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Note that it is not necessary to have $n$ divisible by $p$ to get nonzero $e_p(n).$ All that is necessary is that $n \geq p,$ because we are not factoring $n,$ we are factoring $n!$
Here's a different approach I found while thinking in terms of relating the sum of digits of consecutive numbers written in base $$p$$. Part of the appeal of this approach is you might have at one point learned that there is a connection but don't remember what, this gives a quick way to reconstruct it.
Let's consider $$s_p(n)$$, the sum of digits of $$n$$ in base $$p$$. How does the sum of digits change if we add $$1$$ to it? It usually just increments the last digit by 1, so most of the time,
$$s_p(n+1) = s_p(n)+1$$
But that's not always true. If the last digit was $$p-1$$ we'd end up carrying, so we'd lose $$p-1$$ in the sum of digits, but still gain $$1$$. So the formula in that case would be,
$$s_p(n+1) = s_p(n) + 1 - (p-1)$$
But what if after carrying, we end up carrying again? Then we'd keep losing another $$p-1$$ term. Every time we carry, we leave a $$0$$ behind as a digit in that place, so the total number of times we lose $$p-1$$ is exactly the number of $$0$$s that $$n+1$$ ends in, $$v_p(n+1)$$.
$$s_p(n+1) = s_p(n) + 1 - (p-1)v_p(n+1)$$
Now let's rearrange to make the telescoping series and sum over,
$$\sum_{n=0}^{k-1} s_p(n+1) - s_p(n) = \sum_{n=0}^{k-1}1 - (p-1)v_p(n+1)$$ $$s_p(k) = k - (p-1) \sum_{n=0}^{k-1}v_p(n+1)$$
Notice that since $$v_p$$ is completely additive so long as $$p$$ is a prime number,
$$s_p(k) = k - (p-1) v_p\left(\prod_{n=0}^{k-1}(n+1)\right)$$
$$s_p(k) = k - (p-1) v_p(k!)$$
This rearranges to the well-known Legendre's formula,
$$v_p(k!) = \frac{k-s_p(k)}{p-1}$$
• WOW! Not bad. Though I would recommend you to add the parentheses in order to show what are the terms in the summations. It's a bit confusing. BTW. It's funny that you gave the answer to this old question almost at the same time as I did. – LRDPRDX Jun 6 '20 at 10:31 | {
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Walking down the street I realized how to proof the formula in a very simple way. The key idea is to know how many numbers contain a given prime in the exact $$q$$-th power.
Let us denote the set of all numbers up to $$n$$ (inclusive) that are divisible by $$q$$-th power of prime $$p$$ but not divisible by the greater ($$q+1, q+2, ...$$) power of that prime: $$M^p_q(n) := \{ m : m \le n, p^q | m, p^{q+1} \nmid m\}$$ How many numbers in that set? Well, obviously, the number of those that are divisible by $$p^q$$ minus the number of those which are divisible by $$p^{q+1}$$: $$|M^p_q(n)| = \bigg\lfloor \frac{n}{p^q}\bigg\rfloor - \bigg\lfloor \frac{n}{p^{q+1}}\bigg\rfloor$$ OK. Each of the numbers in $$M^p_q$$ gives us the $$q$$-th power of $$p$$ in $$n!$$ so we need just add them all: $$s_p(n!)=\sum_{q=1}^{\infty}q|M^p_q(n)|$$ Which is the desired result. | {
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Vavg Formula | {
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h interface. Multiply by 3600 and you get watt-seconds, which is also known as Joules. 955*Vs which isn't what i get from the simulation from multisim. 3 m/s upwards. Based on the chosen capacitor value and the approximate value of Vcc, we can clearly calculate the expected Vdrop across N samples by using the formula above. A formula for the number of possible permutations of k objects from a set of n. Thinking they might prove useful as tools or templates to others, it is my pleasure to provide them freely to the scientific community. This means that equation 7 can be reduced to. Peak or crest voltage can be obtained from RMS value of voltage: Vpeak = 1. Find the volume per organism, assuming a spherical shape. 732 = 115,401W = 115. This web site owner is mathematician Dovzhyk Mykhailo. Fig: Average Velocity, r. Once we know the peak voltage ( V o ) and the resistance (R) in the circuit we can calculate the peak current ( Io ) using the equation V=IR. Dalam gas ideal berlaku 3 hukum tentang kinetika gas yaitu. Vavg — where T is the temperature (in kelvins), M is the molar mass (in kilo- grams per mole), and R = 8. Those two variables are temperature and amount of gas (the last one being measured in moles). vi = initial velocity. Yes, watt-hours is a measure of energy, just like kilowatt-hours. You can use the formula A = C^2 / 4pi to find the area of a circle given the circumference. Мама не дала Росси проехать гонку. Subject Company: ITC Holdings Corp, Commission File No. The next step is determining the pipette’s accuracy, manually or via software. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. The level of a waveform defined by the condition that the area enclosed by the curve above this level is exactly equal to the area | {
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by the condition that the area enclosed by the curve above this level is exactly equal to the area enclosed Vp x. Voltage imbalance is given by the formula: Ring connect screw terminals C. 6m/s2 isn't necessary so what you do is you employ the equation Vavg=deltax/delta t then you certainly re-manage it and get the equation (delta t)*(Vavg)=delta x this is what u want this is displacement so then you certainly plug interior the numbers so it would look like this (6. E = C*Vavg Where E is the energy stored in watt-hours, C is the capacity in amp-hours, and Vavg is the average voltage during discharge. 120v Vrms input so 170 Vpp I get 162 vavg using the. IIT-Madras, Momentum Transfer: July 2005-Dec 2005. Substitute (1) into (6) 2(delta x /. = − =528 0 360 1. The 115V is an RMS voltage. The basic definition of average velocity is: Vavg = Δx/Δt = Xf – X0/tf – t0 This is the change in position divided by the time of travel. Following is the formula for Vpp to Vrms conversion. Fundamentals Of Physics Instructors Solutions Manual | Halliday, Resnick | download | B–OK. Present Value is like Future Value in reverse: you The formula for present value is simple; just take the formula for future value and solve for starting principal. Multiply by 3600 and you get watt-seconds, which is also known as Joules. The term average is from mathematics that basically means the central tendency of a set particular set of data. I can use the formula ΔVc = Vm(1-e-t/RC) I don't know why it didn't click sooner, it will become ΔVc = 19. I want to code this do while loop to do this. A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 17. Chapter 1 – Student Solutions Manual 3. the chemical compound silicon dioxide, also known as silica (from the Latin silex), is an oxide of silicon with a chemical formula of SiO 2 and has been known for its hardness since antiquity. The formula can also be used to calculate the present value of money to be | {
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hardness since antiquity. The formula can also be used to calculate the present value of money to be received in the future. Velocity and Most Probable Velocity – a relationship in equation. (c) Substitute t = 1, 2, and 3 into v(t). ARPU Revenue Formula. Easily share your publications and get them in front of Issuu’s. Enzyme-linked Immunosorbent Assay (ELISA) The concentrations of NF-κB-responsive genes (IL-8, TNF-a and IL-1β) were evaluated in cell culture supernatants according to the manufacturer’s protocol (Joyee Biotechnics, Shanghai, China). So is the. Executive Vice President & Chief Financial Officer. Find books. Формула-1 (Formula-1) 2020. standard value calculated using the following calculation formula. WikiHow recommends the value should be between 99 and 101%. Formula 1® Esports Series is back for its 3rd season! Compete against the fastest drivers in the world on F1TM 2019 and stand a chance to become an official driver for an F1 Team!. 1 A to 20,000 A. Answer / soumya ranjan panigrahi we know form factor is the cause of generating voltage in India is 11kV or it be transmitted as multiples of 11, because of simple to desine. Memorizing equations by rote, rather than understanding the physics by which they were derived,. Therefore, vavg m s m/s. For example it can be used to check the format of the container used by a multimedia stream and the format and type of each media stream contained in. As illustrated in the figure, the slope can be positive, negative, or zero. Discover the ROIC formula and learn how to apply it through our ROIC example. 0 Introduction: Quality Assurance and Certification June 17, 2009 ECMPS Reporting Instructions Quality Assurance and Certification 1. Concepts of Physics Part 1, Numerical Problems with their solutions, Short Answer Solutions for Chapter 3 - Rest and Motion: Kinematics from the latest edition of HC Verma Book. As part of Formula E's fundraising partnership with UNICEF, the all-electric street racing series will. | {
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part of Formula E's fundraising partnership with UNICEF, the all-electric street racing series will. For average velocity, d = displacement (vector). We use lower case p(t) because this is the expression for the instantaneous power at time t. These values will be listed and identified as either. These variables also permit you to use behavioral circuit analysis, modeling, and simulation techniques. , the general formula is: Number of organisms = N = 2(t/30 min) = 2(t/0. since only their ratio enter the equation. A Framework for Mining Association Rules in Data Warehouses 161 Assume that in a fact table there is m-dimensions and the quantity attribute exists. formula and/or breastmilk until 12 months of age. Do While (Vavg >= Vavg_error_lb And Vavg =< Vavg_error_ub) ' Call Sim_rand 'Loop Thank you. Average and RMS value of center-tap full wave rectifier. A user enters the voltage,. Quadratic Formula Calculator. = − =528 0 360 1. why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. The first calculation will involve pickets 1 and 2, the next 2 and 3, etc. Enter the world of Formula 1. 1 A to 20,000 A. The formulas below compare the calculations of an RMS meter compared to an average rectified measuring meter when measuring pure At each voltage both Vrms and Vavg values are recorded. The next step is determining the pipette’s accuracy, manually or via software. Any such data mashup is expressed using the Power Query M Formula Language. VAVG - (-VAVG ) ) in the sum of the two areas, thus resulting in zero average voltage over one To find the peak value from a given average voltage value, just rearrange the formula and divide by the. You can't simply do 30x2 (velocity times time) to get the total distance. Kinetic Theory of Gases & Maxwell Distribution Curve & Gas Velocity Vmp,Vavg,Vrms Formulas(Class XI). Cost can add up quickly, especially if you’re a novice and have never attempted a Stamped Concrete Patios installation before. " | {
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if you’re a novice and have never attempted a Stamped Concrete Patios installation before. " In the case of a set of n values. The pulse waveform is shown in Figure 1. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). The highest fre- quencies of 2 to 4 MHz correspond to. Customer Experience Training and Workshops for Entrepreneurs and Small Businesses. Find books. The RMS voltage of a sinusoid or complex waveform can be determined by two basic methods. Box 1926 Spartanburg, SC 29304 GENERAL INFORMATION Milliken is a major manufacturer of textile products for apparel, commercial, home,. If waveform is something different than pure sinewave, reading between two meters will shift appart. Input a constant (in this case: 1) as your Formula and configure the Output Extents. 1-AUG-2019 (Ver. Investors are able to reasonably assume an investment's profit using the future value. Este/a fórmula de encantamiento se usa en la profesión de Encantamiento. It calculates the RMS voltage based on the above formulas for each. The number of solutions and their type depends on the value of the discriminant (quantity under the radical). The instantaneous velocity at any time is the slope of the positiontime graph at that time. Compounding formulas for discrete payments. shortest If set to 1, force the output to terminate when the shortest input terminates. Attached is the sample Application. Vavg = ΔX / Δt For instantaneous velocity, the time interval Δt is shrunk down to the smallest number that you can imagine that is still greater than zero. Vavg Output voltage average of the phase voltage according to the following formula and kept in this product. Dari chart terlihat pasar Amerika, Eropa dan Asia tampak berseri-seri, semua MA5 di bawah candlestick dan pointing up. For example using the formula above we can state that: Vadj = (Vin / 100) * d. For Boyle's Law to be valid, the other two variables must be held constant. 4, the standard | {
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* d. For Boyle's Law to be valid, the other two variables must be held constant. 4, the standard luminance signal's bandwidth ranges from dc to approxi- mately 4 MHz, whereas audio ranges from dc to 20 kHz. 0 furlongs = ( 4. Several questions might immediately come to mind. In this section, we are going to see mensuration formulas which can be used to find cured surface area, total surface area and volume of 3-D shapes like Volume of cylinder = Πr2h. why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. This RMS Voltage calculator helps to find the RMS voltage value from the known values of either peak voltage, peak-to-peak voltage or average voltage. (e) The horizontal line near the bottom of this x-vs-t graph represents the man standing at x = 0 for 0 ≤ t 300 s and the linearly rising line for 300. Download Solutions Manual - Materials Processing in Manufacturing Demargo. compiled from Merck Source. Download books for free. Calculator solution will show work for real Uses the quadratic formula to solve a second-order polynomial equation or quadratic equation. : The maximum instantaneous value of a function as measured from the zero-volt level. And there are certain formulae that are used for the same. Physics Formula Sheet. s di Formula Group S. Securities Act of 1933 and deemed filed pursuant. The formula is (A)*(V) = (W). 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. Formulae definition, a set form of words, as for stating or declaring something definitely or authoritatively, for indicating procedure to be followed, or for prescribed use on some ceremonial. 9610° 36000. It's a functional, case sensitive language similar to F#. VECTORS OPERATIONS ON VECTORS • A scalar quantity (such as mass or energy) can be fully described by a (signed) number. The term streamline flow is descriptive of the laminar flow because, in laminar flow, layers of water flowing over one another at | {
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of the laminar flow because, in laminar flow, layers of water flowing over one another at different speeds with virtually no mixing between layers, fluid particles move in definite and observable paths or streamlines. The complete formula (EFN) is expressed as: EFN = (A/S) x (Δ Sales) - (L/S) x (Δ Sales) - (PM x FS x (1-d)). compiled from Merck Source. Any such data mashup is expressed using the Power Query M Formula Language. For average velocity, d = displacement (vector). Using the above formula, the average voltage can be calculated as. 637 Vpp = Vp • 2 Vp = Vpp ÷ 2 Vinst. High voltage measurement from 10 V up to 6,400 V. Best Price Roadside Assistance. Average Speed formula. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. A chemical formula may be entered to restrict the search to species which match the formula. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s. 5 kg flies through the air at a low speed, so that the air resistance is negligible, What is the net force acting on the ball while it is in motion? Which components of the ball's. 230K likes. This means that the most recent entry always overwrites the oldest one. V m = maximum value of transformer secondary voltage. The average velocity of the sled is (Vavg). Vatios-hora es una medida de energía, al igual que kilovatios-hora. This means that equation 7 can be reduced to. The average velocity of gas particles is found using the root mean square velocity formula: μ rms = (3RT/M) ½ μ rms = root mean square velocity in m/sec R = ideal gas constant = 8. Best Price Roadside Assistance. Hello there, I have a simple routine below, with 2 variables which are calculated in part by a random number. Rules for formula restriction (step 2) (Back to search). Average Force Formula Questions: 1) A dog that weighs 10 kg chases a car for 12 seconds at a velocity of 5 m/s. The difference between modeled and actual | {
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10 kg chases a car for 12 seconds at a velocity of 5 m/s. The difference between modeled and actual velocities can be observed from the curve. 10Beta) CHANGELOG. Several questions might immediately come to mind. Find books. 7979 × √(RT) Now equating them we get, T = (6. T f = 308 K. 758) + A=AIR(3. Value at risk (VAR or sometimes VaR) has been called the "new science of risk management ," but you don't need to be a scientist to use VAR. Run Automatic Analysis/Auto-Analyzer untuk melihat hasilnya. v o represents the object's initial velocity at the beginning of the time interval. The rms speed is a good approximation of the the typical speed of the molecules in a gas. High voltage measurement from 10 V up to 6,400 V. The average velocity of the ball can be found using these values and the formula: v avg = 0. Hi, How do you code, "not greater than or equal to" and "not less than or equal to" in vb. Hence, Vavg Vdc [(12 1) (8 1)]/2 2 V. Someone tosses a tomato straight up in the air. When the payments are all the same, this can be considered a geometric series with 1+r as the common ratio. The hydrocarbon that has at least one double bond is called as alkene with the general formula C n H 2 n. ffmpeg reads from an arbitrary number o. Therefore, vavg m s m/s. RMS voltage of a half wave rectifier, VRMS = Vm / 2 and Average Voltage VAVG= Vm/π, Vm is the peak voltage. If acceleration is constant the Vavg occurs at 1/2 the total time and. The right hand picture illustrates the same formula. vavg ≡ The instantaneous velocity v is defined as the limit of the ratio Δ x Δ t as Δt approaches zero. 10 s vi =?. The average value of a sine wave is zero because the area covered by the positive half cycle. Because, the positive and negative half cycle is equal in magnitude and thus the total value cancels out on summation. Subject Company: ITC Holdings Corp, Commission File No. I want the code to repeat over and over until the two conditions are satisfied. We use lower case p(t) | {
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want the code to repeat over and over until the two conditions are satisfied. We use lower case p(t) because this is the expression for the instantaneous power at time t. En la categoría Fórmulas de encantamiento. Alternate Formulas or Computational Formulas for Variance. The formula for the future value of an annuity, or cash flows, can be written as. In this section, we are going to see mensuration formulas which can be used to find cured surface area, total surface area and volume of 3-D shapes like Volume of cylinder = Πr2h. Average Velocity We begin by writing down the formula for average velocity. SimScale and Formula Student Germany have joined forces to offer a free workshop about the application of CFD in Formula Student. It may also be referred to as the annualized rate of return or annual percent yield or effective annual rate. WikiHow recommends the value should be between 99 and 101%. It calculates the RMS voltage based on the above formulas for each. (2) Methods: The sample. 175 * 10 -3 * d 4 / (f t c vavg),. If any distances x i and x f with their corresponding time intervals t i and t f are given we use the formula. Formula Simulator. How to prepare for States of matter? This chapter is a part of Physical chemistry. Here, in part 1 of this short series on the topic, we. A ball of mass 0. Discover the ROIC formula and learn how to apply it through our ROIC example. Use MathJax to format equations. This would require an integer, however vavg is a ratio of type double. Calculate the average velocity of your car (in cm/tu) for the entire run using the formula Vavg = total distance covered/total time. Formula One Grand Prix (known as World Circuit in the United States) is a racing simulator released This version of Formula One Grand Prix was designed for personal computers with operating system. For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. repeatlast If set to 1, | {
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(pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. repeatlast If set to 1, force the filter to extend the last frame of secondary streams until the end of the primary stream. The Formula Sim is specifically engineered to help drivers to improve their craft in Formula categories. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. The term streamline flow is descriptive of the laminar flow because, in laminar flow, layers of water flowing over one another at different speeds with virtually no mixing between layers, fluid particles move in definite and observable paths or streamlines. This set focuses on formulas, important numbers, word problems, and linear motion terminology. The Center-Tapped Full-Wave Rectifier A center-tapped rectifier is a type of full-wave rectifier that uses two diodes connected to the secondary of a center-tapped transformer, as shown in Figure (a). What do you want to calculate?. Substitute (3) into (2) (4) Vavg = (0 + v)/2. Thus, the total of the masses of neutrons and protons is the atomic mass. 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. It is akin to Ohm’s Law, but something I was never taught in tech school –I had to learn it myself much later –now I use it at least weekly. 0321-5455195 Home Delivery Service is Also Available. (d) The object hits the ground when its position is s(t) = 0. The average velocity of the ball can be found using these values and the formula: v avg = 0. Silica is most commonly found in nature as sand or quartz, as well as in the cell walls of diatoms. Root mean square is a mathematical term that suggests effective level. Basic Properties and Formulas. The RMS value of a pulse waveform can be easily calculated starting with the RMS definition. Exam 1 November 2012, Questions and answers Exam 31 October 2008, Questions and answers Exam 26 October 2009, Questions Exam 10 December 2010, Questions Final Formula Sheet - | {
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and answers Exam 26 October 2009, Questions Exam 10 December 2010, Questions Final Formula Sheet - Summary Mechanics and Waves Phys 131 Equation Sheet. Regards Dan. At the end there is a do while loop with 2 conditions. Pokhriyal2. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. Plugging that into the avg voltage formula: Vavg = 2x217. For simplicity, we will rewrite the formula. EVM - Miscellaneous Formula - Budget at Completion (BAC) is the total budget allocated to the BAC is also used to compute the TCPI and TSPI. man ffprobe (1): ffprobe gathers information from multimedia streams and prints it in human- and machine-readable fashion. Among them, the only time spent for value added activities is the process time. Average voltage (Vavg). In this problem the initial velocity is 10 mph and the final velocity is 7. Materiales de aprendizaje gratuitos. This means that the most recent entry always overwrites the oldest one. key},quantity). pdf) or read book online for free. Calculate the average velocity of your car (in cm/tu) for the entire run using the formula Vavg = total distance covered/total time. avg speed =√(8RT/πM). What do you calculate for the x component of average velocity? vavg. Express your answer in terms of the given quantities and,. HTflux is using the following equations for this task: for laminar pipe flows: Resistance coefficient / hydraulic gradient (R in kg/m 7 ):. If by "going at a If and only if the acceleration is constant, then you can make use the naive formula for average speed, as. Cone : Formulas. Plugging that into the avg voltage formula: Vavg = 2x217. 00000001 1E-08 3561 CHAPTER 28 To find digit. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave displayed on an oscilloscope screen. Enter Data and Calculate. Bankfull and effective discharge in small mountain streams of British Columbia | {
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Data and Calculate. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. RMS vs Average. 10Beta) CHANGELOG. By combining the first two criteria and plugging in the formula of and Update Vavg, sigmaV End Save Vmax,Vmin,Vavg,sigmaV to “gamd_restart. Average Velocity Formula: vavg = x/ t x is the DISPLACEMENT where x = xf - xi (final position - initial position) and t is the time interval over which the displacement occurs. Solution: Use the instantaneous formulas. 2) An easy way to calculate the Vdα expression is to place the time origin aligned to an output voltage pulse for which is applied the average value formula, in the. Because they both start at zero, the formula to convert between the two very easy (in = cm * 0. Non contact or with integrated conductor or clip-on current measurement from 0. since only their ratio enter the equation. 5) mph/ 2 giving us 8. Excel formula in column H is: Vin1 (volts) Formula 28-31 DATA = Vin1 x (255/5 V) dec. Learn physics formulas chapter 3 with free interactive flashcards. This means that equation 7 can be reduced to. pdf - Free ebook download as PDF File (. This histogram shows a theoretical distribution of speeds of molecules in a sample of nitrogen ( ) gas. V m = maximum value of transformer secondary voltage. In physics, you can calculate power based on force and speed. Since velocity is not constant, we find the avg velocity by doing 1/2(vf+vi) and that Vavg=dt. 5 m/s 2, Vi=0 Vf=????, no T. The formula was created using high-performance. Hukum Boyle : V ∝ 1/P (n dan T tetap) Hukum Charles : V ∝ T ( n dan P konstan). web; books; video; audio; software; images; Toggle navigation. i need the derivations that how these formulae comes. Vavg = 2146 / 10 = 214. 99 for most events). EVM - Miscellaneous Formula - Budget at Completion (BAC) is the total budget allocated to the BAC is also used to compute the TCPI and TSPI. Because, the positive and negative half cycle is equal in | {
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is also used to compute the TCPI and TSPI. Because, the positive and negative half cycle is equal in magnitude and thus the total value cancels out on summation. If final Velocity V and Initial velocity U are known, we make use of the formula. Difference between Vavg and Vrms If my Vp is let say 10V and I was to rectify it using ideal diodes and caps, wouldn't the output be 2*10V/π=6. Silica is most commonly found in nature as sand or quartz, as well as in the cell walls of diatoms. Rank the following gases from highest temperature to lowest temperature. v o represents the object's initial velocity at the beginning of the time interval. You need to integrate overtime for average velocity. Formula 15-4 IT = ÖIR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / Ö R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. t2 −t1 Note that no position values are given in the diagram; you will need to estimate these based on the distance between successive positions of the rockets. Several questions might immediately come to mind. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). the split filter instance has two output pads, and the overlay filter instance two input pads. measurements by: Vavg = (VI +V2)/2. Math 106-350/550 Answers to Homework #4 15 Sept ’06. 166 Pa theoretically, which is quite close to what Fluent gives!. Si multiplicamos por 3600 obtenemos vatios-segundos, que también se conoce como Julios. vavg = average velocity. For example it can be used to check the format of the container used by a multimedia stream and the format and type of each media stream contained in. 230K likes. Alternate Formulas or Computational Formulas for Variance. Dalam gas ideal berlaku 3 hukum tentang kinetika gas yaitu. Suppose we | {
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Formulas for Variance. Dalam gas ideal berlaku 3 hukum tentang kinetika gas yaitu. Suppose we forgot the formula that relates mass to density and volume. The very basic version of a settlement formula in a Florida personal injury case is: Settlement = (Full Value of Damages) x (100% – % Chance of Losing Case) x (100% – Your % of Fault) Need to Know Every Part of Settlement Formula If you don’t know any part of this formula, you settlement calculation will be way off. Shows you the step-by-step solutions using the quadratic formula! This calculator will solve your problems. Learn physics formulas chapter 3 with free interactive flashcards. When evaluating an arithmetic expression, FFmpeg uses an internal formula evaluator, implemented through the libavutil/eval. Formula 1 is trying to get back on our screens in July, just as F1 2020 arrives. Average speed formula? Unanswered Questions. 636 x (1/2) Vs peak = 0. If you know average value of voltage then you can use following formula to get peak voltage: Vpeak = (V. 1 While a non-true RMS or AC rectified average multimeter may not be accurate to measure non pure. IVA 02081070977 - All Rights Reserved - Privacy Policy. Para un flujo de tubería laminar completamente desarrollado, Vavg es la mitad de la velocidad máxima. There are user input variables to decide which dimensions will be used. Other multimeters may be designed to measure just the average level. The RMS value of a pulse waveform can be easily calculated starting with the RMS definition. Vavg = total distance/total time = 60. If you know average value of voltage then you can use following formula to get peak voltage: Vpeak = (V. The (average!) velocity profile in a turbulent flow is more flattened than the parabolic profile in a laminar flow. COMPUTERSC 207 - Fall 2019. What is the formula to calculate p-value?. Pressure and Temperature: A Molecular View. The slope of the line on these graphs is equal to the acceleration of the object. A third measure is the | {
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slope of the line on these graphs is equal to the acceleration of the object. A third measure is the root-mean-square (rms) speed, , equal to the square root of. 28*SQRT(D863*E863*F863*G863)) 1000 20000 0. However, the object’s speed, v, is just s divided by t, so …. It includes information about system configuration, database management, user and run registration, configuration of communications links with other BIS systems, and day to day administration activities. 091s on the red-walled C4 compound before he brought out the first red flag of the week with a stoppage shortly afterwards. 3535 * V pp Where V pp is the peak to peak volatge and V rms is the root mean square voltage. Velocity and Most Probable Velocity – a relationship in equation. Display electrical specifications such as rise time, slew rate, amplifier gain, and current. These values will be listed and identified as either. Acoustic impedance is the product of velocity and density. In a coordinate system with north being the positive x-direction, the car's motion is in the southern direction (see figure). Thus, the total of the masses of neutrons and protons is the atomic mass. Related Topics: More Statistics Lessons. In this equation. To compute V P-P from the average voltage, the average voltage is multiplied by 3. Common Derivatives. The displacement due to acceleration is represented by the green triangle. The heat capacity at constant pressure can be estimated because the difference between the molar Cp and Cv is R; Cp – Cv = R. 0 4/14/03 6:27 PM Page 1. 1-3 can be rewritten: Suppose that a car travels at a constant speed s (in feet per second) down a straight highway (see Fig. Blue dot is position of amplitude for dipping Sand 3 before seismic migration moved amplitude 1100 m laterally and 275 m (200 ms) vertically updip (Vavg-2860 m/s). Exam 1H Rev Ques. In this blog, we shall discuss various Ratio Analysis, the various Ratios Formulae, and their importance. Average acceleration in circular | {
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Ratio Analysis, the various Ratios Formulae, and their importance. Average acceleration in circular motion? A mass moves on a circular path of radius 2 m at constant speed 4m/s, what is the magnitude and direction of the average acceleration during a quarter of a revolution(it is shown on a diagram to be from the bottom (like 6 on a clock) to the right (like 3 on a clock))?. 6591 × √(R) Vavg(He) = √(8RTπ × 4) = 0. Calculate the average speed between each two adjacent picket fence stripes and record them in the table. Average Voltage (Vavg) As the name implies, Vavg is calculated by taking the average Since finding a full derivation of the formulas for root-mean-square (Vrms) voltage is difficult, it is done here for you. you can use the approximate formula in the box in Figure 2. Yes, watt-hours is a measure of energy, just like kilowatt-hours. Velocidad promedio Vavg se define como la velocidad promedio a través de una sección transversal. 20 m/s vi = 3,900 m/s vavg =? 2. Interface composition (i COMPRISING i_ref). why the generating voltage in India is 11kV or why is it be transmitted as multiples of 11. F1, FORMULA ONE, FORMULA 1, FIA FORMULA ONE WORLD CHAMPIONSHIP, GRAND PRIX and related marks are trade marks of Formula One Licensing B. FIA Lurani Trophy for Formula Junior Cars. The formula was created using high-performance. Do While (Vavg >= Vavg_error_lb And Vavg =< Vavg_error_ub) ' Call Sim_rand 'Loop Thank you. t i = Initial time. Pavg=____ B) Let us now consider. 3, 4, & 5 represents the relationship of depth and velocity as derived by velocity modeling. Vavg=Vpeak/pi for half wave rectifier Vavg=2Vpeak/pi for full wave. Algebra Formulas For Class 12. Stamped Concrete Pavers: $13-$20 per sq. This set focuses on formulas, important numbers, word problems, and linear motion terminology. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. formula during a Tp | {
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= (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. formula during a Tp time interval (π/3 radians): A 1 ( ) 1 average value of ( ) 0 0 Area T v t dt T V v t p T d p d not d p = = ∫ ⋅ = ⋅ (13. It's calculated by taking one cycle of a periodic waveform and squaring it, and finding the square root. Yes, watt-hours is a measure of energy, just like kilowatt-hours. displacement, B. It calculates the RMS voltage based on the given equations. The first output pad of split is labelled "L1", the first input pad of overlay is labelled "L2", and the second output pad of split is linked to the second input pad of overlay, which are both unlabelled. Let’s recalculate the trapezoidal signal RMS value with the method shown in Equation 1. Balanced three-phase output: Line voltage 3 1. vavg ≡ (x - x0) / t The definition of average velocity vavg = ½ (v + v0) The equation for average velocity in the case of constant acceleration Since we have two equations for average velocity, vavg, they must be equal to each other. You need to integrate overtime for average velocity. It is akin to Ohm’s Law, but something I was never taught in tech school –I had to learn it myself much later –now I use it at least weekly. These values will be listed and identified as either. This banner text can have markup. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. Input a constant (in this case: 1) as your Formula and configure the Output Extents. Confirmed Page owner: Formula ProSCo Production LLC. Kinetic Theory of Gases & Maxwell Distribution Curve & Gas Velocity Vmp,Vavg,Vrms Formulas(Class XI). This set focuses on formulas, important numbers, word problems, and linear motion terminology. For the second equation, the formula is V^2=Vo^2+2ax, with Vo=0, we can use cancel Vo^2 and thus get V^2=2ax, or V= | {
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equation, the formula is V^2=Vo^2+2ax, with Vo=0, we can use cancel Vo^2 and thus get V^2=2ax, or V= squareroot (2ax), x is same as h I just used different symbol. Sponsored Links. | Vavg - V1-2 | + | Vavg - V1-3 | + | Vavg - V2-3 | 2 x Vavg % voltage imbalance = x 100 The maximum allowable voltage imbalance is 2%. The formula is based on theoretical optics and incorporates both regression and artificial intelligence components to further refine its predictions. Securities Act of 1933 and deemed filed pursuant. You know that vavg = 12. 168 m furlong ) 20. Hi, How do you code, "not greater than or equal to" and "not less than or equal to" in vb. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. Materiales de aprendizaje gratuitos. 637 Vpp = Vp • 2 Vp = Vpp ÷ 2 Vinst. A Framework for Mining Association Rules in Data Warehouses 161 Assume that in a fact table there is m-dimensions and the quantity attribute exists. Now if you read a lot of other literature on Precision and Recall, you cannot avoid the other measure, F1 which is a function of Precision and Recall. The complete formula (EFN) is expressed as: EFN = (A/S) x (Δ Sales) - (L/S) x (Δ Sales) - (PM x FS x (1-d)). You should know that the velocity/time equation is just one important equation using velocity , but others exist. Bankfull and effective discharge in small mountain streams of British Columbia Brayshaw, Drew Devoe 2012. How do I generate a calculation to linearize data that has an inverse relationship? This video shows how to transform data from a Boyle's Law experiment so that the data falls along a straight line. Average Voltage of a Waveform in Analytical Method. Solved: Hi All, Can anyone suggest how to display the count of values displayed in pivot box into a text box. Calculator wich uses trigonometric formula to simplify trigonometric expression. It simply involves taking the total revenue in a given time | {
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to simplify trigonometric expression. It simply involves taking the total revenue in a given time period by the number of users in that time period. Confirmed Page owner: Formula ProSCo Production LLC. 6-8 Months of age: Breastfeed every 3-4 hours or Formula 24-37 ounces. The Reynolds number has Vavg=0. 99 for most events). THX! asked by Tim on December 30, 2015; math. Pursuant to Rule 425 under the. 637 times the amplitude (2*Vp/π), Vrms doesn't sounds right to me. Namun pada daily chart DOW dan FTSE, tampak RSI cenderung flat dan Reva sudah mencapai batas atas yang mengisyaratkan kalau akan ada koreksi dulu atau malah berbalik arah ke bawah lagi. The formula was created using high-performance. Chapter 1 – Student Solutions Manual 3. If you wish to find any term (also known as the {n^{th}} term) in the arithmetic sequence, the arithmetic sequence formula should help you to do so. That's the formula for average acceleration. The beautiful and perhaps mysterious formula of Euler which is the subject of this section is. For power computations, the true RMS multimeter provides the. The instantaneous velocity is just the velocity of an object at a specific point in time. Boyle's Law deals with the relationship between pressure and volume (two of the four variables). Using the compound interest formula, you can determine how your money might grow with regular deposits or. vavg = average velocity. Average Voltage of a Waveform in Analytical Method. 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. Ф1 пополнила ряды футболистом. the peak voltage ( Vo ) = half the peak to peak voltage = 60 / 2 = 30 V. 5 kg flies through the air at a low speed, so that the air resistance is negligible, What is the net force acting on the ball while it is in motion? Which components of the ball's. Harvard University. Enzyme-linked Immunosorbent Assay (ELISA) The concentrations of NF-κB-responsive genes (IL-8, TNF-a and IL-1β) were evaluated in cell culture supernatants | {
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of NF-κB-responsive genes (IL-8, TNF-a and IL-1β) were evaluated in cell culture supernatants according to the manufacturer’s protocol (Joyee Biotechnics, Shanghai, China). Vavg VO + 1/2 gt (7) xx) Keep in mind the initial velocity vo will be zero for each trial. Record this in your data table. A ball of mass 0. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. In calculating Vavg, why are you only integrating the first part from 0 to pi/2? Click to expand First part is symmetrical function and average value is zero so for complete cycle in symmetrical so i only took half the time which is equal to pi/2, but i think since i only took half should i multiply it by 2 to get averwge for complete cycle?. Physics Help! Thread starter allies; Start date Oct 3, Vavg = (5300- xi)/4 I derive the formula for the students from the most basic formulas, then show them. 2 Derive the manometer formula ∆p= (ρm−ρ)gh, where his the manometer reading (height), ρm the density of the manometer liquid and ∆pthe pressure difference across two horizontal points in the fluid of density ρ(U-tube manometer, see Example 2. Consequently, aavg = 2. That seems a little low just on a gut feeling, but maybe it is OK?. Using the given conversion factors, we find (a) the distance d in rods to be d = 4. 7V The back of the book says it is 136V. 0 Introduction: Quality Assurance and Certification June 17, 2009 ECMPS Reporting Instructions Quality Assurance and Certification 1. For the waveform shown above, the peak amplitude and peak value are the same, since the average value of the function is zero volts. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave that is displayed on an oscilloscope screen. 0 furlongs = ( 4. Manufacturing Cycle Efficiency Formula. Calculate the rms Value V(rms value) Related Calculations. vavg[t1 , t2 ] = x(t2 )−x(t1 ). 637 Vp = Vavg ÷. A sled of mass (m) is being pulled horizontally by a constant | {
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t2 ] = x(t2 )−x(t1 ). 637 Vp = Vavg ÷. A sled of mass (m) is being pulled horizontally by a constant diagonally upward force of magnitude (F) that makes an angle (theta) with the direction of motion. Business Information Server for Microsoft Windows Administration Help (7846 0284) This Help provides direction for BIS system administrators. x i = Initial Distance. Pressure and Temperature: A Molecular View. Best Price Roadside Assistance. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. Biblioteca en línea. As illustrated in the figure, the slope can be positive, negative, or zero. The highest fre- quencies of 2 to 4 MHz correspond to. Here, in part 1 of this short series on the topic, we. Louis, MO. With calculus, you can find it when given the displacement function, a function which tells you distance moved. If the object is moving with a velocity of -8 m/s, then the slope of the line will be -8 m/s. November 13 2019 Veton Këpuska 14 Hierarchical definition of 74x138 like. t2 −t1 Note that no position values are given in the diagram; you will need to estimate these based on the distance between successive positions of the rockets. Formula 15-4 IT = ÖIR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / Ö R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. As stated by Investopedia, acceptable solvency ratios vary from. 22 except that it has been offset by the addition of a dc component equal to 2 V. Formula SAE Italy, Formula Electric Italy & Formula Driverless 2020. Find books. Thinking they might prove useful as tools or templates to others, it is my pleasure to provide them freely to the scientific community. Related Topics: More Statistics Lessons. ok | {
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to provide them freely to the scientific community. Related Topics: More Statistics Lessons. ok at first to locate the displacement the -a million. Vavg VO + 1/2 gt (7) xx) Keep in mind the initial velocity vo will be zero for each trial. The average velocity of gas particles is found using the root mean square velocity formula: μ rms = (3RT/M) ½ μ rms = root mean square velocity in m/sec R = ideal gas constant = 8. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. This means that equation 7 can be reduced to. It simply involves taking the total revenue in a given time period by the number of users in that time period. (Since the sign of the downward velocity and the downward acceleration is positive, a negative velocity must be directed upward. It states that the volume of a fixed mass of gas at a constant temperature is inversely proportional to the pressure of the gas. 5 km/h, and Δt = (15min)(1h/60min) =. Find the volume per organism, assuming a spherical shape. The root mean square velocity (RMS velocity) is a way to find a single velocity value for the particles. 3, 4, & 5 represents the relationship of depth and velocity as derived by velocity modeling. November 13 2019 Veton Këpuska 14 Hierarchical definition of 74x138 like. V m = maximum value of transformer secondary voltage. It is possible, however, to have negative Voltage Unbalance readings if the device is a Series 600 Power Meter or a Series 2000 Circuit Monitor. Solution: Use the average velocity formula! Therefore, the average velocity is 8 2) Find the velocity function and the acceleration function for the function s(t) = 2t 3 + 5t - 7. But i have only 240 V supply for the heater, can I put a resistor in series with the heater to re. UserVar = (duser 1, duser 2. Physics Help! Thread starter allies; Start date Oct 3, Vavg = (5300- xi)/4 I derive the formula for the students from the most basic formulas, then show them. Stormblood Patch Items. | {
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the formula for the students from the most basic formulas, then show them. Stormblood Patch Items. CAGR is equivalent to the more generic exponential growth rate when the exponential growth interval is one year. Hence, the ratio is not the same; the maximum is higher in the laminar case. To compute V P-P from the average voltage, the average voltage is multiplied by 3. This oscillator formula can also be used as the basis for program trading and backtesting in StrategyDesk. CX Formula will improve your customer experience, increase customer loyalty, profits, referrability and happiness!. Voltage imbalance is given by the formula: Ring connect screw terminals C. For the second equation, the formula is V^2=Vo^2+2ax, with Vo=0, we can use cancel Vo^2 and thus get V^2=2ax, or V= squareroot (2ax), x is same as h I just used different symbol. Босс Renault за отказ от кибергонок. The mass of oil is to be determined. We conclude that p f = 22 atm. This value can be found using the formula: v rms = [3RT/M] 1/2 where v rms = average velocity or root mean square velocity R = ideal gas constant T = absolute temperature M = molar mass The first step is to convert the temperatures to absolute temperatures. clc clear KB=1. Variation is ma = -mg sin theta - mu*mg cos theta if motion of block is UP slope. The heat capacity at constant pressure can be estimated because the difference between the molar Cp and Cv is R; Cp – Cv = R. The instantaneous velocity is just the velocity of an object at a specific point in time. It makes the textbook definitions more comprehensible! SET M…. For the waveform shown above, the peak amplitude and peak. The ratio t1/T is the pulse signal duty-cycle. 494 m/s North. Several questions might immediately come to mind. Maintaining a body fat percentage from 6 to 9 percent. N = 2(72 h/0. The displacement due to acceleration is represented by the green triangle. 175 * 10 -3 * d 4 / (f t c vavg),. the split filter instance has two output pads, and the | {
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triangle. 175 * 10 -3 * d 4 / (f t c vavg),. the split filter instance has two output pads, and the overlay filter instance two input pads. V0 is the value assigned for the pipette to dispense. vavg=vi+vf2, when the acceleration is constant, where vi. For Boyle's Law to be valid, the other two variables must be held constant. vavg[t1 , t2 ] = x(t2 )−x(t1 ). The units for the volume and pressure can be left in l and atm. I can use the formula ΔVc = Vm(1-e-t/RC) I don't know why it didn't click sooner, it will become ΔVc = 19. Harvard University. Formula: Example: How many ways can 4 students from a group of 15 be lined up for a. Biblioteca en línea. 5 m/s 2, Vi=0 ) D= 6. Issuu is a digital publishing platform that makes it simple to publish magazines, catalogs, newspapers, books, and more online. For a sine wave: Vrms = 0. Depth conversion (domain conversion) of seismic time interpretations and data is a basic skill set for interpreters. Calculate dvava/dT at T = 300 K for 0öygen, which has a molar mass of 0. Area of a Circle Formula. If you have questions about this formula or functionality, please call TD Ameritrade's StrategyDesk free helpline at 800 228-8056, or access the online Help Center via the StrategyDesk application. Acoustic impedance is the product of velocity and density. vavg ≡ (x - x0) / t The definition of average velocity vavg = ½ (v + v0) The equation for average velocity in the case of constant acceleration Since we have two equations for average velocity, vavg, they must be equal to each other. Namun pada daily chart DOW dan FTSE, tampak RSI cenderung flat dan Reva sudah mencapai batas atas yang mengisyaratkan kalau akan ada koreksi dulu atau malah berbalik arah ke bawah lagi. The instantaneous velocity is just the velocity of an object at a specific point in time. New formulas from other definitions: D= Vi*T + ½ *A*T 2 and Vf 2 = Vi 2 + 2*A*D. Ninguna Categoria; Subido por Angie Melissa Fluidos- Frank M. 0 furlongs )( 201. We use cookies to | {
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Ninguna Categoria; Subido por Angie Melissa Fluidos- Frank M. 0 furlongs )( 201. We use cookies to improve your navigation experience. The coefficient of kinetic friction is (uk). V m = maximum value of transformer secondary voltage. Copyright © 2007-2019 Formula S. 54 centimetres. Practical Depth Conversion with Petrel. V2-3 = Voltage between phases 2 and 3. Basic Properties and Formulas. 758) + A=AIR(3. 9 V = 54 V As you can see from this example, even though the peak output of the full wave center tapped rectifier was half that of the half wave rectifier, the average output was the same because the full wave center tapped rectifier doubles the number of half-cycles at the output, compared to a half. The PERIODIC TIME (given the symbol T) is the time, in seconds milliseconds etc. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. And a is basically g however a in this case. It includes information about system configuration, database management, user and run registration, configuration of communications links with other BIS systems, and day to day administration activities. 175 * 10 -3 * d 4 / (f t c vavg),. 0 m above a flat horizontal beach. That seems a little low just on a gut feeling, but maybe it is OK?. Calculate the average speed between each two adjacent picket fence stripes and record them in the table. 3; PA=RA*P; PB=RB*P; % B=CH4(3. Therefore, the average voltage value is 214. Voltage Unbalance is calculated using the following formula: Vunbalance = [(Vavg-Vphase) divided by Vavg]. RMS vs Average. 5 kW motor with 110V, 40W anti condensation heater. When compared to protons or neutrons electrons have so much less mass that they don’t influence the calculation. Formula • ფორმულა, Tbilisi, Georgia. Securities Act of 1933 and deemed filed pursuant. s defines the following functions: fit. D=Vavg*T Vavg= (Vi+Vf)/2 A= (Vf-Vi)/T. Average speed is equal to displacement over time, so the formula would be: | {
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(Vi+Vf)/2 A= (Vf-Vi)/T. Average speed is equal to displacement over time, so the formula would be: vavg = (x2 - x1) / (t2 - t1). 0292 m rod = 160 rods, (b) and that distance in chains to be d = ( 4. 5) World Championship. If a = constant, then vavg = (v + v0)/2 where ∆x = displacement. 637 times the amplitude (2*Vp/π), Vrms doesn't sounds right to me. Answer / soumya ranjan panigrahi we know form factor is the cause of generating voltage in India is 11kV or it be transmitted as multiples of 11, because of simple to desine. Use Equation 5 Vf 2 = Vi 2 + 2*A*D. 168 m furlong ) 20. It may also be referred to as the annualized rate of return or annual percent yield or effective annual rate. Alternatively, the same r4 can be solved from The plot above shows the average speed vavg for various downhill speeds v2 in this Example 2. Maintaining a body fat percentage from 6 to 9 percent. The Peak Voltage calculator calculates the peak Voltage value from either the peak-to-peak voltage, the RMS voltage, or the average voltage. 5 h) Solve for N. Using the formula, we calculate v to be equal to (10+7. In this interactive object, learners determine the Vp, Vp-p, Vrms, Vavg, and the frequency of a sine wave that is displayed on an oscilloscope screen. Therefore, it is acceptable to choose the first quarter cycle, which goes from 0 radians (0°) through π /2 radians (90°). 1 A to 20,000 A. Total Kinetic Energy. So I thought the ripple voltage was approximated by the formula Vr,pp = Vp / fRC for a half-wave rectifier, and Vp/2fRC for a full-wave. Pressure and Temperature: A Molecular View. Filed by ITC Holdings Corp. Copy AFL di bawah dan paste ke Formula Editor di amibroker kemudian save. For power computations, the true RMS multimeter provides the. When evaluating an arithmetic expression, FFmpeg uses an internal formula evaluator, implemented through the libavutil/eval. 00000001 1E-08 3561 CHAPTER 28 To find digit. Calculator solution will show work for real Uses the quadratic | {
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1E-08 3561 CHAPTER 28 To find digit. Calculator solution will show work for real Uses the quadratic formula to solve a second-order polynomial equation or quadratic equation. measurements by: Vavg = (VI +V2)/2. SPARKCHARTSTM. formulae_8&9 (2). vi or rather vi stands for initial velocity. 28*SQRT(D863*E863*F863*G863)) 1000 20000 0. 2 for time, with y = 0, using the quadratic formula (choosing the positive root to yield a positive value for t). Exam 1 November 2012, Questions and answers Exam 31 October 2008, Questions and answers Exam 26 October 2009, Questions Exam 10 December 2010, Questions Final Formula Sheet - Summary Mechanics and Waves Phys 131 Equation Sheet. The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic). Unlike the book, this calculator employs the velocity of light $$c$$ at full precision, resulting in slightly shorter, but more precise lengths. VAVG - (-VAVG ) ) in the sum of the two areas, thus resulting in zero average voltage over one To find the peak value from a given average voltage value, just rearrange the formula and divide by the. Previously Viewed. Make sure you enter. When the payments are all the same, this can be considered a geometric series with 1+r as the common ratio. Formula 15-4 IT = ÖIR2 + IL2 =SQRT(D245^2+E245^2) To find Z in a parallel RL circuit Formula 15-5 Z = R x XL / Ö R2 + XL2 =D248*E248/SQRT(D248^2+E248^2) To find G in a parallel RL circuit One R (ohms) Formula 15-6 G = 1/R =D251/E251 To find inductive susceptance Formula 15-7 BL = 1/XL =D254/E254 Z (ohms) Formula 15-8 Y = 1/Z =D257/E257. Express your answer in terms of the given quantities and,. Cost can add up quickly, especially if you’re a novice and have never attempted a Stamped Concrete Patios installation before. The hydrocarbon that has all single bond is called as alkane with the general formula C n H 2 n + 2. Basic | {
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that has all single bond is called as alkane with the general formula C n H 2 n + 2. Basic Properties and Formulas. As stated by Investopedia, acceptable solvency ratios vary from. As with the Vrms formula, a full derivation for the Vavg formula is given here as well. The average velocity of the sled is (Vavg). Describe another way to find the area of a circle when given the circumference. RMS Voltage (Vrms) The root-mean-square or effective value of a waveform. Substitute in the given equation and find the unknown quantity. Furthermore, the formula for computing the boom length \(L. I want to code this do while loop to do this. If you're in the higher end of this range -- 10 to 12 percent fat -- you'll have a fit appearance and -- once exercise and sensible eating become habitual, you'll be able to maintain these lifestyle choices fairly easily. Profit Margin Formula - Explained. Homework Statement For a particle in simple harmonic motion, show that Vmax = (pi/2)*Vavg where Vavg is the average speed during one cycle of the motion. CAGR is equivalent to the more generic exponential growth rate when the exponential growth interval is one year. The displacement due to acceleration is represented by the green triangle. Rank the following gases from highest temperature to lowest temperature. The critical step is to be able to identify or extract known. 6m/s2 isn't necessary so what you do is you employ the equation Vavg=deltax/delta t then you certainly re-manage it and get the equation (delta t)*(Vavg)=delta x this is what u want this is displacement so then you certainly plug interior the numbers so it would look like this (6. Any such data mashup is expressed using the Power Query M Formula Language. This banner text can have markup. 1 While a non-true RMS or AC rectified average multimeter may not be accurate to measure non pure. The formula for Compound Annual Growth Rate (CAGR) is very useful for investment analysis. | {
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Prove that $n^3+5n$ is always divisible by 6 using induction with mod
So I had to prove this in my calculus exam today
Prove that $$n^3+5n$$ is always divisible by 6.
I know that there have been other people around here having asked this already, but I chose a different approach and would really appreciate to hear whether my train of thought here was right or wrong. So I am especially not looking for the "right way" to prove it (i.e., I don't need a proper solution) I would just appreciate for somebody to assess my solution, if it's completely wrong or possibly not complete etc.
so, what I did:
Initial case prove that $$n^3+5n \bmod 6 = 0$$ for $$n=1$$. $$1^3+5\cdot 1 \bmod 6 = 0 \Leftrightarrow 6 \bmod 6 = 0$$ which is trivially true.
Induction step Prove that for every $$n$$, if the statement holds for $$n$$, then it holds for $$n + 1$$. $$((n+1)^3+5(n+1)) \bmod 6 =0 \\ \Leftrightarrow ((n^3+3n^2+3n+1)+5n+5) \bmod 6 =0 \\ \Leftrightarrow (n^3+5n \bmod 6) + (3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 0+(3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 3n^2+3n+6 \bmod 6 =0.$$
And this is true for all $$n \in \mathbb{N}$$, where from the third to the fourth line, we replaced with the induction hypothesis that $$n^3+5n \bmod 6 =0$$ . Take for instance $$1, 2,3,...$$ it is always divisible by 6 without leaving any rest!!
However, my other friends that took the exam said that this is in need of some extension to show that the last line actually holds true for whatever n you insert. But I cannot grasp what they mean; by the definition of the modulus, it should already trivially be given that the last line is right...
I'm really looking forward to your thoughts, do you think I would at least get partial points for this? :( I really don't see how this can be wrong...
Thanks so much, Lin | {
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Thanks so much, Lin
• You just need to show that $2\mid n(n+1)$, then it follows that $6\mid 3n^2+3n+6$. This is certainly only a partial loss of points. Mar 11, 2021 at 12:43
• Yes, it is "in principle" right. I would give full points for some of the solutions here, and less points for you, because your solution is a bit too long and not every step is clear. Have a look at the duplicate yourself! Mar 11, 2021 at 12:45
• Please, see the edits I did to your post so you learn how to use better mathjax. Mar 11, 2021 at 12:47
• Bitte schön, you are welcome! This site has good solutions, and sometimes this helps to improve the own solution a bit. And the advantage is, you can do this on your own. Mar 11, 2021 at 12:51
• BTW, a simpler way to solve this is to see that $$n^3+5n \equiv n^3-n = (n-1)n(n+1)\pmod 6$$, and of course both $2$ & $3$ must divide $(n-1)n(n+1)$. Mar 11, 2021 at 13:11
In view of the final step, $$3n^2+3n+6\equiv 0\mod 6$$ is equivalent to $$3n^2+3n\equiv 0\mod 6$$, i.e.,
$$3(n^2+n)\equiv 0\mod 6$$.
So it is sufficient to show that
$$2\mid n^2+n$$.
But $$n^2+n = n(n+1)$$ is a product of consecutive integers and one of which will be even. Thus $$n^2+n$$ is a multiple of $$2$$ and hence $$3(n^2+n)$$ is a multiple of $$6$$, as required.
• So you also agree that my induction is right, and only the part you mention is missing? :) Mar 11, 2021 at 12:47
Your friends are right. You need to explain why the last equivalent equation is an identity, meaning it solves for any integer n greater than zero.
To do that you show that $$3n^2+3n+6$$ is divisible both by 2 and 3. Divisibility by 3 is obvious since the expression is a multiple of 3. You show that by rewriting the expression into $$3(n^2+n+2)$$ Divisibility by 2 results from observing that the product $$n(n+1)$$ of two consecutive numbers is always an even number: $$n^2+n+2=n(n+1)+2=2k\cdot(2k\pm1) +2=2[k(2k\pm1)]$$ | {
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• Thanks! But just for the fun of it, out of 12 points in total, how many would you give me for my solution? xP Mar 11, 2021 at 13:02
• Is arguable since you failed to finish to prove p(n)->p(n+1). I guess you could loose 1 or 2p depending on how you presented the induction method. p(1) true and p(n)->p(n+1) then p(n) true for n>0. Depends on the teacher. Mar 11, 2021 at 13:10
• yeyyyyy! that is enough excitement for me :D Mar 11, 2021 at 13:11
This is correct, but as your friends said, not proving the last line may cause some loss of points. To prove this, it's not very "long" or something. Just show that $$3n^2+3n+6\equiv 0\ (\textrm{mod}\ 2)$$ Since it is trivial that it is $$0\ \textrm{mod}\ 3$$, the statement follows.
Hope this helps. Ask anything if not clear :)
• Thanks so much to you too!! Mar 11, 2021 at 12:48
• Thanks @LinShao, but one more thing: accept either my or the other answer here just to ensure that this is answered. Mar 11, 2021 at 12:50
Since your final step implies $$3\mid 3n^2-3n$$, it suffices to show $$2\mid n^2-n$$ for any $$n$$, which can be seen as trivial. That was all you missed in my opinion.
• Thanks for also giving me hope :) After all I'm not a total idiot in math, it seems xD Mar 11, 2021 at 13:36 | {
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Writing english statement into predicate logic using quantifiers.
Below was the exercise problem I was solving from Discrete Mathematics by Kenneth Rosen (for preparation of GATE Exam) and I have doubt in it.
Let S(x) be the predicate that "x is a student", F(x) be the predicate "x is a faculty member", and A(x,y) the predicate "x has asked y a question", where the domain consists of all people associated with your school. Use quantifiers to express each of these statements.
(f)Some student has asked every faculty member a question.
Now my doubt is
it can be framed like there is at least one student such that for all faculty members, he must have asked them a question.
so I wrote my expression as
∃x ( (S(x) ^ ∀y ( F(y) $\rightarrow$ A(x,y) ) )
But in Rosen answer is given as below and I have 2 doubts in it.
Rosen's Ans : ∀y ( (F(y) $\rightarrow$ ∃x ( S(x) v A(x,y) ) )
Doubt 1: I think in above expression we must have "and operator" instead of "or operator" in the second part of expression which is quantified by existential quantifier and so it should be
∀y ( (F(y) $\rightarrow$ ∃x ( S(x) ^ A(x,y) ) )
Doubt 2: What is the difference between my answer and Rosen's answer.Which one is correct.
You're right about the operator: that should definitely be a $\land$, rather than a $\lor$
Otherwise, the difference is this: you took the statement "Some student has asked every faculty member a question." to mean that it was the same student who asked every faculty member a quesrion. And, with that interpretation, your answer is correct, and Rosen's (even with the $\land$ instead of the $\lor$) would be incorrect. | {
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However, English (like all natural languages) is ambiguous, and you can also interpret the statement "Some student has asked every faculty member a question." as saying that for each faculty member there is some student (but this time not necessarily the same one for each faculty member) that asked that faculty member a question. And if that is the intended meaning of the sentence, then Rosen's answer (again, with the $\land$ instead of the $\lor$) is correct.
Now, the latter is certainly not a very intuitive reading of the statement, and I much prefer the former interpretation, and hence your answer! However, some people do mean the latter interpretation when expressing the English statement.
• @Bram28-Thank you so much! :) Jun 13 '18 at 4:29
• @user3767495 you're welcome! :) Jun 13 '18 at 14:12
Rosen's intended answer, $\forall y~ (F(y)\to ∃x~( S(x) \wedge A(x,y) ) )$, claims that for any faculty member there is some student that has asked them a question. They need not be the same students.
• That is: "Every faculty member has been asked a question by some student."
Your answer, $\exists x~\forall y~(S(x)\wedge (F(y)\to A(x,y)))$ claims that some student will have asked every faculty member a question. Which is what was required.
• That is: "Some student has asked every faculty member a question."
It is logical distinction which is oftimes misrepresented in natural language. | {
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# Toronto Math Forum
## APM346-2022S => APM346--Lectures & Home Assignments => Chapter 1 => Topic started by: Weihan Luo on January 14, 2022, 12:38:03 AM
Title: Classification of PDEs
Post by: Weihan Luo on January 14, 2022, 12:38:03 AM
I am a little bit confused about the classifications of PDES. Namely, I have trouble distinguishing between linear equations versus quasi-linear equations.
In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?
Title: Re: Classification of PDEs
Post by: Victor Ivrii on January 14, 2022, 02:45:57 AM
In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?
First, it will be not just quasilinear, but also semilinear. Second, it will also be linear since you can move $c(x,y)u$ to the left
Good job, you mastered some $\LaTeX$ basics. :)
Title: Re: Classification of PDEs
Post by: Weihan Luo on January 14, 2022, 11:28:53 AM
Thank you for your response. | {
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Does it mean that all linear PDEs are also quasilinear/or semilinear? If so, on a quiz, I should classify those PDEs as linear right?
Title: Re: Classification of PDEs
Post by: Victor Ivrii on January 14, 2022, 01:47:15 PM
Yes, all linear are also semilinear and all semilinear are also quasilinear. For full mark you need to provide the most precise classification. So, if equation is linear you say "linear", if it is semilinear but not linear you say "semilinear but not linear" and so on,... "quasilinear but not semilinear" and "non-linear and not quasilinear". | {
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# Finding the radius of a circle... impossible?
#### Bailey232008
##### New member
I’ve been working on this for too long. I almost posted this in the Odds and Ends section because I think it’s going to entail more than just geometry. Give this one a shot. Could be a fun challenge for you vets out there if it’s even possible. If you want the back story behind how I came up with it let me know.
The only information provided is what you see in the picture. Solve by any means necessary...
#### MarkFL
##### Super Moderator
Staff member
By Heron, the area $$A$$ of the triangle is:
$$\displaystyle A=22.5\sqrt{(R+22.5)(R-22.5)}$$
We also know:
$$\displaystyle A=22.5(R-5)$$
This implies:
$$\displaystyle \sqrt{(R+22.5)(R-22.5)}=R-5$$
Solving this, we find:
$$\displaystyle R=53.125$$
#### pka
##### Elite Member
I’ve been working on this for too long. I almost posted this in the Odds and Ends section because I think it’s going to entail more than just geometry. Give this one a shot. Could be a fun challenge for you vets out there if it’s even possible. If you want the back story behind how I came up with it let me know.
The only information provided is what you see in the picture. Solve by any means necessary...View attachment 20083
On it we see $$R=\dfrac{c^2}{8h}+\dfrac{h}{2}$$
From the above diagram $$c=45'~\&~h=5'$$.
#### Jomo
##### Elite Member
Draw a line from where it says 45 to the center of the circle. This gives you a right triangle. What are the lengths of the three sides of this right triangle. Now ask Pythagorous for help.
#### Bailey232008
##### New member
Draw a line from where it says 45 to the center of the circle. This gives you a right triangle. What are the lengths of the three sides of this right triangle. Now ask Pythagorous for help.
You lack the interior angle, so this method doesn’t work
#### Bailey232008
##### New member
Mark and Pka,
great stuff, appreciate the help
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##### New member
Mark and Pka,
great stuff, appreciate the help
#### Dr.Peterson
##### Elite Member
You lack the interior angle, so this method doesn’t work
You misunderstood Jomo's suggestion, which doesn't use any angles. In fact, this is how the formula pka gave is derived!
The sides of the triangle are r-h, c/2, and r (the hypotenuse), and solving the Pythagorean theorem for r yields the formula.
#### Jomo
##### Elite Member
You lack the interior angle, so this method doesn’t work
As already mentioned Pythagorous' theorem does not use angles. you have the 3 sides of the right triangle in terms of R. What seems to be the problem?
#### jonah2.0
##### Junior Member
Beer goggles hallucination follows.
Looked at the picture and saw 45° instead of 45'. Ended up thinking R as roughly 65.69.
#### Jomo
##### Elite Member
You lack the interior angle, so this method doesn’t work
Why couldn't you use the 45o angle that you thought was there?
#### yoscar04
##### Junior Member
Following Jomo's suggestion using straightforward Pythagoras theorem, you should get R=53.125. | {
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# Math Help - Proof that sum is rational number
1. ## Proof that sum is rational number
I have to prove that
$\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}$
is rational number.
I have one idea, I hope it's not crazy!
I have spoted that the sum of first three fractions are equal to 1:
$\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }}=1$
I have then discovered that next five fractions are also equal to 1.
$\frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} = 1$.
Continuing, we have that every odd number of fractions are equal to 1.
Considering that there are 99 fractions in that sum we get that because of $3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}=9$
I don't know how to prove that but it seems to me that there is a rule in that sum which I have described.
2. Originally Posted by OReilly
I have to prove that
$\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}$
is rational number.
I have one idea, I hope it's not crazy!
I have spoted that the sum of first three fractions are equal to 1:
$\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }}=1$
I have then discovered that next five fractions are also equal to 1.
$\frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} = 1$. | {
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Continuing, we have that every odd number of fractions are equal to 1.
Considering that there are 99 fractions in that sum we get that because of $3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99$ whole sum is equal to 9: $\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99} + \sqrt {100} }}=9$
I don't know how to prove that but it seems to me that there is a rule in that sum which I have described.
One word,
rationalize.
3. Originally Posted by ThePerfectHacker
One word,
rationalize.
Bit more words?
4. Originally Posted by OReilly
Bit more words?
Okay whenever you have,
$\frac{1}{\sqrt{x}\pm \sqrt{y}}$ you can change the fraction, called "rationalizing". This is done by mutiplying the numerator and denominator by $\sqrt{x}\mp \sqrt{y}$. This changes the denominator into whole number because you use the difference of two squares.
Thus,
$\frac{1}{\sqrt{x}\pm \sqrt{y}}\cdot \frac{\sqrt{x}\mp \sqrt{y}}{\sqrt{x}\mp \sqrt{y}}=\frac{\sqrt{x}\mp \sqrt{y}}{x-y}$
------------
Hence you have,
$\frac{1}{\sqrt{1}+ \sqrt{2} }+\frac{1}{\sqrt{2}+ \sqrt{3} }+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{ \sqrt{99}+\sqrt{100}}$
Upon rationalizing,
$\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}$
Simplyfy,
$-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{99}+\sqrt{100}$
Each one cancels each other out (such a series is called telescoping), thus,
$-1+\sqrt{100}=-1+10=9$
5. Thanks!
6. Welcomed.
That happens to be a nice problem which I really like. Where did you get it from?
7. Originally Posted by ThePerfectHacker
Welcomed.
That happens to be a nice problem which I really like. Where did you get it from?
From high school book. Author has also wrote book with variety of problems which i find most interesting. | {
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3 vehicles out of 5 randomly showcased, what is the probability of each vehicle being showcased?
I have a basic probability doubt. If I have 5 different vehicles backstage, out of which 3 random can be showcased to the public. What is the probability of each vehicle to get to showcase? Once a car is in a showcase, it will not be returned to backstage.
I thought by doing the calculating the combinations- ${5 \choose 3}=10$. And probability of 3 slots getting filled (let's say $P$) - $P=\frac{1}{5}\times\frac{1}{4}\times\frac{1}{3}$. So, total probability is $10\,P= \frac{1}{6}$.
Or will it be $\frac{3}{5}$?
Or any other solution?
I assume that all cars are equally likely to be chosen.
Suppose without loss of generality the cars are labeled 1 through 5. The probability of not choosing car 1 is (4/5) * (3/4) * (2/3) = 2/5, so the probability of choosing it is 1 - 2/5 = 3/5. Of course, by my assumption above, the same argument applies to any of the cars. The answer 3/5 makes intuitive sense since we're drawing 3 things from a set of 5.
• But what about the possible combinations of C1,C3,C4 or C2,C5,C1 etc. Will the 3/5 probability have that into account as well? – Joe Jul 1 '16 at 18:55
• You can renumber the cars however you like and the argument is the same. – Kodiologist Jul 1 '16 at 18:56
• Thank You! Can you please tell in which scenario will my logic below (the same i described in the question) be used <br>- 5C3=10 And probability of 3 slots getting filled (let say P)- 1/5*1/4*1/3 So, total probability = 10*P= 1/6 – Joe Jul 5 '16 at 13:36
• If we have 60 Cars backstage in place of 5, will he situation become- Probability of not chosing car1 is (59/60)*(58/59)*(57/58) = 57/60, so the probability of chosing it will become 1-(57/60)= 3/60 – Joe Jul 7 '16 at 13:51
• Yes, that's right. – Kodiologist Jul 7 '16 at 14:45
You asked for alternative approaches. Here is one you might find useful. | {
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You asked for alternative approaches. Here is one you might find useful.
Let's begin by stating the obvious: you are implicitly assuming the five probabilities are equal. The expected total in the showcase equals the sum of those probabilities, whence it is five times any one of them. Yet the expected total is the average value of all possible totals, weighted by their chances of occurring. Since by design the possible total is always $3$, its average must be $3$. Therefore each probability is $3/5$.
The power of this reasoning about expectations becomes clear when you generalize the question to $k$ cars in the showcase to be chosen (with equal probabilities) out of $n$ cars backstage. | {
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# Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:
## Presentation on theme: "Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:"— Presentation transcript:
Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:
Definition 2: Matrix Addition Let A = ( ai j ) & B = ( bi j ) be mn matrices. Then C = A + B is an mn matrix with elements ci j ai j + bi j . Example: Properties: (A1) A + B = B + A (A2) (A + B) + C = A + (B + C) (A1) A + 0 = A where 0 is a zero matrix (matrix consisting of only zero elements)
Matrix Operations: scalar multiplication
Definition 3: Scalar Multiplication Let A = ( ai j ) and λ R. Then C = λ A is an mn matrix with elements ci j λ ai j . Properties: (S1) λ A = A λ (S2) λ (A + B) = λ A + λ B (S3) (λ1 + λ2) A = λ1 A + λ2 A (S4) λ1 (λ2 A) =(λ1λ2) A
1.3 Matrix Multiplication
Definition: Matrix Multiplication If A = ( ai j ) is an mn matrix and B = ( bi j ) is an np matrix, Then the product AB is defined to be an mp matrix C = ( ci j ) where ci j = ai 1 b1 j + ai 2 b2 j + … + ai n bn j In other words, ci j = ( Row i of A ) ( Column j of B ) Example: Properties: (M1) (AB)C = A(BC) (M2) A(B + C) = AB + AC (M3) (B + C)A = BA + CA
Matrix multiplication lacks commutativity.
Example 1: Example 2: Other properties that matrix multiplication lacks: 1) AB = 0 doesn’t imply A = 0 or B = 0 2) AB = AC doesn’t imply B = C | {
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1.4 Special Matrices Definition (matrix transpose): Given a matrix A, the transpose of A, denoted by AT and read A-transpose, is obtained by changing all the rows of A into columns of AT while preserving the order. Examples: Properties of the transpose: 1) (AT)T = A 2) (λ A)T = λ AT 3) (A + B)T = AT + BT 4) (A B)T = BT AT A symmetric matrix is a matrix that is equal to its transpose while a skew symmetric matrix is a matrix that is equal to the negative of its transpose.
1.4 Special Matrices: row-reduced form
Definition: A matrix is in row-reduced form if it satisfies four conditions: (R1) All zero rows appear below nonzero rows when both types are present in the matrix. (R2) The first nonzero element in any nonzero row is unity. (R3) All elements directly below (that is, in the same column but in succeeding rows from) the first nonzero element of a nonzero row are zero. (R4) The first nonzero element of any nonzero row appears in a later column (further to the right) than the first nonzero element in any preceding row. Example:
Row-reduced form: examples
In row-reduced form Not in row-reduced form
1.4 Special Matrices: identity matrice
Definition: A square matrix that has 1’s on the main diagonal and 0’s off the main diagonal is called an identity matrix. Example: A 3 × 3 identity matrix. Note: An identity matrix has the property that AI = IA = A.
1.4 Special Matrices: lower (upper) triangular matrices
Definition: A square matrix A=[aij] is called lower triangular if aij=0 for j>i (that is, if all the elements above the main diagonal are zero) and upper triangular if aij=0 for j<i (that is, if all the elements below the main diagonal are zero). Example:
1.6 Vectors x1 x2 v Definition: A vector is a 1 n or n 1 matrix.
Magnitude: If , Is a UNIT vector A nonzero vector is normalized if it is divided by its magnitude. Is a unit vector
1.7 The geometry of vectors
Vector Addition v w v+w Vector Subtraction v-w v w | {
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1.7 The geometry of vectors
Vector Addition v w v+w Vector Subtraction v-w v w
Download ppt "Chapter 1: Matrices Definition 1: A matrix is a rectangular array of numbers arranged in horizontal rows and vertical columns. EXAMPLE:"
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# Is a negative number proper fractions?
I was asked this question by a kid, is -$\frac{4}{7}$ a proper fraction or not? As per my knowledge $\frac{4}{7}$ is a proper fraction. If it has a -ve number does it make any difference? Definition says A number whose numerator is smaller than denominator is called a proper fraction. Can we consider -$\frac{4}{7}$as a proper fraction? If not why not please explain. This is my first question I don't have much idea about tags of mathematics if it is tagged wrongly please edit it.
Thank you
Dibya
• mathworld.wolfram.com/ProperFraction.html So it's neither proper or improper. – Lazar Ljubenović Mar 10 '13 at 12:23
• It is proper: en.wikipedia.org/wiki/… – Dennis Gulko Mar 10 '13 at 12:23
• If I had to define a convention that makes sense in the context of where questions like this have relevance, I would say that -4/7 isn't a fraction at all; it is a numeral consisting of a negative sign - and a (proper) fraction 4/7. – Hurkyl Mar 10 '13 at 12:26
From Wikipedia:
Common fractions can be classified as either proper or improper. When the numerator and the denominator are both positive, the fraction is called proper if the numerator is less than the denominator, and improper otherwise. In general, a common fraction is said to be a proper fraction if the absolute value of the fraction is strictly less than one—that is, if the fraction is between $-1$ and $1$.* [Italics mine]
So $\;-1 < \left(-\dfrac 47\right) < 1$ is considered a proper fraction; (alternatively $\;\;0 < \Big|-\dfrac 47 \Big| = \dfrac 47 < 1$.
I think in terms of mathworld's definition: When using the division algorithm, for example, one requires an integer quotient $\times$ an integer divisor, plus a non-negative integer remainder less than the value of the divisor. So if dividing $-4$ by $7$:
$$-4 = -1\cdot 7 + 3, \;\;q = -1;\;\;r = 3, i.e., -\dfrac 47 = -1 + \dfrac 37$$ which would be a mixed fraction. | {
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So the fractional part would be the proper fraction $\dfrac 37$, the integer part, $-1$. Consistent with this, mathworld may require that a proper fraction occurs only when the quotient is $0$, and the remainder a positive integer less than the divisor, hence the fractional part = $\dfrac rd\; d:$divisor,$\;r:$ the remainder when dividing number by $d$.
• I have edited my question added a link form mathworld given by Lazar in the comment. This is confusing. Which one is correct? – NewUser Mar 10 '13 at 12:29
• It's a matter of convention - By any other name, it'd still smell the same. – Guest 86 Mar 10 '13 at 12:33
• The Wikipedia definition is the standard. I think mathworld left out the clause that the $0 <$ |fraction|$< 1$ – Namaste Mar 10 '13 at 12:34 | {
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# Probability of multivariate normal being positive on each coordinate
How can I find the probability that each coordinate of a specified multivariate normal distribution is positive? I tried the following, which I believed should work
mu = {0, 0, 0};
sigma = {{2, 1, 1}, {1, 2, 1}, {1, 1, 2}};
Probability[
x > 0 && y > 0 && z > 0, {x, y, z} \[Distributed]
MultinormalDistribution[mu, sigma]]
Unfortunately, for the output I just get the last line from the input (with mu and sigma replaced by their actual values). I don't see where the problem could possibly be since the matrix is positive definite. If I replace it by the identity matrix everything works fine (i.e. the output is 1/8).
The general integral does not have a closed form solution, so use NIntegrate:
mu = {0, 0, 0};
sigma = {{2, 1, 1}, {1, 2, 1}, {1, 1, 2}};
NIntegrate[
PDF[MultinormalDistribution[mu, sigma], {x, y, z}],
{x, 0, ∞}, {y, 0, ∞}, {z, 0, ∞}]
(* 0.25 *)
Check:
mu = {0, 0, 0};
sigma2 = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
NIntegrate[
PDF[MultinormalDistribution[mu, sigma2], {x, y, z}],
{x, 0, ∞}, {y, 0, ∞}, {z, 0, ∞}]
(* 0.125 *)
Note that symbolic integration and symbolic probability work for the two-dimensional version of this problem:
mu = {0, 0};
sigma = {{2, 1}, {1, 2}};
Probability[x > 0 && y > 0, {x, y} \[Distributed] MultinormalDistribution[mu, sigma]]
and
Integrate[PDF[MultinormalDistribution[mu, sigma], {x, y}], {x, 0, ∞}, {y, 0, ∞}]
(* 1/3 *)
• An even simpler way is to replace Probability with NProbability (adds one additional character to the OP's code). – DumpsterDoofus Jan 24 '15 at 0:41
• Yep... Simpler. – David G. Stork Jan 24 '15 at 0:42
• Nice, thank you very much – Filanto Renika Jan 24 '15 at 0:52
• The general integral for the OPs $n=3$ case does actually have a closed-form solution: posted below. – wolfies Jan 24 '15 at 16:47 | {
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The probability that the OP seeks is known as the multivariate Normal orthant probability. Correctly, for the $n=3$ cased posed here, the general integral DOES in fact have a closed -form solution, though Mma cannot (currently) obtain it.
In particular, given a zero mean vector and variance-covariance matrix:
$$\Sigma =\left( \begin{array}{ccc} 1 & \rho _{\text{xy}} & \rho _{\text{xz}} \\ \rho _{\text{xy}} & 1 & \rho _{\text{yz}} \\ \rho _{\text{xz}} & \rho _{\text{yz}} & 1 \\ \end{array} \right)$$
$\dots$ the standardised trivariate Normal orthant probability is:
$$P(X>0,Y>0,Z>0) \quad = \quad \frac{1}{8} + \frac{\text{ArcSin}\left[\rho _{\text{xy}}\right]+ \text{ArcSin}\left[\rho _{\text{xz}}\right]+ \text{ArcSin}\left[\rho _{\text{yz}}\right]}{4 \pi }$$
For application and more detail, see, for instance:
• Chapter 6 of our book: Rose and Smith, Mathematical Statistics with Mathematica (Section 6.4C) $\rightarrow$ a free download is available at: http://www.mathstatica.com/book/bookcontents.html , or
• Stuart and Ord (1994), Kendall's Advanced Theory of Statistics (6th edition): section 15.10 and 15.11.
Example
Let $(X,Y,Z)$ have a standardised multivariate Normal with zero mean vector, and variance covariance matrix:
sigma = {{1, 27/34, 22/23}, {27/34, 1, 4/5}, {22/23, 4/5, 1}}
Then, the exact orthant probability is given immediately as:
P3 = 1/8 + (ArcSin[27/34] + ArcSin[22/23] + ArcSin[4/5])/(4 Pi)
... which, to 10 decimal places, is:
N[P3, 10]
0.3732564868
By contrast, the approach using numerical integration can be unreliable here:
NIntegrate[ PDF[MultinormalDistribution[{0, 0, 0}, sigma], {x, y, z}], {x, 0, Infinity}, {y, 0, Infinity}, {z, 0, Infinity}]
NIntegrate::slwcon: Numerical integration converging too slowly ...
0.371907
Numerical integration can sometimes be awkward, unreliable, or slow (as in this example) ... and having an exact closed-form instantaneous solution is a better way to proceed, if possible. | {
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From general to standardised
Given a zero mean vector, what if our variance-covariance matrix is not in a standardised form (with 1's along the main diagonal)? We can easily convert it into standardised form. If, say, our variance-covariance matrix is:
sigma = {{3, 1/3, 3/2}, {1/3, 2/3, -1}, {3/2, -1, 4}}
... then the standardised variance-covariance matrix S is:
A = DiagonalMatrix[Diagonal[sigma]^(-(1/2))];
S = A.sigma.A
All done.
• I eagerly looked up your book because I wanted to see "more detail"... but there isn't any more detail, just a restatement of the same formula and a reference to Stuart and Ord's book. – Rahul Jan 24 '15 at 16:51
• The book also gives the closed-form solution for the bivariate case, and then uses these exact formulae as benchmarks to check the performance/accuracy of Mma's numerical integration routines. In case you are about to eagerly consult Stuart and Ord, I should perhaps caution that you won't find very much more detail there either for the $n=2$ or 3 cases as to results (though more as to derivation)... but they do also provide useful references for higher dimensions. – wolfies Jan 24 '15 at 16:59
• Not quite all done. Let's help the OP by giving him the closed form solution for his particular covariance matrix. – David G. Stork Jan 24 '15 at 19:21
• @Rahul I believe the result can be obtained by a linear change of variables that diagonalizes and normalizes the quadratic form of the pdf (to Exp[k (u^2+v^2+w^2)]). The orthant is transformed to a domain that is subtended by a spherical triangle with angles $\pi/2+ \sin^{-1}\rho$, where $\rho$ runs over the nondiagonal coefficients of $\Sigma$. The result is the proportion of the sphere that is cut out (by the symmetry of the integral). A similar result is true for $n=2$. – Michael E2 Jan 24 '15 at 23:22 | {
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# Finding $\lim\limits_{n \to \infty} \sqrt[n^2]{a_n}$ given that $\lim\limits_{n \to \infty} \frac{a_n a_{n+2}}{(a_{n+1})^2}=L^2$
This is the problem I saw in another question - which was later closed and deleted (the close reason was missing context). Still, the problem seemed interesting to me - at least in the sense that the solution is not immediately obvious.
The problem:
Suppose that $(a_n)$ is a real sequence such that $a_n>0$ for $n>0$ and $$\lim\limits_{n\to\infty} \frac{a_na_{n+2}}{a_{n+1}^2}=L^2.$$ Suppose further that $L\ge0$ Find $\lim\limits_{n\to\infty} \sqrt[n^2]{a_n}$. (In terms of $L$.)
The original problem in the linked question was given with $L=2$. However, I do not think that the problem should change that much for any value of $L$. (Maybe one could be suspicious about $L=0$?)
It is also clear - at least for $L>0$ - that if it is possible to express the second limit using $L$, then the limit should be equal to $L$. It suffices to notice that for $a_n = L^{n^2}$ we have $$\frac{a_na_{n+2}}{a_{n+1}^2}= \frac{L^{n^2+(n+2)^2}}{L^{2(n+1)^2}} = L^{(2n^2+4n+4)-2(n^2+2n+2)} = L^2.$$
So we know what is the candidate for the result, we still need to prove whether this is actually true.
I have posted my attempt as answer. But I will be glad to learn about other solutions. (And of course, also corrections to my approach, if I made some mistakes.)
For the sake of simplicity, let us assume that $L>0$. And later we will check what has to be modified in the case $L=0$. | {
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Let us denote $b_n = \frac{\log a_n}{\log L}$. (I.e., the logarithm at the base $L$.) Then the limit $$\lim\limits_{n\to\infty} \frac{a_na_{n+2}}{a_{n+1}^2}=L^2$$ changes to $$\lim\limits_{n\to\infty} (b_{n+2}-2b_{n+1}+b_n) = 2.$$ And if we put $c_n=b_{n+1}-b_n$ then we have $$\lim\limits_{n\to\infty} (c_{n+1}-c_n)=2.$$ Applying Stolz-Cesaro theorem1 twice we get $$\lim\limits_{n\to\infty} \frac{b_{n+1}-b_n}n = \lim\limits_{n\to\infty} \frac{c_n}n =2$$ and $$\lim\limits_{n\to\infty} \frac{b_n}{\frac{n(n-1)}2}=2.$$
Therefore we get $$\lim\limits_{n\to\infty} \frac{b_n}{n(n-1)}=1 \qquad\text{and}\qquad \lim\limits_{n\to\infty} \frac{b_n}{n^2} = 1.$$ So we have $$\lim\limits_{n\to\infty} \frac{\log{a_n}}{n^2\log L} = \frac{\log{\sqrt[n^2]{a_n}}}{\log L}=1$$ i.e. $$\lim\limits_{n\to\infty} \log \sqrt[n^2]{a_n}=\log L \qquad\text{and}\qquad \lim\limits_{n\to\infty} \sqrt[n^2]{a_n}=L.$$
What happens if $L=0$? Already the first step does make sense, since we cannot take logarithm of $L^2=0$.
Let us now take $b_n=\log a_n$. (To avoid $\log L$ in denominator-- which was there just to simplify the numbers we work with, so that we do not have to write $\log L$ every time.)
In the first step we have just used that $x_n\to L$ implies $\log x_n\to\log L$. If we have $x_n\to0^+$, we still have $\log x_n\to -\infty$. So for $b_n=\log a_n$ we $\lim\limits_{n\to\infty} (b_{n+2}-2b_{n+1}+b_n) = -\infty$.
Stolz-Cesaro theorem is true also for $\pm\infty$. Which means that the following steps are correct if we replace $2$ by $-\infty$.
Finally we arrive at $\lim\limits_{n\to\infty} \log \sqrt[n^2]{a_n}=-\infty$, which gives us $$\lim\limits_{n\to\infty} \sqrt[n^2]{a_n}=0$$ simply by using that $\log x_n\to-\infty$ implies $x_n\to0^+$.
Which means that the same result is true for $L=0$.
1See Wikipedia: Stolz–Cesaro theorem. Some proofs of this result can be found in this question. | {
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1See Wikipedia: Stolz–Cesaro theorem. Some proofs of this result can be found in this question.
• Just one doubt. The question was asked 3 Minutes ago. And you posted such a long answer 3 minutes ago. How is this even possible? – user567182 Jun 28 '18 at 8:38
• @user567182 It is possible to post a question and an answer at the same time. This feature was mentioned on meta around the time when it was introduced: Recently rolled out SE Encyclopedia feature. In which situations it is suitable is another question - but it is definitely encouraged that the poster shows their own attempts when posting the question, posting self-answer might be a way to do this. If you are interested in other past discussions about this, you might have a look at posts tagged (self-answer) on meta. – Martin Sleziak Jun 28 '18 at 8:42
• Perfect answer. +1 I had mentioned the same approach in comments to the original (now closed deleted question). – Paramanand Singh Jun 28 '18 at 14:32
Here is an answer that I posted to the question mentioned in this question, but deleted almost immediately due to the lack of context in that question.
If $$\lim_{n\to\infty}\frac{a_n\,a_{n+2}}{a_{n+1}^2}=L^2\tag1$$ then for any $\epsilon\gt0$, there is an $N_\epsilon$ so that for $n\ge N_\epsilon$, $$\left|\,\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)-2\log(L)\,\right|\le\epsilon\tag2$$ Note that \begin{align} \frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right) &=\frac1{n^2}\sum_{k=0}^{n-1}\log\left(\frac{a_0}{a_k}\frac{a_{k+1}}{a_1}\right)\\ &=\frac1n\log\left(\frac{a_0}{a_1}\right)+\frac1{n^2}\log\left(\frac{a_n}{a_0}\right)\tag3 \end{align} Inspired by the following diagram | {
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We can break down the sum on the left side of $(3)$ as follows \begin{align} \frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right) &=\frac1{n^2}\sum_{k=N_\epsilon}^{n-1}\sum_{j=N_\epsilon}^{k-1}\overbrace{\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)}^\text{within \epsilon of 2\log(L)}\\ &\color{#C00}{+\frac{n-N_\epsilon}{n^2}\sum_{j=0}^{N_\epsilon-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)+\frac1{n^2}\sum_{k=0}^{N_\epsilon-1}\sum_{j=0}^{k-1}\log\left(\frac{a_j\,a_{j+2}}{a_{j+1}^2}\right)}\tag4 \end{align} For any $\epsilon\gt0$, the red terms vanish as $n\to\infty$. The remaining term on the right hand side is $\frac1{n^2}$ times a sum of $\frac{(n-N_\epsilon)^2+(n-N_\epsilon)}2$ terms, all within $\epsilon$ of $2\log(L)$.
Therefore, $$\left|\,\lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)-\log(L)\,\right|\le\frac\epsilon2\tag5$$ Since $(5)$ is true for any $\epsilon\gt0$, we have $$\lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^{n-1}\sum_{j=0}^{k-1}\log\left(\frac{a_n\,a_{n+2}}{a_{n+1}^2}\right)=\log(L)\tag6$$ Taking the limit of the right hand side of $(3)$ as $n\to\infty$, we get $$\lim_{n\to\infty}\frac{\log(a_n)}{n^2}=\log(L)\tag7$$ which implies $$\lim_{n\to\infty}\sqrt[n^2]{a_n}=L\tag8$$
• It's great that you posted your answer here. By the time I saw your answer (to the old deleted question) it was in deleted state. +1 – Paramanand Singh Jun 29 '18 at 8:47
• Not surprising, since it was only there for 18 seconds. – robjohn Jun 29 '18 at 11:50
This is not another solution, but is the essence of your solution worked out by me individually.
If $L>0$,
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If $L>0$,
By two applications of Stolz-Cesaro theorem,
\begin{align*} \lim_{n\to\infty}{{\ln a_n}\over{n^2}} &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{2n+1}}\\ &=\frac{1}{2}\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{n}}\\ &=\frac{1}{2}\lim_{n\to\infty}{\ln{\frac{a_{n+2}}{a_{n+1}}-\ln{\frac{a_{n+1}}{a_n}}}\over1}\\ &=\frac{1}{2}\lim_{n\to\infty}\ln\Bigg(\frac{\big(\frac{a_{n+2}}{a_{n+1}}\big)}{\big(\frac{a_{n+1}}{a_{n}}\big)}\Bigg)\\ &=\frac{1}{2}\lim_{n\to\infty}\ln\frac{a_na_{n+2}}{{a_{n+1}}^2}\\ &=\frac{1}{2}\ln(L^2)\\ &=\ln L\\ \therefore\lim_{n\to\infty}\sqrt[n^2]{a_n}&=L \end{align*}
If $L=0$, \begin{align*} \lim_{n\to\infty}{{\ln a_n}\over{n^2}} &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{2n+1}}\\ &=\lim_{n\to\infty}{\ln{\frac{a_{n+1}}{a_n}}\over{n}}\\ &=\lim_{n\to\infty}{\ln{\frac{a_{n+2}}{a_{n+1}}-\ln{\frac{a_{n+1}}{a_n}}}\over1}\\ &=\lim_{n\to\infty}\ln\Bigg(\frac{\big(\frac{a_{n+2}}{a_{n+1}}\big)}{\big(\frac{a_{n+1}}{a_{n}}\big)}\Bigg)\\ &=\lim_{n\to\infty}\ln\frac{a_na_{n+2}}{{a_{n+1}}^2}\\ &=-\infty\\ \therefore\lim_{n\to\infty}\sqrt[n^2]{a_n}&=0 \end{align*} | {
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# Measuring fractal dimension of natural objects from digital images
This is a useful topic. A college physics lab, medical diagnostics, urban growth, etc. - there is a lot of applications. On this site by Paul Bourke about Google Earth fractals we can get a high resolution images (in this post they are low res - import from source for experiments). For example, around Lake Nasser in Egypt:
img = Import["http://paulbourke.net/fractals/googleearth/egypt2.jpg"]
The simplest method I know is Box Counting Method which has a lot of shortcomings. We start from extracting the boundary - which is the fractal object:
{Binarize[img], iEdge = EdgeDetect[Binarize[img]]}
Now we could partition image into boxes and see how many boxes have at least 1 white pixel. This is a very rudimentary implementation:
MinS = Floor[Min[ImageDimensions[iEdge]]/2];
data = ParallelTable[{1/size, Total[Sign /@ (Total[#, 2] & /@ (ImageData /@
Flatten[ImagePartition[iEdge, size]]))]}, {size, 10, MinS/2, 10}];
From this the slope is 1.69415 which is a fractal dimension that makes sense
line = Fit[Log[data], {1, x}, x]
13.0276 + 1.69415 x
Plot[line, {x, -6, -2}, Epilog -> Point[Log[data]],
PlotStyle -> Red, Frame -> True, Axes -> False]
Benchmark: if I run this on high res of Koch snowflake i get something like ~ 1.3 with more exact number being 4/log 3 ≈ 1.26186.
Question: can we improve or go beyond the above box counting method?
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All approaches are acceptable if they find fractal dimension from any image of natural fractal.
• You have a lot of programs in mathematica to measure fractal dimensions and multi fractal spectrum of an image in the book: Fractal Geography, Andre Dauphine, Wiley, 2012 See the book on the Wolfram Mathematica | Books or Amazon  Oct 16, 2012 at 9:32
• Vitaliy, with all due respect, the ending bullets make the question (actually there isn't a question mark anywhere, right?) rather wide ranging. Perhaps you could focus it somewhat and make it sound more like a real question? Oct 16, 2012 at 12:59
• @SjoerdC.deVries Updated, it is a single question now. Oct 16, 2012 at 15:06
• could you perhaps describe or link some examples of use of fractal dimension? Oct 17, 2012 at 10:27
• @magma: Wikipedia lists some examples here: en.wikipedia.org/wiki/…
– shrx
Jun 14, 2013 at 20:22
You can still use box count, but doing it smarter :)
Counting boxes with at least 1 white pixel from ImagePartition can be done more efficiently using Integral Image, a technique used by Viola-Jones (2004) in their now popular face recognition framework. For a mathematical motivation (and proof), Viola and Jones point to this source.
What Integral Image allows you to do is to compute efficiently the total mass of any rectangle in an image. So, you can define the following:
IntegralImage[d_] := Map[Accumulate, d, {0, 1}];
data = ImageData[ColorConvert[img, "Grayscale"]]; (* img: your snowflake image *)
ii = IntegralImage[data];
Then, the mass (white content) of a region, is
(* PixelCount: total mass in region delimited by two corner points,
given ii, the IntegralImage *)
PixelCount[ii_, {p0x_, p0y_}, {p1x_, p1y_}] :=
ii[[p1x, p1y]] + ii[[p0x, p0y]] - ii[[p1x, p0y]] - ii[[p0x, p1y]];
So, instead of partitioning the image using ImagePartition, you can get a list of all the boxes of a given size by | {
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PartitionBoxes[{rows_, cols_}, size_] :=
Transpose /@ Tuples[{Partition[Range[1, rows, size], 2, 1],
Partition[Range[1, cols, size], 2, 1]}];
If you apply PixelCount to above, as in your algorithm, you should have the same data but calculated faster.
PixelCountsAtSize[{rows_, cols_}, ii_, size_] :=
((PixelCount [ii, #1, #2] &) @@ # &) /@ PartitionBoxes[{rows, cols}, size];
Following your approach here, we should then do
fractalDimensionData =
Table[{1/size, Total[Sign /@ PixelCountsAtSize[Dimensions[ii], ii, size]]},
{size, 3, Floor[Min[Dimensions[ii]]/10]}];
line = Fit[Log[fractalDimensionData], {1, x}, x]
Out:= 10.4414 + 1.27104 x
which is very close to the actual fractal dimension of the snowflake (which I used as input).
Two things to note. Because this is faster, I dared to generate the table at box size 3. Also, unlike ImagePartition, my partition boxes are all of the same size and therefore, it excludes uneven boxes at the edges. So, instead of doing minSize/2 as you did, I put minSize/10 - excluding bigger and misleading values for big boxes.
Hope this helps.
Update
Just ran the algorithm starting with 2 and got this 10.4371 + 1.27008 x. And starting with 1 is 10.4332 + 1.26919 x, much better. Of course, it takes longer but still under or around 1 min for your snowflake image.
Update 2
And finally, for your image from Google Earth (eqypt2.jpg) the output is (starting at 1-pixel boxes)
12.1578 + 1.47597 x
It ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs. | {
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It ran in 43.5 secs in my laptop. Using ParallelTable is faster: around 28 secs.
• There isn't a "snowflake" image. All three images come from the same object (and I guess the fractal dimension should be the same for all of them) Dec 23, 2013 at 21:03
• @belisarius, the PO published a Koch snowflake image as Benchmark (see last part of post) and that's the one I used. I called it 'snowflake' for short. On the other hand, this is still a numerical procedure, so the numbers will be different depending on approximation. The Update2 refers to the eqypt2.jpg image, also made available by the PO, as he asked for an improvement suitable for real-life images. Dec 23, 2013 at 22:40
• sorry, I missed that link. Thanks Dec 24, 2013 at 0:05
• +1 This is very neat @caya, thank you! I will wait "a bit" hoping that someone can implement things like wavelet multi-fractals or similar. Feb 14, 2014 at 20:02
• @VitaliyKaurov, glad you liked it. Note that Viola & Jones rightly pointed out a link between integral image and Haar wavelet basis in their paper; although this wasn't explored further. It is unclear to me the link between the paper you mentioned and fractal dimension, but certainly worth reading as I am interested in these kind of problems. Cheers. Feb 16, 2014 at 12:40
"Can we improve the box-counting method?" Most certainly.
But first of all, it is important to point out that if you are only checking if a box is empty or not you are effectively measuring the $D_0$ (the capacity or in other words the box-counting dimension). By referring to $D_0$ as the fractal dimension you are assuming that your object is a monofractal, i.e. that all its generalized dimensions $D_q$ (ref.1) $$D_q=\lim_{\epsilon \rightarrow \infty}\frac{\frac{1}{1-q}\log{\sum_i^Np_i^q}}{\log{\frac{1}{\epsilon}}}$$
are equal; $D_0$=$D_1$(entropy dimension)=$D_2$(correlation dimension) and so on... | {
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are equal; $D_0$=$D_1$(entropy dimension)=$D_2$(correlation dimension) and so on...
Normally you should refrain from such assumptions unless you know the generating process. However, even if you assume that the object is monofractal, review literature of fractal dimension estimation concludes that the box-counting method is a bad choice. As Thelier puts it in "box-counting algorithm is a poor way to estimate the box-counting dimension" (ref. 2).
So what can we do? The basic principle in fractal dimension estimation is to study the object at different scales. The question is how to sample this continuum of scales. There are two different class of methods. The first one is the fixed-sized methods. These count the number of points in a fixed size region (boxes/spheres). They perform poorly because you never know how to grow your boxes (should I double, triple?). The second school is the fixed-mass methods. These grow their extend by reaching over to their mth nearest neighbor. Thus they always assure that they are covering the same mass (hence the name). even though fixed-mass methods are more robust than fixed-size methods, both suffer from finite-size and edge effect. Thus you need some extra measures to avoid the edges of your object, otherwise at large scales (of mass/size) you start to get this leveling off effect (that you also have in your last figure).
A few years ago we developed a complicated method (non-overlapping barycentric fixed-mass method) to do exactly that (ref. 3). Here is a summary of the fractal dimension estimations with the BFM method:
I first gave it a go with your Koch snowflake and I get all $D_q$s to be more or less equal to $1.25$ as you would expect from a monofractal. | {
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Then I run the Egypt2.jpg file and as you can see all $D_q$s are not exactly equal. This can be used to argue that the coastline might be multifractal (even if slightly). Another thing to notice is that the slope is not the same throughout the whole mass range. If you consider only the range $[10 - 631]$ you would get a different $D_q$ curve. This implies that the object is not self-similar throughout the entire scale. This could also also explains why the previous answer finds $D_0$ something around $1.47$ while you find $1.69$. In other words there is a scaling break.
References:
1. Renyi, A. (1970). Probability Theory. Amsterdam: North-Holland.
2. Theiler, J. (1990). Estimating fractal dimension. Journal of the Optical Society of America A, 7(6), 1055.
3. Kamer, Y., Ouillon, G., & Sornette, D. (2013). Barycentric fixed-mass method for multifractal analysis. Physical Review E, 88(2), 022922. | {
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# How do I translate sentences from English to predicate logic?
This question was taken from the MIT OCW Math for Computer Science course.
Translate the following sentences from English to predicate logic. The domain that you are working over is $X$, the set of people. You may use the functions $S(x)$, meaning that “$x$ has been a student of $6.042$,” $A(x)$, meaning that “$x$ has gotten an ‘$A$’ in $6.042$,” $T(x)$, meaning that “$x$ is a TA of $6.042$,” and $E(x, y)$, meaning that “$x$ and $y$ are the same person.”
(a) [$6$ pts] There are people who have taken 6.042 and have gotten A’s in $6.042$
(b) [$6$ pts] All people who are 6.042 TA’s and have taken 6.042 got A’s in $6.042$
(c) [$6$ pts] There are no people who are 6.042 TA’s who did not get A’s in $6.042$.
(d) [$6$ pts] There are at least three people who are TA’s in $6.042$ and have not taken $6.042$
What I got thus far:
a) $\exists x(S(x)\wedge A(x))$
b) $\forall x\in X((T(x)\wedge S(x))\Rightarrow A(x))$
c) $\nexists x\in X(T(x)\wedge ¬A(x))$
d) $\exists x\exists y\exists z\in X(¬E(x,y)∧¬E(x,z)∧¬E(y,z)∧(T(x)\wedge ¬S(x))∧(T(y)\wedge ¬S(y))∧(T(z)\wedge ¬S(z)))$
I'm not sure about any of my answers, but c) and d) gave me the most trouble. In regards to c),I am not sure about how to express "there are no people" in predicate logic. In regards to d), I am not sure about how to express "there are at least three people" in predicate logic.
Hints are much better appreciated than an explicit answer. In addition, I would like to know how to tag this type of math. Is it considered discrete mathematics as most colleges/universities call it in the U.S.?
• I would leave out the $\in X$ stuff. – André Nicolas Apr 25 '16 at 18:51
• Can you please explain why? Wouldn't have to define what $x$ is? – Cherry_Developer Apr 25 '16 at 19:16
• The domain is $X$, so it is understood that variables range over $X$. Also, $\in$ is not in the specified language. – André Nicolas Apr 25 '16 at 19:46 | {
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Your answers all look okay. Specifically for part c), you did indeed translate the sentence into predicate logic correctly. However, often times it is customary to not leave any negation symbols before the quantifiers. We can pass the negation symbol through the existential/universal quantifier by swapping them. For example
$$\neg\exists x\in X(P(x))\iff\forall x\in X(\neg P(x))$$
and
$$\neg\forall x\in X(P(x))\iff\exists x\in X(\neg P(x)).$$
Can you see how you can use this to simplify your answer for part c)?
Also, I personally think the discrete math tag is okay for a question like this, especially since you also used the predicate logic tag.
• I'm not sure about how to simplify it, but I gave it a shot: $\forall x\in X(¬T(x)\vee A(x))$ I was mostly unsure of how to apply the negation symbol to the predicate, so I typed "not (x and not y)" into Wolfram Alpha and it returned that "not x or y" would be an equivalent statement. – Cherry_Developer Apr 25 '16 at 19:13
• Yes, you simplified correctly. Here are the full steps: $\neg \exists x\in X(T(x)\wedge \neg A(x))\iff\forall x\in X\neg(T(x)\wedge \neg A(x))\iff\forall x\in X(\neg T(x)\vee A(x))$. – ervx Apr 25 '16 at 19:16
• The negation of $\exists x$ which becomes $\forall x$ is clear to me, but applying the negation to the predicate is where I get lost. I feel like I cheated myself by just resorting to Wolfram Alpha to find the simplified form of $(T(x)\wedge ¬A(x))$. Is $(T(x)\wedge ¬A(x))\Leftrightarrow (\neg T(x)\vee A(x))$ something I should just have memorized? Or is there a way for me to derive it? – Cherry_Developer Apr 25 '16 at 19:23
• In general, for expressions $A$ and $B$, we have $\neg(A\vee B)\iff (\neg A\wedge\neg B)$ and $\neg (A\wedge B)\iff (\neg A\vee \neg B)$. You can verify these using truth tables if you like. In general, though, it would be good to memorize them. They are essentially DeMorgan's laws for propositional logic. – ervx Apr 25 '16 at 19:26 | {
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# Number of club members if they are organized into four committees if each member is in exactly two committees and any two committees share one member
A club with $x$ members is organized into four committees such that
$(a)$ each member is in exactly two committees,
$(b)$ any two committees have exactly one member in common.
Then $x$ has
$(A)$ exactly two values both between $4$ and $8$
$(B)$ exactly one value and this lies between $4$ and $8$
$(C)$ exactly two values both between $8$ and $16$
$(D)$ exactly one value and this lies between $8$ and $16$.
My thought is option $B$ and $x = 6$. Please somebody tell me am I right or wrong?
• Can you show how you arrived at $x=6$? – RandomUser Apr 25 '14 at 16:53
• I just see that if $x$=$6$ then the two conditions are satisfied.But if $x$<$6$ or $x$>$6$ the condition $b$ is not satisfied so I think $x$ may be $6$. – liesel Apr 25 '14 at 17:15
• Please improve your title. – Alexander Gruber Apr 25 '14 at 17:22
Claim: $x=6$
Let $S:=[x]\left(=\{1,2,\dots ,x\}\right)$ where $x$ is some fixed natural number. Let $C_i\subset S$ be the $i$ th commitee.Let $f$ be a function from the set of unordered pair of distinct commitees to members of the club defined as, $$f\left( (C_i,C_j)\right)=C_i\cap C_j$$ $f$ is injective by (a). $f$ is surjective by (b).
Hence, $|S|=\binom{4}{2}=6$.
Here is a construction:
• Let $C_1\cap C_2=1$ wlog. Then, $1\notin C_i \quad \forall \, i\in [4]\setminus \{1,2\}$.
• Let $C_2\cap C_3=2$ wlog. Then, $2\notin C_i \quad \forall \, i\in [4]\setminus \{2,3\}$.
• Let $C_3\cap C_4=3$ wlog. Then, $3\notin C_i \quad \forall \, i\in [4]\setminus \{3,4\}$.
• Let $C_1\cap C_3=4$ wlog. Then, $4\notin C_i \quad \forall \, i\in [4]\setminus \{1,3\}$.
• Let $C_1\cap C_4=5$ wlog. Then, $5\notin C_i \quad \forall \, i\in [4]\setminus \{1,4\}$.
• Let $C_2\cap C_4=6$ wlog. Then, $6\notin C_i \quad \forall \, i\in [4]\setminus \{2,4\}$. | {
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• Let $C_2\cap C_4=6$ wlog. Then, $6\notin C_i \quad \forall \, i\in [4]\setminus \{2,4\}$.
Let the committees be $A_1, A_2, A_3, A_4$ then $|A_1\cup A_2 \cup A_3 \cup A_4|=n$. There are $\binom{4}{2}=6$ pairwise intersections each of size $1$ and each intersection of triples must be empty since if not some member would belong to $3$ committees, a contradiction. If we make a $4 \times n$ table where rows are committees and columns are members and mark a $1$ whenever member $x_j$ belongs to committee $A_i$ we notice that the $i^{th}$ row sum is $|A_i|$ and the $j^{th}$ column sum is $2$ (since each member belongs to exactly $2$ committees) so equating row and column sums we get $\sum_{i=1}^{4}|A_i|= 2n.$
Putting this information together and using the inclusion-exclusion formula we get $n =6$ | {
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I would like a better understanding of the famous birthday paradox. "What is the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday?"
I understood the first part, where the probability reaches 100% when the number of people reaches 367 by the pigeonhole principle. But I am not understanding the explanation beyond that. How do they say that the probability is 99.9% with 70 people and 50% with 23 people? And how do you further generalize the answer? And why is it a "paradox"?
• Please, post the explanation you have been given. Apr 2 '16 at 10:00
• en.wikipedia.org/wiki/Birthday_problem May be this is helpful to you Apr 2 '16 at 10:00
• @Mc Cheng it might be a bit more helpful to specify a section rather than referring someone to an article (which is debatable in its reliability) that is quite hard (and lengthy) to sort through Apr 2 '16 at 10:03
• I could've typed an answer but explanations of this paradox are readily available all over the internet. Take a look at the Wikipedia article. Tell me exactly what you cannot understand. Maybe then we'd be able to help. Apr 2 '16 at 10:03
• Apr 2 '16 at 10:11
Let the number of people in the group be $n$.
The probability that a pair of people don't share a birthday is given equal to $\frac{364}{365}$ ignoring leap years.
There are $\binom{n}{2}$ pair of people in a group of $n$ people. No pair of people will share a birthday if each person has a distinct birthday. The probability of this happening is given by
$$\frac{364}{365}\times\frac{363}{365} \dots \times \frac{365 - (n-1)}{365}$$
How did I get this probability? | {
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How did I get this probability?
Assume that all birthdays are equally likely. If the first person was born on day $x_1$ then the second person in the group cannot be born on day $x_1$. The probability for this happening is $364\over 365$. Now let the birthday of the second person be $x_2$. The probability that the third person is not born on $x_1$ nor on $x_2$ is $363\over365$. Similarly we get the probability for the $n^{\text{th}}$ person. Since each event is independent, we multiply all the probabilities.
Thus the probability that at least one pair shares a birthday for a group of $n$ people is given by
$$p = 1 - \left(\frac{364}{365}\times\frac{363}{365} \dots \times \frac{365 - (n-1)}{365}\right)$$
Now you have the probability $p$ as a function of $n$. If you know the RHS, then you simply find for what value of $n$ we get the closest RHS to $p$
It so happens that if $p = 99.9\%$, the $n = 70$
• Thank you Banach Tarski! Am I right in understanding, that the only reason it is a "paradox" is because we'd usually expect the probability to be linear in nature? I don't see any paradox otherwise Apr 2 '16 at 10:11
• @aswa09 yup you are correct. A paradox is a self contradictory statement. The contradiction here is with our common sense and not with the mathematics :) Apr 2 '16 at 10:15
• Is it not a multiplication of $364/365$ not $1/365$? Else your formula gives a $364/365$ chance of just two people sharing the same birthday Apr 2 '16 at 10:21
• Yes, I had made a blunder in computing the probability, I have fixed this now :) Apr 2 '16 at 10:39
The paradox is that the rate of growth doesn't match our common sense. We expect that the way to count the number of possibilities for people to have the same birthday is directly from the number of people. However, in reality it's based on the number of pairs of people, which grows much faster than the number of people. | {
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Banach Tarski shows you the derivation which really is that the numerator is the number of people pairs, but still over 365.
• How do I remove the duplication mark? I have edited the question asking about why it's a paradox, so that it isn't a duplicate question Apr 3 '16 at 7:23 | {
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# Factorial decomposition of integers?
This question might seem strange, but I had the feeling it's possible to decompose in a unique way a number as follows:
if $x < n!$, then there is a unique way to write x as: $$x = a_1\cdot 1! + a_2\cdot 2! + a_3\cdot3! + ... + a_{n-1}\cdot(n-1)!$$ where $a_i \leq i$
I looked at factorial decomposition on google but I cannot find any name for such a decomposition.
example:
If I chose :
(a1,a2) =
• 1,0 -> 1
• 0,1 -> 2
• 1,1 -> 3
• 0,2 -> 4
• 1,2 -> 5
I get all number from $1$ to $3!-1$
ideas for a proof:
The number of elements between $1$ and $N!-1$ is equal to $N!-1$ and I have the feeling they are all different, so this decomposition should be right. But I didn't prove it properly.
Are there proofs of this decomposition? Does this decomposition as a name? And above all is this true ?
• It's true, and the proof is easy by induction on $n$. What have you written down? – Olivier Bégassat Jul 23 '11 at 9:19
• I am not sure about number-theory tag, if someone has a better idea, please retag the question. – Martin Sleziak Jul 23 '11 at 9:21
• @Olivier, I actually thinking of a mapping between n! and all permutation, and I should have used that as a proof. But since I couldn't find anything on the web about that I had the feeling something was wrong ... :D! – Ricky Bobby Jul 23 '11 at 9:28
• @Martin: Possibly combinatorics? – Brian M. Scott Jul 23 '11 at 9:39
• If any of you have a good reason to think (number-systems) doesn't belong, please remove that tag... – J. M. is a poor mathematician Jul 23 '11 at 11:56
You're looking for the factorial number system, also known as "factoradic". Searching should give you more results. | {
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Yes, it's true that such a decomposition is always possible. One way to prove it is as follows: given $x < n!$, consider the $x$th permutation of some ordered set of $n$ symbols. This is some permutation $(s_1, s_2, \dots, s_n)$. Now for $s_1$ you had $n$ choices (label them $0$ to $n-1$) and you picked one, so let $a_{n-1}$ be the choice you made. For $s_2$ you had $n-1$ choices (label them $0$ to $n-2$) and you picked one, so let $a_{n-2}$ be the number of the choice you made. Etc. $a_0$ is always $0$ because you have only one choice for the last element. (This is also known as the Lehmer code.)
• I was actually thinking about permutation when this question came to my mind, I didn't thought about using it to prove the decomposition. thanks a lot for your answer. – Ricky Bobby Jul 23 '11 at 9:17
• Why not try what would seem obvious? Divide your numberx by n!, take the highest multiplr $a_n$ with $a_nn!<x$, then divide the remainder x-$a_n n!$ by (n-1)! and take $a_{n-2}$ as the largest multiple, etc. at the end, you will be left with some amount which is always a multiple of 1=1! – gary Jul 23 '11 at 9:37
• @gary: Yes, something like that would work too, if done correctly. But for me, this map numbering permutations was actually more obvious. :-) (You want to divide by $(n-1)$ the first time, also you need to prove that $a_i \le i$ for all $i$. This is easy, but it's not just one line.) – ShreevatsaR Jul 23 '11 at 9:46
• Shreevatsa: if your coefficient for n! was larger than n, then n(n-1)!=n! , and you could then increase your previous coefficient for n! by (at least )1. But alternative explanations are always useful,helpful. – gary Jul 23 '11 at 10:12 | {
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Your conjecture is correct. There is a straightforward proof by induction that such a decomposition always exists. Suppose that every positive integer less than $n!$ can be written in the form $\sum_{k=1}^{n-1} a_k k!$, where $0 \le a_k \le k$, and let $m$ be a positive integer such that $n! \le m < (n+1)!$. There are unique integers $a_n$ and $r$ such that $m = a_nn! + r$ and $0 \le r < n!$, and since $m < (n+1)! = (n+1)n!$, it’s clear that $a_n \le n$. Since $r < n!$, the induction hypothesis ensures that there are non-negative integers $a_1,\dots,a_{n-1}$ such that $r = \sum_{k=1}^{n-1} a_k k!$, and hence $m = \sum_{k=1}^n a_k k!$.
We’ve now seen that each of the $(n+1)!$ non-negative integers in $\{0,1,\dots,n\}$ has a representation of the form $\sum_{k=1}^n a_k k!$ with $0 \le a_k \le k$ for each $k$. However, there are only $\prod_{k=1}^n (k+1) = (n+1)!$ distinct representations of that form, so each must represent a different integer, and each integer’s representation is therefore unique.
You can also reason as follows : suppose you've shown for some integer $n$ that every integer $\in\lbrace0,\dots,n!-1\rbrace$ has a unique decomposition as you suggest. Take $k\in\lbrace0,\dots,(n+1)!-1\rbrace.$ Write $$k=q\cdot n!+r$$ the euclidean division of $k$ with respect to $n!$. Necessarily you have $0\leq q < n+1$. This gives you an expression you want, for $0\leq r<n!$ has, by hypothesis, an expression involving only factorials up to $(n-1)!$ .
Finally, to show uniqueness, you can again use your hypothesis to deduce that if $k=a_n\cdot n!+ \sum_0^{n-1} a_i\cdot i!,$ then $0\leq \sum\dots< n!-1$ by hypothesis, and this tells you that this decomposition is the euclidean division of $k$ by $n!$. So by uniqueness of the euclidean division, it's the one decomposition we defined earlier. | {
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This generalizes to the representation $$n = \sum_{i\ge 0} a_i b_i$$ where $b_0 = 1$ and $b_{n+1} = b_n c_n$ with $c_n > 1$ and $0 \le a_i < c_i$ - i.e., representing $n$ with digits $a_i$ to a "base" with varying ratios of consecutive digit values (phrased awkwardly, I know, but it is late and I am tired). For a usual base $B$ system, $c_i = B$ for all $i$, so $b_i = B^i$. For this system, $c_i = i$ (or maybe $c_i = i+1$, modulo my tiredness), so $b_i = i!$.
I proved too many years ago that this representation is unique iff the $c_i$ were all integers. I am sure that this was proved many years ago (probably by Euler, if not Fermat) and is in Dickson somewhere. | {
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# Why is a number that is divisible by $2$ and $3$, also divisible by $6$?
Why is it that a number such as $108$, that is divisible by $2$ and $3$, is also divisible by $6$?
Is this true that all numbers divisible by two integers are divisible by their product?
• Think about the factorization of a number that's divisible by 2 and 3. – Ben Longo Mar 11 '17 at 14:20
• Every number can be expressed as the product of primes. $2$ & $3$ are coprime ... Can you get it from there Jason ? – Donald Splutterwit Mar 11 '17 at 14:21
• Provided the numbers are mutually prime. – SchrodingersCat Mar 11 '17 at 14:22
• note $108 = 2^2 \cdot 3^3 = 3 \cdot (2\cdot 3) \cdot (2 \cdot 3) = 3 \cdot (6) \cdot (6)$ – Dando18 Mar 11 '17 at 14:23
• What about 30, that's divisible by 10 and 15? – Ethan Bolker Mar 11 '17 at 14:23
It is certainly not true that if $x$ is divisible by both $y$ and $z$, then $x$ is divisible by $yz$. For example, $2$ is divisible by $2$ and $2$, but is not divisible by $2\cdot 2$. As a less silly example, $120$ is divisible by both $15$ and $6$, but is not divisible by $90$.
But if you think about the examples above, you'll notice that the factors involved in each overlap: e.g. both $2$ and $2$ are even, and both $15$ and $6$ are divisible by $3$. Meanwhile, $2$ and $3$ don't overlap - they share no factors in common.
It turns out that this is the crucial property:
If $x$ is divisible by $y$ and $z$, and $y$ and $z$ have no factors in common besides $1$, then $x$ is divisible by $yz$.
Proving this is a good challenge. HINT: think about factorization into primes . . .
• There is perhaps something to add here, since there are divisors that fulfil these rules and DO have factors in common, but still divide some number, e.g. in the case $2$ divides $y$ and $z$, and $4$ divides $x$ – samerivertwice Mar 11 '17 at 14:27 | {
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It is not true that for any two integers $m$, $n$, if $x$ is divisible by $m$ and $n$, it is divisible by $mn$. For example $4$ is divisible by $4$ and $2$, but it is not divisible by $8$.
However, this is true if $m$ and $n$ are relatively prime, i.e. the greatest common divisor of $m$ and $n$ is $1$. Thus, since $2$ and $3$ are relatively prime, any number divisible by $2$ and $3$ will also be divisible by $6$.
More generally, if $x$ is divisible by $m$ and $n$, then it must be divisible by $\operatorname{lcm}(m,n)$ (least common multiple). Case of $m$ and $n$ being relatively prime is a special case of this, since $$\operatorname{lcm}(m,n) = \frac{mn}{\operatorname{gcd}(m,n)}$$
where $\operatorname{gcd}$ stands for the greatest common divisor and thus $\operatorname{lcm}(m,n) = mn$, when $m$ and $n$ are relatively prime.
Let $a$ be divisible by $2$ and $3$. So $$a=2^{r_{0}}\cdot p^{r_{1}}_{1}\cdot p^{r_{2}}_{2}\cdot\ldots=3^{s_{0}}\cdot q^{s_{1}}_{1}\cdot q^{s_{2}}_{2}\cdot\ldots$$ where $p_{i}$ and $q_{j}$ are distinct prime numbers, $r_{i}$ and $s_{j}$ are possibly $0$ for $1\leqslant i,j<\infty$ and $r_{0},s_{0}\geqslant 1$. We know from the fundamental theorem that these two must be unique factorisations (up to re-ordering). Hence there must be some $p_{k}^{r_{k}}=3^{s_{0}}$ and some $q_{\ell}^{s_{\ell}}=2^{r_{0}}$. Therefore $a$ has a factor of $6$: \begin{align*} a&=2^{r_{0}}\cdot p_{1}^{r_{1}}\ldots\cdot p_{k}^{r_{k}}\cdot\ldots\\ &=2^{r_{0}}\cdot p_{1}^{r_{1}}\ldots\cdot3^{s_{0}}\cdot\ldots\\ &=6\cdot(2^{r_{0}-1}\cdot3^{s_{0}-1}\cdot p_{1}^{r_{1}}\cdot\ldots). \end{align*} (The argument works on the other side as well). This might be a little explicit but I hope it helps.
As the comments point out, and as you may suspect, it is not true in general that if a number is divisible by two integers, then it is divisible by their product. What we can say is the following: | {
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If $x,y,z$ are integers such that $z$ is divisible by both $x$ and $y,$ then $z$ is divisible by the least common multiple of $x$ and $y.$
This isn't actually much of a surprise, given the way that least common multiple is often defined:
Given two integers $x$ and $y,$ the least common multiple of $x$ and $y$ is the nonnegative integer $z$ such that (i) $z$ is divisible by both $x$ and $y,$ and (ii) every common multiple of $x$ and $y$ is divisible by $z.$
Proving that such a multiple exists isn't exactly straightforward, though. It is slightly easier to prove the existence of another integer:
Given two integers $x$ and $y,$ the greatest common factor of $x$ and $y$ is the positive integer $z$ such that (i) both $x$ and $y$ are divisible by $z,$ and (ii) $z$ is divisible by every common factor of $x$ and $y.$
One can, for example, use the Euclidean algorithm to prove that $\operatorname{gcf}(x,y)$ exists, regardless of $x,y.$ At that point, one can show that $\frac{|xy|}{\operatorname{gcf}(x,y)}$ turns out to be the least common multiple of $x$ and $y.$ As an immediate consequence of the identity $$\operatorname{lcm}(x,y)=\frac{|xy|}{\operatorname{gcf}(x,y)},$$ together with the result mentioned at the beginning, we find the following:
If $x,y,z$ are integers such that $z$ is divisible by both $x$ and $y,$ and if $\operatorname{gcf}(x,y)=1,$ then $z$ is divisible by $xy.$
Since $\operatorname{gcf}(2,3)=1,$ then any number divisible by both $2$ and $3$ is necessarily divisible by $6.$
The general result is this:
If a number $n$ is divisible by $a$ and $b$, then it is divisible by $\DeclareMathOperator{\lcm}{lcm}\lcm(a,b)$. | {
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Indeed, let $\;d=\gcd(a,b)$, $\;m=\lcm(a,b)$, $\;a'=\dfrac ad$, $\;b'=\dfrac bd$. These numbers satisfy the relation $$md=ab, \enspace\text{so}\quad m=\frac{ab}d=a'b=ab'.$$ By hypothesis, we can write $$n=qa=qda', \qquad n=rb=rb'd$$ We deduce that $\;qa'=rb'$. So $a'$ divides $rb'$. As $a'$ and $b'$ are coprime, Gauß' lemma ensures $a'$ divides $r$, i.e. we have $r=r'a'$, and $$n=(r'a')b'd=r'(a'b)=r'm.$$ | {
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# changeIntegrationVariable
Integration by substitution
## Description
example
G = changeIntegrationVariable(F,old,new) applies integration by substitution to the integrals in F, in which old is replaced by new. old must depend on the previous integration variable of the integrals in F and new must depend on the new integration variable. For more information, see Integration by Substitution.
When specifying the integrals in F, you can return the unevaluated form of the integrals by using the int function with the 'Hold' option set to true. You can then use changeIntegrationVariable to show the steps of integration by substitution.
## Examples
collapse all
Apply a change of variable to the definite integral ${\int }_{\mathit{a}}^{\mathit{b}}\mathit{f}\left(\mathit{x}+\mathit{c}\right)\text{\hspace{0.17em}}\mathit{dx}$.
Define the integral.
syms f(x) y a b c
F = int(f(x+c),a,b)
F =
${\int }_{a}^{b}f\left(c+x\right)\mathrm{d}x$
Change the variable $\mathit{x}+\mathit{c}$ in the integral to $\mathit{y}$.
G = changeIntegrationVariable(F,x+c,y)
G =
${\int }_{a+c}^{b+c}f\left(y\right)\mathrm{d}y$
Find the integral of $\int \mathrm{cos}\left(\mathrm{log}\left(\mathit{x}\right)\right)\mathit{dx}$ using integration by substitution.
Define the integral without evaluating it by setting the 'Hold' option to true.
syms x t
F = int(cos(log(x)),'Hold',true)
F =
$\int \mathrm{cos}\left(\mathrm{log}\left(x\right)\right)\mathrm{d}x$
Substitute the expression log(x) with t.
G = changeIntegrationVariable(F,log(x),t)
G =
$\int {\mathrm{e}}^{t} \mathrm{cos}\left(t\right)\mathrm{d}t$
To evaluate the integral in G, use the release function to ignore the 'Hold' option.
H = release(G)
H =
$\frac{{\mathrm{e}}^{t} \left(\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)\right)}{2}$
Restore log(x) in place of t.
H = simplify(subs(H,t,log(x)))
H =
$\frac{\sqrt{2} x \mathrm{sin}\left(\frac{\pi }{4}+\mathrm{log}\left(x\right)\right)}{2}$ | {
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$\frac{\sqrt{2} x \mathrm{sin}\left(\frac{\pi }{4}+\mathrm{log}\left(x\right)\right)}{2}$
Compare the result to the integration result returned by int without setting the 'Hold' option to true.
Fcalc = int(cos(log(x)))
Fcalc =
$\frac{\sqrt{2} x \mathrm{sin}\left(\frac{\pi }{4}+\mathrm{log}\left(x\right)\right)}{2}$
Find the closed-form solution of the integral $\int \mathit{x}\text{\hspace{0.17em}}\mathrm{tan}\left(\mathrm{log}\left(\mathit{x}\right)\right)\mathit{dx}$.
Define the integral using the int function.
syms x
F = int(x*tan(log(x)),x)
F =
$\int x \mathrm{tan}\left(\mathrm{log}\left(x\right)\right)\mathrm{d}x$
The int function cannot find the closed-form solution of the integral.
Substitute the expression log(x) with t. Apply integration by substitution.
syms t
G = changeIntegrationVariable(F,log(x),t)
G =
The closed-form solution is expressed in terms of hypergeometric functions. For more details on hypergeometric functions, see hypergeom.
Compute the integral ${\int }_{0}^{1}{\mathit{e}}^{\sqrt{\mathrm{sin}\left(\mathit{x}\right)}}\mathit{dx}$ numerically with high precision.
Define the integral ${\int }_{0}^{1}{\mathit{e}}^{\sqrt{\mathrm{sin}\left(\mathit{x}\right)}}\mathit{dx}$. A closed-form solution to the integral does not exist.
syms x
F = int(exp(sqrt(sin(x))),x,0,1)
F =
${\int }_{0}^{1}{\mathrm{e}}^{\sqrt{\mathrm{sin}\left(x\right)}}\mathrm{d}x$
You can use vpa to compute the integral numerically to 10 significant digits.
F10 = vpa(F,10)
F10 = $1.944268879$
Alternatively, you can use the vpaintegral function and specify the relative error tolerance.
Fvpa = vpaintegral(exp(sqrt(sin(x))),x,0,1,'RelTol',1e-10)
Fvpa = $1.944268879$
The vpa function cannot find the numerical integration to 70 significant digits, and it returns the unevaluated integral in the form of vpaintegral.
F70 = vpa(F,70)
F70 = $1.944268879138581167466225761060083173280747314051712224507065962575967$ | {
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F70 = vpa(F,70)
F70 = $1.944268879138581167466225761060083173280747314051712224507065962575967$
To find the numerical integration with high precision, you can perform a change of variable. Substitute the expression $\sqrt{\mathrm{sin}\left(\mathit{x}\right)}$ with $\mathit{t}$. Compute the integral numerically to 70 significant digits.
syms t;
G = changeIntegrationVariable(F,sqrt(sin(x)),t)
G =
${\int }_{0}^{\sqrt{\mathrm{sin}\left(1\right)}}\frac{2 t {\mathrm{e}}^{t}}{\sqrt{1-{t}^{4}}}\mathrm{d}t$
G70 = vpa(G,70)
G70 = $1.944268879138581167466225761060083173280747314051712224507065962575967$
## Input Arguments
collapse all
Expression containing integrals, specified as a symbolic expression, function, vector, or matrix.
Subexpression to be substituted, specified as a symbolic scalar variable, expression, or function. old must depend on the previous integration variable of the integrals in F.
New subexpression, specified as a symbolic scalar variable, expression, or function. new must depend on the new integration variable.
collapse all
### Integration by Substitution
Mathematically, the substitution rule is formally defined for indefinite integrals as
$\int f\left(g\left(x\right)\right)\text{\hspace{0.17em}}g\text{'}\left(x\right)\text{\hspace{0.17em}}dx={\left(\int f\left(t\right)\text{\hspace{0.17em}}dt\right)|}_{t=g\left(x\right)}$
and for definite integrals as
$\underset{a}{\overset{b}{\int }}f\left(g\left(x\right)\right)\text{\hspace{0.17em}}g\text{'}\left(x\right)\text{\hspace{0.17em}}dx=\underset{g\left(a\right)}{\overset{g\left(b\right)}{\int }}f\left(t\right)\text{\hspace{0.17em}}dt.$ | {
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# On the integral $\int_0^\pi\sin(x+\sin(x+\sin(x+\cdots)))\,dx$
This question came into my head when I did a course on Fourier series. However, this is not an infinite sum of sines, but an infinite recurrence of sines in a sum.
Consider $$f_1(x)=\sin(x)$$ and $$f_2(x)=\sin(x+f_1(x))$$ such that $$f_n$$ satisfies the relation $$f_n(x)=\sin(x+f_{n-1}(x)).$$ To what value does $$L:=\lim_{n\to\infty}\int_0^\pi f_n(x)\,dx$$ converge?
Since it is impossible to evaluate the integrals directly, we begin by considering the first few values of $$n$$. A pattern clearly emerges. $$I_1=\int_0^\pi f_1(x)\,dx=2\quad\quad\quad I_2=1.376527...\\I_3=2.188188...\quad\quad\quad\quad\quad I_4=1.625516...\\ I_5=2.179090...\quad\quad\quad\quad\quad I_6=1.732942...\\ I_7=2.155900...\quad\quad\quad\quad\quad I_8=1.927035...$$
For odd values of $$n$$, $$I_n$$ decreases monotonically (except $$n=1$$) and for even values of $$n$$, $$I_n$$ increases monotonically. These two observations have led me to claim that $$L=I_1=2$$.
Is it possible to prove/disprove this claim?
• I feel like a numerical DE approach: let $y=\sin(x+\sin(x+\cdots))$ so that $y=\sin(x+y),$ differentiating, and using Mathematica's NDSolve command should produce a result, but it has numerical instabilities, unfortunately. – Adrian Keister Jul 11 '19 at 17:53
Outline:
• Use the inverse function of $$y=x-\sin x$$ to express $$f_\infty(x)$$.
• Use integral of inverse functions and dominated convergence theorem to prove $$L=2$$.
Claim:$$L=2.$$
Proof: Obviously $$y=t-\sin t$$ is injective on $$t\in[0,\pi]$$.
Define $$y=\operatorname{Sa}(t)$$ as the inverse function of $$y=t-\sin t$$ on $$t\in[0,\pi]$$. Therefore, $$t-\sin t =x \implies t=\operatorname{Sa}(x).$$ | {
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Assume $$f_\infty(x)$$ exists (see 1. the first integral), then we have \begin{align*} f_\infty&=\sin(x+f_\infty)\\ \underbrace{(x+f_\infty)}_{t}-\sin\underbrace{(x+f_\infty)}_{t}&=x\\ x+f_\infty&=\operatorname{Sa}(x)\\ f_\infty(x)&=-x+\operatorname{Sa}(x). \end{align*}
Since $$0-\sin 0 =0\implies \operatorname{Sa}(0)=0$$ and $$\pi-\sin \pi =\pi\implies \operatorname{Sa}(\pi)=\pi$$, \begin{align*} \int_0^\pi f_\infty(x)\,\mathrm dx&=\int_0^\pi -x+\operatorname{Sa}(x)\,\mathrm dx\\ &=\int_0^\pi -x\,\mathrm dx+\int_0^\pi \operatorname{Sa}(x)\,\mathrm dx\\ &=-\frac{\pi^2}2+\left(\pi \operatorname{Sa}(\pi)-0 \operatorname{Sa}(0)-\int_{\operatorname{Sa}(0)}^{\operatorname{Sa}(\pi)}y-\sin y\,\mathrm dy\right)\\ &=-\frac{\pi^2}2+\left(\pi^2-\int_0^\pi y-\sin y\,\mathrm dy\right)\\ &=-\frac{\pi^2}2+\left(\pi^2-\left[\frac{y^2}2+\cos y\right]^\pi_0\right)\\ &=2. \end{align*}
Here we used integral of inverse functions: $$\int_c^df^{-1}(y)\,\mathrm dy+\int_a^bf(x)\,\mathrm dx=bd-ac.$$
Note: Since $$|f_n(x)|\le 1$$ and $$1$$ is integrable on $$[0,\pi]$$, we could interchange limit sign and integral sign from dominated convergence theorem, that is, $$L:=\lim_{n\to\infty}\int_0^\pi f_n(x)\,\mathrm dx=\int_0^\pi \lim_{n\to\infty}f_n(x)\,\mathrm dx=\int_0^\pi f_\infty(x)\,\mathrm dx=2.$$
• What a marvelous answer! – Szeto Oct 6 '18 at 22:29
• That is amazing! I've posted a similar question, but this time with multiplication. Do you have any thoughts on it? Cheers. – TheSimpliFire Oct 7 '18 at 9:19 | {
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1. ## taylor series
Can anyone help me with how to get the Taylor series of a function(for example sinx)?
Furthur how to get a value for sin(for example pi/4) with an particular error?
Thank You
2. Originally Posted by Dili
Can anyone help me with how to get the Taylor series of a function(for example sinx)?
Furthur how to get a value for sin(for example pi/4) with an particular error?
Thank You
The Taylor series for a function $f(x)$ (at least I presume you are talking about a single variable function) to N + 1 terms about a point x = a has the form:
$f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + \frac{1}{2!}f^{\prime \prime}(a)(x - a)^2 + ~ ... ~ + \frac{1}{N!}f^{(N)}(a)(x - a)^N$
or in summation notation:
$f(x) \approx \sum_{k = 0}{N}\frac{1}{k!}f^{(k)}(a)(x - a)^k$
For the error estimate I will refer you here as I have forgotten which one is "standard" (if any.)
So as an example, let $f(x) = sin(x)$ and let us expand the series about the point $x = \frac{\pi}{4}$.
We have:
$f(x) = sin(x) \implies f \left ( \frac{\pi}{4} \right ) = sin \left ( \frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$
$f^{\prime}(x) = cos(x) \implies f^{\prime} \left ( \frac{\pi}{4} \right ) = cos \left ( \frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$
$f^{\prime \prime}(x) = -sin(x) \implies f^{\prime \prime} \left ( \frac{\pi}{4} \right ) = -sin \left ( \frac{\pi}{4} \right ) = -\frac{\sqrt{2}}{2}$
$f^{\prime \prime \prime}(x) = -cos(x) \implies f^{\prime \prime \prime} \left ( \frac{\pi}{4} \right ) = -cos \left ( \frac{\pi}{4} \right ) = -\frac{\sqrt{2}}{2}$
So the Taylor series to 4 terms looks like:
$sin(x) \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \left ( x - \frac{\pi}{4} \right ) - \frac{\sqrt{2}}{4} \left ( x - \frac{\pi}{4} \right )^2 - \frac{\sqrt{2}}{12} \left ( x - \frac{\pi}{4} \right )^3$
-Dan
3. ## lagrange's remainder
Thank you | {
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-Dan
3. ## lagrange's remainder
Thank you
But what if we take the point x=0, where you get the familiar Taylor series expansion of sinx? Is it the same as taking any value(pi/4)?
And can you explain with the Lagrange's remainder?
4. Originally Posted by Dili
Thank you
But what if we take the point x=0, where you get the familiar Taylor series expansion of sinx? Is it the same as taking any value(pi/4)?
And can you explain with the Lagrange's remainder?
I'll have to let someone else explain the remainder (I'm just not up on it myself, which is why I posted the link. )
Yes, you could use the Maclaurin expansion (the Taylor series expansion about the point x = 0), but note that $\frac{\pi}{4} \approx 0.785398$ is not particularly close to 0. The further x is from 0, the greater the error in the approximation.
On the other hand using 3 terms in the expansion I get
$sin(x) \approx x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5$
Gives me
$sin \left ( \frac{\pi}{4} \right ) \approx 0.707143$
which is only off by $5.129 \times 10^{-3}$ %, which is pretty close by most people's judgment.
-Dan
5. Theorem: Let $f(x)$ be an infinitely differenciable function on $(a,b)$ (with $a<0). Let $T_n(x)$ represent the $n$-th degree Taylor polynomial around the origin for the point $x\in (a,b)$ (with $x\not = 0$). And let $R_{n+1}(x) = f(x) - T_{n}(x)$ be the remainder term. Then there exists a number $y$ strictly between $0$ and $x$ so that,
$R_{n+1}(x) = \frac{f^{(n+1)}(y)}{(n+1)!}\cdot x^{n+1}$.
So given $f(x) = \sin x$ let us work on the interval $(a,b)$ where $a=-\infty$ and $b=+\infty$. This function is infinitely differenciable so the above results apply. You can to approximate $T_n \left( \frac{\pi}{4} \right)$. Now by the theorem we know that,
$R_{n+1}\left( \frac{\pi}{4} \right) = \frac{f^{(n+1)} \left( \frac{\pi}{4} \right)}{(n+1)!} \cdot \left( \frac{\pi}{4} \right)^{n+1}$ for some $0. | {
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Notice that $f^{n+1}$ is one of these: $\sin x,\cos x,-\sin x,-\cos x$. Thus, $|f^{n+1}|\leq 1$. And also notice that $\frac{\pi}{4} \leq 1$ thus, $\left(\frac{\pi}{4}\right)^{n+1} \leq 1$.
This means,
$\left| R_{n+1}\left( \frac{\pi}{4}\right) \right| \leq \frac{1}{(n+1)!}$.
This approximation might be greatly improved all I did was place an approximation that works and furthermore converges rapidply.
Remark) The theorem applys in more general to $(n+1)$-differentiable functions but there was no need to use it here. | {
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# Calculate the probability that Juniors are the majority of random selection.
In a business school at ABC University, 1/6 of the enrollment are freshmen, 3/10 are sophomores, 1/3 are juniors and 1/5 are seniors. They compete with several other schools within their region in an annual financial case competition.
Three students were randomly selected to form a team for this year’s competition.
Calculate the probability that the juniors would make up the majority of the team.
I thought that this probability is equal to the probability that exactly 2 out 3 selected are juniors plus the probability that all 3 are juniors. Therefore, I computed the following:
$$P(exactly~2~juniors) = \frac13 \times \frac13 \times \frac23 = \frac2{27}$$ $$P(3~juniors) = (\frac13)^3$$ Therefore, $$P(juniors~make~up~the~majority)= \frac1{27} + \frac2{27} = \frac3{27}$$
My answer is wrong as I have only partially computed the probability. I would appreciate some explanation as what I am missing as I am having trouble figuring it out. Thank you
Your formula $\frac{1}{3}\cdot\frac{1}{3}\cdot \frac{2}{3}$ to pick the two juniors is the probability of first selecting two juniors, and then the non-junior. You can also first pick a junior, then a non-junior, and then a junior again, the probability of which is: $\frac{1}{3}\cdot\frac{2}{3}\cdot \frac{1}{3}$. And, of course, you can first pick the non-junior, and then the two juniors, with a probability of $\frac{2}{3}\cdot\frac{1}{3}\cdot \frac{1}{3}$.
All these give you 2 juniors and 1 non-junior, so the probability of picking exactly 2 juniors is:
$\frac{1}{3}\cdot\frac{1}{3}\cdot \frac{2}{3}+\frac{1}{3}\cdot\frac{2}{3}\cdot \frac{1}{3}+\frac{2}{3}\cdot\frac{1}{3}\cdot \frac{1}{3} = \frac{6}{27}=\frac{2}{9}$ | {
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• I thought that the order in which I choose the students doesn't matter. It obviously does per your solution (thank you!). – the boy 88 Jul 13 '17 at 20:59
• Can I think of my sample space as all combinations of 3-student tuples? And the event I want is choosing exactly 2 juniors and 1 non-junior so the outcomes that satisfy the event are { JJSo, JJF, and JJSe}? So = sophomore, Se = senior, F= freshman. – the boy 88 Jul 13 '17 at 21:07
• @Kovs95 Yes, you can. And you can see them as unordered triples (basically sets with 3 elements), or as ordered triples. If seeing them as unordered, you need to multiply by 3 which is what sds did.. if you see them as ordered you get the formula I got. Same result of course! Now, there is really no need to distinguish between freshmen and sophomores and senior ... you just need to distinguish between junior (1/3 chance) and non-juniors (2/3 chance) – Bram28 Jul 13 '17 at 21:22
You can select the non-junior in 3 different ways (1st, 2nd, 3rd), so
$$P(exactly~2~juniors) = \frac13 \times \frac13 \times \frac23 \times 3 = \frac29$$
Thus
$$P(juniors~ make~ up~ the~ majority)=\frac29+\frac1{27}=\frac{7}{27}$$
• yes sorry I fixed the 2/3 probability of not selecting a junior. – the boy 88 Jul 13 '17 at 20:43
• yes, but this is not the only problem - see "times 3"! – sds Jul 13 '17 at 20:44 | {
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# What are the last two digits of 63*35*37*82*71*41?
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What are the last two digits of 63*35*37*82*71*41?
(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
Kudos for a correct solution. | {
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(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
Kudos for a correct solution.
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
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13 Apr 2015, 06:27
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Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?
(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:
We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.
Remainder of (63*35*37*82*71*41)/ 100
Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:
Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.
So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.
We need the
Remainder of (63*7*37*41*71*41*5*2)/10*5*2
Remainder of (63*7*37*41*71*41)/10
Now using concept 2, let’s write the numbers in form of multiples of 10
Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10 | {
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Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10
Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.
When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.
Answer (D)
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
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07 Apr 2015, 07:49
1
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?
(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
Kudos for a correct solution.
63*35 = 05 *37 = 85 * 82 = 70 * 71 = 70*41 = 70
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
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07 Apr 2015, 10:36
4
First, multiplied 82 and 35, since 2 and 5 will give a 0 in the end. Got last 2 digits 70.
no need to multiply 70 by 41 and 71, since they wont change the last 2 digits (try if you want to, but effective multiplication in case of 41 and 71 is only with 1).
Now, 70 * 3(of 63) gives : last 2 digits as 10.
lastly, 10 * 7(of 37) gives : last 2 digits 70.
Thus D.
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
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09 Apr 2015, 12:31
63*35*37*82*71*41=
We have to focus on the last two digits only, so 63*35=05*37=85*82=70
71*41=81 therefore 81*70=70
Hence Answer is D | {
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
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19 Jul 2016, 10:10
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1
Bunuel wrote:
What are the last two digits of (63)(35)(37)(82)(71)(41)?
(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
(63)(35)(37)(82)(71)(41) = (63)(37)(71)(41)(35)(82)
Multiply the units digits in the red product to get: (#####1)(35)(82)
Rewrite blue product as follows: (#####1)(7)(5)(2)(41)
Rearrange: (#####1)(7)(41)(5)(2)
The units digit of (7)(41) is 7, so we get: (#####1)(##7)(10)
= (####7)(10)
= #####70
Answer:
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
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04 Feb 2018, 23:10
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?
(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
Kudos for a correct solution.
The equation can be written as $$7*9*7*5*37*41*2*71*41$$
$$49*9*10*37*41*71*41$$ = $$401*10*37*41*41*71$$
As you can see, the last digit will be 0, and except for 37 all the digits end with 1.
So, last two Digit will be 70.
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
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07 Feb 2018, 10:30
Bunuel wrote:
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?
(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:
We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.
Remainder of (63*35*37*82*71*41)/ 100
Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note: | {
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Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.
So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.
We need the
Remainder of (63*7*37*41*71*41*5*2)/10*5*2
Remainder of (63*7*37*41*71*41)/10
Now using concept 2, let’s write the numbers in form of multiples of 10
Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10
Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.
When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.
Answer (D)
Hi I thinks this
"Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70."
same as 1470/100 = 70 right ?
also Remainder 3*-3*-3*1*1*1/10 seem more efficient than 3*7*7*1*1*1/10 ? I hope this way is correct too ?
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
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### Show Tags
07 Feb 2018, 12:56
Bunuel wrote:
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?
(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:
We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.
Remainder of (63*35*37*82*71*41)/ 100
Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:
Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.
So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.
We need the
Remainder of (63*7*37*41*71*41*5*2)/10*5*2
Remainder of (63*7*37*41*71*41)/10
Now using concept 2, let’s write the numbers in form of multiples of 10
Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10
Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70. | {
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When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.
Answer (D)
Hi bunuel ..
Can you enlighten how to approach this kind more other questions like to find last three digit..etc etc..
And in that solution to find last two digit how to approach if nothing is there in divisor means if 10 also divides over there...
Sent from my BND-AL10 using GMAT Club Forum mobile app
Manager
Joined: 02 Jan 2016
Posts: 117
Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]
### Show Tags
09 Feb 2018, 07:37
viv007 wrote:
Bunuel wrote:
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?
(A) 10
(B) 30
(C) 40
(D) 70
(E) 80
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:
We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.
Remainder of (63*35*37*82*71*41)/ 100
Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:
Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct. | {
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So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.
We need the
Remainder of (63*7*37*41*71*41*5*2)/10*5*2
Remainder of (63*7*37*41*71*41)/10
Now using concept 2, let’s write the numbers in form of multiples of 10
Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10
Remainder of 3*7*7*1*1*1/10
Remainder of 147/10 = 7
Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.
When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.
Answer (D)
Hi bunuel ..
Can you enlighten how to approach this kind more other questions like to find last three digit..etc etc..
And in that solution to find last two digit how to approach if nothing is there in divisor means if 10 also divides over there...
Sent from my BND-AL10 using GMAT Club Forum mobile app
Hi Please see this
https://gmatclub.com/forum/compilation- ... ml#p650870
Re: What are the last two digits of 63*35*37*82*71*41? &nbs [#permalink] 09 Feb 2018, 07:37
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# What are the last two digits of 63*35*37*82*71*41?
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# Zero sum game, constant sum game
Given any bilateral zero-sum game G, show that strategy profile σ is a Nash equilibrium for G if, and only if, it is a Nash equilibrium for the constant-sum game G' obtained from G by adding any fixed amount "d" to the payoffs of both players. Is the conclusion affected if the fixed amount, call it now $d_i$ for each i = 1 , 2 , differs between the two players?
Suppose $(\sigma_1^*, \sigma_2^*)$ is the Nash Equilibrium of the game $G$ consisting of players $\{1,2\}$ having strategy sets $A_1$ and $A_2$ and the payoff functions $u_1:A_1\times A_2 \rightarrow \mathbb{R}$ and $u_2:A_1\times A_2 \rightarrow \mathbb{R}$. Consider the game $G'$ in which everything else is same as $G$ except that the payoff functions are $v_1:A_1\times A_2 \rightarrow \mathbb{R}$ and $v_2:A_1\times A_2 \rightarrow \mathbb{R}$ defined as: $v_1(a_1, a_2) = u_1(a_1, a_2) + d_1$ and $v_2(a_1, a_2) = u_2(a_1, a_2) + d_2$. We will now show that $(\sigma_1^*, \sigma_2^*)$ is also the Nash Equilibrium of $G'$.
Since $(\sigma_1^*, \sigma_2^*)$ is the Nash Equilibrium of $G$, \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*)) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2)) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}
The above can also be rewritten as \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*)) + d_1 \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*)) + d_1 & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*)) + d_2 \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2)) + d_2 & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*} | {
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Because $d_1$ and $d_2$ are constants, we can take them inside the expectation and rewrite the above inequalities as: \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*) + d_1) \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*) + d_1) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*) + d_2) \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2) + d_2) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}
By definition of $v_1$ and $v_2$,
\begin{eqnarray*} \mathbb{E}(v_1(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(v_1(\sigma_1, \sigma_2^*)) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(v_2(\sigma_1^*, \sigma_2^*)) \geq \mathbb{E}(v_2(\sigma_1^*, \sigma_2)) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}
Therefore, $(\sigma_1^*, \sigma_2^*)$ is also the Nash Equilibrium of $G'$.
Let me borrow from my answer here.
The short answer is that adding constants values, even ones that differ across players, will not change the set of Nash equilibria.
This is easy to see from the definition of Nash equilibrium.
Let $u_i$ represent player $i$'s payoffs as a function of all the players' strategies. A strategy profile $\sigma= (\sigma_i , \sigma_{-i})$, where $\sigma_i$ is $i$'s strategy, and $\sigma_{-i}$ is a vector of all the other players' strategies, is a Nash equilibrium if, for each player $i$,
$$u_i (\sigma_i , \sigma_{-i}) \ge u_i (\sigma_i^\prime ,\sigma_{-i})$$
for any other strategy $\sigma_i^\prime$ of player $i$.
Let $v_i = u_i + d_i$, for each $i$, and let $\sigma$ be a Nash equilibrium strategy profile under $u_i$. Since adding a constant to both sides will preserve the inequality above, it must also be the case that
$$v_i (\sigma_i , \sigma_{-i}) \ge v_i (\sigma_i^\prime ,\sigma_{-i})$$
for all $i$ and $\sigma_i^\prime$, so $\sigma$ is a Nash equilibrium strategy profile under the modified set of payoff functions as well. | {
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In fact, it is a standard assumption in game theory that payoff functions represent preferences that satisfy the von Neumann-Morgenstern axioms. This implies that preference orderings are preserved by affine transformations of utility functions. (In this context, an affine transformation is just multiplication by a positive number and adding arbitrary constants to payoff functions.)
• Why answer the exact same question on two SE sites, instead of flagging one for being crossposted? – Giskard May 27 '17 at 6:44
Other answers took the game theoretic approach, I will take the utility approach. Adding a real number $d_i$ to all payoffs of a player is an affine transformation of said payoffs. This does not change the player's preferences even when dealing with expected payoffs. Thus the desirability of outcomes is unchanged from the player's perspective hence best responses remain the same. | {
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