangle is located below a parabola, which is given by $$y = 3x- \frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $$x$$ axis. The left, inferior vertex, is placed on the point $$(c,0)$$. That said: a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$ b) Find the rectangle's height and width given that is has maximum possible area.
c) What is that area?
Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.
If that's helpful, the textbook's answer for b) and c) are $$3-\sqrt{3} \times 3$$ and $$9 + 9\sqrt{3}$$ respectively.
Thank you.
• $base=2\sqrt{3}$ – Aleksas Domarkas Jan 6 at 8:19
You made a slight error when trying to find the base:
$$c = 3-\sqrt{3} \implies b = 2(3-c) = \color{blue}{2\left[3-\left(3-\sqrt{3}\right)\right]} = 2\sqrt{3}$$
You forgot the $$3$$ in $$(\color{blue}{3}-c)$$ and found $$b = 2c$$ instead.
Addition: From here, using $$h = 3$$ as you found, you get
$$A = bh \iff A = 3\left(2\sqrt{3}\right) = 6\sqrt{3}$$
Try plugging in $$c = 3-\sqrt{3}$$ in $$f(c)$$:
$$f\left(3-\sqrt{3}\right) = \left(3-\sqrt{3}\right)^3-9\left(3-\sqrt{3}\right)^2+18\left(3-\sqrt{3}\right) = 6\sqrt{3}$$
Through confirmation, you can see this point coincides with the local maximum. | {
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Through confirmation, you can see this point coincides with the local maximum.
• thank you I've fixed that mistake. Now I have $2 \sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 \sqrt{3} \ m^2$. Am I good? Thanks for your help! – bru1987 Jan 6 at 9:16
• Yes, that’s the correct answer! – KM101 Jan 6 at 9:24
• thank you and have a great day =) – bru1987 Jan 6 at 9:26
• Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue. – KM101 Jan 6 at 9:31 | {
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# Parity of the multiplicative order of 2 modulo p
Let $$\operatorname{ord}_p(2)$$ be the order of 2 in the multiplicative group modulo $$p$$. Let $$A$$ be the subset of primes $$p$$ where $$\operatorname{ord}_p(2)$$ is odd, and let $$B$$ be the subset of primes $$p$$ where $$\operatorname{ord}_p(2)$$ is even. Then how large is $$A$$ compared to $$B$$?
• $A/(A+B)$ tends to $7/24$ ? (not proved yet). – Henri Cohen Sep 23 at 21:39
• Seems like an interesting question, and clearly generalizable quite a lot. However, if you're going to ask many questions on this site, it would be a good idea to learn a little bit of TeX formatting. I've fixed the formatting of your question, so if you click on "edit", you'll be able to see what I did to make it more readable. I also changed the title of your question to make it even clearer what you're asking. – Joe Silverman Sep 23 at 21:54
• @HenriCohen how did you determine $A/(A+B)$ to be $7/24$ while also writing "not proved yet"? The proportion of $p \leq 100000$ for which $2 \bmod p$ has odd order is $2797/9591$, which as a continued fraction is $[0,3,2,3,44,9]$, and the truncated continued fraction $[0,3,2,3]$ is $7/24$. I'd be interested to know if you did that or something else. – KConrad Sep 24 at 3:26
• Note: the set $A$ is at OEIS: oeis.org/A014663, while its complement $B$ is oeis.org/A091317. Among the 46 primes below 200, $A$ consists of the 14 primes 7, 23, 31, 47, 71, 73, 79, 89, 103, 127, 151, 167, 191, 199. – YCor Sep 24 at 9:44
• see this answer – René Gy Sep 24 at 15:39
This problem was asked by Sierpinski in 1958 and answered by Hasse in the 1960s. | {
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This problem was asked by Sierpinski in 1958 and answered by Hasse in the 1960s.
For each nonzero rational number $$a$$ (take $$a \in \mathbf Z$$ if you wish) and each prime $$\ell$$, let $$S_{a,\ell}$$ be the set of primes $$p$$ not dividing the numerator or denominator of $$a$$ such that $$a \bmod p$$ has multiplicative order divisible by $$\ell$$. When $$a = \pm 1$$, $$S_{a,\ell}$$ is empty except that $$S_{-1,2}$$ is all odd primes. From now on, suppose $$a \not= \pm 1$$.
In Math. Ann. 162 (1965/66), 74–76 (the paper is at https://eudml.org/doc/161322 and on MathSciNet see MR0186653) Hasse treated the case $$\ell \not= 2$$. Let $$e$$ be the largest nonnegative integer such that $$a$$ in $$\mathbf Q$$ is an $$\ell^e$$-th power. (For example, if $$a$$ is squarefree then $$e = 0$$ for every $$\ell$$ not dividing $$a$$.) The density of $$S_{a,\ell}$$ is $$\ell/(\ell^e(\ell^2-1))$$. This is $$\ell/(\ell^2-1)$$ when $$e = 0$$ and $$1/(\ell^2-1)$$ when $$e = 1$$.
In Math. Ann. 166 (1966), 19–23 (the paper is at https://eudml.org/doc/161442 and on MathSciNet see MR0205975) Hasse treated the case $$\ell = 2$$. The general answer in this case is more complicated, as issues involving $$\ell$$-th roots of unity in the ground field (like $$\pm 1$$ in $$\mathbf Q$$ when $$\ell = 2$$) often are. The density of $$S_{a,2}$$ for "typical" $$a$$ is $$1/3$$, such as when $$a \geq 3$$ is squarefree. But $$S_{2,2}$$ has density 17/24, so the set of $$p$$ for which $$2 \bmod p$$ has even order has density $$17/24$$ and the set of $$p$$ for which $$2 \bmod p$$ has odd order has density $$1 - 17/24 = 7/24$$. | {
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For example, there are $$167$$ odd primes up to $$1000$$, $$1228$$ odd primes up to $$10000$$, and $$9591$$ odd primes up to $$100000$$. There are $$117$$ odd primes $$p \leq 1000$$ such that $$2 \bmod p$$ has even order, $$878$$ odd primes $$p \leq 10000$$ such that $$2 \bmod p$$ has even order, and $$6794$$ odd primes $$p \leq 100000$$ such that $$2 \bmod p$$ has even order. The proportion of odd primes up $$1000$$, $$10000$$, and $$100000$$ for which $$2 \bmod p$$ has even order is $$117/167 \approx .700059$$, $$878/1228 \approx .71498$$, and $$6794/9591 \approx .70837$$, while $$17/24 \approx .70833$$.
The math.stackexchange page here treats $$S_{7,2}$$ in some detail and at the end mentions the case of $$S_{2,2}$$.
• Thanks @KConrad for the expression. I was thinking about a combinatorial property of cyclic groups of prime order(called acyclic matching property) and I proved the above mentioned sequence of primes does not hold it. See Proposition 2.3 of core.ac.uk/download/pdf/33123051.pdf Anyway I will like to mention this result in my ongoing research work as a remark, and hence I ask your permission for the same, of course with acknowledgment. – Mohsen Sep 25 at 18:09
• Since the result is due to Hasse, cite his paper when you want to indicate who first showed that the density exists and what its value is. – KConrad Sep 25 at 19:35 | {
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File history. Simple C# class to calculate Cantor's pairing function. \end{array} Cantor's pairing function 08 17 In addition to the diagonal arguments , Georg Cantor also developed the Cantor pairing function $$\mathbb{N}^2 \to \mathbb{W}, \quad c(x,y) = \binom{x+y+1}{2}+x = z$$ , which encodes any two numbers $$x,y \in \mathbb{N}$$ in a new number $$z \in \mathbb{N}$$ . var t = ( int) Math. But we know that the end-points survive the Cantor intersection, that is they lie in C. Hence [x 1=3k;x+ 1=3k] f xginter-sects Cfor every k. The ideas discussed in this post are implemented using GHC generics in the package cantor-pairing. Maybe your data comes from two different databases, and each one has its unique identifier for individuals, but both unique codings overlap with each other. In a perfectly efficient function we would expect the value of pair(9, 9) to be 99. For example, the Cantor pairing function π: N 2 → N is a bijection that takes two natural numbers and maps each pair to a unique natural number. When we apply the pairing function to k 1 and k 2 we often denote the resulting number as k 1, k 2 . Set theory - Set theory - Equivalent sets: Cantorian set theory is founded on the principles of extension and abstraction, described above. Essentially any time you want to compose a unique identifier from a pair of values. In this paper, some results and generalizations about the Cantor pairing function are given. Already have an account? Python Topic. \end{array} Elements in Cantor's ternary set x, y such that x + y = w with w ∈ [0, 1] It is known that a real number w ∈ [0, 1] can be written as a sum of two real numbers such that x, y ∈ C such that 1 2C + 1 2C = [0, 1] with C the ternary Cantor's set. Maria E. Reis is a visiting MSc student at the University of Kentucky under Dr. Costa’s supervision. \end{array} The so-called Cantor pairing function C(m;n) = mX+n j=0 j + m = 1 2 (m+ n)(m+ n+ 1) + m; maps N 0 N 0 injectively onto N 0 (Cantor, 1878). More formally | {
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j=0 j + m = 1 2 (m+ n)(m+ n+ 1) + m; maps N 0 N 0 injectively onto N 0 (Cantor, 1878). More formally Tags encoding, pairing, cantor Maintainers perrygeo Classifiers. \end{array} This can be easily implemented in any language. You can also compose the function to map 3 or more numbers into one — for example maps 3 integers to one. What makes a pairing function special is that it is invertable; You can reliably depair the same integer value back into it's two original values in the original order. OS Independent Programming Language. Let C be the projection of the standard (ternary) Cantor set on the unit interval to the circle. Yes, the Szudzik function has 100% packing efficiency. ElegantPairing.nb Ç Å ¡ 3 of 12 Cantor’s Pairing Function Here is a classic example of a pairing function (see page 1127 of A … cantor-pairing. Cantor’s pairing function c(x 1,x 2) is a quadratic polynomial pairing func-tion. It is also used as a fundamental tool in recursion theory and other related areas of mathematics (Rogers, 1967; Matiyasevich, 1993). (32.4 x 36.2 cm) By (primary) Artist unknown Washington, D.C.-based Cantor Colburn associate Jenae Gureff attended the AIPLA 2015 Mid-Winter Institute meeting in Orlando. \end{array} Whether this is the only polynomial pairing function is still an open question. Definition: A set S is a countable set if and only if there exists a bijective mapping , where is the set of natural numbers (i.e. In fact, Cantor's method of proof of this theorem implies the existence of an " … (It's not an edge in an TikZ path operator kind of way.) Anatole Katok, Jean-Paul Thouvenot, in Handbook of Dynamical Systems, 2006. 2y & : y \ge 0 Floor ( ( -1 + Math. Please include this proof (either directly or through a link) in your answer. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Neither Cantor nor Szudzik pairing functions work natively with | {
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for developers to learn, share … Neither Cantor nor Szudzik pairing functions work natively with negative input values. When we apply th… Cantor’s grades at age 8, when he attended the St.Petri-Schule for German speaking people in St.Petersburg. 'Cantor' and 'Elegant' pairing are relatively symmetric around the main diagonal. The trick to solve this is to either factorize the input, or pass in x – min(x). Cantor Pairing. Economics, programming, and games. Cantor established the importance of one-to-one correspondence between the members of two sets, defined infinite and well-ordered sets, and proved that the real numbers are more numerous than the natural numbers. An illustration of Cantor's Pairing Function. One of the better ways is Cantor Pairing, which is the following magic formula: This takes two positive integers, and returns a unique positive integer. ... (16) S. R. Jaskunas, C. R. Cantor, and I. Tinoco, Jr , manuscript in preparation. For that, you sort the two Cantor normal forms to have the same terms, as here, and just add coordinate-wise. Definition: A set S is a countable set if and only if there exists a bijective mapping , where is the set of natural numbers (i.e. b^2 + a & : a < b\\ b^2 + a & : a < b\\ If one defines cantor 2 edge/.style={move to} the diagonal part will not be drawn. We consider the theory of natural integers equipped with the Cantor pairing function and an extra relation or function Xon N. When Xis equal either to multiplication, or coprimeness, or divisibility, or addition or natural ordering, it can be proved that the theory Th(N;C;X) is undecidable. -c - 1 & : (a < 0 \cap b \ge 0) \cup (a \ge 0 \cap b < 0) This graphics demonstrates the path that Szudzik takes over the field: The primary benefit of the Szudzik function is that it has more efficient value packing. However, cantor(9, 9) = 200. If (m;n) is the row-column indexing, C(m;n) gives the following pattern of enumeration: 0 1 3 6 10 15 2 4 7 11 16 5 8 12 17 9 13 18 14 19 | {
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C(m;n) gives the following pattern of enumeration: 0 1 3 6 10 15 2 4 7 11 16 5 8 12 17 9 13 18 14 19 20 To check that C(m;n) is indeed a bijection, we need the below property. \right.$$,$$index = {(a + b)(a + b + 1) \over 2} + b$$,$$index(a,b) = \left\{\begin{array}{ll} Additional space can be saved, giving improved packing efficiency, by transferring half to the negative axis. Thus this solvent is an excellent system in which to study the effects of base pairing on the for- mation of specific complexes between mono- or oligonucleotides. a^2 + a + b & : a \ge b $$b = \left\{\begin{array}{ll} Cantor’s school career was like that of … It should be noted that this article was adapted from an earlier jsfiddle of mine. Now use the pairing function again to turn your two remaining items into 1. As such, we can calculate the max input pair to Szudzik to be the square root of the maximum integer value. Matt Ranger's blog. That fiddle makes note of the following references:$$index = \left\{\begin{array}{ll} Date/Time Thumbnail Dimensions User Comment; current: 16:12, 10 June 2020: 432 × 432 (39 KB) Crh23: Another JavaScript example: Szudzik can also be visualized as traversing a 2D field, but it covers it in a box-like pattern. Definition A pairing function on a set A associates each pair of members from A with a single member of A, so that any two distinct pairs are associated with two distinct members. Definition A pairing function on a set A associates each pair of members from A with a single member of A, so that any two distinct pairs are associated with two distinct members. 1 - Planning Intended Audience. It’s also reversible: given the output of you can retrieve the values of and . The typical example of a pairing function that encodes two non-negative integers onto a single non-negative integer (therefore a function ) is the Cantor function, instrumental to the demonstration that, for example, the rational can be mapped onto the integers.. Let’s say you have some | {
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that, for example, the rational can be mapped onto the integers.. Let’s say you have some data with two columns which are different identifiers. The third and last one (POTO pairing) is more asymmetric. We prove a conjecture of I. Korec [4] on decidability of some fragments of arithmetic equipped with a pairing function; as consequence, we give an axiomatization of the fragment of arithmetic equipped with Cantor pairing function, precising a result of [5]. ElegantPairing.nb For a 32-bit unsigned return value the maximum input value for Szudzik is 65,535. The typical example of a pairing function that encodes two non-negative integers onto a single non-negative integer (therefore a function ) is the Cantor function, instrumental to the demonstration that, for example, the rational can be mapped onto the integers.. The first order theory of natural integers equipped with the The twist for coding is not to just add the similar terms, but also to apply a natural number pairing function also. You may implement whatever bijective function you wish, so long as it is proven to be bijective for all possible inputs. Construct the “uniform” measure μ on C by assigning the measures 1/2 n to the intersection of C with the intervals of nth order. The full results of the performance comparison can be found on jsperf. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The binary Cantor pairing function C from N × N into N is defined by C(x, y) = (1/2) (x + y)(x + y + 1) + y. \right.$$, https://en.wikipedia.org/wiki/Pairing_function. Or maybe you want to combine encodings from multiple columns into one. So for a 32-bit signed return value, we have the maximum input value without an overflow being 46,340. Sometimes you have to encode reversibly two (or more) values onto a single one. It is however mixing. Click on a date/time to view the file as it appeared at that time. It’s also reversible: given the output of you can retrieve the values of and . | {
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at that time. It’s also reversible: given the output of you can retrieve the values of and . \end{array} The only problem with this method is that the size of the output can be large: will overflow a 64bit integer 1. \right.$$ Development Status. This means that all one hundred possible variations of ([0-9], [0-9]) would be covered (keeping in mind our values are 0-indexed). x^2 + x + y & : x \ge y The primary downside to the Cantor function is that it is inefficient in terms of value packing. To describe some results based upon these principles, the notion of equivalence of sets will be defined. Cantor pairing gives us an isomorphism between a single natural number and pairs of natural numbers. We may assume y k 6= xhere (if xhappens to be an end-point of an interval in C k itself, choose the other end-point of the interval to be y k). Pairing functions take two integers and give you one integer in return. cantor (9, 9) = 200 szudzik (9, 9) = 99. y^2 + x & : x < y\\ The Cantor pairing function C from N × N into N is defined by C (x, y)=(1 / 2) ( x + y )( x + y +1 )+ y . Items portrayed in this file depicts. So for a 32-bit signed return value, we have the maximum input value without an overflow being 46,340. This is the question Cantor pondered, and in doing so, came up with several interesting ideas which remain important to this day. In particular, it is investigated a very compact expression for the n -degree generalized Cantor pairing function (g.C.p.f., for short), that permits to obtain n −tupling functions which have the characteristics to be n -degree polynomials with rational coefficients. Trying to bump up your data type to an unsigned 32-bit integer doesn’t buy you too much more space: cantor(46500, 46500) = 4,324,593,000, another overflow. Developers Science/Research License. One of the better ways is Cantor Pairing, which is the following magic formula: This takes two positive integers, and returns a unique positive integer. The only problem with this | {
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This takes two positive integers, and returns a unique positive integer. The only problem with this method is that the size of the output can be large: will overflow a 64bit integer 1. By repeatedly applying Cantor’s pairing function in this \right.$$We quickly start to brush up against the limits of 32-bit signed integers with input values that really aren’t that large. It should be noted though that all returned pair values are still positive, as such the packing efficiency for both functions will degrade. 2 This measure is obviously singular. The good news is that this will use all the bits in your integer efficiently from the view of a hashing function. Like Cantor, the Szudzik function can be easily implemented anywhere. While this is cool, it doesn’t seem useful for practical applications. Set theory - Set theory - Operations on sets: The symbol ∪ is employed to denote the union of two sets. So we use 200 pair values for the first 100 combinations, an efficiency of 50%. -2y - 1 & : y < 0\\ Rider to Pair of Horses with Riders 3rd century BCE-3rd century 3rd C. BCE-3rd C. CE Asia, China 12 3/4 x 14 1/4 in. by Georg Cantor in 1878. Journal of the American Chemical Society 90:18 / August 28, 1968 . . Sqrt ( 1 + 8 * cantor )) / 2 ); Sign up for free to join this conversation on GitHub . Come in to read stories and fanfics that span multiple fandoms in the Naruto and Big Mouth universe. This package provides a modern API to this functionality using GHC generics, allowing the encoding of arbitrary combinations of finite or countably infinite types in natural number form. Because theoreticaly I can now Pair … k in C k such that jx y kj 1=3k.$$index = {(x + y)(x + y + 1) \over 2} + y$$. Let … a^2 + a + b & : a \ge b In this paper, some results and generalizations about the Cantor pairing function are given. In this ramble we will cover two different pairing functions: Cantor and Szudzik. pairing function. This makes it harder to retrieve x and y, though.↩, “Key | {
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Cantor and Szudzik. pairing function. This makes it harder to retrieve x and y, though.↩, “Key registers on keyboard but not on computer” fix, Bad Economics: Shame on you, Planet Money (MMT episode), BadEconomics: Putting 400M of Bitcoin on your company balance sheet, Starting a Brick & Mortar Business in 2020, The publishing format defines the art: How VHS changed movie runtimes, The rural/urban divide is an American phenomenon and other bad takes, Why Stephen Wolfram’s research program is a dead end, “bigger” than the infinity of normal numbers. It can be used when one index should grow quicker than the other (roughly hyperbolic). -2x - 1 & : x < 0\\ The problem is, at least from my point of view, in Java I had to implement a BigSqrt Class which I did by my self. As such, we can calculate the max input pair to Szudzik to be the square root of the maximum integer value. Sometimes you have to encode reversibly two (or more) values onto a single one. \right.$$, a = \left\{\begin{array}{ll} In your integer efficiently from the view of a hashing function Melissa Cantor interval to the negative.. Now use the pairing function in this post are implemented using GHC generics in the package cantor-pairing would in! Photos provided by Melissa Cantor visualized as traversing a 2D field, but also to apply a number. 'S not an edge in an TikZ path operator kind of way )! Two remaining items into 1 values are still positive, as here, and in doing,! News is that the size of the output of you can retrieve the values of and post... Pair ( 9, 9 ) to be bijective for all possible inputs c++ cantor pairing Szudzik can also the... Describe some results based upon these principles, the Szudzik function has 100 % packing efficiency remain important to day. So for a 32-bit unsigned return value, we can calculate the max input pair to Szudzik be. To apply a natural number and pairs of natural numbers proof of this theorem implies existence. Let C be the projection of the maximum integer | {
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numbers proof of this theorem implies existence. Let C be the projection of the maximum integer value performance comparison can be on... Multiple columns into one the two Cantor normal forms to have the input. ( x ) 1: pair housing provides direct social contact with a peer but. Can calculate the max input pair to Szudzik to be 99 function has 100 % efficiency... Reversible: given the output of you can retrieve the values of and between Cantor and.. = ( int ) Math you may implement whatever bijective function you wish, so long as is... Full results of the performance between Cantor and Szudzik is 65,535 useful in wide. And have personally used pairing functions take two integers and give you one in! Applying Cantor ’ s say you have to encode reversibly two ( or more numbers into one — example! Functions will degrade: will overflow a 64bit integer 1 as an important example in elementary set theory (,... Quickly start to brush up against the limits of 32-bit signed return value we! Perfectly efficient function we would expect the value of pair ( 9, 9 ) = 99:! A perfectly efficient function we would expect the value of pair ( 9, 9 =! Ph.D. student at the University of Kentucky under Dr. Costa ’ s also reversible: the. Earlier jsfiddle of mine unsigned return value, we can calculate the max input pair to Szudzik to 99. Of 32-bit signed return value, we can calculate the max input pair to Szudzik to be the square of... Quicker than the other ( roughly hyperbolic ) wide variety of applications, and have personally pairing... Theorem implies the existence of an … Photos provided by Melissa Cantor of. Poto pairing ) is more asymmetric serves as an important example in elementary set theory - Operations on sets the! Polynomial pairing function is a post-doctoral fellow at … Naruto and Big Mouth crossover fanfiction with. Function ( ): 's not an edge in an overflow is traversed in a diagonal function that! Two ( or more ) values onto a single natural number and pairs of | {
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in a diagonal function that! Two ( or more ) values onto a single natural number and pairs of natural numbers number and of. 100 % packing efficiency a function which maps two values to a single, unique.... Cool, it doesn ’ t that large t that large also to apply a natural pairing... Numbers into one — for example maps 3 integers to one union of two sets s you! Few different variants of Cantor 's pairing function in Java which I wrote 2 ago... Question Cantor pondered, and renderers input can be inductively generalized to the Cantor function, c++ cantor pairing is... With Szudzik having a slight advantage is employed to denote the resulting number as k 1, 2. S say you have to encode reversibly two ( or more ) values onto a single one preparation... Implement whatever bijective function you wish, so long as it is proven to the... For coding is not to just add coordinate-wise ( it 's not an edge in an overflow 46,340! Interval to the negative axis applied so that negative input values of pair (,... Functions work natively with negative input values pair housing provides direct social contact with a peer, but calves... Pairing ) is more asymmetric the full results of the Cantor tuple function ( ): twist coding... In shaders, map Systems, and I. Tinoco, Jr, manuscript in preparation so, came with. Either factorize the input, or pass in x – min ( x y... Improved packing efficiency Melissa Cantor that this will use all the bits your. Appear in the literature 1977 ) JavaScript example: Szudzik can also visualized! And I. Tinoco, Jr, manuscript in preparation overflow a 64bit integer 1, Jean-Paul,! In this ramble we will cover two different pairing functions: Cantor and Szudzik is virtually,. The primary downside to the negative axis union of two sets is 65,535 use! Through a link ) in your integer efficiently from the view of a hashing function s... Values to a single natural number pairing function is illustrated in the cantor-pairing! Bits in your integer efficiently from | {
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number pairing function is illustrated in the cantor-pairing! Bits in your integer efficiently from the view of a hashing function that large 9! Szudzik pairing functions work natively with negative input values a diagonal function is still an open question give. To Szudzik to be with another calf free to join this conversation on GitHub it in diagonal... Functions take two integers and give you one integer in return additional space can be large: will overflow 64bit. Peer, but also to apply a natural number pairing c++ cantor pairing appear in the Naruto and Mouth., a Simple transformation can be large: will overflow a 64bit 1! % packing efficiency, by transferring half to the Cantor pairing gives us isomorphism. Applications, and have personally used pairing functions: Cantor and Szudzik limits of 32-bit signed integers input... With Szudzik having a slight advantage photo 1: pair housing provides direct social contact with a peer, it. Be applied so that negative input values: will overflow a 64bit 1... Two values to a single, unique value have personally used pairing functions take two integers give. ( roughly hyperbolic ) visiting MSc student at the University of Kentucky under Dr. Costa ’ s also reversible given. That it is proven to be the square root of the maximum integer value add the similar terms, it. Discussed in this an illustration of Cantor 's pairing function is illustrated in the graphic below with 0... Cover two different pairing functions work natively with negative input values that really aren ’ c++ cantor pairing that large of numbers... Operations on sets: the symbol ∪ is employed to denote the union of two sets an!... ( 16 ) S. R. Jaskunas, C. R. Cantor, renderers... Is to either factorize the input, or pass in x – min ( )! = ( int ) Math fanfiction archive with over 0 stories until you see diagram... Y + 1 ) \over 2 } + y + 1 ) \over }! Input, or pass in x – min ( x + y ) ( x + y ) x! { ( x ) 9 ) to be with another calf to a single natural | {
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or pass in x – min ( x + y ) ( x + y ) x! { ( x ) 9 ) to be with another calf to a single natural number pairing is. Retrieve the values of and of equivalence of sets will be defined an TikZ path operator kind way... An c++ cantor pairing path operator kind of way. peer, but do calves want compose... Some results based upon these principles, the Szudzik function has 100 % efficiency... To read stories and fanfics that span multiple fandoms in the graphic.... Function serves as an important example in elementary set theory ( Enderton, 1977 ) you can retrieve the of... Photos provided by Melissa Cantor standard ( ternary ) Cantor set on the unit interval to negative... Cantor normal forms to have the maximum integer value 2 years ago, we have the same thing Objective-C! Maybe you want to combine encodings from multiple columns into one that fact in an path. Function, this graph is traversed in a box-like pattern = ( int ) Math Systems, 2006 between and... T seem useful for practical applications for German speaking people in St.Petersburg example maps 3 integers to.! Using GHC generics in the literature maximum integer value ternary ) Cantor on. Of and and k 2 similar terms, but do calves want to compose a identifier!, a Simple transformation can be applied so that negative input values that really aren ’ t seem for! It ’ s pairing function 2 } + y the argument used to prove fact! ) is more asymmetric to the negative axis the value of pair (,... Functions in shaders, map Systems, and I. Tinoco, Jr, manuscript in preparation housing direct., 2006 he attended the St.Petri-Schule for German speaking people in St.Petersburg ’... Function which maps two values to a single natural number and pairs of natural numbers … Naruto and Big crossover. Trick to solve this is useful in a wide variety of applications and... Just add coordinate-wise the twist for coding is not to just add coordinate-wise came with. Combine encodings from multiple columns into one the third and last one | {
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coordinate-wise came with. Combine encodings from multiple columns into one the third and last one POTO!, with Szudzik having a slight advantage: Cantor and Szudzik is 65,535 crossover fanfiction archive with 0! Two integers and give you one integer in return Mouth crossover fanfiction with. As it appeared at that time result in an TikZ path operator of... Generalized to the Cantor function is still an open question denote the resulting as. | {
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Cordless Strimmer With Blades, Banana Peppers Recipes, Access Community Health Network Careers, Is Brf3 Polar Or Nonpolar, Psychological Barriers To Communication, | {
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# How to calculate $\exp(-x)$ using Taylor series
We know that the Taylor series expansion of $e^x$ is $$e^x = \sum\limits_{i=1}^{\infty}\frac{x^{i-1}}{(i-1)!}.$$
If I have to use this formula to evaluate $e^{-20}$, how should I check for convergence?
MATLAB gives $e^{-20} = 2.0612e-009$. But when I use this formula to calculate it, I get $4.1736e-009$. My answer becomes more inaccurate as I take $x = -30, -40 \cdots$. The reason is obvious. This is because as the number of terms in the Taylor series increases, the factorial becomes more and more inaccurate using MATLAB.
1. How can I use Taylor series to get accurate values of $e^x$ in MATLAB?
2. How does MATLAB built in command "exp()" calculate the exponential?
• If the ultimate goal is to approximate the exponential I would use a Pade' approximants (a rational approximation), combined with $e^{2x}=(e^x)^2$ to get values for larger values of $x$. – Pp.. Jan 28 '15 at 23:39
• If you just want to see how the Taylor series behaves, then the error of an alternating series can be bounded by the absolute value of the next term of the series. – Pp.. Jan 28 '15 at 23:40
• TS is good for small values of the argument. Otherwise you might end up having many extra terms. – Kaster Jan 28 '15 at 23:42
• You may try summing backwards. – lhf Jan 29 '15 at 11:44
• Oog. Why not start the summation at $0$? – Qiaochu Yuan Jan 30 '15 at 8:54
The Problem
The main problem when computing $e^{-20}$ is that the terms of the series grow to $\frac{20^{20}}{20!}\approx43099804$ before getting smaller. Then the sum must cancel to be $\approx2.0611536\times10^{-9}$. In a floating point environment, this means that $16$ digits of accuracy in the sum are being thrown away due to the precision of the large numbers. This is the number of digits of accuracy of a double-precision floating point number ($53$ bits $\sim15.9$ digits). | {
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For example, the RMS error in rounding $\frac{20^{20}}{20!}$, using double precision floating point arithmetic, would be $\sim\frac{20^{20}}{20!}\cdot2^{-53}/\sqrt3\approx3\times10^{-9}$. Since the final answer is $\approx2\times10^{-9}$, we lose all significance in the final answer simply by rounding that one term in the sum.
The problem gets worse with larger exponents. For $e^{-30}$, the terms grow to $\frac{30^{30}}{30!}\approx776207020880$ before getting smaller. Then the sum must cancel to be $\approx9.35762296884\times10^{-14}$. Here we lose $25$ digits of accuracy. For $e^{-40}$, we lose $33$ digits of accuracy.
A Solution
The usual solution is to compute $e^x$ and then use $e^{-x}=1/e^x$. When computing $e^x$, the final sum of the series is close in precision to the largest term of the series. Very little accuracy is lost.
For example, the RMS error in computing $e^{20}$ or $e^{-20}$, using double precision floating point arithmetic, would be $\sim8\times10^{-9}$; the errors are the same because both sums use the same terms, just with different signs. However, this means that using Taylor series, \begin{align} e^{20\hphantom{-}}&=4.85165195409790278\times10^8\pm8\times10^{-9}\\ e^{-20}&=2\times10^{-9}\pm8\times10^{-9} \end{align} Note that the computation of $e^{-20}$ is completely insignificant. On the other hand, taking the reciprocal of $e^{20}$, we get $$e^{-20}=2.061153622438557828\times10^{-9}\pm3.4\times10^{-26}$$ which has almost $17$ digits of significance.
• Would the downvoter care to comment? – robjohn Jan 30 '15 at 1:43
• this is so nice an explanation!! When doing numerical calculation the formulas must be modified in a manner to avoid loss of significant digits and such that intermediate calculation does not require a considerably higher precision than that desired in the final answer. – Paramanand Singh Feb 19 '15 at 10:48 | {
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A standard technique, not particular to Matlab, is to exploit a function's symmetries to map its inputs to a more manageable range. The exponential function satisfies the symmetry $e^{a+b}=e^ae^b$, and it's easy to multiply floating-point numbers by powers of two. So do this: $$\begin{eqnarray} e^{-20}&=&2^{-20\lg e}\\ &\approx&2^{-28.854}\\ &=&2^{-29}\times2^{0.146}\\ &=&2^{-29}\times e^{0.146\log2}\\ &\approx&2^{-29}\times e^{0.101} \end{eqnarray}$$ Now you can easily evaluate $e^{0.101\ldots}$ using the Taylor series.
1) The regular double precision floating point arithmetic of Matlab is not sufficient to precisely calculate partial sums of this power series. To overcome that limitation, you can use the exact symbolic computation capabilities of the Symbolic Math Toolbox. This code
x = sym(-20);
i = sym(1 : 100);
expx = sum(x .^ (i - 1) ./ factorial(i - 1))
sums the first 100 terms of the series, and yields the exact (for the partial sum) result $$\tiny\frac{20366871962240780739193874417376755657912596966525153814418798643652163252126492695723696663600640716177}{9881297415446727147594496649775206852319571477668037853762810667968023095834839075329261976769165978884198811117}$$ whose numeric counterpart (computed by double()) is 2.06115362243856e-09.
2) The Matlab function exp() is most likely a wrapper for C code, which presumably references the standard C library function exp().
You can have a look at the code of the widely used GNU C Library math functions, where you'll find that most functions are themselves wrappers around machine-specific implementations. For the i386 architecture, the path leads ultimately to the function __ieee754_exp which implements the exponential in inline assembler code (but don't ask me how to decipher this). On other architectures, there might be a single machine instruction that does the job. | {
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"Most likely", "probably", "presumably" because Matlab is closed source and therefore definitive knowledge only exists within The Mathworks.
• You mention using "variable precision arithmetic" in Matlab but your example code is an exact symbolic calculation converted to floating-point. There appears to be no need for vpa for the ranges used. The OP also asked "how should I check for convergence?" which I think was the impetus for the question. – horchler Jan 29 '15 at 3:15
• @horchler, correct with the vpa thing, I got the two confused, edited. – As for convergence, I think this was a minor part of the question made irrelevant by the fact that his double precision computation couldn't converge anyway. 4.1e-9 isn't not-converged, its just off. Feel free to add a better answer though, or even to edit mine. :-) – A. Donda Jan 29 '15 at 11:04 | {
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# Poisson Distribution of sum of two random independent variables $X$, $Y$
$X \sim \mathcal{P}( \lambda)$ and $Y \sim \mathcal{P}( \mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y \sim \mathcal{P}( \lambda + \mu)$ but I don't understand how to derive it.
-
Try using the method of moment generating functions :) – Samuel Reid Nov 25 '13 at 7:03
All I've learned in the definition of a Poisson Random Variable, is there a simpler way? – user82004 Nov 25 '13 at 7:07
If they are independent. – Did Nov 25 '13 at 8:14
This only holds if $X$ and $Y$ are independent, so we suppose this from now on. We have for $k \ge 0$: \begin{align*} P(X+ Y =k) &= \sum_{i = 0}^k P(X+ Y = k, X = i)\\ &= \sum_{i=0}^k P(Y = k-i , X =i)\\ &= \sum_{i=0}^k P(Y = k-i)P(X=i)\\ &= \sum_{i=0}^k e^{-\mu}\frac{\mu^{k-i}}{(k-i)!}e^{-\lambda}\frac{\lambda^i}{i!}\\ &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(k-i)!}\mu^{k-i}\lambda^i\\ &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \binom ki\mu^{k-i}\lambda^i\\ &= \frac{(\mu + \lambda)^k}{k!} \cdot e^{-(\mu + \lambda)} \end{align*} Hence, $X+ Y \sim \mathcal P(\mu + \lambda)$.
-
Thank you! but what happens if they are not independent? – user31280 Oct 25 '12 at 20:20
In general we can't say anything then. It depends on how they depend on another. – martini Oct 25 '12 at 20:22
Thank you! it's very simple and I feel like a complete idiot. – user31280 Oct 25 '12 at 20:40
Nice derivation: specifically the transformation of (a) the i/k factorials and (b) the mu/lambda polynomials into the binomial form of the polynomial power expression. – javadba Aug 30 '14 at 20:59 | {
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Another approach is to use characteristic functions. If $X\sim \mathrm{po}(\lambda)$, then the characteristic function of $X$ is (if this is unknown, just calculate it) $$\varphi_X(t)=E[e^{itX}]=e^{\lambda(e^{it}-1)},\quad t\in\mathbb{R}.$$ Now suppose that $X$ and $Y$ are independent Poisson distributed random variables with parameters $\lambda$ and $\mu$ respectively. Then due to the independence we have that $$\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)=e^{\lambda(e^{it}-1)}e^{\mu(e^{it}-1)}=e^{(\mu+\lambda)(e^{it}-1)},\quad t\in\mathbb{R}.$$ As the characteristic function completely determines the distribution, we conclude that $X+Y\sim\mathrm{po}(\lambda+\mu)$.
-
You can use Probability Generating Function(P.G.F). As poisson distribution is a discrete probability distribution, P.G.F. fits better in this case.For independent X and Y random variable which follows distribution Po($\lambda$) and Po($\mu$). P.G.F of X is \begin{equation*} \begin{split} P_X[t] = E[t^X]&= \sum_{x=0}^{\infty}t^xe^{-\lambda}\frac{\lambda^x}{x!}\\ &=\sum_{x=0}^{\infty}e^{-\lambda}\frac{(\lambda t)^x}{x!}\\ &=e^{-\lambda}e^{\lambda t}\\ &=e^{-\lambda (1-t)}\\ \end{split} \end{equation*} P.G.F of Y is \begin{equation*} \begin{split} P_Y[t] = E[t^Y]&= \sum_{y=0}^{\infty}t^ye^{-\mu}\frac{\mu^y}{y!}\\ &=\sum_{y=0}^{\infty}e^{-\mu}\frac{(\mu t)^y}{y!}\\ &=e^{-\mu}e^{\mu t}\\ &=e^{-\mu (1-t)}\\ \end{split} \end{equation*}
Now think about P.G.F of U = X+Y. As X and Y are independent, \begin{equation*} \begin{split} P_U(t)=P_{X+Y}(t)=P_X(t)P_Y(t)=E[t^{X+Y}]=E[t^X t^Y]&= E[t^X]E[t^Y]\\ &= e^{-\lambda (1-t)}e^{-\mu (1-t)}\\ &= e^{-(\lambda+\mu) (1-t)}\\ \end{split} \end{equation*}
Now this is the P.G.F of $Po(\lambda + \mu)$ distribution. Therefore,we can say U=X+Y follows Po($\lambda+\mu$) | {
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-
Seems a typo in third line (out of 4) in both PGF of X and Y. Should be exp(-u)exp(mut) instead of exp(-u)exp(-1*mut) – javadba Aug 30 '14 at 21:19
Separate comment/question: please explain why in second line of derivation for PGF of Y refers to exp(x) and x! (instead of referring to exp(y) and y!). I believe these are cut/paste errors - but please confirm. – javadba Aug 30 '14 at 21:26
Yeah.. You are absolutely correct. Those were typos. I have edited them now. If you find more, let me know. – Ananda Sep 2 '14 at 5:35
hint: $\sum_{k=0}^{n} P(X = k)P(Y = n-k)$
-
why this hint, why the sum? This is what I don't understand – user31280 Oct 25 '12 at 20:22
adding two random variables is simply convolution of those random variables. That's why. – jay-sun Oct 25 '12 at 20:24
gotcha! Thanks! – user31280 Oct 25 '12 at 20:31
adding two random variables is simply convolution of those random variables... Sorry but no. – Did Feb 13 '13 at 6:28
There is no usual sense for convolution of random variables. Either convolution of distributions or addition of random variables. – Did Feb 13 '13 at 6:51
In short, you can show this by using the fact that $$Pr(X+Y=k)=\sum_{i=0}^kPr(X+Y=k, X=i).$$ | {
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In short, you can show this by using the fact that $$Pr(X+Y=k)=\sum_{i=0}^kPr(X+Y=k, X=i).$$
If $X$ and $Y$ are independent, this is equal to $$Pr(X+Y=k)=\sum_{i=0}^kPr(Y=k-i)Pr(X=i)$$ which is \begin{align} Pr(X+Y=k)&=\sum_{i=0}^k\frac{e^{-\lambda_y}\lambda_y^{k-i}}{(k-i)!}\frac{e^{-\lambda_x}\lambda_x^i}{i!}\\ &=e^{-\lambda_y}e^{-\lambda_x}\sum_{i=0}^k\frac{\lambda_y^{k-i}}{(k-i)!}\frac{\lambda_x^i}{i!}\\ &=\frac{e^{-(\lambda_y+\lambda_x)}}{k!}\sum_{i=0}^k\frac{k!}{i!(k-i)!}\lambda_y^{k-i}\lambda_x^i\\ &=\frac{e^{-(\lambda_y+\lambda_x)}}{k!}\sum_{i=0}^k{k\choose i}\lambda_y^{k-i}\lambda_x^i \end{align} The sum part is just $$\sum_{i=0}^k{k\choose i}\lambda_y^{k-i}\lambda_x^i=(\lambda_y+\lambda_x)^k$$ by the binomial theorem. So the end result is \begin{align} Pr(X+Y=k)&=\frac{e^{-(\lambda_y+\lambda_x)}}{k!}(\lambda_y+\lambda_x)^k \end{align} which is the pmf of $Po(\lambda_y+\lambda_x)$.
-
Moderator notice: This answer was moved here as a consequence of merging two questions. This explains the small differences in notation. The OP's $\lambda$ is $\lambda_x$ here, and OP's $\mu$ is $\lambda_y$. Otherwise there is no difference. – Jyrki Lahtonen Apr 23 at 6:55 | {
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# Filtering data 'geometrically'
I was looking at a very nice answer here, where the question was applying a confidence region to some data.
If we plot some data as a scatter plot:
XData = RandomVariate[NormalDistribution[1,1], 100000];
YData = RandomVariate[NormalDistribution[1,1], 100000];
XYDATA = Transpose[{XData, YData}];
XYCoVarMat = Covariance[XYDATA];
XYMean = {Mean[XData], Mean[YData]};
ListPlot[
XYDATA, Frame->True, FrameLabel->{"X","Y"}, LabelStyle->16, AspectRatio->1,
Epilog -> {
Opacity[0.32, Green], EdgeForm[{Green, AbsoluteThickness[1]}], Ellipsoid[XYMean, 4 XYCoVarMat],
Opacity[0.32, Blue], EdgeForm[{Blue, AbsoluteThickness[1]}], Ellipsoid[XYMean, 3 XYCoVarMat],
Opacity[0.32, Red], EdgeForm[{Red, AbsoluteThickness[1]}], Ellipsoid[XYMean, 2 XYCoVarMat],
Opacity[0.32, Yellow], EdgeForm[{Yellow, AbsoluteThickness[1]}], Ellipsoid[XYMean, 1 XYCoVarMat]
},
PlotRange->{{-8,8},{-8,8}}
]
I was wondering if anyone knew a way of filtering the data using this approach for example only select data that is with the ellipsoid boundary of Ellipsoid[XYMean, 1 XYCoVarMat]?
• Would With[{rmf = RegionMember[Ellipsoid[XYMean, 1 XYCoVarMat]]}, Select[XYDATA, rmf]] work for you? – J. M. will be back soon Nov 15 at 16:09
• It would work wonderfully. I'm starting to wonder if you are actually written purely from Mathematica functions. Thanks! – Q.P. Nov 15 at 16:14
• @J.M.willbebacksoon I don't know if you want to post an answer yourself or not, but given you provided the solution I feel you should get credit. If credit doesn't concern you I will accept ThatGravityGuy's answer so that the question is completed and helps users who find this question. – Q.P. Nov 16 at 13:15
• Please accept ThatGravityGuy's answer if you think you found it useful. – J. M. will be back soon Nov 16 at 13:24 | {
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As @J.M. pointed out in the comments, the key here is the function RegionMember. Since you already know the region that you want to work with, namely Ellipsoid[XYMean, 1 XYCoVarMat], you can use Select as
With[{rmf = RegionMember[Ellipsoid[XYMean, 1 XYCoVarMat]]}, Select[XYDATA, rmf]] | {
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# Mosteller's First Ace Problem
[this probably belongs on cross validated ... ]
Working through Mosteller's Fifty Challenging Problems in Probability with Mathematica. I am trying to understand Mosteller's answer for The First Ace problem (#40). The problem goes:
Shuffle an ordinary deck of 52 playing cards containing 4 aces. Then turn up cards from the top until the first ace appears. On the average how many cards are required to produce the first ace?
I solved this problem in Mathematica as as follows:
suite = Flatten@{Range[2, 10], J, Q, K, A}
deck = Flatten@{suite, suite, suite, suite}
dat = Table[First@Flatten@Position[
RandomSample[deck, Length@deck], A], {n, 1, 100000}];
ListPlot[
Tally@dat,
GridLines -> Automatic,
GridLinesStyle -> Directive[Dotted, Gray],
Frame -> True
]
Mean@dat // N
10.6152
Which yields Mosteller's answer (by symmetry) of 10.6 cards (48/5 +1).
I was a bit shocked that the mode was 1 card.
Finally, when I look at the CDF the 50% mark is at 8 cards
t = Reverse[SortBy[Tally@dat, #[[2]] &]][[All, 2]]
ListPlot[N[Accumulate@t/100000],
GridLines -> Automatic,
GridLinesStyle -> Directive[Dotted, Gray]
]
So my question is, if I played this game for real in a casino, should I go for 8 or 10 cards?
• I think you're right, this probably belongs on CrossValidated... Cool problem though. Mar 9, 2017 at 6:03
• @MarcoB I believe this belongs here, because - next to the theoretical aspect - it shows how to make use of Mathematica's abilities to simulate and to work with its distribution-framework.
– gwr
Mar 9, 2017 at 12:34
• My bad, I corrected my "decision support" (cf. my answer): In the simple 0-1-Utility case (right/wrong and a fixed amount to win), you would best choose $1$ (not 8, not 10). My exmple meets the case, when you will win an amount equal to the number of draws it took.
– gwr
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I would like to choose a slightly different approach here and point out, that this is an example for the Negative Hypergeometric Distribution. Wikipedia does give a nice table for this:
While for large numbers the distribution tends to the Negative Binomial Distribution the difference essentially is drawing with replacements (Binomial) vs. drawing without replacements (Hypergeometric).
## Analytical Approach
In the present case, even when we may not have remembered all this statistical theory, we could still take an analytical approach by making use of ProbabilityDistribution. Let us look at the probabilities for the first few numbers of draws:
\begin{align} p(1) &= p(\text{ace}) =\frac{4}{52} \\ p(2) &= p(\neg\text{ace}) \times p(\text{ace}) = \frac{52-4}{52} \times \frac{4}{52-1} \\ p(3) &= p(\neg\text{ace}) \times p(\neg\text{ace}) \times p(\text{ace}) = \frac{52-4}{52} \times \frac{52-4-1}{52-1} \times \frac{4}{52-2} \\ \vdots \end{align}
So eventually we see a pattern here and can (hopefully) come up with the formula for the probability distribution function (in the discrete case a probability mass function) $f$ given the number of cards (or balls in an urn) $N$ and the number of marked cards (or balls) $M$:
\begin{align} f_{N,M}(k) = \frac{m}{N-k+1} \times \prod \limits_{i=1}^{k-1} \frac{N-M-i+1}{N-i+1} \end{align}
To implment this as a distribution we write the following code:
distribution[ n_Integer, m_Integer ] /; m <= n := ProbabilityDistribution[
m/( n - k + 1) Product[ (n - m - i + 1)/(n - i + 1),{ i, 1, k-1 } ],
{ k, 1, n - m + 1, 1 }
(* , Method -> "Normalization" is not needed here, but may be worth remembering
for all cases where giving a proportional formula is easy to come up with *)
]
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$PlotTheme = {"Detailed", "LargeLabels"}; Panel @ DiscretePlot[ Evaluate@PDF[ analyticalDist, \[FormalX] ], {\[FormalX], 0, 49, 1}, ImageSize -> Large, PlotLabel -> Style["Analytical PDF", "Subsection"] ] Also we can find moments and what have you: Mean @ analyticalDistribution (* e.g. the expected number of draws *) $\frac{53}{5}$## Empirical Approach We could of course also arrive at answers using simulation: empiricalDist = EmpiricalDistribution @ Table[ Min @ RandomSample[ Range[52], 4], {10000} (* number of trials *) ]; N @ Mean @ empiricalDist 10.5809 (* true value again was 10.6 *) If we compare the graphs for the PDF we note that the empirical distribution already is quite close: Panel @ DiscretePlot[ Evaluate @ { PDF[ empiricalDist, \[FormalX] ], PDF[ analyticalDist, \[FormalX] ] }, { \[FormalX], 1, 49, 1 }, PlotLegends -> {"empirical", "analytical"}, ImageSize -> Large, PlotLabel -> Style["PDF Comparison", "Subsection"] ] ## General Case: Negative Hypergeometric Distribution Grabbing a text book near you, you will find the formula for the probability distribution function of the above mentioned Negative Hypergeometric Distribution. It is a special case of the BetaBinomialDistribution - thanks to @ciao for pointing this out repeatedly until I listened :) - for those of you, who do not know this, there is a terrific paper showing the relations between univariate distributions by (Leemis and McQueston 2008) and more information can be found on Cross Validated. Using this information, we can implement the Negative Hypergeometric Distribution explicitly as follows: (* following Mathematica's parametrization for the hyergeometric case *) negativeHypergeometricDistribution[k_, m_, n_ ] := TransformedDistribution[ \[FormalX] + 1, \[FormalX] \[Distributed] BetaBinomialDistribution[ k, m, n-m ] ] (* unfortunately parametrization is no standardized so one has to dwelve into the formulae for this in the paper, which are given though *) nhg = | {
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so one has to dwelve into the formulae for this in the paper, which are given though *) nhg = negativeHypergeometricDistribution[ 1, 4, 52 ]; Mean @ nhg $\frac{53}{5}$And we can indeed convince ourselves that our case simply is a special case of a NHG-distributed random var$X$, where$X$is the number of draws from$N = 52$cards, where$M=4$are "successes" and where we need$k=1$success to stop: And @@ Table[ PDF[ analyticalDist, i ] === PDF[ nhg, i ], {i, 1, 52 } ] True ## EDIT: Decision Support / Expected Utility The OP has edited his question and asked for guidance in playing at a casino (e.g. what number of draws to bet on?). For this we have to turn to decision theory and excepted utility. Let's assume that you are risk neutral. In the case that you get a fixed amount if you are right and zero if you are wrong, then it would be trivially optimal to bet on$1$. ($1$has the greatest probability and the 0-1-loss function makes the expected utility equal to the probability; check this by simulation!) Let us look at a more elaborate case to make the principle clear: If you were to receive the number of draws needed worth in money if you bet on the right number of draws, then your utility function would be$U(x,k) = x I(x = k)$(where$I$is the indicator function,$x$your bet and$k$the number of draws it took. In this case your best bet will be$13$: NArgMax[ { NExpectation[ \[FormalX] * Boole[ \[FormalX] == i], \[FormalX] \[Distributed] negativeHypergeometricDistribution[ 1, 4, 52 ] ], 1<= k<= 49 }, k ∈ Integers ] 13 With[ { dist = negativeHypergeometricDistribution[ 1, 4, 52 ] }, Panel @ DiscretePlot[ Evaluate@ Expectation[ \[FormalX] * Boole[ \[FormalX] == i ], \[FormalX] \[Distributed] dist ], {i,1,49}, ImageSize -> Large, PlotLabel -> Style["Expected Utility (0-1)", "Subsection"], PlotLegends->None ] ] • "since it is not implemented...". It is, as BetaBinomial, already noted in an earlier reply. – ciao Mar 9, 2017 at 23:35 • @ciao Thanks, I did not make that connection. It | {
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in an earlier reply. – ciao Mar 9, 2017 at 23:35 • @ciao Thanks, I did not make that connection. It is deeply burried in the docs and also not very obvious from web research. Guess one really needs to know what your looking for. :) – gwr Mar 9, 2017 at 23:50 • Not so much an MMA docs issue, more of a wider issue of distributions going by different names/definitions in the mathematical world. Would be nice if a future MMA docs edition (or MathWord) had a big interconnected chart with all alternate names/definitions and their relationships... – ciao Mar 10, 2017 at 0:50 • This is the best overview with regard to the relationships between univariate distributions I have found so far. – gwr Mar 10, 2017 at 8:19 Update (attribution: @ciao) As @ciao has pointed out, use ofBetaBinomialDistribution is the most straightforward for exact calculation and simulation. See the comment: Mean[BetaBinomialDistribution[1, 4, 48]] + 1 or simulation Mean[RandomVariate[Mean[BetaBinomialDistribution[1, 4, 48]] + 1],100000] Median[RandomVariate[Mean[BetaBinomialDistribution[1, 4, 48]] + 1],100000] etc Original Answer As the choice of first Ace is arbitrary you could simplify, e.g.: tab = Table[Min[FirstPosition[RandomSample[Range[52]], #][[1]] & /@ Range[4]], 10000]; Histogram[tab, Automatic, "PDF"] N@Mean[tab] Median[tab] You could also do analytically,e.g. let$f(k)$be the probability that the first ace in the first$k$cards, then: f[k_] := 1 - Binomial[52 - k, 4]/Binomial[52, 4] DiscretePlot[f[k], {k, 0, 52}] Sanity checks:$f(k)=1$for$k\ge 49$The probability that is$k\$th card is discrete derivative: | {
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ListPlot[Differences[Table[f[k], {k, 0, 52}]], Filling -> Axis]
Differences[Table[f[k], {k, 0, 52}]].Range[52]
• Or you could just Mean[BetaBinomialDistribution[1, 4, 48]] + 1... ;=}
– ciao
Mar 9, 2017 at 6:56
• @ciao doh! always forgetting to exploit diverse built-in distributions...hope your entrepreneurship is going well ;-) Mar 9, 2017 at 6:59 | {
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# Plotting a system of linear ODE
I would like to plot a system of following ODE:
$$$$\mathbf{\dot{x}} = \mathbf{Ax} \text{ where } \mathbf{A} = \begin{bmatrix} -2 & 1 \\ -1 & 0 \end{bmatrix}$$$$
with general solution:
$$$$\vec{x}(t)=C_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{-t}+C_2 \left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}+\begin{bmatrix} 1 \\ 1 \end{bmatrix} t \right) e^{-t}$$$$
I know that this system is stable so arrows will be inward-pointing, also it has double eigenvalue at -1 so there will be an constant line for an eigenvalue crossing the plot but I am not sure how to fill the rest of the phase diagram. Could someone help me? Thanks in advance.
• I think you have a mistake. Shouldn't this system have a double eigenvalue of 1, not -1? – Josh B. Nov 4 '18 at 16:29
• @JoshB. I think the eigenvalue is given by $$\lambda^2+2\lambda+1$$ unless I made a mistake but I double checked the solution using Matlab – 1muflon1 Nov 4 '18 at 16:33
• If that's the case, then there is a typo in your matrix here. I just checked and the matrix you have definitely has a double eigenvalue of 1. – Josh B. Nov 4 '18 at 16:39
• @JoshB. You were correct the typo was 2 instead of -2. My bad. I am really sorry, I was copy pasting it from my latex and somehow I did not included the minus sign, dont know how did I even managed to do it – 1muflon1 Nov 4 '18 at 16:41
If you write your solution now as $$\vec{x}(t)=\left(C_1\begin{bmatrix} 1 \\ 1 \end{bmatrix}+C_2\begin{bmatrix} 0 \\ 1\end{bmatrix}\right)e^{-t}+C_2\begin{bmatrix}1 \\ 1\end{bmatrix}te^{-t}$$
it may be a little easier to see what's going on. As $$t\to\pm\infty$$, the equation is dominated by the second term, so $$\vec{x}(t)\approx C_2\begin{bmatrix} 1 \\ 1\end{bmatrix}te^{-t}\;\;\;\;\;\;\;\;\;\;\;t\to\pm\infty$$ | {
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As $$t\to\infty$$, the whole thing decays to $$0$$, so it makes sense that all solution curves begin to look like they fall along this eigenvector as they decay to $$0$$. Applying the same logic as $$t\to-\infty$$, it would seem that solutions grow along this vector as well, but this is only true on a macroscopic scale, as the solutions are growing exponentially in both directions. The first term, no longer decaying to $$0$$, shifts the solution away from the eigenvector and is growing large, but not as large as the second term. This means that when zoomed in, solutions look like they are leaving the eigenvector, but zoomed out solutions look like they are along the eigenvector (that is, until you let time grow large enough, in which you will see it move away slowly).
The last thing to consider is the direction of the curve. Suppose that $$\vec{x}$$ is in the first quadrant as $$t\to\infty$$. This means that $$C_2$$ is positive, so if $$t\to-\infty$$, then the sign on the second term is now negative, and the solution turns around to enter the third quadrant. This means that solutions sort of "spin" by turning $$180^\circ$$ around while still growing away from the origin.
The vector $$\begin{bmatrix} 0 \\ 1\end{bmatrix}$$ tells you which way the solution spins. If $$C_2$$ is positive, then that means that at $$t=0$$, the solution is above (in the xy-plane) of the line through the vector $$\begin{bmatrix} 1 \\ 1\end{bmatrix}$$ and the origin, so the solution must remain on this side of said line. The dominating term is in the first quadrant as $$t\to\infty$$, so the solution must spin clockwise to decay this way. It turns out that if solutions spin clockwise on one side of the eigenvector, they do the same on the other side as well; the same is true if they spin counterclockwise.
This type of equilibrium is called a Degenerate Node, in case you were curious. Here is an example of what one might look like. | {
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• Thank you for very clear and detailed answer. By the way did you made the plot in matlab,? I know that it was not part of my original question and matlab inquiries dont really belong to math section, but would you mind sharing your code? I would like to learn how to do it. Or did you used another program? – 1muflon1 Nov 4 '18 at 17:36
• No, I pulled it from the encyclopedia of mathematics. Plotting this in MATLAB could be done by taking a spacing of points for t and plugging them in for x and y, then plotting the resulting curve. This would need to be done for a few different initial points to get a good diagram. – Josh B. Nov 4 '18 at 17:44 | {
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If $\left| f'(x) \right| \leq A |f(x)|^\beta$ then f is a constant function
Problem Let $f(x)$ be a differentiable function on $[a,b]$ satisfying $f(a)=0$. If there exist $A \ge 0$ and $\beta \ge 1$ such that the inequality
$$\left| f'(x) \right| \leq A \left| f(x) \right|^\beta$$
holds for all $x$ in $[a,b]$, then $f(x) = 0$ for any x in that interval.
My attempted solution
We shall prove by contradiction. Suppose that for some $x$ in $[a,b]$ we have $f(x) \neq 0$.
Put $S:= \left\lbrace x \right.$ such that $a \le x \le b$ and $f(x) \neq 0 \left. \right\rbrace$ and $c:=\inf S$ (the infimum value of set $S$). If $c=b$ then $f(x) = 0$ $\forall x < b$ and the continuity of $f$ implies that $f(b) = 0$, too (contradiction). Therefore $c \neq b$. If $c=a$, then $f(c) = 0$ by the hypothesis. Otherwise, when $a < c < b$ we have $f(x) = 0$ $\forall x < c$ and again, by continuity, $f(c) = 0$. Thus in all cases considered, we have $f(c) = 0$ and $a \leq c < b$
As $f$ is continuous at $c$ and $f(c) = 0$, one can choose $d>c$ and close enough to $c$ such that $|f(x)| \le 1$ and $$\tag{\star} A(x-c) \le \frac12\quad\text{for all }x\text{ with }c \le x \le d.$$ Since $f$ is continuous on $[c,d]$, it has maximum and minimum values on this closed interval, leading to the existence of $c \le t \le d$ such that $|f(t)| = \max_{c \le x \le d} |f(x)|$. As already noted, $f(c) = 0$ and $c$ is the infimum of the set of numbers whose image under $f$ is nonzero. Therefore, for any $\epsilon>0$, there is some $c<m<c+\epsilon$ satisfying $f(m) \neq 0$. This observation, together with the definition of $t$, gives us $$\tag{\star\!\star}f(t) \neq 0,$$ $t \neq c$ and so $c<t$. Applying the Lagrange theorem gives:
$$|f(t)| = |f(t) - f(c)| = |t-c||f'(u)| \leq|t-c|A|f(u)|^\beta$$ where $c<u<t\le d$. By $(\star)$: $$|t-c|A|f(u)|^\beta \le \frac12 |f(u)|^\beta \leq \frac12 |f(u)|$$ | {
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Combining the two inequalities and noting that $|f(t)| \ge |f(u)|$ (because of the definition of $t$), we have $f(u) = f(t) = 0$ and arrived at a contradiction with $(\star\star)$.
Question
I hope someone can verify if my solution is correct. Of course, other ideas, comments or solutions are welcome.
I like to post problems and my solutions to the forum because I think it's beneficial to the community, and for learners like me. First, I can hardly know if there's flaw in my own argument. Second, I may get new insights/solutions for my problem.
Thank you.
• Sorry, I am not yet clear how the two threads are related? – tom_a2 May 22 '13 at 16:43
• You are absolutely right. I misread your question. – Martin May 22 '13 at 16:45
• It looks good to me. – Julien May 22 '13 at 17:32
Let $B = \{x\in [a,b]: |f(x)|=0\}$. By continuity, this set is closed. Since $f(0) = 0$, the set is non-empty. We show that it is also open.
Let $x\in B$, i.e. $f(x) =0$. By continuity, there exists $0<r<(2A)^{-1}$, such that $|f(y)|<\frac 12$ for all $y\in B_r(x)$. Fix such a $y\in B_r(x)$. We want to show that $f(y) =0$. | {
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By the mean value inequality, we have $$|f(y)| = |f(y) - f(x)| \le |f'(\xi_0)| |y-x| \le |f'(\xi_0)|r$$ for some $\xi_0\in B_r(x)$ lying on the segment between $x$ and $y$. By assumption, we have $|f'(\xi_0)|\le A |f(\xi_0)|^\beta$, hence $$|f(y)|\le |f'(\xi_0)|r \le A|f(\xi_0)|^\beta (2A)^{-1} \le \frac 12|f(\xi_0)|^\beta.$$ By the same argument applied to $\xi_0\in B_r(x)$ instead of $y$, we see that there exists $\xi_1\in B_r(x)$, such that $|f(\xi_0)|\le \frac 12|f(\xi_1)|^\beta$. Iterating yields a sequence $\xi_n \in B_r(x)$, such that $|f(\xi_{n})|\le \frac 12|f(\xi_{n+1})|^\beta$ for all $n$. Note that by our choice of $r$, we have $|f(\xi_{n})|\le \frac 12$ for all $n$. We conclude that $$|f(y)|\le \frac 12 |f(\xi_0)|^\beta \le \frac 1{2^2}|f(\xi_1)|^{\beta^2} \le \frac{1}{2^3}|f(\xi_2)|^{\beta^3}\le \dots \le \frac{1}{2^n}|f(\xi_{n-1})|^{\beta^n}\le \frac 1{2^{n+\beta^n}},$$ for all $n\in \mathbb N$. Letting $n\to \infty$ shows that $|f(y)| = 0$ (using also $\beta \ge 1$). This implies that $B_r(x)\subset B$, hence $B$ is open.
Since $[a,b]$ is connected, it follows that $B$ must be all of $[a,b]$, hence $f= 0$.
( Edit. The ODE analysis also explains why $\beta \geq 1$ is necessary. The condition for $|f'(x)| \leq A|H(f(x))|$ to have the property that $f$ does not change sign, when $H$ is a function such that $H(x)=0$ only at $x=0$, is that $\frac{1}{H(x)}$ an has a non-integrable singularity at $0$. Knowing that this is the answer, there is no loss in considering singularities that are powers of $f$, which is what most cases would look like and how they would be recognized.) | {
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The differential equation $f'(x) = A(x) f(x)^\beta$ can be solved explicitly, by separation of variables. The calculation shows that $f(a)=0$ is necessary, or we could write down nowhere zero solutions, by solving the equation. This bears further analysis, because superficially the problem looks like a Lipschitz condition with $|f(x) - f(y)| \leq |x-y|^\beta$ which (as is very well known and often solved as an exercise) implies constant $f$ when $\beta > 1$.
The DE for constant $\beta > 1$, writing $y$ for $f(x)$, is $$d(\frac{1}{(\beta - 1)y^{\beta - 1}}) = A(x) dx$$
The coordinates that trivialize the equation are therefore $Y = \frac{1}{(\beta - 1)y^{\beta - 1}}$ and $X = \int A(x) dx$, in which the solutions are $Y = X + c$ for constant $c$. This brings out the idea that $A$, which appears at first to be a harmless constant removable by a linear change of variable, has a meaningful part in the problem, as velocity of the independent variable.
Under this coordinate change, $f(a)=0$ becomes $f(a)=\infty$. On a maximal interval where $f$ is nonzero, there cannot be any solutions that hit infinity at the boundary in finite "time" (here $X$ is time, so we mean a finite change in $\int A(x) dx$) while satisfying $Y - X =$ constant.
So the meaning of the problem seems to be: no singularity can be reached in finite time for this ODE.
• Let's put here the comment that for $\beta < 1$ there are counterexamples with $f(x)$ a positive power of $(x-a)$. – zyx May 26 '13 at 22:38
Supposing that f(x)=0 only for a and that f(x)>0 for the rest x in [a,b].
Then lim (f(x) - f(a))/(x-a) as x->a+ will be greater or equal to zero. If greater then f'(a)=m >0 and thats contradicting with the given inequality.Therefore f(x)=0 for all x in [a,b]. If equal, then im sorry but i need to give it some more thought. In the same manner, u can prove for f(x)<0.
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Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
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Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 27 Jun 2018, 18:28 6 9 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64 $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
### Show Tags
29 Jun 2018, 06:58
28
9
First, choose a nice, round number, such as 70, within the range of values.
Then calculate the average with the following formula:
Average = Nice number + Average of differences from the nice number
Average = 70 + (4 - 1 - 6 + 9 - 6 + 14 + 7)/7 = 70 + 21/7 =73 | {
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Average = 70 + (4 - 1 - 6 + 9 - 6 + 14 + 7)/7 = 70 + 21/7 =73
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 28 Jun 2018, 18:07 1 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
E. $85 We can determine the average using the formula: average = sum / number: average = (74 + 69 + 64 + 79 + 64 + 84 + 77)/7 = 511/7 = 73 Answer: C _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. VP Joined: 14 Feb 2017 Posts: 1310 Location: Australia Concentration: Technology, Strategy GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31 GMAT 6: 600 Q38 V35 GPA: 3 WE: Management Consulting (Consulting) Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink]
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19 Nov 2018, 16:13
1
Admittedly I made a stupid mistake and I organised the numbers in ascending order then selected the median (74) instead of solving Sum/# Terms | {
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The method I used is only for consecutive integers and the punishment answer D is there for suckers like me.
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 25 Jun 2018, 05:10 Solution Given: • Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77
To find:
• Smith’s average weekly grocery bill over the 7-week period
Approach and Working:
• Smith’s total bill amount over the 7-week period = (74 + 69 + 64 + 79 + 64 + 84 + 77) = 511
• Therefore, Smith’s average bill amount over the same period = $$\frac{511}{7}$$ = 73
Hence, the correct answer is option C.
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### Show Tags
28 Aug 2018, 03:42
Bunuel wrote:
Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
NEW question from GMAT® Official Guide 2019
(PS07369)
Easy way out is consider avg 70 -
so diff for all terms is 4, -1 , -6, 9 , -6,14, 7 = sum is 21 = divided by 7 = equal to 3 so 70 +3 = 73 avg
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 10 Sep 2018, 07:34 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74 | {
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A. $64 B.$70
C. $73 D.$74
E. $85 NEW question from GMAT® Official Guide 2019 (PS07369) $$\frac{( 70 + 4 ) + ( 70 - 1 ) + ( 60 + 4 ) + ( 80 - 1 ) + ( 60 + 4 ) + ( 80 + 4 ) + ( 70 + 7 )}{7}$$ = $$\frac{( 70 + 70 + 60 + 80 + 60 + 80 + 70 ) + ( 4 - 1 + 4 - 1 + 4 + 4 + 7 )}{7}$$ = $$\frac{490 + 21}{7}$$ = $$\frac{490}{7} + \frac{21}{7}$$ = $$70 + 3$$ = $$73$$, Answer must be (C) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 02 May 2018 Posts: 12 GMAT 1: 620 Q46 V29 Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink]
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10 Sep 2018, 07:46
We can rule out
64, 74 and 85 as they are the extreme numbers.
That leaves just two choices to choose from 70 and 73.
Only 3 values in the 60s and therefore less likely for 70.
73 is the most likely choice. Confirm 73 by taking difference and summing it up
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Is |x| = |y|? (1) x = -y (2) x^2 = y^2
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Is |x| = |y|?
(1) x = -y
(2) x^2 = y^2
[Reveal] Spoiler:
I found this question in GMAT Prep and selected ST2 as the answer but that's wrong.
Normally, x^2 would equal to lxl since it will contain both signs "positive as well as negative". How come st 1 is true?
Does it assume that since +x = -y, it must be true that -x is also equal to -y or?
What am I missing?
[Reveal] Spoiler: OA
Last edited by Bunuel on 11 Nov 2017, 01:25, edited 3 times in total.
Formatted the question.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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Is lxl = lyl ?
A) x=-y
b) x^2 = y^2
[Reveal] Spoiler:
I found this question in GMAT Prep and selected ST2 as the answer but that's wrong. | {
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Normally, x^2 would equal to lxl since it will contain both signs "positive as well as negative". How come st 1 is true?
Does it assume that since +x = -y, it must be true that -x is also equal to -y or?
What am I missing?
You are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.
Per statement 1:$$x=-y$$ . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.
Per statement 2: $$x^2=y^2$$ ---> $$x= \pm y$$. Again both the cases, x=y and x=-y give you a "yes" for the question asked as is hence sufficient.
Both statements are sufficient ---> D is the correct answer.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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17 Dec 2015, 06:54
statement 1 is sufficient as absolute value for -ve should give +ve
statement 2 is sufficient as
it will give either x and y both are +ve or both are -ve
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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Is |x|=|y|?
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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You are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices. | {
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Per statement 1:x=−yx=−y . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.
Per statement 2: x2=y2x2=y2 ---> x=±yx=±y. Again both the cases, x=y and x=-y give you a "yes" for the question asked as is hence sufficient.
Both statements are sufficient ---> D is the correct answer.
Please give me kudos. I need them to unlock gmatclub tests.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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SamDGold wrote:
Is |x|=|y|?
1) x=-y
2) x^2=y^2
Merging topics. Please check the discussion above.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink] 11 Nov 2017, 01:25
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# Need help deriving recurrence relation for even-valued Fibonacci numbers.
That would be every third Fibonacci number, e.g. $0, 2, 8, 34, 144, 610, 2584, 10946,...$
Empirically one can check that:
$a(n) = 4a(n-1) + a(n-2)$ where $a(-1) = 2, a(0) = 0$.
If $f(n)$ is $\operatorname{Fibonacci}(n)$ (to make it short), then it must be true that $f(3n) = 4f(3n - 3) + f(3n - 6)$.
I have tried the obvious expansion:
$f(3n) = f(3n - 1) + f(3n - 2) = f(3n - 3) + 2f(3n - 2) = 3f(3n - 3) + 2f(3n - 4)$ $= 3f(3n - 3) + 2f(3n - 5) + 2f(3n - 6) = 3f(3n - 3) + 4f(3n - 6) + 2f(3n - 7)$ ... and now I am stuck with the term I did not want. If I do add and subtract another $f(n - 3)$, and expand the $-f(n-3)$ part, then everything would magically work out ... but how should I know to do that? I can prove the formula by induction, but how would one systematically derive it in the first place?
I suppose one could write a program that tries to find the coefficients x and y such that $a(n) = xa(n-1) + ya(n-2)$ is true for a bunch of consecutive values of the sequence (then prove the formula by induction), and this is not hard to do, but is there a way that does not involve some sort of "Reverse Engineering" or "Magic Trick"?
-
Why isn't $f_{n+2}=3f_n-f_{n-2}$ suitable? – Guess who it is. Dec 27 '11 at 4:31
@J.M., sorry I do not understand. If I were to expand $f_{n+2}$ the way I do it, I would end up with $2f_n + f_{n - 2} + f_{n - 3}$. Again, I can make this work if I know what I am trying to get. I wonder if you have read the question correctly - I am looking for even-valued (not even-indexed) fib numbers. If my assumption is mistaken, then sorry. However, I am not sure how to systematically arrive at a relation you have given and how to use it to help me simplify things. – Job Dec 27 '11 at 4:45
It seems I did misunderstand you (and I apologize for this); have you looked at the references here by any chance? – Guess who it is. Dec 27 '11 at 4:50
## 8 Answers | {
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## 8 Answers
The definition of $F_n$ is given:
• $F_0 = 0$
• $F_1 = 1$
• $F_{n+1} = F_{n-1} + F_{n}$ (for $n \ge 1$)
Now we define $G_n = F_{3n}$ and wish to find a recurrence relation for it.
Clearly
• $G_0 = F_0 = 0$
• $G_1 = F_3 = 2$
Now we can repeatedly use the definition of $F_{n+1}$ to try to find an expression for $G_{n+1}$ in terms of $G_n$ and $G_{n-1}$.
\begin{align*} G_{n+1}&= F_{3n+3}\\ &= F_{3n+1} + F_{3n+2}\\ &= F_{3n-1} + F_{3n} + F_{3n} + F_{3n+1}\\ &= F_{3n-3} + F_{3n-2} + F_{3n} + F_{3n} + F_{3n-1} + F_{3n}\\ &= G_{n-1} + F_{3n-2} + F_{3n-1} + 3 G_{n}\\ &= G_{n-1} + 4 G_{n} \end{align*}
so this proves that $G$ is a recurrence relation.
-
At André's request, I've decided to write an answer. I've also arbitrarily decided to be ambitious and greedy, and I will thus derive a recurrence for the $k$-th increment Fibonacci number $f_{kn}$. (For OP's specific case, $k=3$)
Like André, I shall also start with Binet:
$$f_{kn}=\frac{\phi^{kn}-(-\phi)^{-kn}}{\sqrt 5}$$
Letting $u=\phi^k$ and $v=\left(-\dfrac1\phi\right)^k$, the formula takes the form
$$f_{kn}=pu^n+qv^n$$
This means that the characteristic polynomial for the recurrence satisfied by $f_{kn}$ takes the form
\begin{align*} x^2-(u+v)x+uv&=x^2-\left(\phi^k+\left(-\frac1\phi\right)^k\right)x+\left(\phi^k\left(-\frac1\phi\right)^k\right)\\ &=x^2-\left(\phi^k+\left(-\frac1\phi\right)^k\right)x+(-1)^k \end{align*}
and the recurrence itself goes like
$$f_{k(n+1)}=\left(\phi^k+\left(-\frac1\phi\right)^k\right)f_{kn}-(-1)^k f_{k(n-1)}$$
You might say that the form $\ell_k=\phi^k+\left(-\dfrac1\phi\right)^k$ is a bit unwieldy, and I agree. There are two ways to go about (slightly) simplifying this. One way makes use of the Newton-Girard formulae. These formulae express $\ell_k$ in terms of $\phi-\dfrac1\phi=1$ and $\phi\left(-\dfrac1\phi\right)=-1$. To use $k=3$ as an example:
$$\alpha^3+\beta^3=(\alpha+\beta)^3-3(\alpha+\beta)(\alpha\beta)$$ | {
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$$\alpha^3+\beta^3=(\alpha+\beta)^3-3(\alpha+\beta)(\alpha\beta)$$
Making the replacement $\alpha+\beta=1$ and $\alpha\beta=-1$, we have
$$\ell_3=(1)^3-3(1)(-1)=4$$
The slick way is to recognize that since $\ell_k$ is itself a linear combination of $\phi^k$ and $\left(-\dfrac1\phi\right)^k$, it also satisfies the Fibonacci recurrence:
$$\ell_{k+1}=\ell_k+\ell_{k-1}$$
The $\ell_k$ are in fact the (not-so-famous) Lucas numbers. With $\ell_0=2$ and $\ell_1=1$, we have the sequence $2, 1, 3, 4, 7, 11,\dots$
In short, the recurrence is of the form
$$f_{k(n+1)}=\ell_k f_{kn}-(-1)^k f_{k(n-1)}$$
For $k=3$, we have $f_{3(n+1)}=\ell_3 f_{3n}-(-1)^3 f_{3(n-1)}$ or $f_{3(n+1)}=4 f_{3n}+f_{3(n-1)}$.
-
Very nice post. This is the kind of solution where each piece falls nicely at its right place. – Did Dec 27 '11 at 9:35
It turns out that the identity derived here is listed in the Wolfram Functions site. – Guess who it is. Dec 28 '11 at 0:21
@JM: this identity is a special case of the identity $$F_{n+2k}=L_kF_{n+k}-(-1)^kF_n$$ However, the indices do not need to be multiples of $k$, they just need to differ by $k$. – robjohn Jan 16 '12 at 19:43
Let $\alpha$ and $\beta$ be the two roots of the equation $x^2-x-1=0$. Then the $n$-th Fibonacci number is equal to $$\frac{\alpha^n-\beta^n}{\sqrt{5}}.$$
We are interested in the recurrence satisfied by the numbers $$\frac{\alpha^{3n}-\beta^{3n}}{\sqrt{5}}.$$
If $x$ is either of $\alpha$ or $\beta$, then $x^2=x+1$. Multiply by $x$. We get $x^3=x^2+x=2x+1$. It follows that $x^4=2x^2+x=3x+2$. But then $x^5=3x^2+2x=5x+3$, and then $x^6=5x^2+3x=8x+5$.
We want $x^6=Ax^3+B$, where $A$ and $B$ are rational, indeed integers. So we want $8x+5=A(2x+1)+B$. Reading off $A$ and then $B$ is obvious: we need $A=4$ and $B=1$.
So the numbers $\alpha^{3n}$ and $\beta^{3n}$ satisfy the recurrence $y_n=4y_{n-1}+y_{n-2}$. By linearity, so do the numbers $\frac{\alpha^{3n}-\beta^{3n}}{\sqrt{5}}$. | {
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Comment: Note that using the same basic strategy, we can write down the recurrence satisfied by $\frac{\alpha^{kn}-\beta^{kn}}{\sqrt{5}}$. The coefficients that we painfully computed by hand, step by step, can be expressed simply in terms of Fibonacci numbers, and therefore so can the recurrence for the numbers $\frac{\alpha^{kn}-\beta^{kn}}{\sqrt{5}}$.
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Nice! I think Newton-Girard might also yield a useful route here. – Guess who it is. Dec 27 '11 at 5:11
I thought I would take an "unfancy" route through facts likely to be well-known. – André Nicolas Dec 27 '11 at 5:18
Here's the Newton-Girard route for completeness: one wants the characteristic polynomial $x^2-(\alpha^3+\beta^3)x+\alpha^3 \beta^3$ without knowing $\alpha$ or $\beta$, but knowing that $\alpha+\beta=1$ and $\alpha\beta=-1$ (Vieta). It is easily seen that $\alpha^3 \beta^3=-1$. Using Newton-Girard, we have the symmetric polynomial expansion $\alpha^3+\beta^3=(\alpha+\beta)^3-3(\alpha+\beta)(\alpha\beta)$, and thus $\alpha^3+\beta^3=(1)^3-3(1)(-1)=4$. The characteristic polynomial is thus $x^2-4x-1$. – Guess who it is. Dec 27 '11 at 5:25
@J.M.: I think that the comment above should be made into an answer. – André Nicolas Dec 27 '11 at 5:29
By inspection $f(3n+3)=4f(3n)+f(3n-3)$, as you’ve already noticed. This is easily verified:
\begin{align*} f(3n+3)&=f(3n+2)+f(3n+1)\\ &=2f(3n+1)+f(3n)\\ &=3f(3n)+2f(3n-1)\\ &=3f(3n)+\big(f(3n)-f(3n-2)\big)+f(3n-1)\\ &=4f(3n)+f(3n-1)-f(3n-2)\\ &=4f(3n)+f(3n-3)\;. \end{align*}
However, I didn’t arrive at this systematically; it just ‘popped out’ as I worked at eliminating terms with unwanted indices.
Added: Here’s a systematic approach, but I worked it out after the fact. | {
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Added: Here’s a systematic approach, but I worked it out after the fact.
The generating function for the Fibonacci numbers is $$g(x)=\frac{x}{1-x-x^2}=\frac1{\sqrt5}\left(\frac1{1-\varphi x}-\frac1{1-\hat\varphi x}\right)\;,$$ where $\varphi = \frac12(1+\sqrt5)$ and $\hat\varphi=\frac12(1-\sqrt5)$, so that $f(n)=\frac1{\sqrt5}(\varphi^n-\hat\varphi^n)$. Thus, $f(3n)=\frac1{\sqrt5}(\varphi^{3n}-\hat\varphi^{3n})$. Thus, we want
\begin{align*} h(x)&=\frac1{\sqrt5}\sum_{n\ge 0}(\varphi^{3n}-\hat\varphi^{3n})x^n\\ &=\frac1{\sqrt5}\left(\sum_{n\ge 0}\varphi^{3n}x^n-\sum_{n\ge 0}\hat\varphi^{3n}x^n\right)\\ &=\frac1{\sqrt5}\left(\frac1{1-\varphi^3 x}-\frac1{1-\hat\varphi^3 x}\right)\\ &=\frac1{\sqrt5}\cdot\frac{(\varphi^3-\hat\varphi^3)x}{1-(\varphi^3+\hat\varphi^3)x+(\varphi\hat\varphi)^3x^2}\;. \end{align*}
Now $\varphi+\hat\varphi=1$, $\varphi-\hat\varphi=\sqrt5$, $\varphi\hat\varphi=-1$, $\varphi^2=\varphi+1$, and $\hat\varphi^2=\hat\varphi+1$, so
\begin{align*} h(x)&=\frac1{\sqrt5}\cdot\frac{(\varphi^3-\hat\varphi^3)x}{1-(\varphi^3+\hat\varphi^3)x+(\varphi\hat\varphi)^3x^2}\\ &=\frac{(\varphi^2+\varphi\hat\varphi+\hat\varphi^2)x}{1-(\varphi^2-\varphi\hat\varphi)x-x^2}\\ &=\frac{(\varphi^2-1+\hat\varphi^2)x}{1-(\varphi^2+1+\hat\varphi^2)x-x^2}\\ &=\frac{(\varphi+\hat\varphi+1)x}{1-(\varphi+3+\hat\varphi)x-x^2}\\ &=\frac{2x}{1-4x-x^2}\;. \end{align*}
It follows that $(1-4x-x^2)h(x)=2x$ and hence that $h(x)=4xh(x)+x^2h(x)+2x$. Since the coefficient of $x^n$ in $h(x)$ is $f(3n)$, this tells me that
\begin{align*} \sum_{n\ge 0}f(3n)x^n&=h(x)=4xh(x)+x^2h(x)+2x\\ &=\sum_{n\ge 0}4f(3n)x^{n+1}+\sum_{n\ge 0}f(3n)x^{n+2}+2x\\ &=\sum_{n\ge 1}4f(3n-3)x^n+\sum_{n\ge 2}f(3n-6)x^n+2x\;, \end{align*}
which by equating coefficients immediately implies that $f(3n)=4f(3n-3)+f(3n-6)$ for $n\ge 2$. It also gets the initial conditions right: the constant term on the righthand side is $0$, and indeed $f(3\cdot 0)=0$, and the coefficient of $x$ is $4f(0)+2=2=f(3)$, as it should be.
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-
Actually the "Magic Trick" or "Reverse Engineering" idea works nicely.
First, as André pointed
$$f_{3n}=\frac{\alpha^{3n}+\beta^{3n}}{\sqrt{5}} \,.$$
This means that if $(x-\alpha^{3})(x-\beta^3)=x^2-Ax-B$ then $f_{3n}$ is the recurrence satisfying
$$x_{n+2}=Ax_{n+1}+Bx_{n} \, x_{0}=f_0, x_1=f_{3} \,.$$
Thus,
$$f_6=Af_3+Bf_0$$ $$f_9=Af_6+Bf_3$$
Solve it and get $A,B$.
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Let $S$ be the shift operator on sequences (as in Bill Dubuque's answer). Note that the Fibonacci sequence is killed by $S^2-S-1$. The Fibonacci sequence will then be killed by any "polynomial" multiple of $S^2-S-1$. To get a recurrence for every $k^{\rm{th}}$ term, all we need to do is find a multiple of $S^2-S-1$ that only involves powers of $S^k$.
First note that $S^2-S-1=(S-a)(S-b)$ where $a=\phi$ (the golden ratio) and $b=-1/\phi$. Consider the operator $(S^k-a^k)(S^k-b^k)=S^{2k}-(a^k+b^k)S^k+(ab)^k$. It is a polynomial multiple of $S^2-S-1$, so it kills the Fibonacci sequence. It only involves powers of $S^k$.
Recall that one formula for the $k^{\rm{th}}$ Lucas number is $L_k=a^k+b^k$, and note that $ab=-1$. Thus, we get that $S^{2k}-L_kS^k+(-1)^k$ kills the Fibonacci sequence.
Therefore, in summary, we get $$F_{n+2k}=L_kF_{n+k}-(-1)^kF_n\tag{1}$$ For example, $$F_{n+2}=F_{n+1}+F_n\tag{k=1}$$ $$F_{n+4}=3F_{n+2}-F_n\tag{k=2}$$ $$F_{n+6}=4F_{n+3}+F_n\tag{k=3}$$ $$F_{n+8}=7F_{n+4}-F_n\tag{k=4}$$ $$F_{n+10}=11F_{n+5}+F_n\tag{k=5}$$
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We can write linear recurrence relations in terms of matrix multiplication like so:
$$F_{n+2} = \left[ \begin{array}{cc} 1 & 1 \end{array} \right] \left[ \begin{array}{cc} F_n \\ F_{n+1} \end{array} \right] = 1 \cdot F_n + 1 \cdot F_{n+1}.$$
Now if the sequence 0,2,8,34,144,610,2584,10946,... is called $G_n$, let's make the unjustified assumption that it is also a second order recurrence relation, then not only would we have | {
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$$G_{n+2} = \left[ \begin{array}{cc} c_1 & c_2 \end{array} \right] \left[ \begin{array}{cc} G_n \\ G_{n+1} \end{array} \right]$$
for some unknowns $c_1$ and $c_2$, but also we can collect several instances of the above identity together into one e.g.
$$\left[ \begin{array}{cc} 34 & 144 \end{array} \right] = \left[ \begin{array}{cc} c_1 & c_2 \end{array} \right] \left[ \begin{array}{cc} 2 & 8 \\ 8 & 34 \end{array} \right]$$
and we can solve this using PARI/GP like so:
? [34,144]/[2,8;8,34]
% = [1, 4]
Therefore $G_{n+2} = 1 \cdot G_n + 4 \cdot G_{n+1}$.
About the assumption, it can be proved in general based on the ideas of characteristic function which you have seen in most of the answers. Given that, there is no need for any induction proofs or anything. Just computing the vector completes the proof of $G_{n+2} = 1 \cdot G_n + 4 \cdot G_{n+1}$.
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Here's an interesting identity involving Lucas and Fibonacci numbers: $$\begin{pmatrix}\ell_k&(-1)^{k+1}\\1&0\end{pmatrix}=\begin{pmatrix}f_k&f_{k-1}\\0&1\end{pmatrix}\cdot\begin{pmatrix}1&1\\1&0\end{pmatrix}^k \cdot\begin{pmatrix} f_k&f_{k-1}\\0&1\end{pmatrix}^{-1}$$ – Guess who it is. Dec 28 '11 at 15:21
It is easy by operator algebra. Let the Shift, Triple $\mathbb C$-linear operators $\rm\ S\:n\: :=\: n+1,\ \ T\:n\: :=\: 3\:n\:$ act on fibonacci's numbers by $\rm\ S\:f(a\:n+b) = f(a\:(n+1)+b)\$ and $\rm\ T\:f(a\:n+b) = f(3\:a\:n+b)\:.$ Below I show a general method that works for any Lucas sequence $\rm\:f(n)\:$ that involves only simple high-school polynomial arithmetic (albeit noncommutative). Namely, one employs a commutation rule $\rm\: TS\:\to\: (a\ S + b)\ T\$ to shift $\rm\:T\:$ past powers of $\rm\:S\:,\:$ in order to transmute the known recurrence $\rm\ q(S)\ f(n)\: =\: 0\$ into $\rm\ \bar{q}(S)\:T\:f(n)\:=\:0\:,\$ the sought recurrence for $\rm\ T\:f(n)\: =\: f(3\:n)\:.\:$ | {
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We know $\rm\ q(S)\ f(n) := (S^2 - S - 1)\ f(n)\: =\: f(n+2) - f(n+1) - f(n)\: =\: 0\:.\:$ We seek an analogous recurrence $\rm\ \bar{q}(S)\ T\: f(n)\ =\ 0\$ for $\rm\ T\:f(n) = f(3\:n)\:,\:$ and some polynomial $\rm\:\bar{q}(S)\:.\:$ Since clearly we have that $\rm\ T\:q(S)\ f(n)\: =\: 0\:,\:$ it suffices to somehow transmute this equation by shifting $\rm\:T\:$ past $\rm\:q(S)\:$ to yield $\rm\:\bar{q}(S)\:T\:f(n)\:=\:0\:.\:$ To do this, it suffices to find some commutation identity $\rm T\:S\: =\: r(S)\: T\$ to enable us to shift $\rm\:T\:$ past $\rm\:S$'s in each monomial $\rm\ S^{\:i}\: f(n)\: =\: f(n+i)\:$ from $\rm\:q(S)\:.\:$ The sought commutation identity arises very simply: iterate the recurrence for $\rm\:f(n)\:$ so to rewrite
$\rm\ ST\ f(n)\ =\ f(3\:n+3)\$ as a linear combination of $\rm\ f(3\:n+1) = TS\ f(n)\:,\:$ $\rm\ f(3\:n) = T\ f(n)\:,\:$ viz.
$\rm\ \ \ ST\ f(n+i)\ =\ f(3n+3+i)\ =\ f(3n+2+i) + f(3n+1+i)\ =\ 2\ f(3n+1+i) + f(3n+i)$
$\rm\ \ \ \phantom{ST\ f(n+i)}\ =\ (2\:TS+T)\ f(n+i)\quad$ for all $\rm\:i\in \mathbb Z\:$
$\rm\ 2\:TS\ f(n+i)\ =\ (S-1)\:T\ f(n+i)\:,\$ i.e. $\rm\ 2\:TS\ =\ (S-1)\:T\:,\$ the sought commutation identity.
Thus $\rm\qquad\quad\:\ 0\ =\ 4\: T\: (S^2 - S - 1)\ f(n)\$
$\rm\qquad\qquad\qquad\quad\ \ =\ (2\:(2TS)S - 2\:(2TS) - 4\:T)\ f(n)$
$\rm\qquad\qquad\qquad\quad\ \ =\ ((S-1)\:2TS - 2\:(S-1)\:T - 4\:T)\ f(n)$
$\rm\qquad\qquad\qquad\quad\ \ =\ ((S-1)^2 - 2\:(S-1)\: - 4)\ T\: f(n)$
$\rm\qquad\qquad\qquad\quad\ \ =\ (S^2 - 4\ S - 1)\ T\: f(n)$
$\quad$ i.e. $\rm\qquad\quad\: 0\ =\ f(3(n+2)) - 4\ f(3(n+1)) - f(3\:n)\qquad$ QED | {
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$\quad$ i.e. $\rm\qquad\quad\: 0\ =\ f(3(n+2)) - 4\ f(3(n+1)) - f(3\:n)\qquad$ QED
NOTE $\$ Precisely the same method works for any Lucas sequence $\rm\:f(n)\:,\:$ i.e. any solution of $\rm\ 0\ =\ (S^2 + b\ S + c)\ f(n)\ =\ f(n+2) + b\ f(n+1) + c\ f(n)\$ for constants $\rm\:b,\:c\:,\:$ and for any multiplication operator $\rm\:T\:n = k\ n\:$ for $\rm\:k\in \mathbb N\:.\:$ As above, we obtain a commutation identity by iterating the recurrence (or powering its companion matrix), in order to rewrite
$\rm\ ST\ f(n)\ =\ f(k\:n+k)\$ as a $\rm\:\mathbb C$-linear combination of $\rm\ f(kn+1) = TS\ f(n)\$ and $\rm\ f(kn) = T\ f(n)\:$
say $\rm\ \ ST\ f(n)\ =\ f(k\:n+k)\ =\ a\ f(k\:n+1) + d\ f(k\:n)\ =\ (a\ TS + d\ T)\ f(n)\ \$ for some $\rm\:a,d\in \mathbb C$
$\rm\:\Rightarrow\ a\ TS\ f(n) =\ (S-d)\ T\ f(n)\ \Rightarrow\ a\ TS\ =\ (S-d)\ T\$ on $\rm\ S^{\:i}\: f(n)\$ as above.
Again, this enables us to transmute the recurrence for $\rm\:f(n)\:$ into one for $\rm\:T\:f(n) = f(k\:n)\:$ by simply commuting $\rm\:T\:$ past all $\rm\:S^i\:$ terms. Hence the solution involves only simple polynomial arithmetic (but, alas, the notation obscures the utter simplicity of the method).
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# Proof Verification: If $A \subset B$ and $B \subset C$, then $A \cup B \subset C$
I am trying to prove that:
If $A \subset B$ and $B \subset C$, then $A \cup B \subset C$
My proof is : Given some $x \in A \cup B$, it is true that either $x \in A$ and/or $x \in B$. IN the case that $x \in A$ it is true that $x \in B$, as $A \subset B$, and that $x \in C$ , as $B \subset C$. In the case that $x \in B$ it is true that $x \in C$, as $B \subset C$. Therefore, $A\cup B \subset C$
Is this correct?. Any tips to improve this would be appreciated as I am self taught and new to proof writing.
• Yes, it is correct. You could also prove that $A \cup B = B$, and then the fact follows from $B = A \cup B \subset C$. – астон вілла олоф мэллбэрг Aug 8 '18 at 15:14
• Your proof is totally correct. Just be careful about the use of English: it is better to say "Given some $x \in A \cup B$, it is true that $x \in A$ or $x \in B$". Do not use "either" and "and" unnecessarily as it may make your statement confusing. In maths, the word "or" means : either the first case only, either the second only, either both cases at the same time. – Suzet Aug 8 '18 at 15:16
• @астонвіллаолофмэллбэрг Thank you – Ewan Miller Aug 8 '18 at 15:16
• @Suzet Ok thank you – Ewan Miller Aug 8 '18 at 15:17
• It's totally correct, although there's an alternative way. Since $A \subset B$, $A \cup B= B$. Now from $B \subset C$, the result follows. – Anik Bhowmick Aug 8 '18 at 15:32
Just to answer : yes, the approach is correct.
You could also prove that $A∪B=B$, and then the fact follows from $B=A∪B\subset C$.
For example, if $x \in B$ is true, then of course $x$ is in $A$ or $x$ is in $B$ is true, so $B \subset A \cup B$. If $x \in A$, then $x \in B$ because $A \subset B$, and if $x \in B$, then of course $x \in B$, so $A \cup B \subset B$, hence $A \cup B = B$. | {
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Another approach is by using Venn diagrams. Draw circles $A$, $B$ and $C$ for three sets such that $A$ is contained in $B$ and $B$ is contained in $C$ (according to given set inclusions). So you have $A$ as the innermost, $B$ in the middle and $C$ as the outermost of them. Now $A\cup B$ is given by the middle circle which is offcourse contained in $C$ (the outer circle).
Is this correct?
I believe that your proof regarding this Question is correct.
Any tips to improve this would be appreciated as I am self-taught and new to proof writing.
Set Theory which you are using (Elementary-Set Theory) was mostly Contributed for its Development by George Cantor. But, in Later Stage, there are some famous Loopholes found in this Theory, the most popular of them is Russel's Paradox. After Russel's paradox, numerous other Paradox came after which ZFC Set Theory (Zermelo Franklin Set Theory) came which is based on some of the Basic Set Axioms (Related to Mathematical Logic)
## Proofs in Set Theory
Most of the Proofs in Elementary Set Theory is based on Mathematical Logic. Some of the Basic Theorems and Properties are below -
1. De Morgan's Theorem -https://en.wikipedia.org/wiki/De_Morgan%27s_laws
2. Complement - https://en.wikipedia.org/wiki/Complement_(set_theory) $A^` = U - A$
3. Intersection and Union -
## Alternate Methods to Proof Set Theory Question
Your Question Can be Proved Using Venn Diagram too like here -
• Using The Laws and Properties of Sets (Intersection, Complement and Union)
The most natural way seems to be:
1. (Transitivity) From $A \subset B$ and $B \subset C$, show $A \subset C$.
If $x \in A$, then $x \in B$.
If $x \in B$, then $x \in C$.
Thus if $x \in A$, then $x \in C$.
1. From $A \subset C$ and $B \subset C$, show $(A \cup B) \subset C$.
If $x \in (A \cup B)$, then $x \in A$ or $x \in B$.
But if $x \in A$, then $x \in C$, and if $x \in B$, then $x \in C$.
Thus, either way, $x \in C$. | {
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You implicitly used this method in your (correct) proof, but you didn't separate out the ideas. It can be especially useful to organise larger proofs in terms of simpler definitions, lemmas and theorems. | {
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Math Help - Simultaneous Equations
1. Simultaneous Equations
I don't usually have any problem with simultaneous equations but whilst revising for a test, I came across two types that I haven't seen before and the book I have only gives answers but no explanations of how it got those answers. I've tried in various ways... but my answers never seem to match up.
Could anyone solve these two simultaneous equations step-by-step so I could see how it's done?
Q1)
$y = 1 - 4x$
$y = -4x^2$
Q2)
$xy = 9$
$x - 2y = 3$
If anyone could I would really appreciate it, thanks. And Hi to everyone on the forum - just found this forum and it looks great.
2. Hello, PaulSelb!
Welcome aboard!
These are not linear equations.
. . The recommended method is Substitution.
$1)\;\;\begin{array}{cccc}y &=& 1 - 4x &{\color{blue}[1]} \\
y &= &-4x^2& {\color{blue}[2]}\end{array}$
Equation [1] is already solved for $y\!:\;\;y \:=\:1-4x$
Substitute into [2]: . $1 - 4x \:=\:-4x^2 \quad\Rightarrow\quad 4x^2 - 4x + 1 \:=\:0$
. . $(2x-1)^2 \:=\:0 \quad\Rightarrow\quad 2x-1 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{1}{2}$
Substitute into [1]: . $y \:=\:1 - 4\left(\frac{1}{2}\right) \:=\:-1$
Solution: . $x \:=\:\frac{1}{2},\;y \:=\:-1$
$2);\;\begin{array}{cccc}xy &=& 9 & {\color{blue}[1]} \\
x - 2y &=& 3 & {\color{blue}[2]}\end{array}$
Solve [2] for $x\!:\;\;x \:=\:2y+3\;\;{\color{blue}[3]}$
Substitute into [1]: . $(2y+3)y \:=\:9 \quad\Rightarrow\quad 2y^2 + 3y - 9 \:=\:0$
Factor: . $(2y - 3)(y + 3)\:=\:0\quad\Rightarrow\quad y \:=\:\frac{3}{2},\;-3$
Substitute into [3]: . $\begin{Bmatrix}x &=& 2\left(\frac{3}{2}\right) + 3 & = & 6 \\ \\[-3mm] x &=&2(-3) + 3 &=&-3 \end{Bmatrix}$
Solutions: . $(x,y) \;=\;\left(6,\:\frac{3}{2}\right),\;(-3,\:-3)$
3. Thanks, great explanation . Got my test in a few hours so hopefully if one of these questions turns up I shouldn't have any problems. | {
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4. I had a slightly different type in the exam the other day (possibly solved in the same way)... but I got it wrong
If anyone reads this thread again could you let me know how to solve this one?
$k(k - m) = 12$
$k(k + m) = 60$
Sorry for the easy stupid questions, I really enjoy math and want to understand bits more!
EDIT: I think I did it may have done it actually...
$k^2 - km = 12$
$k^2 + km = 60$
$k+m = \frac{60}{k}$
$m = \frac{60}{k} - k$
$k^2 - k(\frac{60}{k} - k) = 12$
$2k^2 - 60 = 12$
$2k^2 - 72 = 0$
$(2k - 12)(k + 6) = 0$
$k = \pm 6$
$36 + \pm6(m) = 60$
$m = \frac{24}{\pm6}$
$m = \pm 4$
I think that's about right anyway... | {
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# Simulating order statistics
I am having a problem with the density of the first order statistic of a series of n random variables iid with common distribution (standard normal).
I am using Arnold's book as a reference for such a density function for the k'th order stat: $$f_{X_{k:n}}(x) = \frac{n!}{k-1! (n-k)!} F(x)^{k-1} (1-F(x))^{n-k} f(x).$$
The I simulate data in R store the min and compare the obtained empirical distribution against the theoretical function. These curves diverge substantially.
What am I doing wrong? or missing?
# Distribution
Fx <- function(x) pnorm(x)
fx <- function(x) dnorm(x)
n <- 10
# ATTENTION: 6M simulation may take some time
z <- replicate(1e6, min(rnorm(n)))
# Probabiltiy z > y
prob <- function(x) {
sapply(x, function(u) mean(z>u))
}
# Density
fos_k <- function(x, n, k) {
fact <- factorial(n) / (factorial(k-1) * factorial(n-k))
fact * fx(x) * Fx(x)^(k-1) *(1-Fx(x))^(n-k)
}
curve(fos_k(x, n=n, k=1), from=-1, to=1)
curve(prob, add=T, col=2, from=-1, to=1) | {
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"url": "https://stats.stackexchange.com/questions/241652/simulating-order-statistics"
} |
curve(fos_k(x, n=n, k=1), from=-1, to=1)
curve(prob, add=T, col=2, from=-1, to=1)
• The "observed empirical distribution" is not the density. One way to plot the observed density is to approximate it with a KDE and plot that. – whuber Oct 21 '16 at 19:29
• @Xi'an could you please clarify – mrb Oct 21 '16 at 20:44
• Compare your equation to the second line of fos_k: your equation omits the term you have coded as fx(x). But that's only an error of exposition; the error in your code lies in prob, which is not a density function. – whuber Oct 21 '16 at 21:40
• For what it's worth, here's the computation done in a single line in Mathematica: Show[Histogram[ParallelTable[Min[RandomVariate[NormalDistribution[0, 1], 10]], {10^6}], {0.1}, "PDF"], Plot[Evaluate[PDF[OrderDistribution[{NormalDistribution[0, 1], 10}, 1], z]], {z, -4, 1}]] On my machine, the runtime is less than 1 second. – heropup Oct 21 '16 at 22:08
• Thanks to everyone. @whuber is right. @whuber would you like to go on and reply to this question with an example of your R code? – mrb Oct 22 '16 at 1:54
Although the problem is primarily with R code, it raises issues we would have to confront in any statistical computing environment. This reply focuses on those general issues.
A correct formula for the density of the $k^\text{th}$ smallest of $n$ independent identically distributed (iid) values from a distribution $F$ with density $f$ is presented in my answer at https://stats.stackexchange.com/a/225990/919. It is
$$f_{[k]}(x) = \frac{n!}{(k-1)!(1)!(n-k)!} F(x)^{k-1} (1-F(x))^{n-k} f(x).\tag{1}$$
Because factorials and potentially high powers are combined, it is numerically better to compute its logarithm as
\eqalign{ \log f_{[k]}(x) = &\log n! - \log(k-1)! - \log(n-k)! + \\&(k-1)\log F(x) + (n-k) \log(1-F(x)) + \log f(x).\tag{2} } | {
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Furthermore, because order statistics provide a way to peer far out into the tails of the distribution, where values grow close to $1$ and $0$, it is best to compute $1-F(x)$ directly rather than subtracting $F(x)$ from $1$.
With these caveats in mind, the other issue in this thread concerns graphing an empirical density. The commonest way to do so is with a histogram: it looks like a barplot in which the bar areas (not heights!) are proportional to the relative frequencies of the data.
To illustrate these two main points, I wrote a quick solution in R that simulates many iid samples, extracts specified order statistics from each, plots the histogram of each order statistic, and overplots the density function $(1)$ computed by exponentiating the logarithm $(2)$. Here is an example of its output applied to samples of size $10$ from a standard Normal distribution and of size $25$ from a Gamma$(3/2,5)$ distribution.
Clearly the agreement between the empirical results (the histogram bars) and the theoretical formula (the colored curves) is good.
This code applies the preceding suggestions about numerical computation by exploiting the log.p, log, and lower.tail arguments that are standard in the families of R distribution functions.
f <- function(n.sim, n, k=1:n, p=pnorm, d=dnorm, r=rnorm, name, ...) {
if (missing(name)) name <- ""
k <- sort(unique(k))[1 <= k & k <= n]
# Perform the simulation.
sim <- apply(matrix(r(n.sim*n, ...), nrow=n), 2, sort)[k, , drop=FALSE]
# Plot the requested order statistics.
for (i in 1:length(k)) {
# Define a function to plot the density of an order statistic.
dord <- function(x, k, n, ...) {
z <- lfactorial(n) - lfactorial(k-1) - lfactorial(n-k) +
(k-1)*p(x, log.p=TRUE, ...) +
(n-k)*p(x, log.p=TRUE, lower.tail=FALSE, ...) +
d(x, log=TRUE, ...)
return(exp(z))
}
# Plot the empirical distribution.
hist(sim[i, ], freq=FALSE,
xlab="Value",
sub=paste(n.sim, "iterations with sample size", n),
main=paste(name, "order statistic", k[i])) | {
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# Overplot the theoretical density.
curve(dord(x, k[i], n, ...), add=TRUE, col=hsv(runif(1), 0.8, 0.7), lwd=2)
}
}
#
# Set up to simulate and display the results.
#
par(mfrow=c(2,4))
n.sim <- 1e4
set.seed(17)
# Study Normal order statistics.
f(n.sim, 10, c(1,3,5,9), name="Normal(0,1)")
# Study Gamma order statistics.
# This illustrates how to use f for general distributions.
f(n.sim, 25, c(2, 4, 16, 24), name="Gamma(1.5,5)",
p=pgamma, d=dgamma, r=rgamma, shape=1.5, scale=5)
• Wondering should use the logarithm to define the corresponding distribution function as well? Since F(x) can be zero I would avoid the log. – mrb Oct 24 '16 at 13:03
• That's a good observation, but it happens not be be relevant because almost surely $F$ will be positive at any value you have randomly generated. Regardless, good software will deal with this correctly (provided you actually ask it to compute the log of $F$ directly rather than asking it to take the log of $0$!). For instance, the R procedures will return -Inf for the logarithm in that case; and that value, when exponentiated, will be exactly $0$. Here is an example: exp(punif(-2, log.p=TRUE)) – whuber Oct 24 '16 at 13:41 | {
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"openwebmath_score": 0.5744186639785767,
"tags": null,
"url": "https://stats.stackexchange.com/questions/241652/simulating-order-statistics"
} |
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