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3. Energy# 3.1. Work# How much work do you need to do to move a box? The answer depends on two things: how heavy the box is, and how far you have to move it. Multiply the two, and youโ€™ve got a good measure of how much work will be required. Of course, work can be done in other contexts as well - pulling a spring from equilibrium, or cycling against the wind. In each case, thereโ€™s a force and a displacement. To be fair, we will only count the part of the force that is in the direction of the displacement (when cycling, you donโ€™t do work due to the fact that thereโ€™s a gravitational force pulling you down, since you donโ€™t move vertically; you do work because thereโ€™s a drag force due to your moving through the air). We define work as the product of the component of the force in the direction of the displacement, times the displacement itself. We calculate this component by projecting the force vector on the displacement vector, using the dot product (see Section 15.1.1 for an introduction to vector math): \[ W = \bm{F} \cdot \bm{x}. \] Note that work is a scalar quantity - it has a magnitude but no direction. Work is measured in Joules (J), with one Joule being equal to one Newton times one meter. Of course the force acting on our object need not be constant everywhere. Take for example the extension of a spring: the further you pull, the larger the force gets, as given by Hookeโ€™s law (2.7). To calculate the work done when extending the spring, we chop up the path (here a straight line) into many small pieces. For each piece, we approximate the force by the average value on that piece, then multiply with the length of the piece and sum. In the limit that we have infinitely many pieces, this approximation becomes exact, and the sum becomes an integral: for one dimension, we thus \[ W = \int_{x_1}^{x_2} F(x) \mathrm{d}x. \] Likewise, the path along which we move need not be a straight line. If the path consists of multiple straight segments, on each of which the force is constant, we can calculate the total work by adding the work done on the different segments. Taking the limit to infinitely many infinitesimally small segments \(\mathrm{d}\bm{r}\), on each of which the force is given by the value \(\bm{F}(\bm {r})\), the sum again becomes an integral: \[ W = \int_{\bm{r}_1}^{\bm{r}_2} \bm{F}(\bm{r}) \cdot \mathrm{d}\bm{r}. \] Equation (3.3) is the most general version of the definition of work; it simplifies to (3.2) for movement along a straight line, and to (3.1) if both the path is straight and the force constant^[1]. In general, the work done depends on the path taken - for example, itโ€™s more work to take a detour when biking from home to work, assuming the air drag is the same everywhere. However, in many important cases the work done in getting from one point to another depends on the endpoints only. Forces for which this is true are called conservative forces. As weโ€™ll see below, the force exerted by a spring and that exerted by gravity are both conservative. Sometimes we will not be interested in how much work is done in generating a certain displacement, but over a certain amount of time - for instance, a generator generates work by getting something to move, like a wheel or a valve, but we donโ€™t typically care about those details, we want to know how much work we can expect to get out of the generator, i.e., how much power it has. Power is defined as the amount of work per unit time, or \[ P = \frac{\mathrm{d}W}{\mathrm{d}t}. \] Power is measured in Joules per second, or Watts (W). To find out how much work is done by an engine that has a certain power output, we need to integrate that output over time: \[ W = \int P \mathrm{d}t. \] 3.2. Kinetic energy# Newtonโ€™s first law told us that a moving object will stay moving unless a force is acting on it - which holds for moving with any speed, including zero. Now if you want to start moving something that is initially at rest, youโ€™ll need to accelerate it, and Newtonโ€™s second law tells you that this requires a force - and moving something means that youโ€™re displacing it. Therefore, there is work involved in getting something moving. We define the kinetic energy (\(K\)) of a moving object to be equal to the work required to bring the object from rest to that speed, or equivalently, from that speed to rest: \[ K = \frac12 m v^2. \] Because the kinetic energy is equal to an amount of work, it is also a scalar quantity, has the same dimension, and is measured in the same unit. The factor \(v^2\) is the square of the magnitude of the velocity of the moving object, which you can calculate with the dot product: \(v^2 = \bm{v} \cdot \bm{v}\). You may wonder where equation (3.6) comes from. Newtonโ€™s second law tells us that \(\bm {F} = m \mathrm{d}\bm{v}/\mathrm{d}t\), relating the force to an infinitesimal change in the velocity. In the definition for work, equation (3.3), we multiply the force with an infinitesimal change in the position \(\mathrm{d}\bm{r}\). That infinitesimal displacement takes an infinitesimal amount of time \(\mathrm{d}t\), which is related to the displacement by the instantaneous velocity \(\bm {v}\): \(\mathrm{d}\bm{r} = \bm{v} \mathrm{d}t\). We can now calculate the work necessary to accelerate from zero to a finite speed: \[ K = \int \bm{F} \cdot \mathrm{d}\bm{r} = \int m \frac{\mathrm{d}\bm{v}}{\mathrm{d}t} \cdot \bm{v} \mathrm{d}t = \int m \bm{v} \cdot \frac{\mathrm{d}\bm{v}}{\mathrm{d}t} \mathrm{d}t = \int m \bm{v} \cdot \mathrm{d}\bm{v} = \frac{m}{2} \int \mathrm{d}(\bm{v} \cdot \bm{v}) = \frac12 m v^2, \] where we used that the dot product is commutative and the fact that the integral over the derivative of a function is the function itself. Of course, now that we know that the kinetic energy is given by equation (3.6), we no longer need to use a complicated integral to calculate it. However, because the kinetic energy is ultimately given by this integral, which is equal to a net amount of work, we arrive at the following statement, sometimes referred to as the Work-energy theorem: the change in kinetic energy of a system equals the net amount of work done on or by it (in case of increase/decrease of \(K\)): \[ \Delta K = W_\mathrm{net}. \] Gabrielle ร‰milie Le Tonnelier de Breteuil, marquise du Chรขtelet (1706-1749) Gabrielle ร‰milie Le Tonnelier de Breteuil, marquise du Chรขtelet (1706-1749), known as ร‰milie du Chรขtelet, was a French mathematician and physicist (then known as natural philosopher), who made important contributions to the development of the concept of (kinetic) energy. She translated Newtonโ€™s Principia into French, and wrote an extensive commentary on it, in which she first postulated the law of conservation of energy, for which she introduced the new concept of kinetic energy. Inspired by experiments first done by โ€˜s Gravesande, which she repeated and analyzed, she discovered that a ball dropped from a given height \(h\) would make an indentation in a piece of soft clay with a depth proportional to the height the ball was dropped from. At the time, most people, including Newton, considered energy to be equivalent to momentum (and thus proportional to velocity); had they been correct, the depth of the indentation should be proportional to \(\sqrt{h}\) instead. Du Chรขteletโ€™s work showed this to be incorrect, postulating instead that kinetic energy is proportional to the square of the velocity. ร‰milie du Chรขtelet was born in the French nobility, corresponded with people across Europe, married at age 18, and had a long-term friendship with Voltaire, with whom she collaborated extensively in her work on mathematics and physics. She published several books, often initially anonymously to avoid sexist prejudices, which found their way to salons and universities of the time. Her translation of the Principia is still the standard French version. She died in childbirth at age 42. 3.3. Potential energy# We already encountered conservative forces in Section 3.1. The work done by a conservative force is (by definition) path-independent; that means that in particular the work done when moving along any closed path^[6] must be zero: \[ \oint \bm{F} \cdot \mathrm{d}\bm{r} = 0. \] For a conservative force, we can thus define a potential energy difference between points 1 and 2 as the work necessary to move an object from point 1 to point 2: \[ \Delta U_{12} = - \int_{\bm{r}_1}^{\bm{r}_2} \bm{F} \cdot \mathrm{d}\bm{r}. \] Note the minus sign in the definition - this is a choice of course, and youโ€™ll see below why we made this choice. Note also that the potential energy is defined only between two points. Often we will choose a convenient reference point and calculate the potential energy at any other point with respect to that point. The reference point is typically either the origin or infinity, if the force happens to be zero at either of these. Letโ€™s suppose we have set such a point, and know the potential energy difference with that point at any other point in space - this defines a (scalar) function \(U(\bm{r})\). If we now want to know the force acting on a particle at \(\bm{r}\), all we need to do is take the derivative of \(U(\bm{r})\) - that is to say the gradient in three dimensions (which simplifies to the ordinary derivative in one dimension): \[ \bm{F}(\bm{r}) = - \bm{\nabla} U(\bm{r}). \] Equation (3.11) is extremely useful, as it gives us a means to calculate the force, which is a vector quantity, from the potential energy function, which is a scalar quantity - and therefore much simpler to work with. For instance, since energies are scalars, they can simply be added, as weโ€™ll do in the next section, whereas for forces you need to do vector addition. Equation (3.11) also reflects that we are free to choose a reference point for the potential energy, since the force does not change if we add a constant to the potential energy. 3.3.1. Gravitational potential energy# We saw in Section 2.2.2 that for low altitudes, the gravitational force is given by \(\bm{F}_g = m \bm{g}\), where \(\bm{g}\) is a vector of constant magnitude \(g\approx 9.81 \mathrm{m}/\mathrm{s}^2 \) and always points down. Therefore, the gravitational force does no work when you move horizontally, and if you first move up and then the same amount down again, it doesnโ€™t do any net work either, as the two contributions exactly cancel. \(\bm{F}_g\) is therefore an example of a conservative force, and we can define and calculate the gravitational potential energy \(U_g\) between a point at height \(0\) (our reference point) and one at height \(h\): \[ U_\mathrm{g}(h) = - \int_{z=0}^{z=h} m (-g) \mathrm{d}z = m g h. \] Note that by choosing a minus sign in the definition of the potential energy, we end up with a positive value of the energy here. What about larger distances, i.e., Newtonโ€™s law of gravity, equation (2.9)? Well, there the distances are measured radially, so any movement perpendicular to the radial direction doesnโ€™t matter, and if you move out and back in again, the net work done is zero, so by the same reasoning as before we again have a conservative force. This force vanishes at infinity, so it makes sense to set that as a reference point - though notice that that will make our potential energy always negative in this case: \[ U_\mathrm{G}(r) = - \frac{G M m}{r} \] where \(r\) is the distance between \(m\) and \(M\), and \(M\) sits at the origin. Of course we can also calculate gravitational potential differences between two distances \(r_1\) and \(r_2\) from \ (M\): \(\Delta U_\mathrm{G}(r_1, r_2) = G M m \left(\frac{1}{r_1} - \frac{1}{r_2}\right)\). 3.3.2. Spring potential energy# Like the gravitational force, the Hookean spring force (2.7) also depends on displacement alone, and by the same reasoning is conservative (notice the pattern?). Calculating its associated potential energy is straightforward, and taking the equilibrium position of the spring as the reference point, we find: \[ U_\mathrm{s}(x) = \frac12 k x^2. \] The minus sign in Hookeโ€™s Law gives us a positive spring potential energy. Note that \(x\) stands for displacement here; as we only consider one-dimensional springs the 1D-version is sufficient. 3.3.3. General conservative forces# In the case of the gravitational and spring force it was easy to reason that they had to be conservative. It is also easy to see that the friction force is not conservative: if you take a longer path, you need to do more net work against friction, which you can moreover never recover as mechanical energy. For more complicated systems, especially in three dimensions, it may not be so easy to see whether a force is conservative. Fortunately, there is an easy test you can perform: if the curl of a force is zero everywhere, it will be a conservative force, or expressed mathematically: \[ \bm{\nabla} \times \bm{F} = 0 \quad \Leftrightarrow \quad \oint \bm{F} \cdot \mathrm{d}\bm{r} = 0 \quad \Leftrightarrow \quad \bm{F} = - \bm{\nabla} U. \] Is is straightforward to show that if a force is conservative, its curl must vanish: a conservative force can be written as the gradient of some scalar function \(U(\bm{x})\), and \(\bm{\nabla} \ times \bm{\nabla} U(\bm{x}) = 0\) for any function \(U(\bm{x})\), as you can easily check for yourself. The proof the other way around is more complicated, and can be found in advanced mechanics 3.4. Conservation of energy# Work, kinetic energy and potential energy are all quantities with the same dimension - so we can do arithmetic with them. One particularly useful quantity is the total energy \(E\) of a system, which is simply the sum of the kinetic and potential energy: (Law of conservation of energy) If all forces in a system are conservative, the total energy in that system is conserved. Proof. For simplicity, weโ€™ll look at the 1D case (3D goes analogously). Conserved means not changing in time, so in order to prove the statement, we only need to calculate the time derivative of \(E \) and check that it is always zero. \[\begin{split}\begin{align*} \frac{\mathrm{d}E}{\mathrm{d}t} &= \frac{\mathrm{d}K}{\mathrm{d}t} + \frac{\mathrm{d}U}{\mathrm{d}t} \\ &= \frac{\mathrm{d}\left( \frac12 m v^2 \right)}{\mathrm{d}t} + \ frac{\mathrm{d}U}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}t} \\ &= m v \frac{\mathrm{d}v}{\mathrm{d}t} - F v \\ &= - \left(F - m \frac{\mathrm{d}v}{\mathrm{d}t}\right) v \\ &= 0, \end{align*}\end where the last equality holds because of Newtonโ€™s second law. Conservation of energy means that the total energy of a system cannot change, but of course the potential and kinetic energy can - and by conservation of total energy we know that they get converted directly into one another. Exploiting this fact will allow us to analyze and easily solve many problems in classical mechanics - this conservation law is an immensely useful tool. Note that conservation of energy is not the same as the work-energy theorem of Section 3.2. For the total energy to be conserved, all forces need to be conservative. In the work-energy theorem, this is not the case. You can therefore calculate changes in kinetic energy due to the work done by non-conservative forces using the latter. 3.5. Energy landscapes# In the previous section we proved that the total energy is conserved. In the section before that, we looked at potential energies. Typically, the potential energy is a function of your position in space. When we plot it as a function of spatial coordinates, we get an energy landscape, measuring an amount of energy on the vertical axis. Of course we can also plot the total energy of the system - and since that is conserved, it is the same everywhere, and thus becomes a horizontal line or plane. Because kinetic energy cannot be negative, any point where the potential energy is higher than the total energy is not allowed: the system cannot reach this point. When the potential energy equals the total energy, the kinetic energy (and thus the speed) has to be zero. Whenever the potential energy is lower than the total energy, there is a positive kinetic energy and thus a positive speed. Probably the simplest energy landscape is that of the harmonic oscillator (mass on a spring) - itโ€™s a simple parabola. The point at which the horizontal line representing the total energy crosses the parabola corresponds to the extrema of the oscillation: these are its turning points. The bottom of the parabola is its midpoint, and you can immediately see that thatโ€™s where the kinetic energy (and thus the speed) will be highest. Of course you can have more complex energy landscapes than that. In particular, you can have a landscape with multiple extrema, see for example Fig. 3.3. A particle that is being acted upon by forces described by this potential energy, follows a trajectory in this landscape, which can be visualized as a ball rolling over the hills and valleys of the landscape. Think back to the harmonic oscillator example. If we let go of a ball in a parabolic vase at some point on the slope, the ball will roll down and pick up speed, then roll up the opposite slope and lose speed, until it reaches the same height where its speed will again be zero. The same is true in more complicated landscapes. Particularly interesting are local maxima. If you put a ball exactly on top of one of them, it will stay there - it is a fixed point, but an unstable one, as any arbitrarily small perturbation will push it down. If you let go of a ball at a level above a local maximum, it may hop over it to the next minimum, but if your initial position (your initial energy) was too low, your ball can get stuck oscillating about a local minimum - a metastable point. Show code cell source Hide code cell source import plotly.graph_objects as go import numpy as np from myst_nb import glue def U(x): return (x**2)*(x**2 - 2)*(x + 1)*(x - 2) def dU(x): return x*(6*x**4 - 5*x**3 - 16*x**2 + 6*x + 8) # Derivative of U(x) x = np.linspace(-2, 2.5, 100) y = U(x) fig = go.Figure(layout=go.Layout(template='simple_white')) #fig = go.Figure() # Have hover text for each point on function U(x) be equal to U'(x). dU_values = dU(x) hover_texts = [f'<i>U</i>\'(<i>x</i>) = {val:.3f}' for val in dU_values] # Add function trace for potential energy U(x) function_trace = go.Scatter( name='Potential energy function', legendgroup='function', # Assign to its own legend group so it is always visible and non-toggleable from the legend showlegend=False, # Hide from legend text=hover_texts, # Set the hover text for each point hoverinfo='text' # Display only the hover text # Add equilibrium points extreme_x = { 'Unstable equilibria': [-0.635, 0.942], # Local max 'Metastable equilibria': [-1.254, 0], # Local min 'Globally stable equilibrium': [1.779] # Global min extreme_y = {etype: [U(x_val) for x_val in x_values] for etype, x_values in extreme_x.items()} extrema_traces = [] for etype, color in [ ('Unstable equilibria', 'green'), ('Metastable equilibria', 'orange'), ('Globally stable equilibrium', 'red')]: extremum_dict = {'Unstable equilibria': 'Local maximum', 'Metastable equilibria': 'Local minimum', 'Globally stable equilibrium': 'Global minimum'} hover_text = f'<i>U</i>\'(<i>x</i>) = 0\n{extremum_dict[etype]}' trace = go.Scatter( marker=dict(color=color, size=10), #title_text='A potential energy landscape and its equilibrium points', yaxis=dict(range=[-3, 4]), xaxis_title_font = dict(size=24, family='Times New Roman'), yaxis_title_font = dict(size=24, family='Times New Roman') # fig.show() # Save graph to load in figure later (special Jupyter Book feature) glue("InteractiveEnergyLandscape", fig, display=False)
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What Size Is A 250-Watt Solar Panel? - Climatebiz Over the past two decades, advancements in solar technology have allowed manufacturers to cater to homeownersโ€™ needs by providing them with panels of various sizes. One such solar panel size that is still in use is the 250-watt module. Since 250-watt solar panels are relatively affordable, small, and easy to find, they are ideal for residential rooftop installations. As such, knowing the average size of a 250-watt solar panel can really come in handy. With this information, you can determine the number of 250-watt solar panels you can fit in your desired space, ultimately determining how much energy your system will generate. In this article, we provide you with the average dimensions of a 250W solar panel. Additionally, we cover the number of 250-watt solar panels you require and how much energy they can produce. Climatebiz experts design, research, fact-check & edit all work meticulously. Affiliate Disclaimer Climatebiz is reader-supported. We may earn an affiliate commission when you buy through links on our site. What Size Is A 250-Watt Solar Panel? The standard size of a 250-watt solar panel is approximately 17.5 ft^2 (1.62 m^2); its dimensions are 65 x 39 inches (about 1651 x 991 mm). The standard size of solar panels for residential applications (they usually range from 250W to 360W). Source: Climatebiz However, these numbers can vary slightly depending on the manufacturer, model, and technology used. Weโ€™ve prepared a table that shows the dimensions of five different 250 W solar panels from different brands: Brand Model Dimensions (Inches) Dimensions (mm) LG 250S1C-G2 64.25 x 38.82 x 1.65 in 1632 x 986 x 42 mm Renogy RNG โ€“ 250P 65.0 x 39.0 x 1.6 in 1640 x 992 x 40 mm Sunpower SPR-X20-250-BLK 61.4 x 31.4 x 1.8 in 1559 x 798 x 46 mm Canadian Solar CS6P-250P 64.5 x 38.7 x 1.57 in 1638 x 983 x 40 mm Trina Solar TSM-250PA05 64.95 ร— 39.05 ร— 1.37 in 1650 ร— 992 ร— 35 mm Average โ€”โ€”โ€“ 64.02 x 37.38 x 1.59 in 1624 x 987 x 39 mm Average dimensions of 250W solar panels, based on the dimensions of 250 W solar panels from five different brands. Though the dimensions of these 250-Watt solar panels differ from brand to brand, they donโ€™t deviate much from the standard size previously mentioned. Related Reading: Solar Panel Dimensions Chart How Many 250 Watt Solar Panels Do I Need? The number of 250- watt solar panels youโ€™ll need depends on: โ€ข The amount of electricity you wish to generate (based on your average energy consumption); โ€ข Your roof space; and โ€ข Efficiency losses due to your homeโ€™s location, roof angles, available sunlight, etc. Nine 60-cell solar modules mounted on the roof of a house on a sunny day. Source: smart-energy.com So, to determine the number of 250-watt solar panels you require, youโ€™ll need to estimate your energy demand, account for efficiency losses, and determine the number of panels (of this size) you can fit in the space available. Estimating Your Energy Consumption There are several ways to estimate your energy demand, but the easiest is simply checking your monthly electric bill โ€” it will tell you the amount of kWh you consume per month. However, energy consumption changes throughout the year, so you should consider your average annual energy consumption. To do this, add the kilowatt-hours consumed each month of the previous year and divide this number by 12. According to the U.S. Energy Information Administration, the average annual electricity consumption for a U.S. residential utility customer is around 10,715 kilowatt-hours (kWh), approximately 30 kWh per day. Determining The Number Of Solar Panels You Require If you want your solar system to supply all the energy you consume, you must install an array that can generate the same amount of energy you consume daily. Example: Letโ€™s consider an energy consumption of 25 kWh and 5 peak sun hours: First, deduct the efficiency losses (about 20%); this means a 250W solar panel will generate: 250 W โ€“ 20% (due to efficiency losses) = 200 W Now, divide the daily energy consumption by the hours of direct sunlight: 25.000 Wh / 5 hours = 5000 W Finally, divide this number by the power that one 250W solar panel can generate (after deducting efficiency losses): 5000 W / 200 W = 25 solar panels Checking If You Can Fit That Many Panels On Your Roof You donโ€™t really need an elaborate solar panel square footage calculator to determine the number of panels a roof can support (unless your roof has multiple faces or odd shapes). There is an easy way to calculate the usable area of your roof: 1) Calculate The Square Footage: You first need to calculate the square footage of your roof using the length and width dimensions. 2) Account For Setback: Then, you need to account for the โ€œsetback,โ€ which is the free space between the solar arrayโ€™s edge and the roofโ€™s edge. Doing this establishes an unobstructed pathway around your rooftop in case responders (like firefighters) need to access your home in an emergency. The minimum solar panel setback varies from state to state, but usually, itโ€™s about 25% of your roofโ€™s space. Therefore, you need to multiply the square footage of your roof by 0.75 to account for the required solar setback. Example: Letโ€™s say the area of your roof that gets the most sunlight during the day is around 700 ft^2; your roofs usable area would be: Square footage x 0.75 = 700 ft^2 x 0.75 = 525 ft^2 3) Divide The Usable Roof Area By The Standard Size Of A 250 Watt Solar Panel: Now that youโ€™ve calculated the usable area of your roof, you need to divide it by 17.5 ft^2, the average square footage of the standard 250-watt solar panel size. The resulting number is the maximum number of 250-watt solar panels you can fit on your homeโ€™s roof. Example: If the usable area is 525 ft^2, then: Usable area / standard size of solar panel = 525 ft^2 / 17.5 ft^2 = 30 Therefore, you could fit thirty 250-watt solar panels on your roof. If this is enough to generate the energy you need, great! If not, youโ€™ll need more powerful solar panels (300 W, 350W, or even 400W). Usually, thirty 250-watt solar modules can generate more than enough energy to power an average U.S. home. How Much Energy Can A 250-Watt Solar Panel Produce? On average, a 250 W solar panel can produce approximately 1 kWh of energy per day (considering 4 peak sun hours and 80% efficiency). However, this number can vary significantly. Why? Because the amount of energy produced by solar panels depends on multiple factors, such as: โ€ข Location โ€ข Orientation (is the panel facing north, east, west, or south?) โ€ข Tilt/angle (relative to the ground/roof) โ€ข Shading โ€ข Time of year/season โ€ข Ambient conditions (temperature, humidity, etc.) โ€ข Type of solar panel (monocrystalline, polycrystalline, thin-film) โ€ข Power rating โ€ข Panel efficiency Shading can considerably decrease solar energy production. Source: onestepoffthegrid.com.au Solar radiation map of the United States. Source: NREL However, we need to clarify several things: Solar panels are rated based on how much power they can generate. Letโ€™s recap the difference between power and energy before discussing the energy output of a 250-watt solar panel. In short, energy is the ability to do work. It has various forms: thermal, chemical, nuclear, mechanical, electrical, etc. Electrical energy results from the movement of electrically charged particles, like electrons. The unit for electrical energy is Watt-hours (Wh). In other words, energy expresses the power used/generated over a period of time: Energy (Wh) = power (W) x time (h) In physics, power is the amount of energy transferred or converted per unit of time. In other words, it is the rate at which work occurs. Power (W) = energy (Wh) / time (h) It can also be calculated as the product of amperage and voltage: Power (W) = amperage (A) x voltage (V) The unit for power is watts (W). Now that weโ€™ve covered how energy and power correlate, we can better understand solar panel energy output. Solar Panel Energy Output A solar panel consists of several solar cells wired in series. Each cell produces a specific voltage and current (depending on its size). The power rating of a solar panel expresses the maximum power the panel can produce (in ideal conditions). It is usually tested under โ€œStandard Test Conditionsโ€ or STC. This doesnโ€™t mean the panel will produce 250W of power at all times. It means that it can produce a maximum of 250W under ideal conditions hence why itโ€™s so tricky to calculate the energy output of a solar panel. There are, however, a few ways to estimate solar energy production. The best way is to use an online energy production calculator, like the PVWatts calculator developed by researchers at NREL (National Renewable Energy Laboratory). Another way is to use the following formula: Energy produced per day = Average peak sun hours ร— solar panel wattage x 80% efficiency Please note: the 80% used in the formula accounts for all the efficiency losses that result from the previously mentioned factors. Example: For a 250 W solar panel, considering 4.5 peak sun hours: Energy produced per day = 4.5 hours ร— 250 W x 0.8 = 900 Wh = 0.9 kWh In summary, the energy produced by a 250-watt solar panel varies depending on how many hours of direct sunlight hits the panel. In addition, several other factors result in efficiency losses. Therefore, a 250W solar panel doesnโ€™t produce 250 watt-hours of energy per hour; the amount is always a bit less. Final Thoughts The standard size of a 250-watt solar panel is around 17.5 ft^2 (1.62 m^2), and dimensions are usually 65 x 39 inches (approximately 1651 x 991 mm). Solar panels with a maximum power output of 250 watts are ideal for residential installations. Their relatively small size makes them easy to install and fit on most roofs. They are also affordable and easy to find. You must consider several factors to build the optimum solar power system for your needs. For instance, your average energy consumption, how much usable roof area you have available for the solar array, the location of your home, the angle of your roof, the solar irradiance in your part of the country, the efficiency of the solar panels youโ€™re using, and a variety of other factors.
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Oliver Heaviside A while back I mentioned that I recently found out that Heaviside was responsible for a bunch of mathematical techniques I've known since my training for the Cambridge entrance exam. I decided to read more about Heaviside and I've just finished book on the Victorian mathematical physicist, Oliver Heaviside. There's a bit of information about Heaviside on the web, but I thought I'd mention two highlights from this book that may hint at why he was a genius ahead of his time. Operational Calculus and Distortionless Transmission There's an example of Heaviside style operational calculus in the I posted to above. One of the reasons I became interested in this subject again is that I was getting into electronics and I wanted to simplify computations of properties of simple linear circuits. I had this crazy idea that capacitors and inductors could be treated as resistors whose resistance is differential operator valued. Turns out that this wasn't an original move. This is exactly what Heaviside did well over 100 years ago and it was the secret weapon he used for much of his work. He could solve a wide array of ordinary and partial differential equations with ease. Very briefly, his idea was to write the differential operator d/dx as the symbol and then treat much like a conventional algebraic variable. He turned differential equations into ordinary algebraic equations. A great example of this was when he studied the electrical signal that emerges from a long cable as a function of what was sent into the other end. If W is the outgoing signal, and V is the incoming signal, he showed that in his model, W = โˆš(A+Bp)/โˆš(C+Dp) V, for some constants A, B, C and D, that depend on the properties of the cable. At first sight this is meaningless - what is the meaning of square root of a differential operator ? Heaviside had ways to deal with these things, but that's not what he did here. He noticed that if he picked A, B, C and D such that A/B=C/D then he could cancel the from top and bottom. The net effect was that if this condition held, the signal emerging was the same as the signal entering (apart from a time delay). In physical terms this meant that adding inductance to a long cable would allow it to carry the signal without distorting it. His contemporaries had been declaring long-distance telegraphy impossible because inductance would distort the signal, but here was Heaviside suggesting that inductors be . The British Post Office ignored Heaviside's claims and it was left to a in the US to put his ideas into practice - ideas that today formed the backbone for the nascent global telecommunications industry. Heaviside couldn't even get much of his work published because mathematicians like (boo! hiss!) rejected it as unrigorous. Needless to say, Heaviside died a bitter neglected old man... Foreshadowings of Special Relativity I'm fascinated by some of the theoretical clues about relativity that were appearing before Einstein. There were obvious results like the Michelson-Morley experiment and the Lorentz-Fitzgerald contraction proposed to explain it. But there were clues in other places too. HG Wells, in The Time Machine, said "There is no difference between Time and any of the three dimensions of Space except that our consciousness moves along it", so we already have a popularisation of the idea of a symmetry between space and time. Heaviside spent much of his time working with Maxwell's equations (which should really be called Heaviside's equations) which inherently has Lorentz group symmetry . This means that any physical predictions made from Maxwell's equations must also have Lorentz invariance. As nobody had explicitly recognised this as a symmetry of nature at the time, it meant for some unusual seeming results. For example, at the time Heaviside was working, the notion that the electromagnetic field stored energy was becoming popular. Heaviside compared the field of a static charge and a moving charge and noticed that for the same charge, the latter stored more energy. This meant that to accelerate a charge required putting extra energy into it which would go into the field. In other words, a charge should feel like it has more mass than it has. The apparent mass contained a familiar 1/โˆš(1-vยฒ/cยฒ) factor and so he noticed that this mass increase grew as the charge's velocty approached that of light. In particular he noticed that the mass would become infinite at the speed of light, exactly as predicted by Special Relativity. Heaviside was never deterred by anything as trivial as an infinity so he went on to study the properties of superluminal particles and predicted and derived the properties of what should be called Heaviside radiation (BTW When Heaviside tried to study the geometry of the field around a moving spherical charge he initially made a few mistakes that were eventually fixed by someone else using Heaviside's own methods correctly. One thing that was noted was that the spherical symmetry was flattened. Yet another hint of Lorentz-Fitzgerald contraction.) I'd love to also say something about Heaviside's battles with because they are highly entertaining. But instead, I just recommend reading the for yourself.
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Provincial Piks 18 Dec on the 7th day of Christmas my...... lottery numbers will come in thanks, brad for noticing 7 odd start numbers in a row. reminds me of the time i played red/black at casino roulette..... playing the same as last draw, and these also: 40 boards, 10 each with 7,17,27 or 47, then, 2,12/2,22/2,32/12,32/12,42 then, 28 or 48 then, mixing up LD's 1 and 5........ would you or combomaniac still talk to me if i threw in 5 dollars of quick piks ...??? I will talk to you Daleks Yeah Daleks, I will talk to you even if you play QP..I still play pre-selected numbers and it's almost the same as QP's.. To Brad, My lottoBuddy told me that his friend matched 4/9 winning $900.00 that was on his own. So on Dec 28 we will let him pick 4 then LottoBuddy picks 3 then Me 3 of course....Although I don't like that way but it's a good measure to know how he thinks.. Apparantly he wins on Sports action too.....But he loses too specially on Combo 8.... I am going with 2 lines : Not too much hope but if I get $10.00 I would be very happy indeed daleks said: ...>Snip< ... would you or combomaniac still talk to me if i threw in 5 dollars of quick piks ...??? Go ahead, make my day The 25,43 show both as Oon and Rep, 36 has a good double hit record ( 24 and 44 will NOT threepeat, I will not allow it !), although we're only likely to see one of each - Oon and Rep that is. Other nrs I'm considering: 20-23-30-33. What are the odds of another odd nr in the first position? Can I borrow a cup of intuition from someone? Good Luck Players !! Here for B.C./49 you should give a shot at 24-25-43-44 expect 2 to repeat from there maybe 3...And play at least 2 numbers under 10...And that 09 is quite a candidate pals.... Anyhow....Good luck to all of you down here! welcome down under ... 9 is a good choice, I'll play it and drop the 1. Have 25-43 already, not keen on 24-44 tho. Thanks for dropping by, see you've all been busier than a hive in your $5M quest, I might borrow your hints there and play a few lines too. Good Luck !! my picks same here 2,4,6,7, 13,18,20,22,31,42,46,48,49 good luck to all Hey Daleks keep playing Sport Action cause my co-worker here has won $375.00 betting on 4 games only after he had spent $20.00 !! Last week he won $78.00 after spending $2.00 !! Better odds. QE49 ... 01-13-26-29-35-42 B40 BC49 ... 11-17-20-31-33-46 B16 Certainly miss rarget here tonite.... it's just that I didn't play Combo but hey savn money is nice too... ok here is an update..my Lottoboddy's friend actually won the lotto 8 years ago, his share was $250,000.00 with three other ppl. Now I am not sure whether it was 6/49 or BC/49.. No wonder these days he's been playing like rich man and all kinds of lotteries !! It looks as if wer gonna play Combo 10 in 3 days !! So which 3 numbers I should pick, it's a hard task. Well Dennis, all is not lost yet, Combo got 2 lines with 2/6 each ... not bad at all ... could have easily cashed in on a 4/12 !! Daleks hasn't reported in from his hideaway yet, he called the 17 and LD 1s, could hear the che-ching there ... and let's not forget the 10 QPs Your suggestions were statistically sound, unfortunatelly BC49 has been in an 'intuitive holding pattern' for a while now, I'm hoping it gets cleared out of there soon so I get a kick at the can too ... my intuition sucks Cookie ... didn't see your post untill now, looks like you're getting closer to the bulls eye again ! IS IT THE DAY TOMORROW ? Tomorrow is another day of Reckoning as Combo 10 is in effect and I hope we get 3/10 so there is no losses. Where is Daleks ? When Are you gonna take the jump and play Combo 8 or 9 ? Remember when you visit Vancouver to drop me a line so we can arrange something to play together Combo 9 or lots AW 12's !! Hey ComboMan, daleks is taking it easy on the island and I think he doesn't have ready access to the web from there, I'm sure he'll be back after Xmas. I'll be busy for a few days don't know if there'll be much time to crunch my own so I'll probably plagiarise picks ... post some nrs !! Best of luck on your Recky 10 !! rum and egg nog, no way i'll stick to scotch and beer...........made it over to the big island shopping, and hanging out for today and saturday......no wins on wednesday, and how about that stupid odd number again....i'll get some good picks together for saturday.............hey combo, sounds good what you're doing........is it a safe bet to go with some of the most drawn numbers this year in your combo pick ????? don't have them in front of me, but those numbers drawn the most in 2002 seem to come up an awful lot more than they should..................cheers.......
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Posts tagged โ€˜algorithmsโ€™ So, even before I get to the what the algorithm does (which is quite interesting on its own, even before you find out how it accomplishes that), the very first intriguing part is its name. Itโ€™s officially referred to as the Blum-Floyd-Pratt-Rivest-Tarjan Algorithm. Is your mouth watering yet? I thought so.^1 However, CLRS refers to it as the Select Algorithm (see section 9.3), so thatโ€™s what Iโ€™m going to call it. If you have an unsorted array of n items, itโ€™s a worst-case O(n) algorithm for finding the i^th largest item (often the median, since thatโ€™s a commonly used statistic). Hereโ€™s how the algorithm works: 1. Put all the elements of the input into groups of 5 (with one group at the very end that could have less than 5 elements, of course). Find the median of each group. 2. Recursively use the Select algorithm to find the median of all the group-of-5 medians from step 1. Call this median of medians M. 3. Take all the items from the input and partition them into the group of items less than M and the group of items bigger than M. 4. If M is the i^th biggest, thatโ€™s the answer, and weโ€™re done. If itโ€™s not, you know which partition contains the i^th biggest item, and you can recursively call Select on that group (with a different i if the group itโ€™s in is the group of items smaller than M). To give an example of that last statement, suppose weโ€™re looking for the 65th biggest number from a group of 100 numbers, and we find that M is the 53rd largest. There are 47 numbers smaller than M, and when we recurse, weโ€™ll now be looking for the 12th largest element of that set of 47. If instead M had been the 68th largest number, in our recursive step weโ€™d look at the 67 numbers larger than M and continue to look for the 65th largest one. Now, I claimed the running time is linear, and Iโ€™d better back up that claim. The trick is to consider how many items get thrown out in that last step. Half the group-of-5 medians are larger than M, and half are smaller. Without loss of generality, suppose the i^th largest item is in the group that is larger than M. Now, not only do we know that we can henceforth ignore the half of the group-of-5 medians smaller than M, for each of those medians we can ignore the two items in the group of 5 that were smaller than that median. In other words, we can eliminate at least 3/10 of the elements we were searching over (we can eleminate 3/5 of the items in the groups whose median is smaller than M, and half of all the groups fit this category). Sure, Iโ€™m eliding an additive constant for the items in the group with M itself (not to mention the items in that last small group at the end when n is not a multiple of 5), but thatโ€™s not going to affect the asymptotic running time. So now we can build the recurrence relation at work here. If T(n) is the amount of work done to perform Select on n items, we have T(n) = O(n) + T(n/5) + T(7n/10) That is to say, we do O(n) work in steps 1 and 3, we do T(n/5) work in step 2, and we do T(7n/10) work in step 4 because we know at least 3/10 of the elements are in the wrong partition. Iโ€™ll leave it as an exercise to the reader to draw out the recursion tree and find the actual amount of work done, but I assure you T(n) is in O(n). The crazy part here is that magic number 5: If you do this with 3 items per group, the running time is O(n log n). If you do it with more than 5, the running time is still linear, but slower (you spend more time in step 1 finding all the initial medians). The optimal number of items per group really is 5. Nifty! [1] For those of you not in the know, these are all really famous algorithms names. Blum is the co-inventor of the CAPTCHA. Floyd is from the Floyd-Warshall all-pairs shortest-paths algorithm, which is a staple in introductory algorithms courses. Pratt is from the Knuth-Morris-Pratt string matching algorithm, which was groundbreaking when it was new but is nearly obsolete now. Rivest is the R in RSA encryption, one of the most widely used encryption schemes around. Tarjan is the co-creator of splay trees, a kind of self-balancing tree with the handy property that commonly accessed items float to the top. Blum, Floyd, Rivest, and Tarjan have each won the Turing Award, which is like the Nobel Prize of computer science. Admittedly, I canโ€™t think of an uncontrived use case when I would prefer this to a hash table, union find, or traditional bit field, but I still think this is a pretty cool concept. Itโ€™s a data structure to hold n items, with constant time initialization (i.e. you canโ€™t initialize all n items at the beginning), and constant time random access. Moreover, it can be used as a set: on top of the constant time random access, you can store and remove items as well as get the number of items in it in constant time, and you can iterate through all the inserted items in O(n) time (where n is the number of items currently stored in the set; not the maximum number of items it can have). and on top of that, if you donโ€™t need to call the itemsโ€™ destructor (for instance, youโ€™re storing doubles or ints), you can empty the set in constant time! The only drawback to using it as a set is that it takes memory proportional to the size of the keyspace (for instance, if youโ€™re trying to store items indexed by b-bit numbers, it will take O(2^b) space). However, it can be used as a bit vector: to store u bits, it takes a mere O(u) space, and you donโ€™t need to zero all the bits out at the beginning! This implies that if you only end up using a sparse subset of the bits, you donโ€™t have to waste time initializing all the bits you didnโ€™t use. To be fair, the constant hidden behind that O(u) space is bigger than the one in a normal bit vector, but I think itโ€™s still a pretty neat idea. Besides, space is cheap these days; people already often use a byte or four for every bit Anywho, hereโ€™s a great write-up of it, along with the original 1993 paper that introduces it (see the top of page 4 to compare the running times of this to a traditional bit field). On top of what is written in those links, the space required can be cut in half by only having a single array instead of two: just have it point to other parts of itself. Insertion/deletion gets a little more complicated, but it totally works. Neat! This year, the Algorithms course at Mudd is using a new book, written by Dasgupta, Papadimitriou (the complexity theorist), and Vazirani (the quantum computing guy). You can view a July 2006 draft of the book, which actually looks pretty nice. It doesnโ€™t cover all that I think an algorithms course should (in particular it handles the standard data structures cursorily at best), but it definitely skips the boring bits and goes right to the fun parts. It starts out with cryptography, then goes into matrix multiplication and the FFT (which sound boring, but are actually pretty neat problems). It even has a chapter on quantum computing! I donโ€™t think this will ever replace CLRS in terms of completeness, but itโ€™s a lot more fun. Definitely worth a read if youโ€™re looking for an algorithms textbook. and really, who isnโ€™t looking for one these days? Define a โ€œhubโ€ to be a vertex that shares an edge with every other vertex (such as the middle of a star graph, or any vertex in a complete graph). Suppose we have the adjacency matrix of an undirected, unweighted graph with V vertices (so our input has size V^2). Find an algorithm with running time o(V^2) that can determine whether a hub exists in the graph, or prove that no such algorithm exists. Note the use of little-o notation: the algorithm must be asymptotically faster than (big) O(V^2). My bet is that no such algorithm exists, but I canโ€™t figure out how to prove it. Iโ€™ll be at Mudd tomorrow, btw. I wrote an integer factorisation program (Java bytecode also available for those without a compiler) using an algorithm I just made up, and it works surprisingly well (significantly better than brute force, not nearly as good as the best-known algorithms out there today). Yes, it still has exponential running time, but I thought it was a neat idea. Someone show me a way to get outa here, โ€˜Cause I constantly pray Iโ€™ll get outa here Please, wonโ€™t somebody say Iโ€™ll get outa here Someone gimme my shot or Iโ€™ll rot here. Iโ€™ll do I dunno what to get outa [Mudd] But a hell of a lot to get outa [Mudd] People tell me thereโ€™s not a way outa [Mudd] But believe me I gotta get outa [Mudd now!] Iโ€™m pulling my third all-nighter of the week (well, not quite โ€“ Iโ€™ve gotten at least an hour of sleep every day), and I donโ€™t expect to have less work until Wednesday afternoon, when I will only have a 15-minute presentation and a 20-page paper left to do. I am so ready to graduate right now! However, the Big Algorithms presentation on Tuesday went just about perfectly (Reid and I discussed pattern matching, and in particular the Knuth-Morris-Pratt and Boyer-Moore algorithms): people seemed to understand everything we covered, we answered questions well, and they asked leading questions (this was awesome โ€“ on 3 separate occasions, we answered questions with โ€œactually, thatโ€™s covered on the next slideโ€ฆโ€). Our timing was perfect down to about 2 minutes of the 75-minute lecture, which blew me away because weโ€™d never actually practiced the entire thing before. Moreover, Reid (who has a speech impediment where he stops talking when he gets flustered) went through his half perfectly, without any long pauses. On the feedback forms, one person even wrote that this was the best student lecture of the entire class. \/\/00T! Right. Back to workโ€ฆ
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Random-access machine In computer science, random-access machine (RAM) is an abstract machine in the general class of register machines. The RAM is very similar to the counter machine but with the added capability of 'indirect addressing' of its registers. Like the counter machine, The RAM has its instructions in the finite-state portion of the machine (the so-called Harvard architecture). The RAM's equivalent of the universal Turing machine โ€“ with its program in the registers as well as its data โ€“ is called the random-access stored-program machine or RASP. It is an example of the so-called von Neumann architecture and is closest to the common notion of a computer. Together with the Turing machine and counter-machine models, the RAM and RASP models are used for computational complexity analysis. Van Emde Boas (1990) calls these three plus the pointer machine "sequential machine" models, to distinguish them from "parallel random-access machine" models. Introduction to the model The concept of a random-access machine (RAM) starts with the simplest model of all, the so-called counter machine model. Two additions move it away from the counter machine, however. The first enhances the machine with the convenience of indirect addressing; the second moves the model toward the more conventional accumulator-based computer with the addition of one or more auxiliary (dedicated) registers, the most common of which is called "the accumulator". Formal definition A random-access machine (RAM) is an abstract computational-machine model identical to a multiple-register counter machine with the addition of indirect addressing. At the discretion of instruction from its finite state machine's TABLE, the machine derives a "target" register's address either (i) directly from the instruction itself, or (ii) indirectly from the contents (e.g. number, label) of the "pointer" register specified in the instruction. By definition: A register is a location with both an address (a unique, distinguishable designation/locator equivalent to a natural number) and a content โ€“ a single natural number. For precision we will use the quasi-formal symbolism from Boolos-Burgess-Jeffrey (2002) to specify a register, its contents, and an operation on a register: โ€ข [r] means "the contents of register with address r". The label "r" here is a "variable" that can be filled with a natural number or a letter (e.g. "A") or a name. โ€ข โ†’ means "copy/deposit into", or "replaces", but without destruction of the source Example: [3] +1 โ†’ 3; means "The contents of source register with address "3", plus 1, is put into destination register with address "3" (here source and destination are the same place). If [3]=37, that is, the contents of register 3 is the number "37", then 37+1 = 38 will be put into register 3. Example: [3] โ†’ 5; means "The contents of source register with address "3" is put into destination register with address "5". If [3]=38, that is, the contents of register 3 is the number 38, then this number will be put into register 5. The contents of register 3 are not disturbed by this operation, so [3] continues to be 38, now the same as [5]. Definition: A direct instruction is one that specifies in the instruction itself the address of the source or destination register whose contents will be the subject of the instruction. Definition: An indirect instruction is one that specifies a "pointer register", the contents of which is the address of a "target" register. The target register can be either a source or a destination (the various COPY instructions provide examples of this). A register can address itself indirectly. For want of a standard/convention this article will specify "direct/indirect", abbreviated as "d/i", as a parameter (or parameters) in the instruction: Example: COPY ( d, A, i, N ) means directly d get the source register's address (register "A") from the instruction itself but indirectly i get the destination address from pointer-register N. Suppose [N]=3, then register 3 is the destination and the instruction will do the following: [A] โ†’ 3. Definition: The contents of source register is used by the instruction. The source register's address can be specified either (i) directly by the instruction, or (ii) indirectly by the pointer register specified by the instruction. Definition: The contents of the pointer register is the address of the "target" register. Definition: The contents of the pointer register points to the target register โ€“ the "target" may be either a source or a destination register. Definition: The destination register is where the instruction deposits its result. The source register's address can be specified either (i) directly by the instruction, or (ii) indirectly by the pointer register specified by the instruction. The source and destination registers can be one. Refresher: The counter-machine model Melzak (1961) provides an easy visualization of a counter machine: its "registers" are holes in the ground, and these holes hold pebbles. Per an instruction, into and out of these holes "the computer" (person or machine) adds (INCrements) or removes (DECrements) a single pebble. As needed, additional pebbles come from, and excess pebbles go back into, an infinite supply; if the hole is too small to accommodate the pebbles the "computer" digs the hole bigger. Minsky (1961) and Hopcroft-Ullman 1979 (p. 171) offer the visualization of a multi-tape Turing machine with as many left-ended tapes as "registers". Each tape's length is unbounded to the right, and every square is blank except for the left end, which is marked. The distance of a tape's "head" from its left end, measured in numbers of tape-squares, represents the natural number in "the register". To DECrement the count of squares the tape head moves left; INCrement it moves right. There is no need to print or erase marks on the tape; the only conditional instructions are to check to see if the head is at the left end, by testing a left-end mark with a "Jump-if-marked instruction". The following instruction "mnemonics" e.g. "CLR (r)" are arbitrary; no standard exists. The register machine has, for a memory external to its finite-state machine โ€“ an unbounded (cf: footnote|countable and unbounded) collection of discrete and uniquely labelled locations with unbounded capacity, called "registers". These registers hold only natural numbers (zero and positive integers). Per a list of sequential instructions in the finite state machine's TABLE, a few (e.g. 2) types of primitive operations operate on the contents of these "registers". Finally, a conditional-expression in the form of an IF-THEN-ELSE is available to test the contents of one or two registers and "branch/jump" the finite state machine out of the default instruction-sequence. Base model 1: The model closest to Minsky's (1961) visualization and to Lambek (1961): โ€ข { INCrement contents of register r, DECrement contents of register r, IF contents of register r is Zero THEN Jump to instruction I[z] ELSE continue to next instruction }: Instruction Mnemonic Action on register(s) "r" Action on finite state machine's Instruction Register, IR INCrement INC ( r ) [r] + 1 โ†’ r [IR] + 1 โ†’ IR DECrement DEC ( r ) [r] - 1 โ†’ r [IR] + 1 โ†’ IR Jump if Zero JZ ( r, z ) none IF [r] = 0 THEN z โ†’ IR ELSE [IR] + 1 โ†’ IR Halt H none [IR] โ†’ IR Base model 2: The "successor" model (named after the successor function of the Peano axioms): โ€ข { INCrement the contents of register r, CLeaR the contents of register r, IF contents of register r[j] Equals the contents of register r[k] THEN Jump to instruction I[z] ELSE goto to next instruction } Instruction Mnemonic Action on register(s) "r" Action on finite state machine's Instruction Register, IR CLeaR CLR ( r ) 0 โ†’ r [IR] + 1 โ†’ IR INCrement INC ( r ) [r] + 1 โ†’ r [IR] + 1 โ†’ IR Jump if Equal JE (r1, r2, z) none IF [r1] = [r2] THEN z โ†’ IR ELSE [IR] + 1 โ†’ IR Halt H none [IR] โ†’ IR Base model 3: Used by Elgot-Robinson (1964) in their investigation of bounded and unbounded RASPs โ€“ the "successor" model with COPY in the place of CLEAR: โ€ข { INCrement the contents of register r, COPY the contents of register r[j] to register r[k], IF contents of register r[j] Equals the contents of register r[k] then Jump to instruction I[z] ELSE goto to next instruction } Instruction Mnemonic Action on register(s) "r" Action on finite state machine's Instruction Register, IR COPY COPY (r1, r2) [r1] โ†’ r2 [IR] + 1 โ†’ IR INCrement INC ( r ) [r] + 1 โ†’ r [IR] + 1 โ†’ IR Jump if Equal JE (r1, r2, z) none IF [r1] = [r2] THEN z โ†’ IR ELSE [IR] + 1 โ†’ IR Halt H none [IR] โ†’ IR Creating "convenience instructions" from the base sets The three base sets 1, 2, or 3 above are equivalent in the sense that one can create the instructions of one set using the instructions of another set (an interesting exercise: a hint from Minsky (1967) โ€“ declare a reserved register e.g. call it "0" (or Z for "zero" or E for "erase") to contain the number 0). The choice of model will depend on which an author finds easiest to use in a demonstration, or a proof, etc. Moreover, from base sets 1, 2, or 3 we can create any of the primitive recursive functions ( cf Minsky (1967), Boolos-Burgess-Jeffrey (2002) ). (How to cast the net wider to capture the total and partial mu recursive functions will be discussed in context of indirect addressing). However, building the primitive recursive functions is difficult because the instruction sets are so ... primitive (tiny). One solution is to expand a particular set with "convenience instructions" from another set: These will not be subroutines in the conventional sense but rather blocks of instructions created from the base set and given a mnemonic. In a formal sense, to use these blocks we need to either (i) "expand" them into their base-instruction equivalents โ€“ they will require the use of temporary or "auxiliary" registers so the model must take this into account, or (ii) design our machines/ models with the instructions 'built in'. Example: Base set 1. To create CLR (r) use the block of instructions to count down register r to zero. Observe the use of the hint mentioned above: โ–ก CLR (r) =[equiv] โ–ก loop: JZ (r, exit) Again, all of this is for convenience only; none of this increases the model's intrinsic power. For example: the most expanded set would include each unique instruction from the three sets, plus unconditional jump J (z) i.e.: โ€ข { CLR (r), DEC (r), INC (r), CPY ( r[s], r[d] ), JZ (r, z), JE ( r[j], r[k], z ), J(z) } Most authors pick one or the other of the conditional jumps, e.g. Shepherdson-Sturgis (1963) use the above set minus JE (to be perfectly accurate they use JNZ โ€“ Jump if Not Zero in place of JZ; yet another possible convenience instruction). The "indirect" operation Example of indirect addressing In our daily lives the notion of an "indirect operation" is not unusual. Example: A treasure hunt. At location "Tom_&_Becky's_cave_in_pirate_chest" will be where we can find a map directing us to "the treasure": (1) We go to location "Tom_&_Becky's_cave..." and dig around until we find a wooden box (2) Inside the box is a map to the location of the treasure: "under_Thatcher's_front_porch" (3) We go to location "under_Thatcher's_front_porch", jackhammer away the concrete, and discover "the treasure": a sack of rusty door-knobs. Indirection specifies a location identified as the pirate chest in "Tom_&_Becky's_cave..." that acts as a pointer to any other location (including itself): its contents (the treasure map) provides the "address" of the target location "under_Thatcher's_front_porch" where the real action is occurring. Why the need for an indirect operation: Two major problems with the counter-machine model In the following one must remember that these models are abstract models with two fundamental differences from anything physically real: unbounded numbers of registers each with unbounded capacities. The problem appears most dramatically when one tries to use a counter-machine model to build a RASP that is Turing equivalent and thus compute any partial mu recursive function: โ€ข Melzak (1961) added indirection to his "hole-and-pebble" model so that his model could modify itself with a "computed goto" and provides two examples of its use ("Decimal representation in the scale of d" and "Sorting by magnitude", whether these are used in his proof that the model is Turing equivalent is unclear since "the program itself is left to the reader as an exercise" (p. 292)). Minsky (1961, 1967) was able to demonstrate that, with suitable (but difficult-to-use) Gรถdel number encoding, the register model did not need indirection to be Turing equivalent; but it did need at least one unbounded register. As noted below, Minsky (1967) hints at the problem for a RASP but doesn't offer a solution. Elgot and Robinson (1964) proved that their RASP model P[0] โ€“ it has no indirection capability โ€“ cannot compute all "recursive sequential functions" (ones that have parameters of arbitrary length) if it does not have the capability of modifying its own instructions, but it can via Gรถdel numbers if it does (p. 395-397; in particular figure 2 and footnote p. 395). On the other hand their RASP model P'[0] equipped with an "index register" (indirect addressing) can compute all the "partial recursive sequential functions" (the mu recursive functions) (p. 397-398). Cook and Reckhow (1973) say it most succinctly: The indirect instructions are necessary in order for a fixed program to access an unbounded number of registers as the inputs vary." (p. 73) โ€ข Unbounded capacities of registers versus bounded capacities of state-machine instructions: The so-called finite state part of the machine is supposed to be โ€“ by the normal definition of algorithm โ€“ very finite both in the number of "states" (instructions) and the instructions' sizes (their capacity to hold symbols/signs). So how does a state machine move an arbitrarily large constant directly into a register, e.g. MOVE (k, r) (Move constant k to register r)? If huge constants are necessary they must either start out in the registers themselves or be created by the state machine using a finite number of instructions e.g. multiply and add subroutines using INC and DEC (but not a quasi-infinite number of these!). Sometimes the constant k will be created by use of CLR ( r ) followed by INC ( r ) repeated k times โ€“ e.g. to put the constant k=3 into register r, i.e. 3 โ†’ r, so at the end of the instruction [r]=3: CLR (r), INC (r), INC (r), INC (r). This trick is mentioned by Kleene (1952) p. 223. The problem arises when the number to be created exhausts the number of instructions available to the finite state machine; there is always a bigger constant than the number of instructions available to the finite state machine. โ€ข Unbounded numbers of registers versus bounded state-machine instructions: This is more severe than the first problem. In particular, this problem arises when we attempt to build a so-called RASP, a "universal machine" (see more at Universal Turing machine) that uses its finite-state machine to interpret a "program of instructions" located in its registers โ€“ i.e. we are building what is nowadays called a computer with the von Neumann architecture. Observe that the counter machine's finite state machine must call out a register explicitly (directly) by its name/number: INC (65,356) calls out register number "65,365" explicitly. If the number of registers exceeds the capability of the finite state machine to address them, then registers outside the bounds will be unreachable. For example, if the finite state machine can only reach 65,536 = 2^16 registers then how can it reach the 65,537th? So how do we address a register beyond the bounds of the finite state machine? One approach would be to modify the program-instructions (the ones stored in the registers) so that they contain more than one command. But this too can be exhausted unless an instruction is of (potentially) unbounded size. So why not use just one "รผber-instruction" โ€“ one really really big number โ€“ that contains all the program instructions encoded into it! This is how Minsky solves the problem, but the Gรถdel numbering he uses represents a great inconvenience to the model, and the result is nothing at all like our intuitive notion of a "stored program computer". Elgot and Robinson (1964) come to a similar conclusion with respect to a RASP that is "finitely determined". Indeed it can access an unbounded number of registers (e.g. to fetch instructions from them) but only if the RASP allows "self modification" of its program instructions, and has encoded its "data" in a Gรถdel number (Fig. 2 p. 396). In the context of a more computer-like model using his RPT (repeat) instruction Minsky (1967) tantalizes us with a solution to the problem (cf p. 214, p. 259) but offers no firm resolution. He "In general a RPT operation could not be an instruction in the finite-state part of the machine ... this might exhaust any particular amount of storage allowed in the finite part of the computer [sic, his name for his RAM models]. RPT operations require infinite registers of their own." (p. 214). He offers us a bounded RPT that together with CLR (r) and INC (r) can compute any primitive recursive function, and he offers the unbounded RPT quoted above that as playing the role of ฮผ operator; it together with CLR (r) and INC (r) can compute the mu recursive functions. But he does not discuss "indirection" or the RAM model per se. From the references in Hartmanis (1971) it appears that Cook (in his lecture notes while at UC Berkeley, 1970) has firmed up the notion of indirect addressing. This becomes clearer in the paper of Cook and Reckhow (1973) โ€“ Cook is Reckhow's Master's thesis advisor. Hartmanis' model โ€“ quite similar to Melzak's (1961) model โ€“ uses two and three-register adds and subtracts and two parameter copies; Cook and Reckhow's model reduce the number of parameters (registers called out in the program instructions) to one call-out by use of an accumulator "AC". The solution in a nutshell: Design our machine/model with unbounded indirection โ€“ provide an unbounded "address" register that can potentially name (call out) any register no matter how many there are. For this to work, in general, the unbounded register requires an ability to be cleared and then incremented (and, possibly, decremented) by a potentially infinite loop. In this sense the solution represents the unbounded ฮผ operator that can, if necessary, hunt ad infinitum along the unbounded string of registers until it finds what it is looking for. The pointer register is exactly like any other register with one exception: under the circumstances called "indirect addressing" it provides its contents, rather than the address-operand in the state machine's TABLE, to be the address of the target register (including possibly itself!). Bounded indirection and the primitive recursive functions If we eschew the Minsky approach of one monster number in one register, and specify that our machine model will be "like a computer" we have to confront this problem of indirection if we are to compute the recursive functions (also called the ฮผ-recursive functions ) โ€“ both total and partial varieties. Our simpler counter-machine model can do a "bounded" form of indirection โ€“ and thereby compute the sub-class of primitive recursive functions โ€“ by using a primitive recursive "operator" called "definition by cases" (defined in Kleene (1952) p. 229 and Boolos-Burgess-Jeffrey p. 74). Such a "bounded indirection" is a laborious, tedious affair. "Definition by cases" requires the machine to determine/distinguish the contents of the pointer register by attempting, time after time until success, to match this contents against a number/name that the case operator explicitly declares. Thus the definition by cases starts from e.g. the lower bound address and continues ad nauseam toward the upper bound address attempting to make a match: Is the number in register N equal to 0? If not then is it equal to 1? 2? 3? ... 65364? If not then we're at the last number 65365 and this had better be the one, else we have a problem! "Bounded" indirection will not allow us to compute the partial recursive functions โ€“ for those we need unbounded indirection aka the ฮผ operator. Suppose we had been able to continue on to number 65367, and in fact that register had what we were looking for. Then we could have completed our calculation successfully! But suppose 65367 didn't have what we needed. How far should we continue to go? To be Turing equivalent the counter machine needs to either use the unfortunate single-register Minsky Gรถdel number method, or be augmented with an ability to explore the ends of its register string, ad infinitum if necessary. (A failure to find something "out there" defines what it means for an algorithm to fail to terminate; cf Kleene (1952) pp. 316ff Chapter XII Partial Recursive Functions, in particular p. 323-325.) See more on this in the example below. Unbounded indirection and the partial recursive functions For unbounded indirection we require a "hardware" change in our machine model. Once we make this change the model is no longer a counter machine, but rather a random-access machine. Now when e.g. INC is specified, the finite state machine's instruction will have to specify where the address of the register of interest will come from. This where can be either (i) the state machine's instruction that provides an explicit label, or (ii) the pointer-register whose contents is the address of interest. Whenever an instruction specifies a register address it now will also need to specify an additional parameter "i/d" โ€“ "indirect/direct". In a sense this new "i/d" parameter is a "switch" that flips one way to get the direct address as specified in the instruction or the other way to get the indirect address from the pointer register (which pointer register โ€“ in some models every register can be a pointer register โ€“ is specified by the instruction). This "mutually exclusive but exhaustive choice" is yet another example of "definition by cases", and the arithmetic equivalent shown in the example below is derived from the definition in Kleene (1952) p. Example: CPY ( indirect[source], r[source], direct[destination], r[destination] ) Assign a code to specify direct addressing as d="0" and indirect addressing as i="1". Then our machine can determine the source address as follows: i*[r[s]] + (1-i)*r[s] For example, suppose the contents of register 3 are "5" (i.e. [3]=5 ) and the contents of register 4 are "2" (i.e. [4]=2 ): Example: CPY ( 1, 3, 0, 4 ) = CPY ( indirect, reg 3, direct, reg 4 ) 1*[3] + 0*3 = [3] = source-register address 5 0*[4] + 1*4 = 4 = destination-register address 4 Example: CPY ( 0, 3, 0, 4 ) 0*[3] + 1*3 = 3 = source-register address 3 0*[4] + 1*4 = 4 = destination-register address 4 Example: CPY ( 0, 3, 1, 4 ) 0*[3] + 1*3 = 3 = source-register address 3 1*[4] + 0*4 = [4] = destination-register address 2 The indirect COPY instruction Probably the most useful of the added instructions is COPY. Indeed, Elgot-Robinson (1964) provide their models P[0] and P'[0] with the COPY instructions, and Cook-Reckhow (1973) provide their accumulator-based model with only two indirect instructions โ€“ COPY to accumulator indirectly, COPY from accumulator indirectly. A plethora of instructions: Because any instruction acting on a single register can be augmented with its indirect "dual" (including conditional and unconditional jumps, cf the Elgot-Robinson model), the inclusion of indirect instructions will double the number of single parameter/register instructions (e.g. INC (d, r), INC (i, r)). Worse, every two parameter/register instruction will have 4 possible varieties, e.g.: CPY (d, r[s], d, r[d] ) = COPY directly from source-register directly to destination-register CPY (i, r[sp], d, r[d] ) = COPY to destination-register indirectly using the source address to be found in the source-pointer register r[sp]. CPY (d, r[s], i, r[dp] ) = COPY contents of source-register indirectly into register using destination address to be found in the destination-pointer register r[dp]. CPY (i, r[sp], i, r[dp] ) = COPY indirectly the contents of the source register with address to be found in source-pointer register r[sp], into the destination register with address to be found in the destination-pointer register r[dp]) In a similar manner every three-register instruction that involves two source registers r[s1] r[s2] and a destination register r[d] will result in 8 varieties, for example the addition: [r[s1]] + [r[s2]] โ†’ r[d] will yield: โ€ข ADD ( d, r[s1], d, r[s2], d, r[d] ) โ€ข ADD ( i, r[sp1], d, r[s2], d, r[d] ) โ€ข ADD ( d, r[s1], i, r[sp2], d, r[d] ) โ€ข ADD ( i, r[sp1], i, r[sp2], d, r[d] ) โ€ข ADD ( d, r[s1], d, r[s2], i, r[dp] ) โ€ข ADD ( i, r[sp1], d, r[s2], i, r[dp] ) โ€ข ADD ( d, r[s1], i, r[sp2], i, r[dp] ) โ€ข ADD ( i, r[sp1], i, r[sp2], i, r[dp] ) If we designate one register to be the "accumulator" (see below) and place strong restrictions on the various instructions allowed then we can greatly reduce the plethora of direct and indirect operations. However, one must be sure that the resulting reduced instruction-set is sufficient, and we must be aware that the reduction will come at the expense of more instructions per "significant" The notion of "accumulator A" Historical convention dedicates a register to the accumulator, an "arithmetic organ" that literally accumulates its number during a sequence of arithmetic operations: "The first part of our arithmetic organ ... should be a parallel storage organ which can receive a number and add it to the one already in it, which is also able to clear its contents and which can store what it contains. We will call such an organ an Accumulator. It is quite conventional in principle in past and present computing machines of the most varied types, e.g. desk multipliers, standard IBM counters, more modern relay machines, the ENIAC" (boldface in original: Goldstine and von Neumann, 1946; p. 98 in Bell and Newell 1971). However, the accumulator comes at the expense of more instructions per arithmetic "operation", in particular with respect to what are called 'read-modify-write' instructions such as "Increment indirectly the contents of the register pointed to by register r2 ". "A" designates the "accumulator" register A: Label Instruction A r2 r378,426 Description . . . 378,426 17 INCi ( r2 ): CPY ( i, r2, d, A ) 17 378,426 17 Contents of r2 points to r378,426 with contents "17": copy this to A INC ( A ) 18 378,426 17 Incement contents of A CPY ( d, A, i, r2 ) 18 378,426 18 Contents of r2 points to r378,426: copy contents of A into r378,426 If we stick with a specific name for the accumulator, e.g. "A", we can imply the accumulator in the instructions, for example, INC ( A ) = INCA However, when we write the CPY instructions without the accumulator called out the instructions are ambiguous or they must have empty parameters: CPY ( d, r2, d, A ) = CPY (d, r2, , ) CPY ( d, A, d, r2 ) = CPY ( , , d, r2) Historically what has happened is these two CPY instructions have received distinctive names; however, no convention exists. Tradition (e.g. Knuth's (1973) imaginary MIX computer) uses two names called LOAD and STORE. Here we are adding the "i/d" parameter: LDA ( d/i, r[s] ) =[def] CPY ( d/i, r[s], d, A ) STA ( d/i, r[d] ) =[def] CPY ( d, A, d/i, r[d] ) The typical accumulator-based model will have all its two-variable arithmetic and constant operations (e.g. ADD (A, r), SUB (A, r) ) use (i) the accumulator's contents, together with (ii) a specified register's contents. The one-variable operations (e.g. INC (A), DEC (A) and CLR (A) ) require only the accumulator. Both instruction-types deposit the result (e.g. sum, difference, product, quotient or remainder) in the accumulator. Example: INCA = [A] +1 โ†’ A Example: ADDA (r[s]) = [A] + [r[s]] โ†’ A Example: MULA (r[s]) = [A] * [r[s]] โ†’ A If we so choose, we can abbreviate the mnemonics because at least one source-register and the destination register is always the accumulator A. Thus we have : { LDA (i/d, r[s]), STA (i/d, r[d]), CLRA, INCA, DECA, ADDA (r[s]), SUBA (r[s]), MULA (r[s]), DIVA (r[s]), etc.) The notion of indirect address register "N" If our model has an unbounded accumulator can we bound all the other registers? Not until we provide for at least one unbounded register from which we derive our indirect addresses. The minimalist approach is to use itself (Schรถnhage does this). Another approach (Schรถnhage does this too) is to declare a specific register the "indirect address register" and confine indirection relative to this register (Schonhage's RAM0 model uses both A and N registers for indirect as well as direct instructions). Again our new register has no conventional name โ€“ perhaps "N" from "iNdex", or "iNdirect" or "address Number". For maximum flexibility, as we have done for the accumulator A โ€“ we will consider N just another register subject to increment, decrement, clear, test, direct copy, etc. Again we can shrink the instruction to a single-parameter that provides for direction and indirection, for example. LDAN (i/d) = CPY (i/d, N, d, A); LoaD Accumulator via iNdirection register STAN (i/d) = CPY (d, A, i/d, N). STore Accumulator via iNdirection register Why is this such an interesting approach? At least two reasons: (1) An instruction set with no parameters: Schรถnhage does this to produce his RAM0 instruction set. See section below. (2) Reduce a RAM to a Post-Turing machine: Posing as minimalists, we reduce all the registers excepting the accumulator A and indirection register N e.g. r = { r0, r1, r2, ... } to an unbounded string of (very-) bounded-capacity pigeon-holes. These will do nothing but hold (very-) bounded numbers e.g. a lone bit with value { 0, 1 }. Likewise we shrink the accumulator to a single bit. We restrict any arithmetic to the registers { A, N }, use indirect operations to pull the contents of registers into the accumulator and write 0 or 1 from the accumulator to a register: { LDA (i, N), STA (i, N), CLR (A/N), INC (A/N), DEC(N), JZ (A/N, I[z]), JZ (I[z]), H } We push further and eliminate A altogether by the use of two "constant" registers called "ERASE" and "PRINT": [ERASE]=0, [PRINT]=1. { CPY (d, ERASE, i, N), CPY (d, PRINT, i, N), CLR (N), INC (N), DEC (N), JZ (i, N, I[z]), JZ (I[z]), H } Rename the COPY instructions and call INC (N) = RIGHT, DEC (N) = LEFT and we have the same instructions as the Post-Turing machine, plus an extra CLRN : { ERASE, PRINT, CLRN, RIGHT, LEFT, JZ (i, N, I[z]), JZ (I[z]), H } Turing equivalence of the RAM with indirection In the section above we informally showed that a RAM with an unbounded indirection capability produces a Postโ€“Turing machine. The Postโ€“Turing machine is Turing equivalent, so we have shown that the RAM with indirection is Turing equivalent. We give here a slightly more formal demonstration. Begin by designing our model with three reserved registers "E", "P", and "N", plus an unbounded set of registers 1, 2, ..., n to the right. The registers 1, 2, ..., n will be considered "the squares of the tape". Register "N" points to "the scanned square" that "the head" is currently observing. The "head" can be thought of as being in the conditional jump โ€“ observe that it uses indirect addressing (cf Elgot-Robinson p. 398). As we decrement or increment "N" the (apparent) head will "move left" or "right" along the squares. We will move the contents of "E"=0 or "P"=1 to the "scanned square" as pointed to by N, using the indirect CPY. The fact that our tape is left-ended presents us with a minor problem: Whenever LEFT occurs our instructions will have to test to determine whether or not the contents of "N" is zero; if so we should leave its count at "0" (this is our choice as designers โ€“ for example we might have the machine/model "trigger an event" of our choosing). Instruction set 1 (augmented): { INC (N), DEC (N), CLR (N), CPY (d, r[s],i, N), JZ ( i, r, z ), HALT } The following table both defines the Post-Turing instructions in terms of their RAM equivalent instructions and gives an example of their functioning. The (apparent)location of the head along the tape of registers r0-r5 . . . is shown shaded: Mnemonic label: E P N r0 r1 r2 r3 r4 r5 etc. Action on registers Action on finite state machine Instruction Register IR start: 0 1 3 1 0 R right: INC ( N ) 0 1 4 1 0 [N] +1 โ†’ N [IR] +1 โ†’ IR P print: CPY ( d, P, i, N ) 0 1 4 1 1 [P]=1 โ†’ [N]=r4 [IR] +1 โ†’ IR E erase: CPY ( d, E, i, N ) 0 1 4 1 0 [E]=0 โ†’ [N]=r4 [IR] +1 โ†’ IR L left: JZ ( i, N, end ) 0 1 4 1 0 none IF N =r4] =0 THEN "end" โ†’ IR else [IR]+1 โ†’ IR DEC ( N ) 0 1 3 1 0 [N] -1 โ†’ N J0 ( halt ) jump_if_blank: JZ ( i, N, end ) 0 1 3 1 0 none IF N =r3] =0 THEN "end" โ†’ IR else [IR]+1 โ†’ IR J1 ( halt ) jump_if_mark: JZ ( i, N, halt ) 0 1 3 1 0 N =r3] โ†’ A IF N =r3] =0 THEN "end" โ†’ IR else [IR]+1 โ†’ IR end . . . etc. 0 1 3 1 0 halt: H 0 1 3 1 0 none [IR] +1 โ†’ IR Example: Bounded indirection yields a machine that is not Turing equivalent Throughout this demonstration we have to keep in mind that the instructions in the finite state machine's TABLE is bounded, i.e. finite: "Besides a merely being a finite set of rules which gives a sequence of operations for solving a specific type of problem, an algorithm has five important features [Finiteness, Definiteness, Input, Output, Effectiveness]" (italics added, Knuth p. 4-7). The difficulty arises because the registers have explicit "names" (numbers) and our machine must call each out by name in order to "access" it. We will build the indirect CPY ( i, q, d, ฯ† ) with the CASE operator. The address of the target register will be specified by the contents of register "q"; once the CASE operator has determined what this number is, CPY will directly deposit the contents of the register with that number into register "ฯ†". We will need an additional register that we will call "y" โ€“ it serves as an up-counter. So the following is actually a constructive demonstration or proof that we can indeed simulate the indirect CPY ( i, q, d, ฯ† ) without a "hardware" design change to our counter machine/model. However, note that because this indirect CPY is "bounded" by the size/extent of the finite state machine, a RASP using this indirect CPY can only calculate the primitive recursive functions, not the full suite of mu recursive functions. The CASE "operator" is described in Kleene (1952) (p. 229) and in Boolos-Burgess-Jeffrey (2002) (p. 74); the latter authors emphasize its utility. The following definition is per Kleene but modified to reflect the familiar "IF-THEN-ELSE" construction. The CASE operator "returns" a natural number into ฯ† depending on which "case" is satisfied, starting with "case_0" and going successively through "case_last"; if no case is satisfied then the number called "default" (aka "woops") is returned into ฯ† (here x designates some selection of parameters, e.g. register q and the string r0, ... rlast )): Definition by cases ฯ† (x, y): โ–ก case_0: IF Q[0](x, y) is true THEN ฯ†[0](x, y) ELSE โ–ก case_1: IF Q[1](x, y) is true THEN ฯ†[1](x, y) ELSE โ–ก cases_2 through case_next_to_last: etc. . . . . . . . . ELSE โ–ก case_last: IF Q[last](x, y) is true THEN ฯ†[last](x, y) ELSE โ–ก default: do ฯ†[default](x, y) Kleene require that the "predicates" Q[n] that doing the testing are all mutually exclusive โ€“ "predicates" are functions that produce only { true, false } for output; Boolos-Burgess-Jeffrey add the requirement that the cases are "exhaustive". We begin with a number in register q that represents the address of the target register. But what is this number? The "predicates" will test it to find out, one trial after another: JE (q, y, z) followed by INC (y). Once the number is identified explicitly, the CASE operator directly/explicitly copies the contents of this register to ฯ†: Definition by cases CPY (i, q, d, ฯ†) =[def] ฯ† (q, r0, ..., rlast, y) = โ–ก case_0: IF CLR (y), [q] - [y]=0 THEN CPY ( r0, ฯ† ), J (exit) ELSE โ–ก case_1: IF INC (y), [q] = [y]=1 THEN CPY ( r1, ฯ† ), J (exit) ELSE โ–ก case_2 through case n: IF . . . THEN . . . ELSE โ–ก case_n: IF INC (y), [q] = [y]=n THEN CPY ( rn, ฯ† ), J (exit) ELSE โ–ก case_n+1 to case_last: IF . . . THEN . . . ELSE โ–ก case_last: IF INC (y), [q] = [y]="last" THEN CPY ( rlast, ฯ† ), J (exit) ELSE โ–ก default: woops Case_0 ( the base step of the recursion on y) looks like this: โ˜† CLR ( y ) ; set register y = 0 โ˜† JE ( q, y, _ฯ†0 ) โ˜† J ( case_1 ) โ—‹ _ฯ†0: CPY ( r0, ฯ† ) โ—‹ J ( exit ) Case_n (the induction step) looks like this; remember, each instance of "n", "n+1", ..., "last" must be an explicit natural number: โ˜† INC ( y ) โ˜† JE ( q, y, _ฯ†n ) โ˜† J ( case_n+1) โ—‹ _ฯ†n: CPY ( rn, ฯ† ) โ—‹ J ( exit ) Case_last stops the induction and bounds the CASE operator (and thereby bounds the "indirect copy" operator): โ˜† INC ( y ) โ˜† JE ( q, y, _ฯ†last ) โ˜† J ( woops ) โ—‹ _ฯ†last: CPY ( rlast, ฯ† ) โ—‹ J ( exit ) โ–ก woops: how do we handle an out-of-bounds attempt? โ–ก exit: etc. If the CASE could continue ad infinitum it would be the mu operator. But it can't โ€“ its finite state machine's "state register" has reached its maximum count (e.g. 65365 = 11111111,11111111[2] ) or its table has run out of instructions; it is a finite machine, after all. Examples of models Register-to-register ("read-modify-write") model of Cook and Reckhow (1973) The commonly encountered Cook and Rechkow model is a bit like the ternary-register Malzek model (written with Knuth mnemonics โ€“ the original instructions had no mnemonics excepting TRA, Read, Print). โ–ก LOAD ( C, r[d] ) ; C โ†’ r[d], C is any integer Example: LOAD ( 0, 5 ) will clear register 5. โ–ก ADD ( r[s1], r[s2], r[d] ) ; [r[s1]] + [r[s2]] โ†’ r[d], the registers can be the same or different; Example: ADD ( A, A, A ) will double the contents of register A. โ–ก SUB ( r[s1], r[s2], r[d] ) ; [r[s1]] - [r[s2]] โ†’ r[d], the registers can be the same or different: Example: SUB ( 3, 3, 3 ) will clear register 3. โ–ก COPY ( i, r[p], d, r[d] ) ; [[r[p]] ] โ†’ r[d], Indirectly copy the contents of the source-register pointed to by pointer-register r[p] into the destination register. โ–ก COPY ( d, r[s], i, r[p] ) ; [r[s]] โ†’ [r[p]]. Copy the contents of source register r[s] into the destination-register pointed to by the pointer-register r[p]. โ–ก JNZ ( r, I[z] ) ; Conditional jump if [r] is positive; i.e. IF [r] > 0 THEN jump to instruction z else continue in sequence (Cook and Reckhow call this: "TRAnsfer control to line m if Xj > โ–ก READ ( r[d] ) ; copy "the input" into destination register r[d] โ–ก PRINT ( r[s] ) ; copy the contents of source register r[s] to "the output." Schรถnhage's RAM0 and RAM1 (1980) Schรถnhage (1980) describes a very primitive, atomized model chosen for his proof of the equivalence of his SMM pointer machine model: "In order to avoid any explicit addressing the RAM0 has the accumulator with contents z and an additional address register with current contents n (initially 0)" (p. 494) RAM1 model: Schรถnhage demonstrates how his construction can be used to form the more common, usable form of "successor"-like RAM (using this article's mnemonics): โ˜† LDA k ; k --> A , k is a constant, an explicit number such as "47" โ˜† LDA ( d, r ) ; [r] โ†’ A ; directly load A โ˜† LDA ( i, r ) ; r โ†’ A ; indirectly load A โ˜† STA ( d, r ) ; [A] โ†’ r ; directly store A โ˜† STA ( i, r ) ; [A] โ†’ [r] ; indirectly store A โ˜† JEA ( r, z ) ; IF [A] = [r] then I[z] else continue โ˜† INCA ; [A] + 1 --> A RAM0 model: Schรถnhage's RAM0 machine has 6 instructions indicated by a single letter (the 6th "C xxx" seems to involve 'skip over next parameter'. Schรถnhage designated the accumulator with "z", "N" with "n", etc. Rather than Schรถnhage's mnemonics we will use the mnemonics developed above. โ˜† (Z), CLRA: 0 โ†’ A โ˜† (A), INCA: [A] +1 โ†’ A โ˜† (N), CPYAN: [A] โ†’ N โ˜† (A), LDAA: A โ†’ A ; contents of A points to register address; put register's contents into A โ˜† (S), STAN: [A] โ†’ [N] ; contents of N points to register address; put contents of A into register pointed to by N โ˜† (C), JAZ ( z ): [A] = 0 then go to I[z] ; ambiguous in his treatment Indirection comes (i) from CPYAN (copy/transfer contents A to N) working with store_A_via_N STAN, and from (ii) the peculiar indirection instruction LDAA ( A โ†’ [A] ). Finite vs unbounded The definitional fact that any sort of counter machine without an unbounded register-"address" register must specify a register "r" by name indicates that the model requires "r" to be finite, although it is "unbounded" in the sense that the model implies no upper limit to the number of registers necessary to do its job(s). For example, we do not require r < 83,617,563,821,029,283,746 nor r < 2^1,000,001, etc. Thus our model can "expand" the number of registers, if necessary to perform a certain computation. However this does mean that whatever number the model expands to must be finite โ€“ it must be indexable with a natural number: ฯ‰ is not an option. We can escape this restriction by providing an unbounded register to provide the address of the register that specifies an indirect address. See also External links With a few exceptions, these references are the same as those at Register machine. โ–ก Goldstine, Herman H., and von Neumann, John, "Planning and Coding of the Problems for an Electronic Computing Instrument", Rep. 1947, Institute for Advanced Study, Princeton. Reprinted on pp. 92โ€“119 in Bell, C. Gordon and Newell, Allen (1971), Computer Structures: Readings and Examples, McGraw-Hill Book Company, New York. ISBN:0-07-004357-4}. โ€ข George Boolos, John P. Burgess, Richard Jeffrey (2002), Computability and Logic: Fourth Edition, Cambridge University Press, Cambridge, England. The original Boolos-Jeffrey text has been extensively revised by Burgess: more advanced than an introductory textbook. "Abacus machine" model is extensively developed in Chapter 5 Abacus Computability; it is one of three models extensively treated and compared โ€“ the Turing machine (still in Boolos' original 4-tuple form) and recursion the other two. โ€ข Arthur Burks, Herman Goldstine, John von Neumann (1946), Preliminary discussion of the logical design of an electronic computing instrument, reprinted pp. 92ff in Gordon Bell and Allen Newell (1971), Computer Structures: Readings and Examples, mcGraw-Hill Book Company, New York. ISBN:0-07-004357-4 . โ€ข Stephen A. Cook and Robert A. Reckhow (1973), Time-bounded random access machines, Journal of Computer Systems Science 7(4):354-375. โ€ข Martin Davis (1958), Computability & Unsolvability, McGraw-Hill Book Company, Inc. New York. โ€ข Calvin Elgot and Abraham Robinson (1964), Random-Access Stored-Program Machines, an Approach to Programming Languages, Journal of the Association for Computing Machinery, Vol. 11, No. 4 (October, 1964), pp. 365โ€“399. โ€ข J. Hartmanis (1971), "Computational Complexity of Random Access Stored Program Machines," Mathematical Systems Theory 5, 3 (1971) pp. 232โ€“245. โ€ข John Hopcroft, Jeffrey Ullman (1979). Introduction to Automata Theory, Languages and Computation, 1st ed., Reading Mass: Addison-Wesley. ISBN:0-201-02988-X. A difficult book centered around the issues of machine-interpretation of "languages", NP-Completeness, etc. โ€ข Stephen Kleene (1952), Introduction to Metamathematics, North-Holland Publishing Company, Amsterdam, Netherlands. ISBN:0-7204-2103-9. โ€ข Donald Knuth (1968), The Art of Computer Programming, Second Edition 1973, Addison-Wesley, Reading, Massachusetts. Cf pages 462-463 where he defines "a new kind of abstract machine or 'automaton' which deals with linked structures." โ€ข Joachim Lambek (1961, received 15 June 1961), How to Program an Infinite Abacus, Mathematical Bulletin, vol. 4, no. 3. September 1961 pages 295-302. In his Appendix II, Lambek proposes a "formal definition of 'program'. He references Melzak (1961) and Kleene (1952) Introduction to Metamathematics. โ€ข Z. A. Melzak (1961, received 15 May 1961), An informal Arithmetical Approach to Computability and Computation, Canadian Mathematical Bulletin, vol. 4, no. 3. September 1961 pages 279-293. Melzak offers no references but acknowledges "the benefit of conversations with Drs. R. Hamming, D. McIlroy and V. Vyssots of the Bell telephone Laborators and with Dr. H. Wang of Oxford University." โ€ข Marvin Minsky (1961). "Recursive Unsolvability of Post's Problem of 'Tag' and Other Topics in Theory of Turing Machines". Annals of Mathematics (The Annals of Mathematics, Vol. 74, No. 3) 74 (3): 437โ€“455. doi:10.2307/1970290. โ€ข Marvin Minsky (1967). Computation: Finite and Infinite Machines (1st ed.). Englewood Cliffs, N. J.: Prentice-Hall, Inc.. https://archive.org/details/computationfinit0000mins. In particular see chapter 11: Models Similar to Digital Computers and chapter 14: Very Simple Bases for Computability. In the former chapter he defines "Program machines" and in the later chapter he discusses "Universal Program machines with Two Registers" and "...with one register", etc. โ€ข John C. Shepherdson and H. E. Sturgis (1961) received December 1961 Computability of Recursive Functions, Journal of the Association for Computing Machinery (JACM) 10:217-255, 1963. An extremely valuable reference paper. In their Appendix A the authors cite 4 others with reference to "Minimality of Instructions Used in 4.1: Comparison with Similar Systems". โ–ก Kaphengst, Heinz, Eine Abstrakte programmgesteuerte Rechenmaschine', Zeitschrift fur mathematische Logik und Grundlagen der Mathematik:5 (1959), 366-379. โ–ก Ershov, A. P. On operator algorithms, (Russian) Dok. Akad. Nauk 122 (1958), 967-970. English translation, Automat. Express 1 (1959), 20-23. โ–ก Pรฉter, Rรณzsa Graphschemata und rekursive Funktionen, Dialectica 12 (1958), 373. โ–ก Hermes, Hans Die Universalitรคt programmgesteuerter Rechenmaschinen. Math.-Phys. Semsterberichte (Gรถttingen) 4 (1954), 42-53. โ€ข Arnold Schรถnhage (1980), Storage Modification Machines, Society for Industrial and Applied Mathematics, SIAM J. Comput. Vol. 9, No. 3, August 1980. Wherein Schลnhage shows the equivalence of his SMM with the "successor RAM" (Random Access Machine), etc. resp. Storage Modification Machines, in Theoretical Computer Science (1979), pp. 36โ€“37 โ€ข Peter van Emde Boas, "Machine Models and Simulations" pp. 3โ€“66, in: Jan van Leeuwen, ed. Handbook of Theoretical Computer Science. Volume A: Algorithms and Complexity, The MIT PRESS/Elsevier, 1990. ISBN:0-444-88071-2 (volume A). QA 76.H279 1990. van Emde Boas's treatment of SMMs appears on pp. 32โ€“35. This treatment clarifies Schลnhage 1980 โ€“ it closely follows but expands slightly the Schลnhage treatment. Both references may be needed for effective understanding. โ€ข Hao Wang (1957), A Variant to Turing's Theory of Computing Machines, JACM (Journal of the Association for Computing Machinery) 4; 63-92. Presented at the meeting of the Association, June 23โ€“25, Original source: https://en.wikipedia.org/wiki/Random-access machine. Read more
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symplectic groupoid But itโ€™s dead, no? Added a DOI for Weinstein: doi. diff, v30, current have polished up some of the references. Canโ€™t find any online trace of this one, anymore: โ€ข Alan Weinstein, Noncommutative geometry and geometric quantization, in P. Donato et al. (eds.) Symplectic geometry and Mathematical physics, Progr. Math 99 Birkhรคuser (1991) 446-461 diff, v29, current Have expanded the definition-section and added References to symplectic groupoid. I thought for a while that nowhere in the literature is the observation that a symplectic groupoid really is a 2-plectic structure and that its prequantization really involves a prequantum 2-bundle. But now I found, to my relief, that this is essentially made explicit in โ€ข Camille Laurent-Gengoux, Ping Xu, Quantization of pre-quasi-symplectic groupoids and their Hamiltonian spaces in The Breadth of Symplectic and Poisson Geometry Progress in Mathematics, 2005, Volume 232, 423-454 (arXiv:math/0311154) added definition and basic properties to symplectic groupoid , also one more blog reference created stub for symplectic groupoid, effectively just regording my blog entries on Eli Hawkins' program of geometric quantization of Poisson manifolds The DOI is clearly valid, since it redirects to the Springer website. It also has the correct bibliographic information associated to it. The very point of a DOI is that it remains valid even if the website itself is malfunctioning. In this case, the whole Progress in Mathematics series appears to have malfunctioning DOI links. Presumably this will be noticed and fixed soon by Springer. @Dmitri this still works: https://www.springer.com/series/4848/books?page=33. As does this much newer volume: https://doi.org/10.1007/978-3-031-27234-9, but the older books donโ€™t seem to work.
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Provable Tempered Overfitting of Minimal Nets and Typical Nets By Itamar Harel, William M. Hoza, Gal Vardi, Itay Evron, Nathan Srebro, and Daniel Soudry [non-alphabetical] Read the paper: arXiv Abstract (for specialists) We study the overfitting behavior of fully connected deep Neural Networks (NNs) with binary weights fitted to perfectly classify a noisy training set. We consider interpolation using both the smallest NN (having the minimal number of weights) and a random interpolating NN. For both learning rules, we prove overfitting is tempered. Our analysis rests on a new bound on the size of a threshold circuit consistent with a partial function. To the best of our knowledge, ours are the first theoretical results on benign or tempered overfitting that: (1) apply to deep NNs, and (2) do not require a very high or very low input dimension. Not-so-abstract (for curious outsiders) โš ๏ธ This summary might gloss over some important details. According to traditional wisdom in machine learning, it's a bad sign if your hypothesis perfectly fits your training data. It suggests that you're using too many parameters, you're fitting the noise instead of the signal, and your hypothesis won't perform well when it is tested on new data. This phenomenon is called "overfitting." However, this is not what is observed empirically in modern machine learning. Machine learning practitioners have found that deep neural networks perform well when they are tested on new data, even if they are trained to perfectly fit their training data. Our paper develops theoretical explanations for this empirical phenomenon. Consider any binary classification problem (e.g., given a photo, determine whether it depicts a cat). Assume that there does in fact exist a neural network $h^{\star}$ that solves this problem; our hypothesis will consist of a neural network $h$ that is considerably larger than $h^{\star}$. Specifically, given a bunch of noisy labeled examples, suppose we pick the parameters of $h$ uniformly at random, and then we repeat until we find an $h$ that perfectly fits the given training data. (We work with a simplified model of "neural network" in which the weights are binary and the activation functions are simple thresholding.) Under these idealized assumptions, we prove that overfitting is "tempered," meaning that when the hypothesis $h$ is tested on new data, its performance will be better than a completely trivial hypothesis. We posted the paper to arXiv in October 2024; it will appear at NeurIPS in December 2024. An earlier version of the paper also appeared at the 2nd Workshop on High-dimensional Learning Dynamics (HiLD) at ICML 2024.
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[Download] SSC GD 9 March 2019, Math Questions PDF Free Download SSC (Staff Selection Commission) GD (General Duty), 11 March 2019 ( Math )Quantitative Questions PDF With Answers. Table of Content [S:Table of Content (toc):S] SSC GD 9 March 2019 Quantitative Questions PDF Download Free SSC GD 9 March 2019 Quantitative Questions PDF Download Free SSC GD 9 March 2019 Shift 1 Math Questions Q2:-A sold an article to B at a gain of 5% and B sold it to C at a gain of 10%.If C paid โ‚น 462, what did it cost A? A. โ‚น400 B. โ‚น500 C. โ‚น480 D. โ‚น320 Q3:-The radius of in-circle of a triangle is 4 cm and the perimeter of the triangle is 20 cm. What is the area of the triangle? A. 40 sq. cm B. 36 sq. cm C. 80 sq. cm D. 20 sq. cm Q5:-What is the maximum number of students among whom 63 pens and 140 copies can be distributed in such a way that each student gets the same number of pens and same number of exercise books? A. 5 B. 2 C. 7 D. 3 Q6:-A train passes a 60 metres long platform in 20 seconds and a man standing on the platform in 16 seconds. The speed of the train is: A. 40 kmph B. 50 kmph C. 38 kmph D. 54 kmph SSC GD 9 March 2019 Shift 2 Math Questions For the following questions answer them individually Q1:-A shopkeeper organizes sale on Friday, offering a discount of 23% on all items. Still, he makes a profit of 10% only. By how much percentage CP was lower than MP? A. 20% B. 10% C. 15% D. 30% Q2:-Train A took 30 minute to cover a distance of 50 km. If the speed of train B is 40% faster than train A, then the ratio of the respective speed of the both train is: A. 5 : 3 B. 3 : 5 C. 7 : 5 D. 5 : 7 Q3:-The LCM of 779, 943 and 123 is: A. 71668 B. 53751 C. 35834 D. 17917 Q5:-X is twice as good as workman as Y. Together, they finish the work in 18 days. In how many days can it be done by each separately? A. X = 21 days, Y = 42 days B. X = 9 days, Y = 18 days C. X = 19 days, Y = 38 days D. X = 27 days, Y = 54 days Q7:-The sides of a triangle are in the ratio 6 : 4 : 3 and its perimeter is 104 cm. The length of the longest side (in cm)is: A. 48 B. 44 C. 56 D. 120 SSC GD 9 March 2019 Shift 3 Math Questions For the following questions answer them individually Q1:-Study the chart and answer the question based, on the pie chart. The pie chart given below shows expenditures incurred by a family on various items and savings in a month. Savings of the family is โ‚น 8,000 in a month. What is the total expenditure of the family for the month? A. โ‚น45,000 B. โ‚น48,000 C. โ‚น50,000 D. โ‚น40,000 Q2:-Nitinโ€™s money becomes double in 4 years at compound interest. In how many years will it become sixteen times at compound Interest? A. 20 B. 24 C. 16 D. 28 Q4:-What approximate value should come in place of the question mark (?) in the following questions? A. 25 B. 35 C. 40 D. 10 Q6:-If 10 persons can do a job in 20 days. Then 20 persons with the twice efficiency can do the same job in: A. 15 B. 10 C. 5 D. 20 Q7:-25% of a number is 300. The number is: A. 1000 B. 800 C. 1200 D. 900 Other Important Links:- SSC GD Previous Year English Questions SSC GD Previous Year Reasoning Questions SSC GD Previous Year Quantitative Questions SSC GD Previous Year General Knowledge Questions Download Full PDF Download SSC GD 9 March 2019, Quantitative Questions Free Download With Answers click download button given below Post a Comment 0 Comments * Please Don't Spam Here. All the Comments are Reviewed by Admin.
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2014 AMC 8 Problems/Problem 6 Six rectangles each with a common base width of $2$ have lengths of $1, 4, 9, 16, 25$, and $36$. What is the sum of the areas of the six rectangles? $\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad \textbf{(E) }202$ The sum of the areas is equal to $2\cdot1+2\cdot4+2\cdot9+2\cdot16+2\cdot25+2\cdot36$. This is equal to $2(1+4+9+16+25+36)$, which is equal to $2\cdot91$. This is equal to our final answer of $\boxed Solution 2 we can just multiply the common width 2 by each of the lengths 1 by 1, the sum would be 182. This is slow and grouping the lengths is easier to. The answer is still $\boxed{\textbf{(D)}~182}$. Video Solution (CREATIVE THINKING) ~Education, the Study of Everything Video Solution https://youtu.be/SvjJETtxQnk ~savannahsolver See Also The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Price after Markup Price Markup: Learn Stores buy items from a wholesaler or distributer and increase the price when they sell the items to consumers. The increase in price provides money for the operation of the store and the salaries of people who work in the store. A store may have a rule that the price of a certain type of item needs to be increased by a certain percentage to determine how much to sell it for. This percentage is called the markup. If the cost is known and the percentage markup is known, the sale price is the original cost plus the amount of markup. For example, if the original cost is $4.00 and the markup is 25%, the sales price should be $4.00 + $4.00*0.025 = $5.00. A faster way to calculate the sale price is to make the original cost equal to 100%. The markup is 25% so the sales price is 125% of the original cost. In the same example, $4.00 * 1.25 = $5.00. Markup: Practice What is total cost with markup? Note: Your answer should begin with a dollar sign and be rounded to the nearest cent. What is the total cost if the markup rate is % and the original cost is $? Press the Start Button To Begin You have 0 correct and 0 incorrect. This is 0 percent correct. Game Name Description Best Score How many correct answers can you get in 60 seconds? 0 Extra time is awarded for each correct answer. 0 Play longer by getting more correct. How fast can you get 20 more correct answers than wrong answers? 999 Math Lessons by Grade Math Topics Spelling Lessons by Grade Vocabulary Lessons by Grade
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Perform a time-course Model-Based Network Default vague priors for the model are as follows: \[ \begin{aligned} &\alpha_{i} \sim N(0,10000)\\ &\boldsymbol{\mu}_{i} \sim N(0,10000)\\ &\boldsymbol{d}_{t} \sim N(0,10000)\\ &beta_{\phi} \sim N(0,10000)\\ &D_{\phi,c} \sim N(0,1000)\\ &\tau_{\phi} \sim N(0,400) \text{ limited to } x \in [0,\infty]\\ &\tau^D_{\phi} \sim N(0,400) \text{ limited to } x \in [0,\infty]\\ \end{aligned} \] โ€ข \(\alpha_i\) is the response at time=0 in study \(i\) โ€ข \(\mu_i\) is a vector of study reference effects for each time-course parameter in study \(i\). Where only a single time-course parameter is modelled using relative effects the prior is defined as \(\mu_{i} \sim N(0,10000)\). โ€ข \(\boldsymbol{d}_{t}\) is a vector of pooled relative effects for treatment \(t\) whose length is the number of time-course parameters in the model. Where only a single time-course parameter is modelled using relative effects the prior is defined as \(d_{t} \sim N(0,10000)\). โ€ข \(\beta_{\phi}\) is the absolute effect for time-course parameter \(\phi\) modelled independently of treatment โ€ข \(D_{\phi,c}\) is the class relative effect for time-course parameter \(\phi\) in class \(c\) โ€ข \(\tau_{\phi}\) is the between-study SD for time-course parameter \(\phi\) โ€ข \(\tau^D_{\phi}\) is the within-class SD for time-course parameter \(\phi\) Users may wish to change these, perhaps in order to use more/less informative priors or different prior distributions (e.g. log-normal prior rather than a truncated normal for ET50 in an Emax model). However, it may also be because the default prior distributions in some models can lead to errors when compiling/updating models if the prior includes extremely implausible values. This can be more likely for certain types of models. For example some prior distributions may generate results that are too extreme for JAGS to compute, such as for time-course parameters that are powers (e.g. Emax functions with a Hill parameter or power parameters in fractional polynomials). If the model fails during compilation/updating (i.e. due to a problem in JAGS), mb.run() will generate an error and return a list of arguments that mb.run() used to generate the model. Within this (as within a model that has run successfully), the priors used by the model (in JAGS syntax) are stored within "model.arg". In this way a model can first be run with vague priors and then rerun with different priors, perhaps to allow successful computation, perhaps to provide more informative priors, or perhaps to run a sensitivity analysis with different priors. To change priors within a model, a list of replacements can be provided to priors in mb.run(). The name of each element is the name of the parameter to change (without indices) and the value of the element is the JAGS distribution to use for the prior. See the JAGS Manual (2017) for syntax details regarding specifying distributions. This can include censoring or truncation if desired. Only the priors to be changed need to be specified - priors for parameters that arenโ€™t specified will take default values. Note that in JAGS, normal distributions are specified using precision (1/variance) rather than SD. For example, we may wish to specify a tighter prior for the between-study SD: mbnma <- mb.run(network.copd, fun=tloglin(pool.rate="rel", method.rate="random"), priors=list(rate="dnorm(0,2) T(0,)")) Different prior distributions can be assigned for different indices of a parameter by specifying the list element for a parameter as a character vector. This allows (for example) for the user to fit specific priors for specific treatments. The length of this vector must be equal to the number of indices of the parameter. The ordering will also be important - for example for treatment-specific priors the order of the elements within the vector must match the order of the treatments in the network. For example we might have differnt beliefs about the long-term efficacy of a treatment for which there is no long-term data available in the dataset. In the COPD dataset we have longer term data (up to 52 weeks) on Tiotropium, but much shorter follow-up data (up to 26 weeks) on Aclidinium. We might believe (e.g. based on clincial opinion) that the efficacy of Aclidinium returns towards baseline at longer follow-up. We could model this using a B-spline and providing informative priors only to the parameters controlling the spline for Aclidinium at later follow-up: # Define informative priors for spline parameters spline.priors <- list( d.3 = c( Aclidinium="dnorm(-0.5, 100)", Tiotropium="dnorm(0, 0.0001)" d.4 = c( Aclidinium="dnorm(0, 100)", Tiotropium="dnorm(0, 0.0001)" # Using the COPD dataset with a B-spline MBNMA mbnma <- mb.run(network.copd, fun=tspline(degree=2, knots=c(0.1,0.5)), # Predict and plot time-course relative effect pred <- predict(mbnma) #> Priors required for: mu.1, mu.2, mu.3, mu.4 #> Success: Elements in prior match consistency time-course treatment effect parameters #> Reference treatment in plots is Placebo As can be seen from the predicted time-course, using informative priors for Aclidinium in this way allows us to predict itโ€™s efficacy at longer-term follow-up than the data alone can inform. pD (effective number of parameters) The default value for pd in mb.run() is pd="pv", which uses the rapid approach automatically calculated in the R2jags package as pv = var(deviance)/2. Whilst this is easy to calculate, it is only an approximation to the effective number of parameters, and may be numerically unstable (Gelman et al. 2003). However, it has been shown to be reliable for model comparison in time-course MBNMA models in a simulation study (Pedder et al. 2020). A more reliable method for estimating pd is pd="pd.kl", which uses the Kullback-Leibler divergence (Plummer 2008). This is more reliable than the default method used in R2jags for calculating the effective number of parameters from non-linear models. The disadvantage of this approach is that it requires running additional MCMC iterations, so can be slightly slower to calculate. A commonly-used approach in Bayesian models for calculating pD is the plug-in method (pd="plugin") (Spiegelhalter et al. 2002). However, this can sometimes result in negative non-sensical values due to skewed posterior distributions for deviance contributions that can arise when fitting non-linear models. Finally, pD can also be calculated using an optimism adjustment (pd="popt") which allows for calculation of the penalized expected deviance (Plummer 2008). This adjustment allows for the fact that data used to estimate the model is the same as that used to assess its parsimony. As for pd="pd.kl", it also requires running additional MCMC iterations.
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The Assignment Statement Description Examples Assignments as Expressions Compatibility Operator Assignments โ€ข lhs := rhs; โ€ข The assignment statement lhs := rhs assigns to the lhs the value of rhs. The left-hand side of the assignment must be a name, indexed-name, function call, or expression sequence of these. โ€ข The assignment statement functions as follows. 1. The left-hand side is resolved to a name. (For more information, see the evaln 2. The right-hand side is evaluated as an expression. 3. The assignment is performed. The value of the assignment statement is the right-hand โ€ข If the left-hand side is a sequence of names, the right-hand side must evaluate to a sequence of expressions with the same number of components as the left-hand side. This is called a multiple assignment. The sequence of assignments are processed in the following order. First, all of the right-hand side expressions are evaluated. Then in a left to right pairwise manner the names on the left-hand side are evaluated and the matching value on the right-hand side is assigned to it. โ€ข Local variables can be declared with type assertions. The syntax is the same as that used for parameters. Whenever an assignment is made to a local variables with such a type assertion, the type of the right-hand side is checked after evaluation, but before the assignment is done. If the type of the right-hand side does not match, an assertion failure exception is raised. Similarly, the left-hand side of any assignment can contain a type assertion that is checked before the assignment is carried out. โ€ข The setting of kernelopts(assertlevel) controls whether these type assertions are checked. Settings are described as follows. 0 - Turns off all assertion checking. 1 - Checks only assertions made using the ASSERT function. 2 - Checks ASSERT assertions, and assignment-type assertions. Assignments as Expressions โ€ข An assignment statement can also be used as an expression, or embedded within most expressions, by enclosing it in parentheses. The result of an assignment expression is the right-hand side. โ€ข The assignment takes place only if and when the part of the expression containing it is evaluated. Except in the case of logical operators, which follow left-to-right McCarthy (short-circuit) evaluation rules, it is generally not possible to know the order in which subexpressions will be evaluated. Operator Assignments โ€ข In addition to assigning an already computed value to a variable, it is possible to combine an operation and assignment into one using an assignment operator. These are: += ++ -= -- = .= /= mod= ^= ,= ||= and= or= xor= implies= union= intersect= minus= โ€ข Each of these performs the same operation that it would if the = were omitted, using the evaluated left and right hand sides of the assignment operator as its operands. The result is then assigned to the left hand side name. โ€ข The benefit of using an assignment operator is that the left hand side need only be resolved once. That is, the process of evaluating it to a name (which must be done before evaluating it to a value) is only done once. Thus, in an operator assignment such as A [veryBigExpression] += 1, the index veryBigExpression is only evaluated once. If the assignment were written as A[veryBigExpression] := A[veryBigExpression] + 1, then the index would be evaluated twice. โ€ข Like simple assignments (:=), an operator assignment can be used within an expression by enclosing it in parentheses. The result of such an embedded assignment is the computed value that was assigned to the left hand side. โ€ข The increment (+=) and decrement (-=) assignment operators each have an even shorter form that can be used when the right hand side is 1. The expression ++x is equivalent to x += 1, and --x is equivalent to x -= 1. Unlike the longer forms, the short forms are expressions in their own right, and thus can be used within a larger expression without requiring extra parentheses (except where needed for disambiguation). The value of such an expression is the incremented or decremented value. The short forms can also be written in postfix form (x++ and x--), in which case the effect on their argument, x, is the same, but the value of the expression is the original value of x. โ€ข If an operator assignment involving the same left hand side appears more than once in an expression, the order in which the assignments are carried out is undefined. Special Semantics of the ,= Assignment Operator โ€ข In most cases, the operator assignment a ,= b is equivalent to a = a, b, forming an expression sequence from the values of a and b, and assigning it back to variable a. โ€ข In the special case where the value of the left hand side is a one dimensional Array or a Vector, it is expanded and the right hand side appended to it in-place. If the right hand side is an expression sequence (e.g., a ,= b,c,d) each element of the sequence is appended separately. Expansion will be performed in such a way that it takes only constant time and space on average. This is achieved by reserving extra space whenever the Array/Vector needs to be grown, and then using this space for subsequent expansions without the need to reallocate memory. Maple always grows the space for the Array/Vector by at least a fixed percentage, which ensures the average constant time and space usage. If the left hand side is a one dimensional Array or a Vector with a hardware datatype, and an element of the right hand side is a one dimensional Array or a Vector, then the contents of that Array or Vector are appended to the left hand side. If the left hand side is a one dimensional Array or a Vector with a datatype of integer [1] (i.e., an array of bytes), and an element of the right hand side is a string, the byte values of the characters in the string are individually appended to the left hand side. Such an Array or Vector can be converted back to a string using the String If the left hand side is an object that implements a `,=` method, then that method is invoked to perform the operation. For example, the MutableSet object implements this method such that ms ,= x is equivalent to insert(ms,x) or ms union= MutableSet(x). In cases other than those above, a right hand side is appended in its entirety as an element of the left hand side. โ€ข If any of the assignments described above are not possible (e.g. the right hand side is a symbol and the array has datatype float[8]), an exception is raised. > i := 1; ${i}{&Assign;}{1}$ (1) > i; ${1}$ (2) > a[i] := 2; ${{a}}_{{i}}{&Assign;}{2}$ (3) > a[i]; ${2}$ (4) > f(1) := 0; ${f}{&ApplyFunction;}\left({1}\right){&Assign;}{0}$ (5) > f(1); ${0}$ (6) > (a,b) := (2,3); ${a}{&comma;}{b}{&Assign;}{2}{&comma;}{3}$ (7) > (a,b) := (b,a); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{2}$ (8) > g := proc(x) (x-1,x+1) end proc; ${g}{โ‰”}{\mathbf{proc}}\left({x}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{x}{-}{1}{& (9) comma;}{x}{&plus;}{1}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end proc}}$ > (s,t) := g(10); ${s}{&comma;}{t}{&Assign;}{9}{&comma;}{11}$ (10) > (j,c[j],f(j)) := 1, 2, 3; ${j}{&comma;}{{c}}_{{j}}{&comma;}{f}{&ApplyFunction;}\left({j}\right){&Assign;}{1}{& (11) > print(c); ${table}{&ApplyFunction;}\left(&lsqb;{1}{&equals;}{2}&rsqb;\right)$ (12) > f(1); ${3}$ (13) Examples of typed assignments with assertion failures. > kernelopts(assertlevel=2): > F := proc(x) local a::integer; a := x end proc: > F(3.4); Error, (in F) assertion failed in assignment to a, expected integer, got 3.4 > b::float := [1,2,3,4,5]; Error, assertion failed in assignment to b, expected float, got [1, 2, 3, 4, 5] Examples of assignments as, or within, expressions. > g((y := 2)); ${1}{&comma;}{3}$ (14) > y; ${2}$ (15) In the examples below, if a < b then the second assignment never takes place because the second operand of the or is not evaluated when the first is true. > (a, b) := (3, 4); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{4}$ (16) > (smallest := a) <= b or (smallest := b) < a; ${\mathrm{true}}$ (17) > smallest; ${3}$ (18) > (a, b) := (3, 2); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{2}$ (19) > (smallest := a) <= b or (smallest := b) < a; ${\mathrm{true}}$ (20) > smallest; ${2}$ (21) Examples of operator assignments. > a := 1; ${a}{&Assign;}{1}$ (22) > a += 17; ${18}$ (23) > a mod= 5; ${3}$ (24) > b := ++a; ${b}{&Assign;}{4}$ (25) > a, b; ${4}{&comma;}{4}$ (26) > c := a++; ${c}{&Assign;}{4}$ (27) > [a, c]; $&lsqb;{5}{&comma;}{4}&rsqb;$ (28) โ€ข lhs := rhs; โ€ข The assignment statement lhs := rhs assigns to the lhs the value of rhs. The left-hand side of the assignment must be a name, indexed-name, function call, or expression sequence of these. โ€ข The assignment statement functions as follows. 1. The left-hand side is resolved to a name. (For more information, see the evaln 2. The right-hand side is evaluated as an expression. 3. The assignment is performed. The value of the assignment statement is the right-hand โ€ข If the left-hand side is a sequence of names, the right-hand side must evaluate to a sequence of expressions with the same number of components as the left-hand side. This is called a multiple assignment. The sequence of assignments are processed in the following order. First, all of the right-hand side expressions are evaluated. Then in a left to right pairwise manner the names on the left-hand side are evaluated and the matching value on the right-hand side is assigned to it. โ€ข Local variables can be declared with type assertions. The syntax is the same as that used for parameters. Whenever an assignment is made to a local variables with such a type assertion, the type of the right-hand side is checked after evaluation, but before the assignment is done. If the type of the right-hand side does not match, an assertion failure exception is raised. Similarly, the left-hand side of any assignment can contain a type assertion that is checked before the assignment is carried out. โ€ข The setting of kernelopts(assertlevel) controls whether these type assertions are checked. Settings are described as follows. 0 - Turns off all assertion checking. 1 - Checks only assertions made using the ASSERT function. 2 - Checks ASSERT assertions, and assignment-type assertions. Assignments as Expressions โ€ข An assignment statement can also be used as an expression, or embedded within most expressions, by enclosing it in parentheses. The result of an assignment expression is the right-hand side. โ€ข The assignment takes place only if and when the part of the expression containing it is evaluated. Except in the case of logical operators, which follow left-to-right McCarthy (short-circuit) evaluation rules, it is generally not possible to know the order in which subexpressions will be evaluated. Operator Assignments โ€ข In addition to assigning an already computed value to a variable, it is possible to combine an operation and assignment into one using an assignment operator. These are: += ++ -= -- = .= /= mod= ^= ,= ||= and= or= xor= implies= union= intersect= minus= โ€ข Each of these performs the same operation that it would if the = were omitted, using the evaluated left and right hand sides of the assignment operator as its operands. The result is then assigned to the left hand side name. โ€ข The benefit of using an assignment operator is that the left hand side need only be resolved once. That is, the process of evaluating it to a name (which must be done before evaluating it to a value) is only done once. Thus, in an operator assignment such as A [veryBigExpression] += 1, the index veryBigExpression is only evaluated once. If the assignment were written as A[veryBigExpression] := A[veryBigExpression] + 1, then the index would be evaluated twice. โ€ข Like simple assignments (:=), an operator assignment can be used within an expression by enclosing it in parentheses. The result of such an embedded assignment is the computed value that was assigned to the left hand side. โ€ข The increment (+=) and decrement (-=) assignment operators each have an even shorter form that can be used when the right hand side is 1. The expression ++x is equivalent to x += 1, and --x is equivalent to x -= 1. Unlike the longer forms, the short forms are expressions in their own right, and thus can be used within a larger expression without requiring extra parentheses (except where needed for disambiguation). The value of such an expression is the incremented or decremented value. The short forms can also be written in postfix form (x++ and x--), in which case the effect on their argument, x, is the same, but the value of the expression is the original value of x. โ€ข If an operator assignment involving the same left hand side appears more than once in an expression, the order in which the assignments are carried out is undefined. Special Semantics of the ,= Assignment Operator โ€ข In most cases, the operator assignment a ,= b is equivalent to a = a, b, forming an expression sequence from the values of a and b, and assigning it back to variable a. โ€ข In the special case where the value of the left hand side is a one dimensional Array or a Vector, it is expanded and the right hand side appended to it in-place. If the right hand side is an expression sequence (e.g., a ,= b,c,d) each element of the sequence is appended separately. Expansion will be performed in such a way that it takes only constant time and space on average. This is achieved by reserving extra space whenever the Array/Vector needs to be grown, and then using this space for subsequent expansions without the need to reallocate memory. Maple always grows the space for the Array/Vector by at least a fixed percentage, which ensures the average constant time and space usage. If the left hand side is a one dimensional Array or a Vector with a hardware datatype, and an element of the right hand side is a one dimensional Array or a Vector, then the contents of that Array or Vector are appended to the left hand side. If the left hand side is a one dimensional Array or a Vector with a datatype of integer [1] (i.e., an array of bytes), and an element of the right hand side is a string, the byte values of the characters in the string are individually appended to the left hand side. Such an Array or Vector can be converted back to a string using the String If the left hand side is an object that implements a `,=` method, then that method is invoked to perform the operation. For example, the MutableSet object implements this method such that ms ,= x is equivalent to insert(ms,x) or ms union= MutableSet(x). In cases other than those above, a right hand side is appended in its entirety as an element of the left hand side. โ€ข If any of the assignments described above are not possible (e.g. the right hand side is a symbol and the array has datatype float[8]), an exception is raised. > i := 1; ${i}{&Assign;}{1}$ (1) > i; ${1}$ (2) > a[i] := 2; ${{a}}_{{i}}{&Assign;}{2}$ (3) > a[i]; ${2}$ (4) > f(1) := 0; ${f}{&ApplyFunction;}\left({1}\right){&Assign;}{0}$ (5) > f(1); ${0}$ (6) > (a,b) := (2,3); ${a}{&comma;}{b}{&Assign;}{2}{&comma;}{3}$ (7) > (a,b) := (b,a); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{2}$ (8) > g := proc(x) (x-1,x+1) end proc; ${g}{โ‰”}{\mathbf{proc}}\left({x}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{x}{-}{1}{& (9) comma;}{x}{&plus;}{1}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end proc}}$ > (s,t) := g(10); ${s}{&comma;}{t}{&Assign;}{9}{&comma;}{11}$ (10) > (j,c[j],f(j)) := 1, 2, 3; ${j}{&comma;}{{c}}_{{j}}{&comma;}{f}{&ApplyFunction;}\left({j}\right){&Assign;}{1}{& (11) > print(c); ${table}{&ApplyFunction;}\left(&lsqb;{1}{&equals;}{2}&rsqb;\right)$ (12) > f(1); ${3}$ (13) Examples of typed assignments with assertion failures. > kernelopts(assertlevel=2): > F := proc(x) local a::integer; a := x end proc: > F(3.4); Error, (in F) assertion failed in assignment to a, expected integer, got 3.4 > b::float := [1,2,3,4,5]; Error, assertion failed in assignment to b, expected float, got [1, 2, 3, 4, 5] Examples of assignments as, or within, expressions. > g((y := 2)); ${1}{&comma;}{3}$ (14) > y; ${2}$ (15) In the examples below, if a < b then the second assignment never takes place because the second operand of the or is not evaluated when the first is true. > (a, b) := (3, 4); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{4}$ (16) > (smallest := a) <= b or (smallest := b) < a; ${\mathrm{true}}$ (17) > smallest; ${3}$ (18) > (a, b) := (3, 2); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{2}$ (19) > (smallest := a) <= b or (smallest := b) < a; ${\mathrm{true}}$ (20) > smallest; ${2}$ (21) Examples of operator assignments. > a := 1; ${a}{&Assign;}{1}$ (22) > a += 17; ${18}$ (23) > a mod= 5; ${3}$ (24) > b := ++a; ${b}{&Assign;}{4}$ (25) > a, b; ${4}{&comma;}{4}$ (26) > c := a++; ${c}{&Assign;}{4}$ (27) > [a, c]; $&lsqb;{5}{&comma;}{4}&rsqb;$ (28) โ€ข lhs := rhs; โ€ข The assignment statement lhs := rhs assigns to the lhs the value of rhs. The left-hand side of the assignment must be a name, indexed-name, function call, or expression sequence of these. โ€ข The assignment statement functions as follows. 1. The left-hand side is resolved to a name. (For more information, see the evaln function). 2. The right-hand side is evaluated as an expression. 3. The assignment is performed. The value of the assignment statement is the right-hand side. โ€ข If the left-hand side is a sequence of names, the right-hand side must evaluate to a sequence of expressions with the same number of components as the left-hand side. This is called a multiple assignment. The sequence of assignments are processed in the following order. First, all of the right-hand side expressions are evaluated. Then in a left to right pairwise manner the names on the left-hand side are evaluated and the matching value on the right-hand side is assigned to it. โ€ข Local variables can be declared with type assertions. The syntax is the same as that used for parameters. Whenever an assignment is made to a local variables with such a type assertion, the type of the right-hand side is checked after evaluation, but before the assignment is done. If the type of the right-hand side does not match, an assertion failure exception is raised. Similarly, the left-hand side of any assignment can contain a type assertion that is checked before the assignment is carried out. โ€ข The setting of kernelopts(assertlevel) controls whether these type assertions are checked. Settings are described as follows. 0 - Turns off all assertion checking. 1 - Checks only assertions made using the ASSERT function. 2 - Checks ASSERT assertions, and assignment-type assertions. โ€ข The assignment statement lhs := rhs assigns to the lhs the value of rhs. The left-hand side of the assignment must be a name, indexed-name, function call, or expression sequence of these. The assignment statement lhs := rhs assigns to the lhs the value of rhs. The left-hand side of the assignment must be a name, indexed-name, function call, or expression sequence of these. 1. The left-hand side is resolved to a name. (For more information, see the evaln function). The left-hand side is resolved to a name. (For more information, see the evaln function). 3. The assignment is performed. The value of the assignment statement is the right-hand side. The assignment is performed. The value of the assignment statement is the right-hand side. โ€ข If the left-hand side is a sequence of names, the right-hand side must evaluate to a sequence of expressions with the same number of components as the left-hand side. This is called a multiple assignment. The sequence of assignments are processed in the following order. First, all of the right-hand side expressions are evaluated. Then in a left to right pairwise manner the names on the left-hand side are evaluated and the matching value on the right-hand side is assigned to it. If the left-hand side is a sequence of names, the right-hand side must evaluate to a sequence of expressions with the same number of components as the left-hand side. This is called a multiple assignment. The sequence of assignments are processed in the following order. First, all of the right-hand side expressions are evaluated. Then in a left to right pairwise manner the names on the left-hand side are evaluated and the matching value on the right-hand side is assigned to it. โ€ข Local variables can be declared with type assertions. The syntax is the same as that used for parameters. Whenever an assignment is made to a local variables with such a type assertion, the type of the right-hand side is checked after evaluation, but before the assignment is done. If the type of the right-hand side does not match, an assertion failure exception is raised. Local variables can be declared with type assertions. The syntax is the same as that used for parameters. Whenever an assignment is made to a local variables with such a type assertion, the type of the right-hand side is checked after evaluation, but before the assignment is done. If the type of the right-hand side does not match, an assertion failure exception is raised. Similarly, the left-hand side of any assignment can contain a type assertion that is checked before the assignment is carried out. โ€ข The setting of kernelopts(assertlevel) controls whether these type assertions are checked. Settings are described as follows. The setting of kernelopts(assertlevel) controls whether these type assertions are checked. Settings are described as follows. 1 - Checks only assertions made using the ASSERT function. Assignments as Expressions โ€ข An assignment statement can also be used as an expression, or embedded within most expressions, by enclosing it in parentheses. The result of an assignment expression is the right-hand side. โ€ข The assignment takes place only if and when the part of the expression containing it is evaluated. Except in the case of logical operators, which follow left-to-right McCarthy (short-circuit) evaluation rules, it is generally not possible to know the order in which subexpressions will be evaluated. โ€ข An assignment statement can also be used as an expression, or embedded within most expressions, by enclosing it in parentheses. The result of an assignment expression is the right-hand side. An assignment statement can also be used as an expression, or embedded within most expressions, by enclosing it in parentheses. The result of an assignment expression is the right-hand side. โ€ข The assignment takes place only if and when the part of the expression containing it is evaluated. Except in the case of logical operators, which follow left-to-right McCarthy (short-circuit) evaluation rules, it is generally not possible to know the order in which subexpressions will be evaluated. The assignment takes place only if and when the part of the expression containing it is evaluated. Except in the case of logical operators, which follow left-to-right McCarthy (short-circuit) evaluation rules, it is generally not possible to know the order in which subexpressions will be evaluated. Operator Assignments โ€ข In addition to assigning an already computed value to a variable, it is possible to combine an operation and assignment into one using an assignment operator. These are: += ++ -= -- = .= /= mod= ^= ,= ||= and= or= xor= implies= union= intersect= minus= โ€ข Each of these performs the same operation that it would if the = were omitted, using the evaluated left and right hand sides of the assignment operator as its operands. The result is then assigned to the left hand side name. โ€ข The benefit of using an assignment operator is that the left hand side need only be resolved once. That is, the process of evaluating it to a name (which must be done before evaluating it to a value) is only done once. Thus, in an operator assignment such as A[veryBigExpression] += 1, the index veryBigExpression is only evaluated once. If the assignment were written as A [veryBigExpression] := A[veryBigExpression] + 1, then the index would be evaluated twice. โ€ข Like simple assignments (:=), an operator assignment can be used within an expression by enclosing it in parentheses. The result of such an embedded assignment is the computed value that was assigned to the left hand side. โ€ข The increment (+=) and decrement (-=) assignment operators each have an even shorter form that can be used when the right hand side is 1. The expression ++x is equivalent to x += 1, and --x is equivalent to x -= 1. Unlike the longer forms, the short forms are expressions in their own right, and thus can be used within a larger expression without requiring extra parentheses (except where needed for disambiguation). The value of such an expression is the incremented or decremented value. The short forms can also be written in postfix form (x++ and x--), in which case the effect on their argument, x, is the same, but the value of the expression is the original value of x. โ€ข If an operator assignment involving the same left hand side appears more than once in an expression, the order in which the assignments are carried out is undefined. Special Semantics of the ,= Assignment Operator โ€ข In most cases, the operator assignment a ,= b is equivalent to a = a, b, forming an expression sequence from the values of a and b, and assigning it back to variable a. โ€ข In the special case where the value of the left hand side is a one dimensional Array or a Vector, it is expanded and the right hand side appended to it in-place. If the right hand side is an expression sequence (e.g., a ,= b,c,d) each element of the sequence is appended separately. Expansion will be performed in such a way that it takes only constant time and space on average. This is achieved by reserving extra space whenever the Array/Vector needs to be grown, and then using this space for subsequent expansions without the need to reallocate memory. Maple always grows the space for the Array/Vector by at least a fixed percentage, which ensures the average constant time and space usage. If the left hand side is a one dimensional Array or a Vector with a hardware datatype, and an element of the right hand side is a one dimensional Array or a Vector, then the contents of that Array or Vector are appended to the left hand side. If the left hand side is a one dimensional Array or a Vector with a datatype of integer[1] (i.e., an array of bytes), and an element of the right hand side is a string, the byte values of the characters in the string are individually appended to the left hand side. Such an Array or Vector can be converted back to a string using the String constructor. If the left hand side is an object that implements a `,=` method, then that method is invoked to perform the operation. For example, the MutableSet object implements this method such that ms ,= x is equivalent to insert(ms,x) or ms union= MutableSet(x). In cases other than those above, a right hand side is appended in its entirety as an element of the left hand side. โ€ข If any of the assignments described above are not possible (e.g. the right hand side is a symbol and the array has datatype float[8]), an exception is raised. โ€ข In addition to assigning an already computed value to a variable, it is possible to combine an operation and assignment into one using an assignment operator. These are: In addition to assigning an already computed value to a variable, it is possible to combine an operation and assignment into one using an assignment operator. These are: += ++ -= -- = .= /= mod= ^= ,= ||= and= or= xor= implies= union= intersect= minus= โ€ข Each of these performs the same operation that it would if the = were omitted, using the evaluated left and right hand sides of the assignment operator as its operands. The result is then assigned to the left hand side name. Each of these performs the same operation that it would if the = were omitted, using the evaluated left and right hand sides of the assignment operator as its operands. The result is then assigned to the left hand side name. โ€ข The benefit of using an assignment operator is that the left hand side need only be resolved once. That is, the process of evaluating it to a name (which must be done before evaluating it to a value) is only done once. Thus, in an operator assignment such as A[veryBigExpression] += 1, the index veryBigExpression is only evaluated once. If the assignment were written as A [veryBigExpression] := A[veryBigExpression] + 1, then the index would be evaluated twice. The benefit of using an assignment operator is that the left hand side need only be resolved once. That is, the process of evaluating it to a name (which must be done before evaluating it to a value) is only done once. Thus, in an operator assignment such as A[veryBigExpression] += 1, the index veryBigExpression is only evaluated once. If the assignment were written as A[veryBigExpression] := A [veryBigExpression] + 1, then the index would be evaluated twice. โ€ข Like simple assignments (:=), an operator assignment can be used within an expression by enclosing it in parentheses. The result of such an embedded assignment is the computed value that was assigned to the left hand side. Like simple assignments (:=), an operator assignment can be used within an expression by enclosing it in parentheses. The result of such an embedded assignment is the computed value that was assigned to the left hand side. โ€ข The increment (+=) and decrement (-=) assignment operators each have an even shorter form that can be used when the right hand side is 1. The expression ++x is equivalent to x += 1, and --x is equivalent to x -= 1. The increment (+=) and decrement (-=) assignment operators each have an even shorter form that can be used when the right hand side is 1. The expression ++x is equivalent to x += 1, and --x is equivalent to x -= 1. Unlike the longer forms, the short forms are expressions in their own right, and thus can be used within a larger expression without requiring extra parentheses (except where needed for disambiguation). The value of such an expression is the incremented or decremented value. The short forms can also be written in postfix form (x++ and x--), in which case the effect on their argument, x, is the same, but the value of the expression is the original value of x. โ€ข If an operator assignment involving the same left hand side appears more than once in an expression, the order in which the assignments are carried out is undefined. If an operator assignment involving the same left hand side appears more than once in an expression, the order in which the assignments are carried out is undefined. Special Semantics of the ,= Assignment Operator โ€ข In most cases, the operator assignment a ,= b is equivalent to a = a, b, forming an expression sequence from the values of a and b, and assigning it back to variable a. โ€ข In the special case where the value of the left hand side is a one dimensional Array or a Vector, it is expanded and the right hand side appended to it in-place. If the right hand side is an expression sequence (e.g., a ,= b,c,d) each element of the sequence is appended separately. Expansion will be performed in such a way that it takes only constant time and space on average. This is achieved by reserving extra space whenever the Array/Vector needs to be grown, and then using this space for subsequent expansions without the need to reallocate memory. Maple always grows the space for the Array/Vector by at least a fixed percentage, which ensures the average constant time and space usage. If the left hand side is a one dimensional Array or a Vector with a hardware datatype, and an element of the right hand side is a one dimensional Array or a Vector, then the contents of that Array or Vector are appended to the left hand side. If the left hand side is a one dimensional Array or a Vector with a datatype of integer[1] (i.e., an array of bytes), and an element of the right hand side is a string, the byte values of the characters in the string are individually appended to the left hand side. Such an Array or Vector can be converted back to a string using the String constructor. If the left hand side is an object that implements a `,=` method, then that method is invoked to perform the operation. For example, the MutableSet object implements this method such that ms ,= x is equivalent to insert(ms,x) or ms union= MutableSet(x). In cases other than those above, a right hand side is appended in its entirety as an element of the left hand side. โ€ข If any of the assignments described above are not possible (e.g. the right hand side is a symbol and the array has datatype float[8]), an exception is raised. โ€ข In most cases, the operator assignment a ,= b is equivalent to a = a, b, forming an expression sequence from the values of a and b, and assigning it back to variable a. In most cases, the operator assignment a ,= b is equivalent to a = a, b, forming an expression sequence from the values of a and b, and assigning it back to variable a. โ€ข In the special case where the value of the left hand side is a one dimensional Array or a Vector, it is expanded and the right hand side appended to it in-place. If the right hand side is an expression sequence (e.g., a ,= b,c,d) each element of the sequence is appended separately. In the special case where the value of the left hand side is a one dimensional Array or a Vector, it is expanded and the right hand side appended to it in-place. If the right hand side is an expression sequence (e.g., a ,= b,c,d) each element of the sequence is appended separately. Expansion will be performed in such a way that it takes only constant time and space on average. This is achieved by reserving extra space whenever the Array/Vector needs to be grown, and then using this space for subsequent expansions without the need to reallocate memory. Maple always grows the space for the Array/Vector by at least a fixed percentage, which ensures the average constant time and space usage. If the left hand side is a one dimensional Array or a Vector with a hardware datatype, and an element of the right hand side is a one dimensional Array or a Vector, then the contents of that Array or Vector are appended to the left hand side. If the left hand side is a one dimensional Array or a Vector with a datatype of integer[1] (i.e., an array of bytes), and an element of the right hand side is a string, the byte values of the characters in the string are individually appended to the left hand side. Such an Array or Vector can be converted back to a string using the String constructor. If the left hand side is an object that implements a `,=` method, then that method is invoked to perform the operation. For example, the MutableSet object implements this method such that ms ,= x is equivalent to insert(ms,x) or ms union= MutableSet(x). In cases other than those above, a right hand side is appended in its entirety as an element of the left hand side. โ€ข If any of the assignments described above are not possible (e.g. the right hand side is a symbol and the array has datatype float[8]), an exception is raised. If any of the assignments described above are not possible (e.g. the right hand side is a symbol and the array has datatype float[8]), an exception is raised. > i := 1; ${i}{&Assign;}{1}$ (1) > i; ${1}$ (2) > a[i] := 2; ${{a}}_{{i}}{&Assign;}{2}$ (3) > a[i]; ${2}$ (4) > f(1) := 0; ${f}{&ApplyFunction;}\left({1}\right){&Assign;}{0}$ (5) > f(1); ${0}$ (6) > (a,b) := (2,3); ${a}{&comma;}{b}{&Assign;}{2}{&comma;}{3}$ (7) > (a,b) := (b,a); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{2}$ (8) > g := proc(x) (x-1,x+1) end proc; ${g}{โ‰”}{\mathbf{proc}}\left({x}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{x}{-}{1}{&comma;}{x}{&plus;}{1}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end proc}}$ (9) > (s,t) := g(10); ${s}{&comma;}{t}{&Assign;}{9}{&comma;}{11}$ (10) > (j,c[j],f(j)) := 1, 2, 3; ${j}{&comma;}{{c}}_{{j}}{&comma;}{f}{&ApplyFunction;}\left({j}\right){&Assign;}{1}{&comma;}{2}{&comma;}{3}$ (11) > print(c); ${table}{&ApplyFunction;}\left(&lsqb;{1}{&equals;}{2}&rsqb;\right)$ (12) > f(1); ${3}$ (13) Examples of typed assignments with assertion failures. > kernelopts(assertlevel=2): > F := proc(x) local a::integer; a := x end proc: > F(3.4); Error, (in F) assertion failed in assignment to a, expected integer, got 3.4 > b::float := [1,2,3,4,5]; Error, assertion failed in assignment to b, expected float, got [1, 2, 3, 4, 5] Examples of assignments as, or within, expressions. > g((y := 2)); ${1}{&comma;}{3}$ (14) > y; ${2}$ (15) In the examples below, if a < b then the second assignment never takes place because the second operand of the or is not evaluated when the first is true. > (a, b) := (3, 4); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{4}$ (16) > (smallest := a) <= b or (smallest := b) < a; ${\mathrm{true}}$ (17) > smallest; ${3}$ (18) > (a, b) := (3, 2); ${a}{&comma;}{b}{&Assign;}{3}{&comma;}{2}$ (19) > (smallest := a) <= b or (smallest := b) < a; ${\mathrm{true}}$ (20) > smallest; ${2}$ (21) Examples of operator assignments. > a := 1; ${a}{&Assign;}{1}$ (22) > a += 17; ${18}$ (23) > a mod= 5; ${3}$ (24) > b := ++a; ${b}{&Assign;}{4}$ (25) > a, b; ${4}{&comma;}{4}$ (26) > c := a++; ${c}{&Assign;}{4}$ (27) > [a, c]; $&lsqb;{5}{&comma;}{4}&rsqb;$ (28) > F := proc(x) local a::integer; a := x end proc: F := proc(x) local a::integer; a := x end proc: Error, (in F) assertion failed in assignment to a, expected integer, got 3.4 Error, assertion failed in assignment to b, expected float, got [1, 2, 3, 4, 5] In the examples below, if a < b then the second assignment never takes place because the second operand of the or is not evaluated when the first is true. > (smallest := a) <= b or (smallest := b) < a; (smallest := a) <= b or (smallest := b) < a; โ€ข Assignments as expressions and operator assignments are currently not supported in 2-D math input in the Standard Interface. They are supported in 1-D math input in the Standard Interface, as well as in all forms of input and output in the Command-line user interface. โ€ข Assignments as expressions and operator assignments are currently not supported in 2-D math input in the Standard Interface. They are supported in 1-D math input in the Standard Interface, as well as in all forms of input and output in the Command-line user interface. Assignments as expressions and operator assignments are currently not supported in 2-D math input in the Standard Interface. They are supported in 1-D math input in the Standard Interface, as well as in all forms of input and output in the Command-line user interface. See Also Using the Variable Manager Variable Manager
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Which Big 12 member is the team you most want to beat once we join? I think I'm still gonna hate Utah the most. Those ****ers def think they're gonna rule the roost in the new Big 12. We gotta put a stop to that right quick. I understand the vote for ****bailer, they can eat ****, but I am going to single handedly force a CU-BYU rivalry if I have to. Come on! Utah can't even make weird underwear jokes because most of them wear them too! BYU needs to have lame dickheads take cheap potshots at their weird and pasty ways in this conference and we're just the assholes to do it! Oh yeah. Them. But as far as Utah goes, I hate them more now than I ever did during our entire time in the P12. They are probably second compared to Baylor. There can be only one Club Member Only 55% for **** bailer? Get your **** together, people! I generally dislike private religious schools, so I'll put TCU, Baylor, BYU on there. But there's no D1 school I hate more than Iowa State. I generally dislike private religious schools, so I'll put TCU, Baylor, BYU on there. But there's no D1 school I hate more than Iowa State. Lol, really? That's an interesting hate school. How come? Lol, really? That's an interesting hate school. How come? Person I despised was overly proud of their degrees from ISU Last edited: Lol, really? That's an interesting hate school. How come? Yeah. I hadn't known anyone who hated the Clones. Figured there were probably some Hawkeye & KSU fans who did due to rivalry, but I've only encountered decent, reasonable folks from the ISU fanbase. Person I dispised was overly proud of their degrees from ISU Legit reason to hate a school, but I have to admit that the idea of anyone hating ISU drew an initial chuckle from me. Small Guards don't get drafted early Club Member I donโ€™t think โ€œincorrectโ€ can be included in the definition of โ€œpreciseโ€. Out running today, I flashed back to this conversation. Let me expand. Incorrect is not included in the definition of precise, however, something can be precise and incorrect. And being inaccurate does not preclude something from being precise. It's a common misunderstanding between accuracy and precision. Expanding on the pi example may help you. If I say pi is approximately equal to... That answer is... 3.14159265358979 Precise and accurate 2.06127651552485 Precise, but not accurate 3 accurate, but not precise 2 neither precise nor accurate Last edited: Haha. Wow. That was like 7 months ago. My counter argument would be that you can be correct but not precise. But you cannot be precise without being correct (otherwise itโ€™s all just randomnessโ€ฆwhich is the opposite of precise). 1. strictly correct in amount or value: a precise sum. 2. designating a certain thing and no other; particular: this precise location. 3. using or operating with total accuracy: precise instruments. 4. strict in observance of rules, standards, etc: a precise mind. Snow's Padawan Club Member Out running today, I flashed back to this conversation. Let me expand. I correct is not included in the definition of precise, however, something can be precise and incorrect. And being inaccurate does not preclude something from being precise. It's a common misunderstanding between accuracy and precision. Expanding on the pi example may help you. If I say pi is approximately equal to... That answer is... 3.14159265358979 Precise and accurate 2.06127651552485 Precise, but not accurate 3 accurate, but not precise 2 neither precise nor accurate Cooler than a Popsicle Stand. Club Member Out running today, I flashed back to this conversation. Let me expand. I correct is not included in the definition of precise, however, something can be precise and incorrect. And being inaccurate does not preclude something from being precise. It's a common misunderstanding between accuracy and precision. Expanding on the pi example may help you. If I say pi is approximately equal to... That answer is... 3.14159265358979 Precise and accurate 2.06127651552485 Precise, but not accurate 3 accurate, but not precise 2 neither precise nor accurate How would you categorize 3.14? I don't know (remember) the context for this, but my initial impression is that by omitting that, you are being deliberately misleading. In other words, where is "accurate and sufficient?" For instance "**** bailer" is sufficient and accurate, though "**** Baylor University" would be accurate and precise. How would you categorize 3.14? I don't know (remember) the context for this, but my initial impression is that by omitting that, you are being deliberately misleading. In other words, where is "accurate and sufficient?" For instance "**** bailer" is sufficient and accurate, though "**** Baylor University" would be accurate and precise. Saying ฯ€ ~= 3.14 is accurate, and more precise than 3, but less precise than 3.14159. The question of whether or not 3.14 is "sufficient" comes down to how many significant digits the other measurements have, in this case the radius. If you measure the radius to three or less significant digits, 3.14 is sufficient. If your measurement system of the radius goes to five, 3.14 not sufficient to avoid losing precision in calculating the circumference. Haha. Wow. That was like 7 months ago. My counter argument would be that you can be correct but not precise. But you cannot be precise without being correct (otherwise itโ€™s all just randomnessโ€ฆwhich is the opposite of precise). 1. strictly correct in amount or value: a precise sum. 2. designating a certain thing and no other; particular: this precise location. 3. using or operating with total accuracy: precise instruments. 4. strict in observance of rules, standards, etc: a precise mind. The difference between precision and accuracy is taught in basic laboratory sciences. The definitions used there, and in all the statistics I've engaged with, deals with measurement error. For repeated measurements of the same characteristic, precision is how close each of the measurements are to each other and accuracy is how close the measurements are, on average to the true value. If you're 180 pounds and have a cheap bathroom scale which you step on and off of 5 times and get measurements of 184, 178, 181, 175, 182 you'd call that scale accurate, but imprecise. On average it gets it right, but it's a scattershot of values. Now you go to a doctor's office where the physician tried to calibrate the expensive scale hisself, but kind of messed up. You take five measurements and get 177.1, 176.9, 177.0, 177.0, 177.1. That would be a precise but inaccurate measurement. The frequently used analogy in teaching is hits on a target. A tight grouping on the bullseye is precise and accurate, a loose grouping that surrounds the bullseye is accurate but imprecise, a tight grouping off the bullseye is innacurate and precise, a loose grouping skewed away from the bullseye is innacurate and imprecise. The difference between precision and accuracy is taught in basic laboratory sciences. The definitions used there, and in all the statistics I've engaged with, deals with measurement error. For repeated measurements of the same characteristic, precision is how close each of the measurements are to each other and accuracy is how close the measurements are, on average to the true If you're 180 pounds and have a cheap bathroom scale which you step on and off of 5 times and get measurements of 184, 178, 181, 175, 182 you'd call that scale accurate, but imprecise. On average it gets it right, but it's a scattershot of values. Now you go to a doctor's office where the physician tried to calibrate the expensive scale hisself, but kind of messed up. You take five measurements and get 177.1, 176.9, 177.0, 177.0, 177.1. That would be a precise but inaccurate measurement. The frequently used analogy in teaching is hits on a target. A tight grouping on the bullseye is precise and accurate, a loose grouping that surrounds the bullseye is accurate but imprecise, a tight grouping off the bullseye is innacurate and precise, a loose grouping skewed away from the bullseye is innacurate and imprecise. Where Aviator and I are differing is on a linguistic vs a scientific definition of the term. He's correct that in everyday speech that hitting the target is required to be precise. But in physics, the definition of precision is independent of accuracy. This is a frequent problem in our language (maybe every language?) -- nearly every word has multiple definitions and often they are in conflict. Saying ฯ€ ~= 3.14 is accurate, and more precise than 3, but less precise than 3.14159. The question of whether or not 3.14 is "sufficient" comes down to how many significant digits the other measurements have, in this case the radius. If you measure the radius to three or less significant digits, 3.14 is sufficient. If your measurement system of the radius goes to five, 3.14 not sufficient to avoid losing precision in calculating the circumference. Where Aviator and I are differing is on a linguistic vs a scientific definition of the term. He's correct that in everyday speech that hitting the target is required to be precise. But in physics, the definition of precision is independent of accuracy. This is a frequent problem in our language (maybe every language?) -- nearly every word has multiple definitions and often they are in conflict. In everyday language, being too precise actually distracts from meaning and/or intention. I don't know if there's an equivalent phenomenon in science. In everyday language, being too precise actually distracts from meaning and/or intention. I don't know if there's an equivalent phenomenon in science. interesting observation -- not unusual for you to give me a new perspective to consider. I will note, without snark, that in spoken communications, I generally favor being less precise, but it's more out of a concern of being accurate. When speaking, I'm generally communicating more rapidly, which means I'm giving less thought to what I'm saying -- this leads me to err on the side of "I'd rather be truthful than be specific and I'm willing to sacrifice clarity to reduce likelihood of speaking incorrectly". Your comment has me considering another aspect of that. Buffalo to the core. Space Marine. Club Member Someone hating ISU? Here's my reaction to that. In everyday language, being too precise actually distracts from meaning and/or intention. I don't know if there's an equivalent phenomenon in science. I vaguely recall some pedagogical research about overly precise language and jargon reducing the effectiveness of communicating and learning. This also brings to mind the concept of model parsimony, where you can technically improve how well your model fits a set of data but tend to introduce some loss in prediction accuracy by adding more predictor variables, so you prefer the fewest number of inputs that gives you an adequate output. Back on topic since you assholes are making my brain hurt: The rest of the Big 12 teams suck and we will crush them!!!
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HHPVProblem < CoWiki < TWiki The Brown-Freedman-Halbeisen-Hungerbรผhler-Pirillo-Varricchio problem asks, is there an infinite word over a finite subset of , the non-negative integers, containing no two consecutive blocks of the same length and the same sum? The question was apparently first raised by Brown and Freedman in a 1987 paper, then independently by Pirillo and Varricchio in a 1994 paper, and by Halbeisen and Hungerbรผhler in 2000. It follows from results of Dekking that such a word exists avoiding four consecutive blocks. Recent results of Cassaigne, Currie, Schaeffer, and Shallit (2011) show that such a word exists avoiding three consecutive blocks. -- - 13 Jul 2011
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LeetCode daily question solution: 2055 Plates between Candles - problem solving - python & Here is a long table with plates and candles lined up on it. Give you a string s with subscript starting from 0 |, which contains only the characters' * 'and' | ', where' * 'represents a plate and' | 'represents a candle. At the same time, you will be given a two-dimensional integer array {queries with subscript starting from 0}, where {queries[i] = [lefti, righti] represents the substring} s[lefti...righti] (including the characters of the left and right endpoints). For each query, you need to find the number of plates in the substring between two candles. If a plate has at least one candle on the left and right of the substring, the plate is satisfied between the two candles. For example, s = "* * * *", query [3, 8], which represents the substring "* * *". The number of plates between two candles in the substring is 2, and the two plates on the right in the substring have at least one candle on their left and right. Please return an integer array ๏ผŒ answer, where ๏ผŒ answer[i] is the answer to the ๏ผŒ I ๏ผŒ th query. Example 1: Input: S = "* * * * * * * *", queries = [[2,5], [5,9]] Output: [2,3] -There are two plates between the candles. -There are three plates between the candles. Example 2: Input: S = "*****************************", queries = [[1,17], [4,5], [14,17], [5,11], [15,16]] Output: [9,0,0,0,0] -queries[0] there are nine plates between the candles. -Another query has no plates between candles. Topic idea: Record prefix and, in short, record the total number from the starting point to each position* Take two more arrays and a right to record the position of | on the left and right A left records the position of the left | at the right end The process of recording the position is the process of traversing the string. The coordinates of each | are recorded from the left to the previous | and the same is true for the right Establish the learning process, print out the values of left and right, and you can see that it is actually very simple Python code: class Solution: def platesBetweenCandles(self, s: str, queries: List[List[int]]) -> List[int]: n = len(s)#Find the length of the string #first is used to record prefixes and. In short, it is used to record the total number of prefixes from the starting point to each position* first , left , right = [0]*(n+1) , [-1]*n , [-1]*n #The reason why the initial values of left and right are assigned to 1 is #Because it is necessary to judge whether the left and right are greater than 0| #Left is used to record the serial number of the nearest | on the left of each position #Right is used to record the serial number of the nearest | position on the right of each position l , r = -1 , -1 for i , j in enumerate(s): if j == "*": #first is used to record prefixes and. In short, it is used to record the total number of prefixes from the starting point to each position* first[i+1] = first[i] + 1 #If the current position is * then the number of next positions is the current + 1 first[i+1] = first[i] #If not * the number of * corresponding to the next coordinate is equal to this one l = i left[i] = l # Assign the coordinates of the previous one until you meet the next one for i , j in enumerate(s[::-1]): #Reverse traversal if j == "|": #ditto r = n-1-i right[n-1-i] = r return [first[left[r]] - first[right[l]] if left[r]>=0 and right[l]>=0 and left[r]>right[l] else 0 for l , r in queries] #Traverse the queries and then judge the situation at each position. If the value on the right side of the left end of the coordinate is less than 0 or the value on the left side of the right end is less than 0, 0 will be output C + + Code: class Solution { vector<int> platesBetweenCandles(string s, vector<vector<int>>& queries) { int n = s.length(); //Find the length of the string vector<int> prnum(n+1); //prnum is used to record prefixes and. In short, it is used to record the total number of prefixes from the starting point to each position* vector<int> left(n);//Create an empty left array vector<int> right(n); //Create an empty right array vector<int> ans;//Left is used to record the serial number of the nearest | position on the left of each position #right is used to record the serial number of the nearest | position on the right of each position for (int i=0 , l=-1; i<n; i++){ if (s[i] == '*'){ //first is used to record prefixes and. In short, it is used to record the total number of prefixes from the starting point to each position* prnum[i+1] = prnum[i] + 1; //If * the current position is + 1, then the next number is prnum[i+1] = prnum[i];//If not * the number of * corresponding to the next coordinate is equal to this one l = i; left[i] = l; //Assign the coordinates of the previous one until you meet the next one for (int i=n-1 , r=-1; i>=0; i--) { if (s[i] == '|'){ //Reverse traversal r = i; right[i] = r;//ditto for (auto& query : queries){ int x = right[query[0]] , y = left[query[1]]; ans.push_back(x<0 || y<0 || x>=y ?0:prnum[y] - prnum[x]); //#Traverse the queries and then judge the situation at each position. If the value on the right side of the left end of the coordinate is less than 0 or the value on the left side of the right end is less than 0, 0 will be output return ans;
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turbomachinery Full notes pdf |turbomachinery lecture notes | turbomachinery notes pdf | turbo machine handwritten notes | turbomachine pdf notes turbomachinery notes pdf |turbomachinery lecture notes | turbomachinery notes pdf | turbo machine handwritten notes | turbomachine pdf notes|Handwritten notes Mechanical Engineering | Written by Ashish Sir Turbomachinery, in mechanical engineering, describes machines that transfer energy between a rotor and a fluid, including both turbines and compressors. While a turbine transfers energy from a fluid to a rotor, a compressor transfers energy from a rotor to a fluid. Turbomachinery notes Syllabus Unit I: Energy transfer in turbo machines: application of first and second laws of thermodynamics to turbo machines, moment of momentum equation and Euler turbine equation, principles of impulse and reaction machines, degree of reaction, energy equation for relative velocities, one dimensional analysis only. Unit II: Steam turbines: impulse staging, velocity and pressure compounding, utilization factor, analysis for optimum U.F Curtis stage, and Rateau stage, include qualitative analysis, effect of blade and nozzle losses on vane efficiency, stage efficiency, analysis for optimum efficiency, mass flow and blade height. Reactions staging: Parsonโ€™s stages, degree of reaction, nozzle efficiency, velocity coefficient, stator efficiency, carry over efficiency, stage efficiency, vane efficiency, conditions for optimum efficiency, speed ratio, axial thrust, reheat factor in turbines, problem of radial equilibrium, free and forced vortex types of flow, flow with constant reaction, governing and performance characteristics of steam turbines. Unit III: Water turbines: Classification, Pelton, Francis and Kaplan turbines, vector diagrams and work-done, draft tubes, governing of water turbines. Centrifugal Pumps: classification, advantage over reciprocating type, definition of mano-metric head, gross head, static head, vector diagram and work done. Performance and characteristics: Application of dimensional analysis and similarity to water turbines and centrifugal pumps, unit and specific quantities, selection of machines, Hydraulic, volumetric, mechanical and overall efficiencies, Main and operating characteristics of the machines, cavitations. Unit IV: Rotary Fans, Blowers and Compressors: Classification based on pressure rise, centrifugal and axial flow machines. Centrifugal Blowers Vane shape, velocity triangle, degree of reactions, slip coefficient, size and speed of machine, vane shape and stresses, efficiency, characteristics, fan laws and characteristics. Centrifugal Compressor โ€“ Vector diagrams, work done, temp and pressure ratio, slip factor, work input factor, pressure coefficient, Dimensions of inlet eye, impeller and diffuser. Axial flow Compressors- Vector diagrams, work done factor, temp and pressure ratio, degree of reaction, Dimensional Analysis, Characteristics, surging, Polytrophic and isentropic efficiencies. Unit V: Power Transmitting turbo machines: Application and general theory, their torque ratio, speed ratio, slip and efficiency, velocity diagrams, fluid coupling and Torque converter, W.E.F. July 2017 Academic Session 2017-18 characteristics, Positive displacement machines and turbo machines, their distinction. Positive displacement pumps with fixed and variable displacements, Hydrostatic systems hydraulic intensifier, accumulator, press and crane. Turbomachinery Hand Written Notes Unit I: Energy transfer in turbo machines: application of first and second laws of thermodynamics to turbo machines, moment of momentum equation and Euler turbine equation, principles of impulse and reaction machines, degree of reaction, energy equation for relative velocities, one dimensional analysis only Unit II: Steam turbines: impulse staging, velocity and pressure compounding, utilization factor, analysis for optimum U.F Curtis stage, and Rateau stage, include qualitative analysis, effect of blade and nozzle losses on vane efficiency, stage efficiency, analysis for optimum efficiency, mass flow and blade height. Reactions staging: Parsonโ€™s stages, degree of reaction, nozzle efficiency, velocity coefficient, stator efficiency, carry over efficiency, stage efficiency, vane efficiency, conditions for optimum efficiency, speed ratio, axial thrust, reheat factor in turbines, problem of radial equilibrium, free and forced vortex types of flow, flow with constant reaction, governing and performance characteristics of steam turbines. Unit III: Water turbines: Classification, Pelton, Francis and Kaplan turbines, vector diagrams and work-done, draft tubes, governing of water turbines. Centrifugal Pumps: classification, advantage over reciprocating type, definition of mano-metric head, gross head, static head, vector diagram and work done. Performance and characteristics: Application of dimensional analysis and similarity to water turbines and centrifugal pumps, unit and specific quantities, selection of machines, Hydraulic, volumetric, mechanical and overall efficiencies, Main and operating characteristics of the machines, cavitations. Unit IV: Rotary Fans, Blowers and Compressors: Classification based on pressure rise, centrifugal and axial flow machines. Centrifugal Blowers Vane shape, velocity triangle, degree of reactions, slip coefficient, size and speed of machine, vane shape and stresses, efficiency, characteristics, fan laws and characteristics. Centrifugal Compressor โ€“ Vector diagrams, work done, temp and pressure ratio, slip factor, work input factor, pressure coefficient, Dimensions of inlet eye, impeller and diffuser. Axial flow Compressors- Vector diagrams, work done factor, temp and pressure ratio, degree of reaction, Dimensional Analysis, Characteristics, surging, Polytrophic and isentropic efficiencies. Unit V: Power Transmitting turbo machines: Application and general theory, their torque ratio, speed ratio, slip and efficiency, velocity diagrams, fluid coupling and Torque converter, W.E.F. July 2017 Academic Session 2017-18 characteristics, Positive displacement machines and turbo machines, their distinction. Positive displacement pumps with fixed and variable displacements, Hydrostatic systems hydraulic intensifier, accumulator, press and crane. Are You Want More Stuffs !! Leave a Reply Cancel reply Author: engineering Related posts
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HSC Agriculture 1st Paper Question Solution 2023 - (เฆฆเง‡เฆ–เงเฆจ เฆเฆ–เฆพเฆจเง‡) HSC Agriculture 1st Paper Question Solution 2023 โ€“ all board HSC Agriculture 1st Paper Question Solution 2023 is now available here. recently has completed the HSC education examination some time ago today. After completing the test every test is now searching to see the solution of this test question. If you are one of them then you can download Key Education First Paper Question Solution from our register today. 17th August HSC Exam is held accordingly today 17th September from 10 am to 1 pm HSC Agriculture exam was held. The examinees have now completed the examination and returned home. After returning home now they are anxious to see the solution of the question. HSC Agriculture 1st Paper Question Solution 2023 Are you looking for solutions to HSC education questions? Then you have entered the right place, you will get the solution of the question from here today. A total of eight education boards of Bangladesh have held some education exams today. A total of 8,500 examinees of Bangladesh have participated in the Key Education Examination. Testing has already been done on each board. After the completion of each board exam we have collected the questions of each board. After collecting the question paper now he has prepared the question paper solution. Solving this question will be very important for every candidate. Because by solving this question, the candidates will understand how they are going to do in the exam. Dhaka board HSC Agriculture 1st Paper Question answar Dear friends, are you HSC exam candidate of Dhaka Board. Then of course you have participated in the educational examination that was held today. This Agricultural Education Examination of the Department of Humanities was held today from 10 am to 1 pm. Already finished an education test. After completing the exam every candidate has now told us that they have some problem with their question paper. They will now search through MCQ question solutions online. After thinking about them, I collected the question paper after completing the exam. collecting the question papers, I made the experienced teachers. After downloading this question paper, you will get the solution of Dhaka Board HSC Agricultural Education exam. And through the solution of this question I published the solution of the mcq question. Rajshahi Board HSC Agriculture 1st Paper Question answar Rajshahi Board is one of the most necessary and important polls in Bangladesh. Almost lakhs of candidates participate in HSC exam every year under Rajshahi board. According to this year, this year, about 33 thousand 482 candidates have participated from here. Candidates participating in the examination must have participated in the education examination today. Today on September 17 HSC education examination was held. Candidates have already completed the exam. The test was held for a total of three hours. The first 30 minutes of the examination was for answering MCQ questions. And the remaining 2 hours and 30 minutes of written examination. Every candidate has completed the exam well. But after completing the test some number of tests informed us through message. There are some problems with their mcq question paper. And let us provide the solutions to those questions. Through this registration today, I have released the HSC Agriculture 1st Paper Question Solution 2023 of all boards to you. Hope now all the candidates have got the question solutions from our registration. The solutions of the questions that I have provided are completely correct, you can download the solution of this question without any problem. The solutions to these questions have been created by the experienced teachers. So you can download the solutions of these questions without any problem. And thank you very much for reading this my friend carefully so far.
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How do you find the slope of a line parallel to the line that passes through points: (4,0) and (3.8,2)? | HIX Tutor How do you find the slope of a line parallel to the line that passes through points: (4,0) and (3.8,2)? Answer 1 See the solution process below: Since all parallel lines have the same slope, we can find the slope of any line that is parallel to the line in the problem by finding its slope. The slope can be found by using the formula: #m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))# Where #m# is the slope and (#color(blue)(x_1, y_1)#) and (#color(red)(x_2, y_2)#) are the two points on the line. Changing the values from the problem's points yields: #m = (color(red)(2) - color(blue)(0))/(color(red)(3.8) - color(blue)(4)) = 2/-0.2 = -10# Every line parallel to the problem line and the line in the problem have the same slope, which is" #m = 10# Sign up to view the whole answer By signing up, you agree to our Terms of Service and Privacy Policy Answer 2 To find the slope of a line passing through two points (xโ‚, yโ‚) and (xโ‚‚, yโ‚‚), you use the formula: slope = (yโ‚‚ - yโ‚) / (xโ‚‚ - xโ‚). So, for the points (4,0) and (3.8,2), the slope is: (2 - 0) / (3.8 - 4) = 2 / (-0.2) = -10. Therefore, any line parallel to this one will have the same slope of -10. Sign up to view the whole answer By signing up, you agree to our Terms of Service and Privacy Policy Answer from HIX Tutor When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some Not the question you need? HIX Tutor Solve ANY homework problem with a smart AI โ€ข 98% accuracy study help โ€ข Covers math, physics, chemistry, biology, and more โ€ข Step-by-step, in-depth guides โ€ข Readily available 24/7
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University Digital Conservancy :: Browsing by Subject "Lossless source coding" Browsing by Subject "Lossless source coding" Now showing 1 - 1 of 1 โ€ข Results Per Page โ€ข Sort Options โ€ข Information theory of random trees induced by stochastic grammars. Previous research has been done on the information theory of binary random rooted tree models in which every nonleaf vertex in a tree has exactly two children. Let ฮฑ be a positive integer parameter > 2. The main contribution of this thesis is to extend the information theory results for binary random tree models to ฮฑ-random tree models, which are random tree models in which up to ฮฑ children per each nonleaf vertex are allowed. An ฮฑ-random random tree model consists of a sequence {Tn : n = 2, 3, ยท ยท ยท }, in which each Tn is an ฮฑ-random tree having n leaves, and there is a context-free grammar describing how each Tn is randomly generated. In particular, balanced ฮฑ-random tree models are examined, meaning that the context-free grammars which are employed satisfy a type of balance condition. We obtain three types of results for ฮฑ-random tree models, described as follows. First, given a general ฮฑ-random tree model {Tn}, the entropy H(Tn) of each tree Tn is expressed as a recursive formula relating H(Tn) to the entropies of its subtrees. This recursion can be explicitly solved for some random tree models, and we show how this can be done for the binary random search tree model. Our second set of results, which are our main results, concern the balanced ฮฑ-random tree model {Tn}. Defining the entropy rate of Tn as the ratio H(Tn)/n, we examine the asymptotic behavior of the entropy rate as n grows without bound. This asymptotic behavior is described via a continuous non-constant periodic function P#11; โ†’ [0,โˆž), having period one, satisfying the property that H(Tn)/n = P#11;(log#11; n) for every n. The graph of P#11; is seen as a fractal and its fractal properties along with its fractal dimension are investigated. We develop a direct and indirect way of describing P#11;. The direct way expresses the graph of P#11; as the attractor of an iterated function system which is explicitly given. The indirect way obtains P#11; via a change of variable on a hierarchical entropy function discovered by J. Kieffer. Our third and final set of results concern the development of compression methods for some ฮฑ-random tree models {Tn}. In the case in which each random tree Tn is not equiprobable over the ensemble of trees in which it takes it values, we develop a encoding method via which Tn is uniquely represented by a variable-length binary codeword, so that the expected codeword length is roughly equal to the entropy H(Tn). In the case of the balanced ฮฑ-random tree model, each Tn is equiprobable, which allows us to develop encoding methods via which each Tn is uniquely represented by a fixed-length binary codeword of minimal length โŒˆH(Tn)โŒ‰. One of the encoding methods discussed employs an one-to-one mapping of the ensemble of Tn into one of the hierarchical type classes discovered by J. Kieffer, allowing Kiefferโ€™s method for encoding hierarchical type classes to be adapted to an encoding method for balanced ฮฑ-random trees.
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Peter Lynch's Formulas for Valuing a Stock's Growth - Stockezy Peter Lynchโ€™s Formulas for Valuing a Stockโ€™s Growth Why are we discussing Peter Lynch all of a sudden? Well, Peter Lynch is an American investor and a Mutual Fund Manager. As the Magellan Fund manager at Fidelity Investments between 1977 and 1990, Lynch averaged a 29.2% annual return, consistently more than double the S&P 500 stock market index and making it the best-performing mutual fund in the world. During his 13-year tenure, assets under management increased from US$18 million to $14 billion. Source: Wikipedia I hope that last line got you hooked, taking 18 million dollars and turning it into 14 billion dollars. Yes, thatโ€™s a billion with โ€œB,โ€ and there is no typo. Now that youโ€™re aware that we are talking about one of the legendary investors and fund managers, it is important to discuss his investment philosophy. Although we wonโ€™t cover every one of his investment philosophies, but we will divulge into one of my favorites. In his book โ€œOne Up on Wall Streetโ€ (If you havenโ€™t read this masterpiece, then I highly recommended you do so), Lynch gives a straightforward explanation about one of his go-to metrics for valuing a The P/E ratio of any company thatโ€™s fairly priced will equal its growth rate. If the P/E of Coca-Cola is 15, youโ€™d expect the company to be growing at about 15 percent a year, etc. But if the P/E ratio is less than the growth rate, you may have found yourself a bargain. A company, say, with a growth rate of 12 percent a year and a P/E ratio of 6 is a very attractive prospect. On the other hand, a company with a growth rate of 6 percent a year and a P/E ratio of 12 is an unattractive prospect headed for a comedown. In general, a P/E ratio thatโ€™s half the growth rate is very positive, and one thatโ€™s twice the growth rate is very negative. Source: One Up on Wall Street Letโ€™s Understand this in our simple language: The price-earning ratio (P/E ratio) is the ratio of a companyโ€™s share price to its earnings per share. P/E ratio is used for valuing companies to find out whether they are overvalued or undervalued. For example, a share trading at a P/E ratio of 10, is trading at 10X itโ€™s annual earning. Later, Lynch goes on to offer a different approach to the same basic concept: โ€œA slightly more complicated formula enables us to compare growth rates to earnings while also taking the dividends into account. Find the long-term growth rate (say, Company Xโ€™s is 12 percent), add the dividend yield (Company X pays 3 percent), and divide by the p/e ratio (Company Xโ€™s is 10). 12 plus 3 divided by 10 is 1.5.โ€ โ€œLess than a 1 is poor, and a 1.5 is okay, but what youโ€™re really looking for is a 2 or better. A company with a 15 percent growth rate, a 3 percent dividend, and a p/e of 6 would have a fabulous Source: One Up on Wall Street The P/E Ratio vs. The PEG Ratio vs. The Dividend Adjusted PEG Ratio What does all of this mean? Well, later in his book, Lynch introduced his reader to two new concepts that he developed for measuring a companyโ€™s valuation and performance, the PEG ratio and the Dividend-Adjusted PEG ratio. These two new metrics are only a variation of the standard P/E ratio, but they offer a deep insight into company performance. Price/Earnings to Growth (PEG) Ratio: P/E ratio is a great metric for shortlisting stocks, but it has its shortcoming. The biggest limitation is that the P/E ratio tells nothing about the companyโ€™s growth. Letโ€™s say two companies have a P/E ratio of 10, how will you differentiate among them? Well, the P/E ratio doesnโ€™t tell us anything about the companyโ€™s performance. On its own, itโ€™s not very helpful because if a company has no growth and earnings stayed the same, then your investments would not yield good returns. Now what Lynch did to solve the shortcoming of the P/E ratio is to factor in the projected growth rate of future earnings. So now, if two companies are trading at 10x their earnings, and one of them is growing at 5% but the other at 10%, then you can identify the latter as a better bargain that will most likely make you more money. The formula is: PEG ratio = P/E ratio / companyโ€™s earnings growth rate If the result of the PEG ratio is one or lower than that stock is at par or undervalued based on its growth rate. If the result is greater than one, then the stock is overvalued relative to its growth rate. Many investors believe the PEG ratio gives a more complete picture of a companyโ€™s value than a P/E ratio does. The Dividend-Adjusted PEG Ratio: Going even further, lynch developed another ratio called the dividend-adjusted PEG ratio. The problem with the PEG ratio was that it didnโ€™t factor in the companyโ€™s dividends, which make up a big part of the total return of most of the blue-chip stocks. So the Dividend-Adjusted PEG Ratio is a modified version of the PEG ratio that accounts for dividend income. Reinvested Dividends, especially during the Stock market crash, can create a โ€œreturn accelerator,โ€ drastically shortening the time it takes to recover losses. If you buy a stock at 20x earnings that are growing at only 7%, it may look expensive. However, if it distributes a sustainable 10% dividend, thatโ€™s clearly a much better deal. The formula is: Dividend-adjusted PEG ratio = P/E ratio / (earnings growth + dividend yield) Letโ€™s say you invested in a company ABC, which is currently trading at 100 Rs per share. Its earnings were 12 Rs per share over the past year. This is how you can calculate the stock P/E ratio: ABC P/E ratio: 100/12 = 8.3 Now letโ€™s say you find that the company ABC is projected to grow its earnings by 7% over the next three years. You can read the annual reports of the company to find the projected growth. This is how you can calculate PEG ratio: ABC PEG ratio: 8.3/7 = 1.18 Finally, letโ€™s factor in the ABC dividend yield of 3.2% and calculate the dividend-adjusted PEG ratio: ABC dividend-adjusted PEG ratio: 8.3 / ( 7 + 3.2) = 0.81 When comparing the results, you should see that, after adjusting for dividends, ABCโ€™s stock is cheaper than you might think. Here are some of the stocks who offer good dividend yields. Happeningโ€™s Around the Stock Market (24-05-2021) NSE halts all trading on technical glitch: NSE has multiple telecom links with two service providers to ensure redundancy. We have received communication from both the telecom service providers that there are issues with their links due to which there is an impact on NSE system. โ€” NSEIndia (@NSEIndia) February 24, 2021 โ€ข Sebi has asked for a detailed report from NSE on shutdown over a tech glitch. โ€ข Rupee surges 11 paise to close at 72.35 against US dollar. โ€ข D-Street experts feel the same way about bitcoin as our legendary investor Rakesh Jhunjhunwala. Stock in News (24-05-2021) โ€ข Sanofi India board has approved a final dividend of Rs 125 per share and a special dividend of Rs 240 per share. โ€ข US FPA denies application from SPARC for cancer drug Taclantis. โ€ข Heranba Industries IPO subscribed 84% on day 1. Heranba Industries shares are available in a price band of Rs 626-627 per share. โ€ข Alkem Labs received US FDA nod for generic of antibiotic drugs Omnicef and Suprax. โ€ข Mazagon Dock signed MoU with Mumbai Port Trust. โ€ข Tata Power raises Rs 900 crore via non-convertible debentures (NCDs). โ€ข SBI Card raises Rs 550 crore through bonds. โ€ข Gujart issue closure notice to the UPL Jhagadia plant saying that operating the plant is a safety risk. โ€ข RailTel Corporation of India finalised the allotment of its IPO. Check your status here. โ€ข Pfize COVID-19 vaccine has received full approval from Brazilโ€™s health regulatory agency. โ€ข NTPC inks pact to buy GAILโ€™s 25.51% stake in Ratnagiri Gas and Power Pvt Ltd (RGPPL). โ€ข Coal India board to consider the second interim dividend for FY21. Finally, want to read an interesting article about Tesla and Bitcoin, well here it is: Bubbles, bubbles bound for trouble?
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Convert Scientific Notation to Decimal Easy Conversion: Easily Convert Scientific Notation to Decimal In the field of numerical representation and computation, scientific notation provides a concise way of expressing both very large and very small numbers. Introducing Intuitive Scientific to Decimal Converter, a powerful tool designed to seamlessly convert scientific notation to decimal form. With a simple paste of scientific numbers and the click of a button, users can get their decimal equivalents instantly. In this article, we throw light on the importance of this user-friendly converter and how it empowers users to work seamlessly with numbers in their preferred format. 1. Streamlined Conversion Process: Simple Scientific to Decimal Converter simplifies the process of converting scientific notation to decimal form. Users can easily paste their scientific numbers, click on convert button and get the corresponding decimal representation instantly. 2. Smooth operation of large and small numbers: Scientific notation is especially useful for handling numbers with many digits or decimals. The converter ensures that both very large and very small numbers are accurately converted into easy-to-understand decimal format. 3. User Friendly Interface: The intuitive design of the tool requires no technical expertise. With a straightforward form and a conversion button, users can easily switch between scientific and decimal representations without any hassle. 4. Fast and Accurate Conversion: The online nature of the converter ensures fast and accurate conversion from any device with internet access. Users can perform conversions in real time, which can increase their efficiency in working with numbers. 5. Versatility in Application: Effortless Scientific to Decimal Converter finds applications in a variety of fields from scientific research and engineering to finance and data analysis. Its versatility enables users to work with numbers in the format that best suits their needs. Embrace the simplicity and power of the Simple Scientific to Decimal Converter to seamlessly switch between scientific and decimal notation. By converting scientific numbers to decimal form with one click, users can enhance their understanding of numerical data and calculations. Your numerical endeavors will be successful as you harness the capabilities of this intuitive converter, enabling you to work with numbers more effectively and efficiently. Embrace the convenience of this tool and experience the ease of converting decimal to scientific notation, empowering you to navigate the world of numbers with confidence and clarity. Popular Tools
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Meditations on Mathematics right turns. It was difficult working out the pattern of the green maze, especially the upper right corner. See the MoMath Mazes Puzzles and Problems: MoMath Square Wheels Aperiodical website presented by Adam Atkinson: โ€œThere have been various stories in the Italian press and discussion on a Physics teaching mailing list Iโ€™m accidentally on about a question in the maths exam for science high schools in Italy last week. The question asks students to confirm that a given formula is the shape of the surface needed for a comfortable ride on a bike with square wheels. What do people think? Would this be a surprising question at A-level in the UK or in the final year of high school in the US or elsewhere?โ€ I had seen videos of riding a square-wheeled bicycle over a corrugated surface before, but I had never inquired about the nature of the surface. So I thought it would be a good time to see if I could prove the surface (cross-section) shown would do the job. See Square Wheels. (Update 9/14/2023) Square Bridge That Rolls! This is an incredible application of the rolling square wheels idea described on Matt Parkerโ€™s Stand-up Maths Youtube website. It also demonstrates the difference between engineering and pure math. The engineers had to solve some challenging problems to adapt the theoretical math to a practical application. And such solutions are always required under tight time constrictions. Engineering certainly is a noble profession.
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Assignment-07-C Epicycloid Evolute-sehenry Working on this was a bit tricky, but once I understood the concepts and structure behind making these shapes, it became easier. The hardest part for me was using the equations from the website and implementing it into my code. I had to browse through a few posts of other students to get a rough idea of what to do. From the start I wanted to do an epicycloid evolute because it looked really appealing and looked similar to a flower. //Seth Henry //Tuesdays at 10:30 //Assignment: Project 7 Composition with Curves (Epicycloid Evolute) //Global Variables var nPoints = 400; var conX; var scale; var n=10 function setup() { createCanvas(400, 400); function draw() { background(100,50,mouseX); //changes background color based on the mouse position fill(mouseX,100,mouseY); //changes the epicycloid color based on the mouse position var a = 150.0 //radius a var b = 50.0 //radius b var angle = map(conX,0,width,0,6*TWO_PI); //rotate around the constraint (conX) conX = constrain(mouseX, 0, width); //constrain around mouseX and mouseY scaleA = map(conX,0,width,0,3); rotate(angle); //rotate clockwise scale(scaleA,scaleA); //change the size of the epicycloid outer portion //Epicycloid Outer for (var i=0; i<200; i++){ var theta = map(i,0,nPoints,0, 4*TWO_PI); x=(a/(a+2*b))*(a+b)*cos(theta)+b*cos(((a+b)/b)*theta); //xpetal of epicycloid y=(a/(a+2*b))*(a+b)*sin(theta)+b*sin(((a+b)/b)*theta); //ypetal of epicycloid rotate(-angle); //rotate the opposite way of the outer epicycloid //No Rotate //Epicycloid Inner for (var i=0; i<200; i++){ var theta = map(i,0,nPoints,0, 4*TWO_PI); x=(a/(a+2*b))*(a+b)*cos(theta)+b*cos(((a+b)/b)*theta); //xpetal of epicycloid y=(a/(a+2*b))*(a+b)*sin(theta)+b*sin(((a+b)/b)*theta); //ypetal of epicycloid rotate(angle); //rotate same direction of epicycloid beginShape(); //The evolute portion of the flower for (var i=0; i<200; i++){ var theta = map(i,0,nPoints,0, 5*TWO_PI); var petalX = a * (((n-1)*cos(theta)+cos((n-1)*theta))/n) //Xpetal of evolute var petalY = a * (((n-1)*sin(theta)+sin((n-1)*theta))/n) //ypetal of evolute rect(petalX-5,petalY-5,30,30); //draws the inside petals You must be logged in to post a comment.
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Up to 100s Calculation Game - Maths for Kids Up to 100s Calculation Game What to do in this Calculation Game? You just have to calculate. You will get a number when the game starts, then you can get more number clicking the ball, and then choose one that brings you close to your target number. Keep doing it, and finally you can click W (only if needed) when you are too close to complete the number.
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Technical Knowledge Base Types of P-Delta analysis On this page: Simply supported beam P-ฮด P-ฮด is a local effect associated with axial load on displacement relative to element chord extending between end nodes. Figure 1 illustrates the influence of P-ฮด on a simply supported beam. Here, a longitudinal distributed load ฯ‰ correlates with elastic bending-stiffness properties K[E] to induce vertical displacement ฮด. An additional flexural contribution comes from the relationship between this deformed configuration and axial load P. The geometric stiffness properties K[G] which dictate this relationship are discussed further in Dr. Edward L. Wilson's text, Static and Dynamic Analysis of Structures. Values for the maximum flexural response which occurs at element midspan are shown in Figure 1: Figure 1 - P-ฮด applied to a simply supported beam Cantilevered column P-ฮด Now, when observing P-ฮด effect on a cantilevered column, response is shown in Figure 2: Figure 2 - P-ฮด applied to a cantilevered column (single curvature) However, columns seldom displace with single curvature. More commonly, especially with multi-story-building analysis and design, columns deform according to a third-order (cubic) displacement pattern under double curvature. As shown in Figure 3, P-ฮด effect is much less pronounced because an inflection point intersects the element chord near midspan, previously where displacement from chord was Figure 3 - P-ฮด applied to a cantilevered column (double curvature) Cantilevered column P-โˆ† However, what is often of significance, given this loading condition and double-curvature displacement pattern, is P-โˆ† effect. Although displacement deviates from element chord much less, the lateral displacement associated with story drift is significant. With increasing levels of drift, gravity load has a greater effect on mechanical behavior, as shown in Figure 4. P-โˆ† effect should be implemented during design, whether static or dynamic, linear or nonlinear. Figure 4 - P-โˆ† applied to a cantilevered column โ€ข Wilson, E. L. (2004). Static and Dynamic Analysis of Structures (4th ed.). Berkeley, CA: Computers and Structures, Inc.
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Understanding before method - Sylvia Edwards Understanding before method Published: February 26th, 2019 I often wish that schools would not teach children how to solve problems simply by method, often without the understanding upon which method must be based. This is an odd thing to say so letโ€™s explore what I mean. The reason why children in Key Stage 1 do not set out sums in a formal way is because they need to explore and come to understand numbers (up to 100) in an informal way first. In other words, they must understand numbers thoroughly before they are given a formal method for solving number problems. Unfortunately, some children need more time to thoroughly understand the โ€˜inner workingsโ€™ of the numbers and often move into Key Stage 2 without having mastered the required level of understanding. James is one such child who, in Year 4, tries valiantly to remember the method shown by his teacher for subtracting numbers such as 73 โ€“ 38. The โ€˜decompositionโ€™ (changing a ten into units) method of subtraction means little because he has not spent enough time counting forwards and backwards along numberlines to perceive the problem visually. James separates the tens and units โ€“ ending up with 73 โ€“ 38 = 45 instead of 35. He has taken the 3 units from the 8 units because they are less, clearly not recognising the two digit numbers as values that cannot be separated in such a way. This is a common problem. If James had come to understand two digit numbers thoroughly as whole values, he would not be making such mistakes. My plea therefore is for all schools and parents to ensure that children master two digit numbers before they are given formal methods for problem solving. In my view, the best way is for children to devise their own methods rather than being given one. Many children are put off maths quite early on because they struggle to understand what they are doing. Maths is fabulous and fascinating and deserves to be explored. Once children explore numbers thoroughly and come to understand how they work โ€“ most are capable of devising their own methods, all of which leads to more effective and efficient mathematical problem solving. ยซ Back to Blog
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Unique solution - (Intro to Mathematical Economics) - Vocab, Definition, Explanations | Fiveable Unique solution from class: Intro to Mathematical Economics A unique solution in the context of systems of linear equations refers to a scenario where there is exactly one set of values for the variables that satisfies all equations in the system. This situation arises when the equations represent lines that intersect at a single point in a geometric representation, indicating that only one combination of variable values will simultaneously meet all given constraints. congrats on reading the definition of unique solution. now let's actually learn it. 5 Must Know Facts For Your Next Test 1. A system of linear equations has a unique solution if and only if the number of equations is equal to the number of variables and the equations are independent. 2. In graphical terms, a unique solution corresponds to two lines (in two dimensions) that intersect at exactly one point. 3. The determinant of the coefficient matrix must be non-zero for a linear system to have a unique solution, indicating that the lines are not parallel. 4. If a system has more equations than variables, it may still have a unique solution provided that no equation is redundant or contradictory. 5. Systems with unique solutions can often be solved using methods like substitution, elimination, or matrix operations such as row reduction. Review Questions โ€ข What conditions must be met for a system of linear equations to have a unique solution? โ–ก For a system of linear equations to have a unique solution, it must consist of as many independent equations as there are variables. This means that no equation can be derived from another, ensuring that they represent distinct lines or planes in a graphical context. Additionally, the determinant of the coefficient matrix should be non-zero, confirming that the lines are not parallel and will intersect at exactly one point. โ€ข Compare and contrast systems with unique solutions versus dependent systems. How do their graphical representations differ? โ–ก Systems with unique solutions have exactly one intersection point when graphed, indicating a single set of variable values that satisfy all equations. In contrast, dependent systems yield infinitely many solutions because their equations represent the same line or plane, meaning they overlap completely on the graph. The distinction lies in their intersection behavior: unique solutions converge at one point while dependent systems coincide entirely. โ€ข Evaluate how the concept of unique solutions impacts real-world applications such as economics and engineering. โ–ก The concept of unique solutions is vital in fields like economics and engineering where decision-making relies on precise outcomes. For instance, in optimizing resource allocation, having a unique solution ensures that there's a clear optimal strategy without ambiguity. In engineering design, it guarantees that specific parameters lead to functional designs without conflicting results. The implications extend to ensuring stability and predictability in complex systems where multiple variables interact under defined constraints. ร‚ยฉ 2024 Fiveable Inc. All rights reserved. APร‚ยฎ and SATร‚ยฎ are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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calc_contrast_aggregated: Calculate between contrast analysis from aggregated data... in cofad: Contrast Analyses for Factorial Designs Calculate between contrast analysis from aggregated data (means, sds and ns) means numeric vector of mean values for every condition sds numeric vector of standard deviation values for every condition ns numeric vector of sample size values for every condition between factor for the independent variable that divides the data into independent groups lambda_between numeric vector for contrast weights. Names must match the levels of between. If lambda_between does not sum up to zero, this will be done automatically (centering). data optional argument for the data.frame containing all variables except for lambda_between factor for the independent variable that divides the data into independent groups numeric vector for contrast weights. Names must match the levels of between. If lambda_between does not sum up to zero, this will be done automatically (centering). optional argument for the data.frame containing all variables except for lambda_between an object of type cofad_bw, including p-value, F-value, contrast weights, different effect sizes Rosenthal, R., Rosnow, R.L., & Rubin, D.B. (2000). Contrasts and effect sizes in behavioral research: A correlational approach. New York: Cambridge University Press. library(dplyr) furr_agg <- furr_p4 %>% group_by(major) %>% summarize(mean = mean(empathy), sd = sd(empathy), n = n()) lambdas = c("psychology" = 1, "education" = -1, "business" = 0, "chemistry" = 0) calc_contrast_aggregated(mean, sd, n, major, lambdas, furr_agg) For more information on customizing the embed code, read Embedding Snippets.
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See element Cubature methodยถ A cubature method on an element consists in a set of nodes (generally called gauss points) and corresponding loads which define a approximated integration method. In GetFEM it is defined on the reference elements. Degree of freedomยถ The degrees of freedom for a finite element method is the coefficients which multiply the shape functions in order to describe a (scalar or vector) field. Generally, they are the unknowns of the problem in general. An element is a small piece of a domain with a special shape (a segment, a triangle, a quadrilateron, an tetrahedron, a hexahedron or a prism) for dimensions less or equal to three. A mesh is the union of non intersecting elements. Finite element method (fem)ยถ A finite element method is defined on a real element. It consist on a certain number of degrees of freedom linked to the corresponding shape functions and a manner to glue the degrees of freedom from a element to a neighbour element. Integration methodยถ See cubature method. Quadrature methodยถ See cubature method. The mesh is composed of elements. in GetFEM, these elements are often called convexes. A mesh can be composed of elements of different dimensions (triangles, segments, quadrilaters, tetrahedra, hexahedra โ€ฆ). The mesh_fem object is a mesh with a finite element method defined on each element. This represent a finite element space on which a unknown or a data on the considered domain will be described. The mesh_im object is a mesh with a cubature method defined on each element. It is used in assembly procedures. Reference elementยถ A reference element or a convex of reference is a special element on which the elementary computations (integrals) are performed. For instance, the reference segment in GetFEM is the segment [0,1]. The reference triangle is the triangle (0,0), (0,1), (1,0). etc.
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advantages and disadvantages of cronbach alpha In the congeneric condition corrects the underestimation of . This is because the two observations are related over time the closer in time we get the more similar the factors that contribute to error. Fully-functional online survey tool with various question types, logic, randomisation, and reporting for unlimited number of responses and surveys. We are easily distractible. Development of the R language syntax (IT, JA). Table 1. As the duration increases, reliability will increase [ 3, 5, 6 ]. doi: 10.1016/j.jpsychores,.2012.10.010. Is the most common test of neuropsychological function and is well used in research. However, it requires multiple raters or observers. Register a free Taylor & Francis Online account today to boost your research and gain these benefits: Cronbach's Alpha: Review of Limitations and Associated Recommendations, /doi/epdf/10.1080/14330237.2010.10820371?needAccess=true. Schoonheim-Klein M, Muijtens A, Habets L, Manogue M, Van der Vleuten C, Hoogstraten J, et al. This was the result of faculty misunderstanding because it was a first time experience.Footnote 3 This issue was managed with feedback after each exam to avoid these mistakes in future exams. Spearmans rank correlation coefficient is used to assess the strength and direction of a relationship between two variables or to identify and test the strength of a relationship between two sets of data. Alternatively, Cronbachs alpha can also be defined as: $$ \alpha = \frac{k \times \bar{c}}{\bar{v} + (k 1)\bar{c}} $$. This approach also uses the inter-item correlations. In addition, we compute a total score for the six items and use that as a seventh variable in the analysis. 2008;13:47993. Advantages and disadvantages of using alpha-2 agonists in veterinary practice. Its expression is: where x2 is the test variance and tr(Ce) refers to the trace of the inter-item error covariance matrix which it has proved so difficult to estimate. In internal consistency reliability estimation we use our single measurement instrument administered to a group of people on one occasion to estimate reliability. Considering that in practice it is common to find asymmetrical data (Micceri, 1989; Norton et al., 2013; Ho and Yu, 2014), Sijtsma's suggestion (2009) of using GLB as a reliability estimator appears well-founded. The use, distribution or reproduction in other forums is permitted, provided the original author(s) or licensor are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. Despite its theoretical strengths, GLB has been very little used, although some recent empirical studies have shown that this coefficient produces better results than (Lila et al., 2014) and and (Wilcox et al., 2014). J. Multivar. This pilot study was conducted over one semester (FebruaryMay) with 207 year four medical students (the first clinical year after they completed and passed all preclinical courses) as per university law, who took the exam in three groups (in March, April, and May, 2014). Although it is considered a good index for station stability, it has some disadvantages: The measure is affected by exam time and dimensionality. Disadvantages of Python are: Speed. New York: McGraw-Hill; 1994. Advantages: Can compare scores before and after a treatment in a group that receives the treatment and in a group that does not. Psychol. Search for more papers by this author. The second study was the first to discuss the effect of exam duration on the reliability index of the OSCE and reported on the effect of different days of the exam on its validity [7, 15, 16]. The parallel forms estimator is typically only used in situations where you intend to use the two forms as alternate measures of the same thing. There, all you need to do is calculate the correlation between the ratings of the two observers. Probably its best to do this as a side study or pilot study. The present study investigated how ethical ideologies influenced attitude toward animals among undergraduate students. Psychol. The correlations were 0.7, 0.7, and 0.8 (p<0.001) for both Cronbachs alpha and Spearmans rank correlation, which indicated a strong correlation between the checklist score and global rating on all days of the exam. Psychol. All these indexes have been used because no single tool has been considered precise enough. The authors declare that they have no competing interests. Lets assume that the six scale items in question are named Q1, Q2, Q3, Q4, Q5, and Q6, and see below for examples in SPSS, Stata, and R. Note that in specifying /MODEL=ALPHA, were specifically requesting the Cronbachs alpha coefficient, but there are other options for assessing reliability, including split-half, Guttman, and parallel analyses, among others. J. Psychosom. However, when the skewness value increases to 0.50 or 0.60, GLB presents better performance than GLBa. The most commonly used index for this is Pearsons correlation, which is a useful tool for assessing the correlation between the OSCE score and the written exam and has been used in many published articles [1719]. Here, I want to introduce the major reliability estimators and talk about their strengths and weaknesses. McDonald, R. (1999). volume8, Articlenumber:582 (2015) PubMed 22, 209213. Psychometrika 74, 121135. Did you know that with a free Taylor & Francis Online account you can gain access to the following benefits? Effect of Varying Sample Size in Estimation of Coefficients of Internal Consistency. After all, if you use data from your study to establish reliability, and you find that reliability is low, youre kind of stuck. Res. Adv Health Sci Educ Theory Pract. Turning to sample size, we observe that this factor has a small effect under normality or a slight departure from normality: the RMSE and the bias diminish as the sample size increases. This would have been further compounded by the simplicity of calculating this coefficient and its availability in commercial softwares. 1979;13:3954. Cronbach's alpha is a measure used for assessing the dependability and internal consistency of a set of scales and test items. With that new data set active, a Compute command is then . At the end of the semester, the students took the written exam (control exam), consisting of 80 multiple-choice questions. RMSE and Bias with tau-equivalence and congeneric condition for 12 items, three sample sizes and the number of skewed items. The GLB and GLBa coefficients present a lower RMSE when the test skewness or the number of asymmetrical items increases (see Tables 1, 2). doi: 10.1016/S0167-9473(02)00072-5, Ho, A. D., and Yu, C. C. (2014). Is Cronbachs alpha sufficient for assessing the reliability of the OSCE for an internal medicine course? Rstudio: a plataform-independet IDE for R and sweave. Assess. 30, 121144. R syntax to estimate reliability coefficients from Pearson's correlation matrices. 3:34. doi: 10.3389/fpsyg.2012.00034, Sijtsma, K. (2009). Click to reveal You might use the inter-rater approach especially if you were interested in using a team of raters and you wanted to establish that they yielded consistent results. It can also be described simply as a measure of how closely related a set of items are as a collective. Organ. This correlation is known as the test-retest-reliability coefficient, or the coefficient of stability. Strong psychometric properties. 75, 365388. 2011;2:535. Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The test size (6 or 12 tems) has a much more important effect than the sample size on the accuracy of estimates. Measurement errors in multivariate measurement scales. A topic that has attracted particular attention in the psychometric literature is Cronbach's alpha (Cronbach, We would like to acknowledge Dammam University, the Internal Medicine Department, including our chairman Dr. Waleed Albaker, who supports the idea of replacing the long/short cases exam with the OSCE, faculty members, specialists, residents, Mr. Zee Shan, and the medical students who were interested in participating in the OSCE. Scale reliability, cronbach's coefficient alpha, and violations of essential tau- equivalence with fixed congeneric components. If we use Form A for the pretest and Form B for the posttest, we minimize that problem. Advantages Well known neuropsychological measure. For example, Micceri (1989) estimated that about 2/3 of ability and over 4/5 of psychometric measures exhibited at least moderate asymmetry (i.e., skewness around 1). To evaluate whether a single reliability index is enough to assess the OSCE and to ensure fairness among all participants. doi: 10.1111/emip.12100, Headrick, T. C. (2002). Congeneric and (Essentially) Tau-Equivalent estimates of score reliability: what they are and how to use them. 2014;48:62331. The reliability of the written exam was 0.79, and the validity of the OSCE was 0.63, as assessed using Pearsons correlation. Cronbach's alpha, Spearmans rank correlation, and R2 coefficient determinants are reliability indexes and none is considered the best single index. One solution has been to use factorial procedures such as Minimum Rank Factor Analysis (a procedure known as glb.fa). For the test size we generally observe a higher RMSE and bias with 6 items than with 12, suggesting that the higher the number of items, the lower the RMSE and the bias of the estimators (Cortina, 1993). Cronbachs alpha is also not a measure of validity, or the extent to which a scale records the true value or score of the concept youre trying to measure without capturing any unintended characteristics. 66, 930944. If you get a suitably high inter-rater reliability you could then justify allowing them to work independently on coding different videos. Meas. To establish inter-rater reliability you could take a sample of videos and have two raters code them independently. In fact, its possible to produce a high \( \alpha \) coefficient for scales of similar length and variance, even if there are multiple underlying dimensions. Finally, the item option will produce a table displaying the number of non-missing observations for each item, the correlation of each item with the summed index (item-test correlations), the correlation of each item with the summed index with that item excluded (item-rest correlations), the covariance between items and the summed index, and what the \( \alpha \) coefficient for the scale would be were each item to be excluded. 96, 172189. II. GLB and GLBa are found to present better estimates when the test skewness departs from values close to 0. Two computerized approaches were used for estimating GLB: glb.fa (Revelle, 2015a) and glb.algebraic (Moltner and Revelle, 2015), the latter worked by authors like Hunt and Bentler (2015). One way to accomplish this is to create a large set of questions that address the same construct and then randomly divide the questions into two sets. Sheng and Sheng (2012) observed recently that when the distributions are skewed and/or leptokurtic, a negative bias is produced when the coefficient is calculated; similar results were presented by Green and Yang (2009b) in an analysis of the effects of non-normal distributions in estimating reliability. This paper discusses the limitations of Cronbach's alpha as a sole index of reliability, showing how Cronbach's alpha is analytically handicapped to capture important measurement errors and scale dimensionality, and how it is not invariant under variations of scale length, interitem correlation, and sample characteristics. Advantages of a Bogardus Social Distance Scale Some advantages of the Bogardus social distance scale are: Ease of use: The scale is very easy to create and administer. The complication could only arise in the formulating of each option in the distance scale. ABN 56 616 169 021, (I want a demo or to chat about a new project. The requirement for multivariant normality is less known and affects both the puntual reliability estimation and the possibility of establishing confidence intervals (Dunn et al., 2014). Pell G, Fuller R, Homer M, Roberts T. How to measure the quality of the OSCE: a review of metricsAMEE guide no. However, Revelle and Zinbarg (2009) consider that gives a better lower bound than GLB. Cronbach's Alpha 4E - Practice Exercises.doc. 27, 167172. Higher values indicate higher agreement . If all of the scale items you want to analyze are binary and you compute Cronbachs alpha, youre actually running an analysis called the Kuder-Richardson 20. If the internal consistency (as measured by Cronbach's Alpha) is low for a given survey, there are two ways that you can potentially increase it: 1. doi:10.1111/j.1600-0579.2010.00653.x. Graham JM. Educ. The parallel forms approach is very similar to the split-half reliability described below. Data Anal. Bias of coefficient alpha for fixed congeneric measures with correlated errors. In addition, as demonstrated in Table 3, the Cronbach's alpha coefficient was 0.892 with 95% confidence . For legal and data protection questions, please refer to our Terms and Conditions and Privacy Policy. In the case of non-violation of the assumption of normality, is the best estimator of all the coefficients evaluated (Revelle and Zinbarg, 2009). (reverse worded). Both the parallel forms and all of the internal consistency estimators have one major constraint you have to have multiple items designed to measure the same construct. The Cronbach's alpha is the most widely used method for estimating internal consistency reliability. Teach Learn Med. https://doi.org/10.1186/s13104-015-1533-x, DOI: https://doi.org/10.1186/s13104-015-1533-x. In short, youll need more than a simple test of reliability to fully assess how good a scale is at measuring a concept. To measure the validity of the exam, we conducted a Pearsons correlation to compare the results of the OSCE and written exam scores. Psychol. Iramaneerat C, Yudkowsky R, Myford CM, Downing S. Quality control of an OSCE using generalizability theory and many-faceted Rasch measurement. Consider the following syntax: With the /SUMMARY line, you can specify which descriptive statistics you want for all items in the aggregate; this will produce the Summary Item Statistics table, which provide the overall item means and variances in addition to the inter-item covariances and correlations. GLB is recommended when the proportion of asymmetrical items is high, since under these conditions the use of both and as reliability estimators is not advisable, whatever the sample size. doi: 10.1007/s11336-008-9099-3, Green, S. B., and Yang, Y. Plasma noradrenaline and renin concentrations are reduced. Res. This is often no easy feat. Cronbach's , Revelle's , and Mcdonald's H: their relations with each other and two alternative conceptualizations of reliability. Alternatively, you might want to use the option reverse(ITEMS) to reverse the signs of any items/ variables you list in between the parentheses. MHS: Contributed designing the study, analysis and interpretation of data and reviewed the initial draft manuscript. Methodol. In addition, the limitations and strengths of several recommendations on how to ameliorate these problems were critically reviewed. The Kaiser-Meyer-Olkin (KMO) test and Bartlett's chi-square tests were used to test the validity of the questionnaire and whether it was . Cronbach's alpha, a measure of internal consistency, was calculated to test the reliability of the questionnaire. Article Psychometrika 80, 182195. Int J Med Educ. To solve this issue, there must be at least two to three indexes to ensure the reliability of the exam. 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golang convert positive to negative So the code, And, finally it comes to that this is related to constants in Golang. Can I (an EU citizen) live in the US if I marry a US citizen? To make a "black and white negative", first convert a color photo to a black and white here, download result and then select and process it here, on this page, with default settings. . Inside the main() function, we declared an int64 variable with a 10 value. How to copy one slice into another slice in Golang? "math" How to find the index value of specified string in Golang? As you walk, the sun rises higher in the sky and the air temperature increases. The int size is implementation-specific; its either 32 or 64 bits; hence, you wont lose any information when converting from int to int64. Sam Allen is passionate about computer languages. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. fmt.Sprintf (): The fmt.Sprintf () function formats according to a format specifier and returns the resulting string. He is comfortable working in front-end and back-end development. Right click the photo in Win 7 Paint and a menu will launch that has the Invert Color option. Copy that cell with the Ctrl + C. Select the negative numbers you want to convert. How could one outsmart a tracking implant? Making statements based on opinion; back them up with references or personal experience. Anyways, my question is more about how to subtract from unsigned integers when the value (sometimes you want to add and sometimes subtract) is coming from an argument (that must be a signed type) which makes it so that you have to do type conversion before subtracting/adding. ) Step 2 Start an if condition and call Signbit() function in the if condition. The text was updated successfully, but these errors were encountered: You signed in with another tab or window. The math.Abs() function in Golang is used to return the absolute value or We and our partners use cookies to Store and/or access information on a device. Answer (1 of 2): Since potential is just a mathematical trick with no physical meaning, you can convert positive to negative voltage just by naming it differently. VB. SENTENCE [Positive] TRANSFORMED [Negative] Only a fool can say this. These are the two ways to check whether the number is positive or not. Enter the value "-1" in any nearby empty cell. Step 4: Divide num by 10. Right-click on the selection and select Paste Special from the context menu to launch the Paste Special dialog box. acknowledge that you have read and understood our, Data Structure & Algorithm Classes (Live), Full Stack Development with React & Node JS (Live), Data Structure & Algorithm-Self Paced(C++/JAVA), Full Stack Development with React & Node JS(Live), GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, math.Inf() Function in Golang With Examples. 2. If there is no argument ( x - input value) passes to the function, then the compiler will produce an error. Approach to solve this problem. When converting a negative number to an unsigned integer and later adding that value it results in subtracting. Here, id is a variable of type integer. Open your Google Sheets spreadsheet. Solution. Numbers of users = 29 and converted monthly price = -100 then i would need the number of users 29 to be converted to -29 below is an outline of the data i am looking at. I recently needed an implementation of the abs function to solve the Day 20 challenge in Advent of Code 2017. You should remove superfluous parentheses here: a := (uint8)(10) Go language provides inbuilt support for basic constants and mathematical functions to perform operations on the numbers with the help of the math package. 1 ACCEPTED SOLUTION. Golang Itoa: How to use strconv.Itoa() Function, Golang reflect.TypeOf: How to Find Data Type in Go, Golang Max Int: The Largest Possible Value for an Int, Golang min: How to Find Minimum Number in Go. Aug 9, 2012. . I hope the above article on how to convert String to integer in golang is helpful to you. This parameter is required. Asking for help, clarification, or responding to other answers. Asking for help, clarification, or responding to other answers. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Start by focusing on one small area and on how you can approach that in a more positive way. We make use of First and third party cookies to improve our user experience. The strconv.Itoa() function accepts only int32 variables; if you try to use a variable of type int64, you will get the following error. 6 comments Contributor on Dec 21, 2009 removed their assignment In this example, we imported three basic modules. How could magic slowly be destroying the world? Below are two ways as follows. Solution Sage. Why does secondary surveillance radar use a different antenna design than primary radar? "The circle is when the positive and negative . Well occasionally send you account related emails. He has developed a strong foundation in computer science principles and a passion for problem-solving. Share it on Social Media. Given an URL, auto-detect and convert the encoding of the HTML document to UTF-8 if it is not UTF-8 encoded in Go. The signal requirement will be Negative Pulse. Edit for replacing uint32 with int32 for the first situation. @MTCoster yes, because I'm sending packets over a network. Join 6,000 subscribers and get a daily digest of full stack tutorials delivered to your inbox directly.No spam ever. In this tutorial, we will convert an int64 to a string data type. Type: =SIGN (number) into the cell and replace number with the cell . How to Copy an Array into Another Array in Golang? The given string may begin with a leading sign: "+" or "-" to convert as a positive or negative integer. A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. To convert an int to string in Golang, strconv.Itoa (): The strconv.Itoa () function converts an integer base 10 value to an ASCII string. small := float64(3) Positive Scenario. How to perform git tasks with build script? "fmt" In this article, you will learn how to convert a positive number into a negative number in Javascript. For TPS54620, the V DEV (min) is 4.5 V. Hence, the input range of 4.5 to 5.5 V, as given in Table 2, is easily supported by the TPS54620-based inverter circuit in Figure 3. Negative base works similar to positive base. Although it was designed to convert +V to +V, by changing the polarity of assorted components at the secondary site you will be able to generate required negative . For Vista and earlier. Pages are continually updated to stay current, with code correctness a top priority. Manage Settings Check if a number is positive, negative or zero using bit operators in C++, Write a Golang program to check whether a given number is a palindrome or not. Copy the following script into the Script Editor window. In order to convert a positive number into a negative number, you can use the unary negation operator -. Syntax import ( An integer can be int, int64, or int128. @rollin You can multiply by -1 to convert any positive number to negative in PHP, here is code as example: 1 2 3 4 5 6 7 8 9 <?php $number = 5 ; // Make it positive in case if number is negative $negative = abs ($number) * -1 ; // Output: -5 print_r ($negative); 0 | 0 0 1 Rows per page 15 Related Threads: katharina jarod To build a dc-dc converter that can generate stabile negative output voltages you will have to use a transformer .. Take a look at the attached circuit. Note: The unary negation operator - functions by converting any value on its right side into a number and negates it. So be it, and use, confusion about convert `uint8` to `int8`, Microsoft Azure joins Collectives on Stack Overflow. negative :=, package main Have a question about this project? Golang program to find an average of an array of numbers; Golang program to check number is even or odd; Golang program to find the sum of two numbers; Golang Program to remove spaces in a string; Types Conversions examples in Golang. i.e. How to check whether the input number is a Neon Number in Golang? How to rename a file based on a directory name? POSITIVE AND NEGATIVE FEEDBACK MECHANISMS Feedback mechanisms either change a system to a new state or return it to its original state. To convert the int32 variable to string data type, use the strconv.Itoa() function. Already on GitHub? C program to Check Whether a Number is Positive or Negative or Zero? About Archives Categories Tags Authors [Golang] Auto-Detect and Convert Encoding of HTML to UTF-8 October 27, 2018 Edit on Github. Continue with Recommended Cookies. How to check if a number is positive, negative or zero using Python? To learn more about go you can explore these tutorials. if the number is greater than zero, the Number is positive if the number is less than zero, the number is negative else, number=0 is executed, zero value printed to console Here is an example program to check positive or negative numbers. happyume. It is a consequence of not knowing how to implement them efficiently. Submitted by Nidhi, on February 16, 2021 . play.golang.org. How to Convert A Positive Number Into A . In line 14, we initialize the index of the element to delete from the array. Convert the positive and negative integer to an absolute value. An int64 is an immutable value representing signed integers with values ranging from negative. Lets say you have a number variable called a with value 123. That's a total of 4294967296 different integer values (2^32. To do so, you need to import the math module. I would like to use the Mx-22 positive pulse signal into the soon to be released Plantraco buddy box module, (which would handle sending the rf). EFFECTS OF GLOBALIZATION ON BUSINESS MANAGEMENT IN DEVELOPED COUNTRIES. If you would like to change your settings or withdraw consent at any time, the link to do so is in our privacy policy accessible from our home page.. That's a total of 4294967296 different integer values (2^32 i.e. Best, Khuong. I also notice Russ Cox's statement in Golang-nuts The lack of generalized assertions, like the lack of backreferences, is not a statement on our part about regular expression style. Windows 7. What I want is for the loss to be a descending column below the 0 line, and the gain to be a positive column above. I think they are equal. using System ; class Demo { public static void Main ( ) { int [ ] arr = { 10 , - 20 , 30 , - 40 , 50 . The Go programming language has no built in abs function for computing the absolute value of an integer . An int32 can store any integer between -2147483648 and 2147483647. value1 := float64(. This results in an error : prog.go:8: constant 4294967294 overflows int32, How can both result in different results. string representation of a given integer in a chosen base, ranging from 2 to 36). math package of GO provides a Signbit method that can be used to check whether a given number is negative or positive. I need a 'standard array' for a D&D-like homebrew game, but anydice chokes - how to proceed? the value (sometimes you want to add and sometimes subtract) is coming Syntax: math.Lgamma() Function in Golang with Examples, math.Float64bits() Function in Golang With Examples, atomic.AddInt64() Function in Golang With Examples, atomic.StoreInt64 () Function in Golang With Examples, reflect.FieldByIndex() Function in Golang with Examples, strings.Contains Function in Golang with Examples, bits.Sub() Function in Golang with Examples, io.PipeWriter.CloseWithError() Function in Golang with Examples, time.Round() Function in Golang With Examples, reflect.AppendSlice() Function in Golang with Examples. Why are there two different pronunciations for the word Tee? I have try to write one logic is to convert an int32 positive value to a corresponding negative one, i.e., abs(negativeInt32) == positiveInt32. Why did OpenSSH create its own key format, and not use PKCS#8? Golang program to check whether the number is positive or not using Signbit () function in the Math library. Step 2: Define res = 0 variable and start a loop until num becomes 0. math.NaN()) to an absolute value. The two results are different because the first value is typecast to an unsigned int32 (a uint32). Answer link. Method #1 - Multiply by Negative 1 with a Formula The first method is pretty simple. Golang Program to Put Positive and Negative Numbers in a Separate Array In this Go example, we created two separate functions (putPositiveNums and putNegativeNums) that place the positive numbers in the positive and negative numbers in the negative array. How to check the specified rune in Golang String? Step 2 Compare the number with the respective relational operator. We and our partners use data for Personalised ads and content, ad and content measurement, audience insights and product development. Golang Floor (x) returns the greatest integer value less than or equal to x. var a = "codesource"; In order to convert a string to an array, you can use the split method. Find centralized, trusted content and collaborate around the technologies you use most. It's value is 4294967294. ) The given program is compiled and executed successfully on Microsoft Visual Studio. We have 2 ways to Convert Decimal to Binary numbers in Golang. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The output of the above golang program is: 0 strconv.Atoi: parsing "StringTest15": invalid syntax Conclusion. For instance, I have a measure that calculates lost opportunities, intended to be used on a column chart along with a net gain measure. Some of our partners may process your data as a part of their legitimate business interest without asking for consent. By using our site, you Krunal Lathiya Krunal Lathiya is a Software Engineer with over eight years of experience. Change Number Formatting. To understand the below programs, You have the understanding the following features in Go Language. "math" Another possibility would be the freeware IrfanView. Why is water leaking from this hole under the sink? :). Looking to protect enchantment in Mono Black, Attaching Ethernet interface to an SoC which has no embedded Ethernet circuit, Make "quantile" classification with an expression. ) In addition, Krunal has excellent knowledge of Distributed and cloud computing and is an expert in Go Language. Best method to stream results from web service, Redshift returns a []uint8 instead of an integer, converting between them returns incorrect values, Numbered String to Int with leading zeros. How to navigate this scenerio regarding author order for a publication? ) For example in base 2 we multiply bits to 1, 2, 4, 8 and so on to get actual number in decimal. We have 2 ways to convert Decimal to HexaDecimal numbers in Golang. "You may using both int8" - the value I'm subtracting/adding to must be, No it's not a hack, this is the way to go ( idiomatic way). In this example, R5 is set equal to 10 k and V REF to 0.8 V. Consequently, using Equation 2 . func main() { large := float64(40), package main func main() { Just put a "-" or a "0 -" in front. Next, we used the unicode ToUpper function (up := unicode.ToUpper (lwch)) that converts the lowercase to an uppercase character. Please take a look at this one as sample. Python Program to Check if a Number is Positive, Negative or 0, Check whether product of integers from a to b is positive, negative or zero in Python, Golang Program to check whether given positive number is power of 2 or not, without using any branching or loop, Write a Golang program to check whether a given number is prime number or not. How to check whether a number is a prime number or not? Specifies the real number (short int, integer, long, float or double) / infinity / NaN. Method 2: Using Relational Operators In this example, we will use the relational operator >= and < to check whether the number is positive or negative. Related Posts: Create a Sleek Note app With Flutter CRUD Application Using Django and JavaScript This method is using relational operator <, >= with the below syntax. Here are steps to check positive or negative numbers using if-else statements. // C# program to convert negative values //an integer array into positive. What is the difference between int and int64 in Go? Divide by , multiply by 180. Thanks for contributing an answer to Stack Overflow! Christian Science Monitor: a socially acceptable source among conservative Christians? Also, using Equation 2, R4 can be calculated for the desired output voltage. positive to negative voltage. Copy. Theory and Practice. I tried to go around this restriction by converting the integer into a byte, which is then converted to an unsigned integer. Parameters of Inf () Function in Go Language x - Where x is any Valid int Input value (positive or negative). If you want to convert negative to positive and positive to negative I think you can use assign widget and set Value = -Value. Sign up for a free GitHub account to open an issue and contact its maintainers and the community. By using 'no'/'not and the opposite word. Go language provides inbuilt support for basic constants and mathematical functions to perform operations on the numbers with the help of the math package. golang/go Python range() Function Explained With Examples | Tutorial, Disable Connected Experiences In Office 365, Adrianna Vineyard River Stones Malbec 2015, Makeup Obsession All We Have Is Now Eyeshadow Palette. Find centralized, trusted content and collaborate around the technologies you use most. In the following program, we take a string str that has only a numeric value and convert this string value to integer value using strconv.Atoi() function. modulus |x| of a real number x is the non-negative value of x without regard to its sign. Then We convert the "negative" number to a float64 and pass that to the math.Abs method. In the code above: In line 4, we import the fmt package for printing. Problem Solution: In this program, we will read an integer number from the user and check the given number is POSITIVE or NEGATIVE. "math" Let's say you have a 12V battery, because the difference between the terminals is 12V. func main() { Using reflect.TypeOf() function, we can print the data type of a variable in the Go console. Next we introduce the "negative" integer. 'Converts a value from a negative to positive DIM Variable Variable = Abs (SourceFields ("Queryname.Columnname")) CurrentField.Value = Variable. To view the purposes they believe they have legitimate interest for, or to object to this data processing use the vendor list link below. matches any character (except for line terminators) * matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy) $ asserts position at the end of a line. ; In line 10, we initialize an array of type string called originalArray.At the time of declaration, originalArray contains four values. How to define int64 in Golang You check whether the sign of the specified number is negative or negative zero with the help of Signbit () function provided by the math package. He is not an intelligent boy. The original image is not changed. Go provides an int data type that can be declared as anint which will only store signed numeric values. Type 0;"0";0 in the Type box and click the OK button. i.e., we should first get the corresponding value of 1 in int32 type and, then convert it to the two's complementary form as value of -1. How Intuit improves security, latency, and development velocity with a Site Maintenance- Friday, January 20, 2023 02:00 UTC (Thursday Jan 19 9PM Were bringing advertisements for technology courses to Stack Overflow. The two results are different because the first value is typecast to an unsigned int32 (a uint32 ). int positive = 10; int negative = -positive;//-10 int negative2 = -235; int positive2 = -negative2;//235 VB sample Dim positive As Integer = 10 Dim negative As Integer = -positive '-10 Dim negative2 As Integer = -235 Dim positive2 As Integer = -negative2 '235 To subscribe to this RSS feed, copy and paste this URL into your RSS reader. To convert the int64 variable to string data type, use the strconv.FormatInt() function. Could you observe air-drag on an ISS spacewalk? This page was last reviewed on Dec 4, 2021. package main Golang Program to Count Positive and Negative Numbers in an Array using the for loop range. Occasionally, i have learned that why we can not do. By clicking Sign up for GitHub, you agree to our terms of service and value1 :=, package main Go (golang) lets you round floating point numbers. Subsequent firmware updates fixed the issue. Golang program that uses rand.Int. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. Programmer | Writer | bitsized dot me at gmail dot com. You usually call the negative terminal "0V" a. None but a fool can say this. Useful front-end & UX tips, delivered once a week. That's it. Letter of recommendation contains wrong name of journal, how will this hurt my application? ) Is the rarity of dental sounds explained by babies not immediately having teeth? func main() { Then, select the ellipsis button. So, you need to add a math package in your program with the help of the import keyword to access the Inf() function. You can find positive infinity (if sign >= 0) or negative infinity (if sign < 0) with the help of the Inf () function provided by the math package. Hi Esrom, Use can use abs () to convert negative to positive as Nuno said. Select a range of cells. 1. // Change the meaning of n for the implementation below. you could use a schmidt . import ( It takes input a float and returns a bool. [Opposite words: only > none but; always/ever > never; good > bad; right > left/wrong; etc.) "fmt" How can I convert a zero-terminated byte array to string? rev2023.1.18.43170. You can find positive infinity (if sign >= 0) or negative infinity (if sign < 0) with the help of the Inf() function provided by the math package. So SAVE WATER. Which in a number system where one integer less than 0 is 4294967295; 4294967294 happens to be the sum of 0 - 2. import ( This page was last updated on Dec 4, 2021 (edit link). Any number greater than zero is positive, and if it is less than zero, then the negative number. to your account. How to Convert Decimal to HexaDecimal Number with examples. C# Program to check if a number is Positive, Negative, Odd, Even, Zero. Or more simply: uint32(-2). To do it, locate the Value column in the lower-left corner of the Integration Mapping window. Go language provides inbuilt support for basic constants and mathematical functions to perform operations on the numbers with the help of the math package. Signbit() is a function in math library that accept a argument of float type as shown in below syntax. Making statements based on opinion; back them up with references or personal experience. Another way is to use the relation operators using which we can compare a number with zero and predict whether the number is positive or not. Golang math.Abs - Syntax mixed abs(mixed n); Golang math.Abs - Parameter and Retun type Example: #1 - Golang math.Abs Convert the positive and negative integer to an absolute value. Is every feature of the universe logically necessary? Output isif(typeof ez_ad_units!='undefined'){ez_ad_units.push([[250,250],'cloudhadoop_com-box-4','ezslot_3',121,'0','0'])};__ez_fad_position('div-gpt-ad-cloudhadoop_com-box-4-0'); In this post, you will learn the Go language example to check if a given input integer number is positive or negative with the [if else](/2018/11/learn-golang-tutorials-if-else.html) statement, and with the Math signbit function. The strconv package provides functions and utilities that help us convert from and to string values. Well as you have found, we get the value of 4294967294. You can use expression mul ( YOUR_NUMBER ,-1) Please click Accept as solution if my post helped you solve your issue. func RandInt(lower, upper int) int { rand.Seed(time.Now().UnixNano()) rng := upper - lower return rand.Intn(rng) + lower } True random number generation is hard. To avoid the error, always use the int32 variable while working with strconv.Itoa() function. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Not the answer you're looking for? Go Program to Convert Character to Uppercase In this Go Program, to convert the lowercase character to uppercase, we used IsLetter (if unicode.IsLetter (lwch)) to check for alphabets. Agree it has an option for reversing a negative. The best way to convert a float to int in Golang, use the int () function. To convert any exponent from negative to positive you either move the variable from the numerator to the denominator or you move the variable from the denominator to the numerator. If anyone knows of either an existing inverter device or of a schematic which could do this, it would be very much appreciated. If you see the If condition, it checks whether the user-given number is greater than zero. Solution 2 -abs (n) is a really good answer by Tom Karzes earlier because it works whether you know the number is negative or not. For those who come to this problem, I have answered the question myself. Krunal Lathiya is a Software Engineer with over eight years of experience. Solution 1 If you want to force a number to negative, regardless of whether it's initially positive or negative, you can use: -abs (n) Note that 0 will remain 0. This occurs here: uint32(^uint32(int32(2) -1)) How to Check Whether a Number is Krishnamurthy Number or Not in Java? Is this an idiomatic approach or should it be done more explicitly? Using this online tool, you can also make a positive from the negative of photo with all default settings. You'll get a notification every time a post gets published here. Krunal has experience with various programming languages and technologies, including PHP, Python, and JavaScript. Thanks for contributing an answer to Stack Overflow! - The Go Programming Language The given string may begin with a leading sign: "+" or "-" to convert as a positive or negative integer. Two parallel diagonal lines on a Schengen passport stamp, Looking to protect enchantment in Mono Black, Can a county without an HOA or covenants prevent simple storage of campers or sheds, Is this variant of Exact Path Length Problem easy or NP Complete. Dean R. May 30, 2018. Are there developed countries where elected officials can easily terminate government workers? Assert that the Regex below matches. The below program uses Math Signbit function. When converting a negative number to an unsigned integer and later adding that value it results in subtracting. "math" we respect your privacy and take protecting it seriously, CRUD Application Using Django and JavaScript, Build A Desktop Application with Vuejs and Electronjs, Build a Signature Capture Application Using Canvas & Kotlin, Writing cleaner code with higher-order functions, A Comprehensive Roadmap To Web 3.0 For Developers In 2023, How to Build an Animated Slide Toggle in React Native, 5 Best Practices for Database Performance Tuning, From Drawing Board to Drop Date How a Successful App is Developed, How to fix TypeError: numpy.ndarray object is not callable, How to fix the fatal: refusing to merge unrelated histories in Git, How to fix the TypeError: expected string or bytes-like object in Python, How to fix the ImportError: attempted relative import with no known parent package in python, How to fix Crbug/1173575, non-JS module files deprecated. a := (uint8) (10) b := (int8) (-8) fmt.Println (a + (uint8) (b)) // result: 2 Is this an idiomatic approach or should it be done more explicitly? Now if we run the program, the output will be:-5 is a negative number. Here are steps to check positive or negative numbers using if-else statements.if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[320,50],'cloudhadoop_com-medrectangle-3','ezslot_1',117,'0','0'])};__ez_fad_position ('div-gpt-ad-cloudhadoop_com-medrectangle-3-0');if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[320,50],'cloudhadoop_com-medrectangle-3','ezslot_2',117,'0','1'])};__ez_fad_position margin-left:auto!important;margin-right:auto!important;margin-top:7px!important;max-width:100%!important;min-height:50px;padding:0;text-align:center!important}. =, package main have a 12V battery, because I 'm sending packets over a network variable working. Converting the integer into a negative number is no argument ( x - input value positive! Lets say you have a question about this project passion for problem-solving is or... Management in developed COUNTRIES Where elected officials can easily terminate government workers a positive number into negative. 'Ll get a notification every time a Post gets published here - how to copy one slice into another in! Later adding that value it results in an error numbers using if-else statements numeric values implementation of the package... We have 2 ways to convert negative values //an integer array into another array in Golang is helpful you! = -Value the OK button this tutorial, we use cookies to improve user! Only store signed numeric values use data for Personalised ads and content, ad and content, ad and,. Cell with the help of the math module the air temperature increases,! The numbers with the cell can store any integer between -2147483648 and 2147483647.:. Hurt my application? the type box and click the photo in Win 7 Paint and a for. As a part of their legitimate BUSINESS interest without asking for help, clarification or. Function in math library comfortable working in front-end and back-end development time a Post gets published here why... This hole under the sink the sky and the opposite word Monitor a... Or double ) / infinity / NaN is related to constants in Golang 2009 removed their assignment in tutorial... Array to string data type, use the strconv.Itoa ( ): the fmt.sprintf ( ) convert... Another tab or window values //an integer array into another array in Golang Multiply by negative with... Engineer with over eight years of experience knowing how to check whether a number is positive not! Array into another array in Golang V. Consequently, using Equation 2 4294967296 integer. Positive to negative I think you can approach that in a more positive way converted to unsigned. Above: in line 14, we get the value & quot the... ) passes to the math.Abs method it takes input a float to int Golang..., using Equation 2 golang convert positive to negative ; back them up with references or personal experience Decimal. Temperature increases content, ad and content measurement, audience insights and product development Only store numeric! Invert Color option, package main have a question about this project value ( positive or or. To 0.8 V. Consequently, using Equation 2, R4 can be used to check if a number and it... Sending packets over a network experience on our website radar use a antenna... Assign widget and set value = -Value, or responding to other answers -1 ) click! Officials can easily terminate government workers C. select the ellipsis button solution if my Post you... Openssh create its own key format, and not use PKCS #?. Number or not using Signbit ( ) function data as a part their! Occasionally, I have answered the question myself used to check the specified rune in Golang edit on.! Content, ad and content, ad and content, ad and content measurement, audience and. Only store signed numeric values circle is when the positive and negative: = (... One slice into another array in Golang with various programming languages and,. Battery, because I 'm sending packets over a network - how to navigate this scenerio regarding order... Tutorials delivered to your inbox directly.No spam ever your data as golang convert positive to negative part of their BUSINESS. Two different pronunciations for the implementation below account to open an issue and contact its maintainers the! Package for printing Personalised ads and content, ad and content, ad and measurement. ) function in the sky and the community and start a loop until num 0.... Exchange Inc ; user contributions licensed under CC BY-SA =, package main have 12V... Formats according to a float64 and pass that to the math.Abs method in below syntax function in Go process data... Passion for problem-solving how you can use assign widget and set value -Value! Current, with code correctness a top priority Python, and, finally it to! Returns a bool ) to an unsigned integer years of experience golang convert positive to negative this it..., 2018 edit on Github constant 4294967294 overflows int32, how can both result in different results result different... Live in the Go programming language has no built in abs function to solve Day! The air temperature increases submitted by Nidhi, on February 16, 2021 ; quot... Updated successfully, but anydice chokes - how to proceed about this project Writer | bitsized dot at... Short int, integer, long, float or double ) / infinity / NaN an EU citizen ) in. Conservative Christians to Go around this restriction by converting any value on its right side into a byte, is! Declared an int64 variable with a Formula the first value is golang convert positive to negative to an unsigned integer and later that. Consequence of not knowing how to copy one slice into another array Golang! Will launch that has the Invert Color option the meaning of n the... Menu will launch that has the Invert Color option anint which will Only store signed numeric values ; 0V quot!, Sovereign Corporate Tower, we imported three basic modules stay current, with code a! Temperature increases to copy one slice into another slice in Golang in addition, Krunal excellent... Convert an int64 is an expert in Go language provides inbuilt support for basic and... A 'standard array ' for a publication? 2018 edit on Github integer... Of our partners use data for Personalised ads and content, ad and content ad. And select Paste Special from the context menu to launch the Paste Special dialog.... Ctrl + C. select the negative numbers using if-else statements have answered the question myself third party cookies to our. Non-Negative value of x without regard to its original state I have learned that we... To its original state an array of type string called originalArray.At the time of declaration, originalArray contains four.! With all default settings # 1 - Multiply by negative 1 with a 10 value problem-solving! Value = -Value recently needed an implementation of the math module best browsing experience on our.! Context menu to launch the Paste Special from the negative number to an unsigned integer a gets. Implementation below from 2 to 36 ) copy that cell with the cell and replace number with the Ctrl C.... Is greater than zero, then the compiler will produce an error::! Be declared as anint which will Only store signed numeric values Neon number in Javascript if there is no (... To improve our user experience back-end development ' for a publication? type box and the. ] Only a fool can say this to stay current, with correctness. Distributed and cloud computing and is an immutable value representing signed integers with ranging! Useful front-end & UX tips, delivered once a week line 10 we! User-Given number is positive or negative numbers using if-else statements Where x is the non-negative value specified. Integers with values ranging from 2 to 36 ) to open an issue golang convert positive to negative contact its maintainers and air... Inbox directly.No spam ever and our partners may process your data as a part of their legitimate interest... Signbit ( ) function be the freeware IrfanView to its original state result in different.... Best way to convert a golang convert positive to negative number into a negative that accept a argument of float type as shown below... Mtcoster yes, because I 'm sending packets over a network the code and! Is positive, negative or zero golang convert positive to negative Python letter of recommendation contains wrong name journal... Inverter device or of a golang convert positive to negative number x is any Valid int value. Nearby empty cell prog.go:8: constant 4294967294 overflows int32, how can both result in different results best to... Error: prog.go:8: constant 4294967294 overflows int32, how can both result in different results all default settings your. Strconv.Itoa ( ) ) to an unsigned integer and later adding that value it results an. Even, zero live in the type box and click the photo in Win 7 Paint and a menu launch. 10, we will convert an int64 to a string data type, use the strconv.Itoa ( to. Regarding author order for a publication?, ad and content measurement, audience insights and product development directory. On a directory name Win golang convert positive to negative Paint and a passion for problem-solving the Integration Mapping window if-else statements then! Select the ellipsis button to stay current, with code correctness a top.! A question about this project 4294967294 overflows int32, how can both result in different results for. Int32 can store any integer between -2147483648 and 2147483647. value1: = float64 ( into.. Our user experience func main ( ) ) to an unsigned int32 ( uint32! Can both result in different results code 2017, on February 16, 2021 |x|! Type box and click the OK button approach or should it be done more explicitly knowledge. A Signbit method that can be used to check whether a number positive... In Golang one as sample the non-negative value of 4294967294 long, or... Their legitimate BUSINESS interest without asking for help, clarification, or responding to other.! Next we introduce the & quot ; negative & quot ; negative & quot ; in any nearby empty.... William Schur Regis Grandson, Mapanagutang Kilos Loob, Serenity Funeral Home Flint, Mi, Articles G golang convert positive to negative
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Evacuating Robots from a Disk Using Face-to-Face Communication Evacuating Robots from a Disk Using Face-to-Face CommunicationArticle Assume that two robots are located at the centre of a unit disk. Their goal is to evacuate from the disk through an exit at an unknown location on the boundary of the disk. At any time the robots can move anywhere they choose on the disk, independently of each other, with maximum speed $1$. The robots can cooperate by exchanging information whenever they meet. We study algorithms for the two robots to minimize the evacuation time: the time when both robots reach the exit. In [CGGKMP14] the authors gave an algorithm defining trajectories for the two robots yielding evacuation time at most $5.740$ and also proved that any algorithm has evacuation time at least $3+ \frac{\pi}{4} + \sqrt{2} \approx 5.199$. We improve both the upper and lower bound on the evacuation time of a unit disk. Namely, we present a new non-trivial algorithm whose evacuation time is at most $5.628$ and show that any algorithm has evacuation time at least $3+ \frac{\pi}{6} + \sqrt{3} \approx 5.255$. To achieve the upper bound, we designed an algorithm which proposes a forced meeting between the two robots, even if the exit has not been found by either of them. We also show that such a strategy is provably optimal for a related problem of searching for an exit placed at the vertices of a regular hexagon. Volume: vol. 22 no. 4 Section: Distributed Computing and Networking Published on: August 27, 2020 Accepted on: July 31, 2020 Submitted on: March 13, 2020 Keywords: Computer Science - Data Structures and Algorithms Source : OpenAIRE Graph โ€ข Funder: Natural Sciences and Engineering Research Council of Canada
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Advanced Excel Formulas Excel is an invaluable tool for anyone working with data, but its true power lies in its advanced formulas. While basic formulas are essential for simple calculations, mastering advanced formulas can significantly streamline your work, automate complex tasks, and unlock insights hidden within your data. This comprehensive guide will explore some of the most powerful advanced Excel formulas, equipping you with the knowledge to tackle any data challenge. Beyond the Basics: Mastering Advanced Excel Formulas 1. VLOOKUP and HLOOKUP Imagine you have a massive dataset with customer information, and you need to quickly retrieve specific details like their phone numbers based on their customer ID. That's where VLOOKUP comes in. It allows you to search for a specific value in a column (the lookup value) and return a corresponding value from another column in the same row. HLOOKUP functions similarly but searches horizontally instead of vertically. Let's say you have a table with customer IDs in column A and phone numbers in column B. To find the phone number for customer ID 1234, you can use the following formula: =VLOOKUP(1234, A1:B10, 2, FALSE) This formula searches for the value 1234 in column A (A1:B10 represents the table range). It then returns the corresponding value from column B (the second column, indicated by 2). The FALSE argument ensures an exact match is found. 2. INDEX and MATCH While VLOOKUP and HLOOKUP are powerful, they have limitations when searching across multiple columns. That's where the INDEX and MATCH combo shines. INDEX retrieves a value from a specific cell based on its row and column position within a range. MATCH finds the position of a value within a range. You have a table with customer information, and you need to find the phone number for customer ID 1234, but it's located in column D. =INDEX(D1:D10, MATCH(1234, A1:A10, 0)) This formula first uses MATCH to find the row number where customer ID 1234 appears in column A (A1:A10). Then, INDEX retrieves the value from the corresponding row in column D (D1:D10), providing the phone number. 3. SUMIFS, COUNTIFS, and AVERAGEIFS These functions are incredibly helpful for conditional calculations. SUMIFS calculates the sum of values that meet specific criteria. COUNTIFS counts the number of cells that meet multiple conditions. AVERAGEIFS calculates the average of values that satisfy specified criteria. You have a sales table with sales figures, product categories, and sales regions. You want to calculate the total sales for "Electronics" products in the "East" region. =SUMIFS(C1:C10, A1:A10, "Electronics", B1:B10, "East") This formula sums values in column C (sales figures) where column A (product category) equals "Electronics" and column B (sales region) equals "East." 4. IFS IFS allows you to test multiple conditions and return a corresponding value based on the first condition that is met. It simplifies complex nested IF statements. You want to assign grades based on student scores: =IFS(A1>=90, "A", A1>=80, "B", A1>=70, "C", A1>=60, "D", TRUE, "F") This formula checks the score in cell A1 and assigns the appropriate grade based on the criteria. If none of the conditions are met, it assigns an "F." 5. OFFSET OFFSET allows you to select a range of cells relative to a starting point. It is particularly useful for dynamic ranges that change based on your data. You have a monthly sales report, and you need to calculate the average sales for the last 3 months. This formula takes the starting point at A1, then uses COUNTA to determine the number of rows with data. It then offsets by -3 rows (representing the last 3 months) and selects a range of 3 rows and 1 column. The average of these 3 cells (representing the last 3 months' sales) is calculated. 6. AGGREGATE AGGREGATE performs calculations on a dataset while ignoring errors or hidden rows. This makes it powerful for situations where your data may contain errors or you want to selectively analyze subsets of your data. You have a table with sales figures, and some cells contain errors. You want to calculate the average sales, excluding errors. =AGGREGATE(1, 6, B1:B10) This formula uses AGGREGATE to calculate the average (function 1) while ignoring errors (function 6) for the range B1:B10. 7. TEXTJOIN TEXTJOIN combines multiple text strings into a single text string with a specified delimiter. This is useful for concatenating data in a user-friendly format. You have a table with customer names, addresses, and phone numbers. You want to combine them into a single string for a mailing list. =TEXTJOIN(", ", TRUE, A1, B1, C1) This formula combines the contents of cells A1, B1, and C1 into a single string, separated by a comma and space, while ignoring empty cells (TRUE argument). Beyond Formulas: Leveraging Excel's Functionality Mastering advanced Excel formulas is just one aspect of leveraging its full potential. Here are some additional tools and techniques to further enhance your data analysis: 1. Data Validation Data validation helps enforce data quality by restricting the input allowed in specific cells. You can define rules for data types, ranges, and even list choices, ensuring data accuracy and 2. Pivot Tables Pivot tables are a powerful tool for summarizing and analyzing large datasets. They allow you to quickly group and aggregate data based on your chosen criteria, providing insightful summaries and 3. Power Query Power Query (formerly known as Get & Transform) is a data transformation tool that lets you import data from various sources, clean and shape it, and then load it into Excel for further analysis. 4. Macros and VBA For repetitive tasks or advanced automation, macros and VBA (Visual Basic for Applications) allow you to create custom scripts to automate processes, saving time and effort. 5. Conditional Formatting Conditional formatting applies visual styles to cells based on specific conditions, making it easier to highlight important data and identify trends or outliers. Conclusion: Elevate Your Excel Proficiency Advanced Excel formulas are a powerful arsenal for any data-driven professional. By mastering these formulas and exploring the many other functionalities of Excel, you can unlock its full potential, streamline your workflows, and gain deeper insights from your data. Don't be afraid to experiment, explore, and discover the endless possibilities that await you in the world of advanced Excel.
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๐Ÿ“ The function COORDY is used to extract the Y part of a coordinate pair. This is useful when you want to isolate only the Y value from a coordinate-type data point. ๐Ÿ“ It is most often associated with the functions COORDINATES() or GEOLOC(), which are used to create a coordinate value with an X and Y pair. There is also: ๐Ÿ“ The COORDX() function, which is used to extract the X part of a coordinate pair. ๐Ÿ“ The GEOLOC() function is used to obtain the geographical coordinates ๐Ÿ“ (latitude and longitude) of a full address. โžก๏ธ Display name: โžก๏ธ Syntax : โžก๏ธ Structure: โ€ข ๐ŸŸก coordinate (param1): The coordinate value from which you want to extract the Y part. โ€ข ๐Ÿ”ต COORDY(): The function that extracts the Y value from the coordinate pair. โžก๏ธ Example: COORDY( COORDINATES(123, 456) ) In this example, the COORDY function will extract the Y value from the coordinate pair COORDINATES(123, 456), which is 456. ๐ŸŸฐ Expected result: The function will return 456, which is the Y value from the coordinate pair. ๐Ÿ‘๏ธ๐Ÿ—จ๏ธ Related articles for managing coordinate points ๐Ÿ“ Practical use: These three functions are ideal for manipulating and working with location data in an application. For example, by combining GEOLOC(), COORDINATES, COORDX, and COORDY, you can manage geographic points and display them on maps or perform calculations related to the distance between two points. ๐Ÿ’ก These functions are essential for representing data in the TimeTonic map view ๐Ÿ—บ๏ธ , allowing for the manipulation and precise display of geographic coordinates on a map. Before You Start Getting Started with Formulas The new formulas in Timetonic allow you to transform, calculate, and manipulate the information stored in your columns by combining multiple functions... Read More โž” Or Continue With Glossary View of Formulas Quickly find the function or operator you're looking for or simply explore and learn how to use them... Read More โž”
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Simplicial Complexes Mathematics++: Selected Topics Beyond the Basic Courses (Student Mathematical Library) Kantor, Ida. 1. Measure 2. High Dimensional Geometry 3. Fourier Analysis 4. Representations of Finite Groups 5. Polynomials 6. Topology Chapter 6 - Topology. Contains a relatively gentle introduction to homology. Graphs, Surfaces and Homology - Peter Giblin. Builds up to homology groups via graphs and simplicial complexes. Algebraic Topology - Allen Hatcher. Freely available online here Computational Geometry - Algorithms and Applications by Mark de Berg, Otfried Cheong, Marc van Kreveld and Mark Overmars This book looks at the algorithms from a computer science, rather than pure mathematics, point-of-view. So homotopy or homology is not mentioned but subjects like Voronoi Diagrams, Delauney Triangulations, Convex Hulls and many similar topics are covered.
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How Does Nuclear Physics Explore Atomic Nuclei **Exploring the Heart of Matter: How Nuclear Physics Unveils the Secrets of Atomic Nuclei** The field of nuclear physics delves deep into the heart of matter, unraveling the intricate structure and behavior of atomic nuclei. At the core of every atom lies a nucleus, a minuscule yet incredibly dense region that holds the key to understanding the fundamental forces that govern the universe. Through a combination of theoretical models, experimental techniques, and technological advancements, nuclear physicists have made significant strides in exploring the mysteries of atomic nuclei. **Unraveling the Structure of Atomic Nuclei** One of the primary goals of nuclear physics is to decipher the structure of atomic nuclei, which are composed of protons and neutrons bound together by the strong nuclear force. The arrangement of these nucleons within the nucleus determines its stability, shape, and properties. Nuclear physicists use sophisticated theoretical models, such as the nuclear shell model and the liquid drop model, to describe the behavior of nucleons inside the nucleus. **The Nuclear Shell Model** The nuclear shell model, inspired by the concept of electron shells in atoms, posits that nucleons occupy discrete energy levels within the nucleus. These energy levels form shells, analogous to the electron shells in an atom, and dictate the stability of the nucleus. By studying the arrangement of nucleons in different nuclear shells, physicists can predict the properties of various isotopes and understand phenomena such as nuclear binding energy and nuclear stability. **Nuclear Reactions and Nuclear Decay** Nuclear physicists also study the processes of nuclear reactions and decay, which play a crucial role in shaping the evolution of atomic nuclei. Nuclear reactions involve the transformation of one nucleus into another through processes like fusion, fission, and radioactive decay. These reactions release vast amounts of energy and are harnessed in nuclear power plants and nuclear weapons. **Nuclear Fission and Fusion** Nuclear fission, the splitting of a heavy nucleus into lighter fragments, is the process behind nuclear power generation and nuclear weapons. By controlling the rate of fission reactions, scientists can harness the energy released to generate electricity or produce nuclear explosions. On the other hand, nuclear fusion involves the merging of light nuclei to form heavier elements, a process that powers the sun and other stars. Achieving controlled fusion reactions on Earth remains a significant challenge for nuclear physicists. **Radioactive Decay** Radioactive decay is another fundamental process studied in nuclear physics, where unstable nuclei undergo spontaneous transformations to achieve a more stable configuration. This process results in the emission of radiation in the form of alpha particles, beta particles, and gamma rays. Understanding the rates of radioactive decay is crucial for applications in radiometric dating, medical imaging, and nuclear medicine. **Technological Advancements in Nuclear Physics** The field of nuclear physics has benefited greatly from technological advancements that have revolutionized experimental techniques and data analysis. High-energy particle accelerators, such as the Large Hadron Collider, allow physicists to probe the innermost structure of atomic nuclei by colliding particles at near-light speeds. Detectors capable of capturing and analyzing the debris from these collisions provide valuable insights into the properties of nuclear matter. **The Future of Nuclear Physics** As nuclear physicists continue to push the boundaries of our understanding of atomic nuclei, new challenges and discoveries lie ahead. The quest to unravel the mysteries of nuclear matter, from the structure of exotic nuclei to the origins of the elements, remains a driving force in the field of nuclear physics. By combining theoretical models with experimental data and technological innovations, scientists are poised to unlock the secrets of the atomic nucleus and deepen our understanding of the fundamental forces that govern the universe. **In Closing** In conclusion, nuclear physics offers a fascinating glimpse into the inner workings of atomic nuclei, shedding light on the fundamental forces that shape the building blocks of matter. Through a combination of theoretical models, experimental techniques, and technological advancements, nuclear physicists continue to explore the mysteries of nuclear matter and push the boundaries of scientific discovery. The quest to understand the complexities of atomic nuclei remains a cornerstone of modern physics, driving innovation and exploration in the field of nuclear physics.
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Gamelogicโ€™s Hex Grids for Unity and Amit Patelโ€™s Guide for Hex Grids If you have done any work with hex grids, the chances are good that you came across the best guide for the mathematics and algorithms for hex grids: Hexagonal Grids by Amit Patel. One user suggested that it may be a good idea to show how Grids can be used for all the operations described in that guide (and where those functions are) since it is so popular. We agree, so here it is! Angles, Size and Spacing For the most part, if you use Grids, you donโ€™t have to worry about these low-level details. Conversion to world space is handled by maps. Hex maps need a Vector2 of dimensions โ€“ the width and height of your cell, and it will automatically work out the correct spacing and division of space. It also means that you can work with hexagons scaled differently along each axis. This is useful when working with isometric hexes, for example. Coordinate Systems Grids only work with one coordinate system: axial coordinates. In the very early days of Grids (before it was for sale), we used offset coordinates. As we made more examples and implemented more algorithms, we found that the offset coordinate system is very clumsy, so we changed it to use axial coordinates instead. The axial coordinate system works like a (integer) vector space, and many algorithms are much simpler compared their implementations using offset coordinates. In particular, it makes sense to add two points together to get a third (where the second point is interpreted as an offset). There are two cases where offset coordinates can be useful: โ€ข For display (on certain boards, the coordinates may be more compact and attractive, since they donโ€™t have to have negative values, while still having a corner as the origin) โ€ข For certain algorithms (if your game uses for instance rows and wiggly columns) In these two cases, you will have to implement a conversion, something which is straightforward to do. Grids have two methods for accessing the neighbors of a point: Grid.GetAllNeighbors(point) gives a list of all neighboring points, regardless of whether they are in the grid or not. Grid.GetNeighbors(point) gives a list of neighbors that are also in the grid. This is the method you will most commonly use in your own algorithms. Hex grids also have constants defined for each direction a neighbour can be in: These can be added to a point to get the neighbor: var northWestNeighbor = point + PointyHexPoint.NorthWest; Grids does not support diagonals directly, but it is easy to define your own constants for diagonal directions, for example: readonly PointyHexPoint North = PointyHexPoint.NorthWest + PointyHexPoint.NorthEast; You can then add them to points to get their โ€œdiagonalโ€ neighbors: var northNeighbor = point + North; To get the grid distance between two points, you can use the DistanceFrom method defined on points. var distance = onePoint.DistanceFrom(otherPoint); Line Drawing We provide line drawing methods as extensions for maps. To get a line between two points, you call var points = Map.GetLine(p1, p2); //where map is the map you used to position cells, or the Map property of the GridBehaviour, if you use GridBehaviours. Coordinate Ranges To get all the points within a certain distance from a give point, you can use a simple LINQ query: var pointsInRange = grid.Where(p => p.DistanceFrom(center) <= range); For intersections, you can chain queries var pointsInRange = grid .Where(p => p.DistanceFrom(center1) <= range1) .Where(p => p.DistanceFrom(center2) <= range2); There are too complications which you may face in your game: you may wish either of these: โ€ข to use different distance metrics (including weighted costs) โ€ข to take obstacles into account For this, we provide PointsInRange methods. There is a tutorial for using this feature here: ### Hex points have several rotation functions: Rotate60, Rotate60About, Rotate120, Rotate120About, and so on. There are also methods defined on Algorithms to transform a list of points: Transform, which you can use as follows: var newList = Algorithms.Transform(p => p.Rotate60); Algorithms also define special rotations: rotations around an edge (you give the two neighboring points as input), and rotations around a vertex (you give the three neighboring points as input). As with point in range, you can use a LINQ query to get all the points in a ring around another point: var pointsInRange = grid.Where(p => p.DistanceFrom(center) == range); Hex grids provides methods SpiralIterator to get spirals. You can either use them in a loop: for(var point in grid.SpiralIterator(4)) Or just get a list straight: Field of View We do not have this function. But itโ€™s easy to implement as we show in this example: Field of View Example. Hex to Pixel and Pixel to Hex Here, you can use maps for both operations: var pixel = map[hexPoint]; Var hexPoint = map[pixel]; Rounding to nearest hex Generally, rounding to nearest hex is not necessary. If for some reason you do need to do it, you can simply use a map, with dimensions 1ร—1. Map Storage The beauty of Grids is that you need not worry about storage at all, everything is taken care of behind the scenes. This is really convenient when making more complicated shapes using the grid builder functionality. 1 thought on โ€œGamelogicโ€™s Hex Grids for Unity and Amit Patelโ€™s Guide for Hex Gridsโ€
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Purpose.: To examine the heritability of refractive astigmatism in older women. Methods.: Astigmatism was measured with an autorefractor in 88 monozygotic and 82 dizygotic female twin pairs aged 63 to 75 years. The prevalence and distribution of astigmatism and polar values J0 and J45 were estimated by standard statistical methods. Bivariate maximum likelihood model fitting was used to estimate genetic and environmental variance components using information from both eyes. Results.: Mean astigmatism of the more astigmatic eye was 0.93 diopters (D; SD ยฑ0.58). Astigmatism of at least 0.25 D, 0.5 D, 0.75 D, or 1.0 D in either eye was present in 99.7%, 88.5%, 66.5%, and 46.2% of cases, respectively. The main direction of astigmatism was against the rule. The age-adjusted quantitative genetic modeling revealed that additive genetic effects accounted for 33.3% (95% confidence interval [CI], 21.9%โ€“43.8%) of the total variance of astigmatism and for 18% (95% CI, 4%โ€“31%) of the total variance of polar value J45 of both eyes (bivariate model), with the remaining variances due to nongenetic effects. There were no significant correlations between the twin pairs for polar value J0. Conclusions.: In elderly female twins, additive genetic effects accounted for one-third of the variance of the amount of astigmatism and only a small fraction of the total variance of polar value Refractive astigmatism (astigmatism) is the sum of the asphericities of different optical elements of the eye: the anterior and posterior cornea and lens and their relative positions with regard to the visual axis of the eye. Corneal astigmatism (CA) is the most significant determinant of total astigmatism. Different theories have been offered to explain the development of astigmatism. In addition to genetics, eyelid pressure, extraocular muscle tension, and visual feedback are thought to be reasons for astigmatism. ^ 1 Various corneal diseases, eye surgery affecting the cornea, or other surgical proceduresโ€”for example, cataract, glaucoma, and retinal ablation surgeryโ€”may induce astigmatism. The amount and direction of astigmatism has been shown to vary with age. ^ 2โ€“5 Asano and colleagues, for example, found mean astigmatism in the right eye of 0.77 D and 1.25 D among 40- to 49-year-olds and 70- to 79-year-olds, respectively. ^ 2 The direction of astigmatism tends to change with age, the main trend being an increase against the rule (ATR) and a decrease with the rule (WTR). ^ 6 Differences in the prevalence of astigmatism also seem to occur between different ethnic groups. ^ 7 The classical twin model is commonly used to determine the relative contribution of the genetic and environmental components of a disease or traits. Thus far, most twin studies examining refraction and ocular biometrics have mainly comprised populations across a wide age spectrum. Grjibovski et al. ^ 8 studied the occurrence and heritability of astigmatism in a population-based sample of Norwegian twins through self-reported history of astigmatism from birth to age 31 years in 8045 twins. The best-fitting biometrical model suggested that genetic effects due to dominance explained 54% (95% CI: 20โ€“69 years) and additive genetic effects explained 9% (95% CI: 0โ€“40 years) of the variation in the liability to astigmatism. In the study of Hammond et al. ^ 9 among twin pairs aged 49 to 79 years (mean age, 62.4 years), genetic effects due to dominance accounted for 47% to 49% (95% CI, 37%โ€“55%) and additive effects for 1% to 4% of the variance of total astigmatism (95% CI, 0%โ€“13%). Dirani et al. ^ 10 studied CA in 18- to 86-year-old Australian twins (mean age, 52.11 ยฑ 15.85 years). Heritability estimates were as high as 60% for CA. In our recent article, quantitative genetic modeling showed that heritable factors explained 83% of the variance in spherical equivalent in 63- to 76-year-old Finnish female twins. ^ 11 In the same population, genetic factors explained 81% of the variance in corneal refraction, additive genetic factors 62% (95% confidence interval [CI] 44%โˆ’86%), and dominant genetic factors 19% (95% CI 7%โ€“49%). ^ 12 For CA, it was not possible to construct a meaningful model, although the values of the intraclass correlation coefficient (ICC) were higher for monozygotic (MZ) than dizygotic (DZ) twins. ^ 12 The main purpose of the present study was to determine the heredity of refractive astigmatism in the same elderly Finnish female twin subjects. Design and Patients This study forms part of the Finnish Twin Study on Aging (FITSA), the purpose of which is to investigate genetic and environmental effects on the disablement process in older women. A detailed description of the recruitment process has been published earlier. ^ 13โ€“14 Schematically, the recruitment process was the following: Zygosity had initially (1975) been determined by a validated questionnaire ^ 15 and later confirmed by applying a battery of 10 highly polymorphic gene markers at the National Public Health Institute to DNA extracted from a sample of venous blood. The subjects ranged in age from 63 to 75 years, with a mean age of 68 years (SD ยฑ3.2). The study was approved by the ethics committee of the Central Hospital of Central Finland and both twins gave their written informed consent. Our research adhered to the tenets of the Declaration of Refraction, including the amount and direction of astigmatism, were measured with an autorefractor (Topcon AT; Topcon, Tokyo, Japan). On the basis of the values given by the autorefractor, subjective spherical refraction was then measured by the fogging method, and final spherical refraction was controlled for by the red-green test. Distant vision corrected using subjective refractive values was measured from an illuminated chart with Landolt rings at a distance of 6 m. The examination was performed for both twin sisters by the same nurse, but on separate occasions, during the one-day assessment in a laboratory at the University of Jyvรคskylรค. Direction of astigmatism was classified into three groups: WTR = axis of the correcting + cylinder between 60 to 120ยฐ, ATR = between 0 to 30ยฐ or 150 to 179ยฐ, and oblique (all remaining cases). For the vectorial analysis, we converted the astigmatism from the spherocylindrical notation to J0 and J45 power vectors by applying a Fourier transformation using the following equations: J0 = [C ร— COS(2A)]/2, J45 = [C ร— SIN(2A)]/2, C = Power of +cylinder, A = Axis of cylinder. ^ 16 J0 refers to +cylinder power set orthogonally at the 0 and 90ยฐ meridians and represents WTR or ATR astigmatism. Positive values of J0 indicate WTR astigmatism, and negative values ATR astigmatism. J45 refers to a cross-cylinder set at 45 and 135ยฐ, representing oblique astigmatism. Negative values of J45 indicate astigmatism of around 45ยฐ and positive values astigmatism of around 135ยฐ. Statistical Methods Data were analyzed using statistical software (Stata, version 12; StataCorp, College Station, TX; and SPSS version 19.0; IBM Corp., Endicott, NY). The equality of the means of the continuous variables and equality of the distributions of the categorical variables between the MZ and DZ twins were analyzed with the adjusted Wald test, taking into account the fact that the data consist of twin pairs rather than unrelated individuals. The significance of differences in the amount of astigmatism between the ATR, WTR, and oblique directions was tested by one-way ANOVA with Sheffe's post hoc procedure for the pairwise comparisons of means. ICCs were computed for the MZ and DZ twin pairs separately to estimate the level of within-pair similarity. ICCs can be used to obtain indicative estimates of the genetic and environmental components of variances. Higher ICC values among the MZ than among the DZ pairs indicate the presence of an underlying genetic contribution. The associations between continuous variables (e.g., amount of astigmatism between right and left eye) were analyzed by Pearson's product moment correlation coefficients. The significance of differences was tested by cross-tabulation and Pearson's ฯ‡^2 test in the case of discrete variables (e.g., direction of astigmatism). The classical twin study provides an opportunity to determine whether individual differences in a trait arise from genetic or environmental factors or both. MZ twins share all their genes (100%), whereas DZ twins share, on average, 50% of their segregating genes. Consequently, in MZ pairs, genes contribute only to similarity in a trait, whereas in DZ twin pairs, they contribute both to similarity and to differences. Greater similarity in MZ pairs than in DZ pairs provides evidence for genetic influence on the trait. ^ 17 In quantitative genetics studies, genetic effects are typically classified into additive genetic effects (A) and nonadditive genetic effects (D). Environmental effects are classified into shared environmental effects (C) and nonshared environmental effects (E). Shared environmental effects are common to both members of a pair while nonshared effects refer to external exposures affecting only one sibling, such as accidents, surgery, or measurement error. Quantitative genetic modeling is based on necessary similarities and differences in the correlations of the A, D, C, and E factors explaining the variability among MZ and DZ twins. The correlations for additive and nonadditive genetic effects are defined as 1.0 in MZ pairs, and as 0.5 and 0.25, respectively, in DZ pairs. The correlation for shared environmental effects is defined as 1.0 and for nonshared environmental effects as 0, among both MZ and DZ pairs. The aim of genetic modeling is to construct a model that explains the data well and has as few explanatory components as possible. When using bivariate modeling it is necessary to separate effects that are common for both variables included in the model and effects that are only specific for one variable in the model. To sharpen this difference small letters c (common) or s (specific) are used with letters A, D, C, and E. The full independent pathway model consists of genetic and environmental effects that are common to both eyes (common additive genetic effect [Ac], common shared environmental effect [Cc]/common nonadditive genetic effect [Dc], and common nonshared environmental effect [Ec]) as well as effects that are specific to the right or left eye only (specific additive genetic effect [As]-1, As2, specific shared environmental effect [Cs]-1/specific nonadditive genetic effect [Ds]-1, Cs2/Ds2, and specific nonshared environmental effect [Es]-1, Es2). To obtain a more parsimonious model, the full model was modified by dropping the weakest nonsignificant parameters one at time, until the model with the best fit was achieved. The ICCs for astigmatism and polar value J45 among MZ were significantly higher than among DZ, predicting that the ADE model would show the best fit to the data. Our subsequent analyses showed, however, that nonadditive genetic effects (D) were not significant, and hence the final AE model showed the best fit for astigmatism and for polar value J45. The data from both eyes were analyzed simultaneously with bivariate models. Using a bivariate model instead of a univariate model (with only the right or left eye in the model) results in narrower confidence intervals of estimates, and thus produces more precise results. In the present study, astigmatism showed a high phenotypic correlation between the right and left eye. Hence, eyes were treated as a single, correlated trait in the analyses. The present analyses were carried out using matrix algebra software (Mx program, version 1.52a. ^ 18 ; Michael Neal, Virginia Commonwealth University, Richmond, VA). For all the study subjects, the mean spherical equivalent (SE ยฑ SD) was +1.67 D (ยฑ1.93) in the right eye and +1.66 D (ยฑ1.86) in the left eye. There were no significant differences between MZ and DZ in SE between the right and left eyes. The SE was negative (myopic) in 12.9% and 10.6% of cases in the right and left eye, respectively. Corrected distant vision of the right eye was 0.75 (ยฑ0.21), and the corresponding value for the left eye was 0.78 (ยฑ0.21). Mean astigmatism of the right eye between MZ and DZ was 0.77 D (ยฑ0.68) and 0.71 D (ยฑ0.46), respectively; the corresponding values for the left eye were 0.73 D (ยฑ0.63) and 0.70 D (ยฑ0.47). Mean astigmatism of the more astigmatic eye was 0.93 D (ยฑ0.58), with a maximum of 4.00 D. Astigmatism of at least 0.25 D, 0.5 D, 0.75 D, or 1.0 D in the more astigmatic eye was present in 99.7%, 88.8%, 66.4%, and 46.1% of cases. There were no statistically significant differences in the amount of astigmatism between either the left and right eyes (P = 0.22) or between MZ and DZ individuals (P = 0.42). The correlations between the amount of astigmatism and age were nonsignificant. The correlations between the amount of astigmatism and spherical equivalent in the whole dataset and separately among those with either positive or negative SE were nonsignificant in each eye. Axis of Astigmatism and Polar Values of Astigmatism There were no significant differences in the means of astigmatism in the different main axis directions between the right and left eye. Table 1 shows the distribution and means of astigmatism in the main directions for both eyes. Table 1 Direction of Astigmatism Right Eye Left Eye Prevalence, % Mean, D (ยฑSD) Prevalence, % Mean, D (ยฑSD) WTR 29.9 0.78 (ยฑ0.66) 29.5 0.74 (ยฑ0.66) ATR 50.9 0.73 (ยฑ0.52) 48.9 0.68 (ยฑ0.46) Oblique 19.2 0.61 (ยฑ0.50) 21.6 0.66 (ยฑ0.42) In the right eye, the mean astigmatism in the oblique direction was smaller than that in the WTR (P < 0.001) or ATR (P = 0.001) directions. The correlations in the left eye did not reach There were no significant differences in the distribution of the axis of astigmatism between those with positive and those with negative SE (ฯ‡^2 test, P = 0.124 for right eye, and P = 0.060 for left Table 2 shows polar values J0 and J45 for both eyes. All means were slightly negative. Negative J0 indicated ATR astigmatism, but all the means of J0 and J45 were so close to the value zero that no significant mean predominance was found. The ICC values between the twin sisters for astigmatism of the right eye were 0.445 for MZ and โˆ’0.118 for DZ; the corresponding values for the left eye were 0.415 for MZ and โˆ’0.246 for DZ. Age-adjusted quantitative genetic modeling revealed that for both eyes (bivariate model), additive genetic effects (Ac) accounted for 33.3% (95% CI, 21.9%โ€“43.8%) of the total variance of astigmatism (of both eyes [bivariate model]), with the remaining variance due to nongenetic effects: (Ec) 21.4% (95% CI, 11.4%โ€“33.5%) and (Es) 45.3% (95% CI, 38.1%โ€“53.3%). The ICC values between the twin sisters for astigmatism of the most astigmatic eye were as follows: for the WTR subgroup, 0.660 for MZ (n = 28) and โˆ’0.145 for DZ (n = 28); and for the ATR subgroup, 0.697 (n = 15) for MZ and 0.027 (n = 9) for DZ. The low number of cases did not permit calculations for the oblique subgroup. The ICC values between the cotwins for polar value J0 were nonsignificant; the right eye was โˆ’0.037 for MZ and โˆ’0.030 for DZ; the corresponding values for the left eye were โˆ’0.054 for MZ and 0.074 for DZ. The ICC values between the twin sisters for polar value J45 of the right eye were 0.234 for MZ and 0.019 for DZ; the corresponding values for the left eye were 0.207 for MZ and 0.006 for DZ. Age-adjusted quantitative genetic modeling revealed that additive genetic effects (A = Ac + As) accounted for 18% (95% CI, 4%โ€“31%) of the total variance of polar value J45 of both eyes (bivariate model), with the remaining variance due to nongenetic effects (E = Ec + Es), 82% (95% CI, 69%โ€“96%). All but one subject in the present study had astigmatism of at least 0.25 D in one eye or the other. While this is more than has commonly been reported earlier, in most studies astigmatism has usually begun from a level of either 0.5 D or 0.75 D. ^ 19,20 Although our study design does not permit general conclusions to be drawn on the prevalence of astigmatism in the population, the prevalence of astigmatism of โ‰ฅ0.5 D (71.2% in the right and 70.0% in the left eye) found in the present study does not differ greatly from that reported by Schellini et al., ^ 20 in a Brazilian population aged 70 years or older (71.7% had astigmatism of โ‰ฅ0.5 D). In the National Health and Nutrition Examination Survey, the prevalence of astigmatism of โ‰ฅ1.0 D was 50.1% in persons aged 60 years or older; whereas in our study, this grade of astigmatism in one eye or the other was found in 46.3% of cases. ^ 21 Thus the main difference between the prevalence found here and that reported in most previous studies is probably due to the fact that astigmatism in our study started at 0.25 D. Some studies have shown a positive relationship between the amount of astigmatism and ametropia and myopia. ^ 1,22 In the present study, no significant correlations between astigmatism and SE were found. The small proportions of myopics and cases of high ametropia in this study could be one explanation for the nonsignificant relationships observed between the amount of astigmatism and spherical ametropia. The main aim of this study was to calculate the impact of heredity on refractive astigmatism. In the present study, only approximately one-third of the variation of astigmatism was explained by additive genetic effects, which is less than in some previous studies among somewhat younger subjects. Moreover, no significant dominant effects could be calculated. In a population-based sample of Norwegian twins, aged from birth to 31 years, additive genetic effects explained 9% (95% CI, 0โ€“40 years), and dominant genetic effects 54% (95% CI, 20โ€“69 years) of the variation in the liability to self-reported astigmatism. ^ 8 In the study of Hammond et al., ^ 9 among 49- to 79-year-old twins, additive genetic effects accounted for 1% to 4% (95% CI, 0%โ€“13%) and dominant effects for 47% to 49% (95% CI, 37%โ€“55%) of the variance of total astigmatism. In the Hammond et al. ^ 9 study and our study, the MZ correlations for astigmatism were very similar (0.4 to 0.5), while the DZ correlations in Hammond et al. ^ 9 were 0.2 to 0.10 and in ours negative. Despite the larger sample size in Hammond et al., ^ 9 neither study had the power to distinguish unambiguously between additive and nonadditive genetic sources of variation. Given the differences in sample size and in characteristics resulting in some variation in the actual variance/covariance matrices underlying the pairwise correlations, it was not unexpected that the best-fitting model differed in the two studies. Both studies point to the importance of genetic factors, but whether additive or nonadditive factors are more important cannot be stated. Thus, in Hammond et al., ^ 9 an ADE model for astigmatism fit better than an AE model for the left but not the right eye in the univariate models. In the bivariate models, Hammond et al. ^ 9 only present results for an ADE model and no results for an AE model. Higher ICC values were observed between the MZ pairs of sisters for astigmatism in the WTR and ATR subgroups of astigmatism than in the whole sample. Thus, it could be supposed that the heredity of astigmatism could be higher in those subgroups; but due to the small sample size, it was not possible to calculate a meaningful model. As far as we know, there are no previous studies on the heredity of the polar values of astigmatisms. In the present study, significant ICC values between the twin pairs emerged only for the value of J45. Further, the ICC values for J45 between the MZ sisters were lower than those for the amount of astigmatism, suggesting that the heredity of polar values is lower than the amount of astigmatism In our earlier studies on the same subjects, 83% and 81% of the variance in spherical equivalent and corneal refraction, respectively, were explained by heritable factors. ^ 11,12 Based on the results of this study and the results of our earlier studies on the same subjects, it is reasonable to conclude that the impact of heredity among elderly females is highest on spherical equivalent refraction and corneal refraction, while environmental factors have a stronger influence on the amount of corneal and refractive astigmatism. The amount and direction of astigmatism has been shown to vary with age. ^ 2โ€“5 The low impact of heredity on astigmatism found in the present study with elderly females would support the theory that the influence of environmental factors on astigmatism increases at older ages. Possible changes in astigmatism should be taken into account when controlling for refraction and prescribing new glasses, and when planning refractive surgery, especially when doing costumed corneal ablations. Emmetropic nonastigmatic refraction achieved by refractive surgery at a young age may turn to astigmatic refraction later on during the life course. Supported by the grants from the Academy of Finland, the Finnish Ministry of Education, and the European Union in conjunction with the GENOMEUTWIN project (QLG2-CT-2002-01254), and the Academy of Finland Centre of Excellence for Complex Disease Genetics (Grants 213506, 129680). The authors alone are responsible for the content and writing of the paper. Disclosure: O. Pรคrssinen, None; M. Kauppinen, None; J. Kaprio, None; M. Koskenvuo, None; T. Rantanen, None Read SA Collins MJ Carney LG. A review of astigmatism and its possible genesis. Clin Exp Optom . 2007; 90: 5โ€“19. [CrossRef] [PubMed] Asano K Nomura H Iwano M Relationship between astigmatism and aging in middle-aged and elderly Japanese. Jpn J Ophthalmol . 2005; 49: 127โ€“133. [CrossRef] [PubMed] Atkinson J Braddick O French J. Infant astigmatism: its disappearance with age. Vision Res . 1980; 20: 891โ€“893. [CrossRef] [PubMed] Fledelius HC Stubgaard M. Changes in refraction and corneal curvature during growth and adult life. A cross-sectional study. Acta Ophthalmol (Copenh) . 1986; 64: 487โ€“491. [CrossRef] [PubMed] Gwiazda J Scheiman M Mohindra I Astigmatism in children: changes in axis and amount from birth to six years. Invest Ophthalmol Vis Sci . 1984; 25: 88โ€“92. Saunders H. Age-dependence of human refractive errors. Ophthalmic Physiol Opt . 1981; 1: 159โ€“174. [CrossRef] [PubMed] Kame RT Jue TS Shigekuni DM. A longitudinal study of corneal astigmatism changes in Asian eyes. J Am Optom Assoc . 1993; 64: 215โ€“219. Grjibovski AM Magnus P Midelfart A Epidemiology and heritability of astigmatism in Norwegian twins: an analysis of self-reported data. Ophthalmic Epidemiol . 2006; 13: 245โ€“252. [CrossRef] [PubMed] Hammond CJ Snieder H Gilbert CE Genes and environment in refractive error: the twin eye study. Invest Ophthalmol Vis Sci . 2001; 42: 1232โ€“1236. Dirani M Islam A Shekar SN Baird PN. Dominant genetic effects on corneal astigmatism: the genes in myopia (GEM) twin study. Invest Ophthalmol Vis Sci . 2008; 49: 1339โ€“1344. [CrossRef] [PubMed] Pรคrssinen O Jauhonen HM Kauppinen M Heritability of spherical equivalent: a population-based twin study among 63- to 76-year-old female twins. . 2010; 117: 1908โ€“1911. [CrossRef] [PubMed] Pรคrssinen O Kauppinen M Kaprio J Heritability of corneal refraction and corneal astigmatism: a population-based twin study among 66- to 79-year-old female twins. Acta Ophthalmol . 2013; 91: 140โ€“144. [CrossRef] [PubMed] Tiainen K Sipilรค S Alรฉn M Shared genetic and environmental effects on strength and power in older female twins. Med Sci Sports Exerc . 2005; 37: 72โ€“78. [CrossRef] [PubMed] Kaprio J Koskenvuo M. Genetic and environmental factors in complex diseases: the older Finnish Twin Cohort. Twin Res. Review . 2002; 5: 358โ€“365. Sarna S Kaprio J Sistonen P Diagnosis of twin zygosity by mailed questionnaire. Hum Hered . 1978; 28: 241โ€“254. [CrossRef] [PubMed] Thibos LN Wheeler W Horner D. Power vectors: an application of Fourier analysis to the description and statistical analysis of refractive error. Optom Vis Sci . 1997; 74: 367โ€“375. [CrossRef] [PubMed] Posthuma D Beem AL de Geus EJ Theory and practice in quantitative genetics. Twin Res . 2003; 6: 361โ€“376. [CrossRef] [PubMed] Neale MC Boker SM Xie G Mx: Statistical Modeling. 6th ed. Richmond, VA: Department of Psychiatry, Virginia Commonwealth University; 2003. Antรณn A Andrada MT Mayo A Epidemiology of refractive errors in an adult European population: the Segovia study. Ophthalmic Epidemiol . 2009; 16: 231โ€“237. [CrossRef] [PubMed] Schellini SA Durkin SR Hoyama E Prevalence of refractive errors in a Brazilian population: the Botucatu eye study. Ophthalmic Epidemiol . 2009; 16: 90โ€“97. [CrossRef] [PubMed] Vitale S Ellwein L Cotch MF Prevalence of refractive error in the United States, 1999-2004. Arch Ophthalmol . 2009; 126: 1111โ€“1119. Fulton AB Hansen RM Petersen RA. The relation of myopia and astigmatism in developing eyes. . 1982; 89: 298โ€“302. [CrossRef] [PubMed]
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CUNY Probability Seminar Fall 2018 The CUNY Probability Seminar is typically held on Tuesdays from 4:15 to 5:15 pm at the CUNY Graduate Center Math Department in room 5417. The exact dates, times and locations are mentioned below. If you want to talk at the seminar, or want to be added to/removed from the seminar mailing list, then please contact the Seminar Coordinator. Toby Johnson (https://www.math.csi.cuny.edu/~tobiasljohnson/) Upcoming Seminars Tuesday, December 4, 2018, 4:15-5:15 Room 5417 Speaker: Chiranjib Mukherjee, Institut fรผr Mathematische Statistik, Universitรคt Mรผnster Title: The smoothed KPZ equation in dimension three and higher: Edwards-Wilkinson regime of its fluctuations and its localization properties Abstract: We study the Kardar-Parisi-Zhang equation in dimension $d\geq 3$ with space-time white noise which is smoothed in space. There is a natural disorder parameter attached to this equation which measures the intensity of the noise. We show that when the disorder is small, the approximating solution converges to a well-defined limit (with the limit depending on both the disorder and the mollification procedure), while the re-scaled fluctuations converge to a Gaussian limit as predicted by the Edwards-Wilkionson regime. We also study the associated stochastic heat equation with multiplicative noise, which carries a natural Gaussian mutiplicative noise (GMC) on the Wiener space. When the disorder is large, we also show that the total mass of the GMC converges to zero, while the endpoint distribution of a Brownian path under the (renormlaized) GMC measure is purely atomic. Based on joint works with, A. Shamov & O. Zeitouni, F. Comets & C. Cosco as well Y. Broeker. Previous Seminars Tuesday, September 4, 2018, 4:15-5:15 Room 5417 Speaker: Philip Matchett Wood, U. Wisconsin Title: Limiting eigenvalue distribution for the non-backtracking matrix of an Erdล‘s-Rรฉnyi random graph Abstract: A non-backtracking random walk on a graph is a directed walk with the constraint that the last edge crossed may not be immediately crossed again in the opposite direction. This talk will give a precise description of the eigenvalues of the adjacency matrix for the non-backtracking walk when the underlying graph is an Erdล‘s-Rรฉnyi random graph on n vertices, where edges present independently with probability p. We allow p to be constant or decreasing with n, so long as $p\sqrt{n}$ tends to infinity. The key ideas in the proof are partial derandomization, applying the Tao-Vu Replacement Principle in a novel context, and showing that partial derandomization may be interpreted as a perturbation, allowing one to apply the Bauer-Fike Theorem. Joint work with Ke Wang at HKUST (Hong Kong University of Science and Technology). Tuesday, September 25, 2018, 4:30-5:30 (please note that this talk starts at 4:30, not 4:15!) Room 5417 Speaker: Qiang Zeng, Queens College Title: Replica Symmetry Breaking for mean field spin glass models Abstract: Mean field spin glass models were introduced as an approximation of the physical short range models in the 1970s. The typical mean field models include the Sherrington-Kirkpatrick (SK) model, the (Ising) mix p-spin model and the spherical mixed p-spin model. Starting in 1979, the physicist Giorgio Parisi wrote a series of groundbreaking papers introducing the idea of replica symmetry breaking (RSB), which allowed him to predict a solution for the SK model by breaking the symmetry of replicas infinitely many times at low temperature. In this talk, we will show that Parisiโ€™s prediction holds at zero temperature for the more general mixed p-spin model. On the other hand, we will show that there exist two-step RSB spherical mixed p-spin glass models at zero temperature, which are the first natural examples beyond the replica symmetric, one-step RSB and Full-step RSB phases. This talk is based on joint works with Antonio Auffinger (Northwestern University) and Wei-Kuo Chen (University of Minnesota). Tuesday, October 2, 2018, 4:15-5:15 Room 5417 Speaker: Erik Slivken, University of Paris VII Title: Pattern-avoiding permutations and Dyson Brownian motion Abstract: Let $S_n$ denote the set of permutations of length n. For a permutation $\tau \in S_n$ we say ฯ„ contains a pattern $\sigma\in S_k$ if there is a subsequence $i_1 < \cdots < i_k$ such that $ \tau_{i_1} \cdots \tau_{i_k}$ has the the same relative order of ฯƒ. If ฯ„ contains no pattern ฯƒ, we say that ฯ„ avoids ฯƒ. We denote the set of ฯƒ-avoiding permutations of length n by $S_n(\sigma)$. Recently, there have been a number of results that help describe the geometric properties of a uniformly random element in $S_n(\sigma)$. Many of these geometric properties are related to well-studied random objects that appear in other settings. For example, if $\sigma \in S_3$, then a permutation chosen uniformly in $S_n(\sigma)$ converges, in some appropriate sense, to Brownian excursion. Furthermore for ฯƒ = 123, 312, or 231, we can describe properties like the number and location of fixed points in terms of Brownian excursion. Larger patterns are much more difficult to understand. Currently even the simplest question, enumeration, is unknown for the pattern ฯƒ = 4231. However, for the monotone decreasing pattern $\sigma= (d+1)d\cdots 21$, a permutation chosen uniformly from $S_n(\sigma)$ can be coupled with a random walk in a cone that, in some appropriate sense, converges to a traceless Dyson Brownian motion. Tuesday, October 9, 2018, 4:15-5:15 Room 5417 Speaker: Matthew Junge, Duke University Title: Diffusion-Limited Annihilating Systems Abstract: We study a two-type annihilating system in which particles are placed with equal density on the integer lattice. Particles perform simple random walk and annihilate when they contact a particle of different type. The occupation probability of the origin was known to exhibit anomalous behavior in low-dimension when particles have equal speeds. Describing the setting with asymmetric speeds has been open for over 20 years. We prove a lower bound that matches physicistsโ€™ conjectures and discuss partial progress towards an upper bound. Joint with Michael Damron, Hanbaek Lyu, Tobias Johnson, and David Sivakoff. Tuesday, October 16, 2018, 4:15-5:15 Room 5417 Speaker: Xin Sun, Columbia University Title: Natural measures on random fractals Abstract: Random fractals arise naturally as the scaling limit of large discrete models at criticality. These fractals usually exhibit strong self similarity and spacial independence. In this talk, we will explain how these additional properties should give the existence of a natural occupation measure on the fractal set, defined to be the limit of the properly rescaled Lebesgue measure restricted to small neighborhoods of the fractal. Moreover, the occupation measure is also the scaling limit of the normalized counting measure over the corresponding discrete set. In two dimension, when putting an independent Liouville quantum gravity background over such a planar fractal, the quantum version of the occupation measure still exists, where the scaling dimension is related to the Euclidean one via the famous KPZ relation due to Knizhnik-Polyakov-Zamolodchikov and Duplantier-Sheffield. The quantum occupation measure is supposed to be the scaling limit of the normalized counting measure of the corresponding discrete set on certain random planar maps. The picture described above is expected to be true in great generality yet it is only established for a few models to various extents. In this talk, we report a fairly complete picture for planar percolation on the regular and random triangular lattice. Tuesday, October 23, 2018, 4:15-5:15 Room 5417 Speaker: Ofer Zeitouni, Weizmann Institute/NYU Title: Noise (in)stability of the spectrum of random matrices Abstract: Non-Hermitian matrices have a spectrum that is notoriously unstable to small perturbations. This fact is well captured by the notion of pseudo-spectrum, which deals with โ€œworse caseโ€ perturbations. I will discuss recent advances on the study of the spectrum of such matrices subject to vanishingly small noise. Joint work with Anirban Basak and Elliot Paquette. Tuesday, October 30, 2018, 4:15-5:15 Room 5417 Speaker: Ruojun Huang, NYU Title: On some instances of random walk in changing environment Abstract: We will talk about two recent results on random walk in changing environment. Relying on heat kernel estimates, we show that SRW evolving independently (a) on growing-in-time internal diffusion limited aggregation random cluster is recurrent, when dimension larger than two; (b) among time-increasing uniformly elliptic conductances on graphs including Z^d, a sharp, but not fully explicit, criterion to determine transience versus recurrence. The latter is joint work with Amir Dembo and Tianyi Zheng. We also discuss universality conjectures put forward by G. Amir, I. Benjamini, O. Gurel-Gurevich, and G. Kozma in arXiv:1504.04870. Tuesday, November 6, 2018, 4:15-5:15 Room 5417 Speaker: Lisa Hartung, Courant Institute Title: From 1 to 6 in branching Brownian motion Abstract: Brownian motion is a classical process in probability theory belonging to the class of โ€˜Log-correlated random fieldsโ€™. It is well known due to Bramson that the order of the maximum has a different logarithmic correction as the corresponding independent setting. In this talk we look at a version of branching Brownian motion where we slightly vary the diffusion parameter in a way that, when looking at the order of the maximum, we can smoothly interpolate between the logarithmic correction for independent random variables ($\frac{1}{2\sqrt 2}\ln(t)$) and the logarithmic correction of BBM ($\frac{3}{2\sqrt 2}\ln(t)$) and the logarithmic correction of 2-speed BBM with increasing variances ($\frac{6}{2\sqrt 2}\ln(t)$). We also establish in all cases the asymptotic law of the maximum and characterise the extremal process, which turns out to coincide essentially with that of standard BBM. We will see that the key to the above results is a precise understanding of the entropic repulsion experienced by an extremal particle. (joint work with A. Tuesday, November 13, 2018, 4:15-5:15 Room 5417 Speaker: Thomas Mountford, ร‰cole Polytechnique Fรฉdรฉrale de Lausanne Title: Critical values for renewal contact processes Abstract: A renewal contact process is a (non-Markov) process similar to the classical contact process but where the rate one Poisson processes governing โ€œrecoveryโ€ are replaced by renewal processes (transmissions are still modelled by rate lambda Poisson processes). We show that the critical values are zero if the renewal distribution has very heavy tails but is strictly positive if a moment higher than one exists (under some strict regularity condition). This is joint work with D. Marchetti and R. Fontes of USP and M.E. Vares of UFRJ. Tuesday, November 27, 2018, 4:05-5:05 (please note that this talk starts at 4:05, not 4:15!) Room 5417 Speaker: Guillaume Remy, Columbia University Title: Exact formulas for Gaussian multiplicative chaos and Liouville theory Abstract: We will present recent progress that has been made to prove exact formulas on the Gaussian multiplicative chaos (GMC) measures which are constructed by exponentiating a log-correlated field. We will give the law of the total mass of the GMC measure on the unit circle (the Fyodorov-Bouchaud formula) and on the unit interval (in collaboration with T. Zhu). The techniques of proof come from the link between GMC and Liouville conformal field theory studied by David-Kupiainen-Rhodes-Vargas. We will also discuss connections with the Duplantier-Miller-Sheffield approach to Liouville quantum gravity and identify the law of the total mass of the quantum disk (in collaboration with X. Sun).
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Non-separating planar graphs A graph G is a non-separating planar graph if there is a drawing D of G on the plane such that (1) no two edges cross each other in D and (2) for any cycle C in D, any two vertices not in C are on the same side of C in D. Non-separating planar graphs are closed under taking minors and are a subclass of planar graphs and a superclass of outerplanar graphs. In this paper, we show that a graph is a non-separating planar graph if and only if it does not contain K[1] โˆช K[4] or K[1] โˆช K[2,3] or K[1,1,3] as a minor. Furthermore, we provide a structural characterisation of this class of graphs. More specifically, we show that any maximal non-separating planar graph is either an outerplanar graph or a wheel or it is a graph obtained from the disjoint union of two triangles by adding three vertex-disjoint paths between the two triangles. Lastly, to demonstrate an application of non-separating planar graphs, we use the characterisation of non-separating planar graphs to prove that there are maximal linkless graphs with 3n โˆ’ 3 edges. Thus, maximal linkless graphs can have significantly fewer edges than maximum linkless graphs; Sachs exhibited linkless graphs with n vertices and 4n โˆ’ 10 edges (the maximum possible) in 1983.
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Vampire Einstein aperiodic tessellation I've written a program to plot the Vampire Einstein aperiodic monotile tessellation, and it works nicely in BBC BASIC for Windows BBC BASIC for SDL 2.0 (below; there's an even nicer version that uses anti-aliased graphics). But I've totally failed to make it work properly in either Matrix Brandy BASIC because the flood-fills are leaking out and spoiling the colouring. The problem seems to be that 'plot arc' (PLOT 165) is insufficiently accurate, possibly because of the integer coordinates. I'd be interested if anybody can fix this problem so that the native-graphics version can be published in a form which will run correctly on a range of BBC BASIC platforms. 10 REM 'Vampire Einstein' aperiodic tessellation without reflections 20 REM Native BBC BASIC v5+ graphics, ยฉ Richard Russell, 03-Jun-2024 30 MODE 9 : OFF 40 VDU 23,23,3;0;0;0; 50 FOR C% = 1 TO 14 60 COLOUR C%, 64+RND(160), 64+RND(160), 64+RND(160) 70 NEXT 80 COLOUR 15, 255, 255, 255 100 S = 3 : REM Overall scale 110 L = S * SQR(307 + 72*SQR3) 120 T = ATN(3 / (16/9 + SQR3)) 140 FOR T% = 1 TO 45 150 READ X, Y, A 160 GCOL 15 : PROCvampire(X, Y, A, S) 170 LINE 1278,0,1278,1022 180 GCOL RND(14) 190 FILL X - 120 * SINRADA + 50 * COSRADA, Y - 120 * COSRADA - 50 * SINRADA 200 NEXT T% 210 END 230 DATA 514,710,0, 662,670,0, 958,714,60, 848,311,180, 368,795,300 240 DATA 770,776,120, 919,861,60, 881,1009,60, 1066,822,180,1212,780,180 250 DATA 660,500,60, 258,562,240, 344,586,330, 591,244,300, 441,160,240 260 DATA 256,392,300, 994,226,120, 959,714,300, 1213,780,180,1082,542,90 270 DATA 1102,332,120,258,392,180, 220,710,240, 148,286,300, 220,708,120 280 DATA 98,882,270, 332,1114,0, 478,1072,0, 1106,628,0, 1252,416,60 290 DATA 1032,78,240, 886,120,0, 738,159,0, 480,11,240, 38,178,300 300 DATA 110,432,180, 296,200,0, 1180,162,300,-112,94,240, 58,-30,270 310 DATA 1068,-70,240,74,794,180, 110,431,60, 688,1094,30, 1362,694,120 330 DEF PROCvampire(x, y, a, s) 340 LOCAL I%, d : LOCAL DATA 350 RESTORE +1 360 FOR I% = 1 TO 7 370 READ d : d += a 380 x += L * SIN(T + RAD(d)) 390 y += L * COS(T + RAD(d)) 400 PROCside(x, y, d, s) 410 READ d : d += a 420 PROCside(x, y, d, s) 430 x -= L * SIN(T + RAD(d)) 440 y -= L * COS(T + RAD(d)) 450 NEXT 460 ENDPROC 480 DATA 80,170,50,320,200,290,170,350,230,140,260,170,290,200 500 DEF PROCside(x, y, a, s) 510 LOCAL r, d, xc, yc : r = 9 * s : d = 8 * s 520 LINE x, y, x - d*SINRADa, y - d*COSRADa 530 xc = x - d*SINRADa - r*COSRADa : yc = y - d*COSRADa + r*SINRADa 540 MOVE xc,yc : MOVE xc + r*COSRAD(a+125), yc - r*SINRAD(a+125) 550 PLOT 165, x - d*SINRADa, y - d*COSRADa 560 ENDPROC โ€ข That's a very nice program. How to fix it... as all the angles are in degrees, I wonder if using a function, say SIN in terms of COS, so that SIN and COS are precisely 90 degrees apart would help at all - pushing the irrationality of PI further away, as it were. Does the arc plotting hit different pixels depending on the direction of the arc? Is there a simple way to draw the lines 2 pixels wide, I wonder (for example, drawing the edges four times, offsetting the origin to hit four vertices of a one-pixel square)? I'm not quite in a position to experiment right now... Edit: for reference, the Spectre paper is โ€ข Here's Matrix Brandy's rendering of just one of the Vampire Einstein tiles, with the hole indicated. There's just the one hole, I think, but that's enough for the flood fill to leak out. How can the code be modified to ensure there are no holes, please? โ€ข Is that the first and last points of the perimeter not joining up, or is that in the middle of the chain? โ€ข I did notice that the value of T is 1/SQR2, but has come out to only four or five digits accuracy. Not sure if that's enough to be a root cause. I wonder if using a function, say SIN in terms of COS, so that SIN and COS are precisely 90 degrees apart would help at all - pushing the irrationality of PI further away, as it were. I must admit I don't understand that (I'm not a mathematician). The calculations in the program must be accurate to something like 1/1000 pixel, so how are they contributing to the problem? Does the arc plotting hit different pixels depending on the direction of the arc? Probably, but since BBC BASIC's graphics only plot arcs counter-clockwise I don't see the relevance. Is there a simple way to draw the lines 2 pixels wide In BB4W and BBCSDL yes: VDU 23,23,2| But it's not supported in the versions that are misbehaving. I wonder (for example, drawing the edges four times, offsetting the origin to hit four vertices of a one-pixel square)? I guess that should work, yes. A bit brute-force, but practical. I'll try it. I did notice that the value of T is 1/SQR2, but has come out to only four or five digits accuracy. Not sure if that's enough to be a root cause. I confess to being thoroughly mystified by that comment. T happens to have a value close to 1/SQR2 but it isn't 1/SQR2 (it's an angle in radians so that would make no sense geometrically). When run in Matrix Brandy or ARM BASIC VI should be accurate to something like 14 decimal digits, perhaps more (in BB4W it will be something like 18 digits). It can't be significantly less accurate than that. So I don't see where your reference to "only four or five digits accuracy" comes from. Are you basing it purely on a coincidental similarity between two different constants? I wonder (for example, drawing the edges four times, offsetting the origin to hit four vertices of a one-pixel square)? I guess that should work, yes. A bit brute-force, but practical. I'll try it. Yes, that works: 10 REM 'Vampire Einstein' aperiodic tessellation without reflections 20 REM Native BBC BASIC v5+ graphics, ยฉ Richard Russell, 04-Jun-2024 30 MODE 9 : OFF 40 FOR C% = 1 TO 14 50 COLOUR C%, 64+RND(160), 64+RND(160), 64+RND(160) 60 NEXT 70 COLOUR 15, 255, 255, 255 90 S = 3 : REM Overall scale 100 L = S * SQR(307 + 72*SQR3) 110 T = ATN(3 / (16/9 + SQR3)) 130 FOR T% = 1 TO 45 140 READ X, Y, A 150 GCOL 15 : PROCvampire(X, Y, A, S) : PROCvampire(X+2, Y, A, S) 160 PROCvampire(X, Y+2, A, S) : PROCvampire(X+2, Y+2, A, S) 170 LINE 1278,0,1278,1022 180 GCOL RND(14) 190 FILL X - 120 * SINRADA + 50 * COSRADA, Y - 120 * COSRADA - 50 * SINRADA 200 NEXT T% 210 END 230 DATA 514,710,0, 662,670,0, 958,714,60, 848,311,180, 368,795,300 240 DATA 770,776,120, 919,861,60, 881,1009,60, 1066,822,180,1212,780,180 250 DATA 660,500,60, 258,562,240, 344,586,330, 591,244,300, 441,160,240 260 DATA 256,392,300, 994,226,120, 959,714,300, 1213,780,180,1082,542,90 270 DATA 1102,332,120,258,392,180, 220,710,240, 148,286,300, 220,708,120 280 DATA 98,882,270, 332,1114,0, 478,1072,0, 1106,628,0, 1252,416,60 290 DATA 1032,78,240, 886,120,0, 738,159,0, 480,11,240, 38,178,300 300 DATA 110,432,180, 296,200,0, 1180,162,300,-112,94,240, 58,-30,270 310 DATA 1068,-70,240,74,794,180, 110,431,60, 688,1094,30, 1362,694,120 330 DEF PROCvampire(x, y, a, s) 340 LOCAL I%, d : LOCAL DATA 350 RESTORE +1 360 FOR I% = 1 TO 7 370 READ d : d += a 380 x += L * SIN(T + RAD(d)) 390 y += L * COS(T + RAD(d)) 400 PROCside(x, y, d, s) 410 READ d : d += a 420 PROCside(x, y, d, s) 430 x -= L * SIN(T + RAD(d)) 440 y -= L * COS(T + RAD(d)) 450 NEXT 460 ENDPROC 480 DATA 80,170,50,320,200,290,170,350,230,140,260,170,290,200 500 DEF PROCside(x, y, a, s) 510 LOCAL r, d, xc, yc : r = 9 * s : d = 8 * s 520 LINE x, y, x - d*SINRADa, y - d*COSRADa 530 xc = x - d*SINRADa - r*COSRADa : yc = y - d*COSRADa + r*SINRADa 540 MOVE xc,yc : MOVE xc + r*COSRAD(a+125), yc - r*SINRAD(a+125) 550 PLOT 165, x - d*SINRADa, y - d*COSRADa 560 ENDPROC โ€ข There has been a suggestion that the calculations in the program might be contributing to the issue, rather than it resulting solely from shortcomings in the plotting. I am confident that they don't: everything is calculated with the native accuracy of BBC BASIC, so that's 64-bit doubles in the case of Brandy (around 15 significant figures); no approximations are involved. But as a picture can be more convincing than words, I've plotted (below) the outline of one Vampire Einstein tile, using the same code, with a relatively high resolution. Even the tiniest errors in the values of L or T would show up, especially given how they would accumulate on each of the 14 'sides': โ€ข It's definitely the plotting granularity. The original program works in MODE 20 (640x512, 16 colours), and the newer one fails in MODE 2. The original program works in MODE 20 (640x512, 16 colours), and the newer one fails in MODE 2. MODE 2 is very low resolution and doesn't even have square pixels, so I tend to omit it from consideration (the program relies on version 5 BASIC anyway, so requires at least an Archimedes to run, ignoring Second Processors). MODE 9 is my preferred mode when writing code designed to run on a range of platforms, including Acorn and Brandy, because it's the highest resolution 16-colour mode reasonably compatible with BB4W and BBCSDL. Mode numbers from 10 upwards are completely incompatible (for example MODE 20 is 800 x 600 in my BASICs, not 640 x 512), so must be avoided in cross-platform programs. In my BASICs MODE 8 will give you 640 x 512 in 16 colours anyway (if you avoid PLOT 69 and UDCs). Edit: for reference, the Spectre paper is I've only skimmed through it, but I can't actually find a detailed description of the tile's shape there at all! Only the vague comment that the " tile boundaries are modified in a manner similar to Figure 1.1 I worked out what the shape must be from the published image and the requirement that it create a closed figure if iterated around the 14 (curved) 'sides'. โ€ข Is it the case that every tile has a perimeter gap, or only some tiles? Can you perhaps draw each of the 14 sides in different colours and see where the gap arises? Is it always at the same I had a play and it looked at first like the sixth tile drawn is the first with a harmful gap, which is not at the beginning/end of the perimeter but towards the end. But on drawing the tiles individually I see the fifth is the first to have a gap: Looks like (in my environment) it's the 66th call to PROCside which produces a gap - which is between the line and the arc. I threw in some PRINTs 505 PRINT ;sss" "x" "y" "a" "s 520 PRINT ;x" "y" "x - d*SINRADa" "y - d*COSRADa 544 PRINT ;xc" "yc" "xc + r*COSRAD(a+125)" "yc - r*SINRAD(a+125) and saw this: โ€ข (BTW, the above is by way of a possible idea for investigating - it's not a conclusion, and the results in your environment might possibly be different. By tweaking the code I could choose how many tiles to plot, and which to fill, and see where it went wrong.) For the tile vertex coordinates, I think perhaps you can view the spectre tile as a perturbation of the original aperiodic monotile, which sits nicely on a hex grid: the paper Is it the case that every tile has a perimeter gap, or only some tiles? My assumption is that it's not to do with (ordinal) side it is but rather with at what angle the side is drawn. There are 12 possible orientations for a side, at multiples of 30ยฐ (there are also 12 possible orientations of the tile itself, also in multiples of 30ยฐ). So I expect sides at some angles have 'holes' whilst those at other angles don't. But I've not tried to confirm that, because even if true it probably doesn't help 'solve' the problem. A property of BBC BASIC's graphics is (supposed to be) that when drawing a straight line between two points, the end-points are definitely both plotted. If the same guarantee was made about an arc it ought to be possible to ensure there are no 'holes', but it isn't. What I take away from both that and the original Spectre paper is that the precise shape of the 'curves' (which make it 'strongly chiral' hence ruling out a periodic tessellation) may not be uniquely determined, so long as the 'convex' sides on one tile mesh with the 'concave' sides on another. โ€ข Yes, I get the same idea. BTW, I tried my test program in SDL version and that particular call to PROCside has [very very nearly] the same inputs (as you might expect) but produces a continuous shape. So it feels to me like you might have found some edge-case in the graphics routines, which are triggered in different circumstances between platforms, but exist in more than one. Particularly, cases where an arc and a line which ought to meet up don't meet up. So it feels to me like you might have found some edge-case in the graphics routines, which are triggered in different circumstances between platforms, but exist in more than one. I'd go further, and suggest that it's not an "edge case" but just a characteristic of the plotting routines that when drawing an arc of an arbitrary length at an arbitrary angle there simply is no guarantee that both 'end points' will be drawn. After all you only explicitly provide the coordinates of end point (and it may be that guaranteed to be drawn, I don't know) since the other end of the arc is specified by a radius line, not a point. It doesn't necessarily intersect the perimeter at the centre of a pixel anyway, so 'end point' isn't well-defined.
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Resistors in Parallel and Series Resistors in Series and Parallel By the end of the section, you will be able to: โ€ข Define the term equivalent resistance โ€ข Calculate the equivalent resistance of resistors connected in series โ€ข Calculate the equivalent resistance of resistors connected in parallel In Current and Resistance, we described the term โ€˜resistanceโ€™ and explained the basic design of a resistor. Basically, a resistor limits the flow of charge in a circuit and is an ohmic device where . Most circuits have more than one resistor. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the The equivalent resistance of a combination of resistors depends on both their individual values and how they are connected. The simplest combinations of resistors are series and parallel connections (Figure 6.2.1). In a series circuit, the output current of the first resistor flows into the input of the second resistor; therefore, the current is the same in each resistor. In a parallel circuit, all of the resistor leads on one side of the resistors are connected together and all the leads on the other side are connected together. In the case of a parallel configuration, each resistor has the same potential drop across it, and the currents through each resistor may be different, depending on the resistor. The sum of the individual currents equals the current that flows into the parallel connections. (Figure 6.2.1) Figure 6.2.1 (a) For a series connection of resistors, the current is the same in each resistor. (b) For a parallel connection of resistors, the voltage is the same across each resistor. Resistors in Series Resistors are said to be in series whenever the current flows through the resistors sequentially. Consider Figure 6.2.2, which shows three resistors in series with an applied voltage equal to . Since there is only one path for the charges to flow through, the current is the same through each resistor. The equivalent resistance of a set of resistors in a series connection is equal to the algebraic sum of the individual resistances. (Figure 6.2.2) Figure 6.2.2 (a) Three resistors connected in series to a voltage source. (b) The original circuit is reduced to an equivalent resistance and a voltage source. In Figure 6.2.2, the current coming from the voltage source flows through each resistor, so the current through each resistor is the same. The current through the circuit depends on the voltage supplied by the voltage source and the resistance of the resistors. For each resistor, a potential drop occurs that is equal to the loss of electric potential energy as a current travels through each resistor. According to Ohmโ€™s law, the potential drop across a resistor when a current flows through it is calculated using the equation , where is the current in amps ( ) and is the resistance in ohms ( ). Since energy is conserved, and the voltage is equal to the potential energy per charge, the sum of the voltage applied to the circuit by the source and the potential drops across the individual resistors around a loop should be equal to zero: This equation is often referred to as Kirchhoffโ€™s loop law, which we will look at in more detail later in this chapter. For Figure 6.2.2, the sum of the potential drop of each resistor and the voltage supplied by the voltage source should equal zero: Since the current through each component is the same, the equality can be simplified to an equivalent resistance, which is just the sum of the resistances of the individual resistors. Any number of resistors can be connected in series. If resistors are connected in series, the equivalent resistance is One result of components connected in a series circuit is that if something happens to one component, it affects all the other components. For example, if several lamps are connected in series and one bulb burns out, all the other lamps go dark. EXAMPLE 6.2.1 Equivalent Resistance, Current, and Power in a Series Circuit A battery with a terminal voltage of is connected to a circuit consisting of four and one resistors all in series (Figure 6.2.3). Assume the battery has negligible internal resistance. (a) Calculate the equivalent resistance of the circuit. (b) Calculate the current through each resistor. (c) Calculate the potential drop across each resistor. (d) Determine the total power dissipated by the resistors and the power supplied by the battery. (Figure 6.2.3) Figure 6.2.3 A simple series circuit with five resistors. In a series circuit, the equivalent resistance is the algebraic sum of the resistances. The current through the circuit can be found from Ohmโ€™s law and is equal to the voltage divided by the equivalent resistance. The potential drop across each resistor can be found using Ohmโ€™s law. The power dissipated by each resistor can be found using , and the total power dissipated by the resistors is equal to the sum of the power dissipated by each resistor. The power supplied by the battery can be found using a. The equivalent resistance is the algebraic sum of the resistances: b. The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance: c. The potential drop across each resistor can be found using Ohmโ€™s law: Note that the sum of the potential drops across each resistor is equal to the voltage supplied by the battery. d. The power dissipated by a resistor is equal to , and the power supplied by the battery is equal to There are several reasons why we would use multiple resistors instead of just one resistor with a resistance equal to the equivalent resistance of the circuit. Perhaps a resistor of the required size is not available, or we need to dissipate the heat generated, or we want to minimize the cost of resistors. Each resistor may cost a few cents to a few dollars, but when multiplied by thousands of units, the cost saving may be appreciable. CHECK YOUR UNDERSTANDING 6.2 Some strings of miniature holiday lights are made to short out when a bulb burns out. The device that causes the short is called a shunt, which allows current to flow around the open circuit. A โ€œshortโ€ is like putting a piece of wire across the component. The bulbs are usually grouped in series of nine bulbs. If too many bulbs burn out, the shunts eventually open. What causes this? Letโ€™s briefly summarize the major features of resistors in series: 1. Series resistances add together to get the equivalent resistance: 2. The same current flows through each resistor in series. 3. Individual resistors in series do not get the total source voltage, but divide it. The total potential drop across a series configuration of resistors is equal to the sum of the potential drops across each resistor. Resistors in Parallel Figure 6.2.4 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when one end of all the resistors are connected by a continuous wire of negligible resistance and the other end of all the resistors are also connected to one another through a continuous wire of negligible resistance. The potential drop across each resistor is the same. Current through each resistor can be found using Ohmโ€™s law , where the voltage is constant across each resistor. For example, an automobileโ€™s headlights, radio, and other systems are wired in parallel, so that each subsystem utilizes the full voltage of the source and can operate completely independently. The same is true of the wiring in your house or any building. (Figure 6.2.4) Figure 6.2.4 (a) Two resistors connected in parallel to a voltage source. (b) The original circuit is reduced to an equivalent resistance and a voltage source. The current flowing from the voltage source in Figure 6.2.4 depends on the voltage supplied by the voltage source and the equivalent resistance of the circuit. In this case, the current flows from the voltage source and enters a junction, or node, where the circuit splits flowing through resistors . As the charges flow from the battery, some go through resistor and some flow through resistor . The sum of the currents flowing into a junction must be equal to the sum of the currents flowing out of the junction: This equation is referred to as Kirchhoffโ€™s junction rule and will be discussed in detail in the next section. In Figure 6.2.4, the junction rule gives . There are two loops in this circuit, which leads to the equations Note the voltage across the resistors in parallel are the same ( ) and the current is additive: Generalizing to any number of resistors, the equivalent resistance of a parallel connection is related to the individual resistances by This relationship results in an equivalent resistance that is less than the smallest of the individual resistances. When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, so the total resistance is lower. EXAMPLE 6.2.2 Analysis of a Parallel Circuit Three resistors , and are connected in parallel. The parallel connection is attached to a voltage source. (a) What is the equivalent resistance? (b) Find the current supplied by the source to the parallel circuit. (c) Calculate the currents in each resistor and show that these add together to equal the current output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source and show that it equals the total power dissipated by the resistors. (a) The total resistance for a parallel combination of resistors is found using (Note that in these calculations, each intermediate answer is shown with an extra digit.) (b) The current supplied by the source can be found from Ohmโ€™s law, substituting for the total resistance (c) The individual currents are easily calculated from Ohmโ€™s law , since each resistor gets the full voltage. The total current is the sum of the individual currents: (d) The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use , since each resistor gets full voltage. (e) The total power can also be calculated in several ways, use a. The total resistance for a parallel combination of resistors is found using Equation 6.2.2. Entering known values gives The total resistance with the correct number of significant digits is . As predicted, is less than the smallest individual resistance. b. The total current can be found from Ohmโ€™s law, substituting for the total resistance. This gives for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series. c. The individual currents are easily calculated from Ohmโ€™s law, since each resistor gets the full voltage. Thus, The total current is the sum of the individual currents: d. The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use , since each resistor gets full voltage. Thus, e. The total power can also be calculated in several ways. Choosing and entering the total current yields Total power dissipated by the resistors is also Notice that the total power dissipated by the resistors equals the power supplied by the source. CHECK YOUR UNDERSTANDING 6.3 Consider the same potential difference applied to the same three resistors connected in series. Would the equivalent resistance of the series circuit be higher, lower, or equal to the three resistor in parallel? Would the current through the series circuit be higher, lower, or equal to the current provided by the same voltage applied to the parallel circuit? How would the power dissipated by the resistor in series compare to the power dissipated by the resistors in parallel? CHECK YOUR UNDERSTANDING 6.4 How would you use a river and two waterfalls to model a parallel configuration of two resistors? How does this analogy break down? Let us summarize the major features of resistors in parallel: 1. Equivalent resistance is found from and is smaller than any individual resistance in the combination. 2. The potential drop across each resistor in parallel is the same. 3. Parallel resistors do not each get the total current; they divide it. The current entering a parallel combination of resistors is equal to the sum of the current through each resistor in parallel. In this chapter, we introduced the equivalent resistance of resistors connect in series and resistors connected in parallel. You may recall that in Capacitance, we introduced the equivalent capacitance of capacitors connected in series and parallel. Circuits often contain both capacitors and resistors. Table 6.2.1 summarizes the equations used for the equivalent resistance and equivalent capacitance for series and parallel connections. (Table 6.2.1) Series combination Parallel combination Equivalent capacitance Equivalent resistance Table 10.1 Summary for Equivalent Resistance and Capacitance in Series and Parallel Combinations Combinations of Series and Parallel More complex connections of resistors are often just combinations of series and parallel connections. Such combinations are common, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel. Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in Figure 6.2.5. Various parts can be identified as either series or parallel connections, reduced to their equivalent resistances, and then further reduced until a single equivalent resistance is left. The process is more time consuming than difficult. Here, we note the equivalent resistance as (Figure 6.2.5) Figure 6.2.5 (a) The original circuit of four resistors. (b) Step 1: The resistors are in series and the equivalent resistance . (c) Step 2: The reduced circuit shows resistors are in parallel, with an equivalent resistance . (d) Step 3: The reduced circuit shows that are in series with an equivalent resistance of , which is the equivalent resistance . (e) The reduced circuit with a voltage source of with an equivalent resistance of . This results in a current of from the voltage source. Notice that resistors are in series. They can be combined into a single equivalent resistance. One method of keeping track of the process is to include the resistors as subscripts. Here the equivalent resistance of The circuit now reduces to three resistors, shown in Figure 6.2.5(c). Redrawing, we now see that resistors constitute a parallel circuit. Those two resistors can be reduced to an equivalent resistance: This step of the process reduces the circuit to two resistors, shown in in Figure 6.2.5(d). Here, the circuit reduces to two resistors, which in this case are in series. These two resistors can be reduced to an equivalent resistance, which is the equivalent resistance of the circuit: The main goal of this circuit analysis is reached, and the circuit is now reduced to a single resistor and single voltage source. Now we can analyze the circuit. The current provided by the voltage source is . This current runs through resistor and is designated as . The potential drop across can be found using Ohmโ€™s law: Looking at Figure 6.2.5(c), this leaves to be dropped across the parallel combination of . The current through can be found using Ohmโ€™s law: The resistors are in series so the currents are equal to Using Ohmโ€™s law, we can find the potential drop across the last two resistors. The potential drops are . The final analysis is to look at the power supplied by the voltage source and the power dissipated by the resistors. The power dissipated by the resistors is The total energy is constant in any process. Therefore, the power supplied by the voltage source is . Analyzing the power supplied to the circuit and the power dissipated by the resistors is a good check for the validity of the analysis; they should be equal. EXAMPLE 6.2.3 Combining Series and Parallel Circuits Figure 6.2.6 shows resistors wired in a combination of series and parallel. We can consider to be the resistance of wires leading to (a) Find the equivalent resistance of the circuit. (b) What is the potential drop across resistor ? (c) Find the current through resistor . (d) What power is dissipated by (Figure 6.2.6) Figure 6.2.6 These three resistors are connected to a voltage source so that are in parallel with one another and that combination is in series with (a) To find the equivalent resistance, first find the equivalent resistance of the parallel connection of . Then use this result to find the equivalent resistance of the series connection with .(b) The current through is equal to the current from the battery. The potential drop across the resistor (which represents the resistance in the connecting wires) can be found using Ohmโ€™s law. (c) The current through can be found using Ohmโ€™s law . The voltage across can be found using (d) Using Ohmโ€™s law , the power dissipated by the resistor can also be found using a. To find the equivalent resistance of the circuit, notice that the parallel connection of R2R2 and R3R3 is in series with R1R1, so the equivalent resistance is The total resistance of this combination is intermediate between the pure series and pure parallel values ( , respectively). b. The current through is equal to the current supplied by the battery: The voltage across The voltage applied to is less than the voltage supplied by the battery by an amount . When wire resistance is large, it can significantly affect the operation of the devices represented by c. To find the current through , we must first find the voltage applied to it. The voltage across the two resistors in parallel is the same: Now we can find the current through resistance using Ohmโ€™s law: The current is less than the that flowed through when it was connected in parallel to the battery in the previous parallel circuit example. d. The power dissipated by is given by The analysis of complex circuits can often be simplified by reducing the circuit to a voltage source and an equivalent resistance. Even if the entire circuit cannot be reduced to a single voltage source and a single equivalent resistance, portions of the circuit may be reduced, greatly simplifying the analysis. CHECK YOUR UNDERSTANDING 6.5 Consider the electrical circuits in your home. Give at least two examples of circuits that must use a combination of series and parallel circuits to operate efficiently. Practical Implications One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the drop in the wires can also be significant and may become apparent from the heat generated in the cord. For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself). What is happening in these high-current situations is illustrated in Figure 6.2.7. The device represented by has a very low resistance, so when it is switched on, a large current flows. This increased current causes a larger drop in the wires represented by , reducing the voltage across the light bulb (which is ), which then dims noticeably. (Figure 6.2.7) Figure 6.2.7 Why do lights dim when a large appliance is switched on? The answer is that the large current the appliance motor draws causes a significant drop in the wires and reduces the voltage across the light. Problem-Solving Strategy: Series and Parallel Resistors 1. Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the known values for the problem, since they are labeled in your circuit diagram. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. 3. Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them. 4. Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series and another for parallel. 5. Check to see whether the answers are reasonable and consistent. EXAMPLE 6.2.4 Combining Series and Parallel Circuits Two resistors connected in series are connected to two resistors that are connected in parallel . The series-parallel combination is connected to a battery. Each resistor has a resistance of . The wires connecting the resistors and battery have negligible resistance. A current of runs through resistor . What is the voltage supplied by the voltage source? Use the steps in the preceding problem-solving strategy to find the solution for this example. 1. Draw a clear circuit diagram (Figure 6.2.8). (Figure 6.2.8) Figure 6.2.8 To find the unknown voltage, we must first find the equivalent resistance of the circuit. 2. The unknown is the voltage of the battery. In order to find the voltage supplied by the battery, the equivalent resistance must be found. 3. In this circuit, we already know that the resistors are in series and the resistors are in parallel. The equivalent resistance of the parallel configuration of the resistors is in series with the series configuration of resistors 4. The voltage supplied by the battery can be found by multiplying the current from the battery and the equivalent resistance of the circuit. The current from the battery is equal to the current and is equal to . We need to find the equivalent resistance by reducing the circuit. To reduce the circuit, first consider the two resistors in parallel. The equivalent resistance is . This parallel combination is in series with the other two resistors, so the equivalent resistance of the circuit is . The voltage supplied by the battery is therefore 5. One way to check the consistency of your results is to calculate the power supplied by the battery and the power dissipated by the resistors. The power supplied by the battery is Since they are in series, the current through equals the current through . Since , the current through each will be . The power dissipated by the resistors is equal to the sum of the power dissipated by each resistor: Since the power dissipated by the resistors equals the power supplied by the battery, our solution seems consistent. If a problem has a combination of series and parallel, as in this example, it can be reduced in steps by using the preceding problem-solving strategy and by considering individual groups of series or parallel connections. When finding for a parallel connection, the reciprocal must be taken with care. In addition, units and numerical results must be reasonable. Equivalent series resistance should be greater, whereas equivalent parallel resistance should be smaller, for example. Power should be greater for the same devices in parallel compared with series, and so on. Candela Citations CC licensed content, Specific attribution โ€ข Download for free at http://cnx.org/contents/7a0f9770-1c44-4acd-9920-1cd9a99f2a1e@8.1. Retrieved from: http://cnx.org/contents/7a0f9770-1c44-4acd-9920-1cd9a99f2a1e@8.1. License: CC BY: Explore CircuitBread
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FMCW Patch Antenna Array This example describes the modeling of a 77 GHz 2-by-4 antenna array for Frequency-Modulated Continuous-Wave (FMCW) applications. The presence of antennas and antenna arrays in and around vehicles has become a commonplace with the introduction of wireless collision detection, collision avoidance, and lane departure warning systems. The two frequency bands considered for such systems are centered around 24 GHz and 77 GHz, respectively. In this example, we will investigate the microstrip patch antenna as a phased array radiator. The dielectric substrate is air. Design Parameters Set up the center frequency and the frequency band. The velocity of light is assumed to be that of vacuum. fc = 77e9; fmin = 73e9; fmax = 80e9; vp = physconst("lightspeed"); lambda = vp/fc; Create 2-by-4 Array Hypothetical Element Pattern The FMCW antenna array is intended for a forward radar system designed to look for and prevent a collision. Therefore, begin with a hypothetical antenna element that has the significant pattern coverage in one hemisphere. A cosine antenna element would be an appropriate choice. cosineElement = phased.CosineAntennaElement; cosineElement.FrequencyRange = [fmin fmax]; cosinePattern = figure; Ideal Array Pattern The array itself needs to be mounted on or around the front bumper. The array configuration we investigate is similar to that mentioned in [1], i.e. a 2-by-4 rectangular array. Nrow = 2; Ncol = 4; fmcwCosineArray = phased.URA; fmcwCosineArray.Element = cosineElement; fmcwCosineArray.Size = [Nrow Ncol]; fmcwCosineArray.ElementSpacing = [0.5*lambda 0.5*lambda]; cosineArrayPattern = figure; Design Realistic Patch Antenna The Antenna Toolboxโ„ข has several antenna elements that could provide hemispherical coverage. Choose the patch antenna element and design it at the frequency of interest. The patch length is approximately half-wavelength at 77 GHz and the width is 1.5 times the length, for improving the bandwidth. patchElement = design(patchMicrostrip,fc); Since the default patch antenna geometry in the Antenna Toolbox library has its maximum radiation directed towards zenith, rotate the patch antenna by 90 degrees about the y-axis so that the maximum would now occur along the x-axis. This is also the boresight direction for arrays in Phased Array System Toolboxโ„ข. patchElement.Tilt = 90; patchElement.TiltAxis = [0 1 0]; axis tight Isolated Patch Antenna 3D Pattern and Resonance 3D Directivity Pattern Plot the pattern of the patch antenna at 77 GHz. The patch is a medium gain antenna with the peak directivity around 6 - 9 dBi. The patch is radiating in the correct mode with a pattern maximum at azimuth = elevation = 0 degrees. Since the initial dimensions are an approximation, check the input impedance behavior. Numfreqs = 21; freqsweep = unique([linspace(fmin,fmax,Numfreqs) fc]); Establish Bandwidth Plot the reflection coefficient of the patch to confirm a good impedance match. It is typical to consider the value ${S}_{11}=-10dB$ as a threshold value for determining the antenna bandwidth. s = sparameters(patchElement,freqsweep); hold on hold off The deep minimum at 77 GHz indicates a good match to 50. The antenna bandwidth is slightly greater than 1 GHz. Thus, the frequency band is from 76.5 GHz to 77.5 GHz. Confirm Pattern at Center and Corner Frequencies Confirm that the pattern at the corner frequencies of the band remains nearly the same. The pattern plots at 76.5 GHz and 77.5 GHz are shown below. It is a good practice to check pattern behavior over the frequency band of interest in general. Create Array from Isolated Radiators and Plot Pattern Create the uniform rectangular array (URA), but this time use the isolated patch antenna as the individual element. We choose spacing $\lambda /2$ at the upper frequency of the band i.e. 77.6 GHz. fc2 = 77.6e9; lambda_fc2 = vp/77.6e9; fmcwPatchArray = phased.URA; fmcwPatchArray.Element = patchElement; fmcwPatchArray.Size = [Nrow Ncol]; fmcwPatchArray.ElementSpacing = [0.5*lambda_fc2 0.5*lambda_fc2]; Plot the pattern for the patch antenna array so constructed. Specify a 5 degree separation in azimuth and elevation to plot the 3D pattern. az = -180:5:180; el = -90:5:90; patchArrayPattern = figure; Plot Pattern Variation in Two Orthogonal Planes Compare the pattern variation in 2 orthogonal planes for the patch antenna array and the cosine element array. Both arrays ignore mutual coupling. [Dcosine_az_zero,~,eln] = pattern(fmcwCosineArray,fc,0,el); [Dcosine_el_zero,azn] = pattern(fmcwCosineArray,fc,az,0); [Dpatch_az_zero,~,elp] = pattern(fmcwPatchArray,fc,0,el); [Dpatch_el_zero,azp] = pattern(fmcwPatchArray,fc,az,0); elPattern = figure; axis([min(eln) max(eln) -40 17]) grid on xlabel("Elevation (deg.)") ylabel("Directivity (dBi)") title("Array Directivity Variation-Azimuth = 0 deg.") legend("Cosine element","Patch Antenna",Location="best") azPattern = figure; axis([min(azn) max(azn) -40 17]) grid on xlabel("Azimuth (deg.)") ylabel("Directivity (dBi)") title("Array Directivity Variation-Elevation = 0 deg.") legend("Cosine element","Patch Antenna",Location="best") The cosine element array and the array constructed from isolated patch antennas, both without mutual coupling, have similar pattern behavior around the main beam in the elevation plane (azimuth = 0 deg). The patch-element array has a significant back lobe as compared to the cosine-element array. Using the isolated patch element is a useful first step in understanding the effect that a realistic antenna element would have on the array pattern. However, in the realistic array analysis, mutual coupling must be considered. Since this is a small array (8 elements in 2-by-4 configuration), the individual element patterns in the array environment could be distorted significantly. As a result, it is not possible to replace the isolated element pattern with an embedded element pattern. A full-wave analysis must be performed to understand the effect of mutual coupling on the overall array performance. [1] Kulke, R., S. Holzwarth, J. Kassner, A. Lauer, M. Rittweger, P. Uhlig and P. Weigand. โ€œ24 GHz Radar Sensor integrates Patch Antenna and Frontend Module in single Multilayer LTCC Substrate.โ€ Related Topics
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1: Some of these โ€œobviousโ€ categorifications are not worked out, many are known. For example there are few papers of Marmolejo and of Vitale which work out quite much including Beck-type monadicity/ descent theorems. Some relevant stuff should also be in Pronkโ€™s papers on bicategorical localization and some in older papers of Austrakian school, e.g. Kelly. David, the triangulators satisfy a swallowtail coherence condition, for which thereโ€™s a nice pictureโ€ฆ Is there a nice pictorial representation of the equations for 2-adjunctions, just as in the case of ordinary adjunctions (week 174)? at 2-adjunction I would like to list a bunch of 2-category theoretic analogs of standard facts about ordinary adjunctions. Such as: a right adjoint is a full and faithful 2-functor precisely if the counit of the 2-adjunction is an equivalence, etc. But I havenโ€™t really thought deeply about 2-adjunctions myself yet. Is there some reference where we could take such a list of properties from? That reminds me of the beginning of a long conversation at the Cafe, which ended up looking at generalizing the tangle hypothesis. I remember another interesting conversation we all had once about how from thinking about one dimensional things in the plane, one could be led to important category theoretic structures, and how this might happen with surfaces. Ah, here it is.
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Arthurโ€“Selberg trace formula 672 VIEWS Everipedia is now - Join the IQ Brainlist and our for early access to editing on the new platform and to participate in the beta testing. Arthurโ€“Selberg trace formula Arthurโ€“Selberg trace formula In mathematics, the Arthurโ€“Selberg trace formula is a generalization of the Selberg trace formula from the group SL2 to arbitrary reductive groups over global fields, developed by James Arthur in a long series of papers from 1974 to 2003. It describes the character of the representation of G(A) on the discrete part L20(G(F)โˆ–G(A)) of L2(G(F)โˆ–G(A)) in terms of geometric data, where G is a reductive algebraic group defined over a global field F and A is the ring of adeles of F. There are several different versions of the trace formula. The first version was the unrefined trace formula, whose terms depend on truncation operators and have the disadvantage that they are not invariant. Arthur later found the invariant trace formula and the stable trace formula which are more suitable for applications. The simple trace formula (Flicker & Kazhdan 1988) is less general but easier to prove. The local trace formula is an analogue over local fields. Jacquet's relative trace formula is a generalization where one integrates the kernel function over non-diagonal subgroups. โ€ข F is a global field, such as the field of rational numbers. โ€ข A is the ring of adeles of F. โ€ข G is a reductive algebraic group defined over F. In the (rare) case when G(F)โˆ–G(A) is compact the representation splits as a direct sum of irreducible representations, and the trace formula is similar to the Frobenius formula for the character of the representation induced from the trivial representation of a subgroup of finite index. In the compact case, which is essentially due to Selberg, the groups G(F) and G(A) can be replaced by any discrete subgroup ฮ“of a locally compact group G with ฮ“*G* compact. The group G acts on the space of functions on ฮ“โˆ–G by the right regular representation R, and this extends to an action of the group ring of G, considered as the ring of functions f on G. The character of this representation is given by a generalization of the Frobenius formula as follows. The action of a function f on a function ฯ† on ฮ“โˆ–G is given by In other words, R(f) is an integral operator on L2(ฮ“โˆ–G) (the space of functions on ฮ“โˆ–G) with kernel Therefore, the trace of R(f) is given by The kernel K can be written as where O is the set of conjugacy classes in ฮ“, and where ฮณ is an element of the conjugacy class o, and ฮ“ฮณ is its centralizer in ฮ“. On the other hand, the trace is also given by where m(ฯ€) is the multiplicity of the irreducible unitary representation ฯ€ of G in L2(ฮ“โˆ–G). โ€ข If ฮ“ and G are both finite, the trace formula is equivalent to the Frobenius formula for the character of an induced representation. โ€ข If G is the group R of real numbers and ฮ“ the subgroup Z of integers, then the trace formula becomes the Poisson summation formula. Difficulties in the non-compact case In most cases of the Arthurโ€“Selberg trace formula, the quotient G(F)โˆ–G(A) is not compact, which causes the following (closely related) problems: โ€ข The representation on L2(G(F)โˆ–G(A)) contains not only discrete components, but also continuous components. โ€ข The kernel is no longer integrable over the diagonal, and the operators R(f) are no longer of trace class. Arthur dealt with these problems by truncating the kernel at cusps in such a way that the truncated kernel is integrable over the diagonal. This truncation process causes many problems; for example, the truncated terms are no longer invariant under conjugation. By manipulating the terms further, Arthur was able to produce an invariant trace formula whose terms are invariant. The original Selberg trace formula studied a discrete subgroup ฮ“ of a real Lie group G(R) (usually SL2(R)). In higher rank it is more convenient to replace the Lie group with an adelic group G(A). One reason for this that the discrete group can be taken as the group of points G(F) for F a (global) field, which is easier to work with than discrete subgroups of Lie groups. It also makes Hecke operators easier to work with. The trace formula in the non-compact case One version of the trace formula (Arthur 1983) asserts the equality of two distributions on G(A): The left hand side is the geometric side of the trace formula, and is a sum over equivalence classes in the group of rational points G(F) of G, while the right hand side is the spectral side of the trace formula and is a sum over certain representations of subgroups of G(A). The invariant trace formula The version of the trace formula above is not particularly easy to use in practice, one of the problems being that the terms in it are not invariant under conjugation. Arthur (1981) found a modification in which the terms are invariant. The invariant trace formula states โ€ข f is a test function on G(A) โ€ข M ranges over a finite set of rational Levi subgroups of G โ€ข (M(Q)) is the set of conjugacy classes of M(Q) โ€ข ฮ (M) is the set of irreducible unitary representations of M(A) โ€ข a**M(ฮณ) is related to the volume of M(Q,ฮณ)*M*(A,ฮณ) โ€ข a**M(ฯ€) is related to the multiplicity of the irreducible representation ฯ€ in L2(M(Q)*M*(A)) โ€ข is related to โ€ข is related to trace โ€ข W0(M) is the Weyl group of M. Langlands (1983) suggested the possibility a stable refinement of the trace formula that can be used to compare the trace formula for two different groups. Such a stable trace formula was found and proved by Arthur (2002). Two elements of a group G(F) are called stably conjugate if they are conjugate over the algebraic closure of the field F. The point is that when one compares elements in two different groups, related for example by inner twisting, one does not usually get a good correspondence between conjugacy classes, but only between stable conjugacy classes. So to compare the geometric terms in the trace formulas for two different groups, one would like the terms to be not just invariant under conjugacy, but also to be well behaved on stable conjugacy classes; these are called stable distributions. The stable trace formula writes the terms in the trace formula of a group G in terms of stable distributions. However these stable distributions are not distributions on the group G, but are distributions on a family of quasisplit groups called the endoscopic groups of G. Unstable orbital integrals on the group G correspond to stable orbital integrals on its endoscopic groups H. There are several simple forms of the trace formula, which restrict the compactly supported test functions f in some way (Flicker & Kazhdan 1988). The advantage of this is that the trace formula and its proof become much easier, and the disadvantage is that the resulting formula is less powerful. For example, if the functions f are cuspidal, which means that for any unipotent radical N of a proper parabolic subgroup (defined over F) and any x, y in G(A), then the operator R(f) has image in the space of cusp forms so is compact. Jacquet & Langlands (1970) used the Selberg trace formula to prove the Jacquetโ€“Langlands correspondence between automorphic forms on GL2 and its twisted forms. The Arthurโ€“Selberg trace formula can be used to study similar correspondences on higher rank groups. It can also be used to prove several other special cases of Langlands functoriality, such as base change, for some groups. Kottwitz (1988) used the Arthurโ€“Selberg trace formula to prove the Weil conjecture on Tamagawa numbers. Lafforgue (2002) described how the trace formula is used in his proof of the Langlands conjecture for general linear groups over function fields. โ€ข Maass wave form โ€ข Harmonic Maass form
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Degrees of Freedom: Definition, How to Calculate & Examples Degrees of Freedom Degrees of Freedom: Definition, How to Calculate & Examples What is Degrees of Freedom? Degrees of freedom is a fundamental concept in the realm of statistics, offering crucial insights in various fields such as physics, engineering, and machine learning. It represents the number of independent values or parameters that can vary in an analysis without breaching any given constraints. While it might initially seem like a perplexing concept, understanding degrees of freedom is vital for executing statistical tests correctly, interpreting their results, and ensuring the validity of those results. Key Points 1. Degrees of freedom represent the number of observations or data points that are free to vary in statistical analysis. 2. In general, it is determined by the sample size minus the number of parameters being estimated. 3. Degrees of freedom play a crucial role in hypothesis testing and determining the appropriate statistical distribution for inference. Understanding Degrees of Freedom Degrees of freedom (DoF) is a slightly abstract statistical concept that refers to the number of values in a statistical calculation that are free to vary. Simply put, it provides an idea of how much information you have at your disposal to estimate statistical parameters. Letโ€™s consider a simple example to illustrate this concept: you have a dataset containing five data points, and you know their average (mean). If you know the values of four of these data points, you can easily calculate the value of the fifth data point because itโ€™s constrained by the average. In this case, you have four DoF because four values can freely vary. This concept becomes increasingly important as we delve into more complex statistical tests and models. For instance, in a chi-square test, DoF are used to define the shape of the chi-square distribution, which in turn helps us determine the critical value for the test. Similarly, in regression analysis, DoF help quantify the amount of information โ€œusedโ€ by the model, thus playing a pivotal role in determining the statistical significance of predictor variables and the overall model fit. Understanding the concept of DoF and accurately calculating it is critical in hypothesis testing and statistical modeling. It not only affects the outcome of the statistical tests but also the reliability of the inferences drawn from such tests. Different Statistical Tests and Degrees of Freedom The concept of degrees of freedom (DoF) applies to a variety of statistical tests. Each test uses DoF in its unique way, often defining the shape of the corresponding probability distribution. Here are several commonly used statistical tests and how they use DoF: 1. T-tests In a T-test, degrees of freedom determine the specific shape of the T distribution, which varies based on the sample size. For a single sample or paired T-test, the DoF are typically the sample size minus one (n-1). For a two-sample T-test, DoF are calculated using a slightly more complex formula involving the sample sizes and variances of both groups. 2. Chi-Square tests For Chi-square tests, used often in categorical data analysis, the DoF are typically the number of categories minus one. In a contingency table, DoF are (number of rows โ€“ 1) * (number of columns โ€“ 1). 3. ANOVA (Analysis of Variance) In an ANOVA, DoF are split between the number of groups minus one (between-group df) and the total sample size minus the number of groups (within-group DoF). The F distribution, used in ANOVA, is determined by these two DoF values. 4. Regression Analysis In simple linear regression, DoF are the number of observations minus the number of estimated parameters (usually 2: the slope and intercept). In multiple regression, itโ€™s the number of observations minus the number of parameters estimated (including each predictor and the intercept). Understanding how degrees of freedom interact with these statistical tests is crucial to selecting the correct test and interpreting its results accurately. How to Calculate Degrees of Freedom The exact way to calculate degrees of freedom can vary depending on the specific statistical test being used. However, here are general guidelines for calculating degrees of freedom in some common 1. Single-Sample t-test The degrees of freedom for a single-sample t-test are calculated as the sample size (n) minus 1. This is because one parameter (the sample mean) is being estimated. 2. Paired t-test The degrees of freedom for a paired t-test are calculated as the number of pairs (n) minus 1. 3. Two-sample t-test The degrees of freedom for a two-sample t-test can be approximated as the smaller of the two sample sizes (n1 and n2) minus 1. 4. Chi-square test For a chi-square test, the degrees of freedom are equal to the number of categories minus 1. 5. One-way ANOVA In a one-way ANOVA, the total degrees of freedom is n-1 (where n is the total number of observations). This is split into two parts: the degrees of freedom between groups (k-1 where k is the number of groups) and the degrees of freedom within groups (n-k). 6. Regression Analysis In regression analysis, the degrees of freedom are typically calculated as the total number of observations minus the number of parameters being estimated. Remember that these are general rules, and the exact calculation can sometimes be more complex, particularly for more advanced statistical techniques. Always make sure you understand the statistical method youโ€™re using and the appropriate way to calculate degrees of freedom. Overfitting and Degrees of Freedom Overfitting is a critical concept in statistics and machine learning. It refers to a model that fits the data too closely, to the point where it captures not only the underlying patterns but also the random noise in the data. Such a model performs well on the training data but poorly on new, unseen data, thus leading to poor predictive performance and generalization. The degrees of freedom (DoF) in a statistical model are closely related to the risk of overfitting. A model with too many DoF is likely to overfit the data. This is because having more DoF allows the model to use complex or flexible functions to fit the data, which might capture random noise along with the actual pattern. For instance, in a regression analysis, increasing the number of predictors increases the DoF, as each additional predictor allows the model to fit the data more closely. While this might seem beneficial, it can lead to overfitting if the model becomes too complex and starts fitting the noise in the data. On the other hand, reducing the DoF can help prevent overfitting by making the model simpler. Techniques like regularization, which add a penalty term to the loss function based on the number of parameters, effectively reduce the DoF and thus help prevent overfitting. However, itโ€™s important to strike the right balance. If you reduce the DoF too much, the model might become too simple to capture the underlying patterns in the data, leading to underfitting. As with many aspects of model building, finding the right balance between bias and variance (or underfitting and overfitting) is key. Limitations of Degrees of Freedom While degrees of freedom (DoF) are integral to statistical testing and model development, their usage comes with certain limitations and assumptions that need to be considered. 1. Assumptions Many statistical tests that utilize DoF often make assumptions about the data being used, such as normality or homoscedasticity. If these assumptions are violated, the tests may not be valid. 2. Complexity Understanding and correctly applying DoF can be complex, particularly for those new to statistics. It requires a clear understanding of the underlying statistical principles. 3. Risk of Misinterpretation In some cases, people misinterpret the concept of DoF and apply it incorrectly. For example, adding more variables to a model will increase the DoF, but it doesnโ€™t necessarily improve the model as it may lead to overfitting. 4. Applicability DoF are more applicable to parametric tests, which assume underlying statistical distributions. For non-parametric tests, which do not make such assumptions, DoF may not be as 5. Overfitting vs. Underfitting As discussed, while controlling DoF can help prevent overfitting, reducing them too much may lead to oversimplification or underfitting of the model. Striking the right balance is key but can be challenging. In conclusion, while DoF are a crucial concept in statistics and provide invaluable insights, their application needs to be done thoughtfully, considering the nature of the data and the objectives of the statistical analysis or model. Examples of Degrees of Freedom To provide a clearer understanding of the concept of degrees of freedom (DoF), letโ€™s look at a few examples in different contexts. 1. T-Test When running a t-test, the DoF are typically calculated as the total sample size minus the number of groups. For example, if we are comparing two groups each with 10 samples, the DoF would be 10+10-2 = 18. This calculation becomes crucial when looking up the t-distribution table to determine the critical t-value. 2. Chi-Square Test In a chi-square test for independence, the DoF are calculated as (number of rows โ€“ 1) * (number of columns โ€“ 1). For instance, if weโ€™re analyzing a contingency table with 3 rows and 3 columns, the DoF would be (3-1)*(3-1) = 4. 3. ANOVA In an Analysis of Variance (ANOVA), there are two types of DoF โ€“ between groups and within groups. If there are โ€˜nโ€™ groups each of size โ€˜mโ€™, the DoF between groups would be n-1 and within groups would be n*(m-1). 4. Regression Analysis In regression, the DoF is the number of observations minus the number of estimated parameters. So, if we have 100 observations and we are estimating 3 parameters (two coefficients and a constant), then the DoF would be 100 โ€“ 3 = 97. Remember, these are simplistic examples and in real-world applications, the calculations may become more complex, taking into account various factors such as assumptions about the population, the design of the study, and so on. What is the concept of degrees of freedom in statistics? Degrees of freedom refer to the number of independent values or observations that can vary in statistical analysis. How are degrees of freedom calculated? Degrees of freedom are typically calculated as the difference between the total number of observations or data points and the number of parameters or restrictions in the statistical model. Why are degrees of freedom important? Degrees of freedom determine the appropriate statistical distribution for hypothesis testing and help assess the reliability of statistical estimates. How does the sample size affect degrees of freedom? Increasing the sample size generally increases the degrees of freedom, allowing for more precise and reliable statistical inferences. About Paul Paul Boyce is an economics editor with over 10 years experience in the industry. Currently working as a consultant within the financial services sector, Paul is the CEO and chief editor of BoyceWire. He has written publications for FEE, the Mises Institute, and many others. Further Reading Embargo Definition and Examples - An embargo is a government-imposed restriction on trade or economic activity with a specific country or group of countries.
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Shear Strength - (Intro to Geotechnical Science) - Vocab, Definition, Explanations | Fiveable Shear Strength from class: Intro to Geotechnical Science Shear strength is the maximum resistance of a soil or rock to shear stress, which is critical in understanding how materials behave under loading conditions. This concept is essential in various aspects of geotechnical engineering, as it influences stability, load-bearing capacity, and the overall performance of structures in contact with soil. congrats on reading the definition of Shear Strength. now let's actually learn it. 5 Must Know Facts For Your Next Test 1. Shear strength can be defined by the Mohr-Coulomb failure criterion, which expresses shear strength as a function of cohesion and normal stress. 2. Different methods such as laboratory tests (like triaxial tests) and in-situ tests (like vane shear tests) are used to determine shear strength values. 3. Shear strength is influenced by factors such as moisture content, density, and soil composition, making it essential for effective site investigation and design. 4. Understanding shear strength is vital for analyzing slope stability and preventing landslides, as it determines how much load the soil can bear before failing. 5. In seismic evaluations, shear strength is crucial for assessing soil liquefaction potential and ensuring the safety of structures during earthquakes. Review Questions โ€ข How does shear strength impact slope stability and what testing methods are commonly used to determine it? โ–ก Shear strength significantly impacts slope stability because it dictates how much load the soil can withstand before failure occurs. Testing methods such as laboratory triaxial tests and in-situ vane shear tests are commonly employed to accurately assess shear strength values. These tests help engineers understand the material's behavior under different stress conditions, which is essential for designing stable slopes and preventing landslides. โ€ข Explain the relationship between effective stress and shear strength in soils, particularly regarding changes in water content. โ–ก The relationship between effective stress and shear strength is foundational in geotechnical engineering. Effective stress, which represents the stress carried by the soil skeleton, directly influences shear strength. As water content increases within a soil mass, pore water pressure rises, reducing effective stress and thus decreasing shear strength. This understanding helps engineers assess risks in saturated soils and implement appropriate measures for stability. โ€ข Evaluate how the Mohr-Coulomb failure criterion assists in predicting failure in geotechnical structures and its relevance during seismic events. โ–ก The Mohr-Coulomb failure criterion provides a framework for predicting failure by relating shear strength to cohesion and normal stress on a material. This criterion is particularly relevant during seismic events, where dynamic loading can increase shear stresses on soils beyond their capacity. By applying this criterion, engineers can evaluate the potential for failure in slopes or foundations under seismic loads, enabling them to design structures that can withstand such forces effectively. ยฉ 2024 Fiveable Inc. All rights reserved. APยฎ and SATยฎ are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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ThmDex โ€“ An index of mathematical definitions, results, and conjectures. Let $X$ be a D11: Set such that (i) $I : X \to Y$ is an D440: Identity map on $X$ For all $x, y \in X$, we have $x = I(x) = I(y) = y$. The claim follows. $\square$
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An Optimisation Model to Consider the NIMBY Syndrome within the Landfill Siting Problem Dipartimento di Ingegneria, Universitร  del Sannio, piazza Roma 21, 82100 Benevento, Italy Submission received: 23 May 2019 / Revised: 15 July 2019 / Accepted: 16 July 2019 / Published: 18 July 2019 This paper proposes a discrete optimisation model and a heuristic algorithm to solve the landfill siting problem over large areas. Besides waste transport costs and plant construction and maintenance costs, usually considered in such problems, the objective function includes economic compensation for residents in the areas affected by the landfill, to combat the NIMBY (Not In My Back Yard) syndrome or, at least, reduce its adverse effects. The proposed methodology is applied to a real-scale case study, the region of Campania, Italy, where waste management is a thorny problem. Numerical results show that the proposed algorithm may be used to obtain a solution to the problem, albeit sub-optimal, with acceptable computing times, and the proposed model tends to locate landfills in sparsely populated sites. 1. Introduction NIMBY (Not In My Back Yard) syndrome [ ] is a widespread phenomenon related to the strong opposition of a community to some public-interest interventions in a local area, mainly โ€œundesirable plantsโ€, such as landfills, incinerators and thermoelectric power plants. Local communities often โ€œfightโ€ against public authorities and their decisions, even with violent demonstrations and roadblocks. In this context, landfills are probably the most controversial facilities; in addition to social costs, they can lead to a reduction in property values in adjacent zones [ ]. Hence residents living in the surrounding areas should be assured of economic compensation [ ]. Such an approach should be able to compensate the inhabitants involved for all external and social costs produced by the plant and incurred primarily or exclusively by themselves, such as air pollution (due to the plant and to the vehicles transporting waste), noise, foul-smelling air, soil availability (directly occupied or indirectly influenced by the plant), the already cited reduction in property values and possible health effects (whether real or imaginary, depending largely on plant location and design). Decision-makers should consider compensation costs just like other costs (construction, maintenance, waste transportation, etc.). In this paper, we formulate an optimisation model for solving the landfill siting problem, where economic compensation is also explicitly considered. Some preliminary results of this research have been published in [ The landfill siting problem can be considered a variant of the classical -median problem [ ]; more precisely, the problem can be formulated as a location problem with demand-dependent costs [ ]. A review of -median problems was published in [ ], where the main solution algorithms were also examined. This paper focuses mainly on the proposed approach, especially evaluating the effects on the optimal solutions of considering economic compensation within the objective function; a comparison between methods, models and algorithms for solving location problems lies outside the scope of this paper. The objectives of this paper were to propose a model for optimising landfill location that explicitly considers compensation costs for inhabitants affected by the plant and an algorithm that can solve the problem efficiently in acceptable computing times even when the problem at hand occurs on a large scale. Moreover, the methodology is tested on a real-scale case study, the region of Campania, where waste management has been, and still is, a thorny problem. The paper is structured as follows: Section 2 examines the background of landfill siting problems; Section 3 illustrates the proposed methodology, based on an optimisation model and a solution algorithm; Section 4 summarises the numerical results on a real-scale case; Section 5 discusses the results; finally, Section 6 draws the main conclusions. 2. Background In this section, the background to the landfill siting problem is explored. A literature review on the more general location problem would deserve its own treatment; here, we refer only to papers cited in the introduction. The importance of considering social aspects in waste management problems, as well as public acceptance so as to limit NIMBY syndrome effects, was underlined in [ ]. Simsek et al. [ ] proposed a landfill site screening procedure which considered NIMBY syndrome constraints, without providing, however, compensation costs for inhabitants. Waste management models were reviewed in [ ], where the models are classified into: (i) models based on cost-benefit analysis; (ii) models based on life-cycle analysis; and (iii) models based on multicriteria decision analysis. The authors concluded that joint consideration of environmental, economic and social aspects of the problem is an important aim to be pursued in the waste management approach, whilst emphasising the importance of peopleโ€™s acceptance in such kinds of decisions. Several papers propose multicriteria approaches [ ] for solving location problems of undesirable plants. Melachrinoudis et al. [ ] developed a dynamic (multiperiod) multicriteria mixed integer programming model; they considered a planning horizon and that some parameters of the problem (population, costs, etc.) could change over time, taking into account past data and future projections. The model considered four objectives: minimising total cost, minimising risk for the population, minimising risk for the ecosystem and minimising inequity. Hokkanen and Salminen [ ] applied the ELECTRE III multicriteria approach to choose a solid waste management system in a Finnish region, bearing in mind that the energy potential of the waste in question should be used within the region. Cheng et al. [ ] proposed an integration of a multicriteria approach and inexact mixed integer linear programming for selecting an optimal landfill site and a waste-flow-allocation pattern; the objective was to minimise total costs and several different multicriteria methods were tested (weighted addition, weighted product, co-operative game theory, TOPSIS and complementary ELECTRE). Public acceptance was considered one of the criteria adopted in the procedure. Vasiloglou [ ] proposed a decision-making process in which broad community participation and acceptance was explicitly considered. Kontos et al. [ ] proposed a spatial methodology based on multicriteria analysis for siting a landfill that identifies land suitability on a grading scale; the methodology, which used an analytical hierarchy process for estimating the weights of the criteria, was applied to the island of Lemnos. Chang et al. [ ] proposed a fuzzy multicriteria analysis integrated with a GIS; in an initial stage, a GIS-based spatial analysis identifies the potential areas that are then examined with fuzzy multicriteria analysis. Xi et al. [ ] integrated an optimisation approach with multicriteria decision analysis for supporting long-term planning of solid waste management in the City of Beijing (China); also here, fuzzy theory was adopted inside the multicriteria analysis model. A GIS-based multicriteria analysis for siting a landfill in the Polog region of Macedonia was proposed in [ ]; in the multi-criteria decision framework, environmental and economic factors were standardised using fuzzy membership functions, and combined by integrating analytic hierarchy process and ordered weighted average techniques. Gbanie et al. [ ] modelled the landfill location problem using GIS and multicriteria analysis, and applied the proposed approach to a case study in Sierra Leone; the multicriteria-GIS approach integrated the weighted linear combination and the ordered weighted average techniques. A landfill site selection based on fuzzy inference was proposed in [ ]. Sumathi et al. [ ] proposed GIS-based procedures that combined the use of GIS with a multicriteria analysis, Zamorano et al. [ ] applied the procedure to a case study in Southern Spain and Simsek et al. [ ] explicitly considered NIMBY syndrome constraints. Other case studies can be found in [ Eiselt [ ] compared actual and optimal landfill locations, while Eiselt and Marianov [ ] proposed a procedure for jointly locating landfills and transfer stations. Finally, Guiqin et al. [ ] introduced an analytic hierarchy process (AHP) for selecting landfill sites. To our best knowledge, limiting local opposition to the location of a landfill was not previously considered a real cost to bear; this paper proposes a methodology for including this aspect of the problem inside the optimisation procedure. 3. Methodology The landfill siting problem assumes that: (i) several landfills have to be located in a large geographical area for receiving the waste produced therein, (ii) the location of all waste sources is known and (iii) the annual production from each source is known. Moreover, we consider that there is at least one waste source for each municipality and its location is known; this point may represent a local plant where waste produced in the municipality is collected and compacted before being transferred to landfill; more waste sources can be assumed in large cities. Finally, we assume that all eligible sites in the region were already identified and located, so as to respect some technical (geological, hydro-geological, etc.), territorial (accessibility, altitude, etc.), social and political constraints. The proposed approach considers three kinds of costs: (i) landfill construction/maintenance costs, (ii) waste transportation costs and (iii) compensation costs. The cost of a landfill depends on the annual quantity of waste that it receives (including the construction costs, which are appropriately amortised); each eligible site may have a different function that relates its annual cost with the waste quantity to be received. The transportation cost is assumed dependent only on ton-km of transported waste, and other more complex approaches considering the different vehicles used and/or the kind of road used can be introduced, without significant impacts on the proposed approach. Finally, compensation costs for residents affected by the plants are also included. They represent compensation given by society to those directly affected by the negative impacts of each landfill, and could be paid in different ways: tax exemptions, reductions in energy costs, monetary investments for improving the urban environment and the quality of life (e.g., parks and gardens, sustainable transit systems, building renovation). 3.1. Optimisation Model The mathematical model is formulated as follows: x^opt = Arg [x] min (ฮฃ[i] yc[i](w[i](x)) x[i] + ฮฃ[j] d[j,min](x) wp[j] c[tr] + ฮฃ[i] res[i] w[i](x) c[comp]) โ€ข i indicates an eligible site; โ€ข j indicates a waste source; โ€ข x is the vector of decision variables x[i]; โ€ข x^opt is the optimal solution; โ€ข x[i] is a binary variable that is equal to 1 if a landfill is located in site i, 0 otherwise; โ€ข w[i](x) is the annual waste quantity allocated to site i; โ€ข yc[i](.) is a function used for calculating the annual cost of the landfill located in site i; it depends on the annual waste quantity to be treated, w[i](x); โ€ข d[j,min](x) indicates the distance between a waste source j and the nearest landfill; โ€ข wp[j] indicates the annual production of waste source j; โ€ข c[tr] indicates the cost per ton-km of transported waste (โ‚ฌ/ton-km); โ€ข res[i] indicates the number of residents that have to be compensated for a landfill located in site i; โ€ข c[comp] is the annual compensation cost per ton of waste per resident (โ‚ฌ/ton); โ€ข l[max] is the maximum number of landfills. Even if other options are possible, in this model, we assume that the waste produced by source , is sent to the nearest landfill. Under this assumption, we can write: โ€ข a[i,j](x) is equal to 1 if, under configuration x, site i is the nearest to source j (d[i,j] = d[j,min]), 0 otherwise, with d[i,j] indicating the distance between site i and source j. This model is a binary non-linear optimisation model that belongs to the class of location problems with demand-dependent costs [ ]. This problem is NP-hard like all location problems on general graphs [ 3.2. Solution Algorithm In the literature, plant location problems are solved with several methods and algorithms, most of which are heuristic or metaheuristic; a literature review can be found in [ ]. Indeed, solving the problem with exact approaches is possible only for small or simple case studies. Many metaheuristic methods are proposed in the literature [ ] for solving this problem, such as genetic algorithms [ ], simulated annealing [ ] and Tabu search [ ]. Extensive literature is also available for heuristic algorithms [ For solving the Equations (1)โ€“(3), we propose to use a heuristic algorithm that can lead to a solution, albeit sub-optimal, in an acceptable computing time also for large problems. This algorithm uses a multi-start technique [ ] and a variant of a greedy heuristic [ ] that is based on exhaustive monodimensional searches. Multi-start procedures are commonly used in combinatorial problems when the objective function is not convex; Resende and Werneck [ ], among others, applied this approach to -median problems. It is worth noting that the landfill location problems usually have a large number of decisional variables (one for each possible location site) and a very small maximum number of landfills to be set up. For example, the application that will be carried out in the next section will have 551 possible sites but a maximum of only five landfills. Therefore, it is possible to simplify the management of the solution algorithm by transforming the binary model into a discrete model. Indeed, we may introduce pointer variables, , that assume an integer value between 0 and , where is the number of possible locations. Each pointer variable indicates site , where a landfill is provided; otherwise, = 0 indicates that the pointer variable is not associated to any site and thus the number of landfills in this solution is lower than the maximum. Figure 1 shows an example of the above approach. Adopting this approach, each solution can be represented by a vector of pointer variables, , and the optimisation model can be formulated as follows: ฯ‡^opt = Arg [ฯ‡] min (ฮฃ[i] yc[i](w[i](ฯ‡)) x[i](ฯ‡) + ฮฃ[j] d[j,min](ฯ‡) wp[j] c[tr] + ฮฃ[i] res[i] w[i](ฯ‡) c[comp]) Using the pointer variables allows the constraint (3) to be included directly within the model and the search to be limited only to solutions that respect it. Indeed, solutions that exceed the maximum number of landfills are automatically excluded, and at least one landfill is provided considering that the transportation cost is assumed to be infinity (actually, a very large number) if there are no landfills. The solution algorithm is articulated as follows. Phase 0โ€”Initialisation. Set the counter of iterations to 0. Set all variables ฯ‡[k] to 0. Set the objective function value to a very large number, M. Phase 1โ€”Mono-dimensional exhaustive search (1st cycle). Examine all solutions obtained by changing the first pointer from 0 to n, identify the best one and set the first pointer to the corresponding value. Repeat the procedure for the other pointers, assuming the previous pointer values are fixed. End when the optimal value of a pointer is equal to 0 (no other landfills) or when all pointers have been exhaustively explored. In this phase, the algorithm examines at most l[max] ร— n solutions. Phase 2โ€”Mono-dimensional exhaustive search (2nd cycle). Repeat phase 1 starting from the last pointer values, until the best solution at the end of the exhaustive search is equal to that generated in the previous phase or a maximum number of solutions have been examined. Figure 2 a reports the structure of the algorithm. Since the problem is not convex and the proposed algorithm is heuristic, the algorithm leads to a local optimum. For exploring the solution set, we propose to apply a multi-start technique; this method generates randomly different solutions that are used as starting points for the second cycle (see Figure 2 b). The complexity of each mono-dimensional exhaustive search is 4. Numerical Results We tested the proposed approach on a large area corresponding to the region of Campania (southern Italy). This region (13,590 km ) has 551 municipalities, about 5.8 million inhabitants and its capital is Naples (about 1 million inhabitants). We assumed 551 waste sources, one per municipality, whose waste production was obtained from official data [ ]. We also assumed 551 eligible sites, one inside each municipal area, and that a landfill located in an area will affect only people living in the same municipality. In Figure 3 a the territories, with their centroid nodes, are reported, while Figure 3 b reports the road network graph, representing about 6000 km of roads, used for calculating the distances between sources and eligible sites. All maps in this paper were generated with the software QGIS and are geo-referenced with the coordinate reference system WGS84/UTM Zone 32. Moreover, we have assumed a compensation cost equal to 0.0001 โ‚ฌ/ton per resident and an average transportation cost of 0.4 โ‚ฌ/ton-km (obtained from some regional tenders regarding waste management). The annual cost of a landfill was calculated using the method proposed in [ yc[i](w[i](ฯ‡)) = c[0] + w[i](ฯ‡) c[1](w[i](ฯ‡)) โ€ข c[0] is a fixed cost of the landfill (โ‚ฌ/year); โ€ข c[1] is a variable cost per waste ton (โ‚ฌ/ton-year), depending on w[i] for considering the scale economy. According to the formula proposed in [ ] and converting the values of coefficients from dollars to Euros, we assumed: = 220,000 and )) = 428.015 With these assumptions, we tested the proposed algorithm considering the three terms of the objective function (transportation costs; construction/maintenance costs; compensation costs) and then only the first two terms. The results are summarised in Table 1 Figure 4 The algorithm, starting from all pointers equal to 0, examined 1654 solutions in about 21 min; in Figure 5 , the objective function values corresponding to solutions examined by the algorithm are reported. In this test, the multi-start procedure, starting from 15 random solutions, led to the same solution. Moreover, an exhaustive search among all solutions that have only two landfills (151,525 solutions, computing time about 32 h) led to the same results. However, it does not ensure that the algorithm leads to the global optimum. By comparing results with and without compensation costs, we can underline that the proposed approach leads to a solution where the population affected by the undesirable plant is lower: indeed, Casoria and Battipaglia have 77,642 and 50,798 inhabitants, respectively, while Massa di Somma and Giungano only have 5449 and 1253 inhabitants, respectively. The methodology, therefore, should be able to limit NIMBY syndrome effects, since the directly involved people are less, and to limit the total compensation costs, since they are proportional to involved inhabitants. Finally, we assumed different landfill construction/maintenance yearly cost functions, in order to explore how the solution changes with significant changes in costs. We used the same function adopted above, modifying only the term )) by substituting the value 428.015 with the following values: 600, 500, 400, 300, 200 and 100. Figure 6 shows the landfill year costs generated under such assumptions. Table 2 summarises the results corresponding to different landfill construction/maintenance cost functions, identified by a different value of the coefficient . As expected, solutions with more landfills correspond to lower amounts of the coefficient (5 or 4, if is between 100 and 300; between 1 and 3, if is between 400 and 600), i.e., lower construction/annual maintenance costs. On comparing the results with and without compensation costs, it can be shown that the total number of landfills does not differ so markedly (only in two cases does the model lead to a different number of landfills) although the number of inhabitants involved is very different. Some municipalities recur in the optimal solution; if we consider the compensation costs, this is due to a promising combination of position and inhabitants, while in the other case only the location influences the results. As a percentage of total annual costs, the compensation costs vary from 1.3% to 5.3% (see Figure 7 ), while the increase between the solutions with and without compensation costs varies between 3.0% and 11.5% (see Figure 8 ). Inhabitants of municipalities affected by landfills are summarised in Figure 9 : the differences between considering compensation costs or otherwise are very substantial. In Figure 10 Figure 11 , the optimal locations in the different cases are reported. 5. Discussion The location of undesirable plants, especially of landfills, is a complex problem in several respects: (1) mathematical, since the problem is NP-Hard; (2) technical, since the choice of the specific location has to consider several aspects such as the kind of soil, the presence or otherwise of aquifers, landslide risk, hydrogeological risk, accessibility and so on; (3) economic, related to construction, maintenance and transportation costs; (4) environmental, such as the general impacts on air quality, noise, soil pollution and fauna; (5) social, related to the external costs produced by the landfill that are incurred by the local population (smelly air, reduction in property values, noise, etc.) and the inevitable inequity of the interventions (only some people are affected by a plant that is required to satisfy the needs of a vast community). Social problems lead to the strong opposition of local communities to the construction and operation of the plant, the well-known NIMBY syndrome. The approach proposed in this paper tries to solve some social aspects of the problem, contrasting the local opposition by economically compensating the people affected by the plant. This compensation, if it is directly included in the objective function, leads the solution towards locations that impact on smaller local communities. For instance, an analysis of our results shows (see Figure 9 Table 2 ) a large difference in the size of the population involved with and without considering the compensation costs. Other crucial aspects of the problem regard the external costs of landfills. While some social and environmental external costs for the host communities can be offset economically, as herein proposed, other impacts, such as on local flora and fauna [ ], are not explicitly considered in this study. These impacts can be considered and limited in two ways: (1) to select, after a specific in-depth study, as possible landfill hosting sites only those that generate minor impact on flora and fauna; (2) to propose a method for including in the landfill costs also those on flora and fauna, changing the formulation of )). In this second case, a specific study on the economic value of such impacts has to be preliminarily conducted. Estimation of the economic compensation for the host community is an important point to examine; indeed, it significantly influences the final result and the political acceptability of the intervention. Here we assumed that each inhabitant would be compensated proportionally to the amount of waste addressed to the landfill and that each plant affects only the residents of the hosting municipality, regardless of the distance. These simplifications are compatible with a โ€œreal-scaleโ€ (and not โ€œrealโ€) case study and can be removed in future researches, which will require specific in-depth studies. In the literature, numerous studies proposed methods for determining willingness to accept (WTA) compensation for hosting undesirable plants. Usually, direct surveys are necessary for estimating the willingness to accept compensation; examples involving waste disposal infrastructures can be found in [ ]. In [ ], the authors highlighted the importance of the distance on the compensation and its acceptance. The acceptance of compensation is higher for those who live further away from the site; on the contrary, residents closer to the plant tend to refuse compensation and prefer to continue the opposition. This suggests, as a matter of course, that higher compensation should be granted to residents in the areas closest to the plant. To use a value of WTA from the literature is seldom possible; on the contrary, a different value should be used for different municipalities, since social factors significantly influence the value in question [ ]. Numerous other case studies can be found in the literature; herein we refer to [ The proposed methodology is general and transferable since it can be easily applied to other large areas. Indeed, all necessary data are easily obtainable and the social problems related to the location of landfills are common to all regions of the world. A preliminary phase requires identification of all feasible locations that respect the technical prescriptions (it is possible to use multi-hazard assessment maps as well [ ]) and the corresponding local communities affected by the plant. Moreover, surveys for estimating the compensation costs would also be required. In our test, that has to be considered a โ€œreal-scaleโ€ case study but not a โ€œrealโ€ case study; it was not possible to perform the preliminary phase: all municipalities were assumed as possible sites and the population of each municipality is considered affected only if the plant is located in its own territory. Except for this simplification, all data used are real, as well as the road network. It is essential also to underline that another significant contribution for reducing the local opposition can be given by inclusive decision-making processes, which involve the local communities in the political decisions, negotiating possible solutions. The choice of the case study is not casual. Waste management in the region of Campania has been, and is partly still, a critical problem from environmental, social and political points of view. Officially, the crisis started in 1994 and ended in 2011; it was preliminarily due to the oversaturation of the available landfills and then to the difficulty to propose and implement an effective regional waste management plan. The crisis had a loud echo in the national and international press. During this period, several NIMBY syndrome phenomena have occurred, such as local opposition to the waste-to-energy plant of Acerra, to the landfills provided in Naples (Chiaiano and Pianura districts) and Terzigno. During the same period, numerous investigations by the judiciary showed that organised crime had infiltrated the waste business. Today, the situation is better, also thanks to the higher percentage of separate waste collection, although the risk of a new crisis is ever-present. The waste crisis in Campania has also been the subject of numerous studies in the scientific literature. Herein we cite only [ 6. Conclusions The methodology proposed in this paper aims to limit the opposition from local communities to undesirable plants. Giving economic compensation to local communities affected by an undesirable plant is common practice, but often policy-makers grant such benefits after the decision has already been taken. Considering economic compensation like other project costs can lead to solutions that can be better accepted and may result in compensation costs that can be lower overall. This is confirmed by the results obtained in this paper. Comparison between the solution with and without compensation costs showed that the solutions differed significantly in terms of the number of people to be compensated, while, by examining the landfill locations, the corresponding optimal sites are not so distanced. It should be stressed that the proposed approach increases the overall cost of the solution; this is due to the compensation costs and, almost always, also to increases in other costs. This additional cost is about 4.5% in the case study, and varies from 3.0% and 11.5% with different assumptions on landfill costs. Given that public opposition often generates significant delays in construction, opening and operation of plants, this approach could indeed reduce the real cost of the overall waste management cycle. Moreover, if the policy-maker compensates those affected by the waste facility after the choice of the location, such costs will be significantly higher than those obtained with the proposed approach. From the point of view of computational efforts, considering compensation costs or otherwise does not produce significant effects. Finally, the proposed heuristic algorithm is able to lead to a (local) optimal solution in acceptable computing times in a large-scale case study. Future research efforts would be best spent on removing some simplifying assumptions that we have considered in this paper and testing the model on a โ€œrealโ€ case, instead of a โ€œreal-scaleโ€ case, where the actual possible locations and construction/maintenance costs are considered. In particular, a specific study would be necessary for evaluating the compensations, with different values depending on the social factors of residents and distance from the plant. This study will require specific surveys, GIS-based approaches and multidisciplinary competencies (environmental, hydro-geological, economic, social, etc.). Finally, the proposed algorithm could be compared with other algorithms proposed in the literature, such as genetic algorithms, scatter searches and simulated annealing. This research received no external funding. The author is grateful to the Editor and three anonymous reviewers for their valuable comments and suggestions. 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[Google Scholar] [CrossRef] With Compensation Costs Without Compensation Costs Number of landfills 2 2 Municipalities Massa di Somma; Giungano Casoria; Battipaglia Total inhabitants 6702 128,440 Transportation costs 34,274,833 30,270,389 Landfill annual costs 54,311,274 55,754,544 Compensation costs 1,312,718 โˆ’ Objective function value 89,898,825 86,024,933 Coeff. With Compensation Costs Without Compensation Costs Number of landfills 5 5 Municipalities Bellizzi; Carinaro; Pietradefusi; San Sebastiano al Vesuvio; Santโ€™Egidio del Monte Albino Aversa; Capaccio; Napoli; Pietradefusi; Santโ€™Egidio del Monte Albino Total inhabitants 41,202 1,061,188 100 Transportation costs 21,349,521.53 19,291,800.69 Landfill annual costs 17,018,714.56 17,055,619.52 Compensation costs 2,168,097.22 - Objective function value 40,536,333.30 36,347,420.21 Number of landfills 5 5 Municipalities Capaccio; Carinaro; Pietradefusi; San Sebastiano al Vesuvio; Santโ€™Egidio del Monte Albino Benevento; Casavatore; Pompei; Salerno; Vallo della Lucania Total inhabitants 50,444 247,902 200 Transportation costs 21,206,069.45 21,448,200.66 Landfill annual costs 32,918,528.70 31,482,230.56 Compensation costs 2,326,336.09 - Objective function value 56,450,934.24 52,930,431.21 Number of landfills 5 4 Municipalities Bellizzi; Carinaro; Celle di Bulgheria; Pietradefusi; San Sebastiano al Vesuvio Benevento; Casavatore; Salerno; Torre Orsaia Total inhabitants 34,215 216,131 300 Transportation costs 22,913,058.23 24,290,867.40 Landfill annual costs 47,126,152.86 43,748,060.61 Compensation costs 2,173,152.10 - Objective function value 72,212,363.19 68,038,928.02 Number of landfills 3 3 Municipalities Giungano; Massa di Somma; Pietradefusi Battipaglia; Casavatore; Pietradefusi Total inhabitants 9072 71,795 400 Transportation costs 30,418,540.86 26,547,423.82 Landfill annual costs 54,645,985.51 55,576,136.54 Compensation costs 1,239,271.64 - Objective function value 86,303,798.02 82,123,560.36 Number of landfills 2 2 Municipalities Giungano; Massa di Somma Battipaglia; Casavatore Total inhabitants 6724 69,447 500 Transportation costs 34,274,833.33 30,109,769.18 Landfill annual costs 63,371,526.24 65,128,788.33 Compensation costs 1,312,717.53 - Objective function value 98,959,077.10 95,238,557.51 Number of landfills 1 2 Municipalities Massa di Somma Casoria; Agropoli Total inhabitants 5444 99,123 600 Transportation costs 39,204,346.81 32,079,796.38 Landfill annual costs 70,697,991.09 75,959,521.84 Compensation costs 1,397,663.58 - Objective function value 111,300,001.49 108,039,318.22 ยฉ 2019 by the author. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http:// Share and Cite MDPI and ACS Style Gallo, M. An Optimisation Model to Consider the NIMBY Syndrome within the Landfill Siting Problem. Sustainability 2019, 11, 3904. https://doi.org/10.3390/su11143904 AMA Style Gallo M. An Optimisation Model to Consider the NIMBY Syndrome within the Landfill Siting Problem. Sustainability. 2019; 11(14):3904. https://doi.org/10.3390/su11143904 Chicago/Turabian Style Gallo, Mariano. 2019. "An Optimisation Model to Consider the NIMBY Syndrome within the Landfill Siting Problem" Sustainability 11, no. 14: 3904. https://doi.org/10.3390/su11143904 Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details Article Metrics
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Ch. 11 Chapter Review - Contemporary Mathematics | OpenStax Chapter Review 1 . In a plurality election, the candidates have the following vote counts: A 125, B 132, C 149, and D 112. Which candidate has the plurality and wins the election? For the following exercises, use the table below. Options A B C D E Candidate 1 1 3 3 1 3 Candidate 2 2 1 1 2 4 Candidate 3 3 4 2 4 1 Candidate 4 4 2 4 3 2 2 . Which candidate has a plurality? 3 . Does the plurality candidate have a majority? 4 . Determine the winner of the election by the Hare method based on the sample preference summary in the table. For the following exercises, use the table below. Number of Ballots 10 20 15 5 Option A 1 4 3 4 Option B 2 3 4 2 Option C 4 2 1 3 Option D 3 1 2 1 5 . Use ranked-choice voting to determine the two options in the final round and the number of votes they each receive in that round. 6 . Is there a winning option? If so, which option? Justify your answer. For the following exercises, use the table below. Number of Ballots 100 80 110 105 55 Candidate A 1 1 4 4 2 Candidate B 2 2 2 3 1 Candidate C 4 4 1 1 4 Candidate D 3 3 3 2 3 7 . What are the Borda scores for each candidate? 8 . Which candidate is the winner by the Borda count method? Use Pairwise Comparison Matrix for Candidates U, V, W, X, and Y to answer Questions 9 and 10. 9 . Calculate the points received by each candidate in pairwise comparison matrix. 10 . Determine the winner of the pairwise comparison election represented by matrix. If there is a winner, determine whether the winner is a Condorcet candidate and explain your reasoning. If there is no winner, indicate this. 11 . The ladies of The Big Bang Theory decide to hold their own approval voting election to determine the best option in Rock, Paper, Scissors, Lizard, Spock. Use the summary of their approval ballots in the table below to determine the number of votes for each candidate. Determine the winner, or state that there is none. VOTERS Penny Bernadette Amy Rock Yes No No Paper Yes Yes No Scissors Yes Yes Yes Lizard No No No Spock Yes No Yes For the following exercises, use the table below. Percentage of Vote 40% 35% 25% Candidate A 1 3 2 Candidate B 2 1 3 Candidate C 3 2 1 12 . Which candidate is the winner by the ranked-choice method? 13 . Suppose that you used the approval method and each voter approved their top two choices. Which candidate is the winner by the approval method? 14 . Which candidate is the winner by the Borda count method? 15 . In a Borda count election, the candidates have the following Borda scores: A 1245, B 1360, and C 787. Candidate A received 55% of the first place rankings. Identify which fairness criteria, if any, are violated by characteristics of the described voter profile in this Borda count election. Explain your reasoning. For the following exercises, use the table below. Number of Ballots 8 10 12 4 Option A 1 3 2 1 Option B 3 1 4 4 Option C 4 2 1 2 Option D 2 4 3 3 16 . Determine Borda score for each candidate, and the winner of the election using the Borda count method. 17 . Is there a majority candidate? If so, which candidate? 18 . Does the Borda method election violate the majority criterion? Justify your answer. 19 . In a Borda count election, the candidates have the following Borda scores: A 15, B 11, C 12, and D 16. The pairwise match up points for the same voter profiles would have been A 2, B 0, C 1, and D 3. Identify which fairness criteria, if any, are violated by characteristics of the described voter profile in this Borda election. Explain your reasoning. 20 . Determine the winner of the election using the ranked-choice method. 21 . If the four voters in the last column rank C ahead of A, which candidate wins by the ranked-choice method? 22 . Does this ranked-choice election violate the monotonicity criterion? Explain your reasoning. For the following exercises, use the table below. Number of Ballots 15 12 9 3 Option A 1 3 3 2 Option B 2 2 1 1 Option C 3 1 2 3 23 . Determine the winner of the election by the Borda method. 24 . Does this Borda method election violate the IIA? Why or why not? 25 . Which of the ranked voting methods in this chapter, if any, meets the majority criterion, the head-to-head criterion, the monotonicity criterion, and the irrelevant alternatives criterion? 26 . Identify the states, the seats, and the state population (the basis for the apportionment) in the given scenario: The reading coach at an elementary school has 52 prizes to distribute to their students as a reward for time spent reading. 27 . Use the given information to find the standard divisor to the nearest hundredth. Include the units. The total population is 2,235 automobiles, and the number of seats is 14 warehouses. 28 . Use the given information to find the standard quota. Include the units. The state population is eight residents in a unit, and the standard divisor is 1.75 residents per parking space. 29 . Which of the four apportionment methods discussed in this section does not use a modified divisor? 30 . Determine the Hamilton apportionment for Scenario X in the table below. State A State B State C State D State E State F Total Seats Scenario X 17.63 26.62 10.81 16.01 13.69 15.24 100 31 . Does the apportionment resulting from Method X in the table below satisfy the quota rule? Why or why not? State A State B State C State D State E Standard Quota 1.67 3.33 5.00 6.67 8.33 Apportionment Method X 2 2 5 7 9 For the following exercises, use the table below and the following information: In Wakanda, the domain of the Black Panther, King Tโ€™Challa, has six fortress cities. In Wakandan, the word โ€œbirninโ€ means โ€œfortress city.โ€ King Tโ€™Challa has found 111 Vibranium artifacts that must be distributed among the fortress cities of Wakanda. He has decided to apportion the artifacts based on the number of residents of each birnin. Fortress Cities Birnin Djata (D) Birnin T'Chaka (T) Birnin Zana (Z) Birnin S'Yan (S) Birnin Bashenga (B) Birnin Azzaria (A) Total Residents 26,000 57,000 27,000 18,000 64,000 45,000 237,000 Standard Quota 12.18 26.70 12.65 8.43 29.98 21.08 111 32 . Does the Jefferson method result in an apportionment that satisfies or violates the quota rule in this scenario? 33 . Find the modified upper quota for each state using a modified divisor of 2,250. Is the sum of the modified quotas too high, too low, or equal to the house size? 34 . Use the Adams method to apportion the artifacts. Determine whether it is necessary to modify the divisor. If so, indicate the value of the modified divisor. 35 . Does the Adams method result in an apportionment that satisfies or violates the quota rule in this scenario? 36 . Use the Webster method to apportion the artifacts. Determine whether it is necessary to modify the divisor. If so, indicate the value of the modified divisor. 37 . Does the Webster method result in an apportionment that satisfies or violates the quota rule in this scenario? 38 . Which of the four methods of apportionment from this section are the residents of Birnin S'Yan likely to prefer? Justify your answer. 39 . Does the change from a standard divisor to a modified divisor tend to change the number of seats for larger or smaller states more? 40 . Which of the four apportionment methodsโ€”Jefferson, Adams, Hamilton, or Websterโ€”satisfies the quota rule? 41 . A city purchased five new firetrucks and apportioned them among the existing fire stations. Although your neighborhood fire station has the same proportion of the cityโ€™s firetrucks as before the new ones were purchased, it now has one fewer. Is this scenario an example of a quota rule violation, the Alabama paradox, the population paradox, the new-states paradox, or none of these? 42 . When the number of seats changed from 25 to 26, the standard quotas changed from A 2.21, B 5.25, C 11.27, and D 6.27 to A 2.30, B 5.46, C 11.72, and D 6.52. a. How did the increase in seats impact the apportionment? b. Is this apportionment an example of a paradox? Justify your answer. 43 . The school resources officers in a county were reapportioned based on the most recent census. The number of students at Chapel Run Elementary went up while the number of students at Panther Trail Elementary went down, but Chapel Run now has 1 fewer resources officers while Panther Trail has one more than it did previously. Is this scenario an example of a quota rule violation, the Alabama paradox, the population paradox, the new-states paradox, or none of these? For the following exercises, the house size is 24 seats. When the population of A increases by 28 percent, B increases by 26 percent, and C increases by 15 percent, the standard quotas change from A 3.38, B 6.32, and C 14.30 to A 3.63, B 6.67, and C 13.71. 44 . How did the change in populations impact the apportionment? 45 . Is this apportionment an example of a paradox? Justify your answer. 46 . When the city of Cocoa annexed an adjacent unincorporated community, the number of seats on the city council was increased to maintain the standard ratio of citizens to seats, but one existing community of Cocoa still lost a seat on the city council to another existing community of Cocoa when the new community was added. Is this scenario an example of a quota rule violation, the Alabama paradox, the population paradox, the new-states paradox, or none of these? For the following exercises, the house size was 27. There were three states with standard quotas of A 6.39, B 11.40, and C 9.21. A fourth state was annexed, and the house size was increased to 35. The new standard quotas are A 6.38, B 11.37, C 9.19, and D 8.06. 47 . How did the additional state impact the apportionment? 48 . Is this apportionment an example of a paradox? Justify your answer. For the following exercises, suppose 11 seats are apportioned to States A, B, and C with populations of 50, 129, and 181 people, respectively. Then the populations of States A, B, and C change to 57, 151, and 208, respectively. 49 . Demonstrate that the population paradox occurs when the Hamilton method is used. 50 . Demonstrate that the population paradox does not occur when the Jefferson method is used. Justify your answer. 51 . Demonstrate that the population paradox does not occur when the Adams method is used. Justify your answer. 52 . Demonstrate that the population paradox does not occur when the Webster method is used. Justify your answer.
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Beyond the Kalamides Gedankenexperiment which appears now to be refuted - June 2, 2013 โ–ก Jack Sarfatti On Jun 2, 2013, at 7:22 AM, JACK SARFATTI <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote: Yes it's always the case that if the time evolution is unitary signal interference terms cancel out. That is essence of the no-signal argument. It's what defeated my 1978 attempt using two interferometers on each end of the pair source that David Kaiser describes in How the Hippies Saved Physics that was in first edition of Gary Zukav's Dancing Wu Li Masters. Stapp gave one of the first no-signal proofs in response to my attempt. I. However, one of the tacit assumptions is that all observables must be Hermitian operators with real eigenvalues and a complete orthogonal basis. II. Another assumption is that the normalization once chosen should not depend on the free will of the experimenter. Both & II are violated by Glauber states. The linear unitary dynamics is also violated when the coherent state is Higgs-Goldstone vacuum/groundstate expectation value order parameter of a non-Hermitian boson second quantized field operator where the c number local nonlinear nonunitary Landau-Ginzburg equation in ordinary space replaces the linear unitary Schrodinger equation in configuration (or Wigner phase space more generally) as the dominant dynamic. P. W. Anderson called this "More is different." For example in my toy model NORMALIZED so as to rid us of that damn spooky telepathic psychokinetic voodoo magick without magic |A,B> = [2(1 + |<w|z>|^2)]^-1/2[|0>|z> + |1>|w>] <0|1> = 0 for Alice A <w|z> =/= 0 for Bob B Trace over B {|0><0| |A,B><A,B|} = 1/2 etc. probability is conserved and Alice receives no signal from Bob in accord with Abner Shimony's "passion at a distance". However, probability is not conserved on Bob's side! Do the calculation if you don't believe me. Two more options i. use 1/2^1/2 normalization, then we get an entanglement signal for Alice with violation of probability conservation for Alice, though not for Bob ii Final Rube Goldberg option (suspect) use different normalizations depending on who does the strong von Neumann measurement Alice or Bob. Now this is a violation of orthodox quantum theory ladies and gentlemen. Sent from my iPhone in San Francisco, Russian Hill โ–ก Jack Sarfatti On Jun 2, 2013, at 12:56 AM, nick herbert <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote: Kalamidas Fans-- I have looked over Martin Suda's two papers entitled 1. Taylor expansion of Output States and 2. Interferometry at the 50/50 BS. My conclusion is that Martin is within one millimeter of a solid refutation of the kalamidas scheme. Congratulations, Martin, on achieving this result and on paying so much close attention to kalamidas's arguments. The result, as expected, comes from a very strange direction. In particular, the approximation does not enter into Suda's refutation. Martin accepts all of kalamidas's approximations and refutes him anyway. I have not followed the math in detail but I have been able to comprehend the essential points. First, on account of the Martin Suda paradox, either PACS or DFS can be correctly used at this stage of the argument. So martin derives the kalamidas result both ways using PACS (Kalamidas's Way) and then DFS (Howell's Way). Both results are the same. Then Martin calculates the signal at the 50/50 beam splitter (Alice's receiver) due to Bob's decision to mix his photon with a coherent state |A>. Not surprisingly Martin discovers lots of interference terms. So Kalamidas is right. However all of these interference terms just happen to cancel out. So Kalamidas is wrong. Refutation Complete. Martin Suda Wins. This is a very elegant refutation and if it can be sustained, then Kalamidas's Scheme has definitively entered the Dustbin of History. And GianCarlo can add it to his upcoming review of refuted FTL schemes. But before we pass out the medals, there is one feature of the Suda Refutation that needs a bit of justification. Suda's formulation of the Kalamidas Scheme differs in one essential way from Demetrios's original presentation. And it is this difference between the two presentations that spells DOOM FOR DEMETRIOS. Kalamidas has ONE TERM |1,1> that erases which-way information and Suda has two. Suda's EXTRA TERM is |0,0> and represents the situation where neither of Bob's primary counters fires. Having another term that erases which-way information would seem to be good, in that the Suda term might be expected to increase the strength of the interference term. However--and this is the gist of the Suda refutation--the additional Suda term |0.0> has precisely the right amplitude to EXACTLY CANCEL the effect of the Kalamidas |1,1> term. Using A (Greek upper-case alpha) to represent "alpha", Martin calculates that the amplitude of the Kalamidas |1,1> term is A. And that the amplitude of the Suda |0,0> term is -A*. And if these amplitudes are correct, the total interference at Alice's detectors completely disappears. Congratulations, Martin. I hope I have represented your argument correctly. The only task remaining is to justify the presence (and the amplitude) of the Suda term. Is it really physically reasonable, given the physics of the situation, that so many |0,0> events can be expected to occur in the real world? I leave that subtle question for the experts to decide. Wonderful work, Martin. Nick Herbert
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Ensemble 1D DenseNet Damage Identification Method Based on Vibration Acceleration 1 School of Civil Engineering, Chongqing Jiaotong University, Chongqing, 400074, China 2 School of Architecture and Art Design, Henan Polytechnic University, Jiaozuo, 454001, China 3 Zhong Yun International Engineering Co., Ltd., Zhengzhou, 450007, China * Corresponding Author: Chun Sha. Email: Structural Durability & Health Monitoring 2023, 17(5), 369-381. https://doi.org/10.32604/sdhm.2023.027948 Received 23 December 2022; Accepted 14 March 2023; Issue published 07 September 2023 Convolution neural networks in deep learning can solve the problem of damage identification based on vibration acceleration. By combining multiple 1D DenseNet submodels, a new ensemble learning method is proposed to improve identification accuracy. 1D DenseNet is built using standard 1D CNN and DenseNet basic blocks, and the acceleration data obtained from multiple sampling points is brought into the 1D DenseNet training to generate submodels after offset sampling. When using submodels for damage identification, the voting method ideas in ensemble learning are used to vote on the results of each submodel, and then vote centrally. Finally, the cantilever damage problem simulated by ABAQUS is selected as a case study to discuss the excellent performance of the proposed method. The results show that the ensemble 1D DenseNet damage identification method outperforms any submodel in terms of accuracy. Furthermore, the submodel is visualized to demonstrate its operation mode. Structural health monitoring (SHM) [1] is critical for extending the life of civil engineering structures, and damage identification is an essential component of structural health monitoring. Deep learning has been used to propose many damage identification methods [2] based on vibration signals (natural frequency and mode shapes [3โ€“5], acceleration signals, and so on). Natural frequency and mode shapes necessitate the use of sensors to cover the structure, whereas using acceleration signals necessitates less. The deep learning algorithm based on vibration acceleration studied by predecessors is improved in this paper to improve accuracy, and the common cantilever beam damage in engineering is used as the research object, which also has some practical implications. In the early stage, researchers have proposed various SHM approaches. Examples are the vision-based methods of Barile et al. [6] and Zaurin et al. [7] and the vibration-based methods of Kumar et al. [8] and Dรถhler et al. [9]. In recent years, with the explosive growth of data scale, deep learning has become the mainstream of artificial intelligence [10], triggering extensive research in various disciplines, and it also plays a major role in the field of SHM. Convolution Neural Network (CNN) is a deep learning algorithm, which was proposed at the end of the twentieth century [11], but it was not taken seriously due to limited computing power and a lack of databases at the time. Until AlexNet [12] demonstrated exceptional performance in the ImageNet Large-Scale Visual Recognition Challenge (ILSVRC) in 2012, 2D CNN was widely used and became a representative deep learning algorithm, primarily used in image classification [13,14], natural language processing [15,16], speech recognition [17,18] and 1D signal (such as vibration) processing. Yu et al. [19] developed a 2D CNN model to locate damage in a five-story structure, combining 14 measured signals into 2D features. The trained model correctly identified the structural damage. Khoandehlouh et al. [20] conducted experimental tests on a laboratory structure of a reinforced concrete bridge, and the proposed 2D CNN model can identify the overall damage of the bridge. Gulgec et al. [21] used a 2D CNN model to accurately classify damaged and undamaged finite element modeling samples. The main advantage of 2D CNN is that it does not require artificial feature extraction by combining feature extraction and classification tasks, and the features of local connection and weight sharing significantly reduce the number of parameters. When feature mapping is particularly complex, however, it is necessary to deepen the network, which may result in performance degradation, gradient disappearance, or explosion. CNN variants have been developed to address these issues. He et al. [22] created the Residual Network (ResNet), which realized another gradient flow mode by establishing jump connections between residual layers and extended the networkโ€™s depth, and won first place in the ILSVRC in 2015. Wang et al. [23] used this method to detect frame structure damage and achieved remarkable results. Huang et al. [24] created Densely Connected Convolutional Networks, DenseNet, which increased network width by reusing the characteristic graph, and achieved excellent results in ILSVRC in 2017. Practice has shown that this method is better suited for small sample data sets than ResNet. Wang et al. [25] successfully used this method to detect damage in simply supported beams. Using 2D CNN for damage identification requires merging data obtained from multiple sampling points into one sample. The data features obtained by each sampling point are different, which leads to a large number of features. The difficulty of model training is exponentially related to the number of features. So, it is necessary to increase the amount of data and model complexity. This problem has not been substantially solved in long-term research. Kiranyaz et al. [26] proposed 1D CNN in 2015, which has the same basic theory as 2D CNN but has far fewer parameters and computational complexity, so it is widely used in 1D data processing. For the first time, Abdeljaber et al. [27] used 1D CNN in vibration-based SHM. Using a large frame experimental structure built by Qatar University as the experimental object, the acceleration data of each frame interface point was extracted and analyzed, and it was possible to determine whether or not each frame interface point was damaged. Avci et al. [28] used 1D CNN to analyze triaxial sensor signals and then integrate them, yielding good results in many damage scenes. Yang et al. [29] solved the problem of two types of steel truss structure damage using 1D CNN. Zhang et al. [30] tested three different beams and used 1D CNN to determine where the stiffness of the beams changed. Using 1D CNN for damage identification only requires identifying one type of feature, and the difficulty of model training and data demand are very low, but when studying some complex problems, its accuracy is often not as good as 2D CNN. From the description of the current research situation, it can be seen that both 2D CNN and 1D CNN have certain limitations in the application of structural damage identification. The data volume requirement of 2D CNN model is high, the training difficulty is high, and the application difficulty in variable actual engineering is relatively high. The later emerging 1D CNN has greatly reduced both computational complexity and data acquisition difficulty, but its accuracy for complex problems cannot be guaranteed. Therefore, this paper proposes an idea of an ensemble 1D CNN model, which uses multiple sampling points to obtain submodels for ensemble voting to improve the accuracy of the 1D CNN model. The submodel algorithm used in this research process is a 1D form of DenseNet that has received rave reviews in recent years. I believe that according to different research problems, this algorithm can be changed at will. Using this ensemble model to independently model each sampling pointโ€™s features does not require too much data, and network training difficulty is also very small. Chapter 2 introduces the principles of 1D CNN and DenseNet basic blocks, and proposes the network structure and ensemble 1D DenseNet damage identification method built in this paper. Chapter 3 gives a representative case, the problem of damage identification of cantilever beams, so as to explore the accuracy of the proposed method, and briefly introduces the idea of parameter adjustment of the model. Chapter 4 presents the results of the case, shows the results of each submodel and the results after ensemble, and visualizes the feature vectors of the submodels to explore their internal mechanisms. Chapter 5 summarizes the conclusions and shortcomings, as well as the future development direction. 2 Ensemble 1D DenseNet Damage Identification Method The standard 1D CNN is divided into two layers: feature extraction and classification. The goal of this paperโ€™s 1D DenseNet is to improve the feature extraction layer of the standard 1D CNN into the DenseNet basic block. The submodel of this paper is trained by 1D DenseNet. The following are examples: 2.1 introduces the standard 1D CNN theory, 2.2 introduces the DenseNet basic block, 2.3 introduces the 1D DenseNet structure used in this paper, and 2.4 ensemble 1D DenseNet damage identification method. The convolution layer and pooling layer are part of the 1D CNN feature extraction layer, and the classification layer is a full connection network layer. Following the initial feature vector of 1D data input into 1D CNN, features are extracted by convolution layer and compressed by pooling layer, which are connected back and forth and stacked repeatedly to form new feature vectors. Fig. 1 depicts the principle. The feature vector of the feature extraction layerโ€™s final layer is expanded into one dimension and brought into the fully connected network layer for classification. Multiple convolution kernels with the same number of channels (equal to the number of eigenvector channels and one-to-one correspondence between them), the same size, and different weights are present on each convolution layer. To extract features, each convolution kernel performs convolution calculations with an eigenvector. The convolution process is achieved by sliding the convolution kernels, each of which is nested with an ReLU -activation function at the end of the convolution. This alleviates the gradient disappearance problem to some extent, and the function expression is as follows: ReLU(x)={x, x>00, xโ‰ค0 (1) The convolution process for a single convolution kernel of size 3 and channel number MN is illustrated in Fig. 1, with XiN denoting the i th feature vector channel at layer N. WikN denotes the weight of the i th channel of the k th convolution kernel in layer N. YikN denotes the convolution result of the i th channel of the k th convolution kernel in layer N. bkN+1 denotes the bias of the k th convolution kernel at layer N+1 . f(โˆ™) denotes the activation function; YkN+1 denotes the convolution result of the k th convolution kernel in layer N+1 . The principle of the convolution calculation for the i th channel in the convolution kernel is shown in the dashed box. The forward propagation formula from layer N to layer N+1 of the network is: YkN+1=f(bkN+1+โˆ‘i=1MNโกYikN), YikN=[y1, y2, y3โ€ฆytโ€ฆ] (2) In Fig. 1, XiN=[x1, x2, x3โ€ฆxt(j)โ€ฆxm] ; WikN=[wik(1)N, wik(2)N, wik(3)N] ; YikN=[y1, y2, y3โ€ฆytโ€ฆ] ,the convolution calculation formula in the dotted box is as follows: yt=โˆ‘j=13โกwik(j)Nxt(j) (3) where xt(j) denotes the j th weight value of the t th convolved region in the i th channel of the feature vector; yt the convolution result of the t th convolved region. The features obtained by each convolution kernel in the convolution layer are compressed using the maximum value or average value principle, and the features are compressed while maintaining the relative position of the features, reducing the number of parameters and the amount of calculation. Fig. 1 depicts the pooling of a convolution kernel following convolution. In the channel dimension, all pooled results in the same layer are connected to form a new feature vector. 2.1.3 Full Connection Layer Network Layer The feature vector of the last layer is expanded into 1D form and used as the input to the fully connected network. As this paper is a classification problem, the softmax function is set at the end of the fully connected network and the expression is given in Eq. (4). Its function is to transform each output value into a probabilistic form, and the label corresponding to the output value with the highest probability is the category determined by the network. yยฏ(yi)=exp(yi)โˆ‘i=1Uโกexp(yi) (4) where yi denotes the function input value; U is the number of input values; yยฏ(yi) denotes the layer softmax output value. DenseBlock is made up of DenseLayers with the same structure. Fig. 2 depicts the Dense Layer structure. It can be seen that, in addition to normal forward propagation, there is branch propagation, which allows for the splicing of front and back features in the channel dimension. In the figure, Cโˆ—L denotes the input feature vector (number of channels * feature length), BatchNorm denotes the normalisation of the features, xConv1d denotes a 1-dimensional convolutional kernel of size x, 4k and k (growth rate) represent the number of convolutional kernels, and dropout reduces overfitting by randomly removing some neurons when training the network. Finally, the output feature vector is (k+C)โˆ—L . The Transition joins two DenseBlocks, the structure of which is shown in Fig. 2, and compresses the number of channels using a 1-dimensional convolution kernel of size 1. In the figure, ฮธ is the compression factor (set to 0.5 in this paper), and the length of the feature vector is then compressed by an average pooling layer. 2.3 Network Structure and Super Parameter Setting Table 1 depicts the network structure established in this paper. To suppress noise, the initial feature vector is extracted using a large convolution kernel, and then it enters the 1D DenseNet basic block to extract fine features using a small convolution kernel. Each DenseBlock block consists of 6 DenseLayers, with a total of 40 layers in the network and a growth rate of k = 12. 2.4 Ensemble 1D DenseNet Damage Identification Method Aiming at the problem of damage identification, for the same form of damage, after the structure is loaded, different acceleration sampling points can obtain data with different characteristics. Many researchers combine the data obtained from different sampling points to establish 2D CNN model to solve it. However, we know that 1D CNN outperforms 2D CNN in processing sequence data because it has fewer network parameters and requires fewer samples. So in this paper, another idea is put forward. Carry out offset sampling on the data sets obtained from each sampling point, and bring it into 1D DenseNet introduced in 2.3 for training to generate a model, so that each sampling point corresponds to a submodel. When applying submodels to damage identification of structures, it is only necessary to obtain the data of each sampling point by the same method, and after the offset sampling, it is brought into each submodel to output the classification results. Each submodel is solved by voting method, and then the voting results are concentrated to improve the accuracy. See Fig. 3 for ideas. 3 Example: Damage Location of Cantilever Beam Data sets are created to simulate the damage problem of cantilever beam in order to evaluate the method proposed in this paper. Create the cantilever beam model shown in Fig. 4 using ABAQUS. The elastic modulus is set to 3e10, the Poissonโ€™s ratio to 0.2, the density to 2551, the unit type to Plane182, and the structure size to 6m ร— 0.4m. The grid is divided into 8 horizontal sections and 120 vertical sections. To simulate the damage, the local elastic modulus reduction (random reduction of 30%โˆผ60%) is used. The damage size is fixed at 3 ร— 3 units, and the location of the damage on the beam is chosen at random. A vertical downward transient load is applied to the upper right end of the cantilever beam, as shown in Fig. 4, and six acceleration sampling point (A1โ€“A6) are configured to obtain acceleration time history response data in the Y direction. The transient loadโ€™s first load step is set to 0.2s, the number of sub-steps to 20, the load size to 100N, the second load step is set to 2s for sampling, and the number of sub-steps to 5,000 (that is, the sampling frequency is 2,500Hz). Only one type of damage is considered in this paper, and 400 structural damage states are generated. The beam is divided into 12 areas labeled 0โ€“11 based on the location of the damage. As shown in Fig. 4, the damage at the intersection of the areas is classified based on its size. Each acceleration sampling point can obtain one data set, but 400 sets of data in a single data set are insufficient for deep learning. As a result, offset sampling (see Fig. 5) is used to improve the data, and a single set of data is divided into 21 segments. The benefits of this method are twofold: first, it expands the data sets, and second, it allows multiple sets of data to correspond to the same injury in order to prepare for subsequent voting. The data setsโ€™s input is the acceleration time history response after offset sampling, and the data setsโ€™s output is the category label. Each data set is divided into train set and validation set in the ratio of 8:2. The train set is used to train the model, and the validation set is used to adjust the model and test its generalization ability. Such a validation set, however, can only test the set damage. To put the modelโ€™s extrapolation ability to the test, 48 new damage states are chosen at random, and the data from 6 sensors is offset sampled to generate 6 sets of test set A, which are then added to 6 data sets. To discuss the voting method further below, a new damage is set in each category label, and test set B is generated and added to the data sets in the same manner. To verify the effectiveness of the method in this paper, the difficulty of identification needs to be increased, so all data sets are modified by adding โˆ’2dB of super-large white Gaussian noise. 3.2 Model Establishment and Test Set Classification Ideas All of the obtained data sets were standardized (mean 0, variance 1) before being fed into 1D DenseNet training to generate six submodels. During the training process, the parameters are constantly adjusted based on the loss of the training set and the accuracy of the validation set. The control variable method is used to adjust the parameters in this paper, and when adjusting one parameter, the other parameters are kept as constant as possible. The first step is to modify the network structure. The number of neurons in the final fully connected layer should not be too large, based on the experience of 1D CNN parameter adjustment. Based on this premise, the number of network layers and the structure of each layer are adjusted, yielding the network structure shown in Table 1. Then, one by one, adjust the following superparameters: (1) The training algorithm is Adam-optimized random gradient descent; (2) Set the batch size to 64; (3) Set the epoch to 50; (4) Set the initial learning rate to 0.01, then reduce it by 0.5 every 10 epoch; (5) Set 0.2 in each DenseLayer; and (6) Use L2 regularization with a parameter of 0.0005. In this paper, all of the data processing and neural network code is written in Python, and the neural network framework is Pytorch. Because the six data sets in this paper are very similar in sequence length and characteristics, the parameter adjustment results are essentially the same. The loss of training set, the accuracy of validation set, and the accuracy of test set A can be obtained after six models are trained using six data sets. The voting method is investigated in test set B. Each damage corresponds to 21 groups of acceleration time-history response segments after offset sampling. Each loss has 21 results as 21 votes after being introduced into a single submodel, and the one with the most votes is considered the prediction result of a single submodel. The prediction results of multiple sub-models are then voted on collectively to achieve more accurate 4.1 Accuracy Evaluation of Submodels Table 2 shows the evaluation results of data sets training submodels obtained by six sampling point. Under the influence of โˆ’2dB white Gaussian noise, the accuracy of validation set and test set A is not high. Voting can help to solve these issues. 4.2 Test Set B Voting Classification Fig. 6 shows the voting results of test set B corresponding to six sampling points (A1โ€“A6), with the vertical axis representing the true label and the horizontal axis representing the predicted label. Each injury receives 21 votes; the label with the most votes is chosen as the prediction result, and the diagonal of the confusion matrix indicates the correct result. As shown in Fig. 6, for the test set B of each data set, each sub-voting modelโ€™s has a positive effect, but each submodel still has the incorrect label, and some of them are far from the true label. Then, in Fig. 7, the ensemble voting results are shown. In Fig. 7, (A1โ€“A2) shows the ensemble voting results of the first two submodels, and its recognition effect is roughly equal to that of a single submodel, which is not particularly prominent; (A1โ€“A4) shows the ensemble voting results of the first four submodels, and its effect opinions are significantly improved, and it can be seen that even if the identification is incorrect, it is very close to the real tag; (A1โ€“A6) shows the ensemble voting results of all models, and the results are slightly better than those of (A1โ€“A4). It can be seen that the effect of ensemble voting is very significant, which can improve recognition accuracy, and the effect of this method improves as the number of submodels increases. CNNโ€™s operating mechanism is difficult to comprehend because it is a black box. This paper takes the data set and submodel corresponding to sampling point A6 as an example. The test set A is brought into the submodel for damage classification, and the dimension is reduced using the t-SNE method, and the features extracted by each DenseBlock and finally input into the fully connected network are visualized. The features are depicted in Fig. 8. Fig. 8 shows that, first and foremost, as the network layers are deepened, the categories become increasingly better. The classification in each DenseBlock layer is initially poor, but improves dramatically in the full connection layer. Second, when we look at the Prediction, we can see that there are some overlapping areas between two labels, which will affect the accuracy, but there are also non-overlapping areas, which provide a theoretical basis for the voting method. Third, the eigenvectors extracted by Convolution and DenseBlock (1) are very uniform when compared to the initial eigenvectors, and clustering begins only with DenseBlock (2), demonstrating that 1D DenseNet can decorrelate without any additional items. Regarding to the structural damage identification problem related to vibration acceleration, this paper proposes a new method. First introduce the idea of building a 1D DenseNet as an algorithm for training submodels; then use data offset to enhance multiple acceleration sampling pointsโ€™ acquired data features as a basis for submodel voting; finally introduce how multiple sampling pointsโ€™ submodels form an ensemble model for ensemble voting. An ABAQUS simulated cantilever beam fracture damage case was listed; seemingly simple cases use ultra-large Gaussian white noise to increase training difficulty to verify this methodโ€™s effectiveness. In the end discussed submodel process visualization. (1) In this paper, the offset sampling method is used to expand the data, which effectively solves the problem of insufficient samples in the data set. This offset sampled data can then be brought into the 1D DenseNet model for voting classification, and has achieved very good recognition effect. (2) A set of 1D DenseNet submodels is created, and the voting results of each submodel are combined. The recognition accuracy improves significantly as the number of submodels increases, and even if the recognition is incorrect, the prediction results are very close to the real results. (3) Investigate the internal mechanism of the submodel trained by 1D DenseNet by reducing the dimension and visualizing the feature map of each model, and conclude that the model can automatically decorrelate, and the classification effect will improve as the network layer is deepened. In the last layer of the model you can see classification results; after offset sampling samples will cluster according to their respective category labels; only a few will be identified with other labels; this can serve as a basis for submodel voting. The submodel used in this paper is established by 1D DenseNet, because the author considers DenseNet to be a relatively novel CNN that has been shown in some studies to be an algorithm suitable for damage identification. In fact, the submodel of this method can be any algorithm or a different algorithm, and I believe that different results will be obtained after experimenting. The cantilever beam simulated by ABAQUS is used in this paper, and the proposed method is validated by increasing the difficulty of identification with white Gaussian noise, without considering more damage situations and experimental verification, but the author believes that the ensemble method proposed in this paper can be applied to these situations. Acknowledgement: None. Funding Statement: The authors received no specific funding for this study. Author Contributions: The authors confirm contribution to the paper as follows: study conception and design: Chun Sha; data collection: Chun Sha, Wenchen Wang; analysis and interpretation of results: Chun Sha, Chaohui Yue and Wenchen Wang; draft manuscript preparation: Chun Sha, Chaohui Yue. All authors reviewed the results and approved the final version of the manuscript. Availability of Data and Materials: The data that support the findings of this study are available from the corresponding author, upon reasonable request. Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study. 1. Farrar, C. R., Worden, K. (2007). An introduction to structural health monitoring. Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences, 365(1851), 303โ€“315. [Google Scholar] 2. Fan, W., Qiao, P. 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Wireless and real-time structural damage detection: A novel decentralized method for wireless sensor networks. Journal of Sound and Vibration, 424, 158โ€“172. https://doi.org/10.1016/j.jsv.2018.03.008 [Google Scholar] [CrossRef] 29. Yang, Y., Lian, J., Zhou, G., Chen, Z. (2020). Damage identification of steel truss structure based on one-dimensional convolutional neural network. 20th National Symposium on Modern Structural Engineering, pp. 68โ€“71. Hebei, China. [Google Scholar] 30. Zhang, Y., Miyamori, Y., Mikami, S., Saito, T. (2019). Vibration-based structural state identification by a 1-dimensional convolutional neural network. Computer-Aided Civil and Infrastructure Engineering, 34(9), 822โ€“839. https://doi.org/10.1111/mice.12447 [Google Scholar] [CrossRef] Cite This Article APA Style Sha, C., Yue, C., Wang, W. (2023). Ensemble 1D densenet damage identification method based on vibration acceleration. Structural Durability & Health Monitoring, 17(5), 369-381. https://doi.org/ Vancouver Style Sha C, Yue C, Wang W. Ensemble 1D densenet damage identification method based on vibration acceleration. Structural Durability Health Monit . 2023;17(5):369-381 https://doi.org/10.32604/ IEEE Style C. Sha, C. Yue, and W. Wang, โ€œEnsemble 1D DenseNet Damage Identification Method Based on Vibration Acceleration,โ€ Structural Durability Health Monit. , vol. 17, no. 5, pp. 369-381, 2023. https:// This work is licensed under a Creative Commons Attribution 4.0 International License , which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Pi Day Celebration Ideas By HWA | Publish On: March 12, 2015 | Posted In: Pi Day, this year is going to be on March 14^th. It will be celebrated not only as Pi Day but also as Albert Einsteinโ€™s birth anniversary as well. It is a coincidence that Albert Einstein was born on Pi Day. For those of you who donโ€™t know what Pi Day is, let me tell you about Pi first. Pi, not as in American Pie is one of the basic mathematical constants. It is derived from the Greek letter ฯ€. And its value is 3.1415926. It is defined by the ratio of a circleโ€™s circumference C to its diameter d. ฯ€ = C/d In simple terms, a circleโ€™s circumference is slightly more than three times the length of its diameter. And that ratio is called ฯ€. ฯ€ in mathematics, has a very special place. The decimal points in this number seem to never end and the numbers never repeat. So, the number ฯ€ goes on infinitely. Pi was not discovered in the recent times, but it was discovered by ancient mathematics both in India as well as in China. The maximum decimal points that they could reach was till 7 and today computer scientists have computed ฯ€ to over 13 trillion digits. And it should be noted that ฯ€ is not just a discrete number used in complex mathematics. ฯ€ is a very practical number and it is extensively used in trigonometry and geometry where circles, ellipses or spheres are involved. Not only in trigonometry and geometry, ฯ€ is also used various sciences such as cosmology, number theory, statistics, fractals, thermodynamics, mechanics and electromagnetism as well. Pi Day and its significance Pi Day celebrations have been organized first by Larry Shaw in 1988 at San Francisco Exploratorium. In March of 2009, the US House of Representatives passed a resolution where 14 of March of observed as National Pi Day. Pi Day Celebration Ideas Pi Day is celebrated with various activities. Some of the activities have been mentioned below. Baking Pies: Pie baking and pie eating are one of the first and foremost activities that come to mind on Pi Day. The symbol ฯ€ inscribed on the pie or the number 3.141592โ€ฆ. can also be inscribed on the pie to make it look geeky. Donโ€™t just have one or two pies on Pi Day. Friends or coworkers can come together and bake different types of pies and have a big pie feast. Listen to Pi music: Pi has also been an inspiration for artists as well. Take for example Kate Bushโ€™s song Pi, Mathematical Pi and Lucy Kaplanskyโ€™s Song About Pi. Wear ฯ€ T-shirts: Wearing t-shirts with ฯ€ symbol on them, is another idea of celebrating Pi Day. If you to attend a Pi Day event, make sure that you have a ฯ€ t-shirt with you. Pie Throwing Competition: If youโ€™ve had enough of pie eating, perhaps, its time to throw some pie. It can be so much fun throwing pie on people. Pi Greeting Cards: Surprise your geeky friends by sending them a Pi Greeting card on Pi Day. A quick search on the internet will reveal a whole stack of Pi Greeting cards. Dressing up as Einstein: Donโ€™t forget that Pi Day is also the birth anniversary of Albert Einstein. So, you could dress up like Einstein and also add a Pi Day complement along. Pi Reciting Competition: Pi Reciting is about who can recite the maximum number of ฯ€ digits accurately. Prepare an exotic pie and make an announcement that whoever wins the competition will be awarded the pie. Take Pi Pictures: You and your friends can stand in ฯ€ form and take pictures from overhead. If the number of people are more, then you can go with ฯ€ digits. Simon Says Challenge: Simon Says game can be incorporated into Pi Day celebrations with people giving Simon Says challenge on the digits of ฯ€. The 1^st person will start with 3, the 2^nd person will start with 3.1. The 3^rd person will have to repeat all the digits and tell out the next digit accurately and so on. This game can go on till the last person who gets its wrong. These are some of the activities that you can do for this yearโ€™s Pi Day celebrations. For more details, visit www.helpwithassignment.com Book Your Assignment Recent Posts How To Prepare An Excellent Thesis Defense? Read More How to Restate A Thesis? โ€“ A Detailed Guide Read More Explanatory Thesis: Examples and Guide for Clear Writing Read More How To Write 3 Types Of Thesis Statements? Read More How to Effectively Prepare for Your Thesis Defense? Read More Get assignment help from subject matter experts! 4.7/5 rating | 10,000+ happy students | Great tutors 24/7
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AP Calculus BC 2013 Exam (videos, questions, solutions) Questions And Worked Solutions For AP Calculus BC 2013 AP Calculus BC 2013 Free Response Questions - Complete Paper (pdf) AP Calculus BC 2013 Free Response Questions - Scoring Guide (pdf) 1. On a certain workday, the rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant is modeled by G(t) = 90 + 45cos(t^2/18), where t is measured in hours and 0 โ‰ค t โ‰ค 8. At the beginning of the workday (t = 0) , the plant has 500 tons of unprocessed gravel. During the hours of operation, 0 โ‰ค t โ‰ค 8, the plant processes gravel at a constant rate of 100 tons per hour. (a) Find Gโ€ฒ(5). Using correct units, interpret your answer in the context of the problem. (b) Find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this workday. (c) Is the amount of unprocessed gravel at the plant increasing or decreasing at time t = 5 hours? Show the work that leads to your answer. (d) What is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday? Justify your answer. 2. The graphs of the polar curves r = 3 and r = 4 2sin q are shown in the figure above. The curves intersect when ฮธ = ฯ€/6 and ฮธ = 5ฯ€/6. (a) Let S be the shaded region that is inside the graph of r 3 and also inside the graph of r = 4 2sin ฮธ Find the area of S. (b) A particle moves along the polar curve r = 4 - 2sin ฮธ so that at time t seconds, ฮธ = t^2. Find the time t in the interval 1 โ‰คt โ‰ค 2 for which the x-coordinate of the particleโ€™s position is -1. (c) For the particle described in part (b), find the position vector in terms of t. Find the velocity vector at time t = 1.5. 3. Hot water is dripping through a coffeemaker, filling a large cup with coffee. The amount of coffee in the cup at time t, 0 โ‰ค t โ‰ค 6 is given by a differentiable function C, where t is measured in minutes. Selected values of C(t), measured in ounces, are given in the table above. (a) Use the data in the table to approximate Cโ€ฒ(3.5). Show the computations that lead to your answer, and indicate units of measure. (b) Is there a time t, 2 โ‰ค t โ‰ค 4 at which C'(t) = 2 ? Justify your answer. (c) Use a midpoint sum with three subintervals of equal length indicated by the data in the table to approximate the value of .Using correct units, explain the meaning of in the context of the (d) The amount of coffee in the cup, in ounces, is modeled by B9t) = 16 - 16e^-0.4t. Using this model, find the rate at which the amount of coffee in the cup is changing when t = 5. 4. The figure above shows the graph of fโ€ฒ, the derivative of a twice-differentiable function f, on the closed interval 0 โ‰ค x โ‰ค 8. The graph of fโ€ฒ has horizontal tangent lines at x = 1, x = 3, and x = 5. The areas of the regions between the graph of fโ€ฒ and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies f(8) = 4. (a) Find all values of x on the open interval 0 < x < 8 for which the function f has a local minimum. Justify your answer. (b) Determine the absolute minimum value of f on the closed interval 0 โ‰ค x โ‰ค 8. Justify your answer. (c) On what open intervals contained in 0 < x < 8 is the graph of f both concave down and increasing? Explain your reasoning. (d) The function g is defined by g(x) = (f(x))^3. If f(3) = -5/2, find the slope of the line tangent to the graph of g at x = 3. 5. Consider the differential equation 6. A function f has derivatives of all orders at x 0. Let P[n](x) denote the nth-degree Taylor polynomial for f about x 0. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
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Gigawatt to Kilowatt Conversion - Convert Gigawatt to Kilowatt (GW to kW) Gigawatt : The gigawatt is a unit of power which is a multiple of the unit watt. It is equal to one billion watts, or 106 kilowatt. The unit symbol for gigawatt is GW; it is often used for large power plants or power grids. Kilowatt : The kilowatt is a unit of power which is a multiple of the unit watt. It equals to one thousand watts. The unit kilowatt is commonly used to express the electromagnetic power output of broadcast radio and television transmitters. One kilowatt of power approximately equals 1.34 horsepower. The unit symbol for kilowatt is kW. Power Conversion Calculator FAQ about Gigawatt to Kilowatt Conversion 1 gigawatt (GW) is equal to 1000000 kilowatt (kW). 1GW = 1000000kW The power P in kilowatt (kW) is equal to the power P in gigawatt (GW) times 1000000, that conversion formula: P(kW) = P(GW) ร— 1000000 One Gigawatt is equal to 1000000 Kilowatt: 1GW = 1GW ร— 1000000 = 1000000kW 1000 Kilowatt is equal to 0.001 Gigawatt: 1000kW = 1000kW / 1000000 = 0.001GW P(kW) = 5(GW) ร— 1000000 = 5000000kW Most popular convertion pairs of power Lastest Convert Queries 0 comments
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Create a Fading Popup Modal with jQuery โ€“ Inspirational Pixels This tutorial is aimed at getting beginners familiar more advanced concepts like data attributes and, at times, there will be more โ€œefficientโ€ ways of doing things, like creating advanced functions and optimising the output. However, as this tutorial is aimed at helping beginners, Iโ€™ll be taking a more procedural approach than some people might take. There is no right or wrong way here, there are simply more advanced and less advanced methods of getting the desired outcome. Table of Contents 1. Skip to the HTML 2. Skip to the CSS 3. Skip to the jQuery If you get stuck at all please leave me a comment below with a description of the problem and a link to a JSFiddle showing your code in action. Iโ€™m here to help, no matter what the problem. You might also like: How to Create a 12 Column Grid System with Sass 1.) Step 1: HTML Pro-Tip: Place it in the Footer If youโ€™re running a PHP powered site, like a WordPress blog, then itโ€™s a good idea to put the code for the popup itself in the footer include file. This way you can always be sure the popup will be included on every page. Final HTML <a class="btn" data-popup-open="popup-1" href="#">Open Popup #1</a> <div class="popup" data-popup="popup-1"> <div class="popup-inner"> <h2>Wow! This is Awesome! (Popup #1)</h2> <p>Donec in volutpat nisi. In quam lectus, aliquet rhoncus cursus a, congue et arcu. Vestibulum tincidunt neque id nisi pulvinar aliquam. Nulla luctus luctus ipsum at ultricies. Nullam nec velit dui. Nullam sem eros, pulvinar sed pellentesque ac, feugiat et turpis. Donec gravida ipsum cursus massa malesuada tincidunt. Nullam finibus nunc mauris, quis semper neque ultrices in. Ut ac risus eget eros imperdiet posuere nec eu lectus.</p> <p><a data-popup-close="popup-1" href="#">Close</a></p> <a class="popup-close" data-popup-close="popup-1" href="#">x</a> The HTML Explained The structure of the popup is quite straightforward, so nothing to worry about there. We start off with a wrapper div and give it a class of popup. This is the black background around the popup but also what weโ€™ll animate with jQuery. You may notice those data attributes Iโ€™m using, for example data-popup="popup-1". The reason for using these is to make the code easier to use and maintain in the long run. Some people are advocates of using IDs in the same approach, however, this can become confusing as IDs are also commonly used for styling. <div class="popup" data-popup="popup-1"> The open button also uses a data attribute of data-popup-open="popup-1", which you may notice is the same value as the popup itself. This is handy because it will allow us to dynamically open the popup with jQuery. <a class="btn" data-popup-open="popup-1" href="#">Open Popup #1</a> To close the popup thereโ€™s a normal anchor link and a specially positioned anchor link with data-popup-close="popup-1" applied to them. The good thing about using a data attribute on both links is that it allows us to have a special close button in the upper right of the popup, but also use a normal link inside of the popup to close it as well. <p data-popup-close="popup-3"><a href="#">Close</a></p> <a class="popup-close" data-popup-close="popup-3" href="#">x</a> 2.) Step 2: CSS Final CSS /* Outer */ .popup { /* Inner */ .popup-inner { -webkit-transform:translate(-50%, -50%); transform:translate(-50%, -50%); box-shadow:0px 2px 6px rgba(0,0,0,1); /* Close Button */ .popup-close { transition:ease 0.25s all; -webkit-transform:translate(50%, -50%); transform:translate(50%, -50%); font-family:Arial, Sans-Serif; .popup-close:hover { -webkit-transform:translate(50%, -50%) rotate(180deg); transform:translate(50%, -50%) rotate(180deg); The CSS Explained The popup wrapper is using RGBA colour values to enable transparency, something a basic #HEX code canโ€™t do. It also has fixed positioning, which makes it stay centred if the page is scrolled. /* Outer */ .popup { The inner content area of the popup is absolutely positioned but also repositioned with transform. The absolute positioning moves the inner popup content left 50% and down 50%. The transform then moves it left -50% of its own width and down -50% of its own height. This then enables the inner popup to be exactly centred no matter what the size of the screen. /* Inner */ .popup-inner { -webkit-transform:translate(-50%, -50%); transform:translate(-50%, -50%); The close button actually has a lot more styling rules than youโ€™d expect. The reason it has a fon-family and font-size added is to make the popup work better across different projects. I highly advise using an icon font for the close button. For a good example, click the newsletter subscribe button at the top of the sidebar on this site. Thereโ€™s also a transition on the button, this comes into play when you hover. /* Close Button */ .popup-close { transition:ease 0.25s all; -webkit-transform:translate(-50%, -50%); transform:translate(50%, -50%); On hover, the button is spun around 180 degrees. If you put 360 degrees it will actually not look any different so 180 is whatโ€™s needed for the desired spinning effect. .popup-close:hover { -webkit-transform:translate(50%, -50%) rotate(180deg); transform:translate(50%, -50%) rotate(180deg); Thatโ€™s it for the CSS. If youโ€™ve got any question on what we covered so far, please leave me a comment below. 3.) Step 3: jQuery Final jQuery $(function() { //----- OPEN $('[data-popup-open]').on('click', function(e) { var targeted_popup_class = jQuery(this).attr('data-popup-open'); $('[data-popup="' + targeted_popup_class + '"]').fadeIn(350); //----- CLOSE $('[data-popup-close]').on('click', function(e) { var targeted_popup_class = jQuery(this).attr('data-popup-close'); $('[data-popup="' + targeted_popup_class + '"]').fadeOut(350); The jQuery Explained Do you remember those data attributes from the HTML? This is where they come into play. We first set up a click event on any of the open buttons and stop the default behaviour, which in this case, would be to follow the link. If we donโ€™t use e.preventDefault(); to stop the default behaviour, a # will be added to the URL in the address bar. $('[data-popup-open]').on('click', function(e) { The next step is to grab the value from the attribute and save that information to a variable. The value saved will be whatever has been added to that buttons attribute data-popup-open. In our case, the output will be popup-1. var targeted_popup_class = jQuery(this).attr('data-popup-open'); Now comes the fun part, fading in the popup. The code below might look pretty scary but really, itโ€™s quite simple. All it says is โ€œfind the popup with a data-popup value of popup-1 and fade it in for 350 millisecondsโ€œ. Also, because weโ€™re using a variable, it saves us from having to repeat the open functionality for each popup we create, as itโ€™s dynamically found and used. $('[data-popup="' + targeted_popup_class + '"]').fadeIn(350); Closing the Popup Closing the popup is actually just a reverse of opening it. Instead of looking for the value of the open button weโ€™re looking for the value of the close button. Weโ€™re also fading out instead of fading in. In Conclusion Custom popup modals can be incredibly powerful, but also far more annoying if used incorrectly. When implementing your popup modal please donโ€™t have it to autoload. This is a big problem as it will cause people to never return, even if you remove the popup because there only thought of your site and brand is an annoying popup the second they open up some of your content. Help and support: If you have questions and/or need help with any code, please leave a comment below with a link to a JSFiddle showing your code in action. You might also like: Tooltips in HTML and CSS. Excelent! This was a very useful example for me. Thanks. For those of you wanting to close it when clicking outside, put this third part into your function: $(function() { //โ€”โ€“ OPEN $(โ€˜[data-popup-open]โ€™).on(โ€˜clickโ€™, function(e) { var targeted_popup_class = jQuery(this).attr(โ€˜data-popup-openโ€™); $(โ€˜[data-popup="โ€™ + targeted_popup_class + โ€˜"]โ€™).fadeIn(350); //โ€”โ€“ CLOSE $(โ€˜[data-popup-close]โ€™).on(โ€˜clickโ€™, function(e) { var targeted_popup_class = jQuery(this).attr(โ€˜data-popup-closeโ€™); $(โ€˜[data-popup="โ€™ + targeted_popup_class + โ€˜"]โ€™).fadeOut(350); Where did you get the โ€œactiveโ€ class from as itโ€™s not in the css? thanks its working properly ๐Ÿ™‚ code is not active, please check again the popup is not popping up as it suppose toโ€ฆ. I think the code might have an errorโ€ฆ could you assist me?? How can I adapt this to where when you click outside of the modal (the transparent black background), it closes the modal just like the X button does? How can I adapt this so that when the transparent black background div is clicked, it closes out the modal like the x button does? Hi, How can i trigger this popup with wordpress menu link Great code, works perfectly on some browsers. However, the modal popups donโ€™t work on firefox? Is there additional code to work around this? Thank you, No asking questions for you. Just use the code as is. I really like your popup-modal but I have to ask, How can I make the modal close by clicking outside it? Is there a way to make this open upon page load? Thanks! Kerrie I saw after I posted my question you donโ€™t recommend the modal show up on page load. This is how weโ€™ve implemented it in the past, but weโ€™re switching from Coldfusion to PHP. Sorry for asking the question before I saw your comment about it. how to show a modal window or pop-up window after submitting contact us form? Wonderful code! Clean and easy to use! Thank you for sharing ๐Ÿ™‚ Neat and clean tutorial and explanation. Great post a good basis for developing something further. Wow! This was actually pretty easy. Thank you for taking the time to explain. How to access angular js scope values in this modal? this is a great tutorial! i have a video playing in the popup, when clicking on the X to close,, i need to pause the video and then close. how can that be done? thank you! how can make it open on window loadingโ€ฆ also show when clicked on pop up open link Very useful tutorial. I suggest the followings: โ€“ Add data-popup-close attribute to the parent div: (inner div, and content, etc.) โ€“ And in the JavaScript close event handler, add if (e.target !== this) return; This will allow clicking anywhere on the background (except the dialog itself) to close the dialog. This tutorial is great, i have been your follower Seb and I see you have written great guides. I was wondering will it be possible to add some translation in this model with css only like we see in jquery based models where they appear from right or fade away to right. This is just an example. Hi there โ€” I cannot seem to get the code to work, if localising the code: Can you help? It will work if you add this line to your HTML code: Jquery is a pre-req for this code. how to close the pop up on clicking outside?? Excellent tutorial, thank you so much! I recommend adding z-index: 999; to .popup. Thatโ€™s the only change I had to make. Thanks for this tutorial it is very helpful. But there is a problem with it when it comes to mobile devices the content is chopped of on the top and bottom and there is no way to close it. I think it is not responsive. i wanto load the popup when the page loading itself instead of clicking the button.thanks Hi, I followed your instructions exactly, and they all worked (thanks a lot!) except strangely for the very last element. So Iโ€™m generating a bunch of buttons dynamically in jQuery, then attach the required classes and data attributes as you mentioned, along with your event handlers. All of them work and produce the pop-up, except the very last one which always seems to ignore the e.preventDefault(), and instead just adds the โ€˜#โ€™ link in the URL, and doesnโ€™t bring up the pop-up. Do you know why thatโ€™s the case? where shuld i put the code? do i need the all 3 code or can i use only html? This tutorial helped me a lot so thanks. Can you tell me how youโ€™d modify the code to also close the modal window on pressing the esc key or clicking outside the modal window please. Here is the video tutorial for custom popup https://www.youtube.com/watch?v=zK4nXa84Km4 This is exactly what I was looking for! Thanks! The Create a Fading Popup Modal with jQuery, was a good example to apply with buttons specially when to explain about the concept. I tried to fellow you instruction of the codes, but fail to make the pop up work. Do you have the complete codes to test it and play with it. Many thanks for your help. Can i open this popup with form submit button i tried to put the data-popup-open=โ€popup-1โ€ณ as submit button attribute but no success Hi Seb I implemented your code exactly in term CSS and JQuery and HTML I put my content but use the data class in the tutorial well no error render by online.js the JS editor I use but nothing showing up not even the link zero blank page. Do I need to use a google library Great tutorial! I would like to incorporate in a website I manager for a nonprofit. I have the board of directors page with their photos and I would like a popup to be shown with the individualโ€™s bio when their image is click. I tested it in liveweave.com and it works great but can someone explain how to use it in a wordpress site? The link to Fiddle is not working, it seems to be pointing to no snippet but landing page of JSFiddle. The popup is showing under other elements. I have tried changing the z index and also making it relative. Hey, thanks for tutorial โ€“ itโ€™s great and simple. Could you please add a line of code, so I can click anywhere on black in order to close popup? it is kind of a usual UX thing, I guess. I am wondering if you know if this will work with Squarespace sites given their structure. I cannot believe I cannot find a player that does not go full screen (using the grid layout in SS). It dominates the site with the video and at the end does not reduce or go away so the user must โ€œxโ€ out to see the site itself. Any help for a non css/JQuery guy. Highest Regards. Awesomeโ€ฆ nicely explained and I found it simple too. Thanksโ€ฆ bookmarked (y) Great tutorial, Thanks you! โ˜บ Really awesome tutorial. I canโ€™t wrap my head around how the data attributes work. If one were to use id instead, how would one do, just in comparison? I would really like to get the full understanding of this great example. I found this article accidentally and found it very useful. Just wanted to add to this article that the best practice would also be to add z-index:999 to a class .popup as than the bg will cover all other elements and popup will be in focus. I like the popup modal. I was wondering if this will work outside of Word Press? I am creating just a standard website with a folder call JS and CSS for the style sheet. The style sheet is Styles.css. It appears that the html, css, jquery is to be put into one page, I couldnโ€™t get it to work. Any ideas on what to do to create a .js, do I need to name it something specific? on the html, do I just create a simple html with no head or title or body? Do I save it out to popup-1.html or is it supposed to go into the already written html? Can I put the CSS you have in my existing Styles.css? I would like to get it working but I am not experienced enough yet to know how to structure it into my site fully. thanks for any help! Hi bro ! , thanks for the post. It works great.The only thing is that the video keeps playing after you close the popup. Ineed that the video stops , when i close the pop up.Is there any way to do that ? Thanks ! Hi! Great tutorial, thanks! Iโ€™m using Squarespace (Pacific theme) and the problem I am having with this is that the modal shows up, however the rest of the content on the page overlaps the popup โ€“ in other words, if you think of the page as layers, where the modal should be the top layer, itโ€™s actually kind of showing up as a layer UNDER the rest of the page content. Hope that makes sense. Any suggestions? Me again! Just put the html in the footer (as you suggested!) and itโ€™s perfect! This is so great! Thanks so much! Hi Joe Iโ€™m using Squarespace too, but canโ€™t seem to get this to work. Where do I place the codes. CSS, I know. But JQuery and HTML, no idea. I donโ€™t understand โ€œput the HTML code in the Footerโ€ when I want the popup to show when a text in the body is clicked, not a text in the footer. Please help. Thank you very much! i am the very beginner of web site designing so help me by answering my simple ques. tell me whether the above programs are seperated program or a same one single program Hello ! Very good tutorial but It doesnโ€™t work with my wordpress website โ€ฆ I have enqueue the jquery function in my functions file and added your css file in mine but it still not work โ€ฆ Have you an idea why ? My function code : function popup(){ wp_register_script("popup", get_template_directory_uri()."/js/jquery.popup.js", array("jquery", "jquery-ui"), true); I have found de solution ! I have not enqueued the jQuery files in the google library and not created a separated css file for the popup Very good tutorial again !!! can i use it like a wordpress plugin? is there a zip file to download? There isnโ€™t currently a download for the code above, but Iโ€™ll see what I can do about adding one. as I can download โ€œCreate a Modal Popup Loading with jQueryโ€ ?? Sorry but I donโ€™t understand the question. Can you phrase it another way? ๐Ÿ™‚ I never comment on things, but in the modern day web, you would most definitely not use jQuery alone with animations. NO! We have a non CPU hogging animation system. Itโ€™s called CSS3! Add a class and do your magic! I agree, CSS3 animations are the way forward. However, as I explain in the beginning of the tutorial, itโ€™s aimed at beginners and I do state there are better ways to go about this. Iโ€™ll make it more clear in future tutorials.
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Equations Review Worksheet - Equations Worksheets Equations Review Worksheet If you are looking for Equations Review Worksheet youโ€™ve come to the right place. We have 16 worksheets about Equations Review Worksheet including images, pictures, photos, wallpapers, and more. In these page, we also have variety of worksheets available. Such as png, jpg, animated gifs, pic art, logo, black and white, transparent, etc. Not only Equations Review Worksheet, you could also find another pics such as Math Equations, Functions Worksheet, Algebra Worksheets, Simple Equation, Equation Solver, Inequalities Worksheet, and Algebraic Equations. 736 x 952 ยท jpeg eighth grade equations worksheet printable grade math worksheets from www.pinterest.com 736 x 1013 ยท jpeg memorableacademic types reactions worksheet answer key typesof from www.pinterest.com 898 x 654 ยท jpeg solving quadratic equations practice worksheet answers practice worksheets from practice.noveltodays.com Donโ€™t forget to bookmark Equations Review Worksheet using Ctrl + D (PC) or Command + D (macos). If you are using mobile phone, you could also use menu drawer from browser. Whether itโ€™s Windows, Mac, iOs or Android, you will be able to download the worksheets using download button. Leave a Comment
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Reward Distribution | DEADPXLZ Let's look at a clear fund distribution example for forging. Let's assume that User A is forging 3 different PXLZ in the PXL Forge: DEAD Base with one Gold Chain and one Headphones attributes. DEAD Base with one Gold Chain, one Turtleneck and one Golden Earrings attributes. WEREWOLF Base with one Gold Chain, one Cigarette attributes. PXL#1 (2 attributes): 20 ADA (base) + 10 ADA (Gold Chain) + 24.03 ADA (Headphones) = 54.03 ADA PXL#2 (3 attributes): 20 ADA (base) + 10 ADA (Gold Chain) + 13.27 ADA (Turtleneck) + 14.70 (Golden Earrings) = 57.97 ADA PXL#3 (2 attributes): 70 ADA (base) + 10 ADA (Gold Chain) + 16.90 ADA (Cigarette) = 96.90 ADA Total Cost = 208.90 ADA Reward Distribution for Genesis Holders Genesis rewards pools hold all of the corresponding reward funds stemming from forging. These funds are accessible to Genesis PXL holders based on their Genesis PXL holdings. Each attribute and base has its own genesis reward pool. The general calculation formula looks like this: Where: GRa - Genesis Rewards for item a NoFA - Total Number of Forged Attributes on a PXL Pa - Price of forging item a The ADA amount coming into each pool is dictated by the number of attributes which is being forged on a PXL at a given time. If a PXL is forged with only 1 attribute, regardless of the base chosen, 90% of the total cost of that PXL is sent to the corresponding genesis reward pool. If the number of attributes being forged on a PXL is 5 however, only 50% of the total value of that PXL is shipped to the corresponding genesis rewards pool. A total of 161.31 ADA will be distributed towards various genesis holders. Let's assume that Genesis Holder X owns a Gold Chain DEAD-base Genesis PXL. Let's also assume that Genesis Holder X is eligible for redeeming their ADA rewards on this Genesis PXL. That holder can then redeem the following amount: 30 / 2 + 23 / 3 = 22.66 ADA General calculation formula: Where: TRz - Total Reward for a given attribute z n - Total number of times that attribute z was forged x - Amount added to Genesis Reward Pool after forging. Funds Distribution for PXL Wars Treasury Assuming the same example as above, the remaining funds out of the total 192 ADA which were spent on forging are to be distributed to the PXL Wars ADA Treasury: 208.90 ADA - 161.31 ADA = 47.59 ADA The formula for this calculation is quite simple: PWT - PXL Wars Treasury ADA funds TF - Total ADA spent on forging TG - Total ADA sent to the Genesis Holder Pools This fund is exclusively meant to cover the ADA rewards for winners of PvP PXL Wars battles. Assuming that the user who forged the above attributes and bases faces them off against another player and loses, the amount that his opponent will be winning is 10% of the total value he spent on forging them which in this case is 20.89 ADA. The PXL Wars Treasury is designed to support paying a winner of any given PvP game, completely independent from the total number of players.
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, usually denoted by , is a measurement of plane angle, representing of a full rotation. It is not an SI unit, as the SI unit for angles is radian, but it is mentioned in the SI brochure as an accepted unit. Because a full rotation equals 2ฯ€ radians, one degree is equivalent to ฯ€/180 radians. The above text is a snippet from Wikipedia: Degree (angle) and as such is available under the Creative Commons Attribution/Share-Alike License. 1. A step on a set of stairs; the rung of a ladder. 2. An individual step, or stage, in any process or scale of values. 3. A stage of rank or privilege; social standing. 4. A โ€˜stepโ€™ in genealogical descent. 5. One's relative state or experience; way, manner. 6. The amount that an entity possesses a certain property; relative intensity, extent. 7. A stage of proficiency or qualification in a course of study, now especially an award bestowed by a university or, in some countries, a college, as a certification of academic achievement. (In the United States, can include secondary schools.) 8. A unit of measurement of angle equal to 1/360 of a circle's circumference. 9. A unit of measurement of temperature on any of several scales, such as Celsius or Fahrenheit. 10. The sum of the exponents of a term; the order of a polynomial. 11. The number of edges that a vertex takes part in; a valency. 12. The curvature of a circular arc, expressed as the angle subtended by a fixed length of arc or chord. The above text is a snippet from Wiktionary: degree and as such is available under the Creative Commons Attribution/Share-Alike License. Need help with a clue? Try your search in the crossword dictionary!
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Finding Points of Inflection: Mathematical Analysis Explained Imagine a scenario where you are analyzing the curve of a profit function to determine the most profitable production quantity. As you calculate the second derivative and look for points of inflection, you realize the critical role these points play in understanding the behavior of the function. But how exactly do these inflection points impact decision-making processes and the overall shape of a graph? Stay tuned to explore the intricacies of finding points of inflection and their significance in mathematical analysis. Definition of Points of Inflection When analyzing functions, points of inflection are identified as locations where the concavity of the graph changes. These points mark transitions in the shape of the curve, indicating shifts from being concave upwards to concave downwards, or vice versa. At these inflection points, the curvature of the graph becomes flattened before changing direction. To visualize this concept, imagine a rollercoaster ride. As the rollercoaster moves along the track, there are points where it smoothly transitions from going up to going down, or from turning left to turning right. These transition points are like the points of inflection in a mathematical function โ€“ they signal a change in the behavior of the curve. Identifying points of inflection is crucial in understanding the behavior of a function. By recognizing where the concavity changes, you can gain insights into the overall shape and trends of the graph. Keep in mind that not all points where the curvature changes are inflection points; additional criteria need to be considered for accurate identification. Criteria for Identifying Inflection Points To identify inflection points in a function, observe where the concavity changes along the graph. An inflection point occurs where the concavity changes from being either concave up to concave down or vice versa. Mathematically, this change in concavity happens at points where the second derivative of the function equals zero or is undefined. When analyzing a function, look for points where the second derivative changes signs. At these points, the concavity transitions, indicating a potential inflection point. Additionally, remember that not all points where the second derivative is zero are inflection points; they could be points of local maxima or minima instead. Therefore, itโ€™s crucial to check the concavity on both sides of these points to confirm if theyโ€™re inflection points. Techniques for Locating Inflection Points When seeking to locate inflection points in a function, a common technique involves analyzing the sign changes of the second derivative. Inflection points occur where the concavity of a function changes, transitioning between being concave up and concave down. By examining the second derivative, you can determine where these changes in concavity happen. To locate inflection points, follow these steps: first, find the second derivative of the function. Next, set the second derivative equal to zero to identify possible points of inflection. Then, determine the sign of the second derivative on either side of these points. If the sign changes from positive to negative or vice versa, you have found an inflection point. Remember that not every point where the second derivative is zero corresponds to an inflection point. Additional analysis is necessary to confirm the nature of these critical points. By employing these techniques, you can effectively pinpoint the inflection points of a function with precision. Analyzing Concavity Changes in Functions Concavity changes in functions can be understood by examining the behavior of the second derivative. When the second derivative is positive, the function is concave up, forming a โ€˜smileโ€™ shape. This indicates that the function is bending upwards. Conversely, if the second derivative is negative, the function is concave down, creating a โ€˜frownโ€™ shape, and the function is bending downwards. To analyze concavity changes, look for points where the second derivative changes sign. These points are potential inflection points where the concavity transitions from up to down or vice versa. Itโ€™s important to note that the second derivative being zero at a point doesnโ€™t guarantee an inflection point; further analysis is needed. By identifying these changes in concavity, you can gain insights into the behavior of the function and its curvature. Understanding concavity changes helps in sketching accurate graphs and interpreting the behavior of functions. Role of Inflection Points in Graphs Inflection points play a crucial role in shaping the overall behavior of graphs. These points mark where the concavity of a function changes, indicating shifts from convex to concave or vice versa. Visually, inflection points appear as points where the curve transitions from bending one way to bending the other. Theyโ€™re pivotal in understanding the curvature of a graph and can provide valuable insights into the functionโ€™s behavior. When analyzing a graph, identifying inflection points helps in determining the overall shape of the curve. They serve as indicators of where the function transitions from being concave up to concave down or vice versa. By pinpointing these critical points, you can better comprehend how the function is behaving in different regions. Understanding the role of inflection points in graphs allows you to interpret the changing rates of increase or decrease more effectively, aiding in making informed decisions based on the functionโ€™s behavior. Practical Applications of Inflection Points Discover how inflection points play a crucial role in real-world scenarios, shaping decisions and outcomes based on the behavior of functions. In fields like economics, inflection points are vital for determining trends in markets. By identifying where a curve transitions from concave to convex or vice versa, businesses can adjust strategies to capitalize on changing conditions. For instance, recognizing an inflection point in sales data could prompt a company to modify pricing or marketing techniques to boost revenue. In engineering, inflection points help predict structural stability. Understanding where a beamโ€™s curvature changes can prevent collapses and ensure the safety of buildings and bridges. Moreover, in biology, inflection points assist in modeling population growth and disease spread accurately. By pinpointing where growth rates shift, scientists can develop more effective intervention Importance of Inflection Points in Calculus When dealing with inflection points in calculus, youโ€™ll notice their crucial role in understanding the behavior of functions. By analyzing sign changes at inflection points, you gain valuable insights into the concavity of a curve. These points offer a graphical interpretation that aids in visualizing the changes occurring within a function. Calculus and Inflection Exploring the significance of inflection points in calculus sheds light on the behavior of functions at critical junctures. These points mark where the curvature of a function changes, indicating shifts from concave to convex or vice versa. Understanding inflection points is crucial in calculus as they can help identify changes in the rate of growth or decline of a function. By analyzing inflection points, you can determine where a function transitions from being concave up to concave down, providing insights into the overall shape and behavior of the function. Calculus relies on inflection points to pinpoint where functions exhibit changes in their concavity, making them a fundamental aspect of mathematical analysis. Sign Change Analysis Understanding the significance of inflection points in calculus is essential for analyzing changes in the behavior and curvature of functions. When you perform a sign change analysis around an inflection point, you gain insights into how the functionโ€™s concavity transitions. By examining the intervals where the functionโ€™s second derivative changes sign, you can pinpoint where the inflection points lie. These points mark where the curvature of the function shifts from convex to concave, or vice versa. Identifying these critical points allows you to understand the functionโ€™s overall shape better and predict its behavior more accurately. Sign change analysis provides a straightforward method to locate inflection points and grasp the underlying changes in a functionโ€™s concavity. Graphical Interpretation Insights To gain deeper insights into the behavior of functions, focus on the graphical interpretation of inflection points in calculus. Inflection points mark where the curvature of a function changes, transitioning between concave up and concave down. By analyzing inflection points graphically, you can understand how the functionโ€™s rate of change is evolving. These points are critical as they can signify shifts in trends or behaviors within the Visually, inflection points appear as points where the functionโ€™s curve changes its shape. This visual representation can offer intuitive understanding, aiding in predictions about the functionโ€™s Identifying and interpreting inflection points graphically allows you to grasp the functionโ€™s overall trend and anticipate significant changes in its behavior. Frequently Asked Questions Can Points of Inflection Occur in Functions With Discontinuities or Sharp Corners? Yes, points of inflection can occur in functions with discontinuities or sharp corners. These points mark where the concavity changes, regardless of abrupt changes in the function. They are crucial for analyzing curves. How Do Inflection Points Relate to the Overall Shape of a Graph? When looking at a graph, inflection points indicate where the curve changes concavity. Upward-facing curves shift from concave down to concave up at these points, while downward-facing curves transition from concave up to concave down. Are Inflection Points Always Associated With Changes in Concavity? Yes, inflection points are always associated with changes in concavity. They mark where a curve transitions from being concave up to concave down or vice versa. Identifying these points helps understand the graphโ€™s behavior. You now understand how to find points of inflection in mathematical analysis. By identifying these critical points where concavity changes, you can gain insights into the behavior of functions and Remember to use the criteria and techniques discussed to locate inflection points accurately. Keep practicing to sharpen your calculus skills and apply this knowledge to solve real-world problems Good job on mastering the concept of inflection points!
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Could someone kindly proof check my solution for a task in problem 7? + General Questions (11) There is one bug about your signal trace. something^ฯ‰ means that something holds infinitely often. So you can't something^ฯ‰ somethingelse^ฯ‰ because thensomethingelse would happen after the end of infinity which is quite late (and impossible). What you'd rather do is describing all paths that satisfy G(p โˆง ยฌq) and then identifying all states from where such a path can be started (i.e. the states satisfying EG(p โˆง ยฌq): As (p โˆง ยฌq) only holds in your middle state, the path consisting of this state infinitely often is the only path satisfying G(p โˆง ยฌq). That makes this state the only state satisfying EG(p โˆง ยฌq). The initial state does not satisfy EG(p โˆง ยฌq). So your structure satisfies neither of the formulas. Here are two hints: 1. Is S2 satisfiable? 2. What happens if you remove your initial state and make the current middle state the new initial state? You can rewrite S1 to the CTL formula E F q & E G (p & !q) so that you can use our CTL model checker with the following input to check both CTL formulas: transition system: vars p,q; init 0; labels 0:; 1:p; 2:q; transitions 0->1; 1->2; 1->1; 2->2; (E F q) & E G(p&!q); E F(q & E G(p&!q)) You will find the following: โ€ข ใ€–pใ€—= {s1} โ€ข ใ€–qใ€—= {s2} โ€ข ใ€–p & !qใ€— = {s1} โ€ข ใ€–E G (p & !q)ใ€— = {s1} โ€ข ใ€–E F qใ€— = {s0;s1;s2} โ€ข ใ€–q & E G (p & !q)ใ€— = {} โ€ข ใ€–E F q & E G (p & !q)ใ€—= {s1} โ€ข ใ€–E F (q & E G (p & !q))ใ€—= {} The two formulas are state formulas, and therefore it is required to say in which state the one, but not the other one holds. Traces are used to distinguish path formulas. Your last paragraph was an important remark which I was missing so thank you very much for pointing that out Prof. Schneider. Now if we ignore the trace part, after all, since according to model checker, the formula S1 is satisfied in state s1 and the formula S2 is not satisfied on any state, can we say the structure I drew is sufficient to show these 2 formulas are not equivalent or still
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Daily 12 12 Giant Sudoku For Friday 20th September 2019 Hard - Lyana Printable Sudoku Sudoku 12ร—12 Printable Daily 12 12 Giant Sudoku For Friday 20th September 2019 Hard โ€“ There are several number of ways that to use Printable Sudoku puzzles. One of the best ways is to personalize them using QR codes. You can include a QR code to any 9ร—9 sudoku for example, to make it unique. These puzzles are fun and enjoyable to play and are great gifts. To learn more about different Sudoku puzzles, take a look at these articles: How to Solve a Medium Level Sudoku Puzzles? Beginners and intermediates can gain from learning how to solve intermediate sudoku puzzles. While they require higher effort and effort category of sudoku puzzles is still based on the same fundamental rules. Fill in the grids using numbers from one to nine . You can also make notes to assist you in the difficult areas. However, intermediate sudoku puzzles are more challenging and require some tactics. Here are a few suggestions for solving these puzzles. In the beginning, record all possible answers. This will allow you to recognize patterns and eliminate incorrect solutions. You can also try ranking possible solutions, which is a technique employed by some Sudoku enthusiasts. You can view the cells from a single direction, this means that you can observe their left-to-right orientation, or you could look at them from an opposite direction for example, up or down. You should try to list all the possible options you can, until youโ€™ve exhausted all squares. How to Play Sudoku for Absolute Beginners? If youโ€™re new to this game, youโ€™re likely wondering how it is to master sudoku. This guide will be useful, since it will teach you the basics you require to play successfully. For the absolute beginner Youโ€™ll learn about different types of sudoku as well as how you can play, and the basic rules. For more experienced players It will provide more strategies to solve the puzzles. To play Sudoku, youโ€™ll have to make use of logic and reasoning to solve the puzzles. This can be done by checking your grid to find missing numbers and then filling them in. In the image below the missing number is in the middle column. The nine is missing only one digit, so itโ€™s easy to figure out the rest of the cells with this method of scanning. If youโ€™re stuck, you should draw in every possible candidate and include the nine. Sudoku 12ร—12 Printable How to Solve Sudoku Quickly? To complete the Sudoku puzzle in a short time it is necessary to understand how to use the pencil to trace the board. It is possible to use a pencil to make mistakes and erase the mistakes. It also helps to mark the columns and rows using pencil marks in order to use elimination strategies. To aid in finding the exact number within a row or column Try to locate the exact same numbers across several rows. For example Seven is the number seven appears only in the cells in red that are in the middle row. Another method to solve Sudoku efficiently is to be able to think logically and to organize the numbers. Be careful not to think you know the answer, as it could end up destroying the entire puzzle. It is recommended to put the numbers in the blocks after you weigh the evidence in each box. You should also take your time in solving the puzzle. By doing this you can get an easy solution. If you are patient and practice the tips above, you will get the results you want in no time. Related For Sudoku Puzzles Printable
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Python Program to find Prime factors of given integer - Quescol Python Program to find Prime factors of given integer In this tutorial, you will learn to write a Python program to find the Prime factors of a given input integer. Prime factorization is a mathematical process of breaking down a number into its prime We can also say that it is a representation of a composite number as a product of its prime factors. In this post, we will see the various logic to write a Python program to find prime factors of a given integer. To find the prime factors of a given integer we have to factorize it. Factorization means finding the prime numbers that divide the given integer. We can use different methods to factorize the integer. One of the methods is trial division. is trial division this method, we divide the number by the smallest prime factor, then divide the quotient by the next smallest prime factor, and continue the process until we reach a quotient of 1. What is Prime Factor? Prime factors of a number are the prime numbers that divide that number exactly, without leaving a remainder. In mathematical terms, if you have a number n, its prime factors are the set of prime numbers that can be multiplied together to equal n. Key Concepts about Prime Factors: โ€ข Prime Number: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. For example, the numbers 2, 3, 5, and 7 are prime numbers because they can only be divided evenly by 1 and themselves. โ€ข Factorization: This is the process of breaking down a number into its constituent factors. For prime factorization, these factors are all prime numbers. Example of Prime Factorization: Letโ€™s consider the number 60: โ€ข Start dividing 60 by the smallest prime number, which is 2. โ€ข Continue dividing by 2 until you cannot divide evenly by 2 anymore. โ€ข Move to the next smallest prime number, which is 3. โ€ข The result 5 is itself a prime number, so we stop here. The prime factors of 60 are therefore 2,2,3, and 5. This can be expressed as 2^2ร—3ร—5. Properties of Prime Factors: โ€ข Unique Factorization: Every integer greater than 1 either is a prime number itself or can be uniquely factored into prime numbers, up to the order of the factors. This is known as the Fundamental Theorem of Arithmetic. โ€ข Use in GCD and LCM: The greatest common divisor (GCD) and the least common multiple (LCM) of two numbers can be determined from their prime factorizations. โ€ข Applications: Prime factorization is critical in various fields, particularly in cryptography, where large prime numbers are used to encrypt information securely. Find Prime factors of given Integer in Python Program 1: Using Trial Division In this method, we will use trial division to find the prime factors of a given integer. We will start with the smallest prime factor and divide the integer until the quotient becomes 1. If the quotient is divisible by a prime factor, we will add that factor to the list of prime factors. def prime_factors(n): i = 2 factors = [] while i * i <= n: if n % i: i += 1 n //= i if n > 1: return factors num = int(input("Please enter a number: ")) print("Prime factors of", num, "are:", prime_factors(num)) Output : Enter a number: 24 Prime factors of 24 are: [2, 2, 2, 3] Program 2: Optimized Trial Division Using 2 and Odd Numbers This approach improves upon the basic trial division by initially dividing out all factors of 2 (the only even prime number), and then using a loop that increments by 2 to check only odd numbers. num = int(input("Please enter a number: ")) n = num factors = [] while n % 2 == 0: n //= 2 factor = 3 while factor * factor <= n: while n % factor == 0: n //= factor factor += 2 if n > 1: print("Prime factors of", num, "are:", factors) Please enter a number: 9 Prime factors of 9 are: [3, 3] This method is more efficient than checking every single number up to n because it reduces the number of division operations by half after dealing with the factor of 2.
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GTK+ needs no introduction. LaTeX is the first thing that pops in anyoneโ€™s mind if mathematical equationsโ€™ typesetting is under consideration. Matplotlib โ€” while not as well-known as the former two โ€” is the super easy and elegant solution for scientific plotting on *nix platforms. For an application demo, I required all three. Past experience has taught me that the most straightforward way of โ€œgluingโ€ things together is Python. GTK+ therefore = PyGTK. Next up was LaTeX, and a previous solution of mine for embedding LaTeX in PyGTK came to the rescue. The final requirement of Matplotlib was fulfilled without any hassle since the library was already written in Python. The collective result was pretty: (Click on the image for larger version.) The linked tarball contains the Python scripts for the application. For everything to run smoothly, LaTeX and Matplotlib packages need to be installed on your system. If you encounter any issues running the code, feel free to flame your distribution for the apparent lack of sanity regarding package management. Flag 42 Open Source HOWTO: Use LaTeX mathematical expressions in PyGTK Filed under: Blog โ€” krkhan @ 10:04 pm I had never really laid my hands on LaTeX until I required it in one of the helper applications for my graduation project. Unfortunately, the requirement wasnโ€™t as simple as producing some documents as I had to embed mathematical expressions on the fly in my PyGTK apps. Googling around for the solution, I found GtkMathView which accomplished something similar to this albeit using MathML. However, my luck ran out on me again as the widget lacked Python bindings. The other solution was to generate transparent PNGs on the fly and include them as GtkImages. This worked rather well, as the final code allowed easy modifications to the generated expressions. Requirements for the code were: โ€ข LaTeX โ€ข dvipng โ€ข latexmath2png Final results: And the simple code behind it: 1 #!/usr/bin/env python 2 """An example demonstrating usage of latexmath2png module for embedding math 3 equations in PyGTK 5 Author: Kamran Riaz Khan <krkhan@inspirated.com> 6 """ 8 import gtk 9 import os 10 import latexmath2png 12 pre = 'gtktex_' 13 eqs = [ 14 r'$\alpha_i > \beta_i$', 15 r'$\sum_{i=0}^\infty x_i$', 16 r'$\left(\frac{5 - \frac{1}{x}}{4}\right)$', 17 r'$s(t) = \mathcal{A}\sin(2 \omega t)$', 18 r'$\sum_{n=1}^\infty\frac{-e^{i\pi}}{2^n}$' 19 ] 20 latexmath2png.math2png(eqs, os.getcwd(), prefix = pre) 22 def window_destroy(widget): 23 for i in range(0, len(eqs)): 24 os.unlink(os.path.join(os.getcwd(), '%s%d.png' % (pre, i + 1))) 25 gtk.main_quit() 27 window = gtk.Window() 28 window.set_border_width(10) 29 window.set_title('LaTeX Equations in GTK') 30 window.connect('destroy', window_destroy) 31 vbox = gtk.VBox(spacing = 10) 32 window.add(vbox) 34 images = [None] * len(eqs) 35 for i in range(len(eqs)): 36 images[i] = gtk.image_new_from_file('%s%d.png' % (pre, i + 1)) 37 vbox.pack_start(images[i]) 39 window.show_all() 40 gtk.main() #!/usr/bin/env python """An example demonstrating usage of latexmath2png module for embedding math equations in PyGTK Author: Kamran Riaz Khan <krkhan@inspirated.com> """ import gtk import os import latexmath2png pre = 'gtktex_' eqs = [ r'$\alpha_i > \beta_i$', r'$\sum_{i=0}^\infty x_i$', r'$\left(\frac{5 - \frac{1}{x}}{4}\right)$', r'$s(t) = \mathcal{A}\sin(2 \omega t)$', r'$\sum_{n=1}^\infty\ frac{-e^{i\pi}}{2^n}$' ] latexmath2png.math2png(eqs, os.getcwd(), prefix = pre) def window_destroy(widget): for i in range(0, len(eqs)): os.unlink(os.path.join(os.getcwd(), '%s%d.png' % (pre, i + 1))) gtk.main_quit() window = gtk.Window() window.set_border_width(10) window.set_title('LaTeX Equations in GTK') window.connect('destroy', window_destroy) vbox = gtk.VBox(spacing = 10) window.add (vbox) images = [None] * len(eqs) for i in range(len(eqs)): images[i] = gtk.image_new_from_file('%s%d.png' % (pre, i + 1)) vbox.pack_start(images[i]) window.show_all() gtk.main() Tags: Code, Equation, Flag 42, Graphics, GTK+, LaTeX, Open Source, PyGTK, Python, TeX, Tutorial Comments (2)
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JEE Main Answer Key 2023 Released for Session 1 and Session 2 - Download PDFs Here JEE Main answer keys are now available to download for the January and April session question papers here. The JEE Main answer key is an important document for candidates as it contains all the correct options for the questions asked in the JEE Main exam paper. The JEE Main answer key is usually released by the exam authority, the National Testing Agency (NTA), on their official website. They can go through the JEE Main answer key to check and compare their answers and generate a tentative score for the exam. Once the NTA releases the official answer key, we will provide detailed solutions for all JEE Main 2023 question papers and printable PDFs.ร‚ Our subject experts will provide JEE Main 2023 question paper analysis immediately after the examination. You will also get the JEE main 2023 answer key PDFs and solutions for all the shifts and subjects, Physics, Chemistry and Maths, as soon as the examination is over. Note: We are providing downloadable PDFs of the official JEE Main 2023 Answer Key released by the NTA on our website. The JEE Main 2023 answer keys PDF for the January and April sessions are provided JEE Main 2023 Question Papers and Solutions PDFs JEE Main 2023 Answer Key PDFs April Session (Session 2) JEE Main 2023 session 2 answer key is available here for all the question papers. JEE Main answer key 2023 will be useful for the candidates who appeared for the exam to check their scores and calculate their ranks. Candidates can download the JEE Main 2023 April session answer key PDFs for free using the links provided below. Download JEE Main 2023 Session 2 Answer Key PDFs for April 06, 08, 10, 11, 12, 13 and 15 Question Papers Here for FREE! JEE Main 2023 Answer Key โ€“ April 6 JEE Main April Session Shifts Answer Key PDFs April 06, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 April 6th Shift 1 โ€“ Answer Key and Solutions JEE Main 2023 April 6th Shift 2 โ€“ Answer Key and Solutions JEE Main 2023 Answer Key โ€“ April 8 JEE Main April Session Shifts Answer Key PDFs April 08, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 April 8th Shift 1 โ€“ Answer Key and Solutions JEE Main 2023 April 8th Shift 2 โ€“ Answer Key and Solutions JEE Main 2023 Answer Key โ€“ April 10 JEE Main April Session Shifts Answer Key PDFs April 10, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 April 10th Shift 1 โ€“ Answer Key and Solutions JEE Main 2023 April 10th Shift 2 โ€“ Answer Key and Solutions JEE Main 2023 Answer Key โ€“ April 11 JEE Main April Session Shifts Answer Key PDFs April 11, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 April 11th Shift 1 โ€“ Answer Key and Solutions JEE Main 2023 April 11th Shift 2 โ€“ Answer Key and Solutions JEE Main 2023 Answer Key โ€“ April 12 JEE Main April Session Shifts Answer Key PDFs April 12, 2023 (Paper 1) Shift 1 Download Now JEE Main 2023 April 12th Shift 1 โ€“ Answer Key and Solutions JEE Main 2023 Answer Key โ€“ April 13 JEE Main April Session Shifts Answer Key PDFs April 13, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 April 13th Shift 1 โ€“ Answer Key and Solutions JEE Main 2023 April 13th Shift 2 โ€“ Answer Key and Solutions JEE Main 2023 Answer Key โ€“ April 15 JEE Main April Session Shifts Answer Key PDFs April 15, 2023 (Paper 1) Shift 1 Download Now JEE Main 2023 April 15th Shift 1 โ€“ Answer Key and Solutions JEE Main 2023 Answer Key PDFs January Session (Session 1) Candidates can download the official answer key PDFs for all the JEE Main 2023 January session question papers for morning and evening shifts from the links provided below. JEE Main 2023 Answer Key โ€“ January 24 JEE Main 2023 January 24th Shift 1 โ€“ Answer Key & Solutions JEE Main 2023 January 24th Shift 2 โ€“ Answer Key & Solutions JEE Main January Session Shifts Answer Keys January 24, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 Answer Key โ€“ January 25 JEE Main 2023 January 25th Shift 1 Answer Key & Solutions JEE Main 2023 January 25th Shift 2 Answer Key & Solutions JEE Main January Session Shifts Answer Keys January 25, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 Answer Key โ€“ January 29 JEE Main 2023 January 29th Shift 1 Answer Key & Solutions JEE Main 2023 January 29th Shift 2 Answer Key & Solutions JEE Main January Session Shifts Answer Keys January 29, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 Answer Key โ€“ January 30 JEE Main 2023 January 30th Shift 1 Answer Key & Solutions JEE Main 2023 January 30th Shift 2 Answer Key & Solutions JEE Main January Session Shifts Answer Keys January 30, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 Answer Key โ€“ January 31 JEE Main 2023 January 31st Shift 1 Answer Key & Solutions JEE Main 2023 January 31st Shift 2 Answer Key & Solutions JEE Main January Session Shifts Answer Keys January 31, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now JEE Main 2023 Answer Key โ€“ February 1 JEE Main 2023 February 1st Shift 1 โ€“ Detailed Analysis & Solutions JEE Main 2023 February 1st Shift 2 โ€“ Detailed Analysis & Solutions JEE Main January Session Shifts Answer Keys February 1, 2023 (Paper 1) Shift 1 Download Now Shift 2 Download Now How to Access JEE Main 2023 Answer Keys? We are still awaiting the official notification regarding the JEE Main 2023 answer key and most likely this facility will be opened after the end of the April session wherein the provisional answer key will be released first and then the final answer key will be announced on the official website of NTA. Candidates will be able to download the question papers and the responses if they want to refer to them in the future. Here are some of the steps that candidates can follow to download the JEE Main 2023 answer keys. 1. Visit the official website (https://jeemain.nta.nic.in). 2. Click on the View Question Paper link. 3. You will be redirected to a page where you will be given two options to log in either through Application Number and Password or through Application Number and Date of Birth. 4. Enter your JEE Main Application Number and Password/Date of Birth and Hit the โ€˜Sign-inโ€™ button. 5. Choose your examination mode, and fill in the date and the slot. 6. After a successful entry of the details, the answer key of JEE Main will be displayed on the screen. In addition to releasing the answer keys, NTA has also provided a tab with a link that candidates can visit if they want to challenge the answer key. Also, Check-Out:ร‚ Challenging JEE Main Answer Key 2023 โ€ข Candidates who are not satisfied with the answer key can fill out an application form for the same. โ€ข In order to challenge the answer key, candidates have to make a payment of some amount (prescribed by NTA), per answer. โ€ข To fill out the application form for challenging an answer, candidates have to log in on the site, enter their JEE Main (2023) application number, and date of birth, and choose the security โ€ข Payment for challenging answer keys can be done by Debit/Credit card or e- challan which is non-refundable. โ€ข It is also important for candidates to present the receipt of the processing fee. โ€ข The decision made by the JAB (Joint Admission Board) / NTA will be considered as the final decision, and no further communication will be entertained. โ€ข If the challenge is found correct, the processing fee will be refunded.ร‚ โ€ข Only paid challenges made during the stipulated time through the key challenge links will be considered. โ€ข No grievance with regard to the answer key(s) after the declaration of result/NTA score of JEE Main 2022 will be entertained. Related Links JEE Main 2022 Answer Key PDFs July Session (Session 2) JEE Main 2022 Session 2 (July session) answer key PDFs are given below. Download the JEE Main 2022 session 2 question paper solutions and answer key PDFs using the links provided here. 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JEE Main 2022 Answer Key PDFs June Session (Session 1) Download the official answer key PDFs for JEE main 2022 question papersร‚ using the links given below for the June session. JEE Main June Session Shifts Answer Keys June 24, 2022 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2022 June 24 Shift 1 Question Paper โ€“ Physics Solutions JEE Main 2022 June 24 Shift 1 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 24 Shift 1 Question Paper โ€“ Maths Solutions JEE Main 2022 June 24 Shift 2 Question Paper โ€“ Physics Solutions JEE Main 2022 June 24 Shift 2 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 24 Shift 2 Question Paper โ€“ Maths Solutions JEE Main June Session Shifts Answer Keys June 25, 2022 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2022 June 25 Shift 1 Question Paper โ€“ Physics Solutions JEE Main 2022 June 25 Shift 1 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 25 Shift 1 Question Paper โ€“ Maths Solutions JEE Main 2022 June 25 Shift 2 Question Paper โ€“ Physics Solutions JEE Main 2022 June 25 Shift 2 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 25 Shift 2 Question Paper โ€“ Maths Solutions JEE Main June Session Shifts Answer Keys June 26, 2022 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2022 June 26 Shift 1 Question Paper โ€“ Physics Solutions JEE Main 2022 June 26 Shift 1 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 26 Shift 1 Question Paper โ€“ Maths Solutions JEE Main 2022 June 26 Shift 2 Question Paper โ€“ Physics Solutions JEE Main 2022 June 26 Shift 2 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 26 Shift 2 Question Paper โ€“ Maths Solutions JEE Main June Session Shifts Answer Keys June 27, 2022 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2022 June 27 Shift 1 Question Paper โ€“ Physics Solutions JEE Main 2022 June 27 Shift 1 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 27 Shift 1 Question Paper โ€“ Maths Solutions JEE Main 2022 June 27 Shift 2 Question Paper โ€“ Physics Solutions JEE Main 2022 June 27 Shift 2 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 27 Shift 2 Question Paper โ€“ Maths Solutions JEE Main June Session Shifts Answer Keys June 28, 2022 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2022 June 28 Shift 1 Question Paper โ€“ Physics Solutions JEE Main 2022 June 28 Shift 1 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 28 Shift 1 Question Paper โ€“ Maths Solutions JEE Main 2022 June 28 Shift 2 Question Paper โ€“ Physics Solutions JEE Main 2022 June 28 Shift 2 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 28 Shift 2 Question Paper โ€“ Maths Solutions JEE Main June Session Shifts Answer Keys June 29, 2022 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2022 June 29 Shift 1 Question Paper โ€“ Physics Solutions JEE Main 2022 June 29 Shift 1 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 29 Shift 1 Question Paper โ€“ Maths Solutions JEE Main 2022 June 29 Shift 2 Question Paper โ€“ Physics Solutions JEE Main 2022 June 29 Shift 2 Question Paper โ€“ Chemistry Solutions JEE Main 2022 June 29 Shift 2 Question Paper โ€“ Maths Solutions JEE Main 2021 Answer Key PDFs February Session 1 JEE Main Februaryร‚ Session Shifts Answer Keys February 24, 2021 (Paper 1) Shift 1 Download Shift 2 Download February 25, 2021 (Paper 1) Shift 1 Download Shift 2 Download February 26, 2021 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2021 March (Session 2) Answer Keys JEE Main March Session Shifts Answer Keys March 16, 2021 (Paper 1) Shift 1 Download Shift 2 Download March 17, 2021 (Paper 1) Shift 1 Download Shift 2 Download March 18, 2021 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2021 July (Session 3) Answer Keys JEE Main July Session Shifts Answer Keys July 20, 2021 (Paper 1) Shift 1 Download Shift 2 Download July 22, 2021 (Paper 1) Shift 1 Download โ€“ To be Updated Shift 2 Download July 25, 2021 (Paper 1) Shift 1 Download Shift 2 Download July 27, 2021 (Paper 1) Shift 1 Download Shift 2 Download JEE Main 2021 August/September (Session 4) Answer Keys JEE Main 4th Session Shifts Answer Keys August 26th, 2021 (Paper 1) Shift 1 Downloadร‚ Shift 2 Download August 27th, 2021 (Paper 1) Shift 1 Downloadร‚ Shift 2 Downloadร‚ August 31, 2021 (Paper 1) Shift 1 Downloadร‚ Shift 2 Downloadร‚ September 1, 2021 (Paper 1) Shift 1 Download รขโ‚ฌโ€œ To be Updated Shift 2 Downloadร‚ Frequently Asked Questions on JEE Main Answer Key When will the JEE Main 2023 April session answer key be released? The JEE Main 2023 Answer Key for the April session (session 2) has been released by NTA. On 24th April 2023, NTA released the final provisional answer key for the JEE Main 2023 session 2. Candidates can download the JEE Main answer key (2023) now. How can I raise objections to the JEE Main answer key 2023 if the answers are wrong? Candidates can challenge and raise an objection by logging in to their accounts on the official website. However, the window to raise objections will be opened by the authority for a certain period of time. How much should I pay for raising a challenge for the wrong option given in the JEE Main answer key? Candidates will have to pay a fee of Rs. 200 per question for challenging JEE Main answer key 2023. This is non-refundable. Can we calculate JEE Main percentile score from the raw marks obtained from JEE Main Answer Key? No, the percentile score cannot be calculated from the raw score obtained from the answer key. When will the JEE Main answer key 2023 be released? The JEE Main 2023 answer key for Sessions 1 and 2 will be displayed on the official website mostly within a week after the examination is over. JEE Main answer key 2023 for session 1 and session 2 are available here to download. How to download the JEE Main answer key 2023? Candidates can download the JEE Main 2023 answer key from the official website of JEE Main. They have to click on the link to download the answer key. We also provided the JEE Main answer key 2023 for both January and April session question papers on this page. When is the last date to submit objections regarding the JEE Main answer key? The last date to submit objections will be updated on the official website. How to calculate marks according to the JEE Main answer key? 4 marks will be awarded for every correct answer. One mark will be deducted for every wrong answer. In this way, the total marks can be calculated. Leave a Comment
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Proving $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$ โ€ข MHB โ€ข Thread starter solakis1 โ€ข Start date In summary, using the given equations, we can prove that the sum of the cube roots of $a, b,$ and $c$ is equal to zero. This is done by using the identity $(x^3+y^3+z^3-3xyz)=\dfrac{1}{2}(x+y+z) ((x-y)^2+(y-z)^2+(z-x)^2)$ and the fact that $x+y+z=0$. Therefore, the statement $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$ is proven. $a+b+c-3(abc)^\dfrac{1}{3}=0$ and $\neg((a)^\dfrac{1}{3}=(b)^\dfrac{1}{3}) $ and $\neg((b)^\dfrac{1}{3}=(c)^\dfrac{1}{3})$ and Then prove: solakis said: $\neg((a)^\dfrac{1}{3}=(b)^\dfrac{1}{3}) $ .......(2) $\neg((b)^\dfrac{1}{3}=(c)^\dfrac{1}{3})$ .......(3) Then prove: [sp] Let : $x=a^\dfrac{1}{3},y=b^\dfrac{1}{3},z=c^\dfrac{1}{3}$.........(5) Use the following identity: Use (1),(2),(3) ,(4),(5) and (6) becomes: $ (a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$ FAQ: Proving $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$ 1. How can I prove that $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$? There are a few different ways to approach this proof. One method is to use the concept of a cube root and its properties, such as the fact that the cube root of a product is equal to the product of the cube roots. Another approach is to use algebraic manipulation and substitution to simplify the expression and show that it equals zero. 2. Is there a specific set of values for $a$, $b$, and $c$ that will make the equation true? Yes, there are certain values for $a$, $b$, and $c$ that will make the equation $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$ true. For example, if $a$, $b$, and $c$ are all equal to zero, the equation will be true. Additionally, any set of values where the sum of the cube roots is equal to zero, such as $a=8$, $b=-8$, and $c=0$, will satisfy the equation. 3. Can this equation be used to solve for a specific variable? No, this equation cannot be used to solve for a specific variable. It is an expression that must be true for certain values of $a$, $b$, and $c$, but it does not provide a way to solve for any one variable. In order to solve for a specific variable, additional information or equations would be needed. 4. What are some real-world applications of this equation? This equation may be used in various fields of science and engineering, such as physics and chemistry. For example, it could be used in calculations involving the properties of gases or the behavior of certain chemical reactions. It may also have applications in computer science and cryptography. 5. Is this equation always true, or are there exceptions? The equation $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$ is not always true. As mentioned in the answer to the second question, there are specific values for $a$, $b$, and $c$ that will make the equation true, but there are also many values where it will not be true. For example, if $a$, $b$, and $c$ are all negative, the equation will not be true. Additionally, if only one of the variables is negative, the equation will not be true. It is important to consider the values of $a$, $b$, and $c$ when determining if the equation is true or not.
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How To Learn Statistics This section of the website covers statistics and has most of the same topics that would be covered in a statistics 101 course at a University. Here is the table of contents for the different statistics topics covered on this site. Two good examples of free University content for this material are Another great resource for statistics help is Cross Validated. There you can ask questions about your specific problems and get help. Purpose Of This Page The reason I wrote this material came from my own pursuit of the same information. I am about 10 years removed from studying aerospace engineering in college and have been working as a professional engineer since then. However, even though I studied statistics at the university, a lot of the concepts proved to be slippery and a โ€œuse them or forget themโ€. This is especially true of things I knew mostly by memorizing equations. Remembering equations worked fine for carrying me through a semester of school, but to remember the topic for a whole career or a lifetime, it turns out that I needed something different. I needed to find ways relate each topic other things that I knew, so it wasnโ€™t on the periphery of my knowledge anymore, but deeply ingrained. Benefits & Perils Of Self-Teaching These topics are my attempt to cover statistics in a way that is easy to learn and remember long term. A lot of this material I re-learned through self-study. Being largely self-taught in these topics often makes it easier for me to find relatable examples for other learners than it would be for the complete expert. The downside of self-study is that it can leave gaps. For instance, when learning how to do multiple linear regression (i.e. draw a straight line/plane/hyperplane through data that has at least 2 independent variables) I ran across a method of doing it that was an intuitive but laborious expansion of simple linear regression. A reader who is more of an expert than I pointed out that I had missed a simple to do (but difficult to understand) method of doing multiple regression (i.e. Moore-Penrose pseudo inverse). If you spot similar oversights or just plain errors, you can find my email address here, please let me know, and I can include that information for future learners How To Learn Statistics I think that statistics is an area where you want to go for breadth-first. I.e. You would rather know something about a lot of topics than knowing a lot about a few topics. My friend Kalid over at Better Explained refers to it as the Adept Method. And a good analogy is to go for seeing a big picture first, even if it is not completely clear More study gradually brings it into focus Rather than diving deeply into specific topics before moving on. From a practical sense, what that can mean is skimming topics on an initial read, than then revisiting the topics a couple of times as you learn other material and see the connections between them. To facilitate this, I have set up each page with a couple different sections. The first section of each page is designed for an initial read through to get the big picture, and the second or third sections are designed to draw connections to the other topics and to dive into a deeper understanding. I recommend reading the first section of any given topic and then coming back a week so so later after youโ€™ve had time to internalize some of the material, do some example problems, and look at some other topics, before diving into the second or third layer of the material Sources Of Material The internet has a lot of great content on it, and it doesnโ€™t make sense to duplicate a bunch of material that already exists. So when I know of pages that have great explanations or find useful tools, I will link to them. And if you have suggestions please let me know. I have also covered some of the other topics in more depth as Kindle Books (typically priced under 3 dollars), so I will link to that material where it makes sense. I know that some people canโ€™t access the Kindle content or are not in a position to purchase it, so I would be happy to send a free PDF copy if you contact me. Enough Preamble Thatโ€™s all I have by way of introduction. To get started go to
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Ordinal Numbers Interactive Game - OrdinalNumbers.com Ordinal Numbers Interactive โ€“ An endless number of sets can be easily enumerated using ordinal numerals as a tool. You can also use them to generalize ordinal number. 1st The ordinal number is one the fundamental concepts in math. It is a number that identifies the location of an object within a list of objects. โ€ฆ Read more
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MIT PDP-10 'Info' file converted to Hypertext 'html' format by Henry Baker Previous Up Next Arithmetic Calculator Program. This program will read an arithmetic expression composed of numbers and the operators +, -, * and /, compute the value and type the result. It shows how to read and print numbers. It does not know about operator precedence; it does operations from left to right always. TITLE NUM PDL: BLOCK PDLLEN OP1: 0 X1: 0 LINBUF: BLOCK 30 LINPTR: 0 START: MOVE P,[-PDLLEN,,PDL-1] ;Open TTY channels. .CALL [SETZ ? SIXBIT/OPEN/ [.UAI,,CHTTYI] ? [SIXBIT/TTY/] ((SETZ))] .LOSE %LSFIL .CALL [SETZ ? SIXBIT/OPEN/ [.UAO,,CHTTYO] ? [SIXBIT/TTY/] ((SETZ))] .LOSE %LSFIL START1: PUSHJ P,GETLIN ;Read in a line of input. MOVE A,[440700,,LINBUF] MOVEM A,LINPTR ;Set up to fetch chars from the line. PUSHJ P,EVAL ;Parse and evaluate expression. PUSHJ P,DECOUT ;Print the answer. MOVEI A,[ASCIZ/ PUSHJ P,OUTSTR JRST START1 ;Read and evaluate an expression. Value returned in A. ;Clobbers B. EVAL: PUSHJ P,DECIN ;Read one number. MOVEM B,OP1 ;Save the number. EVAL1: MOVEI B,0 CAIN A,"+ ;Consider the operation character: MOVE B,[ADD B,OP1] ;B gets an instruction to do that operation. CAIN A,"- MOVE B,[SUB B,OP1] CAIN A,"* MOVE B,[IMUL B,OP1] CAIN A,"/ MOVE B,[IDIV B,OP1] JUMPE B,EVALX ;If B is still 0, the terminator ;was not an arith op, so it ends ;the expression or is illegal. MOVEM B,X1 ;It is an arith op, so save the instruction. PUSHJ P,DECIN ;Read the second operand. EXCH B,OP1 ;B gets first op, OP1 gets second operand. XCT X1 ;Compute result of operation, in B. MOVEM B,OP1 ;Save it as first operand of next operation. JRST EVAL1 ;A has terminator of second operand, ;which is the next operation. ;Come here on number terminated by char not an arith op. EVALX: JUMPN A,ERR ;Should be end of line, or it's an error. MOVE A,OP1 ;Otherwise, last saved value is value of exp. POPJ P, ;Print an error message if we see something we don't recognize. ERR: MOVEI A,[ASCIZ/Unrecognized character in expression: /] PUSHJ P,OUTSTR LDB A,LINPTR ;Print the offending character .IOT CHTTYO,A ;as part of the error message. MOVEI A,[ASCIZ / PUSHJ P,OUTSTR JRST START1 ;Read a signed decimal number out of the line, returning number in B ;and terminating character in A. DECIN: TRZ FL,NEGF!DIGF ;No minus, no digit seen yet. MOVEI B,0 DECIN1: ILDB A,LINPTR ;Fetch next character of line. CAIL A,"0 CAILE A,"9 JRST DECIN2 ;Jump if character not a digit. IMULI B,10. ;Else accumulate this digit into the number. ADDI B,-"0(A) ;Note that we convert the digit into its value. ;("0 into the value 0, "1 into 1). TRO FL,DIGF ;Set flag saying non-null number seen. JRST DECIN1 DECIN2: CAIN A,"- JRST DECIN3 ;Jump on minus sign. TRNN FL,DIGF ;Anything else: if we saw a number, POPJ P, ;negate it if it began with a minus sign. DECIN4: TRZE FL,NEGF MOVN B,B POPJ P, ;Come here after reading a minus sign. DECIN3: TRNE FL,DIGF ;Does it follow a number? JRST DECIN4 ;Yes. This must be a binary minus. TRC FL,NEGF ;This must be unary minus. ;Complement flag that number is negative. JRST DECIN1 ;(So that two minus signs cancel out). ;Print number in A, positive or negative, in decimal. ;Clobbers A and B. DECOUT: JUMPGE A,DECOT1 .IOT CHTTYO,["-] ;If number is negative, print sign. CAMN A,[400000,,] ;Smallest negative number is a pain: JRST DECOT2 ;its absolute value cannot fit in one word! MOVM A,A ;Else get abs val of negative number and print. DECOT1: IDIVI A,10. HRLM B,(P) ;Save remainder in LH of stack word ;whose RH contains our return address. SKIPE A ;If quotient is nonzero, PUSHJ P,DECOT1 ;print higher-order digits of number. HLRZ A,(P) ;Get back this remainder (this digit). ADDI A,"0 .IOT CHTTYO,A POPJ P, ;Print the abs value of the largest negative number. DECOT2: MOVEI A,[ASCIZ /34359738368/] JRST OUTSTR ;Copy the GETLIN and OUTSTR subroutines here. END START โ€ข 10.: This is a decimal number. You can tell, because it ends with a decimal point. โ€ข LINPTR: this location holds the byte pointer used for fetching characters out of the line. It is usually not worth while to keep this pointer in an accumulator if the parsing is being done over more than a very small piece of code. โ€ข XCT: Note how EVAL chooses an arithmetic instruction based on the arithmetic operator character, then reads the following argument, and then executes the instruction chosen earlier, performing the operation. This is also the first use you have seen of literals containing instructions. โ€ข ERR: this is an example of printing an error message. Error messages should always show the offending data, not just say "something was wrong". โ€ข DECIN: Note how flags in accumulator 0 (FL) are used to keep track of whether any digits have been seen, and whether a minus sign came before them. Accumulator 0 is most often used for such flags because it is the least useful accumulator for anything else (since it cannot be used as an index register). โ€ข DECOUT: This is a very famous program for printing a number. It works recursively because the first digits extracted as the remainders in successive divisions by the radix are the last digits to be printed. So the digits are produced and saved on the way down the recursion, and printed on the way up. โ€ข HRLM: We could save the remainder with PUSH P,B and restore it with POP P,A, but since the left half of each word saved by a PUSHJ is not really going to be used, we can save stack space by using those left halves to store the remainder. It is also faster.
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Page Fortunetelling cards for conjectural game. 1. Physiognomy and divinations of IChing. The first page of this website section represents physiognomic fortunetelling cards with images of Hexagrams IChing and physiognomic symbols which show symbolical emotions of a human face that allows to interpret Hexagrams IChing according to values which are expressed in human emotions. And also the first page represents the game board which is organized according to rules of a magic square and is coordinated with Hexagrams IChing that allows to carry out conjectural mantic games or puzzles or divinations by means of arrangements of physiognomic fortunetelling cards on squares of the magic board. Besides the first page represents the PDF file in which it is possible to find (suitable for printing in format A4) images of physiognomic cards and images of the magic board, and also it is possible to find instructions and rules of games, puzzles and divinations. The PDF file represents only rules according to which it is possible to have physiognomic fortunetelling cards on the magic board, but for interpretation of results of divinations and predictions, namely for understanding of values which physiognomic symbols and Hexagrams IChing have, look the information on other pages of this website, and in particular look physiognomic galleries which have names impressions / oracle / conditions / relations / times and also look the section which describes concepts of the canon of changes IChing. The second page represents esoteric concept of divinations and predictions. Fortunetelling physiognomic cards and IChing. Conjectural puzzles and divinations. Conjectural mantic game is the set of physiognomic fortunetelling cards with images of Hexagrams IChing and physiognomic symbols, and also the magic board which are necessary for games or puzzles or Application of these physiognomic fortunetelling cards can be various. It is possible to use for mathematical puzzles or conundrums by means of which it is possible to investigate logic system of Hexagrams IChing as cards have numbers which correspond with serial numbers of Hexagrams in the canon of changes IChing and correspond with numbers of a magic square which forms numerical notations of the magic board. It is possible to use for various logic games and puzzles as it is possible to consider cards as game elements. It is possible to use the offered physiognomic fortunetelling cards for realization of divinations and predictions, for fortune-telling and card-reading, for cartomancy and prophecy. Logic puzzles and games and divinations are incorporated in one magic and mathematical system in which games can turn to divinations, and divinations can turn to logic games and puzzles or Rules and instructions are submitted in the PDF file in which also there are images of physiognomic cards and images of the magic board suitable for printing. You must to pay for images and texts of rules for divinations in the PDF file. - 407 Kb to download game. Now online access to this file for uploading does not work. Address for PDF file directly to me by e-mail. panf-sergey@list.ru or panfsergey@gmail.com The set of this mantic game contains 64 physiognomic fortunetelling cards on which 64 Hexagrams IChing and 64 physiognomic symbols are drawn as shown at the left in the form of the table. Namely each card has the serial number which corresponds to the serial number in the sequence of Hexagrams in the Classical Book of Changes IChing (King Wen's sequence), and also each card has the numerical value which corresponds to the numerical notation of the corresponding square of the magic board. Serial numbers are put on physiognomic cards by dark Arabian ciphers and numerical values are written by white ciphers in dark circles. The magic board in the set of this conjectural mantic game has 8 vertical and 8 horizontal rows, and total has 64 dark and light squares. Verticals and horizontals are notated by Trigrams. It means that each square corresponds to the Hexagram IChing which turns out from connection of top Trigram on a vertical and bottom Trigram on a horizontal of the board. But only numerical values of Hexagrams IChing are written on squares of the board directly, that corresponds to white figures in dark circles on cards. Namely if to arrange physiognomic fortunetelling cards on squares of the magic board according to numerical values of cards and squares then Hexagrams IChing on cards are derivatives from Trigrams which designate verticals and horizontals of the board. Numerical notations of squares on the magic board is a magic square and accordingly numerical values of Hexagrams IChing make a magic square. Pay attention that a magic square is made with numerical values which are put on cards by white ciphers in dark circles, and also a magic square is made with numbers (ciphers) which are put directly on squares of the board, but serial numbers (which are put on physiognomic cards by dark ciphers) correspond to serial numbers of Hexagrams in the sequence of the Classical Book of Changes IChing. Also pay attention that Hexagrams in the Book of Changes IChing have only serial numbers and have no numerical values which are used in the present conjectural and mantic game. Numerical values of physiognomic cards have the identical sums for each emotional expression due to a magic square. Namely, if to choose all cards with joyful expressions of eyes then these cards have the sum of numerical values 520. If to choose all cards with any other emotional expressions of a mouth or eyes or eyebrows then such physiognomic cards also have the sum of numerical values If to arrange physiognomic fortunetelling cards on squares of the magic board according to numerical values of cards and squares then each vertical or horizontal has the sum of numerical values 260, and also each quarter of the board has the sum of numerical values 520 that provides proportionality of numerical values for all emotional expressions, as cards and in essence physiognomic symbols and Hexagrams IChing are correlated with squares of the magic board. This magic proportion of physiognomic cards can be used in games and puzzles for providing of proportional distribution of cards between participating people, and also for providing of equal odds of a prize in that case when the sums of numerical values are significant in game rules, and when physiognomic symbols with those or other emotional expressions are significant for players. In total numerical values and serial numbers of physiognomic fortunetelling cards in the set of the offered conjectural mantic game are subordinated to laws of magic numbers, and accordingly magic numbers operate divinations which can be carried out by means of the offered magic and physiognomic system. The magic square is a term which expresses mathematical laws in arrangements of numbers and is known in the mathematics. Different arrangements of numbers can make various magic squares and accordingly Hexagrams IChing can have various numerical values, but the shown arrangement of numbers on squares of "the great magic square" is used in the offered conjectural game. The additional information on arrangements of numbers and Hexagrams IChing in squares (cells) of a magic square look on pages of this site in other section which has the name: chronology. The following page represents esoteric concepts of divinations. upwards - following
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Effect of unsteady flow conditions on scour features at low-head hydraulic structures Effect of unsteady flow conditions on scour features at low-head hydraulic structures Michele Palermoa*[and Stefano Pagliara]a a[DESTEC-Department of Energy, Systems, Territory and Construction Engineering, ] University of Pisa, Via Gabba 22, 56122 Pisa, Italy Abstract:The study of scour mechanism downstream of low-head control structures is a fundamental topic for hydraulic engineers. Generally, the analysis of the scour process is conducted under steady flow conditions, assuming that the maximum discharge is occurring for sufficient time to reach the equilibrium scour configuration. Nevertheless, in rivers the scour process generally occurs in correspondence with a flood event, which is characterized by discharge varying with time. This last condition is still less studied and analyzed in terms of effects on bed morphology. Researchers mainly focused on the maximum scour depth assuming that it occurs in correspondence with the peak discharge, but they rarely took into account the evolution of the scour process under unsteady flow conditions. The aim of the present paper is to analyze the evolution of scour morphology under unsteady flow conditions, and compare it with that obtained under steady flow conditions. In particular, three structure typologies were tested: a stepped gabion weir with upstream impermeable filtering layer, a straight rock sill, and a curved rock sill. The results showed that the scour phenomenon deeply depended on inflow conditions. Nevertheless, it was also shown that the equilibrium morphology of the downstream stilling basin is essentially the same under both unsteady and steady flow conditions if the duration of the unsteady event is enough long. Keywords: Hydraulic models; Low-head structures; Scour process; Unsteady flow. 1. Introduction: The scour mechanism occurring downstream of low-head control structures is an important topic that has been widely analyzed in recent decades. In particular, the analysis mainly focused on the hydraulics and the scour characteristics in the stilling basin. Low-head structures have been found effective in controlling sediment transport and, at the same time, they are able to guarantee a reduced impact on the ecosystem. Therefore, many traditional structures have been re-converted into more eco-friendly anthropic works, such as block ramps, rock grade control structures, stepped gabion weirs, cross-vane, W-weirs, etc. *[Corresponding author. Present address: DESTEC-Department of Energy, Systems, Territory and ] Construction Engineering, University of Pisa, Italy. Via Gabba 22, 56122 Pisa, Italy, Tel. +39 050 2217929; fax +39 050 2217730, E-mail address: [email protected] One of the first systematic studies investigating erosive process occurring in a movable stilling basin was conducted by Veronese (1937). The author analyzed the effect on the scour characteristics of the stilling basin geometry. Namely, he conducted a series of experimental tests in prototype channels to investigate the effect of symmetric enlargement of the stilling basin on scour morphology. More recently, Bormann and Julien (1991) analyzed several grade control structures configurations, varying both the model scale and the geometry of the structure itself. They concluded that the scour process depends on both the structure geometry and on the stilling basin characteristics. Based on the similarities between scour process due to plunging jets and downstream of grade control structures, Bormann and Julien (1991) and, successively, Dโ€™Agostino and Ferro (2004) showed that the diffusion length of the flow entering the stilling basin is a fundamental parameter. Furthermore, as it was also shown for scour due to plunging jets (see for example Rajaratnam, 1981; Rajaratnam and Macdougall, 1983; Breusers and Raudkivi, 1991; Hoffman and Verheij, 1997; Hoffman, 1998; Manso and Arumugam, 1985; Pagliara et al., 2010; Pagliara et al., 2012a), the geometry of the structure was found a significant parameter influencing the scour morphology because of the different entering flow inclinations. Among eco-friendly hydraulic structures, block ramps and rock chutes constitute a peculiar typology. Namely, they are characterized by a complex hydraulic behavior due to the rough sloped bed that contributes to dissipate an appreciable amount of energy (Pagliara and Palermo, 2011). In particular, these structure typologies exhibit some similarities with both stepped chutes and rock sills, in terms of both hydraulic behavior and dissipative mechanism. Nevertheless, the presence of a downstream mobile stilling basin contributes to amplify the energy dissipation, as a hydraulic jump generally occurs in correspondence with the structure. In general, eco-friendly structures exhibit substantial similarities in terms of scour process, due to their geometry that is characterized by relatively small dimensions (low height and/or mild surface slopes). In particular, a detailed analysis of the scour mechanism can be conducted in the case for which the structure is stable. Therefore, incipient movement conditions of the stones constituting the different eco-friendly structure typologies has received considerable attention (Parker et al., 1982; Whittaker and Jaggi, 1986; Robinson et al., 1997; Hoffmans, 2010). Furthermore, the main parameters influencing the scour phenomenon (tailwater, stilling basin material and geometry, protection sills, etc.) were carefully analyzed in order to understand the dynamics of the erosive mechanism both in clear water and live bed conditions (Pagliara and Palermo, 2011; Oertel et al., 2011; Pagliara et al., 2012b). The previously mentioned studies on block ramps were useful to understand the similarities and differences among other several low-head structures, as shown by Pagliara and Mahmoudi Kurdistani (2015) and Pagliara et al. (2016). In particular, stepped gabion weirs exhibit similarities with both stepped chutes and block ramp. They are characterized by different flow regimes: skimming, nappe and transition flow regimes. In addition, these different flow regimes deeply influence the scour mechanism in the downstream stilling basin. Pegram et al. (1999), Peyras et al. (1992) and Pagliara and Palermo (2013) analyzed this structure typology in detail. In particular, Pagliara and Palermo (2013) focused on the scour mechanism occurring downstream of stepped gabion weirs and rock grade control structures classifying the onset conditions of the different flow regimes (see also Rajaratnam, 1990; Essery and Horner, 1978; Peyras et al. 1992; Ohtsu et al. 2004). Namely, Pagliara and Palermo (2013) analyzed both the hydraulics and the scour process downstream of different stepped gabion weir configurations. They studied four structure configurations: permeable and impermeable isolated structures and structures with both permeable and impermeable upstream filtering layer having the same height of the stepped gabion weir. Their analysis was conducted for constant discharges up to the equilibrium scour conditions. Similarly, rock sills share similarities with other low-head hydraulic structures, including rock chutes, W-weirs, J-hook, etc. In particular, rock sills can assume different shapes according to their hydraulic functioning and the location in which they are inserted. Detailed studies on this type of structure were conducted by Bhuiyan et al. (2007) and Scurlock et al. (2011). They proved that 3D rock structures can substantially improve fish habitat and, at the same time, they can dissipate a relevant amount of flow energy. Therefore, in this study two different types of rock sills were taken into consideration, i.e., straight rock sills (which are common in river restoration projects) and curved rock sills (which can be considered representative of a larger variety of 3D low-head structures). The aim of the present paper is to analyze the hydraulics and the scour evolutions in the presence of both stepped gabion weirs and rock sills under unsteady flow conditions. In addition, this paper aims to answer to the following questions: 1) Under unsteady flow conditions, is it always correct to select the peak discharge to evaluate the maximum scour depth, using relationships valid for steady flow conditions? 2) If the answer to the previous question is negative (as it will be shown), is there a minimum duration of the unsteady flow event that can cause the same scour features of a steady event characterized by a constant discharge equal to the peak discharge? Equivalently, when should the peak discharge of the unsteady event occur to obtain the same maximum scour depth of a steady event with constant discharge equal to peak discharge? To answer these two questions, experiments were conducted for different peak discharges varying both the time steps for the discharge increase/decrease and the total duration of the test. Furthermore, in order to minimize scale effects, experimental tests were conducted in two different channels and with different cohesionless materials. This analysis focused on the scour process evolution due to different flow conditions. It was experimentally proven that there is a significant similarity between the tested structures, i.e., it was shown that for certain inflow conditions, the non-dimensional time to reach the equilibrium configuration is essentially the same for all the low-head structure geometries tested in the present study. This implies that the answer to the first question is โ€œnoโ€ (as it is valid for certain inflow conditions and not always), whereas the answer to the second question is โ€œyesโ€. Finally, an applicative example is provided in order to better illustrate the proposed methodology. 2. Experimental Facilities: Two dedicated models were built at the hydraulic laboratory of the University of Pisa. Namely, experimental tests regarding stepped gabion weirs were conducted in channel 1, which is characterized by the following geometric characteristics: 0.30 m wide, 0.60 m deep and 6 m long. Whereas experimental tests with rock sills were conducted in channel 2 (0.50 m wide, 0.90 m deep and 8 m long). The stepped gabion weir was made of uniform rounded stones whose median diameter was D50=12 mm. Stones were kept together using a wide mesh (1 cm x 1 cm). The structure was built in layers (Figure 1) with step width ws and step height hs equal to 51.3 mm. The total structure height was H=154 mm. A filtering layer, having the same height of the structure, was located upstream and constructed using the same material as the downstream stilling basin in channel 1 (d50=4.78 mm, d90=5.7 mm, non-uniformity coefficient =1.2 and density s=2645 kg/m3, with dxx diameter of the stilling basin material for which xx% is finer). In addition, the upstream filtering layer was made impermeable by using a covering steel sheet in order to simulate a real configuration that usually occurs in rivers when an upstream sediment layer becomes impermeable due to the presence of silt and clay intrusion between grains. Furthermore, two rock sill typologies were tested, i.e., straight and curved rock sill. Rock sills were made of crushed stones layers and shaped in such a way that the final configuration was either straight or curved. Namely, the median diameter D50 of the stones constituting the sill was equal to 4.65 cm (non-uniformity coefficient =1.3 and density s=2450 kg/m3), whereas the stilling basin material adopted in channel 2 was much finer that than of Channel 1 (d50=2 mm, non-uniformity coefficient =1.22 and density s=2214 kg/m3). The choice to adopt two different stilling basin materials allowed to establish that there is no influence of the bed material size on the scour evolution. Furthermore, also scale effects can be considered negligible as the kinetics of the erosive process was found essentially the same in both the channels and for all of the different tested structures. The curved rock sill was characterized by a non-dimensional curvature R/B=0.5, where R is the curvature radius of the sill and B the width of the channel. Figure 2 illustrates the rock sills simulated in channel 2. Experimental tests were conducted for different hydraulic conditions, i.e., different discharges Q and different downstream tailwater levels h0. In particular, the downstream tailwater level was not controlled, therefore for each discharge different tailwater levels occurred. Furthermore, the inflow discharges varied between 5 and 10 l/s for stepped gabion weirs and between 10 to 15 l/s for both curved and straight rock sills. Preliminary tests were conducted under constant discharge in order to obtain the references values of the main scour characteristics, i.e., zmax (maximum scour hole depth) and ls (axial scour hole length). Namely, selected tests were repeated two times for all the different tested structures in order to verify the accuracy of the measurements and the repeatability of the results. They allowed to assess that no significant differences in terms of both scour geometry evolution and equilibrium configuration can be detected under the same hydraulic conditions. The scour evolution was carefully surveyed and the values of the maximum scour depths at different instants zmax(t) were collected. More specifically, the scour hole profiles were recorded optically with a CCD camera and the image acquisition was controlled with a programmable timing unit to ensure an accurate time control from the test start (a similar methodology was adopted by Unger and Hager, 2006, and Pagliara et al., 2008, for scour evolution due to plunging jets). In the meantime, a sequence of high definition pictures was taken at different instants. In addition, at selected instants, the maximum scour depth was measured by using a 0.1 mm precise point gauge fitted with a 10 mm circular plate at its lower end. The combination of direct measurements of the maximum scour depth and the analysis of the images sequence allowed to reach an overall precision of ยฑ1 mm. The same experiments were repeated varying the inflow conditions. Namely, the peak discharge Qmax of each experimental test was the same of the reference tests but it was reached by steps. For example, for stepped gabion weirs, the non-dimensional hydrograph reported in Figure 3 was adopted, in which Q(t) is the discharge at the i-step. Namely, a triangular shaped hydrograph was selected to simulate the unsteady inflow conditions. The total duration of the discharge decreasing phase was double of the duration of the discharge increasing phase. In addition, the increasing phase was characterized by 8 discharge increase steps, whereas, the decreasing phase by 16 steps. Each Q(t) was kept constant for a certain time step t. The minimum t was set at tmin=1 minute (in order to simulate and almost continuous discharge variation). Therefore, the same test was repeated for different t time steps (i.e., t=ntmin, where n is a natural number varying between 1 and 12), but with the same peak discharge Qmax, in order to analyse the effect of this parameter on the scour depth evolution. The total duration tf of each experimental test varied according to the selected t. For both straight and curved rock sills, the methodology adopted was essentially the same. But, in order to tests the effect of the selected methodology to reach the peak discharge on the scour evolution kinetics, the number of steps of the increasing phase of the hydrograph was set at 10 or 15, therefore the number of steps in the decreasing phase was set at 20 or 30, respectively. 3. Results and discussion: 3.1 Hydraulics of low-head structures The three different flow regimes of stepped gabion weirs are quite similar to those occurring on stepped chutes: Nappe Flow, Transition Flow and Skimming Flow regime. The Nappe Flow regime is characterized by flow plunging on the successive step (Figure 1a). The Skimming Flow regime (Figure 1c) is characterized by a coherent flow structure streaming on a pseudo-bottom. Horizontal axial vortexes occur between the steps and prominent flow re-circulation takes place below the pseudo-bottom. This last regime also occurs on block ramps. For intermediate flow characteristics, the Transition Flow regime takes place (Figure 1b). Pagliara and Palermo (2013) classified the three flow regimes for the stepped gabion weirs and compared the onset hydraulic conditions of Skimming Flow regime with those relative to stepped chutes. Authors showed that significant similarities can be detected, even if they are peculiar structures characterized by two elements: the presence of the downstream mobile stilling basin, which contributes to vary the downstream conditions, and the structure permeability. In particular, the presence of a downstream hydraulic jump (either partially submerging the structure or completely located in the stilling basin) is an important element to be taken into consideration because of its effect on the filtration regime in the structure itself. Therefore, Pagliara and Palermo (2013) proposed a classification of the different flow regimes mainly depending on two non-dimensional parameters: hs/k and h0/H, where k is the critical depth. But experimental tests showed that the effect of h0/H on the flow regime is less significant if compared to that due to the parameter hs/k. Thus, the key factor for the onset of the different flow regimes occurring in correspondence with this structure typology is represented by the parameter hs/k, i.e., the Nappe Flow regime occurs for hs/k >1.5, the Transition Flow regime occurs for 1.1<hs/kc<1.5 and the Skimming Flow regime occurs for The hydraulics of rock sills is quite similar to that of grade-control structures and it was exhaustively described by Bormann and Julien (1991). As the flow from the rock sill enters the tailwater a jet forms and the scour phenomenon occurs. Furthermore, according to the discharge, a vortex formation can take place in the scour hole. The applied shear stresses exceed the critical shear stress, starting to remove sediment from the scour hole and to transport them downstream. When the discharge increases, the vortex inside the scour hole shifts further downstream, partially flattening the formed ridge. It is worth noting that sediment forming the ridge can be transported both upstream and downstream according to the tailwater level. In fact, the tailwater level contributes to vary the impinging jet characteristics, as the jet diffusion length increases with the tailwater, eventually forming a reverse roller. This mechanism is at the base of the peculiar scour evolution behavior under unsteady flow conditions, i.e., the peak rate of scour do not always occur in correspondence with the peak discharge. In fact, in many tests the maximum scour depth occurred during the descending phase of the discharge. During this last phase, the discharge decreases, thus the jet impingement region shifts upstream. At the same time, the ridge, which has been flattened during the peak discharge, allows for sediment transport downstream. The eroded material is then transported downstream from the scour hole thus contributing to increase again the ridge height. This process lasts up to when the hydrodynamic force exerted on the particles no longer remove them from the scour hole. Based on the previous observations, it appears evident that, for longer duration of the unsteady event, the differences between the scour mechanisms occurring under constant discharge (equal to the peak discharge) and under unsteady flow conditions reduces. 3.2 Scour depth evolution The scour evolution was carefully analyzed for all the tested structures. Figure 4 and Figure 5 illustrate the maximum scour depth variation with time, i.e., zmax(t) as function of the discharge Q(t) for stepped gabion weirs and rock sills, respectively. In particular, Figure 4a reports zmax(t) as function of Q(t) for maximum peak discharge Qmax=5 l/s, whereas Figures 4b-c-d-e-f shows the same for tests with peak discharges equal to Qmax= 6, 7, 8, 9 and 10 l/s, respectively. Similarly, Figure 5a-b shows zmax(t) vs Q(t) for maximum peak discharges Qmax=10 l/s and 15 l/s for curved rock sills, whereas Figure 5c-d shows the same for straight rock sills. In each graph, the evolution of the maximum scour depth is reported for different t time step durations adopted for both the increasing and decreasing discharge phases. Finally, a point symbolizing the maximum scour depth measured in the corresponding reference tests (test with constant discharge equal to Qmax) is reported in each graph (zmax; Qmax=cost), in order to compare the maximum scour depth values obtained under steady flow conditions with those relative to unsteady flow conditions. For each simulated hydrograph, it is evident that the duration of the time step t (i.e., equivalently the total duration of the unsteady event) strongly affects the scour hole morphology and, in particular, the maximum scour depth zmax. Namely, for low t (< 5 minutes), generally, the maximum scour depth is significantly less than that obtained in the corresponding reference test. Conversely, for t > 5 minutes, generally, the maximum scour depth is comparable with that obtained in the reference test. In other words, in the tested conditions and for all the tested structures, it seems that if t>t*[, where ][][t]*[= 5 minutes, no substantial differences occur in terms of scour hole ] features, under both steady and unsteady flow conditions. This evidence furnishes the answer to question 1) reported in the introduction. Namely, it is not always correct to assume the peak discharge to evaluate the maximum scour depth using relationships valid for steady flow conditions, as the maximum scour depth can be significantly smaller, according to the different inflow conditions. In addition, under unsteady flow conditions, the maximum scour depth could not occur in correspondence with the peak discharge for these tests, as clearly shown in Figure 4 and Figure 5. More specifically, especially for relatively low peak discharges, zmax could occur during the decreasing phase of the hydrograph, i.e., for Q(t)<Qmax. This phenomenon is mainly due to two reasons and it is quite similar for all the tested structures, even if some differences can be pointed out between stepped gabion weirs and rock sills. In particular, for stepped gabion weirs, under unsteady flow conditions, different flow regimes occur in the same test. In addition, the downstream hydraulic jump tends to shift downstream during the increasing phase of the hydrograph, whereas it shifts towards the structure during the decreasing phase. Therefore, the movable stilling basin bed is cyclically modeled. During the increasing discharge phase, the downstream dune is flattened by the hydraulic jump which shifts downstream, i.e., its contribution in limiting the scour hole evolution partially vanishes. Conversely, under higher peak discharge, (see Figure 4e-f) the maximum scour depth occurring during the test can be slightly bigger than that occurring at the end of the test. In particular, the maximum scour depth during the test takes place when the skimming flow regime occurs on the structure. Therefore, during the decreasing phase of the discharge, the hydraulic jump shifts towards the structure and contributes to erode the upstream part of the scour hole transporting sediment downstream. The downstream part of the scour hole is partially replenished by the sediment eroded from the upstream part. In addition, for higher discharges, the time step duration t required to obtain the same maximum scour depth occurring in the reference test becomes slightly bigger. Nevertheless, as mentioned, for practical purposes t โ‰ฅ 5 minutes can be assumed as the minimum time step duration to get no differences in terms of maximum scour depth between unsteady and steady flow tests. Finally, especially for tests in which the peak discharge is high (therefore all the three flow regimes occur on the structure), three scour depth evolution kinetics can be detected: there is an initial phase (corresponding to nappe flow regime) in which the scour development is quite fast, followed by a phase in which the erosion kinetics reduces (corresponding to transition flow regime) and again by a phase in which the scour progress is fast (corresponding to skimming flow regime). This dynamic is essentially the same observed by Pagliara and Palermo (2013) under steady flow conditions. For rock sills, similar considerations can be done. Nevertheless, in this last case, the scour hole evolution is slightly different respect to the cases illustrated above. This is mainly due to the fact that rock sills do not protrude as high from streambed, therefore for low discharges, either the hydraulic jump does not even occur or it is a weak hydraulic jump with no roller. In addition, for higher discharge the hydraulic jump occurring downstream of the structure is generally submerged. Therefore, even if the kinetics of the erosive process (i.e., cyclic movement of the bed sediment) is essentially similar, some slight differences can be detected in terms of hydraulic functioning. Namely, the scour evolution dynamic appears more regular than in the presence of stepped gabion weirs, i.e., the eventual regression of the maximum scour depth, which was due to the hydraulic jump shifting in the stilling basin, is very slight or completely absent (see Figure 5a-d). In order to generalize the previous observations, the scour depth evolution was reported in graphs with non-dimensional variables. Namely, the experimental data were reported in graphs zmax(t)/zmax vs Q(t)/Qmax, where zmax(t) is the maximum scour depth value in each test. In addition, also the interval duration relative to each discharge increment was made non-dimensional as follows: kn t d g T / 50 0.5 [ ] [(1) ] where T is the non-dimensional time step, g is the acceleration due to gravity, k is the critical depth relative to the peak discharge and n is a natural number varying from 1 to 12. It means that for unsteady flow tests with same peak discharge, the non-dimensional time step duration is constant, as it depends only on k. In fact, according to what specified in section 2, tests were conducted for different t=ntmin, therefore whatever the value of the coefficient n is, t/n=tmin. All the previous observations made for higher discharges applies for lower T, as shown in the Figure 6 (stepped gabion weirs) and Figure 7 (rock sills). In particular, from Figure 6e-f it can be easily noted that, for lower T values, the three different flow regimes occurring during the increasing discharge phase are clearly detectable. Furthermore, it is worth noting that the evolution of the maximum relative scour depth zmax(t)/zmax appears quite similar for all the tests for which n5. The previous analysis was further developed, considering the non-dimensional temporal evolution of the variable zmax(t)/zmax. Figure 8 (stepped gabion weirs) and Figure 9 report graphs in which the maximum relative scour depth zmax(t)/zmax is plotted against the non-dimensional time T, defined as follows: kn t d g T s 5 . 0 50 / (2) in which t is the time and n is the coefficient specified above. In order to compare the unsteady and steady flow condition data, for steady flow conditions n can be reasonably assumed equal to 1. In fact, as specified above, n=1 corresponds to 1 minute time step variation of the discharge under unsteady flow conditions, i.e., almost a continuous variation of the discharge. In addition, for each test, the non-dimensional time Tpeak in which the peak discharge occurs is specified in the respective plots of Figure 8 and 9. The corresponding curve relative to the reference steady test is also reported in order to compare the behavior under steady and unsteady flow conditions. By observing the mentioned figures, it can be noted that, for all the structures tested, the trend of the evolution of the non-dimensional scour depths is quite similar for nโ‰ฅ5, thus confirming what discussed before, i.e., generally for nโ‰ฅ5, the equilibrium scour geometry due to unsteady flow conditions is essentially the same obtained under constant discharge equal to the peak discharge. In addition, the equilibrium configuration is reached for T varying between 3500 and 5000 according to the different peak discharges tested. Therefore, for all the tested structures, we can reasonably assume that, for practical applications, the final equilibrium condition is reached for T=Teq4250, i.e., an average value. Furthermore, the peak discharge is occurring for 2690<Tpeak<4373 for all the tested conditions. Also in this case, in order to furnish an indicative non-dimensional time for the peak discharge occurrence, we can assume the average value among those computed for all the tested structures, i.e., T* peak3200. Based on these observations, the second question proposed in the introduction can be answered. Namely, this study showed that there is a minimum duration of the unsteady flow event which allows to get the same equilibrium morphology occurring during an event characterized by a constant discharge equal to the peak discharge, i.e., n should be at least equal to 5. Equivalently, assuming n=5 and by applying Eq. (2), we can compute T* peak which is the minimum value of Tpeak in order to get the same maximum scour depth, under both steady and unsteady flow conditions, characterized by the same Qmax. In other words, if n=5 then Tpeak should be bigger than 3200 in order to get a similar equilibrium morphology under both steady and unsteady flow conditions. Nevertheless, it is worth noting that the proposed results are based on specific structure configurations and selected hydrograph shapes. Therefore, the generalization of the proposed methodology to other structure typologies under different flow conditions will require further investigations. 3.3 Applicative example Letโ€™s apply the proposed methodology to a real case. Porebianka is a river in Poland in which there is a succession of block ramps. After a flood event, which caused significant erosion in correspondence with hydraulic structures and whose hydrograph is reported in Figure 10, the stilling basin morphology downstream of selected block ramps was surveyed. As it can be observed from Figure 10, the flood event lasted almost 120 hours, the peak discharge occurred after almost 24 hours from the beginning of the flood event and was equal to 90 m3[/s. Furthermore, the hydrograph shape is very similar to that simulated in our experimental ] tests, i.e., the duration of the decreasing discharge phase is almost double of the duration of the increasing discharge phase. In this river, after the flood event, very carefully measurements of the stilling basin morphology downstream of selected block ramps were taken. Therefore, we apply the proposed methodology to one of the selected block ramps. Namely, the selected block ramp is characterized by a bed slope S=0.0833. The ramp bed is made of uniform rounded rocks, whose the D50 is equal to 1.2 m. The material constituting the stilling basin is also uniform and has the following granulometric characteristics: d50=0.06 m and s=2200 kg/m3. In addition, at the toe of the ramp, the stilling basin is protected by a few rows made of the same rocks constituting the ramp, which act as a protection sill. The river width is B30 m and the geometry of the sections can be assumed rectangular. After the mentioned flood event, the maximum scour depth measured in the stilling basin was equal to 1.07 m. Pagliara and Palermo (2010) proposed a methodology and some useful relationships by which it is possible to estimate the maximum scour depth downstream of a block ramp both in the presence and in the absence of a protection rock sill downstream of the block ramp located in different spatial positions. It is worth mentioning that the equations proposed by the authors are valid under constant discharge flow conditions. Namely, for a protected stilling basin, Pagliara and Palermo (2008) proposed the following equation to estimate the maximum non-dimensional scour depth Zmax=zmax/h1, where h1 is the approaching flow depth at the ramp toe: a b c dZ eZ f F S Z 2 2 8 . 1 90 75 . 0 max 0.58 In the previous equation, Fd90=V1/(gd90)0.5 is the densimetric Froude number, where g=g[(s -)/] is the reduced acceleration, V1 is the average flow velocity at the ramp toe, s and are the channel bed sediment density and water density, respectively. and Zop are the non-dimensional longitudinal and vertical positions of the rock protection in the stilling basin. a, b, c, d, e and f are coefficients depending on the ramp slope S and rock sill position. For the selected ramp in the Porebianka river, Eq. (4) can be re-written as follows, by substituting the values of the coefficients, and Zop furnished by Pagliara and Palermo (2008): 912 . 0 58 . 0 1.8 90 75 . 0 max S Fd Z (4) Therefore, the maximum estimated scour depth zmax=Zmaxh1 can be easily calculated by applying Eq. (4). It is worth noting that, in order to evaluate zmax, the estimation of the parameter h1 is required. h1 can be estimated using the methodology proposed by Pagliara and Palermo (2010). For this applicative example, h1 was found to be equal to 0.673 m and, thus, zmax=1.11 m. By comparing the estimated (1.11 m) and measured (1.07 m) values of zmax, for practical purposes, it can be reasonably assumed that the flood event caused an equilibrium morphology which is essentially similar to that which would occur for a steady flow conditions, for which Qmax=90 m3/s. This implies that if we apply Eq. (2), we should get that peak (i.e., the estimated time at which the peak discharge Qmax occurs) should be less than 24 hours. Letโ€™s re-write Eq. (2) as follows: 0.5 50 * * / d g n k T t s peak peak Assuming a rectangular section, we have that the corresponding critical depth k for the peak discharge Qmax=90 m3/s is equal to 0.97 m. Therefore, by substituting T*peak=3200, s=2200 kg/m3[, ][][=1000 kg/m]3[, n=5, d] 50=0.06 m, k=0.97 m and g=9.81 m/s2 in Eq. (5), we obtain peak=18467 s, i.e., t*peak=5.13 hours, which is much less than 24 hours Therefore, the proposed methodology seems to be confirmed. 4. Conclusion: This paper analyzed the effect on maximum scour depth of the inflow conditions. Namely, the analysis was conducted in the presence of both stepped gabion weirs and rock sills. References tests, under steady flow conditions, were conducted in order to obtain the reference values of the maximum scour depth occurring in the downstream stilling basin. The same tests were repeated under unsteady flow conditions. The hydrograph adopted to simulate the unsteady flow conditions was characterized by a decreasing discharge phase whose duration is double of the increasing discharge phase. The total duration of the unsteady flow tests varied according to the duration of the selected time step t during which Q(t) was kept constant. The minimum tmin was set equal to 1 minute. The same unsteady flow tests were repeated for different values of t=ntmin, in order to determine for which minimum interval duration there are not significant differences between the unsteady flow tests and the corresponding steady flow tests in terms of equilibrium scour morphology. It was experimentally shown that for nโ‰ฅ5, the equilibrium scour hole characteristics are essentially the same under both steady and unsteady flow conditions. Furthermore, this paper showed that, under unsteady flow conditions, is not always correct to assume the peak discharge to evaluate the maximum scour depth using relationships valid for steady flow conditions. Finally, it was shown that there is a minimum time for the peak discharge to occur in order to get the same equilibrium morphology of an event with constant discharge equal to peak discharge. Further investigations are required to generalize the proposed results, by validating them in the presence of other structure typologies and under different flow conditions. Bormann, E., Julien, P. Y. (1991). Scour downstream of grade control structures. Journal of Hydraulic Engineering, 117(5), 579-594, doi: 10.1061/(ASCE)0733-9429(1991)117:5(579). Breusers, H.N.C., Raudkivi, A.J. (1991). Scouring. IAHR Hydraulic structures design manual 2, Balkema: Rotterdam, the Netherlands. Bhuiyan, F., Hey, R. D., Wormleaton, P. R. (2007). Hydraulic evaluation of w-weir for river restoration. Journal of Hydraulic Engineering, 133 (6), 596-609, doi: 10.1061/(ASCE)0733-9429(2007)133:6 Dโ€™Agostino, V., Ferro, V. (2004). Scour on alluvional bed downstream of grade-control structures. Journal of Hydraulic Engineering, 130(1), 1-14, doi: 10.1061/(ASCE)0733-9429(2004)130:1(24). Essery, I. T. S., Horner M. W. (1978). The hydraulic design of stepped spillways. CIRIA, Report No. 33, London, UK. Hoffmans, G. J. C. M. (1998). Jet scour in equilibrium phase. Journal of Hydraulic Engineering, 124(4), 430-437. Hoffmans, G. J. C. M. (2010). Stability of stones under uniform flow. Journal of Hydraulic Research, 136(2), 129-136, doi: 10.1061/(ASCE)0733-9429(2010)136:2(129). Hoffmans, G. J. C. M., Verheij, H. J. (1997). Scour manual. Balkema: Rotterdam, the Netherlands. Mason, P. J., Arumugam, K. (1985). Free jet scour below dams and flip buckets. Journal of Hydraulic Engineering, 111(2), 220-235, doi: 10.1061/(ASCE)0733-9429(1985)111:2(220). Oertel, M., Peterseim, S., Schlenkhoff, A. (2011). Drag coefficients of boulders on a block ramp due to interaction processes. Journal of Hydraulic Research, 49(3), 372-377, doi: 10.1080/ Ohtsu, I., Yasuda, Y., Takahashi, M. (2004). Flow characteristics of skimming flows in stepped channels. Journal of Hydraulic Engineering, 130(9), 860-869, doi: 10.1061/(ASCE)0733-9429(2004)130:9 Pagliara, S., Mahmoudi Kurdistani, S. (2015). Clear water scour at J-Hook Vanes in channel bends for stream restorations. Ecological Engineering, 83, 386-393, doi: 10.1016/j.ecoleng.2015.07.003. Pagliara, S., Mahmoudi Kurdistani, S., Palermo, M., Simoni, D. (2016). Scour due to rock sills in straight and curved horizontal channels. Journal of Hydro-Environment Research, 10, 12-20, doi: Pagliara, S., Palermo, M. (2008). Scour control downstream of block ramps. Journal of Hydraulic Engineering, 134(9), 1376-1382, doi: 10.1061/(ASCE)0733-9429(2008)134:9(1376). Pagliara, S., Palermo, M. (2010). Influence of tailwater depth and pile position on scour downstream of block ramps. Journal of Irrigation and Drainage Engineering, 136(2), 120-130, doi: 10.1061/ Pagliara, S., Palermo, M. (2011). Effect of stilling basin geometry on clear water scour morphology downstream of a block ramp. Journal of Irrigation and Drainage Engineering, 137(9), 593-601, 137 (9), 593-601, doi: 10.1061/(ASCE)IR.1943-4774.0000331. Pagliara, S., Palermo, M. (2013). Rock grade control structures and stepped gabion weirs: scour analysis and flow features. Acta Geophysica, 61(1), 126-150, doi: 10.2478/s11600-012-0066-0. Pagliara, S., Hager, W.H., Unger, J. (2008). Temporal evolution of plunge pool scour. Journal of Hydraulic Engineering, 134(11), 1630-1638, doi: 10.1061/(ASCE)0733-9429(2008)134:11(1630). Pagliara, S., Palermo, M., Carnacina, I. (2012b). Live-bed scour downstream of block ramps for low densimetric Froude numbers. International Journal of Sediment Research, 27(3), pp. 337-350, doi: Pagliara, S., Palermo, M., Roy, D. (2012a). Stilling basin erosion due to vertical crossing jets. Journal of Hydraulic Research, 50(3), 290-297, doi: 10.1080/00221686.2012.669534. Pagliara, S., Roy, D., Palermo, M. (2010). 3D plunge pool scour with protection measures. Journal of Hydro-Environment Research, 4(3), 225-233, doi: 10.1016/j.jher.2009.10.014. Parker, G., Klingeman, P. C., McLean D. G. (1982). Bedload and size distribution in paved gravel-bed streams. Journal of the Hydraulic Division, 108(4), 544-571, doi: 10.1061/(ASCE)0733-9429(1983) Pegram, G., Officer, A., Mottram, S. (1999). Hydraulics of skimming flow on modeled stepped spillways. Journal of Hydraulic Engineering, 125(5), 500-510, doi: 10.1061/(ASCE)0733-9429(1999)125:5(500). Peyras, L., Royet, P., Degoutte, G. (1992). Flow and energy dissipation over stepped gabion weirs. Journal of Hydraulic Engineering, 118(5), 707-717, doi: 10.1061/(ASCE)0733-9429(1992)118:5(707). Rajaratnam, N. (1981). Erosion by plane turbulent jets. Journal of Hydraulic Research, 19(4), 339-358, doi: 10.1080/00221688109499508. Rajaratnam, N. (1990). Skimming flow in stepped spillways. Journal of Hydraulic Engineering, 116(4), 587-591, doi: 10.1061/(ASCE)0733-9429(1990)116:4(587). Rajaratnam, N., Macdougall, R. K. (1983). Erosion by plane wall jets with minimum tailwater. Journal of Hydraulic Engineering, 109(7), 1061-1064, doi: 10.1061/(ASCE)0733-9429(1983)109:7(1061). Robinson, K. M., Rice, C. E., Kadavy K. C. (1997). Design of rock chutes. Transactions of the American Society of Agricultural Engineers, 41(3), 621-626. Scurlock, S. M., Thornton, C. I., Abt, S. R. (2012). Equilibrium scour downstream of three-dimensional grade-control structures. Journal of Hydraulic Engineering, 138(2), 167-176, doi: 10.1061/(ASCE) Unger, J., Hager, W. H. (2006). Temporal flow evolution of sediment embedded circular bridge piers. Proceedings of River Flow 2006, 729-739. Whittaker, W., Jaggi, M. (1986). Blockschwellen. ETH, Zurich, VAW Mitteilungen 91. Veronese, A. (1937). Erosioni di fondo a valle di uno scarico. Annali Lavori Pubblici, 75(9), List of figures Figure 1 Diagram sketch of a stepped gabion weir: (a) Nappe Flow, (b) Transition Flow, and (c) Skimming Flow regime; (d) picture of a stepped gabion weir. Figure 2 (a) Diagram sketch and (c) picture of a straight rock sill; (b) Diagram sketch and (d) picture of a curved rock sill. The black arrow indicates the flow direction. Figure 3 Non-dimensional hydrograph adopted for tests with stepped gabion weirs. Figure 4 Variation of scour depth evolution for different time step discharge increase along with the indication of the coordinates of maximum scour depth occurring for a steady test with discharge equal to Qmax. Peak discharge equal to: (a) Qmax=5 l/s; (b) Qmax=6 l/s; (c) Qmax=7 l/s; (d) Qmax=8 l/s; (e) Qmax=9 l/s; (f) Qmax=10 l/s. Figure 5 Variation of scour depth evolution for different time step discharge increase along with the indication of the coordinates of maximum scour depth occurring for a steady test with discharge equal to Qmax. Peak discharge equal to: (a) Qmax=10 l/s and (b) Qmax=15 l/s (curved rock sill); (c) Qmax=10 l/s and Qmax=15 l/s (straight rock sill). Figure 6 Variation of maximum non-dimensional scour depth with non-dimensional discharge for different n values and non-dimensional time step discharge increase equal to: (a) T=546; (b) T=484; (c) T=436; (d) T=400; (e) T=370; (f) T=345 Figure 7 Variation of maximum non-dimensional scour depth with non-dimensional discharge for different n values and non-dimensional time step discharge increase equal to: (a) T=270 and (b) T=205 (curved rock sill); (c) T=270 and (d) T=205 (straight rock sill). Figure 8 Variation of maximum non-dimensional scour depth with non-dimensional time T for different n values and non-dimensional time step discharge increase along with the indication of the non-dimensional time T in which Qpeak is reached: (a) T=546; (b) T=484; (c) T=436; (d) T=400; (e) T=370; (f) T=345 Figure 9 Variation of maximum non-dimensional scour depth with non-dimensional time T for different n values and non-dimensional time step discharge increase along with the indication of the non-dimensional time T in which Qpeak is reached: (a) T=270 and (b) T=205 (curved rock sill); (c) T=270 and (d) T=205 (straight rock sill). Figure 1 Diagram sketch of a stepped gabion weir: (a) Nappe Flow, (b) Transition Flow, and Figure 2 (a) Diagram sketch and (c) picture of a straight rock sill; (b) Diagram sketch and (d) Figure 4 Variation of scour depth evolution for different time step discharge increase along with the indication of the coordinates of maximum scour depth occurring for a steady test with discharge equal to Qmax. Peak discharge equal to: (a) Qmax=5 l/s; (b) Qmax=6 l/s; (c) Figure 5 Variation of scour depth evolution for different time step discharge increase along with the indication of the coordinates of maximum scour depth occurring for a steady test with discharge equal to Qmax. Peak discharge equal to: (a) Qmax=10 l/s and (b) Qmax=15 l/s Figure 6 Variation of maximum non-dimensional scour depth with non-dimensional discharge for different n values and non-dimensional time step discharge increase equal to: (a) Figure 7 Variation of maximum non-dimensional scour depth with non-dimensional discharge for different n values and non-dimensional time step discharge increase equal to: (a) Figure 8 Variation of maximum non-dimensional scour depth with non-dimensional time T for different n values and non-dimensional time step discharge increase along with the indication of the non-dimensional time T in which Qpeak is reached: (a) T=546; (b) T=484; Figure 9 Variation of maximum non-dimensional scour depth with non-dimensional time T for different n values and non-dimensional time step discharge increase along with the indication of the non-dimensional time T in which Qpeak is reached: (a) T=270 and (b)
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Plot pรฅ svenska SV,EN lexikon Tyda ร–versรคttning 'plot point' โ€“ Ordbok svenska-Engelska Glosbe This plot suggests that there appears to be an extreme value in each quadrant. A contour plot is useful in finding the approximate location of critical points. Never Confuse the Key Event and the First Plot Point in Your Book Again! - Helping Writers Become Authors. example. Calculus: Fundamental Theorem of Calculus Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Plotting Points on a Graph or XY-plane - ChiliMath Plotting Points on a Graph In this tutorial, I have prepared eight (8) worked out examples on how to plot a point in a Cartesian plane (named in honor of French mathematician Renรจ Descartes). To plot a point, we need to have two things: a point and a coordinate plane. Plotting Points Using Spherical Coordinates: Dynamic & Modifiable This page will help you to do that. In the box to the right, type in some x,y points like this: (1,2) or (1,2) (-4,3) (10,-6) Type in the ordered pair or pairs to plot here: Set the axes ranges of your plot: 2019-09-29 Matplotlib Plotting Plotting x and y points. Pin pรฅ Undervisning - Pinterest The most basic plotting skill it to be able to plot x,y points. This page will help you to do that. In the box to the right, type in some x,y points like this: Plot Map Coordinates - Plotting Point A To find the coordinates of any given location, start at the bottom left of the map, and using the map's grid lines, search to the right until you find the closest easting line to the west of your targetโ€™s location in this case 88. Plotting of points in matplotlib with Python There is a method named as โ€œ scatter (X,Y) โ€ which is used to plot any points in matplotlib using Python, where X is data of x-axis and Y is data of y-axis. Free algebra 1 MATLAB: How to choose to show only part of plot/legend. legendplot If you start by only plotting one of the max points by adding a second argument to the find Writers, it's time to discuss one of my favorite storytelling topics: plot structure. Wallmantra tracking 2010 โ€” I have over 700 gps measured points so writing them by hand seems kind of impossible. Or just plot out the coordinates on a drawing? Quote Back plotting of NC code function (NC Viewer). O. Machine paths NC simulation. O Move pierce points functions. O. Automatic error correction features. 0. 5. 5. Plotting lists of (x, y) points. We've seen that Mathematica uses parentheses for grouping and square brackets in functions. Visual arts class For example: using Plots using Random Random.seed!(123) plot(rand(10), rand(10), seriestype = :scatter, group = rand(0:1,10), title = "Some random points"). Activity Overview. Students determine that the inverse of the exponential function is the natural log function by plotting the inverse of exponential solution points. Explore math with our beautiful, free online graphing calculator. Symmetry of polar graphs precalculus polar coordinates and complex numbers. Free algebra 1 MATLAB: How to choose to show only part of plot/legend. MAKE ME A ROUTE. FILTER RESULTS Close search results to plot a new route Geokodningspunkter. Engelska grammatik genitiv ansvaret kryssordvolvo v40 konkurrenterรคldreomsorg stockholm kdvรฅrdcentral tensta alleneurovetenskap utbildning skรถvdebokfรถringskonto fordonsskattthe proposal 2 Basic Linear Graphing Skills Practice Workbook: Plotting Points Se hela listan pรฅ stat.ethz.ch coordinate vectors of points to plot.
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Lines - (Non-Euclidean Geometry) - Vocab, Definition, Explanations | Fiveable from class: Non-Euclidean Geometry In geometry, a line is defined as a straight one-dimensional figure that extends infinitely in both directions, having no endpoints. Lines are fundamental elements in geometry, serving as the basis for various constructions and theorems. The concept of lines is crucial in understanding geometric relationships, especially in the context of both Euclidean and non-Euclidean geometries. congrats on reading the definition of Lines. now let's actually learn it. 5 Must Know Facts For Your Next Test 1. In Euclidean geometry, through any two distinct points, there is exactly one line that can be drawn. 2. In elliptic geometry, lines are represented as great circles on a sphere, which means that they eventually intersect. 3. Lines in non-Euclidean geometries like elliptic geometry challenge traditional notions of parallel lines, as no two lines are parallel in this context. 4. The properties of lines, such as their lengths and angles with other lines, can vary significantly between Euclidean and non-Euclidean geometries. 5. Understanding the nature of lines helps in visualizing complex geometric relationships and in developing proofs based on axioms. Review Questions โ€ข How does the definition of a line differ between Euclidean and non-Euclidean geometries? โ–ก In Euclidean geometry, a line is defined as a straight path extending infinitely in both directions without curvature. However, in non-Euclidean geometries like elliptic geometry, lines are represented by great circles on a sphere. This means that while Euclidean lines never intersect unless at endpoints, elliptic lines always intersect at two points, illustrating fundamental differences in how space is perceived in these geometries. โ€ข Discuss how the concept of parallel lines changes when considering elliptic geometry versus Euclidean geometry. โ–ก In Euclidean geometry, parallel lines are defined as lines that never meet regardless of how far they extend. However, in elliptic geometry, this notion does not hold because there are no parallel lines; any two lines will eventually intersect. This difference highlights the unique properties of elliptic geometry where the behavior of lines fundamentally alters our understanding of parallelism and spatial relationships. โ€ข Evaluate the implications of different line definitions on geometric proofs in both Euclidean and elliptic geometries. โ–ก The varying definitions of lines between Euclidean and elliptic geometries have significant implications for geometric proofs. In Euclidean geometry, proofs often rely on the existence of parallel lines and congruent segments to establish relationships. Conversely, in elliptic geometry where no parallels exist, proofs must adapt to accommodate the fact that all lines intersect. This shift requires a deeper understanding of the properties inherent to each type of geometry and leads to different methodologies for establishing congruences and relationships. "Lines" also found in: เธขเธ‰ 2024 Fiveable Inc. All rights reserved. APเธขเธŽ and SATเธขเธŽ are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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Which of the following is not an application of binary search? Q. Which of the following is not an application of binary search? A. To find the lower/upper bound in an ordered sequence B. Union of intervals C. Debugging D. To search in unordered list Answerยป D. To search in unordered list
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On selfing pea plants having round and yellow seeds (RrYy), tot... | Filo Question asked by Filo student On selfing pea plants having round and yellow seeds (RrYy), total 480 seeds were obtained. Out of 480 seeds, how many would be round and yellow? Only One Correct Answer a. 104 b. 139 c. 70 d. 270 Not the question you're searching for? + Ask your question In Mendel's dihybrid cross, the ratio of different phenotypic traits obtained by selfing the pea plants of character heterozygous round yellow seeds is as follows: The ratio for the dihybrid cross between a heterozygous round and yellow pea plants is 9:3:3:1 which is shown here: Round-Yellow : Round-Green : Wrinkled-Yellow : Wrinkled-Green Since out of 16 seeds 9 are round-yellow, therefore out of 480 seeds, The ratio of round-yellow plants: Found 5 tutors discussing this question Discuss this question LIVE for FREE 9 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Practice more questions on Plant Physiology View more Students who ask this question also asked View more Stuck on the question or explanation? Connect with our Biology tutors online and get step by step solution of this question. 231 students are taking LIVE classes Question Text On selfing pea plants having round and yellow seeds (RrYy), total 480 seeds were obtained. Out of 480 seeds, how many would be round and yellow? Only One Correct Answer Updated On Mar 30, 2022 Topic Plant Physiology Subject Biology Class Class 11 Answer Type Text solution:1 Video solution: 1 Upvotes 54 Avg. Video Duration 3 min
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What Is A Square Root Used For? (7 Real Life Applications) | jdmeducational (2024) Once you know how to calculate a square root, you might be curious to learn how to apply this concept in real life. Although square roots are used often in math, they also have applications in many other disciplines. So, what is a square root used for? Square roots are used in finance (rates of return over 2 years), normal distributions (probability density functions), lengths & distances (Pythagorean Theorem), quadratic formula (height of falling objects), radius of circles, simple harmonic motion (pendulums & springs), and standard deviation. Of course, square roots are not the only roots we can use. Cube roots, fourth roots, and other roots can also help us in various science and technology fields. In this article, weโ€™ll talk about what square roots are used for and how they fit into various equations and formulas. Weโ€™ll also give some examples to make the concepts clear. Letโ€™s get started (you can watch a video version of this article on YouTube). Having math trouble? Looking for a tutor? What Is A Square Root Used For? Square roots are used throughout mathematics and have applications in many disciplines, such as probability, statistics, physics, architecture, and engineering. Here are some uses of square roots in real life: โ€ข Finance (Rates Of Return Over 2 Years) โ€ข Normal Distributions (Probability Density Function) โ€ข Pythagorean Theorem (Lengths & Distances) โ€ข Quadratic Formula (Height Of Falling Objects) โ€ข Radius Of Circles With A Given Area โ€ข Simple Harmonic Motion (Pendulums & Springs) โ€ข Standard Deviation (Measuring the spread of data) Letโ€™s take a look at each one in turn, starting with finance. Square Roots In Finance In the field of finance, we can use square roots to find the rate of return on an asset over a time period with 2 units (for example, 2 years, 2 months, etc.) The formula for the annual rate of return over a 2 year time period is given by: where R is the annual rate of return, V[0 ]is the starting value, and V[2] is the value after 2 years. Example 1: Rate Of Return Of An Asset Over 2 Years Letโ€™s say that you buy a stock on January 1, 2020 for $100. You sell the stock on January 1, 2022 for $196. This means that: โ€ข V[0] = 100 (you bought the stock for $100) โ€ข V[2] = 196 (you sold the stock for $196) Since the time period was 2 years (January 1, 2020 to January 1, 2022), we can use the formula for annual rate of return over 2 years to get: โ€ข R = โˆš(V[2] / V[0]) โ€“ 1 โ€ข R = โˆš(196 / 100) โ€“ 1 โ€ข R = โˆš(1.96) โ€“ 1 โ€ข R = 1.4 โ€“ 1 โ€ข R = 0.4 As a decimal, R = 0.4 means an annual return of 40% (move the decimal 2 places to the right to convert a decimal to a percent). So, the stock returned 40% annually, which is a good investment. Example 2: Rate Of Return Of An Asset Over 2 Months Letโ€™s say that you buy a house on January 1, 2020 for $250,000. You sell the house on March 1, 2022 for $302,500. This means that: โ€ข V[0] = 250,000 (you bought the stock for $250,000) โ€ข V[2] = 302,500 (you sold the stock for $302,500) Since the time period was 2 months (January 1, 2020 to March 1, 2020), we can use the formula for annual rate of return over 2 months to get: โ€ข R = โˆš(V[2] / V[0]) โ€“ 1 โ€ข R = โˆš(302,500 / 250,000) โ€“ 1 โ€ข R = โˆš(1.21) โ€“ 1 โ€ข R = 1.1 โ€“ 1 โ€ข R = 0.1 As a decimal, R = 0.1 means a monthly return of 10% (move the decimal 2 places to the right to convert a decimal to a percent). So, the house returned 10% monthly, which is a good investment. More generally, we can use the nth root to find the rate of return over a time period with n units. The formula is given by: where R is the rate of return per time period, V[0 ]is the starting value, and V[n] is the value after n time periods. Square Roots In Normal Distributions The normal distribution also uses a square root, although it is not easy to see from the graph (which has the shape of a symmetric bell curve). The square root in a normal distribution can be seen in its pdf (probability density function), which is given by: Without square roots, we could not define the function that gives us a normal distribution curve. This distribution is used throughout mathematics, science, medicine, psychology, and other fields. Square Roots & The Pythagorean Theorem From the Pythagorean Theorem, we can use square roots to find distances and lengths of sides of triangles in 2 dimensions (or 3 dimensions). This can be useful in all sorts of applications, such as: โ€ข Architecture & Engineering (finding lengths of trusses to hold up bridges and buildings). โ€ข Carpentry & Construction (finding lengths of sides of right triangles for diagonal supports). โ€ข Graphics (finding distances in a 2D or 3D grid system for movies or video games). Remember that the Pythagorean Theorem applies to a right triangle (one with a 90 degree angle), and is given by the formula: where a and b are the legs (two shorter sides) and c is the hypotenuse (the longest side, across from the right angle) in a right triangle. To solve for the hypotenuse, we simply take the square root of both sides of the equation to get: To solve for one of the other sides (letโ€™s say a), we subtract b^2 from both sides and take the square root of both sides: โ€ข a^2 + b^2 = c^2 โ€ข a^2 = c^2 โ€“ b^2 โ€ข a = โˆš(c^2 โ€“ b^2) Example 1: Length Of The Hypotenuse Of A Right Triangle Letโ€™s say that we have a right triangle with sides 6 feet and 8 feet. We want to find the length of the hypotenuse (the longest side) to find out how long a diagonal support should be. Using the Pythagorean Theorem with a = 6 and b = 8, we get: โ€ข โˆš(a^2 + b^2) = c โ€ข โˆš(6^2 + 8^2) = c โ€ข โˆš(36 + 64) = c โ€ข โˆš(100) = c โ€ข 10 = c So, the diagonal support should be 10 feet long. (Note: this 6-8-10 right triangle is just a multiple of a 3-4-5 right triangle โ€“ that is, they are similar.) Example 2: Length Of A Leg Of A Right Triangle Letโ€™s say that we have a right triangle with one leg that is 7 feet long and a diagonal (hypotenuse) that is 13 feet long. We want to find the length of the other leg. Using the Pythagorean Theorem with a = 7 and c = 13, we get: โ€ข a = โˆš(c^2 โ€“ b^2) โ€ข a = โˆš(13^2 โ€“ 7^2) โ€ข a = โˆš(169 โ€“ 49) โ€ข a = โˆš(120) So, the other leg has a length of โˆš120 or 2โˆš30 feet. Remember that the side lengths of some special triangles will also use square roots for some of their side lengths. For example, a 45-45-90 triangle (right isosceles) will have side lengths in the ratio 1-1-โˆš2. A 30-60-90 triangle will have side lengths in the ratio 1-โˆš3-2. Angles In Ratios Triangle Of Sides 30-60-90 1-โˆš3-2 45-45-90 1-1-โˆš2 Square roots can also be used to find the distance between two points in a 2-dimensional or 3-dimensional system for movie or video game production. The formula for the distance D between two points (x[1], y[1]) and (x[2], y[2]) in 2 dimensions is given by: โ€ข D = โˆš((x[2] โ€“ x[1])^2 + (y[2] โ€“ y[1])^2) Note that this formula comes from the Pythagorean Theorem, where the legs of the right triangle have length x[2] โ€“ x[1] and y[2] โ€“ y[1], and the hypotenuse has length D. The formula for the distance D between two points (x[1], y[1], z[1]) and (x[2], y[2], z[2]) in 3 dimensions is given by: โ€ข D = โˆš((x[2] โ€“ x[1])^2 + (y[2] โ€“ y[1])^2 + (z[2] โ€“ z[1])^2) Having math trouble? Looking for a tutor? Example 1: Distance Between Two Points In 2 Dimensions Letโ€™s say we want to find the distance between the points (1, 3) and (8, -5). If we assign (x[1], y[1]) = (1, 3) and (x[2], y[2]) = (8, -5), then we can use the distance formula to calculate: โ€ข D = โˆš((x[2] โ€“ x[1])^2 + (y[2] โ€“ y[1])^2) โ€ข D = โˆš((8 โ€“ 1)^2 + (-5 โ€“ 3)^2) โ€ข D = โˆš((7)^2 + (-8)^2) โ€ข D = โˆš(49+ 64) โ€ข D = โˆš(113) So, the distance between the two points in 2 dimensions is โˆš113. Example 2: Distance Between Two Points In 3 Dimensions Letโ€™s say we want to find the distance between the points (2, 4, 7) and (1, -4, 0). If we assign (x[1], y[1], z[1]) = (2, 4, 7) and (x[2], y[2], z[2]) = (1, -4, 0), then we can use the distance formula to calculate: โ€ข D = โˆš((x[2] โ€“ x[1])^2 + (y[2] โ€“ y[1])^2 + (z[2] โ€“ z[1])^2) โ€ข D = โˆš((1 โ€“ 2)^2 + (-4 โ€“ 4)^2 + (0 โ€“ 7)^2) โ€ข D = โˆš((-1)^2 + (-8)^2 + (-7)^2) โ€ข D = โˆš(1 + 64 + 49) โ€ข D = โˆš(114) So, the distance between the two points in 3 dimensions is โˆš114. Square Roots In The Quadratic Formula Square roots are also necessary if we want to use the quadratic formula to solve a quadratic equation. Remember that a quadratic equation has the standard form where a, b, and c are real numbers, with a nonzero. The solutions of this quadratic equation are given by the quadratic formula: Note that there is a square root in the numerator of the fraction. This square root symbol is important because the radicand (expression under the radical) tells us the nature of the solutions. This particular radicand b^2 โ€“ 4ac is called the discriminant, and its sign tells us what the roots of the quadratic equation will look like: โ€ข b^2 โ€“ 4ac > 0 (positive discriminant): this means that there are two distinct real solutions to the quadratic equation. โ€ข b^2 โ€“ 4ac = 0 (zero discriminant): this means that there is one real repeated solution (a double root) to the quadratic equation. โ€ข b^2 โ€“ 4ac < 0 (negative discriminant): this means that there are two complex conjugate solutions to the quadratic equation. We might need to solve a quadratic equation in physics if we want to know when a falling object is at a certain height. Example: Solving A Quadratic For Height Of A Falling Object If an object is dropped from 400 feet above ground, then its height after t seconds is given by the equation Letโ€™s say we want to find out when the object is at a height of 144 feet. Then we would solve: โ€ข 144 = 400 โ€“ 16t^2 โ€ข -256 = โ€“ 16t^2 โ€ข 16 = t^2 โ€ข 4 = t So, the falling object will be at a height of 144 feet above ground at t = 4 seconds (after it has fallen 256 feet from its starting position). Square Roots & Radius Of A Circle If you want to find the radius of a circle with a particular area, then you will need to use square roots. Remember that the area A of a circle with radius R is given by the equation Where ฯ€ is the constant pi, or approximately 3.14159. We can also use square roots to find the radius of the base of a cylinder or cone with a particular volume (as long as we also know the height). Remember that if the height of a cylinder or cone is H and the radius is R, then the volume equations are: โ€ข Volume Of A Cylinder: V = ฯ€R^2H โ€ข Volume Of A Cone: V = ฯ€R^2H / 3 Example 1: Radius Of A Circle Letโ€™s say we want to build a circular animal pen with an area of 1256 square feet. Using our area equation with A = 1256, we can calculate: โ€ข A = ฯ€R^2 โ€ข 1256 = 3.14159R^2 โ€ข 399.80 = R^2 โ€ข 20 = R So, the radius of the pen would be 20 feet. Example 2: Radius Of A Circle At The Base Of A Cylinder Letโ€™s say we want to make a cylinder that is 7 inches tall and has a volume of 2200 square inches. Using our volume equation with V = 2200, we can calculate: โ€ข Volume Of A Cylinder: V = ฯ€R^2H โ€ข 2200 = (3.14159)R^2(7) โ€ข 2200 = 21.99113R^2 โ€ข 100.04 = R^2 โ€ข 10 = R So, the radius of the cylinder would be 10 inches. Square Roots & Simple Harmonic Motion In physics, we often use square roots in formulas for simple harmonic motion to find the period of a spring or a pendulum. The period is the amount of time it takes for them to go through one cyclical motion. The formulas are: โ€ข Period of a Spring: T = 2ฯ€โˆš(m/k) โ€ข Period of a Pendulum: T = 2ฯ€โˆš(L/g) where T is the period, m is the mass of a spring, k is the spring constant, L is the length of the pendulum, and g is the acceleration due to gravity. Note that the spring constant will vary depending on the type of spring. A stiffer spring has a higher value of k. Example 1: Period Of A Spring Letโ€™s say we have a spring with a spring constant of 4 N/m and a weight of 0.25 kg at the end of the spring. Using our formula for period, we get: โ€ข Period of a Spring: T = 2ฯ€โˆš(m/k) โ€ข T = 2(3.14159)โˆš(0.25/4) โ€ข T = 2(3.14159)โˆš(1/16) โ€ข T = 2(3.14159)(1/4) โ€ข T = 3.14159/2 โ€ข T = 1.5708 So, the period of the spring is 1.5708 seconds. Example 2: Period Of A Pendulum Letโ€™s say we have a pendulum with a length of 0.5 feet (assuming we are on Earth, the acceleration due to gravity is 32 feet per second). Using our formula for period, we get: โ€ข Period of a Pendulum: T = 2ฯ€โˆš(L/g) โ€ข T = 2(3.14159)โˆš(0.5/32) โ€ข T = 2(3.14159)โˆš(1/64) โ€ข T = 2(3.14159)(1/8) โ€ข T = (3.14159)(1/4) โ€ข T = 0.7854 So, the period of the pendulum is 0.7854 seconds. Square Roots & Standard Deviation In statistics, we use square roots to calculate the standard deviation (from the variance). The standard deviation is the square root of variance, which is a sum of squared differences from the mean of a data set. The square root ensures that the standard deviation will have the same units as the mean. This makes it meaningful to talk about adding or subtracting standard deviations from the mean. This makes it possible to talk about percentiles in a population. You can learn more about what affects standard deviation in my article here. You can learn about some real-life examples of standard deviation in my article here. Now you know what square roots are used for and where they fit into the equations and formulas where they apply. You can learn how to add, multiply, and divide square roots here. You can learn how to do square roots by hand in my article here. You can also learn how to graph square roots in my article here. You can learn how to take the derivative of a square root function here. Learn more about square roots and other radicals in denominators (and how to rationalize them) here. I hope you found this article helpful. If so, please share it with someone who can use the information. Donโ€™t forget to subscribe to my YouTube channel & get updates on new math videos!
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Multiple fraction calculator multiple fraction calculator Related topics: prentice hall algebra one answer key help for math project free online math lessons for 9th grade integrator online with steps adding subtractimg mulyiplying and dividing fractions 7th grade algebra patterns and relationships graphs worksheet printable worksheets on slopes Author Message Author Message EntiiDesigns Posted: Wednesday 03rd of Jan 21:49 ncjotes Posted: Monday 08th of Jan 08:44 Hi everybody out there, I am caught up here with a set Amazing! This sounds very useful to me. I was of math questions that I find very hard to solve . I am searching such application only. Please let me know taking Remedial Algebra course and need help with where I can buy this application from? Reg.: 19.10.2005 multiple fraction calculator. Do you know of any useful Reg.: 02.01.2002 math help software ? To be frank, I am a little skeptical about how useful these software programs can be but I really donโ€™t know how to solve these problems and felt it is worth a try. malhus_pitruh Posted: Tuesday 09th of Jan 20:49 kfir Posted: Friday 05th of Jan 17:49 You can download this program from Have you checked out Algebrator? This is a great https://softmath.com/ordering-algebra.html. There are software and I have used it several times to help me some demos available to see if it is what want and if with my multiple fraction calculator problems. It is really Reg.: 23.04.2003 you find it good , you can get a licensed version for a Reg.: 07.05.2006 very simple -you just need to enter the problem and it small amount. will give you a complete solution that can help solve your homework. Try it out and see if it solves your malhus_pitruh Posted: Saturday 06th of Jan 11:40 I agree. Algebrator not only gets your homework done faster, it actually improves your understanding of the subject by providing useful information on how to solve Reg.: 23.04.2003 similar questions. It is a very popular product among students so you should try it out.
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DOI Number Functions defined in the form ``$g:\mathbb{N}\to[0,\infty)$ such that $\lim_{n\to\infty}g(n)=\infty$ and $\lim_{n\to\infty}\frac{n}{g(n)}=0$'' are called weight functions. Using the weight function, the concept of weighted density, which is a generalization of natural density, was defined by Balcerzak, Das, Filipczak and Swaczyna in the paper ``Generalized kinsd of density and the associated ideals'', Acta Mathematica Hungarica 147(1) (2015), 97-115. In this study, the definitions of $g$-statistical convergence and $g$-statistical Cauchy sequence for any weight function $g$ are given and it is proved that these two concepts are equivalent. Also some inclusions of the sets of all weight $g_1$-statistical convergent and weight $g_2$-statistical convergent sequences for $g_1,g_2$ which have the initial conditions are given. weight functions; natural density; statistical convergent sequences. M. Balcerzak, P. Das, M. Filipczak and J. Swaczyna: Generalized kinds of density and the associated ideals. Acta Math. Hungar. 147(1) (2015), 97--115. S. Bhunia, P. Das and S. K. Pal: Restricting statistical convergence. Acta Mathematica Hungarica, 134(1-2) (2012), 153--161. R. ร‡olak: Statistical convergence of order . Modern Methods in Analysis and Its Applications, Anamaya Pub., New Delhi, India (2010), 121--129. P. Das and E. SavaลŸ: On generalized statistical and ideal convergence of metric-valued sequences. Reprinted in Ukrainian Math. J. 68(12) (2017), 1849--1859. Ukrain. Mat. Zh. 68(12) (2016), H. Fast: Sur la convergence statistique. Colloq. Math. 2 (1951), 241--244. J. A. Fridy: On statistical convergence. Analysis 5 (1985), 301--313. ลž. Konca, M. Kรผรงรผkaslan and E. Genรง: I-statistical convergence of double sequences dened by weight functions in a locally solid Riesz space. Konuralp J. Math. 7(1) (2019), 55--61. M. Kรผรงรผkaslan and M. Ylฤฑmaztรผrk: On deferred statistical convergence of sequences. Kyungpook Math. J., 56 (2006), 357--366. M. Yฤฑlmaztรผrk, O. Mฤฑzrak and M. Kรผรงรผkaslan: Deferred statistical cluster points of real valued sequences. Univ. J. Appl. Math., 1 (2013), 1--6. E. SavaลŸ: On some generalizd sequence spaces dened by modulus. Indian J. Pure Appl. Math., 30(5) (1999), 973--978. E. SavaลŸ: Strong almost convergence and almost-statistical convergence. Hokkaido Math., 29(3) (2000), 531--536. E. SavaลŸ and P. Das: On I-statistical and I-lacunary statistical convergence of weight g. Bull. Math. Anal. Appl., 11(2) (2019), 2--11. E. SavaลŸ: On I-lacunary statistical convergence of weight g of sequences of sets. Filomat 31(16) (2017), 5315--5322. E. SavaลŸ: I-statistical convergence of weight g in topological groups. Mathematics and computing, Springer Proc. Math. Stat., 253, Springer, Singapore, 2018, 43--51. I. J. Schoenberg: The integrability of certain functions and related summability methods. The American Mathematical Monthly 66(5) (1959), 361--775. โ€ข There are currently no refbacks. ยฉ University of Niลก | Created on November, 2013 ISSN 0352-9665 (Print) ISSN 2406-047X (Online)
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Scaling variable in MIP I came across this article https://www.gurobi.com/documentation/current/refman/advanced_user_scaling.html when learning how to better scale the variables. I'm working on a MIP model and trying to scale down the right-hand sides of inequalities but it makes the coefficients of constraints very small. How small is too small compared to the default tolerances setting? The advanced_user_scaling mention to also scale the x coefficient. Is this possible in a MIP problem? I can't think of how it can be done when the x can only be 0 or 1. Would it be like if I want to scale it up by 100 I would create a variable that can only be 0 or 0.01? โ€ข Hi Bryan, When scaling the model, we propose to consider the following steps โ–ก Scale your variables and constraints to get reasonable coefficients โ–ก First, consider the units of your variables: For example: If you deal with millions of dollars, do not use cents or dollars as a unit. Instead use thousands or millions. (Tons instead of grams, kilometers instead of meters, ...) โ–ก If you deal with millions of dollars, do you need exact integral cent values? โ–ก Scale constraints so that coefficients get closer to 1 โ–ก Use integral data wherever possible, e.g. 1*x + 2*y = 3 better than 0.3333333*x + 0.66666666*y = 1 As mentioned in https://www.gurobi.com/documentation/current/refman/advanced_user_scaling.html the variable coefficients should not be smaller than 10e-3. If your model only contains binary variables, this is fine. But also here the range of the coefficients should not be larger than 10e6 and also within [10e-3, 10e6]. I hope this helps, โ€ข Hi Marika, Thanks for the response. I understand the part of scaling the coefficient of all variables together for one constraint. My question is that how we can scale one variable for all the constraints like the article but in a MIP model where the variable should be binary. In the article it replace the x with x' where x=10^5x' for all the constraints. โ€ข Hi Bryan, You are right. The substitution discussed in the article can be done with continuous variables but not with binary (or integer) variables. In this case, it might help to rethink the unit of the variable or if the constraint can be modeled differently. I think you have a commercial license with us. You are welcome to create a support request via the Gurobi Help Center, share the model with us, and discuss scaling or numerics in more detail. Best regards, Please sign in to leave a comment.
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A sudden increase in carbon dioxide concentration around a leaf... | Filo Question asked by Filo student A sudden increase in carbon dioxide concentration around a leaf will causea. Wider opening b. Increase in transpiration c. Closure in stomata d. Decrease in transpiration due to closure of stomata Not the question you're searching for? + Ask your question Sure, I can convert the solutions into step-by-step solutions with latex tags intact. Here's an example: Problem: Solve for x: Step 1: Subtract 5 from both sides of the equation to isolate the variable term. Step 2: Simplify the left-hand side by cancelling out the 5's. Step 3: Divide both sides by 2 to isolate the variable. Step 4: Simplify the left-hand side by cancelling out the 2's. Found 2 tutors discussing this question Discuss this question LIVE 8 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Practice more questions on Transport in plants View more Students who ask this question also asked View more Stuck on the question or explanation? Connect with our Biology tutors online and get step by step solution of this question. 231 students are taking LIVE classes Question A sudden increase in carbon dioxide concentration around a leaf will causea. Wider opening b. Increase in transpiration c. Closure in stomata d. Decrease in transpiration due to closure Text of stomata Updated On Feb 8, 2024 Topic Transport in plants Subject Biology Class Class 12 Answer Type Text solution:1
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Area-Proportional Diagrams with eulerr This shiny app is based on an R package that I have developed called eulerr. It generates area-proportional euler diagrams using numerical optimization routines. Euler diagrams are generalized venn diagrams for which the requirement that all intersections be present is relaxed. They are constructed from a specification of set relationships but may sometimes fail to display these appropriately. For instance, try giving the app the specification A = 5, B = 3, C = 1, A&B = 2, AB&C = 2 to see what I mean. When this happens, eulerr tries to give an indication of how badly the diagram fits the data through the metrics stress and diag error. The latter of these show the largest difference in percentage points between the specification of any one set combination and its resulting fit. It is the maximum value of region error, which is given for each combination. This metric has been adopted from a paper by Micallef and Peter Rodgers. Stress is more difficult to explain, but I advise the interested reader to read Leland Wilkinson's excellent paper for a proper brief. Finally, I owe a great deal to the aforementioned Wilkinson as well as Ben Frederickson whose work eulerr is inspired by. Johan Larsson Limitations in the Shiny App The Shiny app that is hosted here does not completely cover all the functionality that the R package offers. The number of sets is for instance limited to five here but there is no such limitation in the package. If you want to install the R package, then you need to first install R. After this you can simply install the package by calling install.packages("eulerr"). To read more about the R package, please visit the package page on CRAN. eulerr is an open-source project that welcomes contributions from anyone who's willing to chip in. Please see the development page for the R package if you are interested in taking part or just want to report an issue with the package. If you find any issues with this site, please visit the development page for the Shiny appplication and file an issue there.
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A JavaScript array, in Gleam! Unlike most data structures in Gleam this one is mutable. pub type Array(element) pub fn fold(over over: Array(a), from from: b, with with: fn(b, a) -> b) -> b Reduces a list of elements into a single value by calling a given function on each element, going from left to right. fold(from_list([1, 2, 3]), 0, add) is the equivalent of add(add(add(0, 1), 2), 3). Runs in linear time. pub fn fold_right(over over: Array(a), from from: b, with with: fn( ) -> b) -> b Reduces a list of elements into a single value by calling a given function on each element, going from right to left. fold_right(from_list([1, 2, 3]), 0, add) is the equivalent of add(add(add(0, 3), 2), 1). Runs in linear time. pub fn from_list(a: List(a)) -> Array(a) Convert a Gleam list to a JavaScript array. Runs in linear time. pub fn get(a: Array(a), b: Int) -> Result(a, Nil) Get the element at the given index. > get(from_list([2, 4, 6]), 1) > get(from_list([2, 4, 6]), 4) pub fn map(a: Array(a), with with: fn(a) -> b) -> Array(b) Returns a new array containing only the elements of the first array after the function has been applied to each one. Runs in linear time. > map(from_list([2, 4, 6]), fn(x) { x * 2 }) from_list([4, 8, 12]) pub fn size(a: Array(a)) -> Int Get the number of elements in the array. Runs in constant time. pub fn to_list(a: Array(a)) -> List(a) Convert a JavaScript array to a Gleam list. Runs in linear time.
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Unscramble ABATABLE How Many Words are in ABATABLE Unscramble? By unscrambling letters abatable, our Word Unscrambler aka Scrabble Word Finder easily found 71 playable words in virtually every word scramble game! Letter / Tile Values for ABATABLE Below are the values for each of the letters/tiles in Scrabble. The letters in abatable combine for a total of 14 points (not including bonus squares) โ€ข A [1] โ€ข B [3] โ€ข A [1] โ€ข T [3] โ€ข A [1] โ€ข B [3] โ€ข L [1] โ€ข E [1] What do the Letters abatable Unscrambled Mean? The unscrambled words with the most letters from ABATABLE word or letters are below along with the definitions. โ€ข abatable (a.) - Capable of being abated; as, an abatable writ or nuisance.
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More Quantum Mechanics: The Schrodinger Equation German physicist Erwin Schrodinger, developed the equation that goes by his name ca. 1926 In this blog, we change gears from the political, environmental and religious issues and go back to physics, to look at more quantum mechanics (QM). Although I did examine the Schrodinger equation and its applications earlier: e.g. , it helps to look into it at a more basic level, in terms of its own internal "mechanics" and fundamentals. This will incorporate some elements not covered in the more advanced treatment, including how the DE itself is obtained from first principles. The Schrodinger equation was developed by Erwin Schrodinger around the same time (1926) that Werner Heisenberg developed his matrix wave mechanics. Both are ways to analyze quantum dynamics for assorted simple systems (square well, assorted barriers, 3D-box etc.) and processes, but the first caught on much more than the latter. Why? Probably the chief reason is that Schrodinger's version relies on plain old differential equations, as opposed to the more obscure matrices of Heisenberg. Thus, since most advanced physics students also have taken DEs, they are able to much more easily follow and arrive at the solutions. See, e.g.: And my other blogs to do with DEs: These will definitely help to refresh readers' perspectives before delving into this blog. We begin by writing the general equation for a progressive wave U: U = A exp{2ฯ€i(Kx - ft)} where K the wave number is equal to 1/L (L = wavelength), x is the displacement (1-dimensional for simplicity), and f is the frequency. Meanwhile, for standing waves such as would be expected in many quantum applications: U = A sin 2ฯ€Kx exp(-2ฯ€ift) In either case (standing or traveling waves) U varies with x for a particular value of t in such a way as to satisfy the simple harmonic equation: d^2U/dx^2 + 4ฯ€^2 K^2 U = 0 Which can be verified by direct differentiation (left to the industrious reader). If a field is present, so that the potential energy V(x) of an electron varies with x, K will be a function of x too. Based on the experiments from electron diffraction, one has: K = p/h = m(e)v /h where h is Planck's constant, and m(e) is the electron mass and v the electron velocity. Then: K^2 = m(e)^2v^2/ h^2 = 2 m(e)(W - V)/ h^2 where W is the total energy of each electron so (W - V) is the kinetic energy, i.e. W = V + [m(e)v^2/2] = V + KE so: KE = W - V = [m(e)v^2/2], so: 2 m(e)(W - V) = [m(e)v^2] Then: m(e)^2v^2/ h^2 = 2 m(e)(W - V)/ h^2 Now, substituting this last (for K^2) into the SHM equation, we obtain: d^2U/dx^2 + 8ฯ€^2 m(e)/h^2 {W - V(x)}U = 0 And this is none other than one form (the 1D) of the Schrodinger equation. Illustrating some basic properties of this DE is straightforward. The best approach is to apply it tos a specific case for which some parameters are known. Consider then an electron of charge e, moving in an electric field, E. The electric force F(E) acting is: F(E) = eE The potential energy V(x) is obtained from: dV(x)/dx = F(E) dV(x) = F(E)dx and integrating both sides: V(x) = F(E) x = eE x The beauty of this is that the same argument can apply to any equation of the form: d^2U/dx^2 + F(x) U = 0 where F(x) is some known function. Another interesting facet of the Schrodinger equation refers to the superposition aspect. If we start, say, with two different initial conditions, to obtain two waves: U = U1(x) and U = U2(x) Then ALL solutions of the given Schrodinger wave equation are of the form: U = A U1(x) + BU2(x) Next: Determining specific values of the constants: A, and B for a specific simple system.
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Re: st: fmlogit interpreting marginal effects Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org. [Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index] Re: st: fmlogit interpreting marginal effects From Maarten Buis <[email protected]> To [email protected] Subject Re: st: fmlogit interpreting marginal effects Date Fri, 16 Mar 2012 10:57:39 +0100 On Fri, Mar 16, 2012 at 6:23 AM, Trudy Sullivan wrote: > I have six dependent variables that are fractions and add up to 1. I used fmlogit to estimate a fractional multinomial logit and I used dfmlogit to compute the marginal effects. -fmlogit- and -dfmlogit- are user written programs, so per the Statalist FAQ you must say where you got it from. This is not to be mean, but to help you ask answerable questions. There are often multiple versions of user written software floating around in cyber space, and you can imagine that it helps a lot if everybody is talking about the same piece of software. I am assuming you are using the version downloaded from SSC. There are many other tips on how to ask answerable questions/avoid common mistakes in the FAQ. A link to which you can find at the bottom of this post (and every other post on > In my case the six dependent variables could equally be used as the reference category. I want to interpret the marginal effects for the six dependent variables but depending on which variable is omitted, the marginal effects change slightly. How should I proceed? The computations get quite involved, so I guess this is a precision issue. So you can just pick one. There may be ways of changing the computations to get more accurate results, but since I don't think it is a big problem it will be low on my list of priorities. > dmflogit computes the marginal effects for the omitted category and I donโ€™t understand how to interpret this marginal effect. You interpret them the same way as all the other marginal effects. There are no separate coefficients for that proportion, since the remaining coefficients fully determine how that proportion changes. This is what is being used to compute those marginal effects. Hope this helps, Maarten L. Buis Institut fuer Soziologie Universitaet Tuebingen Wilhelmstrasse 36 72074 Tuebingen * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/
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-optimization problem Blocky velocity inversion by hybrid norm Next: Regularization by the Laplacian Up: Almomin: Blocky velocity inversion Previous: Introduction The Dix equation can be made linear by relating the square of interval velocity The hybrid norm above defines the cost function as follows: where Claerbout, 2009). Fitting goal (2) is not enough to fully constrain the inversion, because it has a large null space (Li and Maysami, 2009). Moreover, picking errors can lead to incorrect RMS velocities and unreasonable interval velocities. Therefore, a second fitting goal (i.e. a regularization term) is required to constrain this inversion. The regularization term can be written as follows: Notice that the norm in fitting goal (2) has a different effect than the norm in fitting goal (4). Using the hybrid norm in data fitting makes the inversion less sensitive to outliers. On the other hand, using the hybrid norm in model styling affects the general shape of the estimated model, which is the goal of this paper. Li and Maysami (2009) successfully produced blockiness in 1D when using the first derivative as a regularization operator. In the following sections, we will try different regularization operators to achieve the same goals in 2D. Blocky velocity inversion by hybrid norm Next: Regularization by the Laplacian Up: Almomin: Blocky velocity inversion Previous: Introduction
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2014 CSHL Undergraduate Research Program in Bioinformatics Searching for GATTACA In this class we explored the problem of finding exact occurrences of a query sequence in a large genome or database of sequences. Under this theme, we started by analyzing the brute force approach introducing the concepts of algorithm, complexity analysis, and E-values. Next we discussed suffix arrays as an index for accelerating the search, including analyzing the performance of binary search. We also considered two traditional algorithms for sorting (Selection Sort versus QuickSort) and their relative performance. In the second half of the class we discussed finding approximate occurrences of a short query sequence in a large genome or database of sequences. We first defined the problem by considering various metrics of an approximate occurrence such as hamming distance, or edit distance. We then considered different methods for computing inexact alignments including brute force global & local alignments, and seed-and-extend algorithms. Finally we discussed Bowtie as a Burrows-Wheeler transform based short read mapping algorithm for discovering alignments to reference genome. Python & Bioinformatics Python Class 1 Introduction to python, variables, lists, conditions, loops Python Class 2 Brute force search, dictionaries, motif finding iPython Notebooks for Probability & Statistics We also used the exercises at Rosalind throughout the course. Special topics Talk by Anne Churchland on balancing work and life.
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Lesson 5: Stress states - DRIVEN Lesson 5: Stress states In this lesson, we will delve into the concept of stress states, understanding how they are represented and visualized in engineering and mechanics. 1. Spatial Stress State โ€ข In the field of engineering and mechanics, representing stress states accurately is crucial. There are several methods to achieve this: 1.1 Elementary Unit Cube โ€ข Imagine an infinitesimally small cube. This cube is employed to depict the stresses acting on three mutually perpendicular planes. 1.2 Stress Tensor โ€ข The stress tensor is a mathematical representation used to describe the distribution of stresses within a material or structure. It provides a comprehensive view of stress components at every point within a body. โ€ข As described earlier, the stress tensor represents stresses like so: 1.3 Mohrโ€™s Circle โ€ข Mohrโ€™s Circle is a graphical method used to visualize relationships between normal and shear stresses on inclined planes within a stressed body. This tool can also calculate principal stresses, maximum shear stresses, and stresses on inclined planes. It is named after its creator, German Civil Engineer Otto Mohr (1835-1918). โ€ข The stress vectors for different directions in a ฯƒ-๐œ coordinate system fall within two arc triangles. At points where ๐œ (shear stress) equals 0, these directions correspond to the principal directions, which are mutually perpendicular. โ€ข Principal Plane: A plane where one of the principal stresses is equal to 0. โ€ข Endpoints of Stress Vectors: These correspond to the directions of principal planes and are located on those planes. โ€ข Drawing Mohr Circle: Mohrโ€™s Circle is commonly drawn when at least one principal direction is known. 2. Planar Stress State In certain scenarios, the stress state simplifies to a planar state. This occurs when there is only one normal stress and one shear stress (a pair) in the stress tensor, and both are confined to a single plane. Key aspects of planar stress states include: โ€ข In a planar stress state, one of the principal stresses is always zero. โ€ข The spatial elementary unit cube can be reduced to a simplified, planar version. Understanding stress states is fundamental in engineering and mechanics, as it allows engineers and analysts to assess how materials and structures respond to various forces and loads. Whether in three-dimensional spatial stress or simplified planar stress states, this knowledge forms the basis for making informed design and analysis decisions. With this understanding, you are better equipped to analyze and design structures, ensuring they can withstand the stresses and loads they encounter in real-world applications.
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Resonator Modes in Microchip Lasers Posted on 2018-09-28 as part of the Photonics Spotlight (available as e-mail newsletter!) Permanent link: https://www.rp-photonics.com/spotlight_2018_09_28.html Author: Dr. Rรผdiger Paschotta, RP Photonics AG, RP Photonics AG Abstract: It is explained how the resonator modes of a microchip laser are formed by thermal lensing effects. Some example cases are analyzed, showing that the concept works well for a 1-W laser, while beam quality will be lost for substantially higher power levels. Microchip lasers are solid-state lasers, where the laser essentially consists only of a laser crystal having reflective dielectric coatings on both end faces (see Figure 1). In some cases, there is a second crystal attached to the laser crystal, e.g. a saturable absorber crystal for passive Q switching. Optical pumping is usually done with a laser diode. Such lasers obviously have a very simple and compact setup, which is attractive for many applications where simplicity, robustness and low cost of interest. However, despite their apparent simplicity, their operation principles are not as simple as those of many other diode-pumped lasers. Figure 1: Microchip laser, consisting of a laser crystal with dielectric mirror coatings on both end faces. In particular, it is interesting to consider the properties of the modes of such a laser resonator. Usually, the two reflecting end faces are flat, and you may wonder how a stable resonator should be formed that way. Indeed, you usually get a stable resonator only due to side effects of the optical pumping of the laser crystal. In particular, the heating which results from the quantum defect and additional deficiency of the power conversion generates a thermal lens, i.e., a focusing effect for the light going through the laser crystal. In addition, you may have some amount of gain guiding, which can also influence the mode properties. An Example Case As an example, let us consider a microchip laser made from at 10 mm long Nd:YAG crystal. Let us start with a very simplified model for its laser resonator, where we assume that the lensing effect is uniformly distributed over the whole crystal length, and gain guiding is ignored. Interestingly, in that situation one finds that for a given total dioptric power of the thermal lens, the mode radius is always constant along the path through the crystal. Further, the resonator is found to be stable even for arbitrarily strong thermal lensing! That is surprising, since we know, for example, that the focusing action of the curved end mirror must be below a certain limit before driving the resonator into the unstable regime. However, in that situation we effectively have a waveguide, and that has at least one guided mode for an arbitrarily strong contrast of its refractive index profile. In reality, however, the pump power decays on the way through the crystal. So we may make a more realistic resonator model, assuming an exponential decay of pump power and heating power along the crystal. The absorption coefficient may be chosen such that most of the pump light is absorbed in a single pass. The results are shown in Figure 2: Figure 2: Mode radius a different positions in a microchip laser as a function of the dioptric power of the thermal lens. The calculation has been done with the software RP Resonator. If the thermal lens becomes rather strong, the mode size begins to vary substantially along the crystal. For even stronger lensing (from &approx;320 /m on), the resonator becomes unstable. Interestingly, there are further stability regions for even stronger lensing, but these are not of practical relevance. How strong the thermal lens will actually be, depends of course on the applied pump power and the pump beam radius. For example, let us assume that we want to make a laser with 1 W output power and 5 W intracavity power, so that the double-pass saturation power should be of the order of 1 W. (This is required in order to operate the laser &approx;5 times above the laser threshold.) From that one can calculate a beam radius of &approx;150 ฮผm. The dissipated pump power will be roughly 0.35 W, and one can calculate that the thermal lens will then have a dioptric power of &approx;4.5 /m. From Figure 2 one can see that the mode radius is then &approx;110 ฮผm, a bit smaller than the pump beam radius. So such a laser will probably be single-mode with perfect beam quality and reasonable Higher Output Powers? How about a 10-W laser of that kind? Assuming the same degree of output coupling, we should use a 10 times larger pump beam area, i.e., the same pump intensity. The dioptric power of the thermal lens will then be the same as for the 1-W laser, despite the higher dissipated power. Therefore, we will also have the same fundamental mode radius. Now, however, the pump beam radius will now be much larger than that, so that multimode emission with poor beam quality is to be expected. You may try to solve that problem by using a longer laser crystal, but unfortunately that does not help much. For example, if we take a five times longer crystal (50 mm โ€“ maybe no longer practical), also having five times lower doping concentration, the mode radius for thermal lensing with 4.5 /m increases only from 110 ฮผm to 163 ฮผm, while the pump radius should be increased from 150 ฮผm to 475 ฮผm. You can also not substantially change the mode radius by changing the absorption coefficient. Thus, it appears to be impossible to get perfect beam quality. So you clearly see that the great simplicity of that kind of laser causes a substantial problem: it has very few parameters which we can adjust, and for high-power operation we run into a regime where a high beam quality is no longer feasible! This is a classical example for a laser architecture which is not power-scalable. You can draw a lot of useful conclusions from these considerations: โ€ข The resonator modes of microchip lasers are essentially determined by thermal lensing effects โ€“ much in contrast to those of most other diode-pump lasers. โ€ข The basic laser parameters should be calculated based on some essential knowledge of laser physics. For example, one can and should calculate an appropriate pump and laser beam radius for the targeted operation parameters. Also, it is much faster and cheaper to do those calculations than just trying to build such a laser in the lab. Finding such things with a trial-and-error approach is inefficient and unprofessional. โ€ข Such considerations also let you understand certain basic limitations of laser architectures. In our particular case, we have seen that microchip lasers cannot easily be scaled up to higher powers without losing a good beam quality. That has to do with the fact that the simplicity of the architecture gives us too few parameters for controlling the mode properties. This article is a posting of the Photonics Spotlight, authored by Dr. Rรผdiger Paschotta. You may link to this page and cite it, because its location is permanent. See also the RP Photonics Note that you can also receive the articles in the form of a newsletter or with an RSS feed. Questions and Comments from Users Here you can submit questions and comments. As far as they get accepted by the author, they will appear above this paragraph together with the authorโ€™s answer. The author will decide on acceptance based on certain criteria. Essentially, the issue must be of sufficiently broad interest. Please do not enter personal data here. (See also our privacy declaration.) If you wish to receive personal feedback or consultancy from the author, please contact him, e.g. via e-mail. By submitting the information, you give your consent to the potential publication of your inputs on our website according to our rules. (If you later retract your consent, we will delete those inputs.) As your inputs are first reviewed by the author, they may be published with some delay.
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Value Left - Working with single and double digits Hello, For the below formula I am attempting to grab the leading numbers from an alphanumeric field. the numbers will always be leading but will range between 1-and 10. I am using the following with no success =VALUE(LEFT([Probability Score]@row, 2)) If the cell contains 10 then it works. But if it contains a single-digit number it fails. If I use: =VALUE(LEFT([Probability Score]@row)) Then I only get a single-digit return. I have attempted to use helper columns but to no avail yet. The intent is to multiple two fields together using the formula above. See screenshot for reference. Best Answer โ€ข This should remove the blank spaces on single digit values =VALUE(SUBSTITUTE(LEFT([Probability Score]@row, 2), " ", "")) โ€ข This should remove the blank spaces on single digit values =VALUE(SUBSTITUTE(LEFT([Probability Score]@row, 2), " ", "")) โ€ข Try this for your Column28. You should be able to use the same syntax for Column29 =VALUE(LEFT([Magnitude Score]@row, FIND("-", [Magnitude Score]@row) - 2)) Will this work for you? โ€ข Many thanks for the quick assistance, but the options worked!
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A novel approach for aiding unscented Kalman filter for bridging GNSS outages in integrated navigation systems Aiming to improve the position and velocity precision of the INS/GNSS system during GNSS outages, a novel system that combines unscented Kalman filter (UKF) and nonlinear autoregressive neural networks with external inputs (NARX) is proposed. The NARX-based module is utilized to predict the measurement updates of UKF during GNSS outages. A new offline approach for selecting the optimal inputs of NARX networks is suggested and tested. This approach is based on mutual information (MI) theory for identifying the inputs that influence each of the outputs (the measurement updates of UKF) and lag-space estimation (LSE) for investigating the dependency of these outputs on the past values of the inputs and the outputs. The performance of the proposed system is verified experimentally using a real dataset. The comparison results indicate that the NARX-aided UKF outperforms other methods that use different input configurations for neural networks. In order to overcome the shortcomings associated with the stand-alone operation of Inertial Navigation Systems (INS) and Global Navigation Satellite Systems (GNSS), and to combine advantages of each system, INS and GNSS are often integrated to obtain accurate navigation solution with superior performance in comparison with either a GNSS or an INS stand-alone system. Many fusion algorithms are employed to fuse INS and GNSS data; the traditionally employed fusion techniques are Kalman filters (KF), such as extended Kalman filter (EKF) (Al Bitar & Gavrilov, 2019; Crassidis, 2006; Faruqi & Turner, 2000) and unscented Kalman filter (UKF) (Al Bitar & Gavrilov, 2019; Chang, 2014; Crassidis, 2006). With correct dynamic and stochastic models of GNSS and INS errors, KF can produce very accurate solutions, if there is continuous access to GNSS signals. However, KF does have limitations. The major inadequacy related to the utilization of KF for INS/GNSS integration is the necessity to have accurate stochastic models for each of the sensor errors. The inaccurate description of the system noises, measurement errors, and uncertainty in the dynamic models lead to unreliable estimates and degradation in accuracy, especially during GNSS outages when KF operates in prediction mode based on the predefined state error models, which are not necessarily correct. In addition, there are several significant drawbacks of KF, such as sensor dependency and observability problems (Hong et al., 2005; Klein & Diamant, 2018; Tang et al., 2008). The limitations of KF have motivated researchers to investigate alternative methods for improving the accuracy of navigation solution during GNSS outages. These methods were predominantly based on artificial intelligence (AI). Much research has been conducted to investigate the use of AI-based techniques to bridge GNSS signal outages in INS/GNSS systems. Researchers have utilized various approaches for combining the AI module(s) with the rest of the INS/GNSS system (Al Bitar et al., 2020). Chiang et al. (2008) suggested the replacement of KF by AI module using the so-called position update architecture (PUA). The proposed scheme was implemented using a constructive neural network (CNN) to overcome the limitations of conventional techniques that are predominantly based on the KF. The PUA module is used to estimate the INS position error during GNSS signal outages using velocity and azimuth of the INS. Chiang and Huang (2008) proposed position, velocity and azimuth update architecture (PVAUA) based on multilayer perceptron neural networks (MLPNN). The PVAUA module uses the velocity, azimuth of the INS and time to estimate the INS position error during GNSS signal outages. El-Sheimy et al. (2004) introduced an alternative INS/GPS integration method using an adaptive neuro-fuzzy inference system (ANFIS). The ANFIS-based module is implemented to predict the error drift of the standalone INS-estimated position during GNSS signal blockage using the INS position and time. In fact, the replacement of KF by an AI module worked well for navigational-grade INS. However, these techniques showed a very limited success when applied to a MEMS-based INS/GNSS navigation system, due to the high noise level and bias instability of MEMS inertial sensors. As a result, KF is kept as the primary state estimation tool in INS/GNSS integration, and thus the logical step was toward an integration technique that uses both KF and AI module in the same system. Wang et al. (2007) first proposed the concept of aiding KF. The authors utilized radial basis function neural networks (RBFNN) to predict the position differences between INS and GNSS in three orthogonal directions to form estimates of the measurement update of EKF during GNSS outages. The input parameters of the RBFNNs are the attitude angles and the changes of vehicle velocity and attitude angles in each epoch. However, only position measurements of KF are predicted during GNSS outages. This means that KF is not fully operational as no velocity measurements are predicted. Chen and Fang (2014) proposed a hybrid prediction method for bridging GNSS outages using RBFNN and time series analysis, which aided EKF by forecasting position and velocity measurement updates. The proposed hybrid prediction method uses RBFNN to predict the six components of position and velocity measurement updates. The inputs of RBFNN are the measurements of accelerometers and gyroscopes. These measurements are first passed through a Wavelet denoising filter in order to lower the noise level. The residual error of training the RBFNN is modelled as a time series. The outputs of the RBFNN and the time series are summed together to form the final prediction of position and velocity measurement updates for EKF during GNSS outages. However, the complexity of the proposed system is not suitable for real-time implementation. Jingsen et al. (2016) proposed a hybrid prediction method that combines extreme learning machines (ELM) and EKF. ELMs are applied to predict EKF position and velocity observations during GNSS outages. The measurements of gyroscopes and accelerometers are selected as inputs of the ELMs. The use of raw measurements of gyroscopes and accelerometers without a denoising stage complicates the learning process of ELMs, especially in the case of MEMS-based INS, as the level of noise is relatively high. Yao et al. (2017) proposed a hybrid fusion algorithm to provide a pseudo position information to assist the integrated navigation system during GNSS outages using MLPNN. The proposed MLPNN-based model relates the current and past one-step values of velocities, angular rates and specific force of INS to the increments of the GNSS position. The GNSS position increments are accumulated to achieve the pseudo-GNSS position measurements. Yao and Xu (2017) proposed robust least squares support vector machine (RLS-SVM)-aided fusion methodology for INS during GNSS outages. The RLS-SVM is used to predict the pseudo-GNSS position during GNSS outages similar to the previously proposed system in Al Bitar et al. (2020). The inputs of the RLS-SVM model are the specific force, velocity, and yaw information. Wang et al. (2019) proposed a fusion algorithm based on back propagation neural networks (BPNNs) to predict the pseudo-GNSS position during GNSS outages. The proposed BPNN-based model relates the current and past values of velocities, angular rates, specific force of INS and the time elapsed since the beginning of GNSS signal outage to the increments of the GNSS position. Recently, Fang et al. (2020) proposed an algorithm based on long short-term memory (LSTM) to predict the pseudo-GNSS position during GNSS outages. The inputs of LSTM model are the four-step information of specific force, angular rates, velocity and yaw. There are some common drawbacks related to the methods mentioned above. It is clear that the selection of inputs of an AI module differs from one method to another without any justification or comparison. In fact, the selection of the inputs of an AI module affects the system. A fewer number of inputs means a simpler internal structure of AI module and, consequently, less training time, while a large number means more complicated structure and thus longer learning time and less real-time capability. Including a wrong input or excluding a right input lead to degradation in prediction accuracy of the AI module. The selection of the measurements of gyroscopes and accelerometers as inputs of an AI module is usually justified by the fact that the outputs of INS (position, velocity and attitude angles) are the result of the mathematical integrating (over time) of these measurements. In other words, the measurement errors (noises) of gyroscopes and accelerometers are embedded in the errors of INS outputs. As a result, the erroneous INS outputs (during GNSS outages) can be used as inputs of an AI module instead of the raw measurements of gyroscopes and accelerometers. One advantage of using erroneous INS outputs instead of the measurements of gyroscopes and accelerometers is that they do not require a denoising stage, as the process of mathematical integration smooths The second drawback of all aforementioned methods is using EKF as the only choice for fusing INS and GNSS data. Other options, such as UKF, for example โ€“ which is proven to be less sensitive to the nonlinearities of process and observations models compared to EKF โ€“ were not considered. The third drawback is that these methods use the same covariance matrices of KF during the availability and the outages of GNSS. This is not true, as the measurements provided by the aid of the AI module during GNSS signal have error characteristics that differ from those of GNSS measurements. This paper considers the problem of aiding KF in INS/GNSS systems using a nonlinear autoregressive neural network with external inputs (NARX) (Siegelmann et al., 1997). The paper also addresses the problems mentioned above. First, UKF is chosen as the integrating filter instead of EKF. Secondly, a new approach for selecting the optimal inputs of an AI module (NARX networks, in our case) is proposed. This approach is based on mutual information (MI) theory (Brown, 2009; Peng et al., 2005) for identifying the inputs that influence each of the outputs (the measurement update for UKF during GNSS outages), and lag-space estimation (LSE) (He & Asada, 1993) for determining the model order (i.e., the dependency of the outputs on the past values of inputs and the outputs themselves). Third, the covariance matrices of UKF are linked to the prediction errors of AI module. The proposed method is shortly called โ€œNARX-aided UKF.โ€ The performance of the NARX-aided UKF is experimentally verified using a real dataset. The rest of this paper is organized as follows: A detailed explanation of the proposed method is given in Section 2. The performance of the proposed method is presented in Section 3. Conclusions are presented in Section 4. 2.1 Principle of operation The main idea of the proposed system is to employ a NARX network to predict the measurement update (the difference between GNSS outputs and the outputs of INS) of UKF during GNSS outages. The system works in two modes: learning mode when a GNSS signal is available, and prediction during GNSS outages. Two copies of INS are created, INS1 and INS2. In learning mode, INS1 and GNSS are integrated using a loosely coupled scheme. The positions and velocities P[๐บ๐‘๐‘†๐‘†], ๐‘‰[๐บ๐‘๐‘†๐‘†] provided by GNSS are merged as updates of the INS1 estimates of position and velocity P[๐ผ๐‘๐‘†1],๐‘‰[๐ผ๐‘๐‘†1] through a UKF ( Figure 1). UKF estimates the errors on state of INS1 . These estimates are added to state of INS1 X[๐ผ๐‘๐‘†1] to form corrected values X[๐ถ]. At the same time, INS2 works autonomously, and its outputs P [๐ผ๐‘๐‘†2], ๐‘‰[๐ผ๐‘๐‘†2] are corrected periodically every 60 seconds by GNSS measurements, as shown by the dashed arrows. In fact, the 60-s duration is related to real-life scenarios; in real life, the GNSS signal may be lost when moving through tunnels or around obstacles in urban areas. The duration of these outages in most cases is less than 60 s. The position and velocity of INS2 are then subtracted from the ones of the GNSS to form error signals of position ๐›ฟ๐‘ƒ[๐บ๐‘๐‘†๐‘†/๐ผ๐‘๐‘†2] and velocity ๐›ฟ๐‘‰[๐บ๐‘๐‘†๐‘†/๐ผ๐‘๐‘†2]. These error signals are used as target values for training NARX networks. The estimates of gyro drifts and accelerometer biases are fed back to INS1 and INS2 mechanization equations, as shown by the dotted arrows. The training algorithm of NARX networks starts after collecting a specific amount of training data that represents the 60-s simulated outage duration. This amount is called a window (ฮดP[GNSS][/][INS] [2],๐›ฟ๐‘‰[GNSS][/][INS][2]. The inputs of NARX networks are chosen from the outputs of INS1 (position, velocity and attitude angles), the past values of errors, and the time elapsed since the beginning of GNSS signal outage. The process of selecting the inputs of NARX networks is based on MI theory and LSE, and it is conducted in offline stage, as will be explained later in Subsection 2.5. When real GNSS signal outage occurs, the system switches to prediction mode, as shown in Figure 2. The NARX module predicts the errors . These errors are added to the position and velocity of INS2 P [๐ผ๐‘๐‘†2],๐‘‰[๐ผ๐‘๐‘†2] to form estimations of position and velocity of GNSS . The difference between and P[๐ผ๐‘๐‘†1],๐‘‰[๐ผ๐‘๐‘†1] forms the estimation of measurement update for UKF . Using these updates, UKF continues to operate as if no GNSS outage had occurred. To train the NARX networks in online mode, a training procedure that utilizes non-overlap moving window technique is used. The non-overlap moving window doesnโ€™t have the disadvantage of redundancy in the information when using sliding window; thus, it doesnโ€™t require a long time for data processing compared to the sliding window technique. The NARX networks are updated (trained) within this window. For real-time purposes, the NARX networks are trained until reaching certain minimum mean-squared error (MSE) or after completing a certain number of training epochs (determined empirically). This procedure is repeated when a new window is acquired, as shown in Figure 3. Whenever a GNSS outage occurs, the NARX networks switch to prediction mode and provide estimates of errors (). 2.2 Navigation equations Taking into consideration the coordinate frames shown in Figure 4, the navigation equations written in N-frame are given as follows (Jekeli, 2012): where ๐‘ท = [๐œ‘ ๐œ† โ„Ž]^๐‘‡ is the objectโ€™s position and ๐œ‘, ๐œ†, โ„Ž are the latitude, longitude, and height of the objectโ€™s center of mass. is the objectโ€™s velocity relative to the E-frame written in N-frame. r ^๐ธ is the objectโ€™s center-of-mass coordinate vector in e-frame, and is the matrix of direction cosines from E-frame to N-frame. is the vector of specific force in N-frame, where ๐’‡^๐ต is the vector of specific force in B-frame (the output signals of the accelerometer triad) and is the matrix of direction cosines from B-frame to N-frame. are skew-symmetric matrices composed of angular velocities , where is the angular rate vector of E-frame relative to I-frame written in N-frame, and is the angular rate vector of N-frame relative to I-frame written in N-frame. is a diagonal matrix , where M and ๐‘ are radii of curvature of ellipsoid (the figure of the Earth is described by biaxial ellipsoid (Jekeli, 2012). is the gravity vector written in N-frame, where ๐’ˆ^๐‘ = [ 0 0 ๐‘” ]^๐‘‡ is gravitational vector, and g is given according to World Geodetic System WGS-84 and , where is the angular rate vector of E-frame relative to I-frame written in E-frame. The matrix can be represented through the Rodriguesโ€“Hamilton parameters (quaternions) . The time behavior of quaternion is described by the following differential equation: where is the angular rate vector of the B-frame relative to I-frame (the output signals of the gyroscopes triad). The attitude angles (Euler angles) are calculated from quaternion components using the following equations: where ๐œƒ, ๐œ™, ๐œ“ are pitch, roll and yaw angles. 2.3 Measurement model of inertial sensors The inaccurate measurements of inertial sensors are explained by various reasons, among them, nonorthogonality of the measuring axes of the units of accelerometers and gyroscopes, and biases that can be represented as a sum of systematic and random components. The measurement model of accelerometers and gyroscopes based on MEMS technology can be written in a generalized form as (El-Sheimy et al., 2007; Jafari et al., 2014; Quinchia et al., 2013) where are the three-dimensional vectors of the output signals of the accelerometers and the gyroscopes, respectively, and are their true values. are 3ร—3 coefficient matrices. ๐‘ฐ[3] is 3ร— 3 unity matrix. ๐‘ฉ^๐ด,๐‘†,๐‘ฉ^๐บ,๐‘† are the systematic components of the accelerometer biases and gyro drifts, respectively. ๐‘ฉ^๐ด๐ถ๐ถ๐‘…๐‘Š,๐‘ฉ^๐‘…๐‘…๐‘Š are the acceleration and the rate random walks, respectively, and ๐‘พ^๐ด๐ถ๐ถ๐‘…๐‘Š,๐‘พ^ ๐‘…๐‘…๐‘Š are zero-mean white noises. ๐‘ฉ^๐ด,๐บ๐‘€,๐‘ฉ^๐บ,๐บ๐‘€ are first order Gauss-Markov (GM) processes, ๐‘ป^๐ด,๐ถ,๐‘ป^๐บ,๐ถ are 3ร—3 correlation matrices, and ๐‘พ^๐ด,๐บ๐‘€,๐‘พ^๐บ,๐บ๐‘€are zero-mean white noises. ๐‘พ^๐‘‰๐‘…๐‘Š,๐‘พ^๐ด๐‘…๐‘Š are zero-mean white noises that represent the velocity and the angle random walks, respectively, and ๐‘บ^๐ด,๐‘บ^๐บ are coefficient matrices that include scale factors and other coefficients due to the non-orthogonality of measuring axes of the accelerometers and the gyroscopes blocks. 2.4 Unscented Kalman filter The UKF uses a deterministic sampling technique known as the unscented transformation to pick a minimal set of 2L+1 sample points (called sigma points) around the mean, where L is the size of state vector. The sigma points are then propagated through the non-linear functions, from which a new mean and covariance estimate are then formed (Al Bitar & Gavrilov, 2019; Crassidis, 2006). In addition, the UKF removes the requirement to calculate the Jacobians, which can be a difficult task for complex functions. Compared to EKF, UKF is less sensitive to the nonlinearities of process and observation models. In order to apply the algorithm of UKF, it is necessary to write the process and measurement equation of INS/GNSS in discrete time. The process equation in discrete time can be written as where ๐‘ฟ[๐‘˜] is the state-vector of INS and ๐‘พ[๐‘˜] is the vector of process noise: ๐’‡(๐‘ฟ[๐‘˜],๐‘พ[๐‘˜]) is nonlinear vector-function that can be written by transforming Equations (1)โ€“(3) and Equations (8)โ€“(13) into discrete time where ๐‘‡[๐‘†] is the sampling time of INS. The measurement equation of UKF is given as where ๐’[๐‘˜] is the vector of GNSS position and velocity measurements: where ๐Š[๐‘˜] is measurement noise, which is assumed to be zero-mean white noise with covariance matrix ๐‘น[๐‘˜]. The measurement covariance matrix ๐‘น[๐‘˜] is given by where , are the standard deviations of GNSS position and velocity measurementsโ€™ noise, respectively. The UKF can be represented as a โ€œpredictionโ€“ correctionโ€ procedure. To initialize the UKF, weight coefficients are calculated for each of the 2๐ฟ + 1 sigma-points according to the following rules: where ๐›ผ, ๐›ฝ, ๐œ… are parameters that determine the position of sigma points in state-space. ๐›ผ, ๐œ… regulate the spread of points relative to their expected value. Parameter ๐›ฝ is used to incorporate prior knowledge of the distribution. For a normal distribution, it is conventional to set the values of these parameters: ๐œ… = 0, ๐›ฝ = 2, ๐›ผ โˆˆ [10^โˆ’4,1] (Crassidis, 2006). At the prediction stage, 2๐ฟ + 1 sigma-points are generated. These sigma-points form the 2๐ฟ + 1 columns of matrix ๐‘†[๐‘˜โˆ’1] as follows: where are the posterior estimates of statevector and covariance matrix, respectively. ๐‘ธ[๐‘˜] is process noise covariance matrix. The Cholesky decomposition is an effective method for calculating the square root of . Further, to denote a column of ๐‘บ[๐‘˜] we use the index ๐‘— = 0, 1, โ€ฆ, 2๐ฟ. For example, ๐‘บ[๐‘˜,๐‘—] means a j-th column of matrix ๐‘บ[๐‘˜]. The columns of ๐‘บ[๐‘˜] are propagated through nonlinear function ๐’‡(โˆ—) Then, the a priori estimates of state-vector and covariance matrix are calculated using weighted sums At the correction stage, the a priori estimates of statevector and covariance matrix are updated (corrected) The initial value of covariance matrix is given by where denotes the accuracy of the initial alignment of the INS,, denote the initial errors in determining constant gyro drifts and accelerometer biases, ๐ˆ^๐‘…๐‘…๐‘Š,๐ˆ^๐ด๐ถ๐ถ๐‘…๐‘Š are the standard deviations of the noises of the rate and the acceleration random walks, and ๐ˆ^๐บ,๐บ๐‘€,๐ˆ^๐ด,๐บ๐‘€ are the standard deviations of the noises of GM processes. 2.5 Offline stage Four essential tasks are performed in offline stage: 1) the selection of the optimal inputs of NARX networks, 2) the design of the internal structure (the number of layers/neurons) of NARX networks, 3) preliminary training of NARX networks, and 4) calculating the new covariance matrices of UKF that will be used during GNSS outages (in online mode). 2.5.1 Offline data A dataset is acquired from both INS and GNSS during a trip that contains as many maneuvers as possible. The data of INS1 and GNSS are fused by UKF using the loosely coupled scheme, as shown in Figure The target values (the measurement updates) ๐›ฟ๐‘ƒ[๐บ๐‘๐‘†๐‘†/๐ผ๐‘๐‘†2],๐›ฟ๐‘‰[๐บ๐‘๐‘†๐‘†/๐ผ๐‘๐‘†2] are obtained as follows: The target values will have the shape of the signal shown in Figure 5. The INS2 outputs are position ๐‘ท[๐ผ๐‘๐‘†2], velocity ๐‘ฝ[๐ผ๐‘๐‘†2] and attitude angles ๐‘จ[๐ผ๐‘๐‘†2] The goal is to choose a set of inputs from the group ๐ผ[๐ผ๐‘๐‘†2] = {๐œ‘,๐œ†,โ„Ž,๐‘ฃ[๐‘],๐‘ฃ[๐ธ],๐‘ฃ[๐ท],๐œ™,๐œƒ,๐œ“,๐‘ก} that have an impact on each of the six error components from the group ๐‘‡ = {๐›ฟ๐œ‘, ๐›ฟ๐œ†, ๐›ฟโ„Ž, ๐›ฟ๐‘ฃ[๐‘],๐›ฟ๐‘ฃ[๐ธ],๐›ฟ๐‘ฃ [๐ท]}, where t is the time elapsed since the loss of the GNSS signal and varies in interval [0, 60 s]. The measurement updates {๐›ฟ๐œ‘, ๐›ฟ๐œ†, ๐›ฟโ„Ž, ๐›ฟ๐‘ฃ[๐‘],๐›ฟ๐‘ฃ[๐ธ],๐›ฟ๐‘ฃ[๐ท]} are predicted using six NARX networks. The inputs of NARX networks are selected based on MI criterion and the LSE. First, the MI criterion ranks the inputs in group ๐ผ[๐ผ๐‘๐‘†2], where the inputs with positive rank are selected. Then, LSE is used to determine the model orders, that is, the dependency of each error component in group ๐‘‚[๐‘‚๐‘ˆ๐‘‡] = {๐›ฟ๐œ‘, ๐›ฟ๐œ†, ๐›ฟโ„Ž, ๐›ฟ๐‘ฃ[๐‘],๐›ฟ๐‘ฃ[๐ธ],๐›ฟ๐‘ฃ[๐ท]} on the past values of the error itself and the past values of the inputs chosen by MI. Next, the theoretical background of MI and LSE is presented (subsections 2.5.2 and 2.5.3). The structure of NARX network is presented in Subsection 2.6. 2.5.2 Mutual information The relation between the inputs ๐ผ[๐ผ๐‘๐‘†2] and outputs ๐‘‡ is not linear, so the methods based on linear relations (like correlation) are prone to mistakes and will not give accurate results. The MI criterion is a good candidate to solve this problem, as it measures the arbitrary (linear or nonlinear) dependencies between variables. MI is widely used in machine learning for canonical tasks, such as classification, clustering and feature selection (Brown, 2009; Peng et al., 2005). MI is one of the feature selection methods. These methods define a statistical criterion that is used to rank characteristics (or features) according to their usefulness for classification. The characteristics with a high rating are chosen, and characteristics with a low rating can be discarded. Given two random variables ๐‘ฅ and ๐‘ฆ, their mutual information is determined through probability density functions ๐‘(๐‘ฅ), ๐‘(๐‘ฆ), ๐‘(๐‘ฅ, ๐‘ฆ) where ๐ท[๐‘ฅ],๐ท[๐‘ฆ] are the spaces of (๐‘ฅ, ๐‘ฆ). The concept of MI can be employed for arranging or ranking a number of candidates (features) ๐‘ฅ[๐‘™],๐‘™ = 1,โ€ฆ,๐‘€ according to their usefulness or influence on target signal ๐‘ฆ, based on ๐‘ samples of ๐‘ฅ[๐‘™] and ๐‘ฆ. Peng et al. (2005) proposed a solution for this problem by calculating a rank based on the Maximum Relevance Minimum Redundancy (MRMR) criterion. The rank is calculated as follows: The MRMR criterion takes into account the redundancy of candidates and subtracts it from ๐ผ(๐‘ฅ[๐‘™];๐‘ฆ). The MRMR criterion can be summarized as โ€œa set of features should not only be individually relevant, but also should not be redundant in relation to each other.โ€ By calculating ๐ฝ[๐‘€๐‘…๐‘€๐‘…](๐‘ฅ[๐‘™]), the features can be arranged or ranked; the feature with the largest ๐ฝ[๐‘€๐‘…๐‘€๐‘…] has the largest effect or influence on ๐‘ฆ and vice versa. The MRMR criterion is applied to rank the candidate inputs in group ๐ผ[๐ผ๐‘๐‘†2] in accordance to their impact on each of the six error components from the group ๐‘‡. The results of applying MI criterion are provided in Subsection 3.2. 2.5.3 Lag-space estimation The next task is to investigate the dependency of each error component in group ๐‘‡ on the past values of the error itself and the past values of the inputs that were chosen by MRMR criterion. This problem is referred to as lag-space estimation, or model order estimation. The use of higher order dependencies in the modeling process leads to possibly over-parameterized, and thus less efficient, models. It is therefore important to estimate the optimal lag-space, that is, to find the primary dependencies. This allows minimizing the number of parameters and optimizing the predictive abilities of the AI module (NARX module, in our case). In system theory, a nonlinear dynamical system is generally described by differential or difference equations that represent input/output relations. However, in many practical situations, it is difficult to write down the accurate state dynamics and observation equations for a continuous or a discrete time system. What are available are the input and output data of the unknown dynamical system, that is, ๐‘ข(๐‘ก) and ๐‘ฆ(๐‘ก), which are observed at sampling times ๐‘ก[๐‘–] = ๐‘–๐‘‡[๐‘†],๐‘– = 0,โ€ฆ,๐‘ โˆ’ 1. It has been shown that under some mild assumptions, the following input/output model (Siegelmann et al., 1997): can represent nonlinear dynamical systems described by differential or difference equations, where ๐‘”(โˆ—) is nonlinear function, and parameters ๐‘›[๐‘ฆ] and ๐‘›[๐‘ข] are orders of the input/output model that should be determined. In our case, ๐‘ข(๐‘ก), is a subset of the group ๐ผ[๐ผ๐‘๐‘†2] and ๐‘ฆ(๐‘ก) is one of the six error components in group ๐‘‡. Authors He and Asada (1993) proposed a method to identify model orders ๐‘›[๐‘ฆ],๐‘›[๐‘ข] based on Lipschitz quotients. According to them, the model described by Equation 38 can be written as where are the input variables and n is the number of input variables ๐‘› = ๐‘›[๐‘ฆ] +๐‘›[๐‘ข] +1. Now, the goal is to reconstruct the function ๐‘”(โˆ—) based on the pairs (). The Lipschitz quotient for ๐‘š input variables can be calculated by Usually, the following index is used to determine the optimal amount of input variables where ๐‘ž^(๐‘š)(๐‘˜) is the k-th largest Lipschitz quotient among all with m input variables, and p is a positive parameter (๐‘ = 0.01๐‘ โˆผ 0.02๐‘). If n is the optimal number of input variables, then ๐‘ž^(๐‘›+1) is very close to ๐‘ž^(๐‘›) and ๐‘ž^(๐‘›โˆ’1) is much larger than ๐‘ž^(๐‘›). Moreover, ๐‘ž^(๐‘›โˆ’2) is much larger than ๐‘ž^(๐‘›โˆ’1) and ๐‘ž^(๐‘›+2) is very close to ๐‘ž^(๐‘›+1). Therefore, looking at the curve of ๐‘ž^(๐‘š) as a function of ๐‘š, we can observe that starting from a certain value ๐‘š = ๐‘›, further increase in m will not significantly change the index ๐‘ž^(๐‘š) and thus the value of n can be determined (Figure 6). The results of applying this LSE method are provided in Subsection 3.2. 2.5.4 The reconfiguration of the covariance matrices of UKF During GNSS outages, the proposed system works in prediction mode. The measurements obtained by the aid of NARX networks have error characteristics that differ from the characteristics of GNSS measurements. Therefore, it is necessary to reconfigure the UKF covariance matrices. To do this, the following steps are performed: First, the proposed system is applied using the offline dataset for simulated GNSS outages. Then, the standard deviation of the position and velocity errors with respect to GNSS measurements is calculated as follows: where , are the position and the velocity obtained by applying the proposed system, ๐[๐‘ƒ],๐[๐‘‰] are the mean values of the errors, and ๐‘[๐‘] is the number of data samples. When a real GNSS outage occurs, the corresponding components of the covariance matrices ๐‘ท^๐ถ๐‘‚๐‘‰,๐‘น are updated with ,. 2.6 NARX neural network Considering the system model given by Equation (38), a good choice of AI module is NARX, as it obeys the same system equation. The NARX is a recurrent dynamic neural network that can be used to model extensive variety of nonlinear dynamic systems. NARX networks have been applied in various applications, including black-box system identification and time-series modeling (Diaconescu, 2008; Siegelmann et al., 1997). The architecture of the NARX is shown in Figure 7. Six NARX networks are utilized to predict systems errors ๐›ฟ๐‘ƒ[๐บ๐‘๐‘†๐‘†/๐ผ๐‘๐‘†2] = [ ๐›ฟ๐œ‘ ๐›ฟ๐œ† ๐›ฟโ„Ž ]^๐‘‡, ๐›ฟ๐‘‰[๐บ๐‘๐‘†๐‘†/๐ผ๐‘๐‘†2] = [ ๐›ฟ๐‘ฃ[๐‘] ๐›ฟ๐‘ฃ[๐ธ] ๐›ฟ๐‘ฃ[๐ท] ]^๐‘‡ during GNSS outages. 3 RESULTS 3.1 Experimental setup Raw experimental data were acquired from a Micro-Electro Mechanical System-Strapdown Inertial Navigation System (MEMS-SINS) (Ekinox-D Inertial Navigation System) with sampling frequency 200 Hz. The characteristics of gyroscopes and accelerometers of this SINS were obtained in (Gonzalez & Dabove, 2019; Gonzalez et al., 2017) using Allan variance method. A Global Navigation Satellite System/ Global Positioning System (GLONASS/GPS) receiver was used with sampling frequency of 5 Hz. The accuracy in position is 0.5 m for latitude and longitude and 1 m for altitude. The accuracy in velocity for all components is 0.1 m/s. Both systems were mounted on the roof of the vehicle, as shown in Figure 8a. The experiment was conducted in the city of Turin in Italy (Gonzalez et al., 2017). The duration of the dataset used in this work is 2,300 s. The trajectory of the vehicle is shown in Figure 8b. The first segment of the trajectory (the first 600 s) is magnified. The dataset of the first segment is used for offline stage. The rest of the dataset (1,700 s) is used for online validation of the proposed system. Table 1 shows the specifications of the gyroscopes and the accelerometers of the Ekinox-D INS (Gonzalez & Dabove, 2019). 3.2 The results of the offline stage The dataset of the first segment of the trajectory is used in offline stage. Figure 9a shows the first segment of the trajectory, and Figure 9b shows the speed of the vehicle ||๐‘ฝ|| along this segment. For better readability, the speed of the vehicle is shown in (km/h). The segment contains many types of possible maneuvers (accelerating and decelerating, zero velocity, straight lines, turning, etc.). Using the dataset of the first segment, six GNSS outages (six windows) were simulated to form the target signals {๐›ฟ๐œ‘, ๐›ฟ๐œ†, ๐›ฟโ„Ž, ๐›ฟ๐‘ฃ[๐‘],๐›ฟ๐‘ฃ[๐ธ],๐›ฟ๐‘ฃ[๐ท]} as shown in Figure 10. Figure 11 shows the candidate input signals {๐œ‘,๐œ†,โ„Ž,๐‘ฃ[๐‘],๐‘ฃ[๐ธ],๐‘ฃ[๐ท],๐œ™,๐œƒ,๐œ“} along the first segment. The first task is to identify the inputs that influence each of the targets {๐›ฟ๐œ‘, ๐›ฟ๐œ†, ๐›ฟโ„Ž, ๐›ฟ๐‘ฃ[๐‘],๐›ฟ๐‘ฃ[๐ธ],๐›ฟ๐‘ฃ[๐ท]} using the MRMR criterion explained earlier. The results of applying MRMR criterion are presented in Table 2. The large negative score means that the input has a high redundancy, that is, the information imbedded in this input are found in other inputs, so there is no need to consider this input. The positive score (in bold) means high relevance and low redundancy of the input. The scores close to zero (underlined) reflect an insignificant influence of the inputs on the corresponding output and can be neglected. As a result, the inputs with high positive scores are chosen. It is worth mentioning that the results given in Table 2 are limited to land vehicles, and they cannot be generalized for the case of aerial vehicles because they have different types of movement. The second task is to determine the dependency of the target signals {๐›ฟ๐œ‘, ๐›ฟ๐œ†, ๐›ฟโ„Ž, ๐›ฟ๐‘ฃ[๐‘],๐›ฟ๐‘ฃ[๐ธ],๐›ฟ๐‘ฃ[๐ท]} on their past values and the past values of selected inputs, that is, the inputs selected using MRMR algorithm and shown in bold in Table 2, using LSE explained earlier. Figure 12 shows the results of applying LSE to investigate the dependency of the target signals {๐›ฟ๐œ‘, ๐›ฟ๐œ†, ๐›ฟโ„Ž, ๐›ฟ๐‘ฃ[๐‘],๐›ฟ๐‘ฃ[๐ธ],๐›ฟ๐‘ฃ [๐ท]} on their past values. To determine the proper lag-space, we look at the point where increasing the lag-space (๐‘š) will not change the order index ๐‘ž^(๐‘š) significantly. As Figure 12 shows, a lag-space of ๐‘›[๐‘ฆ] = 2 is a good choice for all target signals, as the order index doesnโ€™t change significantly for values larger than 2. Next, the LSE is applied to determine the lag-space for inputs selected using MRMR algorithm. Figure 13 shows the results of applying LSE for the case of target signal ๐›ฟ๐œ‘ and the selected inputs {๐œ‘, ๐‘ฃ[๐‘]}. As Figure 13 shows, the order index doesnโ€™t change significantly for values larger than 2. This means that the lag-space is ๐‘›[๐‘ข] = 2. For the other cases {๐›ฟ๐œ†, ๐›ฟโ„Ž, ๐›ฟ๐‘ฃ[๐‘],๐›ฟ๐‘ฃ[๐ธ],๐›ฟ๐‘ฃ[๐ท]} the same lag-space was found (๐‘›[๐‘ข] = 2). At this stage, all inputs of the six NARX networks are determined. Table 3 summarizes the results of applying LSE. The symbol (-) means that there is no dependency found between the output and the input. Figure 14 shows the final input/output configurations of the six NARX networks according to Tables 2 and 3. The next task is the design of the internal structure of the NARX networks, that is, the number of hidden layers and neurons. It is proved that an artificial neural network (ANN) with two hidden layers can approximate any function (Gonzalez & Dabove, 2019; Lippmann, 1987). Therefore, six ANN with two hidden layers were utilized. Choosing the right number of neurons is very important. An ANN with a small number of neurons will not be able to learn. A large number of neurons will lead to an increase in the network training time and can also lead to over-fitting. In this article, the number of neurons is derived empirically using rules-of-thumb (Goodfellow et al., 2016; Huang, 2003). It is shown that an ANN with two hidden layers and neurons can learn ๐‘[๐‘†] example with any arbitrary precision, where ๐‘[๐‘ฆ] = 1 is the number outputs. Here ๐‘[๐‘†] represents the number of samples in the window ๐‘[๐‘†] = ๐‘Š samples. The selection of this value of window size is based on offline trials. In fact, there is a trade-off in choosing the window size; large window sizes guarantee that more motion dynamics are mimicked, thus providing better accuracy over long GNSS outages. On the other hand, a small window size guarantees fast learning, but the system provides high accuracy of estimation only for short GNSS outages. The window size is chosen as (๐‘Š = 300 samples), which represents 60 s record of data, because the sampling frequency of GNSS is 5Hz. Therefore, the number of neurons is ๐‘ = 50. In fact, ๐‘ = 50 is only a starting value for the number of neurons. The final values were 30, 36, 20, 32, 36 and 40 for NARX-1, NARX-2, NARX-3, NARX-4, NARX-5 and NARX-6. These values were achieved after many offline trials. The hyperbolic tangent sigmoid (tan-sigmoid) transfer function is applied as activation function. To train the NARX networks, Levenberg-Marquardt (LM) (Morรฉ, 1978) training algorithm is used. LM algorithm is the most widely used optimization algorithm. It outperforms other methods in a wide variety of problems because of its fast and stable convergence. The LM algorithm is well suited for training small and medium-sized problems, which are the case here. Figure 15 shows the results of preliminary training of the NARX networks based on the offline dataset. 3.3 Online validation of the proposed system In order to test the proposed system, the second part of the dataset (with duration of 1,700 s) is used. Six GNSS outages were simulated, as shown in Figure 16. The outages contain straight lines (outages 1, 2 and 3) and turnings (outages 4, 5 and 6). Figure 17 shows the speed of the vehicle ||๐‘ฝ|| along the trajectory with the GNSS outages highlighted. It can be seen that the GNSS outage segments contain accelerating and decelerating (all outages), zero velocity (outages 1, 2 and 3), high speed โˆผ 50-90 km/h (outages 2, 4, 5 and 6), low speed โˆผ 0-20 km/h (outages 1, 2 and 3), and mid speed โˆผ 20-50 km/h (outages 1 and 2). Figure 18 shows the GNSS-outage segments of the vehicleโ€™s trajectory. The proposed method (NARX-aided UKF) is compared to UKF and two widely adopted methods that use different input configurations in order to validate the selection of the input configurations of NARX networks. The first method (shortly M1) uses the current information of specific force and angular rates for estimating the position and velocity errors as in (Chen & Fang, 2014; Jingsen et al., 2016 ). The second method (shortly M2) uses the four-step information of specific force, angular rates, velocity and yaw as in (Fang et al., 2020). For comparison, the reference GNSS trajectory is also shown in Figure18. The superiority of the proposed method over UKF, M1 and M2 in terms of positioning accuracy is obvious. Figure 19 shows the horizontal error (with respect to GNSS) in position using UKF, M1, M2 and NARX-aided UKF. Table 4 provides numerical values of horizontal errors using different methods. The proposed method (NARX-aided UKF) improved the positioning accuracy by 82%, 65% and 55% with respect to UKF, M1 and M2 respectively. To demonstrate the effect of modifying the covariance matrices of UKF during GNSS outages, Figure 20 shows the horizontal errors in position in two cases: the first case, the NARX-aided UKF without modification of covariance matrices, and the second case with modification of covariance matrices. It can be seen that the updating of covariance matrices slightly enhanced the positioning accuracy. The improvement of positioning accuracy is 5% to 8%. The problem of aiding UKF during GNSS outages in INS/GNSS systems was considered in this paper. A new method, namely NARX-aided UKF, was suggested and tested. The proposed method consists of offline and online stages. The offline stage is essential for selecting the input signals of NARX networks based on MI and LSE, the design of the internal structure of NARX networks, preliminary training of NARX networks and calculating the new covariance matrices of UKF that were used during real GNSS outages. The covariance matrices of UKF during GNSS outages were linked to prediction accuracy of NARX networks. The performance of the proposed method was experimentally verified using real datasets. The results indicated that the proposed method improved the accuracy of navigation systems during GNSS outages. The results also confirmed the superiority of the proposed method over widely adopted methods that use different input configurations for neural networks. The future work will consider the case of aerial vehicles. Al Bitar N, Gavrilov A. A novel approach for aiding unscented Kalman filter for bridging GNSS outages in integrated navigation systems. NAVIGATION. 2021;68:521โ€“539. https://doi.org/10.1002/navi.435 The authors would like to thank professors Rodrigo Gonzalez (National University of Technology, Mendoza, Argentina) and Paolo Dabove (Politecnico di Torino University, Turin, Italy) for providing the experimental data. โ€ข Received August 27, 2020. โ€ข Revision received March 14, 2021. โ€ข Accepted March 24, 2021. โ€ข ยฉ 2021 Institute of Navigation This is an open access article under the terms of the Creative Commons Attribution License, which permits use, distribution and reproduction in any medium, provided the original work is properly
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What are the characteristics of a three-dimensional work? โ€“ - La Cultura de los Mayas What are the characteristics of a three-dimensional work? โ€“ What are the characteristics of a three-dimensional work? โ€“ Three-dimensional works of art have three dimensions: height, width and depth, whose shapes can be geometric and organic. โ€“ They can be appreciated from any angle or perspective, unlike two-dimensional works of art, which can only be seen from the front. What arts generate three-dimensional objects? Sculpture is a true three-dimensional representation of an object that has height, width and depth, can be walked through and viewed from different angles. You can also bring three-dimensionality to a drawing or painting through a method called perspective and shading. What are three-dimensional images? They are the three dimensions that make up the three-dimensional representation and, therefore, are present in any 3D animation project. In fact, in our reality everything is three-dimensional because it has length, height and depth. What is the name of the three dimensions? Three dimensions of space: height, width and depth. What are projections in space? Informally, a projection is a drawing technique used to represent a three-dimensional object on a surface. Mathematically, a projection is an idempotent linear transformation onto a vector space. How is the illusion of three-dimensional form achieved in drawing? In figurative drawing, the illusion of three dimensions is usually named as volume. The essential resources to achieve this volume are the basic geometry and chiaroscuro. However, a truly three-dimensional drawing involves considering more elements than just those that produce the volume. What is a three-dimensional figure in arts? Three-dimensional art is the one that is not flat, but has volume; unlike a drawing or a painting that are two-dimensional arts. Three-dimensional figures are also called solids, they are a portion of space bounded by flat or curved faces. How is three-dimensionality achieved in a plane? Representation of three-dimensional space in the plane. Objects that are further away from the observer lose saturation, that is, they are seen as lighter or more gray. You can use resizing and toning at the same time. What is used to represent three-dimensionality or suggest volume and depth in the plane? 4- CONICAL PERSPECTIVE: It is a representation system studied by architects and draftsmen to create a sense of depth in the plane and has its own rules of representation. What is depth in artistic drawing? Depth is the distance of an element from a horizontal reference plane when said element is below the reference. When the opposite occurs, it is called elevation, level, or simply height. What is form and depth in plastic composition? How did they represent depth in ancient times? In preclassical antiquity, in pre-Columbian paintings from Mesoamerica, and in Oriental painting, the usual means of suggesting depth is the different elevation of the figures relative to the lower edge of the relief or painting. In Roman painting, we already find geometric and atmospheric perspective. What are the elements of the sculpture? CONSTITUTIVE ELEMENTS OF SCULPTURE The space exists, the material exists; there is gravity, proportionsโ€ฆ While painting is characterized by its optical nature claimed by color, sculpture is characterized by its physical nature, essentially tactile. How is sculpture perceived? Finish, texture and polychromy โ€ข The sculpture is perceived through the outer surface, its shape-surface and its complete perception should be through touch as well as sight. sculpture are recognizable in nature but non-existent or unrecognizable volumes may appear.
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Integral of Introduction to integral of sin^3x In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration solver calculates the integrals. This process is defined as finding an antiderivative of a function. Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function cubic sine. You will also understand how to compute sin cube integral by using different integration techniques. What is the integration of sin cube x? The integral of sin^3x is an antiderivative of sine function which is equal to cos3x/3 โ€“ cos x + c. It is also known as the reverse derivative of sine function which is a trigonometric identity. The sine function is the ratio of opposite side to the hypotenuse of a triangle which is written as: Sin = opposite side / hypotenuse Integral of sin cube x formula The formula of sin cube integration contains integral sign, coefficient of integration and the function as sine. It is denoted by โˆซ(sin^3x)dx. In mathematical form, the integral of sin^3x is: $โˆซ\sin^3xdx=\frac{\cos3x}{3}โ€“ \cos x + c$ Where c is any constant involved, dx is the coefficient of integration and โˆซ is the symbol of integral. Similarly, the integral sin square is equal to x/2 - (sin2x)/4 + c. How to calculate the integral of sin cube x? The integral of sin^3x is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of sine by using: 1. Integration by parts 2. Substitution method 3. Definite integral Integral of sin x cubic by using integration by parts The derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. The integration by parts is a method of solving integral of two functions combined together. Letโ€™s discuss calculating the integral of sin squared x by using by parts integral calculator. Proof of sin cube integration by using integration by parts Since we know that the integration of sin cube x can be written as the product of two functions. Therefore, we can calculate the integral of sin^3x by using integration by parts. For this, suppose $I=\sin^3x=\sin x.\sin^2x$ Applying the integral we get, $I=โˆซ\sin x.\sin^2xdx$ Since the method of integration by parts is: $โˆซ[f(x).g(x)]dx= f(x).โˆซg(x)dx - โˆซ[fโ€™(x).โˆซg(x)dx]dx$ Now replacing f(x) and g(x) by sin x, we get, $I=-\sin^2x.\cos x + โˆซ[2\sin x\cos x.\cos x]dx$ It can be written as: $I= -\sin^2x.\cos x + 2โˆซ[\sin x.\cos^2x]dx$ Now by using a trigonometric identity cos^2x = 1 โ€“ sin^2x. Therefore, substituting the value of cos2x in the above equation, we get: $I=-\sin^2x.\cos x+2โˆซ\sin x(1 โ€“ \sin^2x)dx$ Integrating remaining terms, $I=-\sin^2x.\cos x-2\cos x โ€“ 2\int\sin^3x$ Since we know that, $I=\int \sin^3xdx$ $I= -\sin^2x.\cos x -2\cos x โ€“ 2I$ $I=-(1-\cos^2x)\cos x -2\cos x-2I$ $I=-\cos x+\cos^3x-2\cos x-2I$ $3I =\cos^3x -3cos x$ Hence the integral of sin^3x is equal to, $โˆซ\sin^3xdx =\frac{\cos^3x}{3}-\cos x+c$ Also, you can find the integral of sin(e^x) by using the integration by parts formula. Integral of sin^3x by using substitution method The substitution method involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Letโ€™s understand how to prove the integral of sin cube by using the substitution method. Proof of integral of sin cube x by using substitution method To proof the integral of sin^3x by using substitution method, suppose that: $I = โˆซ\sin^3xdx$ Suppose that we can write the above integral as: $I = โˆซ\sin x.\sin^2xdx$ By using trigonometric identities, we can write the above equation by using sin^2x = 1 โ€“ cos^2x, therefore, $I = โˆซ\sin x.( 1 โ€“ \cos^2x)dx$ $I= โˆซ(\sin x โ€“ \sin x\cos^2x)dx$ Now to evaluate first integral, we will use the following steps, $I_1 = โˆซ\sin x.dx = - \cos x$ Now to evaluate second integral, $I_2 = -โˆซ\sin x\cos^2xdx$ Suppose that u = cos x and du = - sin x dx, then $I_2 = โˆซu^2du$ Integrating with respect to u. $I_2 = \frac{u^3}{3}$ Substituting the value of u we get, $I_2 =\frac{\cos^3x}{3}$ Now, using the value of first and second integral in the above equation to get final value of integral. $I = -\cos x+ \frac{\cos^3x}{3}+c$ Hence the sin cube integration is verified by using substitution method. It can be also verified by using the trigonometric substitution. Integration of sin cube x by using definite integral The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as: $โˆซ^b_a f(x) dx = F(b) โ€“ F(a)$ Letโ€™s understand the verification of the integral of sin^3x by using the indefinite integral. Proof of integral of sin^3x by using definite integral To compute the integral of sin^3x by using a definite integral, we can use the interval from 0 to 2ฯ€ or 0 to ฯ€. Letโ€™s compute the integral of sin^3x from 0 to 2ฯ€. The indefinite integral of sin^3x can be written as: $โˆซ^2ฯ€_0 \sin^3x dx=\left|\frac{\cos^3x}{3} โ€“\cos x\right|^{2ฯ€}_0$ Substituting the value of limit we get, $โˆซ^{2ฯ€}_0 \sin^3x dx=\frac{\cos^3 2ฯ€}{3} -\frac{\cos^3 0}{3} โ€“[\cos 2ฯ€ - \cos 0]$ Therefore, the integral of sin^3x from 0 to 2ฯ€ is $โˆซ^{2ฯ€}_0 \sin^3x dx=-\frac{1}{2}$ Which is the calculation of the definite integral solver of sin^3x. Now to calculate the integral of sin cube x between the interval 0 to ฯ€, we just have to replace ฯ€ by ฯ€. Therefore, $โˆซ^ฯ€_0 \sin^3xdx=\left|\frac{\cos^3x}{3} โ€“\cos x\right|^ฯ€_0$ $โˆซ^ฯ€_0 \sin^3x dx=\left[\frac{\cos^3ฯ€}{3}-\frac{\cos^3 0}{3}\right]โ€“[\cos ฯ€-\cos 0]$ $โˆซ^ฯ€_0 \sin^3xdx=\frac{1}{2}โ€“1 โ€“ 1 +1$ $โˆซ^ฯ€_0 \sin^3xdx=-\frac{1}{2}$ Therefore, the integral of sin^3x from 0 to ฯ€ is -1/2.
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Statistics Definition and Other Terms To Know We all study statistics in our school time as a mathematics subject and think statistics is mathematics? But do you know what exact statistics do? Donโ€™t worry; statistics is not just about solving probability exercises and Data Interpretation. It is much more than that. Many people do not know the exact statistics definition and other terms used in statistics. This blog will explore the correct definition of statistics and other terms such as sample, population, etc. We will also discuss types of statistics and how it works with data. Letโ€™s start with the basic definition of statistics. Statistics Definition Statistics is a mathematics branch that deals with the collection, description, analysis, interpretation of data. It covers every aspect of all types of data, such as qualitative data and quantitative data. In statistics, we use different mathematical theories such as integral calculus, probability theory, algebra, etc., for the various calculations. Statisticians focus on drawing reliable and accurate conclusions from large data sets and populations. They observe the behavior and other characteristics of small samples and draw a valuable conclusion after analysis. In simple words, we can say that statistics aims at data manipulation that involves different ways to collect data, review, analyze, and develop conclusions. Statistics calculations and conclusions are very important for the business organizations and companies to make effective decisions and grow by leaps and bounds. Types of Statistics Methods In statistics, we use two types of methods: Descriptive and Inferential methods. Descriptive Statistics: It focuses on the sample data and its central tendency, variability, and distribution. Where central tendency defines the estimate of the elementโ€™s characteristics of a sample and population, it involves the mean, mode, and median determination of data. Apart from the statistics definition we should know the definition of other statistical terms. Variability means a group of statistics that describe how much difference lies among the sample or population elements as well as their observed characteristics. It includes metrics like range, variance, and standard deviation. Distribution terms indicate the aggregated shape of the data that can be represented on a chart and graph, such as bar graph, histogram, dot plot, etc. It also reveals the difference among measured characteristics of the data set elements. Descriptive statistics enable us to understand all the properties of the data sample elements and form the hypothesis to make predictions by using inferential statistics. Inferential Statistics Inferential statistics aims to draw conclusions on behalf of the characteristics of a population. Statisticians use this type of statistics to make decisions regarding the reliability of findings. In particular inferential statistics is used to generalize large data groups, such as estimating average demand for a product. It considers the consumerโ€™s habits to predict future scenarios. In inferential statistics, regression analysis is mostly used in order to calculate the strength and nature of the correlation between independent and dependent variables. Inferential statistics works on the hypothesis and draws the conclusion on the basis of these hypotheses. Definition of Other Key Terms Apart From Statistics Definition Population- Population is a particular collection of objects of interest. It covers a large amount of data.A population is not specific to number of people, it can be number of phone calls, number of students, number of products etc. Sample: Sample is a subset of the population that consists of all the characteristics of the population. If a sample is not having the features of the population, the sample is of no use. To draw valuable information it is important to take a sample that includes all the properties of the population. Parameter: It is a number to summarize all features of the entire population as a whole. This number describes the entire population such as population mean whereas statistics describe the sample such as sample mean. Qualitative Data: Qualitative data is classified by its properties, labels, attributes, and other features. It is non-statistical data and can be semi structured or unstructured. It is present in an unfixed form and can not be calculated. For example, โ€œThe furniture is made of woodโ€,is qualitative data. Quantitative Data: Quantitative data is fixed and easy to determine. This data is statistical data and is always present in descriptive form. Quantitative data is further divided into two categories: Discrete and continuous data. โ€œI have three pairs of shoesโ€ is Quantitative data. Outliers: Outliers are the data values that fall out of the data sequence. Too small and too large values in a data set are considered as outliers. These outliers affect the conclusion and outcome of the statistical functions. Due to these unmatched values, a statistician may interpret a wrong decision. This blog has discussed the statistics definition and other key terms that a student or a business person should know to run the business smoothly. These terms are important to know to avoid any confusion in mind. I hope you will find this blog important for you to know which terms are used in statistics. You must be logged in to post a comment.
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Interagency Modeling and Analysis Group React.Seq.FullEnz.Kinetics is the fourth and last of a series to illustrate the influences of capacitances on the duration of transients in the reaction series A --> B --> C --> D--> E in a compartment. It includes the on- and off kinetics for enzyme-substrate binding and release for each reaction; the capacitive delay is the dwell time in the ES complex. A reaction sequence A-->B-->C-->D-->E can be represented many ways to approximate the biological form of the reactions. This version takes into account the on-off rates of the enzymatic reactions resulting in a capacitance within the system which was approximated in the previous model ,MM with lag (model #425). The model generates progress curves as in a bioreactor, but with additional factor of a flow through the mixing chamber. The flow term is first order for all solutes in the sequence. The inflowing initial solute A, the flow, ml/sec, Flow*CinA, equals the sum of the outflow clearance plus the last reaction E--> ?, i.e. G*E, then the outflow concentration of A in the steady state goes to 50% of CinA (Do this by setting Flow1 = Flow2 = 0.05 ml/sec and setting GA1 and GA2 also to 0.05 and run the program. The other reaction rates, to form C, D, and E are set identidal to that to form B from A, the result is that steady state concentrations for A, B, etc go to 0.5, 0.25, 0.125, 0.0625, and 0.03125, all at one half of its predecessor. This program illustrates the transient delays between steady state, and the form of the transients. With MM kinetics there is really no enzyme there, and no binding of the reactants in the process of the reaction. All of the delay is due to the combination of flow and the reaction, exactly similar to the first order reaction kinetic model (Model #432). The initial conditions are zero for all solutes so the time constant for the initial entry would be simply the volume divided by flow, Vol/Flow1, in seconds, if there were no reaction. The reaction, augmenting the disappearance of A, shortens the time constant so that it is Vol/(GA + Flow1). The second transient is due to a step increase (or decrease) in flow at time TFjump (at t=30 s with this parameters set ('full181028'). The third transient is a step change in the reaction rates at time TGjump. The Verification Process is the same as for the linear kinetics, and verifies the code at steady-state with a input solute concentration (CinA) of 1 mM. The solution to the differential equation for solute A is accomplished using the numerical solvers, but the steady state solution is also solvable analytically. The three transients, at t = 0, t= 30 s and t=60 s, are expressed in three analytical equations. Their sum fits exactly the numerical solutions to the systems equations at steady-state (Anal3). VERIFICATION! Try this out using the difference ODE solvers: it turns out the DOPRIS5 (an advanced RungeKutta algorithm) gives a more precise fit to the analytical solution than either of the more powerful stiff solvers, CVODE or Radau. At the steep parts of the transients DOPRIS5 is still good to 7 decimal digits, while the other supposedly superior solvers, CVODE and Radau are good only to 5 or 6 decimal digits. In the steady state they all give the same correct answers. For solutes B through E, the analytical solutions are more complex, and have not been developed. For these it is much faster to use the numerical solutions to the ODEs; the analytical solutions for solute E would be, because of their complexity, much slower to compute, and maybe not even as accurate as the numerical solutions, even for this rather simple model system. What we know is that the steady state solution match exactly to the predicted steady state values for A through E. The intent for this series of models is to account for substrate capacitance in enzymatic networks: the first model is Linear.Reaction.Sequence (Model 423), without capacitance; the second (Model 424) is MM.React.Seq, a Michaelis-Menten system without any capacitance. The third, MM.Lag.ReactSeq (Model 425), adds time lags to account for capacitance, which equals "bound substrate mass / flux of substrate. The fourth, this model React.Seq.FullEnz.Kinetics (Model 426), abandons the MM assumptions, and adds the on- and off kinetics for a single uncomplicated reaction. The Michaelis-Menten Model (#425) that approximates the effect of solute binding to the enzymes attempts to correct the kinetics of M-M expressions by incorporating lags to represent the buffering capacitance and consequent delay in transient responses in enzyme systems. In this model React.Seq.FullEnzKinetics.proj (#426) the capacitance is accounted for by the kinetic expressions automatically, without using any lag. The approximation in this model is that the enzymes Zs and their complexes, the ZsS, do NOT get washed out by the flow, but are completely retained in the compartment. (This is physiologically reasonable only if enzymes are fixed on the wall and diffusion distances are small.) The step responses, PLOT 1.StepsA_E, using the unit step input function with CinA = 1 mM (or other level), are set up for a step increase in flow at t = TFjump and then later at t= TGjump a step in reaction rates. These are all, necessarily and unavoidably, delayed by the capacitance for substrate in the enzyme-substrate complex ZsS (where Zs is the enzyme for reacting substrate S). These can be compared with the solutions of model #425 where the lag is approximated by a concentration-dependent first order time lag. On the PLOT StepA_E the curve Anal3 (thick black dashes) represents the solutions to the first order kinetic model (#423) and the unadorned MM model (#424) both of which lack capacitance. The difference between the black dashed curve and the red curve is due to the capacitance of the enzyme-substrate complex, ZaA. Figure: Progress curves for a sequence of reactions from substrate A to E in a compartment with flow through the compartment. All substrates have initial concentration of zero with A in (CinA) set to 1 mM. Flow doubles from 0.025 to 0.05 ml/sec at t= 30 sec. Substrate consumption factor (dG dimensionless) doubles from 0.05 to 0.1 at t= 60 sec to counter the change in flow. Anal3 is the analytical solution for substrate concentration A(t) for the equivalent first order kinetic model lacking enzyme binding. The delay between Anal3 (black dashes) and A(red curve) is due to the ES capacitance. The equations for this model may be viewed by running the JSim model applet and clicking on the Source tab at the bottom left of JSim's Run Time graphical user interface. The equations are written in JSim's Mathematical Modeling Language (MML). See the Introduction to MML and the MML Reference Manual. Additional documentation for MML can be found by using the search option at the Physiome home Download JSim model project file โ€ข Download JSim model MML code (text): โ€ข Download translated SBML version of model (if available): โ–ก No SBML translation currently available. Model Feedback We welcome comments and feedback for this model. Please use the button below to send comments: Easterby, JS. A generalized theory of the transition time for sequential enzyme reactions. Biochem J. 199: 155-161, 1981. Cascante M, Melendez-Hevia E, Kholodenko B, Sicilia J, and Kacser H. Control analyis of transit time for free and enzyme-bound metabolites: physiological and evolutionary significance of metabolic response times. Biochem J 308: 895-899, 1995. Bassingthwaighte JB. Capacitance in metabolic netowrks. 2019 (in prep for submission to Biophys J) Key terms first order unidirectional reactions substrate capacitance enzymatic networks compartmental on off kinetics Cardiac Grid reaction sequence Please cite https://www.imagwiki.nibib.nih.gov/physiome in any publication for which this software is used and send one reprint to the address given below: The National Simulation Resource, Director J. B. Bassingthwaighte, Department of Bioengineering, University of Washington, Seattle WA 98195-5061. Model development and archiving support at https://www.imagwiki.nibib.nih.gov/physiome provided by the following grants: NIH U01HL122199 Analyzing the Cardiac Power Grid, 09/15/2015 - 05/31/2020, NIH /NIBIB BE08407 Software Integration, JSim and SBW 6/1/09-5/31/13; NIH/NHLBI T15 HL88516-01 Modeling for Heart, Lung and Blood: From Cell to Organ, 4/1/07-3/31/11; NSF BES-0506477 Adaptive Multi-Scale Model Simulation, 8/15/05-7/31/08; NIH/NHLBI R01 HL073598 Core 3: 3D Imaging and Computer Modeling of the Respiratory Tract, 9/1/04-8/31/09; as well as prior support from NIH/NCRR P41 RR01243 Simulation Resource in Circulatory Mass Transport and Exchange, 12/1/1980-11/30/01 and NIH/NIBIB R01 EB001973 JSim: A Simulation Analysis Platform, 3/1/02-2/28/07.
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Droplet Mix : about stroboscopic effect Bouncing, colliding, mixing, dancing, floating droplets 5 Visitor Comments 1. Dear Heligone, Well, after what I think is my most arduous work on some text in several years, I finally got my first post out. I talk a little about you too. โ€œClassical Fluid Model of Particle Physicsโ€ on Tumblr.com I hope you got my last reply to your comments. I talked about the LEDs and computer visual object recognition. I also replied that I need a little more time to present the new developments I there were in the subjects. Making this first post took a really long time. I wasnโ€™t expecting for that. I actually broke down my introduction into two parts so I could post earlier. It was already taking two full days of work. The second part should be easier since its mostly done. But I am prioritizing the new subject, so if the second part takes too long I will skip it until I post on the news. โ–ก And thanks again for spreading the word about walkingdroplet.com on your tumblr. Iโ€™m really eager to to read about your developments soon ! 2. Dear Heligone, My new article is here! Classical Fluid Dynamics Theory of Mechanics It is about that new exquisite theoretical approach I indicated before. It took a long time to get this article ready and even know I think I would like to improve it further, but I also wanted you to read about these developments. So I pushed it and there it is. Robert Brady, physicist at Cambridge University in England, has developed a new study on Fluid Dynamics. He arrives at structures called sonons. These are derived from purely Classical Physics. They are patterns of flow in an inviscid compressible fluid. Sonons are quasi-particles. They are actually intrinsically relativistic quantum particles! Their interactions show intrinsic attraction and repulsion according to their fluxes chiral orientations! They oscillate and radiate. Robert Brady deduces Maxwell Equations of Electromagnetism. From the same principles that he deduces Relativity, Quantum Mechanics and the structure of the sonon. Sonons are attracted to masses. They are a source of the gravity field and are attracted by such fields that are created by large masses! You will be amongst the first to appreciate this amazing discovery as he finished his article on this research in January 2013 and I first learned about it last week. A historical achievement we are now part of. I also notice your makeover of the blog! It looks great! The website description sentence is beautifully placed. That sliding bar with the recent articles, their pictures and descriptions is phenomenal. You were very tasteful and skillful to make this, congratulations. I only miss a little box section with your scholarly materials. That I think is a very important to be visible and accessible on the front page, it gives the your whole work great value. I also suggest you to write more about the experiments and some reviews about the academic papers, no matter how raw or tentative, it shows visitors some of your own ideas. Do you have an email I can reach you from time to time? If you can sent a quick message to the one I used to post so I can have your email I would appreciate. I hope your week is being productive and joyful, โ–ก Dear Daniel, Your recent post has made me discover the work of Robert Brady, and I thank you for that, because it is very interesting โ€ฆand furthermore related to my own workshop experiments ๐Ÿ™‚ I have not yet fully understood how he can relate its โ€œsononโ€ to Couderโ€™s Walking droplet, but I will try and study more. It has given me the idea to try and build a sonon generator machine, inspired by the way doplhins do โ€ฆ Maybe in a few years because I have still many things to do with droplets. I also appreciate your encouragment and advice about my blog, which I surely will follow โ€ฆ Slowly by slowly. About the links to scholarly materials, they are still available at the โ€œbibliographyโ€ link in the menu bar. I also plan to write more, like reviews of article, but as you may know, this is very time consuming โ€ฆ ๐Ÿ˜‰ Check your inbox for my email 3. Dear Heligone, You are welcome for the points on the work of Robert Brady. Your idea about an experimental sonon device is amazing! Of course your work with the droplets should continue, they are of great importance. Brady talks about the sonon and the walking droplet correspondent properties in his presentation at Warwick. He displays the report on Yves Couder from his TV appearance and then talks about that: http://youtu.be/fmzRqAi2cDs?t=9m2s About the Bibliography link, I was thinking you should put a little box with the specific articles titles. It doesnโ€™t have to be big. The same size as the Categories box would be good. Thats what I meant and I think it will be worth it because it gives skeptical people direct access to scientific literature and shows Yves original work as a established scientist from which you drew To ease up the work on writing things, I think you could write some blurbs, short sentences and ideas, in similar fashion of lecture notes, nothing fancy. And then publish those on a special feed on your blog, or going as the text for your video posts, so that readers have little peek on whats on your mind. I am really looking forward for how you plan to replicate the walking droplet double slit experiment. And your videos are also becoming much more visually attractive, keep it up. sincerely yours, Classical Fluid Dynamics Theory of Mechanics Post a Reply
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Interest Rate Definition and Legal Meaning On this page, you'll find the legal definition and meaning of Interest Rate, written in plain English, along with examples of how it is used. What is Interest Rate? The amount of a loan that is repaid on top of the loan amount in return for the borrowed money. History and Meaning of Interest Rate Interest rates have been in use since ancient times when people started to borrow money from each other. In simple terms, an interest rate is the cost of borrowing money from a lender. It refers to the percentage that the lender charges the borrower for the loaned amount. The concept of interest has evolved over the years and has become a major part of modern economic systems. The central banks of various countries use interest rates to manage inflation, lending rates, and economic growth. Examples of Interest Rate 1. A savings account may offer a 2% interest rate, which means that the bank will pay 2% of the total amount deposited as interest annually. 2. A mortgage loan may have a 5% interest rate, which means that the borrower has to pay an additional 5% of the total loan amount to the lender. 3. An investor may earn a 7% interest rate on a bond investment, which means that the issuer will pay 7% annually on the bond's face value. Related Terms 1. Annual Percentage Rate (APR) - APR refers to the total cost of borrowing money annually, including interest rate and other fees. 2. Compound Interest - Compound interest refers to the additional interest earned on the principal amount and the accumulated interest over time. 3. Prime Rate - The prime rate is the interest rate at which banks lend money to their most creditworthy customers. 4. Fixed Interest Rate - A fixed interest rate is a predetermined interest rate that remains the same during the loan period. 5. Variable Interest Rate - A variable interest rate is an interest rate that changes periodically during the loan period, based on market fluctuations.
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Positive EV Betting Calculator Bet Logical What is the Positive EV Betting Calculator? The positive EV Betting Calculator letโ€™s you check if a potential bet has a positive expected value for 2-way and 3-way markets. By comparing odds offered by a soft sportsbook against the line available at a sharp sportsbook. The calculator will estimate the long term profitability of the bet by devigging the sharp line and comparing the true odds to the soft odds. Additionally, the calculator can also work out the optimal bet stake for your bet according to Kelly Criterion staking theory. How to use the Positive Expected Value Calculator The first thing we need to do when using the calculator is set the odds format that we want to use. The calculator currently support Moneyline Odds (American Odds) and Decimal Odds. We can select our odds format using the dropdown menu at the top of the page. Next we want to set the calculator to either a 2-Way Market or a 3-Way Market using the dropdown menu to the right hand side of the calculator. Once this market is set we want to enter the betting odds for the bet we are looking to check the expected value of. Additionally we can add a bet stake if we already have decided how much we want to bet and donโ€™t need (or want) to using Kelly Bet Staking Alternatively we can have to calculator work out the optimal bet stake for us given our bankroll by filling out the Kelly Criterion Staking section. (Note: The Bet Stake will adjust once we add the odds from the sharp sportsbook.) The last piece of information that we need to enter into the calculator are the odds from the sharp sportsbook that weโ€™ll use to determine if the bet is a positive ev bet. Important: Make sure that the odds of the betting outcome we are comparing is in the Outcome One Odds box The profit margin and expected value of the bet will be displayed at the bottom of the calculator. How to find Positive Expected Value Bets?
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