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10000	https://www.thermopedia.com/de/content/553/	AIR (PROPERTIES OF) Atmospheric air is a mixture of nitrogen and oxygen being the earth atmosphere. Main components of air which are practically the same throughout the globe are nitrogen (78.08 volume per cent) and oxygen (20.95 v.%). Along with them air contains 0.94 v.% of inert gases and 0.03 v.% of carbon dioxide. The air of such a composition is named dry. Its molecular mass is regarded to be M = 28.96 g/mole. In the lower atmosphere strata the air contains also water vapor, its concentration is substantially variable depending on the partial water vapor pressure at the appropriate temperature and relative humidity. For instance at 20°C and relative humidity 80% air contains about 0.02 v.% of water vapor. In the air layers adjacent to the earth surface other components may be present being in most cases of antropogenic origin. At ambient pressure and temperature air can be regarded as a perfect gas, its properties may be described by equations: where v denotes specific volume; u is specific internal energy; R is the gas constant for air. At low temperatures the air is liquified. The normal (at 0.1013 MPa) boiling (condensation) temperature of the oxygen is equal— 183°C, that of the nitrogen -195.8°C. Liquid air at atmospheric pressure behaves practically as an ideal solution following the Raoult's Law. The normal condensation temperature of air is -191.4°C, the normal boiling temperature -194°C. At elevated temperatures air undergoes some physicochemical transformations. The nitrogen reacts with oxygen producing various oxides: N2O, NO, NO2, NO3. Their equilibrium concentration can be derived from the isotherm equations of the respective reactions. At temperatures higher than 2000 K and moderate pressures the nitrogen and oxygen start to dissociate, and at temperatures exceeding 4000 K and atmospheric pressure the ionization of oxygen, nitrogen, and other components becomes evident. This implies the transition of air into the plasma state. The equilibrium dissociation degree can be calculated according to the Saha equation. The thermodynamic properties of air along the saturation curve are given in Table 1; these properties for the liquid and gaseous air—in Table 2. Table 1. Thermodynamic properties of air along the saturation curve Table 2. Thermodynamic properties of liquid and gaseous air The enthalpy is taken as zero at an arbitrary point. The entropy is taken zero for the solid air at 0K. Air is a mixture mainly consisting of diatomic gases. Therefore its heat capacity at close to normal temperatures and pressures may with good accuracy be taken equal to where With increasing temperature the heat capacity slightly increases due to exciting of the vibrational degrees of freedom in the oxygen and nitrogen molecules. Table 3 gives air heat capacity values for a wide range of temperatures and pressures. Table 3. Air heat capacity cp, KJ/kg · K As for all pure substances in the supercritical region, the isobars and isotherms of the heat capacity cp have maximums the steeper the closer to the critical point. The temperature dependence of the viscosity of air is qualitatively the same as for pure substances: in the liquid phase the viscosity decreases with temperature following an approximately exponential function; in the gas phase at low pressures the viscosity increases according to equation: with increasing pressure at constant temperature the viscosity increases. This dependence is most strong in the vicinity of the critical point. Air viscosity values at various temperatures and pressures are given in Table 4. Table 4. Air viscosity η · 107, N · s/m2 The behavior of the thermal conductivity of air is similar to the viscosity: in the liquid phase with growing temperature the heat conductivity decreases whereas in the gas phase-increases. At low pressures the temperature dependence is described by the equation: Along the isotherm with increasing pressure the thermal conductivity increases. In Table 5 the air thermal conductivity is given at various temperatures and pressures. Table 5. Air thermal conductivity λ · 103, W/m · K At low pressures and high temperatures the thermal conductivity sharply increases due to dissociation. With growing temperature the thermal conductivity goes through maximums which are connected with maximum heat transfer by the heats of respective reactions. Thermal conductivities of air at dissociation conditions are given in Table 6. Table 6. Air thermal conductivity at high temperatures λ · 103, W/m · K REFERENCES Additional information about air properties can be found in the following literature: Handbook, edited by V. P. Glushko (1978) "Nauka" Publishing House, Moskow (in Russian). Wassermann, A. A. and Rabinovitch, V. A. (1968) Thermophysical properties of liquid air and its components. Standarts Publishing House, Moscow (in Russian). Handbook Thermophysical Properties of Gases and Liquids, edited by N. B. Vargaftic (1972) "Nauka" Publishing House, Moscow (in Russian). Verweise Verwandte Inhalte in anderen Produkten
10001	https://www.sydney.edu.au/content/dam/students/documents/mathematics-learning-centre/integration-definite-integral.pdf	Mathematics Learning Centre Introduction to Integration Part 2: The Deﬁnite Integral Mary Barnes c ⃝1999 University of Sydney Contents 1 Introduction 1 1.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 Finding Areas 2 3 Areas Under Curves 4 3.1 What is the point of all this? . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.2 Note about summation notation . . . . . . . . . . . . . . . . . . . . . . . . 6 4 The Deﬁnition of the Deﬁnite Integral 7 4.1 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 5 The Fundamental Theorem of the Calculus 8 6 Properties of the Deﬁnite Integral 12 7 Some Common Misunderstandings 14 7.1 Arbitrary constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 7.2 Dummy variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 8 Another Look at Areas 15 9 The Area Between Two Curves 19 10 Other Applications of the Deﬁnite Integral 21 11 Solutions to Exercises 23 Mathematics Learning Centre, University of Sydney 1 1 Introduction This unit deals with the deﬁnite integral. It explains how it is deﬁned, how it is calculated and some of the ways in which it is used. We shall assume that you are already familiar with the process of ﬁnding indeﬁnite inte-grals or primitive functions (sometimes called anti-diﬀerentiation) and are able to ‘anti-diﬀerentiate’ a range of elementary functions. If you are not, you should work through Introduction to Integration Part I: Anti-Diﬀerentiation, and make sure you have mastered the ideas in it before you begin work on this unit. 1.1 Objectives By the time you have worked through this unit you should: • Be familiar with the deﬁnition of the deﬁnite integral as the limit of a sum; • Understand the rule for calculating deﬁnite integrals; • Know the statement of the Fundamental Theorem of the Calculus and understand what it means; • Be able to use deﬁnite integrals to ﬁnd areas such as the area between a curve and the x-axis and the area between two curves; • Understand that deﬁnite integrals can also be used in other situations where the quantity required can be expressed as the limit of a sum. A B C D A B D C Mathematics Learning Centre, University of Sydney 2 2 Finding Areas Areas of plane (i.e. ﬂat!) ﬁgures are fairly easy to calculate if they are bounded by straight lines. The area of a rectangle is clearly the length times the breadth. The area of a right-angled triangle can be seen to be half the area of a rectangle (see the diagram) and so is half the base times the height. Area of rectangle Area of triangle =length × breadth =1 2 area of rectangle = 1 2 length × breadth The areas of other triangles can be found by expressing them as the sum or the dif-ference of the areas of right angled trian-gles, and from this it is clear that for any triangle this area is half the base times the height. Area of △ABC Area of △ABC = area of △ABD = area of △ABD + area of △ACD −area of △ACD Using this, we can ﬁnd the area of any ﬁgure bounded by straight lines, by di-viding it up into triangles (as shown). Areas bounded by curved lines are a much more diﬃcult problem, however. In fact, although we all feel we know intuitively what we mean by the area of a curvilinear ﬁgure, it is actually quite diﬃcult to deﬁne precisely. The area of a ﬁgure is quantiﬁed by asking ‘how many units of area would be needed to cover it?’ We need to have some unit of area in mind (e.g. one square centimetre or one square millimetre) and imagine trying to cover the ﬁgure with little square tiles. We can also imagine cutting these tiles in halves, quarters etc. In this way a rectangle, and hence any ﬁgure bounded by straight lines, can be dealt with, but a curvilinear ﬁgure can never be covered exactly. We are therefore forced to rely on the notion of limit in order to deﬁne areas of curvilinear ﬁgures. To do this, we make some simple assumptions which most people will accept as intuitively obvious. These are:-1. If one ﬁgure is a subset of a second ﬁgure, then the area of the ﬁrst will be less than or equal to that of the second. Mathematics Learning Centre, University of Sydney 3 2. If a ﬁgure is divided up into non-overlapping pieces, the area of the whole will be the sum of the areas of the pieces. Using these assumptions, we can approximate to curved ﬁgures by means of polygons (ﬁgures with straight line boundaries), and hence deﬁne the area of the curved ﬁgure as the limit of the areas of the polygons as they ‘approach’ the curved ﬁgure (in some sense yet to be made precise). x y a b Area y = f(x) x subdivisions y a b y = f(x) x lower sum y a b y a b x upper sum Mathematics Learning Centre, University of Sydney 4 3 Areas Under Curves Let us suppose that we are given a positive function f(x) and we want to ﬁnd the area enclosed between the curve y = f(x), the x-axis and the lines x = a and x = b. (The shaded area in the diagram.) If the graph of y = f(x) is not a straight line we do not, at the moment, know how to calculate the area precisely. We can, however, approximate to the area as follows: First we divide the area up into strips as shown, by dividing the interval from a to b into equal subinter-vals, and drawing vertical lines at these points. Next we choose the least value of f(x) in each subin-terval and construct a rectangle with that as its height (as in the diagram). The sum of the areas of these rectangles is clearly less than the area we are trying to ﬁnd. This sum is called a lower sum. Then we choose the greatest value of f(x) in each subinterval and construct a rectangle with that as its height (as in the diagram opposite). The sum of the areas of these rectangles is clearly greater than the area we are trying to ﬁnd. This sum is called an upper sum. Thus we have ‘sandwiched’ the area we want to ﬁnd in between an upper sum and a lower sum. Both the upper sum and the lower sum are easily calculated because they are sums of areas of rectangles. y x division into narrower strips f(x') a b x y ∆ ∆x y = f(x) x f(x) Mathematics Learning Centre, University of Sydney 5 Although we still can’t say precisely what the area under the curve is, we know between what limits it lies. If we now increase the number of strips the area is divided into, we will get new upper and lower sums, which will be closer to one another in size and so closer to the area which we are trying to ﬁnd. In fact, the larger the number of strips we take, the smaller will be the diﬀerence between the upper and lower sums, and so the better approximation either sum will be to the area under the curve. It can be shown that if f(x) is a ‘nice’ function (for example, a continuous function) the diﬀerence between the upper and lower sums approaches zero as the number of strips the area is subdivided into approaches inﬁnity. We can thus deﬁne the area under the curve to be: the limit of either the upper sum or the lower sum, as the number of subdivi-sions tends to inﬁnity (and the width of each subdivision tends to zero). Thus ﬁnding the area under a curve boils down to ﬁnding the limit of a sum. Now let us introduce some notation so that we can talk more precisely about these con-cepts. Let us suppose that the interval [a, b] is divided into n equal subintervals each of width ∆x. Suppose also that the greatest value of f(x) in the ith subinterval is f(x∗ i ) and the least value is f(x ′ i). Then the upper sum can be written as: f(x∗ 1)∆x + f(x∗ 2)∆x + ... + f(x∗ n)∆x or, using summation notation: n i=1 f(x∗ i )∆x. Similarly, the lower sum can be written as: f(x ′ 1)∆x + f(x ′ 2)∆x + ... + f(x ′ n)∆x or, using summation notation: n i=1 f(x ′ i)∆x. With this notation, and letting A stand for the area under the curve y = f(x) from x = a to x = b, we can express our earlier conclusions in symbolic form. The area lies between the lower sum and the upper sum and can be written as follows: n i=1 f(x ′ i)∆x ≤A ≤ n i=1 f(x∗ i )∆x. Mathematics Learning Centre, University of Sydney 6 The area is equal to the limit of the lower sum or the upper sum as the number of subdivisions tends to inﬁnity and can be written as follows: A = lim n→∞ n i=1 f(x ′ i)∆x or A = lim n→∞ n i=1 f(x∗ i )∆x. 3.1 What is the point of all this? Well, ﬁrstly it enables us to deﬁne precisely what up till now has only been an impre-cise intuitive concept, namely, the area of a region with curved lines forming part of its boundary. Secondly it indicates how we may calculate approximations to such an area. By taking a fairly large value of n and ﬁnding upper or lower sums we get an approximate value for the area. The diﬀerence between the upper and lower sums tells us how accurate this approximation is. This, unfortunately, is not a very good or very practical way of approximating to the area under a curve. If you do a course in Numerical Methods you will learn much better ways, such as the Trapezoidal Rule and Simpson’s Rule. Thirdly it enables us to calculate areas precisely, provided we know how to ﬁnd ﬁnite sums and evaluate limits. This however can be diﬃcult and tedious, so we need to look for better ways of ﬁnding areas. This will be done in Section 5. At this stage, many books ask students to do exercises calculating upper and lower sums and using these to estimate areas. Frequently students are also asked to ﬁnd the limits of these sums as the number of subdivisions approaches inﬁnity, and so ﬁnd exact areas. We shall not ask you to do this, as it involves a great deal of computation. 3.2 Note about summation notation The symbol (pronounced ‘sigma’) is the capital letter S in the Greek alphabet, and stands for ‘sum’. The expression 4 i=1 f(i) is read ‘the sum of f(i) from i = 1 to i = 4’, or ‘sigma from i = 1 to 4 of f(i)’. In other words, we substitute 1, 2, 3 and 4 in turn for i and add the resulting expressions. Thus, 4 i=1 xi stands for x1 + x2 + x3 + x4, 5 i=1 i2 stands for 12 + 22 + 32 + 42 + 52, and 2 i=1 f(x1)∆x stands for f(x1)∆x + f(x2)∆x. Mathematics Learning Centre, University of Sydney 7 4 The Deﬁnition of the Deﬁnite Integral The discussion in the previous section led to an expression of the form A = lim n→∞ n i=1 f(xi)∆x (1) where the interval [a, b] has been divided up into n equal subintervals each of width ∆x and where xi is a point in the ith subinterval. This is a very clumsy expression, and mathematicians have developed a simpler notation for such expressions. We denote them by b a f(x)dx which is read as ‘the integral from a to b of f(x)dx’. The sign is an elongated ‘s’ and stands for ‘sum’, just as the did previously. The diﬀerence is that in this case it means ‘the limit of a sum’ rather than a ﬁnite sum. The dx comes from the ∆x as we pass to the limit, just as happened in the deﬁnition of dy dx. Thus the deﬁnite integral is deﬁned as the limit of a particular type of sum i.e. sums like that given in (1) above, as the width of each subinterval approaches zero and the number of subintervals approaches inﬁnity. 4.1 Notes 1. Although we used the area under a curve as the motivation for making this deﬁnition, the deﬁnite integral is not deﬁned to be the area under a curve but simply the limit of the sum (1). 2. Initially, when discussing areas under curves, we introduced the restriction that f(x) had to be a positive function. This restriction is not necessary for the deﬁnition of a deﬁnite integral. 3. The deﬁnition can be made more general, by removing the requirement that all the subintervals have to be of equal widths, but we shall not bother with such generali-sations here. 4. Sums such as (1) are called Riemann sums after the mathematician Georg Riemann who ﬁrst gave a rigorous deﬁnition of the deﬁnite integral. 5. The deﬁnition of a deﬁnite integral requires that f(x) should be deﬁned everywhere in the interval [a, b] and that the limit of the Riemann sums should exist. This will always be the case if f is a continuous function. c x A(x) y = f(t) t y ∆ t y c x P Q S R x + x f(x) T Q U P S R ∆x f(x') Mathematics Learning Centre, University of Sydney 8 5 The Fundamental Theorem of the Calculus So far, we have deﬁned deﬁnite integrals but have not given any practical way of calcu-lating them. Nor have we shown any connection between deﬁnite integrals and diﬀeren-tiation. Let us consider the special case where f(t) is a con-tinuous positive function, and let us consider the area under the curve y = f(t) from some ﬁxed point t = c up to the variable point t = x. For diﬀerent values of x we will get diﬀerent areas. This means that the area is a function of x. Let us denote the area by A(x). Clearly, A(x) increases as x increases. Let us try to ﬁnd the rate at which it increases, that is, the deriva-tive of A(x) with respect to x. At this point, recall how we ﬁnd derivatives from ﬁrst principles: Given a function f(x), we let x change by an amount ∆x, so that f(x) changes to f(x+∆x). The derivative of f(x) is the limit of f(x + ∆x) −f(x) ∆x as ∆x →0. We shall go through this process with A(x) in place of f(x). When we increase x by ∆x, A(x) increases by the area of the ﬁgure PQRS. That is, (see the diagram) A(x + ∆x) −A(x) = area PQRS. Now that the area PQRS is bounded by a curved line at the top, but it can be seen to lie in between the areas of two rectangles: area PURS < area PQRS < area TQRS. Both of these rectangles have width ∆x. Let the height of the larger rectangle be f(x∗) and the height of the smaller rectangle f(x ′). (In other words, x∗ and x ′ are the values of x at which f(x) attains its maximum and minimum values in the interval from x to x + ∆x.) Thus area PURS = f(x ′)∆x and area TQRS = f(x∗)∆x. So, f(x ′)∆x ≤A(x + ∆x) −A(x) ≤f(x∗)∆x. Mathematics Learning Centre, University of Sydney 9 Now if we divide these inequalities all through by ∆x, we obtain f(x ′) ≤A(x + ∆x) −A(x) ∆x ≤f(x∗). Finally, if we let ∆x →0, both f(x ′) and f(x∗) approach f(x), and so the expression in the middle must also approach f(x), that is, the derivative of A(x), dA dx = f(x). This result provides the link we need between diﬀerentiation and the deﬁnite integral. If we recall that the area under the curve y = f(t) from t = a to t = x is equal to x a f(t)dt, the result we have just proved can be stated as follows: d dx x a f(t)dt = f(x). (2) This is the Fundamental Theorem of the Calculus. In words If we diﬀerentiate a deﬁnite integral with respect to the upper limit of inte-gration, the result is the function we started with. You may not actually use this result very often, but it is important because we can derive from it the rule for calculating deﬁnite integrals: Let us suppose that F(x) is an anti-derivative of f(x). That is, it is a function whose derivative is f(x). If we anti-diﬀerentiate both sides of the equation (2) we obtain x a f(t)dt = F(x) + c. Now we can ﬁnd the value of c by substituting x = a in this expression. Since a a f(t)dt is clearly equal to zero, we obtain 0 = F(a) + c, and so c = −F(a). Thus x a f(t)dt = F(x) −F(a), or, letting x = b, b a f(t)dt = F(b) −F(a). This tells us how to evaluate a deﬁnite integral • ﬁrst, ﬁnd an anti-derivative of the function • then, substitute the upper and lower limits of integration into the result and sub-tract. Note A convenient short-hand notation for F(b) −F(a) is [F(x)]b a. y y = x –2x+5 3 x 1 2 Mathematics Learning Centre, University of Sydney 10 To see how this works in practice, let us look at a few examples: i Find 1 0 x2dx. An anti-derivative of x2 is 1 3x3, so we write 1 0 x2dx = 1 3x3 1 0 = 1 3 (1)3 −1 3 (0)3 = 1 3. ii Find π 0 sin tdt. π 0 sin tdt = [−cos t]π 0 = −cos(π) + cos 0 = −(−1) + 1 = 2. iii Find the area enclosed between the x-axis, the curve y = x3−2x+5 and the ordinates x = 1 and x = 2. In a question like this it is always a good idea to draw a rough sketch of the graph of the function and the area you are asked to ﬁnd. (See below) If the required area is A square units, then A = 2 1 x3 −2x + 5 dx = x4 4 −x2 + 5x 2 1 = (4 −4 + 10) − 1 4 −1 + 5 = 53 4. Exercises 5 1. a. 2x34 2 b. 1 x2 3 1 c. √x 16 9 d. [ln x]4 2 2. a. 9 4 1 √xdx b. π 2 −π 2 cos tdt x y 1 2 1 y = x + 1 2 1 3 y = 1/x x y √ 2 4 u 2 v v = (4 – u) y π 0 y = 2sint 2 t Mathematics Learning Centre, University of Sydney 11 c. 2 1 1 y2dy d. −1 −2 (s2 + 2s + 2)ds 3. Find the area of the shaded region in each of the diagrams below: a. b. c. d. 4. Evaluate a. 1 0 xex2dx b. −1 −2 1 3 −xdx c. π 2 0 sin 2ydy d. 5 1 t 4 + t2dt a c x y b b x y y = g(x) y = f(x) a Mathematics Learning Centre, University of Sydney 12 6 Properties of the Deﬁnite Integral Some simple properties of deﬁnite integrals can be derived from the basic deﬁnition, or from the Fundamental Theorem of the Calculus. We shall not give formal proofs of these here but you might like to think about them, and try to explain, to yourself or someone else, why they are true. a. a a f(x)dx = 0. If the upper and lower limits of the integral are the same, the integral is zero. This becomes obvious if we have a positive function and can interpret the integral in terms of ‘the area under a curve’. b. If a ≤b ≤c, c a f(x)dx = b a f(x)dx + c b f(x)dx. This says that the integral of a function over the union of two intervals is equal to the sum of the integrals over each of the intervals. The diagram opposite helps to make this clear if f(x) is a positive function. c. b a cf(x)dx = c b a f(x)dx for any constant c. This tells us that we can move a constant past the integral sign, but beware: we can only do this with constants, never with variables! d. b a (f(x) + g(x))dx = b a f(x)dx + b a g(x)dx. That is, the integral of a sum is equal to the sum of the integrals. e. If f(x) ≤g(x) in [a, b] then b a f(x)dx ≤ b a g(x)dx. That is, integration preserves inequalities between functions. The diagram opposite explains this result if f(x) and g(x) are positive functions. c a b x y c b – a y = f(x) M m a b x y Mathematics Learning Centre, University of Sydney 13 f. b a cdx = c(b −a). This tells us that the integral of a constant is equal to the product of the constant and the range of integra-tion. It becomes obvious when we look at the diagram with c > 0, since the area represented by the integral is just a rectangle of height c and width b −a. g. We can combine (e) and (f) to give the result that, if M is any upper bound and m any lower bound for f(x) in the interval [a, b], so that m ≤f(x) ≤M, then m(b −a) ≤ b a f(x)dx ≤M(b −a). This, too, becomes clear when f(x) is a positive func-tion and we can interpret the integral as the area un-der the curve. h. Finally we extend the deﬁnition of the deﬁnite integral slightly, to remove the restric-tion that the lower limit of the integral must be a smaller number than the upper limit. We do this by specifying that a b f(x)dx = − b a f(x)dx. For example, 1 2 f(x)dx = − 2 1 f(x)dx. Mathematics Learning Centre, University of Sydney 14 7 Some Common Misunderstandings 7.1 Arbitrary constants When you ﬁrst learned how to ﬁnd indeﬁnite integrals (anti-derivatives), you probably also learned that it was important to remember always to add an arbitrary constant to the answer. There is no arbitrary constant in a deﬁnite integral. If we interpret a deﬁnite integral as an area, it is clear that its value is a ﬁxed number (the number of units of area in the region). There is no ambiguity, and so no need to add an arbitrary constant - in fact, it is wrong to do so. When we apply the Fundamental Theorem of the Calculus to ﬁnding a deﬁnite integral, however, the possibility of an arbitrary constant appears to arise. For example, in calculating 2 1 x2dx, we have to ﬁnd an anti-derivative for x2. The most natural choice would be 1 3x3, but instead of that we could choose 1 3x3 + c, where c is any constant. Then, 1 0 x2dx = 1 3x3 + c 1 0 = 1 3(1)3 + c − 1 3(0)3 + c = 1 3. Note that the constants cancel one another out, and we get the same answer as we did before. Thus we might as well take the simplest course, and forget about arbitrary constants when we are calculating deﬁnite integrals. 7.2 Dummy variables What is the diﬀerence between b a f(x)dx and b a f(t)dt? Let’s work them both out in a special case. 4 2 1 xdx = [ln x]4 2 = ln 4 −ln 2. 4 2 1 t dt = [ln t]4 2 = ln 4 −ln 2. So both integrals give the same answer. It is clear that the value of a deﬁnite integral depends on the function and the limits of integration but not on the actual variable used. In the process of evaluating the integral, we substitute the upper and lower limits for the variable and so the variable doesn’t appear in the answer. For this reason we call the variable in a deﬁnite integral a dummy variable - we can replace it with any other variable without changing a thing. Thus, b a f(x)dx = b a f(y)dy = b a f(t)dt = b a f(θ)dθ. 0 x y ∆x f(x) y = f(x) Mathematics Learning Centre, University of Sydney 15 8 Another Look at Areas We have deﬁned the deﬁnite integral b a f(x)dx as the limit of a particular type of sum, without placing any restrictions on whether the function f(x) is positive or negative. We know that, if f(x) is positive, b a f(x)dx is equal to the area between the curve y = f(x), the x-axis and the ordinates x = a and x = b, (which we refer to as ‘the area under the curve’). The natural question to ask now is: what does b a f(x)dx equal if f(x) is negative? Can we represent it as an area in this case too; perhaps ‘the area above the curve’? If we go back to the deﬁnition of b a f(x)dx as the limit of a sum, we can see clearly that if f(x) is always negative then each of the terms f(xi)∆x will also be negative (since ∆x is positive). So the sum n i=1 f(xi)∆x will be a sum of negative terms and so will be negative too. And when we let n approach inﬁnity and pass to the limit, that will be negative also. Thus, if f(x) is negative for x between a and b, b a f(x)dx will also be negative. Now areas are, by deﬁnition, positive. Remember that, in section 1, we explained that we can measure the area of a region by counting the number of little square tiles (each of unit area) needed to cover it. Since we can’t cover a region with a negative number of tiles (it doesn’t make sense to talk of it) we can’t have a negative area. On the other hand, if we ignore the fact that each of the terms f(x)∆x is negative, and consider its numerical value only, we can see that it is numerically equal to the area of the rectangle shown. And, if we go through the usual process, adding up the areas of all the little rectangles and taking the limit, we ﬁnd that b a f(x)dx is numerically equal to the area between the curve and the x-axis. So to ﬁnd the area, we calculate b a f(x)dx, which will turn out to be negative, and then take its numerical (i.e. absolute) value. x 0 1 –1 Mathematics Learning Centre, University of Sydney 16 To see this more clearly, let’s look at an example. Consider the curve, y = x(x2 −1). This is a cubic curve, and cuts the x-axis at −1, 0 and 1. A sketch of the curve is shown below. Let us ﬁnd the shaded area. First we calculate the deﬁnite integral 1 0 x(x2 −1)dx. 1 0 x(x2 −1)dx = 1 0 (x3 −x)dx = 1 4x4 −1 2x2 1 0 = 1 4 −1 2 −(0 −0) = −1 4. Since x(x2 −1) is negative when x lies between 0 and 1, the deﬁnite integral is also negative, as expected. We can conclude that the area required is 1 4 square units. As a check, let us ﬁnd the area of the other ‘loop’ of the curve, i.e. the area between the curve and the x-axis from −1 to 0. Since x(x2 −1) is positive for this range of values of x, the area will be given by 0 −1 x(x2 −1)dx = 1 4x4 −1 2x2 0 −1 = (0 −0) − 1 4 −1 2 = 1 4. This is the answer we would expect, since a glance at the diagram shows that the curve has ‘point symmetry’ about the origin. If we were to rotate the whole graph through 180◦, the part of the curve to the left of the origin would ﬁt exactly on top of the part to the right of the origin, and the unshaded loop would ﬁt on top of the shaded loop. So the areas of the two loops are the same. Now let us calculate 1 −1 x(x2 −1)dx. 1 −1 x(x2 −1)dx = 1 4x4 −1 2x2 1 −1 = 1 4 −1 2 − 1 4 −1 2 = 0. This makes it very clear that a deﬁnite integral does not always represent the area under a curve. x y 0 1 –1 2 A B Mathematics Learning Centre, University of Sydney 17 We have found that 1. If f(x) is positive between a and b, then b a f(x)dx does represent the area under the curve. 2. If f(x) is negative between a and b, then b a f(x)dx represents the area above the curve, since the value of b a f(x)dx is negative. 3. If f(x) is sometimes positive and sometimes negative between a and b, then b a f(x)dx measures the diﬀerence in area between the part above the x-axis and the part below the x-axis. (In the example above, the two areas were equal, and so the diﬀerence came out to be zero.) Let’s look at another example. Consider the function y = (x + 1)(x −1)(x −2) = x3 −2x2 −x + 2. This is a cubic function, and the graph crosses the x-axis at −1, 1 and 2. A sketch of the graph is shown. The area marked A is given by 1 −1(x3 −2x2 −x + 2)dx = 1 4x4 −2 3x3 −1 2x2 + 2x 1 −1 = 1 4 −2 3 −1 2 + 2 − 1 4 + 2 3 −1 2 −2 = −4 3 + 4 = 22 3. So the area of A is 22 3 square units. The area marked B can be found by evaluating 2 1 x3 −2x2 −x + 2 dx. This works out as −5 12. (The details of the calculation are left to you.) So the area of B is 5 12 square units. y x a c b Mathematics Learning Centre, University of Sydney 18 If we calculate 2 −1(x3 −2x2 −x + 2)dx the answer will be the diﬀerence between the area of A and the area of B, that is, 21 4 square units. (Check it out for yourself.) If we want the total area enclosed between the curve and the x-axis we must add the area of A and the area of B. i.e. 2 2 3 + 5 12 = 3 1 12 square units. WARNING In working out area problems you should always sketch the curve ﬁrst. If the function is sometimes positive and sometimes negative in the range you are interested in, it may be necessary to divide the area into two or more parts, as shown below. The area between the curve and the x-axis from a to b is NOT equal to b a f(x)dx. Instead, it is c a f(x)dx + \| b c f(x)dx\|. Before you can calculate this, you must ﬁnd the value of c, i.e. ﬁnd the point where the curve y = f(x) crosses the x-axis. Exercises 8 1. Find the area enclosed by the graph of y = 3x2(x −4) and the x-axis. 2. i Find the value of 2π 0 sin xdx. ii Find the area enclosed between the graph of y = sin x and the x-axis from x = 0 to x = 2π. 3. Find the total area enclosed between the graph of y = 12x(x + 1)(2 −x) and the x-axis. x y y = f(x) y = g(x) x y y = f(x) y = g(x) f(x') – g(x) i i x y (2,2) (2,–3) (4,–2) (4,–3) Mathematics Learning Centre, University of Sydney 19 9 The Area Between Two Curves Sometimes we want to ﬁnd, not the area between a curve and the x-axis, but the area enclosed between two curves, say between y = f(x) and y = g(x). We can approach this problem in the same way as before by dividing the area up into strips and approx-imating the area of each strip by a rectangle. The lower sum is found by calculating the area of the in-terior rectangles as shown in the diagram. The height of each interior rectangle is equal to the diﬀerence between the least value of f(x), f(x′), and the greatest value of g(x), g(x∗), in the rectangle. The area of the ith rectangle is (f(x′ i) −g(x∗ i ))∆x. The lower sum = n i=1 (f(x′ i) −g(x∗ i ))∆x. The upper sum can be found in the same way. The area enclosed between the curves is sandwiched be-tween the lower sum and the upper sum. When we pass to the limit as ∆x →0, we get Area enclosed between the curves = b a (f(x) −g(x))dx. Note that the height is always f(x)−g(x), even when one or both of the curves lie below the x-axis. For example, if for some value of x, f(x) = 2 and g(x) = −3, the distance between the curves is f(x) − g(x) = 2 −(−3) = 5, or, if f(x) = −2 and g(x) = −3, the distance between the curves is (−2) −(−3) = 1 (see the diagram). So, to ﬁnd the area enclosed between two curves, we must: 1. Find where the curves intersect. 2. Find which is the upper curve in the region we are interested in. Mathematics Learning Centre, University of Sydney 20 3. Integrate the function (upper curve −lower curve) between the appropriate limits. In other words, if two curves f(x) and g(x) intersect at x = a and x = b, and f(x) ≥g(x) for a ≤x ≤b, then Area enclosed between the curves = b a (f(x) −g(x))dx. Exercises 9 (Remember to draw a diagram ﬁrst, before beginning any problem.) 1. Find the area enclosed between the parabola y = x(x −2) and the line y = −x + 2. 2. Find the area enclosed between the two parabolas y = x2 −4x + 2 and y = 2 −x2. 3. Check that the curves y = sin x and y = cos x intersect at π 4 and 5π 4 , and ﬁnd the area enclosed by the curves between these two point. 4. i Sketch the graphs of the function y = 6 −x −x2 and y = x3 −7x + 6. ii Find the points of intersection of the curves. iii Find the total area enclosed between them. \| \| ∆x Area A(x) Pi–1 Pi Pn ∆li P 0 Mathematics Learning Centre, University of Sydney 21 10 Other Applications of the Deﬁnite Integral The problem with which we introduced the idea of the deﬁnite integral was that of ﬁnding the area under a curve. As a result, most people tend to think of deﬁnite integrals always in terms of area. But it is important to remember that the deﬁnite integral is actually deﬁned as the limit of a sum: lim n→∞ n i=1 f(xi)∆x and that any other problem which can be approximated by a similar sum will give rise to a deﬁnite intregral when we take the limit. Examples 1. Volume of a solid If we want to ﬁnd the volume of a solid, we can imagine it being put through a bread slicer, and cut into slices of thickness ∆x. If A(x) is the cross sectional area at distance x along the x-axis, the volume of the slice will be approximately A(x)∆x, and the total volume of the solid will be approximately n i=1 A(xi)∆x. When we pass to the limit as ∆x →0 and n →∞, this becomes the deﬁnite integral b a A(x)dx. 2. Length of a curve We can approximate to the length of a curve by dividing it up into segments, as shown, and approximating the length of each seg-ment by replacing the curved line with a straight line joining the end points. If the length of the ith straight line segment is ∆li, the total length of the curve will be approx-imately n i=1 ∆li. If we take the limit of this sum as the length of each segment approaches zero and the number of segments approaches inﬁnity, we again get a deﬁnite integral. The details are rather complicated and are not given here. Mathematics Learning Centre, University of Sydney 22 3. Mass of a body of varying density Suppose we have a bar, rope or chain whose linear density (mass per unit length) varies. Let the density at distance x along the x-axis be d(x). If we subdivide the object into small sections of length ∆x, the total mass can be approximated by the sum n i=1 d(xi)∆x. When we take the limit as n →∞, we obtain the deﬁnite integral b a d(x)dx. 4. Work done by a variable force In mechanics, the work done by a constant force is deﬁned to be the product of the magnitude of the force and the distance moved in the direction of the force. If the force F(x) is varying, we can approximate to the work by dividing up the distance into small subintervals. If these are small enough, we can regard the force as eﬀectively constant throughout each interval and so the work done in moving through distance ∆x is approximately F(x)∆x. The total work is thus approximately n i=1 F(xi)∆x and when we take the limit as n →∞, we ﬁnd that the work done in moving the force from x = a to x = b is b a F(x)dx. Many other examples could be given, but these four should be suﬃcient to illustrate the wide variety of applications of the deﬁnite integral. Mathematics Learning Centre, University of Sydney 23 11 Solutions to Exercises Exercises 5 1. a. 2(43) −2(23) = 112 b. 1 9 −1 1 = −8 9 c. √ 16 − √ 9 = 1 d. ln 4 −ln 2 = ln 4 2 = ln 2 2. a. 9 4 x−1 2dx = 2x 1 2 9 4 = 2 √ 9 −2 √ 4 = 2 b. π 2 −π 2 cos tdt = [sin t] π 2 −π 2 = sin π 2 −sin −π 2 = 1 −(−1) = 2 c. 2 1 y−2dy = −y−12 1 = −1 2 − −1 1 = 1 2 d. −1 −2 (s2 + 2s + 2)ds = 1 3s3 + s2 + 2s −1 −2 = −1 3 + 1 −2 − −8 3 + 4 −4 = 11 3 3. a. Area = 2 1 (x2 + 1)dx = 1 3x3 + x 2 1 = 8 3 + 2 − 1 3 + 1 = 31 3 b. Area = 3 1 1 xdx = [ln x]3 1 = ln 3 −ln 1 = ln 3 c. Area = 4 0 (4 −u)du = − 4 0 (4 −u) 1 2(−1)du = − 2 3(4 −u) 3 2 4 0 = 51 3 d. Area = π 0 2 sin tdt = [−2 cos t]π 0 = −2 cos π + 2 cos 0 = 4 4. a. 1 0 xex2dx = 1 2 1 0 ex2 · 2xdx = 1 2 ex21 0 = 1 2(e −1) b. −1 −2 1 3 −xdx = − −1 −2 1 3 −x · (−1)dx = −[ln(3 −x)]−1 −2 = −(ln 4 −ln 5) = ln 5 −ln 4 = ln 5 4 c. π 2 0 sin 2ydy = 1 2 π 2 0 sin 2y · 2dy = 1 2 [−cos 2y] π 2 0 = 1 2(−cos π + cos 0) = 1 d. 5 1 t 4 + t2dt = 1 2 5 1 2t 4 + t2dt = 1 2 ln(4 + t2) 5 1 = 1 2 ln 29 5 y x 0 4 x y 0 π 2π x y –1 0 2 B A Mathematics Learning Centre, University of Sydney 24 Exercises 8 1. First, draw a graph. The area is below the x-axis, so we ﬁrst cal-culate 4 0 3x2(x −4)dx. 4 0 3x2(x −4)dx = 4 0 (3x3 −12x2)dx = 3 4x4 −4x3 4 0 = −64. The required area is therefore 64 units. 2. i 2π 0 sin xdx = [−cos x]2π 0 = −cos 2π + cos 0 = −1 + 1 = 0 ii From the graph we see that the area Area = π 0 sin xdx + \| 2π π sin xdx\| = [−cos x]π 0 + \| [−cos x]2π π \| = (−cos π + cos 0) + \| −cos 2π + cos π\| = (−(−1) + 1) + \| −1 + (−1)\| = 4. 3. The graph of the curve cuts the x-axis at −1, 0 and 2. The total area = area A + area B. Area A = \| 0 −1 12x(x + 1)(2 −x)dx\| = \| 0 −1(−12x3 + 12x2 + 24x)dx\| = \| −34 + 4x3 + 12x 0 −1 \| = \|0 −(−3 −4 + 12)\| = \| −5\| = 5. Area B = 2 0 12x(x + 1)(2 −x)dx = −3x4 + 4x3 + 12x22 0 = (−48 + 32 + 48) = 32. x y (–1,3) (2,0) (2,–2) (0,2) y = 2–x 2 y x x y π/4 5π/4 Mathematics Learning Centre, University of Sydney 25 Therefore the total area is 5 + 32 = 37 square units. Exercises 9 1. The curves y = x2 −2x and y = −x + 2 intersect where x2 −2x = −x + 2. i.e. at x = −1 or x = 2. The upper curve is y = −x + 2. Area = 2 −1((−x + 2) −(x2 −2x))dx = 2 −1(2 + x −x2)dx = 2x + 1 2x2 −1 3x3 2 −1 = (4 + 2 −8 3) −(−2 + 1 2 + 1 3) = 41 2. 2. The curves intersect where x2 −4x + 2 = 2 −x2 i.e. 2x2 −4x = 0 i.e. x = 0 or x = 2. The upper curve is y = 2 −x2 (see sketch). Area = 2 0 ((2 −x2) −(x2 −4x + 2))dx = 2 0 (4x −2x2)dx = 2x2 + 2 3x3 2 0 = (8 −2 3 · 8) −0 = 22 3. 3. When x = π 4, sin x = 1 √ 2 and cos x = 1 √ 2. When x = 5π 4 , sin x = −1 √ 2 and cos x = −1 √ 2. So the curves y = sin x and y = cos x inter-sect at π 4 and 5π 4 . Area = 5π 4 π 4 (sin x −cos x)dx = [−cos x −sin x] 5π 4 π 4 x y 0 –3 2 A B Mathematics Learning Centre, University of Sydney 26 = (−cos 5π 4 −sin 5π 4 ) + (cos π 4 + sin π 4 ) = 4 √ 2 = 2 √ 2. 4. (i) and (ii) The curves are easier to sketch if we ﬁrst ﬁnd the points of intersection: they meet where x3 −7x + 6 = 6 −x −x2. That is, x3 + x2 −6x = 0 or x(x −2)(x + 3) = 0. So the points of intersection are (0, 6); (2, 0); and (−3, 0). The ﬁrst curve is an ‘upside-down’ parabola, and the second a cubic. Total area = area A + area B. Area A = 0 −3((x3 −7x + 6) −(6 −x −x2))dx = 0 −3(x3 + x2 −6x)dx = 1 4x4 + 1 3x3 −3x2 0 −3 = 153 4. Area B = 2 0 ((6 −x −x2) −(x3 −7x + 6))dx = 2 0 (6x −x2 −x3)dx = 51 3. ˙ . . the total area = 153 4 + 5 1 3 = 21 1 12 square units.
10002	https://www.annemergmed.com/article/S0196-0644(08)00680-X/fulltext	Dehydration in Infants and Young Children - Annals of Emergency Medicine Skip to Main ContentSkip to Main Menu Login to your account Email/Username Your email address is a required field. E.g., j.smith@mail.com Password Show Your password is a required field. Forgot password? [x] Remember me Don’t have an account? Create a Free Account ACEP Member Login ACEP Members, full access to the journal is a member benefit. Use your society credentials to access all journal content and features. ACEP Member Login If you don't remember your password, you can reset it by entering your email address and clicking the Reset Password button. 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Ok Evidence-based emergency medicine/rational clinical examination abstractVolume 53, Issue 3p395-397 March 2009 Download Full Issue Download started Ok Dehydration in Infants and Young Children Stephen Emond, MD Stephen Emond, MD Affiliations Department of Emergency Medicine, The Permanente Medical Group, Inc., Santa Clara, CA; and the Kaiser-Stanford Emergency Medicine Residency Program, Stanford University School of Medicine, Palo Alto, CA Search for articles by this author Affiliations & Notes Article Info Department of Emergency Medicine, The Permanente Medical Group, Inc., Santa Clara, CA; and the Kaiser-Stanford Emergency Medicine Residency Program, Stanford University School of Medicine, Palo Alto, CA Publication History: Published online May 13, 2008 DOI: 10.1016/j.annemergmed.2008.04.003 External LinkAlso available on ScienceDirect External Link Copyright: © 2008 American College of Emergency Physicians. Download PDF Download PDF Outline Outline Rational Clinical Examination Review Source Objective Data Sources Study Selection Data Extraction and Analysis Main Results Conclusions Rational Clinical Examination Author Contact Commentary: Clinical Implications Take-Home Message EBEM Teaching Point References Article metrics Related Articles Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley bluesky Add to my reading list More More Download PDF Download PDF Cite Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley Bluesky Add to my reading list Set Alert Get Rights Reprints Download Full Issue Download started Ok Previous articleNext article Show Outline Hide Outline Rational Clinical Examination Review Source Objective Data Sources Study Selection Data Extraction and Analysis Main Results Conclusions Rational Clinical Examination Author Contact Commentary: Clinical Implications Take-Home Message EBEM Teaching Point References Article metrics Related Articles [Ann Emerg Med. 2009;53:395-397.] Rational Clinical Examination Review Source This is a rational clinical examination abstract, a regular feature of the Annals' Evidence-Based Emergency Medicine (EBEM) series. Each features an abstract of a rational clinical examination review from the Journal of the American Medical Association and a commentary by an emergency physician knowledgeable in the subject area. The source for this rational clinical examination review abstract is: M Steiner, D DeWalt, J Byerly. The rational clinical examination: is this child dehydrated? JAMA. 2004;291:2746-2754. The Annals' EBEM editors assisted in the preparation of the abstract of this rational clinical examination review, as well as selection of the Evidence-Based Medicine Teaching Points. Objective To systematically review the precision and accuracy of symptoms, signs, and basic laboratory tests for evaluating dehydration in infants and children (aged 1 month to 5 years). Data Sources The authors performed direct searches for potential sources of primary data or reviews with potential background information on the MEDLINE database through the PubMed search engine. They supplemented these searches with the standardized search technique used in the “Rational Clinical Examination” series to yield 1,561 potential articles. Searches of bibliographies of retrieved articles, the Cochrane Library, textbooks, and private collections of experts in the field yielded an additional 42 articles. Study Selection All authors reviewed the titles and available abstracts of the 1,603 potential articles to identify studies for further review. Of 110 retained articles, 26 contained original data on the precision or accuracy of a symptom, sign, or laboratory value for the diagnosis of dehydration in young children. These underwent full quality assessment with a standardized methodologic filter, with group consensus on level of evidence. The difference between the rehydration weight and the acute weight divided by the rehydration weight was chosen as the best available criterion standard of percentage of volume lost. Thirteen studies were assigned level 4 quality or higher and were included in the meta-analysis. Data Extraction and Analysis Two of the 3 authors independently reviewed and abstracted data into 2×2 tables for each test to calculate sensitivity, specificity, and likelihood ratios, with corresponding 95% confidence intervals (CIs). When 2 studies evaluated an individual diagnostic test, a range of values was provided. If more than 2 studies evaluated a test, a random-effects model was used. Significant heterogeneity (a common finding with this model) was demonstrated for most signs. Main Results The authors found few high-quality studies with accurate criterion standards and minimal systematic bias. Tests of dehydration were imprecise, generally showing only fair to moderate agreement among examiners (key signs of dehydration are listed in the Figure; precision of signs on examination is listed in Table 1). Historical points had moderate sensitivity in screening for dehydration, but parental reports of dehydration symptoms were so nonspecific that they had limited clinical utility. Prolonged capillary refill time, abnormal skin turgor, and abnormal respiratory pattern were the best 3 signs of dehydration (Table 2), and groups of signs or the use of clinical scales improved accuracy. Laboratory tests generally were helpful only when results were markedly abnormal, but none were considered definitive for dehydration with the authors' reference standard (Table 3). Figure viewer Figure Key signs in young children with suspected dehydration. \| Finding \| Total No. of Participants \| Range of κ Values \| --- \| Prolonged capillary refill \| 216 \| 0.01-0.65 \| \| Abnormal skin turgor \| 184 \| 0.36-0.55 \| \| Abnormal respiratory pattern \| 184 \| 0.04-0.40 \| \| Extremity perfusion \| 100 \| 0.23-0.66 \| \| Absent tears \| 184 \| 0.12-0.75 \| \| Sunken fontanelle \| 100 \| 0.10-0.27 \| \| Sunken eyes \| 184 \| 0.06-0.59 \| \| Dry mucous membranes \| 184 \| 0.28-0.59 \| \| Weak pulse \| 184 \| 0.15-0.50 \| \| Poor overall appearance \| 184 \| 0.18-0.61 \| Table 1 Precision of signs for dehydration. Open table in a new tab \| Finding \| Total No. Participants \| LR Summary Value (95% CI) or Range \| --- \| Present \| Absent \| \| Prolonged capillary refill \| 478 \| 4.1(1.7-9.8) \| 0.57(0.39-0.82) \| \| Abnormal skin turgor \| 602 \| 2.5(1.5-4.2) \| 0.66(0.57-0.75) \| \| Abnormal respiratory pattern \| 581 \| 2.0(1.5-2.7) \| 0.76(0.62-0.88) \| \| Sunken eyes \| 533 \| 1.7(1.1-2.5) \| 0.49(0.38-0.63) \| \| Dry mucous membranes \| 533 \| 1.7(1.1-2.6) \| 0.41(0.21-0.79) \| \| Cool extremity \| 206 \| 1.5-18.8 \| 0.89-0.97 \| \| Weak pulse \| 360 \| 3.1-7.2 \| 0.66-0.96 \| \| Absent tears \| 398 \| 2.3(0.9-5.8) \| 0.54(0.26-1.13) \| Table 2 Summary test characteristics for clinical findings to detect 5% dehydration. LR, Likelihood ratio. Open table in a new tab \| Laboratory Value \| Total No. of Participants \| LR Summary, Value (95% CI) or Range \| --- \| Present \| Absent \| \| BUN, mg/dL \| \| \| \| \| >8 \| \| 2.1-2.4 \| 0.41-0.76 \| \| >45 \| 168 \| 46.1(2.9-733) \| 0.58(0.49-0.68) \| \| BUN/Cr ratio >40 \| 40 \| 2.1(0.5-8.9) \| 0.87(0.62-1.20) \| \| Bicarbonate, mEq/L \| \| \| \| \| <17 \| 97 \| 3.5(2.1-5.8) \| 0.22(0.12-0.43) \| Table 3 Summary test characteristics for selected laboratory tests assessing dehydration BUN, Blood urea nitrogen; Cr, creatinine. Open table in a new tab Conclusions The authors advocate the approach recommended by the World Health Organization (WHO) and other groups: use the physical examination to classify dehydration as none, some, or severe, and then use this general assessment to guide clinical management. Table 4 summarizes this classification scheme. \| Variable/Sign \| Dehydration \| --- \| \| Mild (4%-5%) \| Moderate (6%-9%) \| Severe (≥10%) \| \| General appearance \| Thirsty, restless, alert \| Thirsty, drowsy, postural hypotension \| Drowsy, limp, cold, sweaty, cyanotic extremities \| \| Radial pulse \| Normal rate and strength \| Rapid and weak \| Rapid, thready, sometimes impalpable \| \| Respirations \| Normal \| Deep, may be rapid \| Deep and rapid \| \| Anterior fontanelle \| Normal \| Sunken \| Very sunken \| \| Systolic blood pressure \| Normal \| Normal or low \| Low \| \| Skin elasticity \| Pinch retracts immediately \| Pinch retracts slowly \| Pinch retracts very slowly \| \| Eyes \| Normal \| Sunken \| Grossly sunken \| \| Tears \| Present \| Absent \| Absent \| \| Mucous membranes \| Moist \| Dry \| Very dry \| Table 4 Dehydration Assessment Scale. Adapted from Steiner et al.1 1. King, C.K. ∙ Glass, R. ∙ Bresee, J.S. ... Managing acute gastroenteritis among children: oral rehydration, maintenance, and nutritional therapy MMWR Recomm Rep. 2003; 52:1-16 PubMed Google Scholar Level of classification may be classified as none, some (mild or moderate dehydration), or severe. Open table in a new tab Rational Clinical Examination Author Contact Darren A. DeWalt, MD, MPH University of North Carolina at Chapel Hill School of Medicine Chapel Hill, NC E-maildewaltd@med.unc.edu Commentary: Clinical Implications Each year in the United States, acute gastroenteritis results in 5% of pediatric outpatient visits and is frequently complicated by some degree of dehydration. Volume depletion accounts for 5% of pediatric hospital admissions and more than 300 deaths each year. In the United States, gastroenteritis accounts for more than 1.5 million pediatric outpatient visits and 200,000 hospitalizations annually.1 1. King, C.K. ∙ Glass, R. ∙ Bresee, J.S. ... Managing acute gastroenteritis among children: oral rehydration, maintenance, and nutritional therapy MMWR Recomm Rep. 2003; 52:1-16 PubMed Google Scholar The goals of treatment of the child with gastroenteritis are prevention of dehydration, rehydration efforts when necessary, rapid refeeding, and pharmacologic treatment whenever appropriate (eg, antinauseants and antimotility agents). Throughout the last decade, randomized controlled trials in both developed and developing countries have confirmed the success rates (90%) of oral rehydration solutions in achieving these treatment goals, as well as their lower complication rates, shorter treatment times, and lower cost compared with intravenous therapy.2,3 2. Fonseca, B.K. ∙ Holdgate, A. ∙ Craig, J.C. Enteral vs intravenous rehydration therapy for children with gastroenteritis: a meta-analysis of randomized controlled trials Arch Pediatr Adolesc Med. 2004; 158:483-490 Crossref Scopus (142) PubMed Google Scholar 3. Hartling, L. ∙ Bellemare, S. ∙ Wiebe, N. ... Oral versus intravenous rehydration for treating dehydration due to gastroenteritis in children Cochrane Database Syst Rev. 2006; 3 CD004390. DOI: PubMed Google Scholar Unfortunately, oral rehydration solution therapy remains underused, particularly in developed countries.4 4. Ozuah, P.O. ∙ Avner, J.R. ∙ Stein, R.E. Oral rehydration, emergency physicians, and practice parameters: a national survey Pediatrics. 2002; 109:259-261 Crossref Scopus (111) PubMed Google Scholar Historically, clinicians who provide care in emergency departments have been more likely to choose intravenous over oral rehydration when vomiting is a major symptom.5 5. Reis, E.C. ∙ Goepp, J.G. ∙ Katz, S. ... Barriers to use of oral rehydration therapy Pediatrics. 1994; 93:708-711 PubMed Google Scholar In one survey, 36% of pediatricians reported that vomiting was a contraindication to oral rehydration5 5. Reis, E.C. ∙ Goepp, J.G. ∙ Katz, S. ... Barriers to use of oral rehydration therapy Pediatrics. 1994; 93:708-711 PubMed Google Scholar ; however, a recent large trial demonstrating safety and efficiency of oral-dissolvable ondansetron in controlling nausea and vomiting in gastroenteritis is likely to increase the use and success rate of oral rehydration.6 6. Freedman, S.B. ∙ Adler, M. ∙ Seshadri, R. ... Oral ondansetron for gastroenteritis in a pediatric emergency department N Engl J Med. 2006; 354:1698-1705 Crossref Scopus (223) PubMed Google Scholar The clinical problem is to distinguish the mild to moderately dehydrated child, in whom adequate fluid intake can be achieved with oral rehydration solution therapy, from the more seriously dehydrated child requiring intravenous hydration and consideration for admission. More aggressive measures may be required if excessive fluid loss or dehydration is evident or if patients are at risk of dehydration (eg, infants, immunosuppressed patients, patients with comorbid disease). Such patients may benefit from hospital admission and intravenous fluids, in addition to oral rehydration solution with isotonic electrolyte solutions containing glucose or starch. The most accurate indicator of the magnitude of dehydration appears to be the percentage of loss of body weight, a measure usually unavailable in the ED, so estimates of the degree of dehydration must be made from clinical data. Diagnostic criteria published by WHO7 7. Gove, S. Integrated management of childhood illness by outpatient health workers: technical basis and overviewThe WHO Working Group on Guidelines for Integrated Management of the Sick Child Bull World Health Organ. 1997; 75:7-24 PubMed Google Scholar and a practice measure on the management of acute gastroenteritis published by the American Academy of Pediatrics (AAP)8 8. American Academy of Pediatrics Practice parameter: the management of acute gastroenteritis in young children Pediatrics. 1996; 97:424-435 PubMed Google Scholar (now retired in deference to the more recent Centers for Disease Control and Prevention guidelines9 9. King, C.K. ∙ Glass, R. ∙ Bresee, R.S. ... Managing gastroenteritis among children MMWR Recomm Rep. 2003; 52:1-16 PubMed Google Scholar ) are widely referenced. These classification systems differ slightly, and categorizing severity can be difficult if the child has signs and symptoms that fit into more than 1 category. The authors reference a strategy using groups of signs and symptoms to predict the percentage of loss of body weight (Table 2) and then treating patients by degree of dehydration. The WHO classification, designed for health care workers in developing countries with limited health care resources and high diarrhea-related morbidity and mortality, has proven to be effective in the triage of patients with acute diarrheal illness,10 10. Santosham, M. Oral rehydration therapy: reverse transfer of technology Arch Pediatr Adolesc Med. 2002; 156:1177-1179 Crossref Scopus (27) PubMed Google Scholar whereas the AAP classification has been less well studied. The authors' classification scheme, although reasonable, has also not been prospectively evaluated. Take-Home Message Prolonged capillary refill time, abnormal skin turgor, and abnormal respiratory pattern are the 3 best signs of dehydration; groups of signs and the use of clinical scales are more accurate in defining extent. Laboratory tests are helpful only when results are markedly abnormal, and none are considered definitive for dehydration. Use of physical findings can help identify children with no, mild, or moderate dehydration, all of whom are likely to respond well to oral rehydration solution therapy. EBEM Commentator Contact Stephen Emond, MD Department of Emergency Medicine Kaiser Santa Clara Medical Center Santa Clara, CA E-mailsteve.emond@kp.org EBEM Teaching Point Random-effects modeling Random-effects modeling is a statistical technique used to combine data from disparate analyses for summary analysis (eg, meta-analysis). Both within-study sampling error (variance) and between-studies variation are included in the assessment of the uncertainty (CI) of the results of a meta-analysis. As the authors of this study reported, whenever there is significant heterogeneity among the results of the included studies, random-effects models give wider CIs than fixed-effect modeling. References 1. King, C.K. ∙ Glass, R. ∙ Bresee, J.S. ... Managing acute gastroenteritis among children: oral rehydration, maintenance, and nutritional therapy MMWR Recomm Rep. 2003; 52:1-16 PubMed Google Scholar 2. Fonseca, B.K. ∙ Holdgate, A. ∙ Craig, J.C. Enteral vs intravenous rehydration therapy for children with gastroenteritis: a meta-analysis of randomized controlled trials Arch Pediatr Adolesc Med. 2004; 158:483-490 Crossref Scopus (142) PubMed Google Scholar 3. Hartling, L. ∙ Bellemare, S. ∙ Wiebe, N. ... Oral versus intravenous rehydration for treating dehydration due to gastroenteritis in children Cochrane Database Syst Rev. 2006; 3 CD004390. DOI: PubMed Google Scholar 4. Ozuah, P.O. ∙ Avner, J.R. ∙ Stein, R.E. Oral rehydration, emergency physicians, and practice parameters: a national survey Pediatrics. 2002; 109:259-261 Crossref Scopus (111) PubMed Google Scholar 5. Reis, E.C. ∙ Goepp, J.G. ∙ Katz, S. ... Barriers to use of oral rehydration therapy Pediatrics. 1994; 93:708-711 PubMed Google Scholar 6. Freedman, S.B. ∙ Adler, M. ∙ Seshadri, R. ... Oral ondansetron for gastroenteritis in a pediatric emergency department N Engl J Med. 2006; 354:1698-1705 Crossref Scopus (223) PubMed Google Scholar 7. Gove, S. Integrated management of childhood illness by outpatient health workers: technical basis and overviewThe WHO Working Group on Guidelines for Integrated Management of the Sick Child Bull World Health Organ. 1997; 75:7-24 PubMed Google Scholar 8. American Academy of Pediatrics Practice parameter: the management of acute gastroenteritis in young children Pediatrics. 1996; 97:424-435 PubMed Google Scholar 9. King, C.K. ∙ Glass, R. ∙ Bresee, R.S. ... Managing gastroenteritis among children MMWR Recomm Rep. 2003; 52:1-16 PubMed Google Scholar 10. Santosham, M. Oral rehydration therapy: reverse transfer of technology Arch Pediatr Adolesc Med. 2002; 156:1177-1179 Crossref Scopus (27) PubMed Google Scholar Figures (1)Figure Viewer Article metrics Related Articles View abstract Open in viewer Dehydration in Infants and Young Children Hide CaptionDownloadSee figure in Article Toggle Thumbstrip Figure Download .PPT Go to Go to Show all references Expand All Collapse Expand Table Authors Info & Affiliations Home Access for Developing Countries Articles & Issues Current issue Articles in Press List of Issues Supplements Collections Editors Free Top Selections Annals CME Articles News and Perspective The Science of Peer Review Equity in Health Care Journal Club FOAM Expert Clinical Management Residents' Perspective Review Articles Safety/Operations EBEM Resources Systematic Review Snapshots Disaster Medicine Change of Shift Case Reports ACEP Clinical Policies Annals Podcast Archives Images For Authors Instructions for Authors Researcher Academy Submit to Annals Visual Abstract Instructions Journal Info Editorial Board Conflict of Interest Policy Diversity Pledge Editorial Board Declarations of Competing Interests The Peer Review Process at Annals Peer Reviewer Information Permission to Reuse Info for Advertisers Journal Access ACEP Login Register Institutions E-mail Alert Free Trial Issue Subscription/Pricing Information ACEP Join ACEP ACEP Now emCareers.org Contact Us Follow Us Twitter Facebook Instagram The content on this site is intended for healthcare professionals. 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10003	https://www.youtube.com/watch?v=gHyCPbgT1hk	Translations in Desmos Classroom! #math #geometry Susan Carriker 734 subscribers 2 likes Description 54 views Posted: 20 Apr 2024 Visit for other cool Desmos geometry activities. . Find a link to this activity and MORE at on the FREE teaching resources page. #maths #iteachmath #mtbos #edtech Transcript:
10004	https://www.youtube.com/watch?v=Jkmwih6e6Kw	Prove that T=2π(L/g) is dimensionally correct. #dimensionanalysis CHAMPION TUTORS ACADEMY 364 subscribers 197 likes Description 15529 views Posted: 25 Mar 2022 12 comments Transcript:
10005	https://www.ck12.org/flexi/cbse-math/rationalize-the-denominator/how-do-you-simplify-cube-root-of-fractions/	How do you simplify cube root of fractions? Flexi Says: To simplify a cube root of fraction, you need to simplify the numerator and denominator separately and then take the cube root of the simplified result. Here's the general process: Simplify the numerator and denominator separately. Take the cube root of the simplified numerator. Take the cube root of the simplified denominator. If possible, simplify the cube roots further. For example, let's simplify the cube root fraction @$\begin{align}\frac {27}{64}:\end{align}@$ Step 1: Simplify the numerator and denominator.@$\begin{align}&27 = 3^3\ &64 = 4^3\end{align}@$ Step 2: Take the cube root of the simplified numerator and denominator.@$\begin{align}&\sqrt{3^3}= 3\ &\sqrt{4^3}= 4\end{align}@$ Step 3: Write the simplified cube roots as the numerator and denominator of the new fraction:@$\begin{align}\frac {3}{4}\end{align}@$ So, the simplified cube root fraction of @$\begin{align}\frac {27}{64}\end{align}@$ is @$\begin{align}\frac {3}{4}.\end{align}@$ Try Asking: Are perfect squares only even numbers?What is equal to the square root of pi?Simplify. Remove all perfect squares from inside the square roots. Assume x and y are positive By messaging Flexi, you agree to our Terms and Privacy Policy
10006	https://brainly.com/question/36789826	[FREE] Find the relative extrema for the function f(x) = x^3 - 3x using the second derivative test. - brainly.com 6 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +69,3k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +41,8k Ace exams faster, with practice that adapts to you Practice Worksheets +5,1k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified Find the relative extrema for the function f(x)=x 3−3 x using the second derivative test. 1 See answer Explain with Learning Companion NEW Asked by bellasmith5941 • 09/04/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 793117 people 793K 0.0 0 Upload your school material for a more relevant answer By finding the first and second derivatives of the given function and applying the second derivative test, we find that the function has a relative maximum at x=-1 and a relative minimum at x=1. Explanation First we need to find the first derivative of the function f(x)=x^3−3x. The derivative is the rate of change at a certain point on the graph. To determine this, take the derivative of f(x) which gives f'(x)=3x^2 - 3. The next step is to set this derivative equal to 0 to find the critical points. Solving this gives x=1 and x=-1. To apply the second derivative test, we need to find the second derivative of the function, which comes out to be f''(x)=6x. Substituting the critical points into the second derivative tells us if the function has a relative max, min, or neither. For x=1, f''(x)=6 and for x=-1, f''(x)=-6. So by the second derivative test, we see that f(x)=x^3−3x has a relative maximum at x=-1, and a relative minimum at x=1. Learn more about Second Derivative Test here: brainly.com/question/34429954 SPJ11 Answered by Tomstark •6.1K answers•793.1K people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 793117 people 793K 0.0 0 Upload your school material for a more relevant answer The function f(x)=x 3−3 x has a relative maximum at x=−1 and a relative minimum at x=1, as found using the second derivative test. Explanation To find the relative extrema of the function f(x)=x 3−3 x, we will use the second derivative test. Here's a step-by-step breakdown: Find the First Derivative: The first derivative of the function is given by: f′(x)=d x d(x 3−3 x)=3 x 2−3. Find Critical Points: Set the first derivative equal to zero to find critical points: 3 x 2−3=0 Simplifying gives: 3(x 2−1)=0 Therefore, the solutions are: x 2−1=0⇒x=−1 and x=1. Find the Second Derivative: Now, we compute the second derivative: f′′(x)=d x d(3 x 2−3)=6 x. Apply the Second Derivative Test: Evaluate the second derivative at the critical points: For x=−1: f′′(−1)=6(−1)=−6 (Since this is less than 0, there is a relative maximum at x=−1) For x=1: f′′(1)=6(1)=6 (Since this is greater than 0, there is a relative minimum at x=1) Conclusion: Thus, the function f(x)=x 3−3 x has a relative maximum at x=−1 and a relative minimum at x=1. Examples & Evidence For example, if we graph the function f(x)=x 3−3 x, we will see that the highest point in the interval around x=−1 is at that point, confirming it as a maximum, while at x=1, we see the lowest point around that interval confirming it as a minimum. The use of the second derivative test is a standard method in calculus to determine points of extrema. This method is well documented in calculus textbooks and online resources detailing derivatives. Thanks 0 0.0 (0 votes) Advertisement bellasmith5941 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer Find the relative extrema of the function f(x)=x3−4x2−16x−6 and justify your answer either using the first or the second derivative test (explain your answer and show your supporting work). Community Answer Find all relative extrema. Use the Second Derivative Test where applicable. (If an answer does not exist, enter DNE.) f(x) = x^3 − 3x^2 + 3 Community Answer 4. Find relative extrema using the Second Derivative Test if applicable: (a) f(x) = 6x – x2 = (b) f(x) = x4 – 4x3 + 2 = Community Answer Find the First Derivative Test to find the relative extrema and the intervals where the function increases and decreases. f(x)=3x3−4x Community Answer Let f(x) = x3 - 3x + 1. Use the first-derivative test to find the relative extrema. Label the relative extrema as a relative maximum or relative minimum. (DO NOT SKETCH THE FUNCTION) (5 points) Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? New questions in Mathematics Simplify the expression. 5(2 x−3)=5(2 x)−5(3)=□x−□ Write an expression equivalent to 5(2 x−3). Use the distributive property to expand the expression. 5(2 x−3)=5(□x)−5(?) Write an expression equivalent to 2 1(16 m+12). Write an expression equivalent to 4(3 x−7). Use the distributive property to expand the expression. 4(3 x−7)=4(□x)−4(?) 7 x−5 y−31=0 16 x+15 y−15=0 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
10007	https://www.educative.io/answers/how-to-implement-falling-factorial-in-python	How to implement falling factorial in Python Falling factorial is a notion used in counting issues to describe the product of decreasing integers up to a given point. It is especially helpful when selecting a certain number of elements from a bigger collection is required, and order is important. Definition of falling factorial The sum of the firstkkkdescending integers fromnnnis the falling factorial ofnnnwith a factor ofkkk, represented as(n)k(n)_k(n)k. In terms of math, it is stated as: Generalization of falling factorial The falling factorial, which we can use to select the topkkkelements out ofnnnelements can be written as a ratio of factorials: Recurrence relation The recurrence connection can be used to relate the falling factorial to the factorial: By using this relation, we may create a recursive definition of the falling factorial by expressing it in terms of a smaller falling factorial. Implementation Let’s implement the falling factorial in Python: Explanation Line 1: Here, we import the math module. math Lines 3–9: We calculate the falling factorial of a given number n with k terms. n k Lines 5–6: We check if the value of n (representing the base) is less than the value of k (representing the number of terms). If n is less than k, it means there aren't enough terms to calculate the falling factorial, so it raises a ValueError with an appropriate message. n k n k ValueError Line 8: We calculate the falling factorial using the formula n! / (n - k)!. It first calculates the factorial of n using math.factorial(n) and then divides it by the factorial of n - k to get the falling factorial. The // operator is used for integer division to ensure the result is an integer. n! / (n - k)! n math.factorial(n) n - k // Line 9: It returns the calculated falling factorial value. Lines 12–15: We print the result is an example of using the function with n=10 and k=3. n=10 k=3 Relevant Answers Explore Courses Free Resources
10008	https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2010/6942723a3a99c1e7cfe0c3210b6f13ca_MIT6_042JF10_chap17.pdf	“mcs-ftl” — 2010/9/8 — 0:40 — page 445 — #451 17 Random Variables and Distributions Thus far, we have focused on probabilities of events. For example, we computed the probability that you win the Monty Hall game, or that you have a rare medical condition given that you tested positive. But, in many cases we would like to more more. For example, how many contestants must play the Monty Hall game until one of them ﬁnally wins? How long will this condition last? How much will I lose gambling with strange dice all night? To answer such questions, we need to work with random variables. 17.1 Deﬁnitions and Examples Deﬁnition 17.1.1. A random variable R on a probability space is a total function whose domain is the sample space. The codomain of R can be anything, but will usually be a subset of the real numbers. Notice that the name “random variable” is a misnomer; random variables are actually functions! For example, suppose we toss three independent1, unbiased coins. Let C be the number of heads that appear. Let M D 1 if the three coins come up all heads or all tails, and let M D 0 otherwise. Every outcome of the three coin ﬂips uniquely determines the values of C and M. For example, if we ﬂip heads, tails, heads, then C D 2 and M D 0. If we ﬂip tails, tails, tails, then C D 0 and M D 1. In effect, C counts the number of heads, and M indicates whether all the coins match. Since each outcome uniquely determines C and M, we can regard them as func-tions mapping outcomes to numbers. For this experiment, the sample space is S D fHHH; HHT; HTH; HT T; THH; THT; T TH; T T T g and C is a function that maps each outcome in the sample space to a number as 1Going forward, when we talk about ﬂipping independent coins, we will assume that they are mutually independent. 1 “mcs-ftl” — 2010/9/8 — 0:40 — page 446 — #452 Chapter 17 Random Variables and Distributions follows: C.HHH/ D 3 C.THH/ D 2 C.HHT / D 2 C.THT / D 1 C.HTH/ D 2 C.T TH/ D 1 C.HT T / D 1 C.T T T / D 0: Similarly, M is a function mapping each outcome another way: M.HHH/ D 1 M.THH/ D 0 M.HHT / D 0 M.THT / D 0 M.HTH/ D 0 M.T TH/ D 0 M.HT T / D 0 M.T T T / D 1: So C and M are random variables. 17.1.1 Indicator Random Variables An indicator random variable is a random variable that maps every outcome to either 0 or 1. Indicator random variables are also called Bernoulli variables. The random variable M is an example. If all three coins match, then M D 1; otherwise, M D 0. Indicator random variables are closely related to events. In particular, an in-dicator random variable partitions the sample space into those outcomes mapped to 1 and those outcomes mapped to 0. For example, the indicator M partitions the sample space into two blocks as follows: HHH T T T „ ƒ‚ … M D 1 HHT HTH HT T THH THT T TH „ ƒ‚ … M D 0 : In the same way, an event E partitions the sample space into those outcomes in E and those not in E. So E is naturally associated with an indicator random variable, IE, where IE.w/ D 1 for outcomes w 2 E and IE.w/ D 0 for outcomes w … E. Thus, M D IE where E is the event that all three coins match. 17.1.2 Random Variables and Events There is a strong relationship between events and more general random variables as well. A random variable that takes on several values partitions the sample space into several blocks. For example, C partitions the sample space as follows: T T T „ƒ‚… C D 0 T TH THT HT T „ ƒ‚ … C D 1 THH HTH HHT „ ƒ‚ … C D 2 HHH „ƒ‚… C D 3 : 2 “mcs-ftl” — 2010/9/8 — 0:40 — page 447 — #453 17.1. Deﬁnitions and Examples Each block is a subset of the sample space and is therefore an event. Thus, we can regard an equation or inequality involving a random variable as an event. For example, the event that C D 2 consists of the outcomes THH, HTH, and HHT . The event C 1 consists of the outcomes T T T , T TH, THT , and HT T . Naturally enough, we can talk about the probability of events deﬁned by proper-ties of random variables. For example, PrŒC D 2 D PrŒTHH C PrŒHTH C PrŒHHT D 1 8 C 1 8 C 1 8 D 3 8: As another example: PrŒM D 1 D PrŒT T T C PrŒHHH D 1 8 C 1 8 D 1 4: 17.1.3 Functions of Random Variables Random variables can be combined to form other random variables. For exam-ple, suppose that you roll two unbiased, independent 6-sided dice. Let Di be the random variable denoting the outcome of the ith die for i D 1, 2. For example, PrŒD1 D 3 D 1=6: Then let T D D1 C D2. T is also a random variable and it denotes the sum of the two dice. For example, PrŒT D 2 D 1=36 and PrŒT D 7 D 1=6: Random variables can be combined in complicated ways, as we will see in Chap-ter 19. For example, Y D eT is also a random variable. In this case, PrŒY D e2 D 1=36 and 3 “mcs-ftl” — 2010/9/8 — 0:40 — page 448 — #454 Chapter 17 Random Variables and Distributions PrŒY D e7 D 1=6: 17.1.4 Conditional Probability Mixing conditional probabilities and events involving random variables creates no new difﬁculties. For example, Pr C 2 j M D 0 is the probability that at least two coins are heads (C 2) given that not all three coins are the same (M D 0). We can compute this probability using the deﬁnition of conditional probability: Pr C 2 j M D 0 D PrŒC 2 \ M D 0 PrŒM D 0 D PrŒfTHH; HTH; HHT g PrŒfTHH; HTH; HHT; HT T; THT; T THg D 3=8 6=8 D 1 2: The expression C 2 \ M D 0 on the ﬁrst line may look odd; what is the set operation \ doing between an inequality and an equality? But recall that, in this context, C 2 and M D 0 are events, and so they are sets of outcomes. 17.1.5 Independence The notion of independence carries over from events to random variables as well. Random variables R1 and R2 are independent iff for all x1 in the codomain of R1, and x2 in the codomain of R2 for which PrŒR2 D X2 > 0, we have: Pr R1 D x1 j R2 D x2 D PrŒR1 D x1: As with events, we can formulate independence for random variables in an equiva-lent and perhaps more useful way: random variables R1 and R2 are independent if for all x1 and x2 PrŒR1 D x1 \ R2 D x2 D PrŒR1 D x1 PrŒR2 D x2: For example, are C and M independent? Intuitively, the answer should be “no”. The number of heads, C, completely determines whether all three coins match; that is, whether M D 1. But, to verify this intuition, we must ﬁnd some x1; x2 2 R such that: PrŒC D x1 \ M D x2 ¤ PrŒC D x1 PrŒM D x2: 4 “mcs-ftl” — 2010/9/8 — 0:40 — page 449 — #455 17.1. Deﬁnitions and Examples One appropriate choice of values is x1 D 2 and x2 D 1. In this case, we have: PrŒC D 2 \ M D 1 D 0 and PrŒM D 1 PrŒC D 2 D 1 4 3 8 ¤ 0: The ﬁrst probability is zero because we never have exactly two heads (C D 2) when all three coins match (M D 1). The other two probabilities were computed earlier. On the other hand, let F be the indicator variable for the event that the ﬁrst ﬂip is a Head, so “F D 1” D fHHH; HTH; HHT; HT T g: Then F is independent of M, since PrŒM D 1 D 1=4 D Pr M D 1 j F D 1 D Pr M D 1 j F D 0 and PrŒM D 0 D 3=4 D Pr M D 0 j F D 1 D Pr M D 0 j F D 0 : This example is an instance of a simple lemma: Lemma 17.1.2. Two events are independent iff their indicator variables are inde-pendent. As with events, the notion of independence generalizes to more than two random variables. Deﬁnition 17.1.3. Random variables R1; R2; : : : ; Rn are mutually independent iff PrŒR1 D x1 \ R2 D x2 \ \ Rn D xn D PrŒR1 D x1 PrŒR2 D x2 PrŒRn D xn: for all x1; x2; : : : ; xn. A consequence of Deﬁnition 17.1.3 is that the probability that any subset of the variables takes a particular set of values is equal to the product of the probabilities that the individual variables take their values. Thus, for example, if R1; R2; : : : ; R100 are mutually independent random variables, then it follows that: PrŒR1 D 7 \ R7 D 9:1 \ R23 D D PrŒR1 D 7 PrŒR7 D 9:1 PrŒR23 D : The proof is based on summing over all possible values for all of the other random variables. 5 “mcs-ftl” — 2010/9/8 — 0:40 — page 450 — #456 Chapter 17 Random Variables and Distributions 17.2 Distribution Functions A random variable maps outcomes to values. Often, random variables that show up for different spaces of outcomes wind up behaving in much the same way because they have the same probability of having any given value. Hence, random variables on different probability spaces may wind up having the same probability density function. Deﬁnition 17.2.1. Let R be a random variable with codomain V . The probability density function (pdf) of R is a function PDFR W V ! Œ0; 1 deﬁned by: PDFR.x/ WWD ( PrŒR D x if x 2 range.R/ 0 if x … range.R/: A consequence of this deﬁnition is that X x2range.R/ PDFR.x/ D 1: This is because R has a value for each outcome, so summing the probabilities over all outcomes is the same as summing over the probabilities of each value in the range of R. As an example, suppose that you roll two unbiased, independent, 6-sided dice. Let T be the random variable that equals the sum of the two rolls. This random variable takes on values in the set V D f2; 3; : : : ; 12g. A plot of the probability density function for T is shown in Figure 17.1: The lump in the middle indicates that sums close to 7 are the most likely. The total area of all the rectangles is 1 since the dice must take on exactly one of the sums in V D f2; 3; : : : ; 12g. A closely-related concept to a PDF is the cumulative distribution function (cdf) for a random variable whose codomain is the real numbers. This is a function CDFR W R ! Œ0; 1 deﬁned by: CDFR.x/ D PrŒR x: As an example, the cumulative distribution function for the random variable T is shown in Figure 17.2: The height of the ith bar in the cumulative distribution function is equal to the sum of the heights of the leftmost i bars in the probability 6 “mcs-ftl” — 2010/9/8 — 0:40 — page 451 — #457 17.2. Distribution Functions 3=36 6=36 x 2 V 2 3 4 5 6 7 8 9 10 11 12 PDFT.x/ Figure 17.1 The probability density function for the sum of two 6-sided dice. 0 1=2 1 x 2 V 0 1 2 3 4 5 6 7 8 9 10 11 12 : : : CDFT.x/ Figure 17.2 The cumulative distribution function for the sum of two 6-sided dice. 7 “mcs-ftl” — 2010/9/8 — 0:40 — page 452 — #458 Chapter 17 Random Variables and Distributions density function. This follows from the deﬁnitions of pdf and cdf: CDFR.x/ D PrŒR x D X yx PrŒR D y D X yx PDFR.y/: In summary, PDFR.x/ measures the probability that R D x and CDFR.x/ measures the probability that R x. Both PDFR and CDFR capture the same information about the random variable R—you can derive one from the other—but sometimes one is more convenient. One of the really interesting things about density functions and distribution func-tions is that many random variables turn out to have the same pdf and cdf. In other words, even though R and S are different random variables on different probability spaces, it is often the case that PDFR D PDFs: In fact, some pdfs are so common that they are given special names. For exam-ple, the three most important distributions in computer science are the Bernoulli distribution, the uniform distribution, and the binomial distribution. We look more closely at these common distributions in the next several sections. 17.3 Bernoulli Distributions The Bernoulli distribution is the simplest and most common distribution func-tion. That’s because it is the distribution function for an indicator random vari-able. Speciﬁcally, the Bernoulli distribution has a probability density function of the form fp W f0; 1g ! Œ0; 1 where fp.0/ D p; and fp.1/ D 1 p; for some p 2 Œ0; 1. The corresponding cumulative distribution function is Fp W R ! Œ0; 1 where: Fp.x/ D 8 ˆ < ˆ : 0 if x < 0 p if 0 x < 1 1 if 1 x: 8 “mcs-ftl” — 2010/9/8 — 0:40 — page 453 — #459 17.4. Uniform Distributions 17.4 Uniform Distributions 17.4.1 Deﬁnition A random variable that takes on each possible value with the same probability is said to be uniform. If the sample space is f1; 2; : : : ; ng, then the uniform distribu-tion has a pdf of the form fn W f1; 2; : : : ; ng ! Œ0; 1 where fn.k/ D 1 n for some n 2 NC. The cumulative distribution function is then Fn W R ! Œ0; 1 where Fn.x/ D 8 ˆ < ˆ : 0 if x < 1 k=n if k x < k C 1 for 1 k < n 1 if n x: Uniform distributions arise frequently in practice. For example, the number rolled on a fair die is uniform on the set f1; 2; : : : ; 6g. If p D 1=2, then an indicator random variable is uniform on the set f0; 1g. 17.4.2 The Numbers Game Enough deﬁnitions—let’s play a game! I have two envelopes. Each contains an in-teger in the range 0; 1; : : : ; 100, and the numbers are distinct. To win the game, you must determine which envelope contains the larger number. To give you a ﬁghting chance, we’ll let you peek at the number in one envelope selected at random. Can you devise a strategy that gives you a better than 50% chance of winning? For example, you could just pick an envelope at random and guess that it contains the larger number. But this strategy wins only 50% of the time. Your challenge is to do better. So you might try to be more clever. Suppose you peek in one envelope and see the number 12. Since 12 is a small number, you might guess that that the number in the other envelope is larger. But perhaps we’ve been tricky and put small numbers in both envelopes. Then your guess might not be so good! An important point here is that the numbers in the envelopes may not be random. We’re picking the numbers and we’re choosing them in a way that we think will defeat your guessing strategy. We’ll only use randomization to choose the numbers if that serves our purpose, which is to make you lose! 9 “mcs-ftl” — 2010/9/8 — 0:40 — page 454 — #460 Chapter 17 Random Variables and Distributions Intuition Behind the Winning Strategy Amazingly, there is a strategy that wins more than 50% of the time, regardless of what numbers we put in the envelopes! Suppose that you somehow knew a number x that was in between the numbers in the envelopes. Now you peek in one envelope and see a number. If it is bigger than x, then you know you’re peeking at the higher number. If it is smaller than x, then you’re peeking at the lower number. In other words, if you know a number x between the numbers in the envelopes, then you are certain to win the game. The only ﬂaw with this brilliant strategy is that you do not know such an x. Oh well. But what if you try to guess x? There is some probability that you guess cor-rectly. In this case, you win 100% of the time. On the other hand, if you guess incorrectly, then you’re no worse off than before; your chance of winning is still 50%. Combining these two cases, your overall chance of winning is better than 50%! Informal arguments about probability, like this one, often sound plausible, but do not hold up under close scrutiny. In contrast, this argument sounds completely implausible—but is actually correct! Analysis of the Winning Strategy For generality, suppose that we can choose numbers from the set f0; 1; : : : ; ng. Call the lower number L and the higher number H. Your goal is to guess a number x between L and H. To avoid confusing equality cases, you select x at random from among the half-integers: 1 2; 11 2; 21 2; : : : ; n 1 2 But what probability distribution should you use? The uniform distribution turns out to be your best bet. An informal justiﬁcation is that if we ﬁgured out that you were unlikely to pick some number—say 50 1 2— then we’d always put 50 and 51 in the envelopes. Then you’d be unlikely to pick an x between L and H and would have less chance of winning. After you’ve selected the number x, you peek into an envelope and see some number T . If T > x, then you guess that you’re looking at the larger number. If T < x, then you guess that the other number is larger. All that remains is to determine the probability that this strategy succeeds. We can do this with the usual four step method and a tree diagram. 10 “mcs-ftl” — 2010/9/8 — 0:40 — page 455 — #461 17.4. Uniform Distributions Step 1: Find the sample space. You either choose x too low (< L), too high (> H), or just right (L < x < H). Then you either peek at the lower number (T D L) or the higher number (T D H). This gives a total of six possible outcomes, as show in Figure 17.3. choices of x number peeked at TDH TDL TDH TDL TDH TDL 1=2 1=2 1=2 1=2 1=2 1=2 L=n .HL/=n .nH/=n result lose win win win win lose probability L=2n L=2n .HL/=2n .HL/=2n .nH/=2n .nH/=2n x too low x too high x just right Figure 17.3 The tree diagram for the numbers game. Step 2: Deﬁne events of interest. The four outcomes in the event that you win are marked in the tree diagram. Step 3: Assign outcome probabilities. First, we assign edge probabilities. Your guess x is too low with probability L=n, too high with probability .n H/=n, and just right with probability .H L/=n. Next, you peek at either the lower or higher number with equal probability. Multi-plying along root-to-leaf paths gives the outcome probabilities. 11 “mcs-ftl” — 2010/9/8 — 0:40 — page 456 — #462 Chapter 17 Random Variables and Distributions Step 4: Compute event probabilities. The probability of the event that you win is the sum of the probabilities of the four outcomes in that event: PrŒwin D L 2n C H L 2n C H L 2n C n H 2n D 1 2 C H L 2n 1 2 C 1 2n The ﬁnal inequality relies on the fact that the higher number H is at least 1 greater than the lower number L since they are required to be distinct. Sure enough, you win with this strategy more than half the time, regardless of the numbers in the envelopes! For example, if I choose numbers in the range 0; 1; : : : ; 100, then you win with probability at least 1 2 C 1 200 D 50:5%. Even better, if I’m allowed only numbers in the range 0; : : : ; 10, then your probability of winning rises to 55%! By Las Vegas standards, those are great odds! 17.4.3 Randomized Algorithms The best strategy to win the numbers game is an example of a randomized algo-rithm—it uses random numbers to inﬂuence decisions. Protocols and algorithms that make use of random numbers are very important in computer science. There are many problems for which the best known solutions are based on a random num-ber generator. For example, the most commonly-used protocol for deciding when to send a broadcast on a shared bus or Ethernet is a randomized algorithm known as expo-nential backoff. One of the most commonly-used sorting algorithms used in prac-tice, called quicksort, uses random numbers. You’ll see many more examples if you take an algorithms course. In each case, randomness is used to improve the probability that the algorithm runs quickly or otherwise performs well. 17.5 Binomial Distributions 17.5.1 Deﬁnitions The third commonly-used distribution in computer science is the binomial distri-bution. The standard example of a random variable with a binomial distribution is the number of heads that come up in n independent ﬂips of a coin. If the coin is 12 “mcs-ftl” — 2010/9/8 — 0:40 — page 457 — #463 17.5. Binomial Distributions f20.k/ 0:18 0:16 0:14 0:12 0:10 0:08 0:06 0:04 0:02 0 k 10 15 20 5 0 Figure 17.4 The pdf for the unbiased binomial distribution for n D 20, f20.k/. fair, then the number of heads has an unbiased binomial distribution, speciﬁed by the pdf fn W f1; 2; : : : ; ng ! Œ0; 1 where fn.k/ D n k ! 2n for some n 2 NC. This is because there are n k sequences of n coin tosses with exactly k heads, and each such sequence has probability 2n. A plot of f20.k/ is shown in Figure 17.4. The most likely outcome is k D 10 heads, and the probability falls off rapidly for larger and smaller values of k. The falloff regions to the left and right of the main hump are called the tails of the distribution. We’ll talk a lot more about these tails shortly. The cumulative distribution function for the unbiased binomial distribution is Fn W R ! Œ0; 1 where Fn.x/ D 8 ˆ < ˆ : 0 if x < 1 Pk iD0 n i 2n if k x < k C 1 for 1 k < n 1 if n x: 13 “mcs-ftl” — 2010/9/8 — 0:40 — page 458 — #464 Chapter 17 Random Variables and Distributions f20;:75.k/ 0:25 0:2 0:15 0:1 0:05 0 k 10 15 20 5 0 Figure 17.5 The pdf for the general binomial distribution fn;p.k/ for n D 20 and p D :75. The General Binomial Distribution If the coins are biased so that each coin is heads with probability p, then the number of heads has a general binomial density function speciﬁed by the pdf fn;p W f1; 2; : : : ; ng ! Œ0; 1 where fn;p.k/ D n k ! pk.1 p/nk: for some n 2 NC and p 2 Œ0; 1. This is because there are n k sequences with k heads and n k tails, but now the probability of each such sequence is pk.1 p/nk. For example, the plot in Figure 17.5 shows the probability density function fn;p.k/ corresponding to ﬂipping n D 20 independent coins that are heads with probability p D 0:75. The graph shows that we are most likely to get k D 15 heads, as you might expect. Once again, the probability falls off quickly for larger and smaller values of k. 14 “mcs-ftl” — 2010/9/8 — 0:40 — page 459 — #465 17.5. Binomial Distributions The cumulative distribution function for the general binomial distribution is Fn;p W R ! Œ0; 1 where Fn;p.x/ D 8 ˆ < ˆ : 0 if x < 1 Pk iD0 n i pi.1 p/ni if k x < k C 1 for 1 k < n 1 if n x: (17.1) 17.5.2 Approximating the Probability Density Function Computing the general binomial density function is daunting when k and n are large. Fortunately, there is an approximate closed-form formula for this function based on an approximation for the binomial coefﬁcient. In the formula below, k is replaced by ˛n where ˛ is a number between 0 and 1. Lemma 17.5.1. n ˛n ! 2nH.˛/ p 2˛.1 ˛/n (17.2) and n ˛n ! < 2nH.˛/ p 2˛.1 ˛/n (17.3) where H.˛/ is the entropy function2 H.˛/ WWD ˛ log 1 ˛ C .1 ˛/ log 1 1 ˛ : Moreover, if ˛n > 10 and .1 ˛/n > 10, then the left and right sides of Equa-tion 17.2 differ by at most 2%. If ˛n > 100 and .1˛/n > 100, then the difference is at most 0:2%. The graph of H is shown in Figure 17.6. Lemma (17.5.1) provides an excellent approximation for binomial coefﬁcients. We’ll skip its derivation, which consists of plugging in Theorem 9.6.1 for the fac-torials in the binomial coefﬁcient and then simplifying. Now let’s plug Equation 17.2 into the general binomial density function. The probability of ﬂipping ˛n heads in n tosses of a coin that comes up heads with 2log.x/ means log2.x/. 15 “mcs-ftl” — 2010/9/8 — 0:40 — page 460 — #466 Chapter 17 Random Variables and Distributions H.’/ 1 0:8 0:6 0:4 0:2 0 ’ 0:4 0:6 0:8 1 0:2 0 Figure 17.6 The Entropy Function probability p is: fn;p.˛n/ 2nH.˛/p˛n.1 p/.1˛/n p 2˛.1 ˛/n D 2n ˛ log. p ˛/C.1˛/ log 1p 1˛ p 2˛.1 ˛/n ; (17.4) where the margin of error in the approximation is the same as in Lemma 17.5.1. From Equation 17.3, we also ﬁnd that fn;p.˛n/ < 2n ˛ log. p ˛/C.1˛/ log 1p 1˛ p 2˛.1 ˛/n : (17.5) The formula in Equations 17.4 and 17.5 is as ugly as a bowling shoe, but it’s useful because it’s easy to evaluate. For example, suppose we ﬂip a fair coin n times. What is the probability of getting exactly pn heads? Plugging ˛ D p into Equation 17.4 gives: fn;p.pn/ 1 p 2p.1 p/n : 16 “mcs-ftl” — 2010/9/8 — 0:40 — page 461 — #467 17.5. Binomial Distributions Thus, for example, if we ﬂip a fair coin (where p D 1=2) n D 100 times, the probability of getting exactly 50 heads is within 2% of 0:079, which is about 8%. 17.5.3 Approximating the Cumulative Distribution Function In many ﬁelds, including computer science, probability analyses come down to get-ting small bounds on the tails of the binomial distribution. In a typical application, you want to bound the tails in order to show that there is very small probability that too many bad things happen. For example, we might like to know that it is very unlikely that too many bits are corrupted in a message, or that too many servers or communication links become overloaded, or that a randomized algorithm runs for too long. So it is usually good news that the binomial distribution has small tails. To get a feel for their size, consider the probability of ﬂipping at most 25 heads in 100 independent tosses of a fair coin. The probability of getting at most ˛n heads is given by the binomial cumulative distribution function Fn;p.˛n/ D ˛n X iD0 n i ! pi.1 p/ni: (17.6) We can bound this sum by bounding the ratio of successive terms. In particular, for i ˛n, n i 1 ! pi1.1 p/n.i1/ n i ! pi.1 p/ni D nŠpi1.1 p/niC1 .i 1/Š.n i C 1/Š nŠpi.1 p/ni iŠ.n i/Š D i.1 p/ .n i C 1/p ˛n.1 p/ .n ˛n C 1/p ˛.1 p/ .1 ˛/p: 17 “mcs-ftl” — 2010/9/8 — 0:40 — page 462 — #468 Chapter 17 Random Variables and Distributions This means that for ˛ < p, Fn;p.˛n/ < fn;p.˛n/ 1 X iD0 ˛.1 p/ .1 ˛/p i D fn;p.˛n/ 1 ˛.1 p/ .1 ˛/p D 1 ˛ 1 ˛=p fn;p.˛n/: (17.7) In other words, the probability of at most ˛n heads is at most 1 ˛ 1 ˛=p times the probability of exactly ˛n heads. For our scenario, where p D 1=2 and ˛ D 1=4, 1 ˛ 1 ˛=p D 3=4 1=2 D 3 2: Plugging n D 100, ˛ D 1=4, and p D 1=2 into Equation 17.5, we ﬁnd that the probability of at most 25 heads in 100 coin ﬂips is F100;1=2.25/ < 3 2 2100. 1 4 log.2/C 3 4 log. 2 3// p 75=2 3 107: This says that ﬂipping 25 or fewer heads is extremely unlikely, which is consis-tent with our earlier claim that the tails of the binomial distribution are very small. In fact, notice that the probability of ﬂipping 25 or fewer heads is only 50% more than the probability of ﬂipping exactly 25 heads. Thus, ﬂipping exactly 25 heads is twice as likely as ﬂipping any number between 0 and 24! Caveat. The upper bound on Fn;p.˛n/ in Equation 17.7 holds only if ˛ < p. If this is not the case in your problem, then try thinking in complementary terms; that is, look at the number of tails ﬂipped instead of the number of heads. In fact, this is precisely what we will do in the next example. 17.5.4 Noisy Channels Suppose you are sending packets of data across a communication channel and that each packet is lost with probability p D :01. Also suppose that packet losses are independent. You need to ﬁgure out how much redundancy (or error correction) to 18 “mcs-ftl” — 2010/9/8 — 0:40 — page 463 — #469 17.5. Binomial Distributions build into your communication protocol. Since redundancy is expensive overheard, you would like to use as little as possible. On the other hand, you never want to be caught short. Would it be safe for you to assume that in any batch of 10,000 packets, only 200 (or 2%) are lost? Let’s ﬁnd out. The noisy channel is analogous to ﬂipping n D 10;000 independent coins, each with probability p D :01 of coming up heads, and asking for the probability that there are at least ˛n heads where ˛ D :02. Since ˛ > p, we cannot use Equa-tion 17.7. So we need to recast the problem by looking at the numbers of tails. In this case, the probability of tails is p D :99 and we are asking for the probability of at most ˛n tails where ˛ D :98. Now we can use Equations 17.5 and 17.7 to ﬁnd that the probability of losing 2% or more of the 10,000 packets is at most 1 :98 1 :98=:99 210000.:98 log. :99 :98/C:02 log. :01 :02// p 2.:98/.1 :98/10000 < 260: This is good news. It says that planning on at most 2% packet loss in a batch of 10,000 packets should be very safe, at least for the next few millennia. 17.5.5 Estimation by Sampling Sampling is a very common technique for estimating the fraction of elements in a set that have a certain property. For example, suppose that you would like to know how many Americans plan to vote for the Republican candidate in the next presidential election. It is infeasible to ask every American how they intend to vote, so pollsters will typically contact n Americans selected at random and then compute the fraction of those Americans that will vote Republican. This value is then used as the estimate of the number of all Americans that will vote Republican. For example, if 45% of the n contacted voters report that they will vote Republican, the pollster reports that 45% of all Americans will vote Republican. In addition, the pollster will usually also provide some sort of qualifying statement such as “There is a 95% probability that the poll is accurate to within ˙4 per-centage points.” The qualifying statement is often the source of confusion and misinterpretation. For example, many people interpret the qualifying statement to mean that there is a 95% chance that between 41% and 49% of Americans intend to vote Republican. But this is wrong! The fraction of Americans that intend to vote Republican is a ﬁxed (and unknown) value p that is not a random variable. Since p is not a random variable, we cannot say anything about the probability that :41 p :49. 19 “mcs-ftl” — 2010/9/8 — 0:40 — page 464 — #470 Chapter 17 Random Variables and Distributions To obtain a correct interpretation of the qualifying statement and the results of the poll, it is helpful to introduce some notation. Deﬁne Ri to be the indicator random variable for the ith contacted American in the sample. In particular, set Ri D 1 if the ith contacted American intends to vote Republican and Ri D 0 otherwise. For the purposes of the analysis, we will assume that the ith contacted American is selected uniformly at random (with replacement) from the set of all Americans.3 We will also assume that every contacted person responds honestly about whether or not they intend to vote Republican and that there are only two options—each American intends to vote Republican or they don’t. Thus, PrŒRi D 1 D p (17.8) where p is the (unknown) fraction of Americans that intend to vote Republican. We next deﬁne T D R1 C R2 C C Rn to be the number of contacted Americans who intend to vote Republican. Then T=n is a random variable that is the estimate of the fraction of Americans that intend to vote Republican. We are now ready to provide the correct interpretation of the qualifying state-ment. The poll results mean that Pr jT=n pj :04 :95: (17.9) In other words, there is a 95% chance that the sample group will produce an esti-mate that is within ˙4 percentage points of the correct value for the overall popu-lation. So either we were “unlucky” in selecting the people to poll or the results of the poll will be correct to within ˙4 points. How Many People Do We Need to Contact? There remains an important question: how many people n do we need to contact to make sure that Equation 17.9 is true? In general, we would like n to be as small as possible in order to minimize the cost of the poll. Surprisingly, the answer depends only on the desired accuracy and conﬁdence of the poll and not on the number of items in the set being sampled. In this case, the desired accuracy is .04, the desired conﬁdence is .95, and the set being sampled is the set of Americans. It’s a good thing that n won’t depend on the size of the set being sampled—there are over 300 million Americans! 3This means that someone could be contacted multiple times. 20 “mcs-ftl” — 2010/9/8 — 0:40 — page 465 — #471 17.5. Binomial Distributions The task of ﬁnding an n that satisﬁes Equation 17.9 is made tractable by observ-ing that T has a general binomial distribution with parameters n and p and then applying Equations 17.5 and 17.7. Let’s see how this works. Since we will be using bounds on the tails of the binomial distribution, we ﬁrst do the standard conversion Pr jT=n pj :04 D 1 Pr jT=n pj > :04 : We then proceed to upper bound Pr jT=n pj > :04 D PrŒT < .p :04/n C PrŒT > .p C :04/n D Fn;p..p 0:4/n/ C Fn;1p..1 p :04/n/: (17.10) We don’t know the true value of p, but it turns out that the expression on the righthand side of Equation 17.10 is maximized when p D 1=2 and so Pr jT=n pj > :04 2Fn;1=2.:46n/ < 2 1 :46 1 .:46=:5/ fn;1=2.:46n/ < 13:5 2n.:46 log. :5 :46/C:54 log. :5 :54// p 2 0:46 0:54 n < 10:81 2:00462n pn : (17.11) The second line comes from Equation 17.7 using ˛ D :46. The third line comes from Equation 17.5. Equation 17.11 provides bounds on the conﬁdence of the poll for different values of n. For example, if n D 665, the bound in Equation 17.11 evaluates to :04978 : : : . Hence, if the pollster contacts 665 Americans, the poll will be accurate to within ˙4 percentage points with at least 95% probability. Since the bound in Equation 17.11 is exponential in n, the conﬁdence increases greatly as n increases. For example, if n D 6;650 Americans are contacted, the poll will be accurate to within ˙4 points with probability at least 1 1010. Of course, most pollsters are not willing to pay the added cost of polling 10 times as many people when they already have a conﬁdence level of 95% from polling 665 people. 21 “mcs-ftl” — 2010/9/8 — 0:40 — page 466 — #472 22 MIT OpenCourseWare 6.042J / 18.062J Mathematics for Computer Science Fall 2010 For information about citing these materials or our Terms of Use, visit:
10009	https://artofproblemsolving.com/wiki/index.php/Circumradius?srsltid=AfmBOoq0qdiMgbTEu3ddrmzqTp6twHhpyboWD9DND39lPmQZVy4emLdt	Art of Problem Solving Circumradius - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Circumradius Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Circumradius The circumradius of a cyclicpolygon is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. Contents [hide] 1 Formula for a Triangle 2 Proof 3 Formula for Circumradius 4 Circumradius, bisector and altitude 5 Euler's Theorem for a Triangle 6 Proof 7 Right triangles 7.1 Theorem 8 Equilateral triangles 9 If all three sides are known 10 If you know just one side and its opposite angle 11 See also Formula for a Triangle Let and denote the triangle's three sides and let denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply . This can be rewritten as . Proof We let , , , , and . We know that is a right angle because is the diameter. Also, because they both subtend arc . Therefore, by AA similarity, so we have or However, remember that . Substituting this in gives us and then simplifying to get and we are done. Formula for Circumradius Where is the circumradius, is the inradius, and , , and are the respective sides of the triangle and is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that . But, if you don't know the inradius, you can find the area of the triangle by Heron’s Formula: Circumradius, bisector and altitude Circumradius and altitude are isogonals with respect bisector and vertex of triangle. Euler's Theorem for a Triangle Let have circumcenter and incenter .Then Proof See Right triangles The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle. This results in a well-known theorem: Theorem The midpoint of the hypotenuse is equidistant from the vertices of the right triangle. The midpoint of the hypotenuse is the circumcenter of a right triangle. Equilateral triangles where is the length of a side of the triangle. If all three sides are known Which follows from the Heron's Formula and . If you know just one side and its opposite angle by the Law of Sines. (Extended Law of Sines) See also Inradius Semiperimeter Retrieved from " Category: Geometry Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
10010	https://math.libretexts.org/Courses/Monroe_Community_College/MTH_165_College_Algebra_MTH_175_Precalculus/00%3A_Preliminary_Topics_for_College_Algebra/0.03%3A_Review_-_Radicals_(Square_Roots)	5 3 4 17 5\|x\|\|y\|2yz−−−√ x y 10\|x\| Skip to main content 0.03: Review - Radicals (Square Roots) Last updated : Aug 16, 2025 Save as PDF 0.02e: Exercises - Whole number exponents 0.03e: Exercises - Square Roots Page ID : 38216 ( \newcommand{\kernel}{\mathrm{null}\,}) Evaluating Square Roots When the square root of a number is squared, the result is the original number. Since 42=1642=16, the square root of 1616 is 44.The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root. In general terms, if aa is a positive real number, then the square root of aa is a number that, when multiplied by itself, gives aa. The square root could be positive or negative because multiplying two negative numbers gives a positive number. However, the √ √ symbol denotes just the non-negative result, or so-called principal square root. The square root obtained using a calculator is the principal square root. The principal square root is the nonnegative number that when multiplied by itself equals aa. The principal square root of aa is written as √aa−−√. The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression. Example 0.3.10.3.1 Does √25=±525−−√=±5? Solution No. Although both 5252 and (−5)2(−5)2 are 2525, the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 2525 is √25=525−−√=5. Definition: Principal Square Root The principal square root of aa is the nonnegative number that, when multiplied by itself, equals aa. It is written as a radical expression √aa−−√, with the symbol called a radical,over the term aa, called the radicand. √aa−−√. Example 0.3.20.3.2: Evaluating Square Roots Evaluate each expression. √100100−−−√ √√1616−−√−−−−√ √25+14425+144−−−−−−−√ √4949−−√-√8181−−√ Solution √100=10100−−−√=10 because 102=100102=100 √√16=√4=216−−√−−−−√=4–√=2 because 42=1642=16 and 22=422=4 √25+144=√169=1325+144−−−−−−−√=169−−−√=13 because 132=169132=169 √49−√81=7−9=−249−−√−81−−√=7−9=−2 because 72=4972=49 and 92=8192=81 Example 0.3.30.3.3: For √25+14425+144−−−−−−−√, can we find the square roots before adding? Solution No. √25+√144=5+12=1725−−√+144−−−√=5+12=17. This is not equivalent to √25+144=√169=1325+144−−−−−−−√=169−−−√=13. The order of operations requires us to add the terms of the radicand together before finding the square root. Try It 0.3.30.3.3 Evaluate each expression. \| \| \| \| \| --- --- \| \| a. √2525−−√ \| b. √√8181−−√−−−−√ \| c. √25−925−9−−−−−√ \| d. √36+√12136−−√+121−−−√ \| Answers : a. 5 b. 3 c.4 d. 17 Using the Product Rule to Simplify Square Roots To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite √1515−−√ as √3×√53–√×5–√. We can also use the product rule to express the product of multiple radical expressions as a single radical expression. The Product Rule For Simplifying Square Roots If aa and bb are nonnegative, the square root of the product abab is equal to the product of the square roots of aa and bb √ab=√a×√b ab−−√=a−−√×b√ How to: Simplify the square root of a product. Factor any perfect squares from the radicand. Write the radical expression as a product of radical expressions. Simplify. Example 0.3.40.3.4: Using the Product Rule to Simplify Square Roots Simplify the radical expression. √300300−−−√ √162a5b4162a5b4−−−−−−√ Solution a. √100×3100×3−−−−−−√ Factor perfect square from radicand. √100×√3100−−−√×3–√ Write radical expression as product of radical expressions. 10√3103–√ Simplify b. √81a4b4×2a81a4b4×2a−−−−−−−−−√ Factor perfect square from radicand √81a4b4×√2a81a4b4−−−−−√×2a−−√ Write radical expression as product of radical expressions 9a2b2√2a9a2b22a−−√ Simplify Try It 0.3.40.3.4 Simplify √50x2y3z50x2y3z−−−−−−−√ Answer : 5\|x\|\|y\|√2yz Notice the absolute value signs around x and y. That’s because their value must be positive! How to: Simplify the product of multiple radical expressions Express the product of multiple radical expressions as a single radical expression. Simplify. Example 0.3.50.3.5: Using the Product Rule to Simplify the Product of Multiple Square Roots Multiply. Simplify the radical expression. a. √12×√312−−√×3–√ b. √6x3y3×√2x36x3y3−−−−−√×2x3−−−√. Solution a. Express the product as a single radical expression: √12×3=√36=6Express the product as a single radical expression: 12×3−−−−−√=36−−√=6 b. Begin by writing as a single radical expression: √12x6y312x6y3−−−−−−√. Determine the square factors of 12,x612,x6 , and y3y3. 12=22⋅3x6=(x3)2y3=y2⋅y}Squarefactors12=22⋅3x6=(x3)2y3=y2⋅y⎫⎭⎬⎪⎪⎪⎪Squarefactors Make these substitutions, and then apply the product rule for radicals and simplify. √12x6y3=√22⋅3⋅(x3)2⋅y2⋅yApplytheproductruleforradicals.=√22⋅√(x3)2⋅√y2⋅√3ySimplify.=2⋅x3⋅\|y\|⋅√3y=2\|x3y\|√3y12x6y3−−−−−−√=22⋅3⋅(x3)2⋅y2⋅y−−−−−−−−−−−−−√Applytheproductruleforradicals.=22−−√⋅(x3)2−−−−−√⋅y2−−√⋅3y−−√Simplify.=2⋅x3⋅\|y\|⋅3y−−√=2\|x3y\|3y−−√ Notice the absolute value bars around x3yx3y. That is because the result of taking a square root is never negative. Try It 0.3.50.3.5 Simplify √50x×√2x50x−−−√×2x−−√ assuming x>0x>0. Answer : 10\|x\| Using the Quotient Rule to Simplify Square Roots Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. THE QUOTIENT RULE FOR SIMPLIFYING SQUARE ROOTS The square root of the quotient abab is equal to the quotient of the square roots of aa and bb , where a≥0a≥0 and b>0b>0 . √ab=√a√b ab−−√=a−−√b√ How to: Simplify the radical of a quotient Simplify the quotient radicand. Write the simplified quotient radical expression as the quotient of two radical expressions. Simplify the numerator and denominator. Example 0.3.60.3.6: Using the Quotient Rule to Simplify Square Roots Simplify the radical expression. \| \| \| --- \| \| a. √536536−−−√ \| b. √18a5b818a5b8−−−−−√ \| Solution a. Write as a quotient of two radical expressions and then simplify the denominator: √5√36=√565–√36−−√=5–√6 b. Begin by determining the square factors of 1818, a5a5, and b8b8. 18=2⋅32a5=a2⋅a2⋅a=(a2)2⋅ab8=b4⋅b4=(b4)2}Squarefactors18=2⋅32a5=a2⋅a2⋅a=(a2)2⋅ab8=b4⋅b4=(b4)2⎫⎭⎬⎪⎪⎪⎪Squarefactors Make these substitutions, apply the product and quotient rules for radicals, and then simplify. √18a5b8=√2⋅32⋅(a2)2⋅a(b4)2Applytheproductandquotientruleforradicals.=√32⋅√(a2)2⋅√2a√(b4)2Simplify.=3a2√2ab4 Try It 0.3.6 Simplify √2x29y4 Answer : \|x\|√23y2 We do not need the absolute value signs for y2 because that term will always be nonnegative. How to: Simplify the quotient of multiple radical expressions Express the quotient of multiple radical expressions as a quotient of a single radical expression. Simplify the radical. Example 0.3.7: Use the Quotient Rule to Simplify the Quotient of Two Square Roots Divide. Simplify the radical expression. \| \| \| --- \| \| a. √234x11y√26x7y \| b. √50x6y4√8x3y \| Solution a. First combine numerator and denominator into one radical expression √234x11y√26x7y=√234x11y26x7yWrite under one radical=√9x4Simplify radicand=3x2Simplify square root b. Write as a single square root and cancel common factors before simplifying. √50x6y4√8x3y=√50x6y48x3yApplythequotientruleforradicalsandcancel.=√25x3y34Simplify.=√25x3y3√4=5\|xy\|√xy2 Try It 0.3.7 Simplify √9a5b14√3a4b5 Answer : b4√3ab From this point on the assumption will be made that all variables represent non-negative real numbers.Therefore using absolute values when simplifying will not be necessary. Adding and Subtracting Square Roots We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. For example, the sum of √2 and 3√2 is 4√2. However, it is often possible to simplify radical expressions, and that may change the radicand. The radical expression √18 can be written with a 2 in the radicand, as 3√2, so √2+√18=√2+3√2=4√2 How to: Simplify a radical expression requiring addition or subtraction of square roots Simplify each radical expression. Add or subtract expressions with equal radicands. Often, we will have to simplify before we can identify the like radicals within the terms. If the radicand and the index are not exactly the same, then the radicals are not similar and we cannot combine them. Example 0.3.8: Add or Subtract Square Roots Perform the indicated operation and simplify. 4√10−5√10 5√12+2√3. 20√72a3b4c−14√8a3b4c Solution a. =(4−5)√10=−1√10=−√10 b. We can rewrite 5√12 as 5√4×3. According the product rule, this becomes 5√4√3. The square root of √4 is 2 , so the expression becomes 5×2√3, which is 10√3. Now we can the terms have the same radicand so we can add. 10√3+2√3=12√3 c. Rewrite each term so they have equal radicands. 20√72a3b4c=20√9√4√2√a√a2√(b2)2√c=20(3)(2)\|a\|b2√2ac=120\|a\|b2√2ac 14√8a3b4c=14√2√4√a√a2√(b2)2√c=14(2)\|a\|b2√2ac=28\|a\|b2√2ac Now the terms have the same radicand so we can subtract. 120\|a\|b2√2ac−28\|a\|b2√2ac=92\|a\|b2√2ac Try It 0.3.8 Add √5+6√20 Subtract 3√80x−4√45x Answers : a. 13√5 b. 0 Example 0.3.9: Adding and Subtracting Square Roots Simplify. 10√5+6√2−9√5−7√2 √32−√18+√50 2a√125a2b−a2√80b+4√20a4b. Solution a. 10√5+6√2−9√5−7√2 =10√5−9√5+6√2−7√2=√5−√2 We cannot simplify any further because √5 and √2 are not like radicals; the radicands are not the same. Caution: It is important to point out that √5−√2≠√5−2. We can verify this by calculating the value of each side with a calculator. √5−√2≈0.82 is not the same as √5−2=√3≈1.73 b. √32−√18+√50 =√16⋅2−√9⋅2+√25⋅2=4√2−3√2+5√2=6√2 At first glance, the radicals do not appear to be similar. However, after simplifying completely, we see that we can combine them. c. Step 1: Simplify the radical expression. Step 2: Combine all like radicals. Remember to add only the coefficients; the variable parts remain the same. 2a√125a2b−a2√80b+4√20a4b=2a√25⋅5⋅a2⋅b−a2√16⋅5⋅b+4√4⋅5⋅(a2)2bFactor.=2a⋅5⋅a√5b−a2⋅4√5b+4⋅2⋅a2√5bSimplify.=10a2√5b−4a2√5b+8a2√5bCombineliketerms.=14a2√5b Try It 0.3.9 Simplify: √20+√27−3√5−2√12. Answer : −√5−√3 Caution: Simplifying Radicals Take careful note of the differences between products and sums within a radical. Assume both x and y are nonnegative. ProductsSums√x2y2=xy√x2+y2≠x+y The property says that we can simplify radicals when the operation in the radicand is multiplication. There is no corresponding property for addition. Simplifying Products of Expressions containing Square Roots Often, there will be coefficients in front of the radicals. Example 0.3.10.1x: Multiply: 3√6⋅5√2 Solution Using the product rule for radicals and the fact that multiplication is commutative, we can multiply the coefficients and the radicands as follows. 3√6⋅5√2=3⋅5⋅√6⋅√2Multiplicationiscommutative.=15⋅√12Multiplythecoefficientsandtheradicands.=15√4⋅3Simplify.=15⋅2⋅√3=30√3 Typically, the first step involving the application of the commutative property is not shown. We will often find the need to subtract a radical expression with multiple terms. If this is the case, remember to apply the distributive property before combining like terms. Example 0.3.10.2x: Use the distributive property with square roots Simplify: (5√x−4√y)−(4√x−7√y)Solution: =5√x−4√y−4√x+7√yDistribute.=5√x−4√x−4√y+7√y=√x+3√y Use the distributive property when multiplying rational expressions with more than one term. Example 0.3.10.3x: Multiply: 5√2x(3√x−√2x). Solution: Apply the distributive property and multiply each term by 5√2x. 5√2x(3√x−√2x)=5√2x⋅3√x−5√2x⋅√2xDistribute.=15√2x2−5√4x2Simplify.=15x√2−5⋅2x=15x√2−10x The process for multiplying radical expressions with multiple terms is the same process used when multiplying polynomials. Apply the distributive property, simplify each radical, and then combine like terms. Example 0.3.10.4x: Multiply: (√x−5√y)2. Solution (√x−5√y)2=(√x−5√y)(√x−5√y) Begin by applying the distributive property. =√x⋅√x+√x(−5√y)+(−5√y)√x+(−5√y)(−5√y)=√x2−5√xy−5√xy+25√y2=x−10√xy+25y The binomials (a+b) and (a−b) are called conjugates. When multiplying conjugate binomials the middle terms are opposites and their sum is zero. Example 0.3.10.5x: Multiply: (√10+√3)(√10−√3). Solution Apply the distributive property, and then combine like terms. (√10+√3)(√10−√3)=√10⋅√10+√10(−√3)+√3(√10)+√3(−√3)=√100−√30+√30−√9=10−√30+√30−3=10−3=7 It is important to note that when multiplying conjugate radical expressions, we obtain a rational expression. This is true in general (√x+√y)(√x−√y)=√x2−√xy+√xy−√y2=x−y Alternatively, using the formula for the difference of squares we have, (a+b)(a−b)=a2−b2Differenceofsquares.(√x+√y)(√x−√y)=(√x)2−(√y)2=x−y Try It 0.3.10x Multiply: (3−2√y)(3+2√y). (Assume y is positive). Answer : 9−4y Rationalizing Denominators When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator. We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical. Monomial Denominators For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is b√c, multiply by √c√c. How To: Rationalize the denominator of an expression with a monomial denominator Multiply the numerator and denominator by the radical in the denominator. Simplify. Sometimes, we will find the need to reduce, or cancel, after rationalizing the denominator. Example 0.3.11: Rationalize a Denominator Containing a Single Term Write in simplest form (rationalize the denominator). \| \| \| \| --- \| a. 2√33√10 \| b. √2√5x \| c. 3a√2√6ab \| Solution a. The radical in the denominator is √10. So multiply the fraction by √10√10. Then simplify. 2√33√10×√10√10=2√3030=√3015 b. The goal is to find an equivalent expression without a radical in the denominator. The radicand in the denominator determines the factors that you need to use to rationalize it. In this example, multiply by 1 in the form √5x√5x. √2√5x=√2√5x⋅√5x√5x=√10x√25x2Simplify.=√10x5x c. In this example, we will multiply by 1 in the form √6ab√6ab. 3a√2√6ab=3a√2√6ab⋅√6ab√6ab=3a√12ab√36a2b2Simplify.=3a√4⋅3ab6ab=6a√3ab6abCancel.=√3abb Notice that b does not cancel in this example. Do not cancel factors inside a radical with those that are outside. Try It 0.3.11 Write in simplest form (rationalize the denominator). \| \| \| --- \| \| a. 12√3√2 \| b. √9x2y \| Answer : a. 6√6 b. 3√2xy2y Binomial Denominators For a denominator containing the sum or difference of rational or irrational terms, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign that connects the two terms in the denominator. If the denominator is a+b√c , then the conjugate is a−b√c. How to: Rationalize the denominator of an expression with a binomial denominator Find the conjugate of the denominator. Multiply the numerator and denominator by the conjugate. Use the distributive property. Simplify. Example 0.3.12: Rationalizing a Denominator Containing Two Terms Write in simplest form (rationalize the denominator). \| \| \| \| \| --- --- \| \| a. 41+√5 \| b. 1√5−√3 \| c. √10√2+√6 \| d. √x−√y√x+√y \| Solution a. Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of 1+√5 is 1−√5. Then multiply the fraction by 1−√51−√5 . 41+√5×1−√51−√54−4√5−4Use the distributive property−1+√5Simplify b. In this example, the conjugate of the denominator is √5+√3. Therefore, multiply by 1 in the form (√5+√3)(√5+√3). 1√5−√3=1(√5−√3)(√5+√3)(√5+√3)Multiplyby"1".=√5+√3√25+√15−√15−√9Simplify.=√5+√35−3=√5+√32 Notice that the terms involving the square root in the denominator are eliminated by multiplying by the conjugate. We can use the property (√a+√b)(√a−√b)=a−b to expedite the process of multiplying the expressions in the denominator. c. Multiply by 1 in the form √2−√6√2−√6. √10√2+√6=(√10)(√2+√6)(√2−√6)(√2−√6)Multiplebytheconjugate.=√20−√602−6Simplify.=√4⋅5−√4⋅15−4=2√5−2√15−4=2(√5−√15)−4=√5−√15−2=−√5−√152=−√5+√152 d. In this example, we will multiply by 1 in the form √x−√y√x−√y. √x−√y√x+√y=(√x−√y)(√x+√y)(√x−√y)(√x−√y)Multiplybytheconjugateofthedenominator.=√x2−√xy−√xy+√y2x−ySimplify.=x−2√xy+yx−y Try It 0.3.12 Write in simplest form (rationalize the denominator). \| \| \| --- \| \| a. 72+√3 \| b. 2√35−√3 \| Answer : a. 14−7√3 b. 5√3+311 0.02e: Exercises - Whole number exponents 0.03e: Exercises - Square Roots
10011	https://www.education.com/worksheet/article/evaluate-expressions-using-order-of-operations-1/	Worksheet Evaluate Expressions Using Order of Operations #1 Get practice evaluating expressions using the order of operations with this mixed operations math worksheet! Tailored for learners in fifth and sixth grades, this worksheet covers a wide range of difficulty, beginning with expressions that require two steps to evaluate and advancing to more intricate expressions that require up to five steps to evaluate. With many of the problems containing exponents, this worksheet also gives students the opportunity to practice evaluating exponents. As a warm-up, consider having students complete the Order of Operations: PEMDAS worksheet. Then, for more practice, check out the next two worksheets in this series: Evaluate Expressions Using Order of Operations #2 and Evaluate Expressions Using Order of Operations #3. Grades: Subjects: View aligned standards Educational Tools Support Connect About IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Wyzant Trusted tutors for 300 subjects Vocabulary.com Adaptive learning for English vocabulary ABCya Fun educational games for kids SpanishDictionary.com Spanish-English dictionary, translator, and learning Emmersion Fast and accurate language certification TPT Marketplace for millions of educator-created resources Copyright © 2025 Education.com, Inc, a division of IXL Learning • All Rights Reserved.
10012	https://pmc.ncbi.nlm.nih.gov/articles/PMC4238837/	Neurological aspects of human parvovirus B19 infection: a systematic review - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Rev Med Virol . 2014 Jan 24;24(3):154–168. doi: 10.1002/rmv.1782 Search in PMC Search in PubMed View in NLM Catalog Add to search Neurological aspects of human parvovirus B19 infection: a systematic review Faraj Barah Faraj Barah 1 Center for Neuroscience and Cell Biology, Faculty of Medicine, University of Coimbra, Coimbra, Portugal Find articles by Faraj Barah 1,, Sigrid Whiteside Sigrid Whiteside 2 Department of Medical Statistics, Education & Research Centre, University Hospital of South Manchester, Manchester, UK 4 Wythenshawe Hospital, Manchester, UK Find articles by Sigrid Whiteside 2,4, Sonia Batista Sonia Batista 3 Neurology Department, Coimbra Hospital and University Centre, Coimbra, Portugal Find articles by Sonia Batista 3, Julie Morris Julie Morris 2 Department of Medical Statistics, Education & Research Centre, University Hospital of South Manchester, Manchester, UK 4 Wythenshawe Hospital, Manchester, UK Find articles by Julie Morris 2,4 Author information Article notes Copyright and License information 1 Center for Neuroscience and Cell Biology, Faculty of Medicine, University of Coimbra, Coimbra, Portugal 2 Department of Medical Statistics, Education & Research Centre, University Hospital of South Manchester, Manchester, UK 3 Neurology Department, Coimbra Hospital and University Centre, Coimbra, Portugal 4 Wythenshawe Hospital, Manchester, UK ✉ Correspondence author: F. Barah, PhD, Center for Neuroscience and Cell Biology, Faculty of Medicine, University of Coimbra, 3004–504 Coimbra, Portugal., E-mail: farajbarah@hotmail.com Received 2013 Oct 22; Revised 2013 Nov 28; Accepted 2013 Nov 29; Issue date 2014 May. © 2014 The Authors. Reviews in Medical Virology published by John Wiley & Sons, Ltd. This is an open access article under the terms of the Creative Commons Attribution-NonCommercial License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited and is not used for commercial purposes. PMC Copyright notice PMCID: PMC4238837 PMID: 24459081 Abstract Parvovirus B19 has been linked with various clinical syndromes including neurological manifestations. However, its role in the latter remains not completely understood. Although the last 10 years witnessed a surge of case reports on B19-associated neurological aspects, the literature data remains scattered and heterogeneous, and epidemiological information on the incidence of B19-associated neurological aspects cannot be accurately extrapolated. The aim of this review is to identify the characteristics of cases of B19-associated neurological manifestations. A computerized systematic review of existing literature concerning cases of B19-related neurological aspects revealed 89 articles describing 129 patients; 79 (61.2%) were associated with CNS manifestations, 41 (31.8%) were associated with peripheral nervous system manifestations, and 9 (7.0%) were linked with myalgic encephalomyelitis. The majority of the cases (50/129) had encephalitis. Clinical characteristic features of these cases were analyzed, and possible pathological mechanisms were also described. In conclusion, B19 should be included in differential diagnosis of encephalitic syndromes of unknown etiology in all age groups. Diagnosis should rely on investigation of anti-B19 IgM antibodies and detection of B19 DNA in serum or CSF. Treatment of severe cases might benefit from a combined regime of intravenous immunoglobulins and steroids. To confirm these outcomes, goal-targeted studies are recommended to exactly identify epidemiological scenarios and explore potential pathogenic mechanisms of these complications. Performing retrospective and prospective and multicenter studies concerning B19 and neurological aspects in general, and B19 and encephalitic syndromes in particular, are required. © 2014 The Authors. Reviews in Medical Virology published by John Wiley & Sons, Ltd. INTRODUCTION Since its discovery in the 1970s of last century 1, human parvovirus B19 (B19) has been linked with a broad spectrum of clinical syndromes, including erythema infectiosum (EI), transient aplastic crisis, persistent infection manifesting as pure red cell aplasia in immunocompromised individuals, nonimmune hydrops fetalis, and arthritis. Less commonly recognized, but receiving increasing attention recently, are the neurological manifestations, a variety of which have been described in patients with either clinically diagnosed or laboratory-confirmed B19 infection. The last 10 years witnessed a surge of case reports on the association of B19 with neurological aspects. However, the literature on B19 infection and its association with neurological aspects continue to be heterogeneous, and epidemiological data on the incidence of B19-associated neurological aspects cannot be accurately extrapolated. Therefore, the role of B19 in neurological diseases remains incompletely described and understood. The pathogenesis of B19 infection is complex and variable, so it is likely that a combination of mechanisms contribute to the development of neurological manifestations 2, although there is a lack of detailed descriptions of autopsy reports. The objectives of this systematic review are to search for cases of B19-related neurological aspects and identify the clinical characteristics of those patients that could be associated with B19 infection. METHODS A computerized search was conducted using all databases included in Web of Knowledge in addition to PubMed database. The search was performed combining the terms (‘human parvovirus’ or ‘parvovirus B19’ or ‘B19’ or ‘erythema infectiosum’) and (‘neurologic complication’ or ‘neurological disorder’ or ‘neurological manifestation’ or ‘central nervous system’ or ‘peripheral nervous system’ or ‘a specific term for a specific neurological disorder’) without language and time restrictions. The specific terms for neurological disorders used in the search were obtained from the website of National Institute of Neurological Disorders and Stroke 3, with a total of 442 disorders and manifestations. In addition, all cited references listed in the identified papers were hand-searched for other relevant articles. An article was considered for inclusion in the systematic review if it reported cases with B19 infection that presented with neurological manifestations. A case was considered eligible for the following reasons: (i) if data of age, sex, immune status, description of manifestations and investigation, treatment, and outcomes were presented and (ii) if B19 infection was diagnosed in the presence of B19 DNA or anti-B19 IgM specific antibodies in the serum or the CSF. Exceptions included cases with neurological manifestations associated with the presence of clinical presentation of EI while laboratory tests were not performed or available. The legitimacy behind that relies on the fact that B19 is the sole agent for EI. In the absence of B19 specific markers, other common B19-related clinical manifestations, such as transient aplastic crisis, persistent infection manifesting as pure red cell aplasia, nonimmune hydrops fetalis, and arthritis, were not considered as indicators of B19 infection because the latter is not their sole etiological agent. Cases of B19-associated neurological manifestations that result from intrauterine infection were also excluded. B19-associated myalgic encephalomyelitis (ME) cases were included because of the neurological classification of ME in the World Health Organization's International Classification of Diseases (ICD G93.3) but classified and labeled separately. Cases that did not fulfill the International Consensus Criteria of ME 4 were excluded. The computerized search was conducted for the last time on 30 June 2013. The preferred reporting items for systematic review and meta-analysis guidelines were followed 5. Data were summarized using percentages and cross tabulations. Comparisons between subgroups were made using Fisher's exact tests. The 95% confidence intervals (CIs) for percentages were calculated using the Wilson method. All statistical analyses used the conventional two-sided 5% significance level and were carried out using spss version 20 and cia version 2.0. RESULTS As shown in Figure 1, the search using Web of Knowledge databases identified 998 publications, whereas PubMed database search identified 903 publications, with a combined search result of 1065 publications. A scrupulous analysis resulted in 89 eligible articles 6–94 describing the history of 129 patients, published between the years 1970 and 2012, which were further evaluated. Figure 1. Open in a new tab Flow diagram of information through the different phases of the review Seventy nine of the eligible cases (61.2%) were associated with CNS manifestations (Table 1), whereas 41 (31.8%) were associated with peripheral nervous system (PNS) manifestations (Table 2), and nine cases (7.0%) were linked with ME (Table 3). Many of the cases (50/129) had encephalitis, encephalopathy, or meningoencephalitis. The patients age ranged from 1 day to 75 years; median age of 12.5 years, with 70 (54.3%) children (<18 years) and 59 (45.7%) adults (≥18 years). The male-to-female ratio was 4 : 5 (55.8% female). One hundred cases (77 · 5%) were immunocompetent, whereas 29 cases (22.5%) were patients with suppressed immune status. Table 1. Cases of B19 infection and manifestations related to the central nervous system Open in a new tab Table 2. Cases of B19 infection and manifestations related to the peripheral nervous system Open in a new tab Table 3. Cases of B19 infection and myalgic encephalomyelitis Open in a new tab Clinical features of B19-associated encephalitis, encephalopathy, and meningoencephalitis cases were subject to comprehensive analysis in this review because many of the B19-associated neurological aspects were cases belonging to this category. For readers who are interested in other B19-related neurological aspects, they should refer to Tables 3 and to the Discussion Section of this review. B19 and encephalitis, encephalopathy, and meningoencephalitis Fifty cases of encephalitic syndromes were found and reviewed 6–39, representing 63.3% of B19-related CNS cases and 38.8% of total B19-related neurological cases currently found in the literature (Table 1). In most of these cases (33/50), B19 was sought after other possible pathogens were proved to be negative, and the etiological cause was not determined. In addition, B19 was investigated in these cases because of the appearance of B19-related symptoms or merely because of a suspicion of B19 infection. This association was confirmed by the detection of at least one specific marker for B19 infection, with the exception of two cases of encephalitis associated with EI prior to the recognition of the etiological role of B19 in this disease 6,7. However, 12 cases were identified during two retrospective studies of 43 and 282 patients, respectively, with etiologically undiagnosed neurological symptoms and with no sufficient clinical information to support the detection of a recent B19 infection [14,16]. Four more cases were detected in another retrospective study that targeted 346 patients with aplastic crisis 19. An additional case was detected during a screening program of 1572 sera from hospitalized pediatric patients with various presentations submitted for viral investigation 8. Analysis of CSF for cell count and protein and glucose concentrations varied according to the cases. From those who were subject to CSF analysis, 19 cases had normal white cells count, whereas 21 cases had raised count, 15 had normal protein concentration whereas 16 had higher concentration, and 23 had normal glucose concentration whereas only two had lower concentrations. The majority of the cases (34, 68.0%) were immunocompetent and 16 (32.0%) were immunocompromised. Typical EI rash was observed in 15 cases (30.0%), 13 cases among children (33.0%, 95% CI 20.6–49.0%) and two cases among adults (20%, 95% CI 5.7–51.0%). Only one of these (no. 33) had suppressed immunity, as might be expected from the immunopathological nature of the rash. All 17 cases detected during screening programs or retrospective studies (except one) were free from the rash 8,14,16,19. The timing of neurological symptoms in relation to the rash varied considerably. B19-associated encephalitis presented prior to the appearance of the rash in five cases, contemporaneously with the appearance of the rash in four cases, or following the appearance of the rash in six cases. There were statistically significant differences between patients with competent and suppressed immune status and symptoms of rash (p = 0.018); rash was observed by 42.4% of patients with competent immune status, compared with only 6.2% with suppressed immune status. Anemia was detected in 21 cases (42.0%), 17 cases among children (43.6%, 95% CI 29.3–59.0%) and four cases among adults (40.0%, 95% CI 16.8–68.7). There were statistically significant differences between patients with competent and suppressed immune status and symptoms of anemia (p = 0.002). The majority of the cases with anemia were observed at reduced immune status (12/16, 75%) comparing with the immunocompetent group (9/34, 26.5%). Arthralgia was present only in three cases among children (7.7%, 95% CI 2.7–20.3%) and one case among adults (10.0%, 95% CI 1.8–40.4). There were no statistically significant differences between patients with competent and suppressed immune status and symptoms of arthralgia (p = 0.289). Neuroimaging studies (Table 4) were performed on 34 cases using computed axial tomography (20 cases), electroencephalogram (20 cases), and/or magnetic resonance imaging (27 cases). Among the computed axial tomography cases, 12 were normal whereas eight showed a range of abnormalities including enlarged ventricles, lesions, frontal, and occipital vasogenic swelling. By electroencephalogram, only three cases were normal whereas 17 showed abnormal activities. Using magnetic resonance imaging, 11 cases were normal whereas 16 showed various abnormalities including enlarged ventricles and increased signal from the white matter. Two cases were normal with all three types of neuroimaging studies. Table 4. Neuroimaging studies for B19-associated encephalitis, encephalopathy and meningoencephalitis cases Open in a new tab Excluding 25 cases in which treatment regimen was not known (12 cases), not specific (10 cases), or the encephalitis presentation resolved spontaneously without medical intervention (three cases), most cases were initially treated with antivirals (11 cases) and/or antibiotics (eight cases) to cover a possibility of unidentified viral and/or bacterial encephalitis, respectively, accompanied sometimes with an addition of antiepileptics (10 cases) to relieve the symptoms. However, when B19 involvement was either suspected or confirmed, 16 cases were treated with intravenous immunoglobulins (IVIGs) and/or steroids. Four cases were treated only with IVIG, two of them showing concomitant clinical improvement. Four other cases were treated only with steroids, two of them showing associated clinical improvement. Eight cases were treated with IVIG and steroids (according to the physicians reports about these cases, two showed improvement most probably due to IVIG, two showed improvement most probably due to steroid treatment, while three showed improvement due to combined IVIG and steroids treatment). None of the cases treated with IVIG and/or steroids died but four of them (no. 24, 32, 37, and 50) treated in later stage with either IVIG or steroids (but not both) had mild neurological sequelae. It should be noted that two of these cases had underlying immunodeficiency (no. 24 and 32) while the other two were treated with IVIG alone (no. 50) or steroids alone (no. 37). In contrast, seven out of nine patients who did not receive IVIG and/or steroids were either slow in their recovery (three cases), had a form of neurological sequelae (three cases), or even died (one case). There were no statistically significant differences between patients with competent and suppressed immune status and the success of the treatment (IVIG and/or steroids) (p = 0.188). The prognosis of the encephalitis associated with B19 appears to vary. Although in some cases, complete recovery without neurological sequelae was achieved, there were seven deaths (14.0%) following the illness, and in 13 cases (26.0%), long-term neurological sequelae were observed, ranging from mild language learning difficulties and slurred speech to mental and motor retardation. DISCUSSION This is the first systematic review that targets the association between B19 infection and neurological aspects. We identified in this review 129 cases, reported in 89 publications including case reports, brief communications, comments or letters to the editors, retrospective studies or screening programs, and follow-up studies. These publications linked the virus with various neurological aspects either confined to CNS (including encephalitis; 50 cases, meningitis; 12 cases (no. 51–62), cerebellar ataxia as isolated neurological event; one case (no. 63), stroke; seven cases (no. 64–70), transverse myelitis; three cases (no. 71–73), seizures; five cases (no. 74–78), Reye's syndrome; one case (no. 79)) or related to PNS (including neuralgic amyotrophy; eight cases (no. 80–87), peripheral neuropathy; 19 cases (no. 88–106), carpal tunnel syndrome (CTS); 10 cases (no. 107–116), Guillain–Barré syndrome (GBS); four cases (no. 117–120)), in addition to nine cases that reported an association between B19 infection and ME (no. 121–129). The most common B19-associated neurological manifestation was encephalitic syndromes, representing 38.8% of the total B19-related neurological cases that are currently in the literature. B19 is not usually investigated during encephalitis episodes, and in most of the cases reported, it was only sought after other pathogens that are commonly involved in encephalitis were proved to be negative, and the etiological cause remained undetermined. In addition, in other cases, B19 was suspected because of the appearance of one of its related symptoms or merely because of the physician anticipation of or interest in B19 infection. This could explain to some extent the rarity of B19 involvement in such a widespread neurological disorder, which occurs at a rate of six to seven cases among 100 000 individuals worldwide 95. That means if B19 is investigated at the same rate as other pathogens that commonly cause encephalitis, more cases could be detected. This is supported by the fact that when B19 was sought retrospectively in encephalitis cases, a 4% detection rate was found in the two retrospective studies performed so far, detecting 12 cases with etiologically undiagnosed encephalitis and with no sufficient information to support the detection of a recent B19 infection 14,16. Therefore, we suggest performing larger multicenter retrospective studies and further prospective studies to support these findings. Because the cost of detecting anti-B19 IgM antibodies and B19 DNA in serum or CSF is relatively low, we recommend at this stage that detection of B19 should be incorporated in the differential diagnosis of encephalitis cases. The represented cases of B19-associated encephalopathy clearly indicate that there are no distinguishing features of B19-associated encephalitis compared with encephalitis caused by many other viral pathogens, except for the presence of rash, anemia, and arthritis in some patients. It is clear from the evidence of B19 infection upon retrospective analyses that there are no clinical clues regarding the diagnosis 14,16. For example, physicians cannot depend only on typical EI rash to confirm the concurrent B19 infection with encephalitis because many case reports, especially those identified retrospectively, failed to identify the coexistence of the rash with the encephalitic episode. In addition, the timing of neurological symptoms in relation to the rash varied considerably without a clear pattern. Furthermore, in immunocompromised individuals, B19-specific rash is usually absent around the time of neurological illness because of the immunopathological nature of the rash. Also, a physician cannot depend on the appearance of other B19-related symptoms, such as arthralgia, which is present in few cases, and anemia, which is detected mostly in cases with reduced immune status because of various conditions other than neurological symptoms. On the other hand, analysis of CSF for cell count and protein and glucose concentrations vary among cases, and significant indications cannot be obtained from these data and thus cannot be used in supporting the association of B19 infection in such cases. In addition, physicians cannot rely on neuroimaging studies in detecting specific abnormalities related to B19-associated encephalitis. Therefore, we conclude that the detection or confirmation of associated B19 infection with encephalopathy should always depend on the presence of B19 specific markers, namely, the anti-B19 IgM antibodies and B19 DNA in serum or CSF. However, B19 DNA may be detectable for extended periods, even in healthy individuals 96. Therefore, the presence of low levels of B19 DNA alone cannot be used to diagnose B19 infection with encephalopathy. Intravenous immunoglobulin are considered the only treatment option for many clinical syndromes associated with B19 infection because it is believed that they include a good source of antibodies to neutralize the virus, although the mechanism of IVIG action is not precisely known. On the other hand, steroid therapy was also suggested for treatment of neurological disorders in general and encephalopathy in particular. When B19 involvement was either suspected or confirmed, 16 cases of encephalopathy were treated with IVIG and/or steroids. Given the clinical cases of encephalopathy presented, we are clearly not in a position to fully support the use of IVIG, steroids, or their combination in encephalitis associated with B19 infection, although 12 cases showed clinical improvement, because favorable outcomes with full recovery seem to be another distinctive feature for many encephalopathy reported cases without IVIG and/or steroids treatment, which may suggest a casual association with IVIG and/or steroids when they are used. In addition, IVIG and steroids were given together in eight cases showing clinical improvement, and therefore, an objective assessment of efficacy of either treatment cannot be obtained. However, with the absence of other effective treatment regimens, the use of combined treatment of IVIG and steroids in B19-associated encephalopathy could be considered. We, therefore, recommend that treatment of severe cases might benefit from a combined regime of IVIG and steroids, until a randomized prospective clinical trial of this regimen can be conducted. There were seven deaths following encephalitis associated with B19, and in 12 cases, long-term neurological sequelae were observed, urging the necessity of rapid diagnosis of B19 infection and swift clinical intervention with combined IVIG and steroids regimen. In contrast to encephalitis cases, prognosis appears to be good in cases of B19-associated meningitis with a high rate of spontaneous cure and no sequelae reported. In addition to one case of ataxia as an individual isolated event in association with B19 50, there are also six cases where cerebellar involvement was additionally suggested either clinically (no. 16, 20, 44, and 50) or pathologically (no. 17 and 60). Although aplastic crisis can be, by itself, a risk factor for stroke, B19 could participate in the latter in patients without aplastic crisis. Isolated events of seizures were also reported, although episodes of seizures seemed more part of a wider neurological picture. Scattered case reports have linked B19 infection with transverse myelitis and Reye's syndrome. We are not in a position to confirm these associations because of the low number of cases reported in the literature. Large-scale retrospective studies are required to confirm the association of B19 infection with these CNS presentations. B19 is generally not regarded to be neurotropic, but direct infection and local replication of the virus could not be ruled out as possible pathogenesis mechanism for B19-associated CNS aspects. There have been controversial reports regarding the detection of B19 DNA in brain tissues. In one study, no evidence of B19 invasion of the brain was observed using relatively insensitive techniques 97. However, in other reports, B19 DNA was detected in a brain biopsy specimen from a 67-year-old woman with severe meningoencephalitis by PCR 15 and in the nucleus of the multinucleated giant cells and solitary endothelial cells for a hydropic fetus by in situ PCR 98. These data were supported by recent large-scale studies that used a highly specific nested PCR and nucleotide sequencing to detect B19 sequences in the dorsolateral prefrontal cortex 99 and cerebellum 100 of postmortem adult human brains. However, concerns should be cast over the fact that B19 DNA could persist in many tissues in detectable levels for years. In these cases, B19 DNA is likely to be existing in the circulation at high levels after infections and which becomes sequested and persists in the tissues as a result. Distinguishing that remnant DNA from the products of an active infection in a tissue has plagued many studies of B19 pathogenesis. B19 DNA persistence in brain tissues, however, could by itself provoke the pathogenic action of the virus through inducing chromosomal defect or damage 2. On the other hand, immune-related pathogenic mechanisms cannot be ruled out, supported by complete resolution of symptoms in cases treated with steroids. In addition, recent reports suggest that some cases of anti-N-methyl-d-aspartate receptor encephalitis could be triggered by B19 35,36,101. Vascular injury, particularly in the cerebellum, could also be involved in the pathogenesis. Therefore, we conclude that complex and variable pathogenesis is likely to contribute to the CNS manifestations. Exact mechanisms of actions through thorough prospective and retrospective pathological studies on sera, CSF, and postmortem tissues are required. The number of B19-associated neuralgic amyotrophy cases does not necessarily reflect the extent of B19 involvement in this neurological disorder of unknown etiology because all were case reports. Therefore, retrospective and prospective studies are required to give a more comprehensive picture of the involvement of B19 in this disorder. The pathological mechanism of brachial plexus neuritis after B19 infection is also not known. However, clinical characteristics of reported cases show some striking similarities: All cases but one concerned adults, whereas the majority of parvovirus infections usually occur in children. All patients but one presented symptoms of brachial plexus neuritis coinciding with or immediately after the appearance of EI rash, which is interesting because the typical rash is believed to be immune mediated and generally coincides with the appearance of viral antibodies. It is likely that either autoantibodies or immune complex deposition are involved in these cases while direct infection and local replication of the virus could be ruled out. Reported B19-associated peripheral neuropathy cases occurred as paraesthesia, dysesthesia, cranial nerve palsy, optic neuropathy, mononeuritis multiplex (MM), or acute autonomic sensory and motor neuropathy (AASM). In general, these cases followed a subacute but progressive course with unpredictable extension and severity ranging from complete recovery with no further neurological symptoms to pure limited sensory disorders and could end with severe multifocal sensory motor axonal loss with marked functional disability. The way in which B19 is able to trigger neuropathy is not fully elucidated. Possible mechanisms may involve necrotizing vasculitis through immune complex deposition or hypersensitivity vasculitis secondary to B19 infection. Persistence of B19 infection may play a role in prolongation of the disorder. When followed up, B19 DNA was present in serum beyond 6 months after onset of neuropathy in at least five cases (no. 94–95 and 99–101). This could provide a rationale for the treatment of such patients with IVIG. However, and as discussed in B19-associated encephalitis cases, its combination with steroids in three patients and the possibility of a spontaneous progressive recovery after acute nerve injury does not allow definite conclusions to be made about the efficacy of IVIG in B19-associated neuropathy. Complete neurological, neurophysiological, and nerve histological examination of B19-associated neuropathy should be performed. Further epidemiological studies are required to confirm the link between B19 and the neuropathy and to assess the frequency of B19-related neuropathy in comparison with other causes of B19-free neuropathy. However, B19 infection should be routinely considered in the etiological assessment, especially in the event of initial sensory symptoms with concurrent rash, as this could lead to an early appropriate treatment with IVIG. There are several causes of CTS, but in many cases, the etiology remains unknown. The assumptions that B19 can be the infectious agent that triggers CTS, and that the coexistence of B19 infection and CTS was not causal in the reported cases, require prospective and follow-up studies that compare B19 markers (detection and quantification) between CTS patients and a CTS-free control group, preferably during an epidemic peak of B19 infection. This is because most CTS cases are usually confirmed long time after laboratory detection of B19 acute infection. B19 is not usually cited as a cause of GBS compared with Campylobacter jejuni and cytomegalovirus that constitute the most frequent bacterial and viral triggers 102. Only four cases linked to B19 infection are reported in the literature, and retrospective studies have yet to prove any association. However, it is worth mentioning that, in a relatively old prospective study, serum samples taken from a group of GBS patients were examined for the presence of a variety of pathogens 103. Anti-B19 IgM antibodies were detected in four patients (4%) but not in controls. Although this was statistically insignificant, this finding suggests that parvovirus B19 may be an important cause of some cases of GBS. These four cases were not included in this study because of insufficient data concerning them. Few case reports and follow-up studies have documented an association between acute B19 infection and ME (no. 121–129), whereas others found no association 104,105. Reasons behind this disparity could be due to various groups promoting different nomenclature, diagnostic criteria, etiologic hypotheses, and treatments for ME, resulting in controversy about many aspects of this disorder. For example, ME is not always classified as a neurological disorder. In addition, although ME may follow acute B19 infection, attribution of a case of ME to B19 infection may be extremely difficult in the absence of serological confirmation of acute infection at fatigue onset. Although the gold standard method of diagnosis is to confirm acute B19 infection through positive anti-B19 IgM antibodies or B19 DNA at the time of onset of fatigue, this would be a rare occurrence, and a more practical method would is needed to detect other B19 markers during the illness. Kerr et al. 106 found out that antibodies against B19 nonstructural protein are associated with chronic and not acute B19 infection and therefore could be used as a marker for B19-assciated ME cases. Because several different infections may be involved in ME, the proportion resulting from any one agent, such as B19, is likely to vary, as evident in various follow-up studies, with regard to sampling strategy, time and place, and its relation to the prevalence of each infection. Two particularly important factors are whether there is an outbreak in progress at a particular location and the selection strategy for the control/comparison groups. These factors should be taken into consideration when attempting to confirm the possible link between B19 and ME. Interestingly, five out of nine patients were improved after IVIG 92,94,107, a matter that should be considered when B19 is suspected or confirmed to be associating with ME cases. This study has various limitations: Firstly, the literature data on this topic are few and heterogeneous in terms of criteria used for characterization of B19-associated neurological symptoms, in addition to the differences in diagnosis, complementary investigation, treatment, and follow-up of these cases. Therefore, epidemiological data on the incidence of B19-associated neurological aspects cannot be accurately extrapolated. Prospective, structured, multicenter studies would be necessary to determine the real epidemiological scenario of these complications that are currently receiving increasing attention. Furthermore, although there are some hypotheses on the pathogenesis of B19-associated neurological aspects, lack of detailed descriptions of autopsy reports render the pathogenesis not completely understood, and therefore, thorough prospective and retrospective pathological studies on postmortem tissues using sensitive immunocytochemistry and in situ hybridization techniques are a priority. In conclusion, pending answers to the questions raised, we recommend that B19 infection should be included in the differential diagnosis of encephalitic syndrome and some PNS manifestations regardless of the age. We recommend that diagnosis of B19-associated neurological aspect should solely depend on the investigation of anti-B19 IgM antibodies and B19 DNA in serum or CSF. We also suggest that severe cases of B19-associated neurological aspects might benefit from a combined regime of IVIGs, and steroids and a randomized controlled trial should be considered. CONFLICT OF INTEREST The authors have no competing interest. Acknowledgments The support by the projects WELCOMEII/6/CNC/1055/2011 and PEst-C/SAU/LA0001/2013-2014 “financiado por Fundos FEDER através do Programa Operacional Factores de Competitividade – COMPETE e por Fundos Nacionais através da FCT – Fundação para a Ciência e a Tecnologia” are gratefully acknowledged. Glossary AASM acute autonomic sensory and motor neuropathy B19 human parvovirus B19 CAT computed axial tomography CIs confidence intervals CTS carpal tunnel syndrome EEG electroencephalogram EI erythema infectiosum GBS Guillain–Barré syndrome IVIGs intravenous immunoglobulins ME myalgic encephalomyelitis MM mononeuritis multiplex MRI magnetic resonance imaging NIHF nonimmune hydrops fetalis PNS peripheral nervous system PRCA persistent infection manifesting as pure red cell aplasia PRISMA preferred reporting items for systematic review and meta-analysis TAC transient aplastic crisis REFERENCES Cossart YE, Field AM, Cant B, Widdows D. Parvovirus-like particles in human sera. Lancet. 1975;1:72–73. doi: 10.1016/s0140-6736(75)91074-0. [DOI] [PubMed] [Google Scholar] Barah F, Vallely PJ, Cleator GM, Kerr JR. Neurological manifestations of human parvovirus B19 infection. Reviews in Medical Virology. 2003;13:185–199. doi: 10.1002/rmv.388. 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[DOI] [PubMed] [Google Scholar] Articles from Reviews in Medical Virology are provided here courtesy of Wiley ACTIONS View on publisher site PDF (6.2 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract INTRODUCTION METHODS RESULTS DISCUSSION CONFLICT OF INTEREST Acknowledgments Glossary REFERENCES Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
10013	http://www.math.utah.edu/~wortman/1060text-loc.pdf	Law of Cosines Suppose we have a triangle with one of its angles, θ, identiﬁed. Suppose further that the length of the side of the triangle that is opposite the angle θ is c. The other two sides of the triangle have length a and b. Then the law of cosines is the formula c2 = a2 + b2 −2ab cos(θ) Problem. Find cos(θ) if θ is the angle shown in the triangle below. Solution. The law of cosines tells us that 22 = 32 + 42 −2(3)(4) cos(θ) 252 C b 0 C b a a n 4-, n 0 Simpliﬁed, we have 4 = 25 −24 cos(θ) We can subtract 25 −21 = −24 cos(θ) and then divide by −24 cos(θ) = −21 −24 = 7 8 Problem. Find c if c is the length of the side of the triangle shown below. Solution. The law of cosines states that c2 = 42 + 52 −2(4)(5) cos π 3 Simpliﬁed, we have c2 = 16 + 25 −40 cos π 3 We know that cos π 3 = 1 2 so that the equation can be simpliﬁed further as c2 = 16 + 25 −20 = 21 Therefore, either c = √ 21 or c = − √ 21. However, c is a length, which must be a positive number, so c = √ 21. That’s the end of the solution. 253 n 4-, n 0 Why the law of cosines is true If we have the triangle then we can draw another line from the top of the triangle to the base in a way that creates a right angle. We’ll say that the length of this new line is d, and that the length of the base to the right of this new line is f. Because the length of the entire base of our original triangle was a, the length of the base to the left of the new line we drew must be a −f. The line we drew cuts our original triangle into two right triangles. From the right triangle on the right we see that sin(θ) = d b (the length of the side opposite θ divided by the length of the hypotenuse) and that cos(θ) = f b (the length of the side adjacent to θ divided by the length of the hypotenuse). 254 C b 0 C b a a n 4-, n 0 C b 0 C b a a These last two equations are equivalent to the equations b sin(θ) = d and b cos(θ) = f, respectively. We’ll use these two equations in applying the Pythagorean theorem to the triangle on the left: c2 = (a −f)2 + d2 = a2 −2af + f 2 + d2 = a2 −2a b cos(θ) + b cos(θ)]2 + [b sin(θ) 2 = a2 −2ab cos(θ) + b2 cos(θ)2 + b2 sin(θ)2 = a2 −2ab cos(θ) + b2 cos(θ)2 + sin(θ)2 We know that cos(θ)2 + sin(θ)2 = 1. This is the Pythagorean Identity. It was one of the identities that we learned in the Sine and Cosine chapter as Lemma 7. Therefore, c2 = a2 −2ab cos(θ) + b2 cos(θ)2 + sin(θ)2 = a2 −2ab cos(θ) + b2 = a2 + b2 −2ab cos(θ) 255 Exercises Find cos(θ). 1.) 3.) 2.) 4.) 256 03 U’ -ç tJ 03 U’ -ç tJ 03 U’ -ç tJ 03 U’ -ç tJ Find c. You can consult the chart on page 227 in the chapter “Sine and Cosine” to ﬁnd the values of cosine that you need to complete these problems. 5.) 8.) 6.) 9.) 7.) 10.) 257 L) 0 ci c) cc) 0 c3D ci N L) 0 ci c) cc) 0 c3D ci N L) 0 ci c) cc) 0 c3D ci N L) 0 ci c) cc) 0 c3D ci N L) 0 ci c) cc) 0 c3D ci N L) 0 ci c) cc) 0 c3D ci N
10014	https://fishingbooker.com/blog/types-of-bass-in-north-america/	Browse Fishing Charters Types of Bass in North America: A Simple Guide for 2025 Updated on Aug 25, 2025 \| 6 minute read \| Written by Albert Comments Written by Albert Reading Time: 6 minutes “Bass fishing” can mean very different things to different people. For some, it brings up memories of lakeside mornings and tournament circuits. For others, it means trolling beachfronts or deep sea adventures. This article breaks down the different types of Bass in North America, with a quick look at how they fit together. There are dozens, possibly hundreds of fish called “Bass” in the world. It’s impossible to cover them all in one place. However, there are two main families of Bass in North America, and a few extras that are well worth mentioning. You probably know most of them, but there might be a couple of surprises hidden along the way. Types of Black Bass For many people, Black Bass are the only Bass. These guys have a multi-billion-dollar industry built up around them. Tournaments are held every week of the year. Bass fishing pros tour the country, battling monster fish as well as each other. Let’s take a look at some of the fish causing all this commotion. Largemouth Bass They say you can never fool a Largemouth the same way twice. They’ll attack fish, insects, and even small birds with incredible aggression, but will avoid the most convincing lures if they came across them before. Their intelligence is probably exaggerated by Black Bass fanatics, but nobody can deny that Largemouth are an amazing game fish. The funny thing is that, to your average non-angler, Largemouth Bass probably look pretty boring. They’re small. They’re round. They have no crazy fins or exciting color patterns. It really is all about the fight with these lake-loving legends. Just beware that, if you’re Bass fishing in Florida, you’re most likely targeting a separate strain altogether! Top Largemouth Bass Fishing Charters in Texas ### Tracking Texas Trophies 4.9 / 5 (93 reviews) Sulphur Springs 21ft 2 people Instant confirmation trips from US $400 See availability ### Reel Texas Fishing Tours 5 / 5 (27 reviews) Austin 21ft 2 people Instant confirmation trips from US $400 See availability ### JDR Fishing Adventures 4.9 / 5 (209 reviews) Alba 22ft 4 people Instant confirmation trips from US $350 See availability ### Lake Fork Premier Guide Service 4.9 / 5 (42 reviews) Quitman 20ft 2 people Instant confirmation trips from US $350 See availability See all fishing charters in Texas Smallmouth Bass Their big-mouthed brothers may get the headlines, but Smallmouth Bass are just as worthy of the limelight. They’re every bit as wily and put up an even better fight pound for pound. At least, that’s what the “Smallie” crowd says. We’ve covered Largemouth and Smallmouth Bass in much more detail elsewhere, but the main differences are size and habitat. Smallmouth are smaller on average. They like colder water and stronger current. You can find both fish in the same place, but one always tends to be more dominant. Spotted Bass “Spotties” are much less common and are unfairly overlooked by many Bass anglers. When people do catch them, they often mistake them for Largemouth because of their similar coloring. The best way to tell the two apart is that the upper jaw on a Spotted Bass doesn’t extend past its eye. In plain English, it doesn’t have such a large mouth as Largemouth. Spotted Bass like water somewhere between what their famous cousins go for. It needs to have some current, like with Smallmouth, but in warm, murky water, where you’d expect to find Largemouth. Essentially, they’re the “baby bear” of the Black Bass family. And More! These are the most common types of Black Bass, but there are plenty more fish in the family. Some only inhabit a single river, like the Guadalupe Bass of Texas. Others have only recently been recognized as a separate species, such as Florida’s Choctaw Bass. There could be even more species just waiting for someone to discover them! Types of Temperate Bass Of course, just because Black Bass are big business, doesn’t mean they’re the only fish out there. The Temperate Bass family includes one of America’s most important sport fish. Let’s take a look at North America’s “other” Bass family. Striped Bass If you’ve ever fished on the East Coast, chances are you’ve at least tried to catch a Striped Bass. Ken Schultz describes Stripers as “one of the most valuable and popular fish in North America” in his Fishing Encyclopedia. Big words, but well deserved. These guys are big, strong, and mean – all the makings of the perfect sport fish. Stripers spend most of their lives in the sea, but head inland to spawn. The problem is that most of them migrate into one place – the Chesapeake Bay. This bottleneck leaves them vulnerable to overfishing. Recently, several states canceled their trophy Striper season to try and avoid this. Whatever you make of the closures, let’s hope it helps to keep the species healthy. White Bass White Bass are Stripers’ smaller, freshwater cousins. Unlike most of the fish on our list, these guys aren’t really considered sport fish by many anglers. They’re less aggressive than Stripers and less wily than Black Bass. They’re perfect for kids and beginners, though. White Bass like large lakes and reservoirs. They prefer clear waters that are at least 10 feet deep. As they hang out in schools, you can really fill the boat when you find them. Opinions vary on White Bass as food. They have a particular taste that some people love and others avoid. The only way to find out which group you’re in is to catch one! Yellow Bass Another rung down the ladder, Yellow Bass are one of the smallest members of the Temperate Bass family. They don’t put up much of a fight, and rarely weigh more than a pound. Even complete beginners will have an easy time reeling one in. To be fair, people catch them more for their meat than their might. They’re supposed to be even tastier than Stripers. What’s more, they’re way less overfished, making them a great choice for sustainably-minded fish lovers. And More! These are the most important Temperate Bass species. The other one worth a mention is White Perch. “Wait, that’s not a Bass!” Actually, it is. It’s in the same family are Stripers. There’s also another branch of Temperate Bass in Europe, which includes European Seabass – every British beach angler’s favorite fish. “Bass” That Aren’t Bass There are fish called “Bass” all over the world. Most of them have little or no connection to each other. Here are the top species of Bass-not-Bass in North America. They all prefer completely different habitats. Each one of them is from a different family. The one thing they’ve got in common is that they’re incredibly popular to catch. Black Seabass These guys are technically a species of Grouper. They show up around reefs and piers from Texas to Maine. Black Seabass are mainly caught for food, but they can also put up a good fight on light tackle. It’s not clear how Black Seabass got their name. They don’t look or act like any other type of Bass. One thing’s for sure: they’re absolutely delicious, and a welcome sight on any fishing trip. White Seabass Is White Seabass related to Black Seabass? No. Does it have anything to do with European Seabass? No. Is it big and tasty? Yes. Sold! Also known as “White Weakfish,” these confusingly-named fish are actually distant cousins of Redfish. They’re most closely related to Geelbek, a popular catch in South Africa. How they ended up in the East Pacific is a mystery, but you won’t hear any complaints from Californian anglers. Peacock Bass People often assume that these brightly-colored brutes are Black Bass. They fight just as hard and look kind of similar, after all. Peacock Bass are actually Cichlids, from the same family as Tilapia. There are around 15 different species of Peacock Bass that we know of. The most common one in the US is Butterfly Peacock Bass. Peacock Bass were introduced in Florida in the ‘80s to battle invasive Tilapia and Oscar. They quickly got to work eating up their relatives, and helped to stabilize the ecosystem. In the process, they became a much-loved sport fish and are now an iconic part of South Florida’s fishing scene. And More! “Bass” is one of those names that gets handed out to all kinds of fish. Some of them are popular table fare such as Chilean Seabass. Others are critically-endangered goliaths like the Giant Seabass. It seems like pretty much any fish can take the name as long as it tastes great, fights hard, or both. We’ve tried to cover the top types of Bass in North America. It’s not an easy list to narrow down, but these are the species you’re most likely to come across. Hopefully, we’ve helped to clear up one of the most confusing fish names out there! Did any of the fish on our list surprise you? Which ones did we miss? Why do you think so many different species are called “Bass”? Let us know your thoughts and theories in the comments below! Albert Share this post with friends Comments (28) Zachary Meeks July 7, 2024 Jul 7, 2024 Recently, I caught a 5.4 lb bass on a swimbait. I’m usually on point at identifying fish, but I couldn’t figure out what it was. It looked like a largemouth at first, but didn’t have the eyes. It also wasn’t as chunky as a largemouth, it was built more like a walleye, so I think it might have been a spotted. Could you figure out what is was? Sorry, I don’t have a photo. I caught it at Riverwalk Park in Bakersfield, CA in 105+ degree weather. Tanya Replied on July 8, 2024 Jul 8, 2024 Hi Zachary, glad to hear your hunt was a successful one! That area is famous for Largemouth Bass and Trout, not so much for Spotted and Smallmouth Bass, for example. Those fisheries are stocked with Crappie and Bluegill as well, but they don’t match your description. It’s hard to tell what species that could have been without looking at the fish 🤔 But maybe one of our fellow anglers and readers was in a similar situation, so they might chip in and share their experience with you. View all replies Marinos Baris November 30, 2023 Nov 30, 2023 Red-Eyed Rock Bass are a rather common catch in many of Maryland’s rivers. I’ve caught them in all branches of the Upper and Lower Gunpowder, as well as the Patapsco River. They’re a fun catch, although they are quite quarrelsome and territorial and sometimes scare away a LOT of the stocked Rainbow and Golden Rainbow Trout in Spring and early Autumn. I’ve done some light snorkeling in these rivers in the summer months, and have run into a small family of them. Either of the parents, usually the more aggressive male will guard their fry and “nest” fiercely, and have no problem going in for a flash-attack against any one threatening them. Funny thing is I’ve seen them totally ignore people’s legs and feet, while they will charge me with great aggressiveness, whenever my face is under water! They’re very beautiful, too. Under water, they take on a very light or dark all-body coloration depending on where they’re located, and their eyes, mouth area, and fins are bright red, and quite aggressive looking, while the female is more subdued and camouflaged as she is protecting their young. Unfortunately, lately I’ve seen a LOT of people walking through the middle of a stream casting seines all around them, and keeping fish of all types regardless of size. Their kids will find schools of minnows and swine them, too, although some of those schools are of larger species whose parents were also caught and taken out. This is destroying the natural species that have taken so long to establish themselves and is causing even more damage to the ecosystem. I’d like to see more enforcement against these types of poachers, including heavy fines for multiple violations or in addition to other crimes. M Baris Rhys Replied on November 30, 2023 Nov 30, 2023 Hi Marinos, Thanks for sharing your insights with us. Red-eyed Rock Bass are certainly an exciting species and have very similar attributes to the Smallmouth Bass. In fact, they often fight for the same food sources, so you can often catch them when going after Smallies! I definitely agree that they’re a species worth protecting and, here at FishingBooker, we always support responsible and ethical fishing – including adhering to all size and bag limits. I fully support your suggestion to get law enforcement agencies and your local Department for National Resources involved in punishing anyone found guilty of harming the species and not respecting regulations. I suggest you reach out to them to see if they can do something. Tight lines, View all replies Jonathan Heu July 17, 2023 Jul 17, 2023 There is also the Wiper, its a hybrid of a Stripper & White Bass. Lets not forget the Saddam Bass, only found in Iraq. Both the Wiper & Saddam Bass are mean. Keep Rippin Lips. #StayBlessed Load more comments Top Charters Nearby #### Chesapeake Bay #### California #### Florida Related posts [### Best Bass Bait: An Angler’s Guide for 2025 Are you ready for an epic Bass fishing trip? Pretty much every Bass species is beloved by anglers wherever they show up. And, like all fish, they respond well to different baits. Artificial and natural lures both have the potential to catch a trophy Bass. 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10015	https://www.caliper.com/glossary/what-is-a-thematic-map.htm?srsltid=AfmBOooiu081jx0Nv4HRRgj47IoOX7GNwfGPdO6RHT71mkJ9_QSo4OKG	What is a Thematic Map? 9 Theme Types Defined Toggle navigation Desktop Maptitude Desktop Overview Features Examples API Map Your Data Data Included Latest Version Free Add-in Tools Brochure (PDF) Redistricting (Electoral, USA) Online Pricing Pricing Purchase/Store Learn Options Training Video Tutorials Free Webinars Tech Tips Learning Portal Glossary Blog Resources Testimonials Blog Case Studies Featured Maps Newsletters Articles News About Us Contact Us Solutions Banking Business Census Data Mapping Geocoding GIS Mapping Government Healthcare Marketing & Sales Microsoft Integration Online Mapping Political Mapping Real Estate Retail Resources & Utilities Territory Mapping Three Dimensional & GPS Mapping Transportation USA Data Australia Data Brazil Data Canada Data UK Data Europe Data Other Country Packages Deutsch English Español Français Italiano Portugêse العربية 中文 עברית Indonesia اردو Call Us 617-527-4700ReviewsFree TrialRequest Demo Mapping Software and GIS Glossary DEFINITION What is a thematic map? 9 Types of Thematic Maps 5 February 2025 A thematic map is a map created to display information about a topic or theme. Themes can be shown with a choropleth (color), heat map, sizes, charts, dot-densities, three-dimensional prisms, cluster (bubble), or cartogram. A bivariate thematic map combines two types of themes (e.g., color and size or color and prism) so that you can display information about two variables. This thematic map uses a color theme (heat map) to illustrate the median home value in ZIP Codes around San Diego. ZIP Codes shown with a darker color have higher home values. This map also uses a chart theme to illustrate housing tenure. The size of the pie chart indicates the total number of housing units in the ZIP Code, and the pie pieces show the share of owner-occupied vs. renter occupied. In this article: Types of Thematic Map Deciding Which Type of Theme to Use GIS Software for Creating Thematic Maps Thematic Mapping Software Reviews What is a choropleth map? What is a Heat Map? What is a size theme map? What are pie and bar chart theme maps? What is a dot-density map? What is a prism map? What is a cluster/bubble map? What is a Cartogram? What is a bivariate map? Frequently Asked Questions Who uses thematic maps? How do you read a thematic map? What is the difference between a thematic map and a physical map? How do you create a thematic map in Excel? What are the most commonly used thematic maps? Are political maps thematic maps? Are topographic maps thematic? Types of Thematic Map In some kinds of thematic maps, the appearance of each feature is based on its characteristics. In other types of thematic maps, charts or symbols on each feature illustrate comparative information. When data are portrayed in a thematic map, the information becomes easier to see, understand, and interpret. Choropleth Theme Symbol Theme Dot-Density Theme Chart Theme Proportional Symbol/Size Theme 3D Theme Deciding Which Type of Theme to Use Thematic maps illustrate the characteristics of map features, so that the characteristics are easy to see and understand. A map theme is a styling of map features according to the values of one or more data fields. Choosing the correct theme can be tricky. The table below helps you choose the correct tool for the data you are using. \| If your data contain... \| For example... \| Then you should... \| --- \| Counts or amounts \| Population, Sales \| Use any kind of theme \| \| Percentages, ratios, or averages \| Mean Income, Percent Hispanic \| Avoid dot-density themes \| \| Negative numbers \| Change in Population \| Use color or symbol themes \| \| Non-numeric information \| Zoning, Sales Territory \| Use color or symbol themes \| GIS Software for Creating Thematic Maps Maptitude Mapping Software gives you all of the tools, maps, and data you need to analyze and understand how geography affects you and your business. Create informative color, chart, scaled-symbol, prism, heat, and dot-density thematic maps that show the characteristics of your area of interest. Free TrialRequest a Demo Thematic Mapping Software Reviews “Maptitude has a wealth of analytical features that satisfy my ongoing needs in the area of demographic analysis. Regular in the business/demographic sphere include thematic mapping, route planning, territory design, trade area analysis and location/allocation modelling.” Nich Nicholas Sydney, Australia “I've made tons of use of the themes (you can color the maps with backgrounds based on the included geodemographics). I've also used the band feature - retailers LOVE it. I recommend it ALL THE TIME!” John Miglautsch Hartland, Wisconsin “All of the tools you need to do spatial analysis are included in the software at the base price. No need to pay for costly add-ons or extensions. This includes state, county, MSA, zip code and census tract level line work and data including all 255 Census fields. The user has the ability to do point maps, thematic maps, drive-time rings and advanced spatial and statistical analysis.” Brett Lucas Spokane, Washington “Thematic maps are truly unique in terms of graphic capabilities and data representation, highly appreciated by customers. With a low cost combined with an excellent training program and support service, Maptitude is an added value for my business!” Giuseppe Staiti Rome, Italy “I've been using Maptitude by Caliper mapping software for over 20 years and love it! The functionality and value are industry leading in the Thematic Mapping space. Check out this software if you are looking for a way to show data in visual ways you didn't know existed. Good stuff! :)” Scott Jablonski Columbus, Ohio What is a choropleth map? Achoropleth map(also called acolor theme) is a thematic map in which points, lines or areas are shaded to represent different data values. In a typical example, administrative areas are colored or shaded according to the range in which the aggregated statistic of interest falls. Choropleth maps are often confused with heat maps. A choropleth map represents distinct data values for geographic features such as states, counties, or postal codes. A heat map shows a continuous representation of density using grid cells. A location that is "hot" and colored red in a heat map could be between and encompass geographic features such as high sales location points. A choropleth map of quantitative data: This map uses color to illustrate the median income by ZIP Code in Miami, Florida. The darker the color, the higher the median income is in that ZIP Code. A choropleth map of nominal data: This map uses color to illustrate how each parcel is zoned. Try an interactive choropleth map now! This interactive choropleth map lets you see the per capita GDP of each U.S. county. Create your own choropleth maps! Try the full-featured version of Maptitude with a 1-month risk-free trial. No credit card is required Choropleth Maps: Best used for: Displaying geographical data by shading or coloring geographic regions according to the value of a particular variable. Advantages: Easy to interpret, allows for a quick comparison of regions, and can easily be used in conjunction with other mapping techniques. Disadvantages: Can be subject to misinterpretation, difficult to accurately convey data when regions are very small. But, the Maptitude thematic mapping Wizard makes it easy to correctly create choropleth maps. Free TrialRequest a Demo What is a Heat Map? Aheat mapis used to visualize and analyze point data by transforming the points into a density grid. Each resulting grid cell is assigned a value that is determined by the proximity of nearby points, optionally weighting each point using a weighting value. Heat maps use color palettes that represent the values from low to high. Geographic heat maps are often confused with choropleth maps, but a heat map does not conform to geographic locations such as states, counties, postal codes, or land parcels. A heat map can illustrate the concentration of stores across a city to help you visualize over and underserved areas. Heat maps are best used for visualizing data that has a geographical component. They are particularly effective for showing the intensity of a variable across a region. For example, map crime data to find hot spots, map vehicle miles traveled to find areas of congestion and higher air pollution, and analyze the pattern of clients around a store weighted by the cost of their purchases to find areas that are underserved. Try an interactive heat map now! This interactive heat map shows where there are larger concentrations of private schools. Create your own heat maps! Try the full-featured version of Maptitude with a 1-month risk-free trial. No credit card is required Heat Maps: Best Used For: Heat maps are best used for visualizing data that has a geographical component. They are particularly effective for showing the intensity of a variable across a region, such as population density or temperature. Advantages: Heat maps are easy to read and interpret, since they rely on color to convey information. They can quickly show patterns and trends, making them an effective tool for data analysis. Disadvantages: Heat maps can be difficult to interpret, since color variations can be subtle and hard to distinguish. Additionally, they can be misleading if they are not properly scaled or interpreted. But, the Maptitude thematic mapping Wizard makes it easy to correctly create heat maps. See also: How do I create a heat map of my locations? What is a size theme map? A size theme (also called a scaled-symbol or graduated-symbol theme) is a type of thematic map that uses symbols of different sizes or lines of different widths to show the value of a data field. Features that have a greater value for the chosen field will be shown larger or with a greater width. A point layer of customers with a size theme on the value of sales. Customers with higher sales are shown with a larger symbol. A line layer of highways with a size theme on the value of vehicle flow. Highways with more traffic are shown wider than those with less traffic. Try an interactive size theme map now! This interactive size theme map lets you see the the locations of hospitals and the number of beds that they have. Create your own size-theme maps! Try the full-featured version of Maptitude with a 1-month risk-free trial. No credit card is required Graduated Symbol or Size Maps: Best Used For: Visualizing data with multiple attributes and showing the distribution of the data over the map. Advantages: Provides an easy way to visualize data and compare multiple attributes. It is an effective way to show the magnitude of geographic features. Disadvantages: It can be difficult to distinguish between different attributes and can be difficult to read. It can be difficult to determine the exact values of the data being represented. But, the Maptitude thematic mapping Wizard makes it easy to correctly create size theme maps. Free TrialRequest a Demo What are pie and bar chart theme maps? Pie and bar chart themes are a type of thematic map that use pie or bar charts on map features to illustrate the data that go with each feature. A pie chart theme where the size of the chart indicates the number of housing units in a Census Tract and the pie wedges indicate the size of the structures that contain those units. A bar chart theme that shows the number of owner-occupied housing units in each ZIP Code within different value ranges. Try an interactive chart theme map now! This interactive chart theme map shows the commute times for workers in each state. Create your own chart-theme maps! Try the full-featured version of Maptitude with a 1-month risk-free trial. No credit card is required What is a dot-density map? A dot-density map is a type of thematic map that uses dots or other symbols on the map to show the values of one or more numeric data fields. Each dot on a dot-density map represents some amount of data. In a dot-density map, areas with many dots indicate high concentrations of values for the chosen field and fewer dots indicate lower concentrations. This Maptitude dot-density map shows the distribution of the Hispanic and Non-Hispanic population. Each blue dot represents 30,000 Non-Hispanic people and each red dot represents 30,000 Hispanic people. Population dense areas have many dots and less populated areas have fewer dots. Areas with higher Hispanic populations stand out in central California, southern Texas, and southern Florida. Dot-Density Maps: Best used for: Displaying population data or other data points with a high density. Advantages: Easy to interpret, allows for a quick comparison of regions, and can easily be used in conjunction with other mapping techniques. Disadvantages: Can be subject to misinterpretation, difficult to accurately convey data when regions are very small. But, the Maptitude thematic mapping Wizard makes it easy to correctly create dot-density maps. Free TrialRequest a Demo What is a prism map? Aprism map is a three-dimensional thematic map that uses prisms to show the relative values of a data field where higher prisms indicate areas with higher values. An example of a prism map where states are depicted as three-dimensional prisms whose heights indicate the median home value. What is a cluster/bubble map? Acluster map (also known as a bubble map) is a type of thematic map that uses clusters of symbols or bubbles to represent and analyze spatial patterns in point data. Each cluster or bubble represents a grouping of items based on proximity and often varies in size to reflect the quantity or magnitude of the data within that group. To create a cluster map, spatial data points are aggregated based on their proximity to one another. As the user zooms in, the clusters break apart to reveal smaller clusters or individual points. Conversely, zooming out combines points into larger clusters. The visual representation typically involves varying the size of the bubbles to indicate the number of points they contain. An example of a cluster map where individual points for all of the customers would be too dense to differentiate so they are combined into clusters based on proximity and the clusters are labeled with the number of customers that they represent Cluster Maps: Best Used For: Cluster maps are ideal for visualizing large datasets with numerous individual points, such as locations of businesses, incidents of crime, or occurrences of a particular event. They help to simplify the display by aggregating points into clusters, making it easier to discern patterns and trends at different scales. Advantages: Simplifies Complex Data: By clustering points, these maps reduce visual clutter and make it easier to process large amounts of data. Scalable Analysis: Cluster maps can be zoomed in and out to show more or less detail, allowing for both macro and micro-level analysis. Quantitative Representation: The size of the bubbles can be scaled to represent the quantity of data points, providing a quick visual indication of density or concentration. Disadvantages: Loss of Specificity: Individual data points are not visible when clustered, which can obscure specific location details. Dependent on Scale: The appearance and interpretation of clusters can change significantly with the scale of the map, potentially leading to different conclusions. Requires Interaction: To fully explore the data, users often need to interact with the map, such as by zooming or clicking on clusters, which may not be ideal for static displays. Try an interactive cluster map now! This interactive cluster map shows where there are larger concentrations of private schools. Create your own cluster maps! Try the full-featured version of Maptitude with a 1-month risk-free trial. No credit card is required Free TrialRequest a Demo What is a Cartogram? A cartogram is a map that is rehsaped to represent data such as population demographics, business data, or natural phenomena. In the map below, for example, the sizes of the states represent the total number of African Americans. A cartogram of African American population by state. States with a large African American population are shown larger than states with a small population. What is a bivariate map? Abivariate map is a type of thematic map that displays two variables by using different colors and/or symbols. The variables in a bivariate map can be on the same layer or separate layers. This bivariate map uses color and size on the same layer to show the number of rental units and median rent in different Census tracts. The size of the dots indicates the number of rental units and the color indicates the rent. Free TrialRequest a Demo Frequently Asked Questions Who uses thematic maps? Thematic maps are used by a wide range of people for a variety of purposes. Some common users of thematic maps include: Geographers and cartographers to study patterns and trends in geographic data, such as population density, migration, or land use Government agencies to plan and make decisions about infrastructure, resource management, and public policy. Businesses to analyze market trends and make business decisions, such as identifying potential customers or choosing new locations for stores or offices Environmental organizations to monitor and protect natural resources such as forests, wetlands, and wildlife habitats. Educators to teach students about geography and spatial data analysis Overall, thematic maps are used by anyone who needs to understand and visualize geographic data in order to make informed decisions or communicate information to others How do you read a thematic map? Reading a thematic map involves understanding the meaning of the symbols, colors, and other visual elements used on the map. Here are some steps to follow when reading a thematic map: Look at the title: The title of the map should give you an idea of what the map is showing and the area it covers. Examine the legend: The legend is a key that explains the symbols and colors used on the map. It should tell you what each symbol or color represents. Identify the map's projection: The projection is the way the map represents the curved surface of the earth on a flat piece of paper. Different projections can affect the way the map looks and the distances and shapes shown on it. Look at the scale: The scale is a numerical representation of the size of the map relative to the actual size of the area it represents. The scale will help you understand the distances shown on the map. Examine the data: Look at the map and identify the patterns and trends shown by the data. Pay attention to the symbols and colors used on the map and refer back to the legend if you are not sure what they mean. Interpret the data: Use the information on the map to draw conclusions and make interpretations about the data it represents. What is the difference between a thematic map and a physical map? the main difference between a thematic map and a physical map is the type of information they represent. Thematic maps focus on a specific theme or subject such as demographic or business characteristics. Physical maps show the physical features of an area, such as mountains, rivers, lakes, and coastlines. How do you create a thematic map in Excel? There are a few steps you can follow to create a thematic map in Excel: Collect and organize your data: Begin by gathering the data you want to represent on your map. Choose a map type: Decide which type of thematic map you want to create, such as a choropleth map or a proportional symbol map. Prepare your data for mapping: Make sure your data is in a format that can be used to create a map. This may involve cleaning or formatting the data, or creating new columns or calculations. Create a base map: Use a tool such as Google Maps or a GIS software such as Maptitude mapping software to create a base map of the area you want to represent. Add your data to the map: Use the Excel mapping features to add your data to the map. You can use features such as data labels, colors, and symbols to represent your data on the map. Customize your map: Use the formatting and styling options to suit your needs, such as by changing the colors, fonts, or symbols used to represent the data. What are the most commonly used thematic maps? There are many types of thematic maps, which are maps that are designed to illustrate a specific theme or subject. The top five types of thematic maps are: Population density maps: These maps show the distribution of population within a given area. They can be used to visualize the distribution of population density across a city, state, or country. Climate maps: These maps illustrate the distribution of temperature, precipitation, and other climatic variables across a given area. They can be used to understand the local climate and how it varies over time. Land use maps: These maps show the different types of land uses in an area, such as residential, commercial, agricultural, and industrial. They can be used to understand the patterns of land use and how they may impact the environment or the economy. Elevation maps: These maps show the elevation of different parts of the land surface. They can be used to understand the topography of an area and how it may affect the local climate or vegetation. Political maps: These maps show the boundaries of different political entities, such as countries, states, and provinces. They can be used to understand the political divisions of a region and how they relate to one another. Are political maps thematic maps? Yes, political maps are a type of thematic map. Thematic maps are maps that are designed to illustrate a specific theme or subject, and political maps are a type of thematic map that show the boundaries of different political entities, such as countries, states, provinces, and other administrative divisions. Political maps can be used to understand the political divisions of a region and how they relate to one another. They can also be used to show the location of important political features, such as capital cities and major cities. Political maps can be created at different scales, from global maps that show the entire world to more detailed maps that show the political divisions of a specific country or region. Political maps can be useful for a variety of purposes, including studying geography, understanding the political landscape of a region, and supporting decision-making in areas such as business, politics, and diplomacy. Are topographic maps thematic? Topographic maps are a type of thematic map that depict the three-dimensional shape and surface features of the earth's surface. They are used to represent the natural and man-made features of an area, such as mountains, valleys, rivers, roads, and buildings. Topographic maps are produced by government agencies and are used for a variety of purposes, including land-use planning, resource management, and recreational activities such as hiking and fishing. They are often used in conjunction with other types of thematic maps, such as maps showing vegetation or land cover, to provide a more complete picture of an area. Topographic maps use contour lines to show the elevation of the land surface. The contour lines are spaced at regular intervals, usually in feet or meters, and are used to show the slope of the land. The closer the contour lines are together, the steeper the slope. Topographic maps also include a legend that explains the symbols and colors used on the map. Learn MoreFree TrialFree for Students/Teachers How to Make a Thematic Map using Thematic Mapping Software This video provides step-by-step instructions for how to create attractive and informative theme maps. Related Articles Maptitude Thematic Mapping Software How Do I Make a Map Look Nice? Maptitude Learning Portal Theme Articles Home Maptitude Mapping Software Learning Glossary Thematic Map 617-527-4700 1172 Beacon St.,Suite 300 Newton MA 02461 USA sales@caliper.com Caliper Home Maptitude Desktop Online Tutorial Videos Learning Pricing & Requirements Solutions TestimonialsTransCAD About Pricing Learning Caliper Online Store Buy Maptitude Buy TransModeler SE Buy Maptitude DataMaptitude for Redistricting About Online TransModeler About Pricing TransModeler SE AboutCompany About Contact Career Opportunities Free Mapping Resources Legal Media Relations Press Releases Publications Products Home\|Products\|Contact\|Secure Store Copyright © Caliper Corporation
10016	https://histologyguide.com/slidebox/04-muscle-tissue.html	Histology Guide virtual microscopy laboratory About Us Contact Us Terms of Use HOME SLIDE BOX GALLERY ELECTRON MICROSCOPY QUIZ INDEX SEARCH HELP Chapter 4 - Muscle Tissue Muscle tissue is composed of cells specialized for contraction. Muscle is classified into three types according to their structure and function: Skeletal muscle cells - striated, voluntary control Cardiac muscle cells - striated, involuntary control Smooth muscle cells - nonstriated, involuntary control Fig 021 Muscle Types Skeletal and cardiac muscle cells are called striated because they show an alternating series of bands. The repeating arrangement of their basic contractile unit, the sarcomere, produces these striations. In all types of muscle, contraction is caused by the movement of myosin filaments along actin filaments. The terms muscle cell and muscle fiber are synonymous. Skeletal Muscle Skeletal muscle fibers are long cylindrical, multinucleated, striated, and under voluntary control. MH 055a Skeletal Muscle (longitudinal and cross-section) H&E MH 055ahr Skeletal Muscle (longitudinal section) H&E UCSF 118 Skeletal Muscle (longitudinal section) Phosphotungstic Acid/Hematoxylin MHS 237 Skeletal Muscle (longitudinal section) H&E MHS 262 Skeletal Muscle (cross-section) H&E Individual skeletal muscle cells can be seen by teasing apart a muscle. MHS 238 Teased Muscle H&E Muscle Bone Junctions Muscles connect to the skeletal framework to enable motion and provide stability. These connections transmit the forces generated by muscle fibers (cells) through two distinct mechanisms: Tendons - muscles taper at their ends into strong, fibrous cords or sheet-like structures that connect muscle to bone Muscle Insertions - collagen fibers of the endomysium (connective tissue surrounding individual muscle cells) extend and interweave with the collagen fibers of the periosteum covering bone. MH 030 Tendon (muscle tendon junction H&E MH 023bx Fascia Lata (muscle tendon junction) Azan MH 029b Bone (muscle insertion) H&E Cardiac Muscle Cardiac muscle are short branching fibers, have a single, centrally located nucleus, show the same striations as skeletal muscle, and are under involuntary control. MH 054 Cardiac Muscle H&E MH 056 Muscle Types (intercalated discs) Phosphotungstic Acid/Hematoxylin Purkinje fibers are modified cardiac muscle cells that convey electrical impulses that coordinate contraction of cardiac muscle. MH 058a Heart Periodic Acid-Schiff Stain MH 074 Heart Phosphotungstic Acid/Hematoxylin Smooth Muscle Smooth muscle cells are spindle-shaped (fusiform), have a single, centrally located nucleus, and are under involuntary control. The uniform, nonstriated appearance gives rise to the name smooth muscle. MH 053 Smooth Muscle H&E MH 024 Mesentery H&E MH 016x Small Intestine H&E MH 130a Gallbladder H&E
10017	https://www.statisticshowto.com/tables/t-distribution-table/	Skip to content T-Distribution Table (One Tail and Two-Tails) For more info on the parts of the t table, including how to calculate them, see: degrees of freedom and alpha level. T-Distribution Table (One Tail) For the T-Distribution Table for Two Tails, Click Here. \| df \| a = 0.1 \| 0.05 \| 0.025 \| 0.01 \| 0.005 \| 0.001 \| 0.0005 \| --- --- --- --- \| \| ∞ \| ta = 1.282 \| 1.645 \| 1.960 \| 2.326 \| 2.576 \| 3.091 \| 3.291 \| \| 1 \| 3.078 \| 6.314 \| 12.706 \| 31.821 \| 63.656 \| 318.289 \| 636.578 \| \| 2 \| 1.886 \| 2.920 \| 4.303 \| 6.965 \| 9.925 \| 22.328 \| 31.600 \| \| 3 \| 1.638 \| 2.353 \| 3.182 \| 4.541 \| 5.841 \| 10.214 \| 12.924 \| \| 4 \| 1.533 \| 2.132 \| 2.776 \| 3.747 \| 4.604 \| 7.173 \| 8.610 \| \| 5 \| 1.476 \| 2.015 \| 2.571 \| 3.365 \| 4.032 \| 5.894 \| 6.869 \| \| 6 \| 1.440 \| 1.943 \| 2.447 \| 3.143 \| 3.707 \| 5.208 \| 5.959 \| \| 7 \| 1.415 \| 1.895 \| 2.365 \| 2.998 \| 3.499 \| 4.785 \| 5.408 \| \| 8 \| 1.397 \| 1.860 \| 2.306 \| 2.896 \| 3.355 \| 4.501 \| 5.041 \| \| 9 \| 1.383 \| 1.833 \| 2.262 \| 2.821 \| 3.250 \| 4.297 \| 4.781 \| \| 10 \| 1.372 \| 1.812 \| 2.228 \| 2.764 \| 3.169 \| 4.144 \| 4.587 \| \| 11 \| 1.363 \| 1.796 \| 2.201 \| 2.718 \| 3.106 \| 4.025 \| 4.437 \| \| 12 \| 1.356 \| 1.782 \| 2.179 \| 2.681 \| 3.055 \| 3.930 \| 4.318 \| \| 13 \| 1.350 \| 1.771 \| 2.160 \| 2.650 \| 3.012 \| 3.852 \| 4.221 \| \| 14 \| 1.345 \| 1.761 \| 2.145 \| 2.624 \| 2.977 \| 3.787 \| 4.140 \| \| 15 \| 1.341 \| 1.753 \| 2.131 \| 2.602 \| 2.947 \| 3.733 \| 4.073 \| \| 16 \| 1.337 \| 1.746 \| 2.120 \| 2.583 \| 2.921 \| 3.686 \| 4.015 \| \| 17 \| 1.333 \| 1.740 \| 2.110 \| 2.567 \| 2.898 \| 3.646 \| 3.965 \| \| 18 \| 1.330 \| 1.734 \| 2.101 \| 2.552 \| 2.878 \| 3.610 \| 3.922 \| \| 19 \| 1.328 \| 1.729 \| 2.093 \| 2.539 \| 2.861 \| 3.579 \| 3.883 \| \| 20 \| 1.325 \| 1.725 \| 2.086 \| 2.528 \| 2.845 \| 3.552 \| 3.850 \| \| 21 \| 1.323 \| 1.721 \| 2.080 \| 2.518 \| 2.831 \| 3.527 \| 3.819 \| \| 22 \| 1.321 \| 1.717 \| 2.074 \| 2.508 \| 2.819 \| 3.505 \| 3.792 \| \| 23 \| 1.319 \| 1.714 \| 2.069 \| 2.500 \| 2.807 \| 3.485 \| 3.768 \| \| 24 \| 1.318 \| 1.711 \| 2.064 \| 2.492 \| 2.797 \| 3.467 \| 3.745 \| \| 25 \| 1.316 \| 1.708 \| 2.060 \| 2.485 \| 2.787 \| 3.450 \| 3.725 \| \| 26 \| 1.315 \| 1.706 \| 2.056 \| 2.479 \| 2.779 \| 3.435 \| 3.707 \| \| 27 \| 1.314 \| 1.703 \| 2.052 \| 2.473 \| 2.771 \| 3.421 \| 3.689 \| \| 28 \| 1.313 \| 1.701 \| 2.048 \| 2.467 \| 2.763 \| 3.408 \| 3.674 \| \| 29 \| 1.311 \| 1.699 \| 2.045 \| 2.462 \| 2.756 \| 3.396 \| 3.660 \| \| 30 \| 1.310 \| 1.697 \| 2.042 \| 2.457 \| 2.750 \| 3.385 \| 3.646 \| \| 60 \| 1.296 \| 1.671 \| 2.000 \| 2.390 \| 2.660 \| 3.232 \| 3.460 \| \| 120 \| 1.289 \| 1.658 \| 1.980 \| 2.358 \| 2.617 \| 3.160 \| 3.373 \| \| 1000 \| 1.282 \| 1.646 \| 1.962 \| 2.330 \| 2.581 \| 3.098 \| 3.300 \| Two Tails T Distribution Table \| df \| a = 0.2 \| 0.10 \| 0.05 \| 0.02 \| 0.01 \| 0.002 \| 0.001 \| --- --- --- --- \| \| ∞ \| ta = 1.282 \| 1.645 \| 1.960 \| 2.326 \| 2.576 \| 3.091 \| 3.291 \| \| 1 \| 3.078 \| 6.314 \| 12.706 \| 31.821 \| 63.656 \| 318.289 \| 636.578 \| \| 2 \| 1.886 \| 2.920 \| 4.303 \| 6.965 \| 9.925 \| 22.328 \| 31.600 \| \| 3 \| 1.638 \| 2.353 \| 3.182 \| 4.541 \| 5.841 \| 10.214 \| 12.924 \| \| 4 \| 1.533 \| 2.132 \| 2.776 \| 3.747 \| 4.604 \| 7.173 \| 8.610 \| \| 5 \| 1.476 \| 2.015 \| 2.571 \| 3.365 \| 4.032 \| 5.894 \| 6.869 \| \| 6 \| 1.440 \| 1.943 \| 2.447 \| 3.143 \| 3.707 \| 5.208 \| 5.959 \| \| 7 \| 1.415 \| 1.895 \| 2.365 \| 2.998 \| 3.499 \| 4.785 \| 5.408 \| \| 8 \| 1.397 \| 1.860 \| 2.306 \| 2.896 \| 3.355 \| 4.501 \| 5.041 \| \| 9 \| 1.383 \| 1.833 \| 2.262 \| 2.821 \| 3.250 \| 4.297 \| 4.781 \| \| 10 \| 1.372 \| 1.812 \| 2.228 \| 2.764 \| 3.169 \| 4.144 \| 4.587 \| \| 11 \| 1.363 \| 1.796 \| 2.201 \| 2.718 \| 3.106 \| 4.025 \| 4.437 \| \| 12 \| 1.356 \| 1.782 \| 2.179 \| 2.681 \| 3.055 \| 3.930 \| 4.318 \| \| 13 \| 1.350 \| 1.771 \| 2.160 \| 2.650 \| 3.012 \| 3.852 \| 4.221 \| \| 14 \| 1.345 \| 1.761 \| 2.145 \| 2.624 \| 2.977 \| 3.787 \| 4.140 \| \| 15 \| 1.341 \| 1.753 \| 2.131 \| 2.602 \| 2.947 \| 3.733 \| 4.073 \| \| 16 \| 1.337 \| 1.746 \| 2.120 \| 2.583 \| 2.921 \| 3.686 \| 4.015 \| \| 17 \| 1.333 \| 1.740 \| 2.110 \| 2.567 \| 2.898 \| 3.646 \| 3.965 \| \| 18 \| 1.330 \| 1.734 \| 2.101 \| 2.552 \| 2.878 \| 3.610 \| 3.922 \| \| 19 \| 1.328 \| 1.729 \| 2.093 \| 2.539 \| 2.861 \| 3.579 \| 3.883 \| \| 20 \| 1.325 \| 1.725 \| 2.086 \| 2.528 \| 2.845 \| 3.552 \| 3.850 \| \| 21 \| 1.323 \| 1.721 \| 2.080 \| 2.518 \| 2.831 \| 3.527 \| 3.819 \| \| 22 \| 1.321 \| 1.717 \| 2.074 \| 2.508 \| 2.819 \| 3.505 \| 3.792 \| \| 23 \| 1.319 \| 1.714 \| 2.069 \| 2.500 \| 2.807 \| 3.485 \| 3.768 \| \| 24 \| 1.318 \| 1.711 \| 2.064 \| 2.492 \| 2.797 \| 3.467 \| 3.745 \| \| 25 \| 1.316 \| 1.708 \| 2.060 \| 2.485 \| 2.787 \| 3.450 \| 3.725 \| \| 26 \| 1.315 \| 1.706 \| 2.056 \| 2.479 \| 2.779 \| 3.435 \| 3.707 \| \| 27 \| 1.314 \| 1.703 \| 2.052 \| 2.473 \| 2.771 \| 3.421 \| 3.689 \| \| 28 \| 1.313 \| 1.701 \| 2.048 \| 2.467 \| 2.763 \| 3.408 \| 3.674 \| \| 29 \| 1.311 \| 1.699 \| 2.045 \| 2.462 \| 2.756 \| 3.396 \| 3.660 \| \| 30 \| 1.310 \| 1.697 \| 2.042 \| 2.457 \| 2.750 \| 3.385 \| 3.646 \| \| 60 \| 1.296 \| 1.671 \| 2.000 \| 2.390 \| 2.660 \| 3.232 \| 3.460 \| \| 120 \| 1.289 \| 1.658 \| 1.980 \| 2.358 \| 2.617 \| 3.160 \| 3.373 \| \| 8 \| 1.282 \| 1.645 \| 1.960 \| 2.326 \| 2.576 \| 3.091 \| 3.291 \| References Beyer, W. (2017). Handbook of Tables for Probability and Statistics 2nd Edition. CRC Press. Comments? Need to post a correction? Please Contact Us.
10018	https://courses.physics.illinois.edu/tam251/su2021/TAM251_Chapter7_TransverseShear_postlecture_Johnson_July16.pdf	Shear Flow in Built-up Beams 12 Consider the built-up beam below where the section is composed of 4 rectangular segments glued to one another. How can we calculate the shear stress in the glued segments? Shear Flow in Built-up Beams 13 Consider the built-up beam below where the section is composed of 4 rectangular segments glued to one another. How can we calculate the shear stress in the glued segments? A beam is made of four planks glued together. Knowing that the vertical shear in the beam is 𝑉𝑉= 500 N, determine the minimum required shear strength 𝜏𝜏𝑔𝑔for the glue. Example 2 14 Example 2 15 Built up beams with fasteners (bolts or nails) 16 Unlike glue, fasteners supply resistance to longitudinal shear forces at fixed internals. Fasteners are typically spaced at a constant interval Δ𝑠𝑠along the length of the beam. If we know the shear flow 𝑞𝑞, how much load does each fastener carry? Δ𝑠𝑠 Δ𝑠𝑠 A beam is made of four planks, nailed together as shown. If each nail can support a shear force of 30 lb, determine the maximum spacing s of the nails at B and at C so that the beam will support the force of 80lb. Example 3 17 A beam is made of four planks, nailed together as shown. If each nail can support a shear force of 30 lb, determine the maximum spacing s of the nails at B and at C so that the beam will support the force of 80lb. Example 3 18 A square box beam is constructed from four planks as shown. Knowing that the spacing between nails is 1.5 in. and the beam is subjected to a vertical shear of magnitude V = 600 lb, determine the shearing force in each nail. = 𝑡𝑡 𝑏𝑏= Example 4 19
10019	https://www.nanosweb.org/opticneuritis/	Optic Neuritis \| North American Neuro-Ophthalmology Society PAY DUESJOINDONATE Log In Reset password About Board of Directors Historical Board Listing Meet NANOS Board Bylaws Committees NANOS Committee Rosters Committee Highlights Dr. Thomas Carlow Thomas and Susan Carlow Young Investigator Award Fund NANOS Awards, Grants & Recognitions NANOS International Community Mission Statement NANOS in the News NANOS in the News Archives NANOS Policies NANOS Code of Conduct NANOS Guest and Children Policy Code of Ethics NANOS Anti-harassment Policy NANOS Whistleblower Policy NANOS Avoidance of Anticompetitive Behavior Guidelines NANOS Clinical Survey and Investigation Request Policy NANOS Co-Sponsored Symposia in Other International Conferences Policy Online Communication Platforms Rules and Terms NANOS Promotion Policy for Medical Meetings Social Media & Professionalism Opinion NANOS Speaker Policy NANOS Strategic Plan Who We Are History Contact Us Online Community Online Community Guide NANOS Online Communities and Committees Directory 2026 MEETING 2025 ON-DEMAND 2025 Meeting Materials 2025 Exhibitor and Supporter Information 2025 Abstract Information 2025 CME Information Events 2024 MEETING 2024 CME Information 2024 Abstract Submission Guidelines NANOS Future Meetings Other Meetings of Interest Past Meetings NANOS Historical Faculty Listing Membership NANOS Spotlight NANOS Spotlight - June 2025 NANOS Spotlight - May 2025 NANOS Spotlight - April 2025 NANOS Spotlight - March 2025 NANOS Spotlight - February 2025 NANOS Spotlight - January 2025 NANOS Spotlight - December 2024 NANOS Spotlight - November 2024 NANOS Spotlight - October 2024 NANOS Spotlight - September 2024 NANOS Spotlight- August 2024 NANOS Spotlight - July 2024 NANOS Spotlight - June 2024 NANOS Spotlight - May 2024 NANOS Spotlight- April 2024 NANOS Spotlight- March 2024 NANOS Spotlight- January 2024 Pay Dues Donate to NANOS Stories Create a Fundraiser View NANOS Donors Get Involved! 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Optic neuritis is inflammation of the optic nerve. In optic neuritis, the covering around the fibers of the optic nerve (myelin) is damaged by inflammation (demyelination),which typically results in blurred or dark vision. What causes optic neuritis? Your immune system normally protects your body from infections. However, it can become confused and attack your own body (autoimmune reaction). Most cases of optic neuritis are caused by an autoimmune reaction to the optic nerve, but optic neuritis can also be directly caused by certain infections. Optic neuritis can be caused by neurologic disease, systemic conditions, infections, or it may not have any known cause (idiopathic). Optic neuritis is frequently a manifestation of MS, and sometimes the very first sign of the disease.Multiple sclerosis is an autoimmune disease that specifically attacks your brain and spinal cord. Optic neuritis can also be caused by other neurological inflammatory diseases such as neuromyelitis optica (NMO) or myelin oligodendrocyte glycoprotein (MOG)-associated disease. How does optic neuritis affect me? Vision Loss In optic neuritis, the eye itself is not affected, but the optic nerve behind the eye is inflamed.This may cause blurred vision, darkened vision, decreased color vision, or decreased peripheral vision.The most common symptom of optic neuritis is decreased vision, which happens suddenly or quickly over the course of several days. Many people describe the blurring and darkening of their vision as “dimming” or like “the brightness is turned down.” Optic neuritis does not affect your glasses prescription. Eye Pain Most people with optic neuritis will have eye pain or discomfort when looking in different directions. When the optic nerve is inflamed and the eye moves, there may be pulling on the optic nerve that can cause irritation. This pain typically goes away after several days. Other Vision Problems Decreased color visionis also common. Colors are often described as “washed out.” Difficulty seeing contrast is also common. Some people will see flashes of light, sparkles, or shifting squares. Some people will have episodes of blurred vision lasting up to an hour triggered by exercise or hot temperatures, which may continue after the episode of optic neuritis has resolved because of residual damage to the optic nerve. Why do I need to see a neuro-ophthalmologist? Neuro-ophthalmologists specialize in diseases of the optic nerve. While a neuro-ophthalmologist may do the same tests as your regular eye doctor, a neuro-ophthalmologist is experienced in sorting out the differences between optic neuritis and other optic nerve diseases. Your doctor may test your visual fieldand may scan your optic nerves using specialized machines in the office. Your doctor will order an MRI of the brain with special views of the orbits (where the majority of the optic nerve runs), which can confirm optic neuritis when the nerve looks bright after intravenous contrast is injected. Your doctor may order other tests, such as blood tests or a chest X-ray, to look for different causes of optic neuritis. If your symptoms and eye exam are typical for optic neuritis, your doctor may choose to order very few tests at first. If your symptoms are unusual, your doctor may order other tests. Why does my doctor say I may have multiple sclerosis (MS)? There has been a lot of research that links the most common form of optic neuritis with multiple sclerosis. Most people with multiple sclerosis will have an episode of optic neuritis in their lifetime. While not everyone who gets optic neuritis will develop multiple sclerosis, people who have optic neuritis have a greater risk for having MS in the future, compared to those who have never had an episode. Because early detection and treatment may keep you from having more severe problems, it is important to follow your doctor’s advice and follow up for subsequent exams An MRI of the brain is recommended in optic neuritis because it can help determine the risk for developing MS in the future, and determine if you need additional treatment. Your doctor may recommend that you see a neurologist to be screened for multiple sclerosis, even if you do not have any other symptoms. Studies have shown that 75% (3 out of 4) people with optic neuritis and additional brain abnormalities on MRI will develop MS within 15 years. In people with a normal MRI at the time of their optic neuritis episode, 25% (1 out of 4) people develop MS within 15 years. Future MRIs may be recommended by your doctor. How is optic neuritis treated? In typical cases of optic neuritis, no treatment is necessary. Research has shown that there is no difference in final vision, or the likelihood of developing MS in the future, between people with optic neuritis who were treated and those who were not treated. However, treatment can speed up recovery; however, the treatment for optic neuritis has many side effects and is therefore not recommended for everyone. Treatment is most likely to be recommended if your vision is severely affected or if your symptoms or test results are unusual. When optic neuritis is treated, treatment consists of intravenous (IV) steroidsor high dose oral steroids for 3-5 days, which may be followed by pills taken by mouth (oral steroids). There is evidence that using oral steroids (prednisone) in doses of 60-80 mg per day without the preceding intravenous course may increase the risk of recurrence of optic neuritis. Depending on your overall health, steroid treatment may not be safe for you. Steroids have many side effects including fluid retention, increased blood pressure, increased blood sugar, weight gain, increased appetite, mood and behavior problems, stomach irritation, and glaucoma. What is my prognosis? Most people with typical cases of optic neuritis will get better with or without treatment.Vision typically begins to recover within a few weeks. 92% of people with typical cases of optic neuritis will recover most of their vision. However,most people will continue to report some mild decreased quality of their vision, even though they may still read the small letters on the 20/20 line in the doctor’s office. These changes are typically related to reduced color vision, distorted vision, or difficulty with contrast (distinguishing shades of light and dark). Up to 35% of people with optic neuritis will have further episodes in the same eye or other eye. As of now, there are no medications, supplements, or other treatments proven to reduce the risk of recurrence unless you have been diagnosed with multiple sclerosis or another neurological disease requiring immune suppression. The prognosis may be worse depending on the cause of the optic neuritis. Infections, such as syphilis, tuberculosis, Lyme disease and certain viruses, and autoimmune diseases, such as neuromyelitis optica (NMO), myelin oligodendrocyte glycoprotein (MOG)-antibody disease, sarcoidosis, and lupus, can also cause optic neuritis. In these cases, the vision may not improve as much or at all. Many of these patients require specific treatments directed at the underlying cause (if an infection) or more aggressive treatment with steroids and other drugs that treat autoimmune inflammation. These can include intravenous immunoglobulin (IVIg) or plasmapheresis. How often do I need to have checkups for optic neuritis? Your doctor will check your eyes and vision when you are first having symptoms (acute phase). After determining if you need treatment, other testing, or a referral to a neurologist, your doctor may recheck your eyes and vision after a few weeks to months. Once your vision has stabilized and as long as you do not have any new episodes, you can have routine eye exams. Additional Reading/Resources Websites Optic Neuritis, by the National Library of Medicine on MedlinePlus ( Optic Neuritis, by the Mayo Clinic ( Optic Neuritis, by the American Academy of Ophthalmology’s EyeSmart Project ( Support Groups National Multiple Sclerosis Society: Multiple Sclerosis Foundation: Optic Neuritis Facebook Support Group: Copyright © 2023. North American Neuro-Ophthalmology Society. All rights reserved. This information was developed collaboratively by the Patient Information Committee of the North American Neuro-Ophthalmology Society. This has been written by neuro-ophthalmologists and has been edited, updated, and peer-reviewed by multiple neuro-ophthalmologists. The views expressed in this brochure are of the contributors and not their employers or other organizations. Please note we have made every effort to ensure the content of this is correct at time of publication, but remember that information about the condition and drugs may change. Major revisions are performed on a periodic basis. This information is produced and made available “as is” without warranty and for informational and educational purposes only and do not constitute, and should not be used as a substitute for, medical advice, diagnosis, or treatment. Patients and other members of the general public should always seek the advice of a physician or other qualified healthcare professional regarding personal health or medical conditions. 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10020	https://en.wikipedia.org/wiki/Algebraic_element	Jump to content Search Contents (Top) 1 Examples 2 Properties 3 See also 4 References 5 Further reading Algebraic element Català Čeština Deutsch Ελληνικά Español Esperanto فارسی Français Galego 한국어 Ido עברית Nederlands 日本語 Polski Português Romnă Українська Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Concept in abstract algebra In mathematics, if A is an associative algebra over K, then an element a of A is an algebraic element over K, or just algebraic over K, if there exists some non-zero polynomial with coefficients in K such that g(a) = 0. Elements of A that are not algebraic over K are transcendental over K. A special case of an associative algebra over is an extension field of . These notions generalize the algebraic numbers and the transcendental numbers (where the field extension is C/Q, with C being the field of complex numbers and Q being the field of rational numbers). Examples [edit] The square root of 2 is algebraic over Q, since it is the root of the polynomial g(x) = x2 − 2 whose coefficients are rational. Pi is transcendental over Q but algebraic over the field of real numbers R: it is the root of g(x) = x − π, whose coefficients (1 and −π) are both real, but not of any polynomial with only rational coefficients. (The definition of the term transcendental number uses C/Q, not C/R.) Properties [edit] The following conditions are equivalent for an element of an extension field of : is algebraic over , the field extension is algebraic, i.e. every element of is algebraic over (here denotes the smallest subfield of containing and ), the field extension has finite degree, i.e. the dimension of as a -vector space is finite, , where is the set of all elements of that can be written in the form with a polynomial whose coefficients lie in . To make this more explicit, consider the polynomial evaluation . This is a homomorphism and its kernel is . If is algebraic, this ideal contains non-zero polynomials, but as is a euclidean domain, it contains a unique polynomial with minimal degree and leading coefficient , which then also generates the ideal and must be irreducible. The polynomial is called the minimal polynomial of and it encodes many important properties of . Hence the ring isomorphism obtained by the homomorphism theorem is an isomorphism of fields, where we can then observe that . Otherwise, is injective and hence we obtain a field isomorphism , where is the field of fractions of , i.e. the field of rational functions on , by the universal property of the field of fractions. We can conclude that in any case, we find an isomorphism or . Investigating this construction yields the desired results. This characterization can be used to show that the sum, difference, product and quotient of algebraic elements over are again algebraic over . For if and are both algebraic, then is finite. As it contains the aforementioned combinations of and , adjoining one of them to also yields a finite extension, and therefore these elements are algebraic as well. Thus set of all elements of that are algebraic over is a field that sits in between and . Fields that do not allow any algebraic elements over them (except their own elements) are called algebraically closed. The field of complex numbers is an example. If is algebraically closed, then the field of algebraic elements of over is algebraically closed, which can again be directly shown using the characterisation of simple algebraic extensions above. An example for this is the field of algebraic numbers. See also [edit] Algebraic independence References [edit] ^ Roman, Steven (2008). "18". Advanced Linear Algebra. Graduate Texts in Mathematics. New York, NY: Springer New York Springer e-books. pp. 458–459. ISBN 978-0-387-72831-5. Further reading [edit] Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556, Zbl 0984.00001 Retrieved from " Category: Algebraic properties of elements Hidden categories: Articles with short description Short description is different from Wikidata Algebraic element Add topic
10021	https://www.geeksforgeeks.org/physics/einsteins-photoelectric-equation/	Published Time: 2022-03-09 12:00:19 Einstein's Photoelectric Equation - GeeksforGeeks Skip to content Tutorials Python Java Data Structures & Algorithms ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps And Linux Software and Tools School Learning Practice Coding Problems Courses DSA to Development Get IBM Certification Newly Launched! 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Light, according to Einstein, is a wave that interacts with matter as a packet of energy or a quantum of energy. The photon was the quantum of radiation, and the equation was known as Einstein's photoelectric equation. When a substance absorbs electromagnetic radiation, electrically charged particles are emitted from or inside it, resulting in the photoelectric effect. The phenomenon of emission of electrons by certain substance (metal), when it is exposed to radiations of suitable frequencies is called as photoelectric effect and emitted electrons are called photoelectrons. The effect is frequently characterized as the ejection of electrons from a metal plate when exposed to light. The radiant energy may be infrared, visible, or ultraviolet light, X-rays, or gamma rays; the substance could be solid, liquid, or gas; and the particles discharged could be ions or electrons. Characteristics of Photoelectric Effect: For each particular photosensitive material, there is a minimal cut-off frequency of the incoming radiation, known as the threshold frequency ν o, below which no photo-electrons are produced. The threshold frequency varies depending on the metal. The photoelectric current is directly proportional to the intensity of incident light for a particular photosensitive material and the frequency of incident radiation (above threshold frequency). Above the threshold frequency ν 0, the maximal kinetic energy of the emitted photo-electrons grows linearly with incident radiation frequency but is independent of incoming radiation intensity. The emission of photo-electrons occurs in an instant. There is no time lag between irradiation of the metal surface and photo-electronic emission. Einstein`s Photoelectric Equation In 1905, Einstein expanded Planck's notion by deriving his own equation that correctly described the features of the photoelectric effect. He presupposed two things. Radiation with frequency ν consists of a stream of discrete quanta or photons, with energy hν, where h is Planck's constant. Photons travel at the speed of light through space. When photons and electrons in the emitter's atoms collide when radiation of frequency ν is incident on a photosensitive surface. During such a collision, the photon's whole energy is transmitted to the electron with no time lag. A photon is not a material particle but quanta of energy. The incident photon's absorbed energy hv by an electron is utilized in two ways. The electron uses some of its energy to break free from the atom. The minimum energy required to free-electron from a given surface is called photoelectric work function φ 0 of the material of the surface. The residual energy (h ν- φ 0) emerges as electron kinetic energy. If the electron does not lose any of its energy in impact with the surface and exits with the greatest possible kinetic energy. ∴ 1 / 2 m v 2 max = hν - φ 0 .....(1) where, m = mass of the electron v max = maximum velocity of electron All the photoelectrons emitted from the metal surfaces do not have the same energy. The photoelectric current becomes zero when the stopping potential is sufficient to repel even the most energetic photoelectrons, with the maximum kinetic energy, so that the stopping potential (in volts) is numerically equal to the maximum kinetic energy of photo-electron in eV. 1 / 2 m v 2 max = ev 0.....(2) where, e = magnitude of the charge on an electron. From (2) we have, 1 / 2 m v 2 max = hν - φ 0 eV0= hν - φ0 This equation is Einstein`s photoelectric equation. With the aid of Einstein's photoelectric equation, we can now describe all of the features of the photoelectric effect: If the frequency of incident radiation is decreased, the kinetic energy of photoelectrons also decreases, and finally, it becomes zero for a particular frequency (say ν). ν is called the threshold frequency. Thus, When ν=ν o, then KE max = 1/2 m V 2 max = 0 Therefore from equation (1), we get 0 = hν 0 - φ 0 hν 0 = φ 0 Therefore, Einstein`s equation can be written as, 1/2 m v 2 max = h(ν - ν 0) .....(3) From the above equation, we can say three points, ν > ν 0, the photoelectrons are emitted with some velocity, ν < ν 0, no photoelectrons are emitted, and ν = ν 0, photoelectrons are just emitted with zero kinetic energy. A more intense beam, according to quantum theory, includes a higher number of photons. As a result, the number of photon-electron collisions increases, and more photoelectrons are released. This explains why the photoelectric current increases with incoming photon intensity. Because the photoelectric work function (φ 0) is constant for every given emitter, equation (3) demonstrates that the KE max of photoelectrons grows with the frequency of incoming radiation but does not rely on intensity. Photo-electrons are emitted as a result of electron-photon collisions. Such collisions occur as soon as the radiation strikes the photosensitive surface, and photoelectrons are released. There is no emission of photoelectrons at the incidence cut-off. As a result, the photoelectric effect occurs instantly. Particle nature of light: Photon The photoelectric effect, therefore, provided proof for the unusual fact that light acted as though it were formed of quanta or packets of energy, each of energy hν. Is the light quantum of energy to be connected with a particle? Einstein discovered that the light quantum may also be related to momentum hv. A definite value of energy and momentum is a strong indication that the light quantum can be connected with a particle. This particle was eventually given the name photon. The particle-like behaviour of light was established in 1924 by A.H. Compton's (1892-1962) experiment on the scattering of X-rays from electrons. Einstein received the Nobel Prize in Physics in 1921 for his contributions to theoretical physics and the photoelectric effect. Millikan received the Nobel Prize in Physics in 1923 for his discoveries on the elementary charge of electricity and the photoelectric effect. Summarized Photon picture of electromagnetic radiation When radiation interacts with matter, it behaves as if it were made up of particles known as photons. Each photon has energy E (= hu) and momentum p (= hν / c) where c is the speed of light. Whatever the intensity of radiation, all photons of light of a specific frequency ν or wavelength λ, have the identical energy E( = hν = hc / λ) and momentum p (= hν / c = h/λ). By raising the intensity of light of a certain wavelength, the number of photons per second crossing a given area increases, with each photon having the same energy. As a result, photon energy is independent of radiation intensity. Photons are electrically neutral and are unaffected by electric or magnetic forces. Total energy and total momentum are preserved in a photon-particle collision (such as a photon electron collision). However, the number of photons in a collision may not be preserved. The photon might be absorbed or a new photon could be produced. Sample Problems Problem 1: An electron is accelerated from rest through a potential difference of 500 volts. Find the speed of the electron. (Given e = 1.6 × 10-19C, m = 19 ×10-31Kg) Solution: Given that, Voltage = 500 V e = 1.6 × 10-19 C m = 19 × 10 Kg KE of emitted electron = eV = 1/2 m v 2 = 2 × (1.6 × 10-19) × (500) / (9.1 × 10-31 ) V = √1.758 × 10 14 = 1.326 × 107m/s Problem 2: A metal whose work function is 4.2 eV is irradiated by radiation whose wavelength is 2000 A°. Find the maximum kinetic energy of the emitted electron. Solution: Given that, φ 0 = 4.2eV = 6.72 × 10-19 J λ = 2000 A° 1/2 m v 2 max = hν - φ 0 KE max = hν - φ 0 = (hc / λ) - φ 0 = (6.63 ×10-34 × 3 × 10) / (2 × 10-7) - (6.72 × 10-10) = (9.945 - 6.72) × 10-19 = 2.015 eV Problem 3: If photo-electrons are to be emitted from potassium surface with a speed of 6 ×105m/s, what frequency of radiation must be used? The threshold frequency for potassium is 4.22 ×1014Hz. Solution: v max = 6 × 10 5 m/s ν 0 = 4.22 × 10 14 Hz KE max = 1/2 mv 2 max = h(ν - ν 0) ν = 1/2 × (mv 2 max / h) + ν 0 = (1/2) × [(9.1 × 10-31 ×(6 × 10 5)2 / 6.63 × 10-34] +4.22 × 10 14 Hz = 2.47 × 10 14+ 4.22 × 10 14 = 6.69 × 1014Hz Problem 4: A photon of wavelength 3310A° falls on a photo-cathode and an electron of energy 3 × 10-19J is ejected. if the wavelength of the incident photon is changed to 5000 A°, the energy of the ejected electron is 9.72 × 10-20J. Calculate the value of Plank`s constant and threshold wavelength of the photon. Solution: λ 1 = 3310 = 3.31 × 10-7 m KE max (1) = 3 × 10-19 J λ 2 = 5000 = 5 × 10-7 m KE max = 9.72 × 10-20 J= 0.972 × 10-19 J KE max (1) = hc/λ 1 - φ 0 KE max (2) = hc/λ 2 - φ 0 KE max (1) - KE max (2) = hc (1/λ 1 - 1/λ 2) 3 × 10-19 - 0.972 × 10-19 = h × 3 × 10 8 [(1/3.31 × 10-7) - (1/5 × 10-7)] 2.028 × 10-19 = h × 3 × 10 15 (1.69 / 16.55) h = 2.028 × 10-19 × 16.55 / 3 × 10 15 × 1.69 = 6.62 × 10-34Js Now, KE max (1) = hc/λ 1 - φ 0 φ 0 = hc/λ 1 - KE max (1) = (6.62 × 10-34 × 3 × 10 8 / 3.31 × 10-7 ) - 3 × 10-19 = 3 × 10-19J Also, φ 0 = hc / λ 0 λ 0 = hc / φ 0 = 6.62 × 10-34 × 3 × 10 8 /3 × 10-19 = 6.62 × 10-7 = 6620 A° Problem 5: The Photo-electric function for a metal surface is 2.4 eV. If the light of wavelength 5000 A° is incident on the surface of the metal, find the threshold frequency and incident frequency, Will there be an emission of photo-electrons or not? Solution: φ 0 = 2.34eV = 3.84 × 10-19 J λ = 5000A° = 5 × 10-7 ν = c / λ = 3 × 10 8/ 5 × 10-7 = 6 × 10 34 Hz ν 0 = φ 0 / h = 3.84 × 10-19 / 6.63 × 10-34 = 5.792 × 10 14 Hz As ν > ν0, photoelectric emission is possible. Comment More info Advertise with us Next Article Einstein's Photoelectric Equation L lunaticgemini2510 Follow 2 Improve Article Tags : School Learning Physics Class 12 Geeks Premier League Geeks-Premier-League-2022 Physics-Class-12 Modern Physics +2 More Similar Reads CBSE Class 12 Physics Notes 2023-24 CBSE Class 12 Physics Notes are an essential part of the study material for any student wanting to pursue a career in engineering or a related field. Physics is the subject that helps us understand our surroundings using simple and complex concepts combined. Class 12 physics introduces us to a lot o 10 min read Chapter 1 - ELECTRIC CHARGES AND FIELDS Electric Charge and Electric Field Electric Field is the region around a charge in which another charge experiences an attractive or repulsive force. Electric Field is an important concept in the study of electrostatics which is the branch of physics. Electric Field despite its invisible nature, powers our homes with electricity, tra 15+ min readElectric Charge Electric Charge is the basic property of a matter that causes the matter to experience a force when placed in a electromagnetic field. It is the amount of electric energy that is used for various purposes. Electric charges are categorized into two types, that are, Positive ChargeNegative ChargePosit 8 min readConductors and Insulators When humans remove synthetic clothing or sweater, especially in dry weather, he or she often sees a spark or hear a crackling sound. 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Columb's Law states that the force between any two charged particles is directly proportional to the product of the charge but is inversely proportional to the square of the distance between t 9 min readForces Between Multiple Charges When our synthetic clothing or sweater is removed from our bodies, especially in dry weather, a spark or crackling sound appears. With females' clothing like a polyester saree, this is almost unavoidable. Lightning, in the sky during thunderstorms, is another case of electric discharge. It is an ele 10 min readElectric Field Electric field is a fundamental concept in physics, defining the influence that electric charges exert on their surroundings. This field has both direction and magnitude. It guides the movement of charged entities, impacting everything from the spark of static electricity to the functionality of ele 14 min readElectric Field Lines Electric field lines are a representation used to visualize the electric field surrounding charged objects. They provide a way to understand the direction and strength of the electric field at different points in space. It helps analyze electric fields in various situations, such as around point cha 5 min readWhat is Electric Flux? Electric flux is a fundamental concept in physics that helps us understand and quantify the electric field passing through a given surface. It provides a means to describe the flow of electric field lines through an area. Electric flux forms the basis of Gauss's Law, to calculate the net charge encl 12 min readElectric Dipole An electric dipole is defined as a pair of equal and opposite electric charges that are separated, by a small distance. An example of an electric dipole includes two atoms separated by small distances. The magnitude of the electric dipole is obtained by taking the product of either of the charge and 11 min readContinuous Charge Distribution Electric charge is a fundamental feature of matter that regulates how elementary particles are impacted by an electric or magnetic field. Positive and negative electric charge exists in discrete natural units and cannot be manufactured or destroyed. There are two sorts of electric charges: positive 7 min readApplications of Gauss's Law Gauss's Law states that the total electric flux out of a closed surface equals the charge contained inside the surface divided by the absolute permittivity. The electric flux in an area is defined as the electric field multiplied by the surface area projected in a plane perpendicular to the field. N 9 min read Chapter 2 - ELECTROSTATIC POTENTIAL AND CAPACITANCE Electric Potential Energy Electrical potential energy is the cumulative effect of the position and configuration of a charged object and its neighboring charges. The electric potential energy of a charged object governs its motion in the local electric field.Sometimes electrical potential energy is confused with electric pot 15+ min readElectric Potential Due to a Point Charge Electric forces are responsible for almost every chemical reaction within the human body. These chemical reactions occur when the atoms and their charges collide together. In this process, some molecules are formed and some change their shape. Electric forces are experienced by charged bodies when t 7 min readElectric Potential Of A Dipole and System Of Charges Electric Potential is defined as the force experienced by a charge inside the electric field of any other charge. mathematically it is defined as the ratio of electric potential energy that is required to take a test charge from infinity to a point inside the electric field of any other charge with 7 min readEquipotential Surfaces When an external force acts to do work, moving a body from a point to another against a force like spring force or gravitational force, that work gets collected or stores as the potential energy of the body. When the external force is excluded, the body moves, gaining the kinetic energy and losing a 9 min readPotential Energy of a System of Charges When an external force works to accomplish work, such as moving a body from one location to another against a force such as spring force or gravitational force, that work is collected and stored as the body's potential energy. When the external force is removed, the body moves, acquiring kinetic ene 11 min readPotential Energy in an External Field When an external force operates to conduct work, such as moving a body from one location to another against a force like spring force or gravitational force, the work is gathered and stored as potential energy in the body. When an external force is removed, the body moves, acquiring kinetic energy a 11 min readElectrostatics of Conductors When an external force is used to remove a body from a situation. Point to another in the face of a force like spring or gravitational force That work is stored in the body as potential energy. When the external environment When a force is eliminated, the body moves, gaining and losing kinetic energ 11 min readDielectrics and Polarisation Have you noticed how many of the insulators are made of wood, plastic, or glass? But why is that? When we utilise wood or plastic, why don't we receive electric shocks? Why do you only get severe shocks from metal wires? We'll look at dielectrics, polarisation, the dielectric constant, and more in t 10 min readCapacitor and Capacitance Capacitor and Capacitance are related to each other as capacitance is nothing but the ability to store the charge of the capacitor. Capacitors are essential components in electronic circuits that store electrical energy in the form of an electric charge. They are widely used in various applications, 11 min readWhat is a Parallel Plate Capacitor? Answer: A Parallel Plate Capacitor is a capacitor with two parallel conducting plates separated by an insulating material and capable of storing electrical charge. Capacitance can be defined in Layman's terms as a physical quantity that indicates the ability of a component or circuit to collect and 8 min readCapacitors in Series and Parallel Capacitors are special devices that can hold electric charges for instantaneous release in an electric circuit. We can easily connect various capacitors together as we connected the resistor together. The capacitor can be connected in series or parallel combinations and can be connected as a mix of 7 min readEnergy stored in a Capacitor Capacitors are used in almost every electronic device around us. From a fan to a chip, there are lots of capacitors of different sizes around us. Theoretically, the basic function of the capacitor is to store energy. Its common usage includes energy storage, voltage spike protection, and signal filt 6 min read Chapter 3 - CURRENT ELECTRICITY Electric Current Electricity has become an essential part of our everyday life, changing the way we live and work. In the past, people depended on fire for light, warmth, and cooking. Today, we can easily turn on lights, heat our homes, and charge our devices with just a switch or button. This is all possible becaus 10 min readElectric Current in Conductors Electric current in conductors is the movement of electric charge through a substance, usually a metallic wire or other conductor. Electric current is the rate at which an electric charge flows past a certain point in a conductor, and it is measured in amperes. When a potential difference (voltage) 8 min readOhm's Law - Definition, Formula, Applications, Limitations According to Ohm's law, the voltage or potential difference between two locations is proportional to the current of electricity flowing through the resistance, and the resistance of the circuit is proportional to the current or electricity travelling through the resistance. V=IR is the formula for O 5 min readDrift Velocity Drift Velocity as the name suggests refers to the slow movement of electrons in the conductor when an Electromotive force(emf) is introduced. Electrons do not move in a straight line in the conductor, but they move randomly in the conductor colliding with the other electrons and atoms exchanging ene 12 min readLimitations of Ohm's Law Ohmâ€™s Law is a relationship between three physical phenomena: current, voltage, and resistance. This relationship was introduced by German physicist George Simon Ohm. That is why the law is well known as Ohmâ€™s law. It states that the amount of steady current through a large number of materials is di 10 min readResistivity Resistance is the physical property of the material which opposes the current flow in the circuit whereas resistivity is the intrinsic property that helps us understand the relation between the dimension of the substance and the resistance offered by it.Â In this article, we will learn about Resista 9 min readTemperature Dependence of Resistance Devices such as batteries, cells, etc. are essential for maintaining a potential difference across the circuit and are referred to as voltage sources. When a voltage source is connected across a conductor, it creates an electric field which causes the charges to move and this causes current. The val 5 min readElectrical Energy and Power Electric energy is the most important form of energy and is widely used in almost all the electrical devices around us. These devices have a rating written on them. That rating is expressed in Watts and intuitively explains the amount of electricity the device will consume. Bigger devices like AC, r 9 min readElectromotive Force Electromotive Force or EMF is the work done by the per unit charge while moving from the positive end to the negative end of the battery. It can also be defined as the energy gain per unit charge while moving from the positive end to the negative end of the battery. The battery or the electric gener 10 min readCombination of Cells in Series and Parallel There are many resistances in complex electrical circuits. There are methods to calculate the equivalent resistances in case multiple resistances are connected in series or parallel or sometimes in a combination of series and parallel. In many situations, batteries or different types of voltage sour 6 min readKirchhoff's Laws Kirchhoff's Laws are the basic laws used in electrostatics to solve complex circuit questions. Kirchhoff's Laws were given by Gustav Robert Kirchhoff who was a famous German Physicist. He gave us two laws Kirchhoffâ€™s Current Law and Kirchhoffâ€™s Voltage Law which are discussed in this article.These l 8 min readWheatstone Bridge Wheatstone bridge is a device that is used to find the resistance of a conductor, in 1842, scientist Wheatstone proposed a theory, which is called the principle of Wheatstone bridge after his name. we can prove or establish the formula for Wheatstone by using Kirchhoff laws. Wheatstone bridge is sim 10 min read Chapter 4 - MOVING CHARGES AND MAGNETISM Magnetic Force on a Current carrying Wire When a charge is moving under the influence of a magnetic field. It experiences forces, which are perpendicular to its movement. This property of charge is exploited in a lot of fields, for example, this phenomenon is used in the making of motors which in turn are useful for producing mechanical for 5 min readMotion of a Charged Particle in a Magnetic Field This has been already learned about the interaction of electric and magnetic fields, as well as the motion of charged particles in the presence of both electric and magnetic fields. We have also deduced the relationship of the force acting on the charged particle, which is given by the Lorentz force 9 min readBiot-Savart Law The Biot-Savart equation expresses the magnetic field created by a current-carrying wire. This conductor or wire is represented as a vector quantity called the current element. Lets take a look at the law and formula of biot-savart law in detail, Biot-Savart Law The magnitude of magnetic induction a 7 min readMagnetic Field on the Axis of a Circular Current Loop Moving charges is an electric current that passes through a fixed point in a fixed period of time. Moving charges are responsible for establishing the magnetic field. The magnetic field is established due to the force exerted by the flow of moving charges. As the magnetic field is established moving 7 min readAmpere's Circuital Law and Problems on It AndrÃ©-Marie Ampere, a French physicist, proposed Ampere's Circuital Law. Ampere was born in Lyon, France, on January 20, 1775. His father educated him at home, and he showed an affinity for mathematics at a young age. Ampere was a mathematician and physicist best known for his work on electrodynamic 5 min readForce between Two Parallel Current Carrying Conductors Moving charges produce an electric field and the rate of flow of charge is known as current. This is the basic concept in Electrostatics. The magnetic effect of electric current is the other important phenomenon related to moving electric charges. Magnetism is generated due to the flow of current. M 8 min readCurrent Loop as a Magnetic Dipole When a charge move it generates an electric field and the rate of flow of charge is the current in the electric field. This is the basic concept in Electrostatics. The magnetic effect of electric current is the other important concept related to moving electric charges. Magnetism is generated due to 11 min readMoving Coil Galvanometer Hans Christian Oersted discovered in 1820 that a current-carrying conducting wire produces a magnetic field around it. His findings from his experiments are as follows: The magnetic compass needle is aligned tangent to an imaginary circle centered on the current-carrying cable.When the current is re 10 min read Chapter 5 - MAGNETISM AND MATTER Magnetism Magnetism in Physics is defined as the property of the material that is responsible for the magnetic behaviour of the magnets. Magnetism is defined as the force that is produced by the moving charge and it attracts or repels other magnets and moving charge. Initially, magnetism is defined as the pro 11 min readBar Magnet Bar Magnet is a magnet that is rectangular in shape and has two poles, the North Pole and South Pole. The magnetic field of a bar magnet is maximum at its pole and minimum at its center. Bar Magnets are made up of Iron, cobalt, or any other Ferromagnetic materials that show magnetic properties. Bar 9 min readGauss's Law Gauss's law is defined as the total flux out of the closed surface is equal to the flux enclosed by the surface divided by the permittivity. The Gauss Law, which analyses electric charge, a surface, and the issue of electric flux, is analyzed. Let us learn more about the law and how it functions so 15+ min readMagnetization and Magnetic Intensity We've all had fun with magnets as kids. Some of us are now even playing with them! What makes them magnetic though? Why aren't there magnetic fields in all materials and substances? Have you ever given it any thought? The subjects of magnetization and magnetic intensity will be covered in this chapt 6 min read CHAPTER 6 - ELECTROMAGNETIC INDUCTION Experiments of Faraday and Henry For a long time, electricity and magnetism were thought to be separate and unrelated phenomena. Experiments on electric current by Oersted, Ampere and a few others in the early decades of the nineteenth century established the fact that electricity and magnetism are inter-related. They discovered th 5 min readMagnetic Flux Magnetic Flux is defined as the surface integral of the normal component of the Magnetic Field(B) propagating through that surface. It is indicated by Ï† or Ï†B. Its SI unit is Weber(Wb). The study of Magnetic Flux is done in Electromagnetism which is a branch of physics that deals with the relation b 6 min readFaradayâ€™s Laws of Electromagnetic Induction Faraday's Law of Electromagnetic Induction is the basic law of electromagnetism that is used to explain the working of various equipment that includes an electric motor, electric generator, etc. Faraday's law was given by an English scientist Michael Faraday in 1831. According to Faraday's Law of El 10 min readLenz's Law Lenz law was given by the German scientist Emil Lenz in 1834 this law is based on the principle of conservation of energy and is in accordance with Newton's third law. Lenz law is used to give the direction of induced current in the circuit. In this article, let's learn about Lenz law its formula, e 7 min readMotional Electromotive Force The process of induction occurs when a change in magnetic flux causes an emf to oppose that change. One of the main reasons for the induction process in motion. We can say, for example, that a magnet moving toward a coil generates an emf, and that a coil moving toward a magnet creates a comparable e 14 min readInductance - Definition, Derivation, Types, Examples Magnetism has a mystical quality about it. Its capacity to change metals like iron, cobalt, and nickel when touched piques children's interest. Repulsion and attraction between the magnetic poles by observing the shape of the magnetic field created by the iron filling surrounding the bar magnet will 13 min readAC Generator - Principle, Construction, Working, Applications A changing magnetic flux produces a voltage or current in a conductor, which is known as electromagnetic induction. It can happen when a solenoid's magnetic flux is changed by moving a magnet. There will be no generated voltage (electrostatic potential difference) across an electrical wire if the ma 7 min read CHAPTER 7 - ALTERNATING CURRENT AC Voltage Applied to a Resistor Alternating Currents are used almost as a standard by electricity distribution companies. In India, 50 Hz Alternating Current is used for domestic and industrial power supply. Many of our devices are in fact nothing but resistances. These resistances cause some voltage drop but since the voltage thi 5 min readPhasors \| Definition, Examples & Diagram Phasor analysis is used to determine the steady-state response to a linear circuit functioning on sinusoidal sources with frequency (f). It is very common. For example, one can use phasor analysis to differentiate the frequency response of a circuit by performing phasor analysis over a range of freq 10 min readAC Voltage Applied to an Inductor Alternating Currents and Voltages vary and change their directions with time. They are widely used in modern-day devices and electrical systems because of their numerous advantages. Circuits in everyday life consist of resistances, capacitors, and inductances. Inductors are devices that store energy 5 min readAC Voltage Applied to a Capacitor Alternating Currents and Voltages vary and change their directions with time. They are widely used in modern-day devices and electrical systems because of their numerous advantages. Circuits in everyday life consist of resistances, capacitors, and inductance. Capacitors are the devices that accumula 6 min readSeries LCR Circuits In contrast to direct current (DC), which travels solely in one direction, Alternating Current (AC) is an electric current that occasionally reverses direction and alters its magnitude constantly over time. Alternating current is the type of electricity that is delivered to companies and homes, and 8 min readPower Factor in AC circuit The power factor is determined by the cosine of the phase angle between voltage and current. In AC circuits, the phase angle between voltage and current is aligned, or in other words, zero. But, practically there exists some phase difference between voltage and current. The value of the power factor 8 min readTransformer A transformer is the simplest device that is used to transfer electrical energy from one alternating-current circuit to another circuit or multiple circuits, through the process of electromagnetic induction. A transformer works on the principle of electromagnetic induction to step up or step down th 15+ min read CHAPTER 8 - ELECTROMAGNETIC WAVES Displacement Current Displacement current is the current that is produced by the rate of change of the electric displacement field. It differs from the normal current that is produced by the motion of the electric charge. Displacement current is the quantity explained in Maxwell's Equation. It is measured in Ampere. Dis 12 min readElectromagnetic Waves A wave is a propagating dynamic disturbance (change from equilibrium) of one or more quantities that is commonly described by a wave equation in physics, mathematics, and related subjects. Electromagnetic waves are a mix of electric and magnetic field waves produced by moving charges. The origin of 11 min readElectromagnetic Spectrum Electromagnetic Spectrum: The sun is our planet's principal source of energy, and its energy travels in the form of electromagnetic radiation. Electromagnetic energy moves across space at the speed of light in the form of waves of electric and magnetic fields with a range of frequencies or wavelengt 11 min read CHAPTER 9 - RAY OPTICS AND OPTICAL INSTRUMENTS Spherical Mirrors Spherical mirrors are generally constructed from glass. A spherical surface is a part cut from a hollow sphere. This curved surface of the glass has a silver coating on one side and a polished surface on the other, where the reflection of light takes place. The term "convex mirror" refers to a mirro 11 min readRefraction of Light Refraction is an important term used in the Ray Optics branch of Physics. Refraction of light is defined as the change in direction or the bending of a wave passing from one medium to another due to the change in speed of the wave. Some natural phenomena occurring in nature where refraction of light 11 min readTotal Internal Reflection In Physics, total internal reflection is the complete reflection of a light ray within the medium (air, water glass, etc). For example, the total internal reflection of rays of light takes place in a Diamond. Since Dimond has multiple reflecting surfaces through which the Total internal reflection t 8 min readImage formation by Spherical Lenses You might have used a microscope in the science lab for magnifying the micro-size object. It basically magnifies tiny objects and we can see the enlarged image of that object. Telescopes are used by scientists to the planets and stars which are far- far away from the earth. You might see the spectac 8 min readDispersion of Light through a Prism Dispersion of Light happens when white light is split into its constituent hues due to refraction. Dispersion of Light can be achieved through various means but the most common way to achieve dispersion of light is through Prism. 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10022	https://www.funwithpuzzles.com/2014/06/hourglass-problem-brain-teaser.html	Hourglass Problem Brain Teaser with Answer Home Puzzles Popular Puzzles Easy Puzzles Medium Puzzles Tough Puzzles Brain Test Quick Puzzles Logic Crack the Code Square Reasoning Circle Reasoning Missing Numbers Lateral Thinking Parking Riddles Number Logic Interview Questions Number Questions Maths Arithmetic Puzzles Algebra Problems Maths Reasoning Sequence Puzzles Logical Equations Triangle Math Puzzles Pyramid Math Puzzles Matchstick Puzzles Maths Riddles Picture Find the Mistake Odd One Out Puzzles Hidden Numbers Jigsaw Puzzles Shadow Riddles Water Tank Puzzles Emoji Puzzles Can You Read this? Spatial Reasoning Optical Illusions Riddles Tricky Riddles Mystery Riddles Funny Riddles English Word Riddles Missing Vowels Quiz What am I? 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Using these hourglasses you have to measure 15 minutes. Let's see if you can solve this time puzzle without wasting any time? Can you solve this Hourglass Brain Teaser? The answer to this "Hourglass Problem Brain Teaser", can be viewed by clicking on the button. Please do give your best try before looking at the answer. View Answer The Answer is.......... First start both the hourglasses. Once 7 minute hourglass finishes, turn it around. Once 11 minute hourglass finishes, 7 minute hourglass has run for 4 minutes. Turn around 7 minutes hourglass to measure 15 minutes. 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10023	https://www.collinsdictionary.com/us/dictionary/english-thesaurus/contradict	English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese More English Italiano Português 한국어 简体中文 Deutsch Español हिंदी 日本語 English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese Definitions Summary Synonyms Sentences Pronunciation Collocations Conjugations Grammar Credits × Synonyms of 'contradict' in American English contradict (verb) Synonyms deny be at variance with belie challenge controvert fly in the face of negate rebut Synonyms of 'contradict' in British English contradict 1 (verb) Definition (of a fact or statement) to suggest that (another fact or statement) is wrong We knew she was wrong, but nobody liked to contradict her.His comments contradict remarks he made earlier that day. Synonyms dispute He disputed the allegations. deny challenge The move was immediately challenged by the opposition. belie The facts of the situation belie his testimony. fly in the face of make a nonsense of gainsay (archaic, literary) There was no-one to gainsay this assertion. be at variance with 2 (verb) Definition to declare the opposite of (a statement) to be true The result appears to contradict a major study carried out last December. Synonyms negate I can neither negate nor affirm this claim. deny She denied the accusations. oppose Mr Taylor was bitter towards those who had opposed him. counter contravene This deportation order contravenes basic human rights. rebut impugn controvert Opposites support , agree , confirm , defend , endorse , affirm , verify , authenticate Copyright © 2016 by HarperCollins Publishers. All rights reserved. Additional synonyms in the sense of belie Definition to show to be untrue The facts of the situation belie his testimony. Synonyms disprove, deny, expose, discredit, contradict, refute, repudiate, negate, invalidate, rebut, give the lie to, make a nonsense of, gainsay (archaic, literary), prove false, blow out of the water (slang), controvert, confute in the sense of challenge Definition to call (a decision or action) into question The move was immediately challenged by the opposition. Synonyms dispute, question, tackle, confront, defy, object to, disagree with, take issue with, impugn (formal), throw down (US, slang) in the sense of contravene This deportation order contravenes basic human rights. Synonyms conflict with, cross, oppose, interfere with, thwart, contradict, hinder, go against, refute, counteract You may also like English Quiz ConfusablesEnglish Word listsLatest Word SubmissionsEnglish GrammarGrammar PatternsLanguage Lover's BlogCollins ScrabbleThe Paul Noble MethodEnglish Quiz ConfusablesEnglish Word listsLatest Word SubmissionsEnglish GrammarGrammar PatternsLanguage Lover's BlogCollins ScrabbleThe Paul Noble MethodEnglish Quiz ConfusablesEnglish Word listsLatest Word Submissions Browse alphabetically contradict contraction contractive contractual contradict contradicting contradiction contradictions All ENGLISH synonyms that begin with 'C' 1 2 ## Wordle Helper ## Scrabble Tools Quick word challenge Quiz Review Question: 1 - Score: 0 / 5 SYNONYMS Select the synonym for: exactly cheerfullyexplicitlyjauntilyerroneously SYNONYMS Select the synonym for: confused devotedenterprisingfashionableperplexed SYNONYMS Select the synonym for: to jump to hopto gulpto seeto sip SYNONYMS Select the synonym for: dinky terrifyinguncomplicatedvacillatingdainty SYNONYMS Select the synonym for: to include to containto maintainto peruseto demonstrate Your score: Study guides for every stage of your learning journey Whether you're in search of a crossword puzzle, a detailed guide to tying knots, or tips on writing the perfect college essay, Harper Reference has you covered for all your study needs. 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10024	https://www.youtube.com/watch?v=eIQndXQ5ujQ	cobb douglas production function \| microeconomics \| Short run \| long run \| econ mafia 1410 subscribers 58 likes Description 6403 views Posted: 13 Jun 2020 In economics and econometrics, the Cobb–Douglas production function is a particular functional form of the production function, widely used to represent the technological relationship between the amounts of two or more inputs (particularly physical capital and labor) and the amount of output that can be produced 3 comments Transcript: 해 o from 도모 뭐 금지 있겠나이까 플렉스 안에서는 porter two to point out into two people are too in possession of us 악보 따고 쌀의 스파이 캠 숨겨진 e wave 앞으로 pores console 옵스 to 시대 안데스 rendered seiwa 션 쌓기 of western 수업에서 애 카메 이렇게 a 라 주어진 붓다 학교 젠트 포함 에 이렇게 아르고 있어 최소 on vero notes on 쏭 같은 목회 거래소 intel 중이었어요 내줄 것을 보아야 os x 에 대해서 포함해 쌓아 교육보다 옵션 벌이 wood 앞으 페이셜케어 wonder live food was open to write a note inverter 몇개 더 웨버 벤 버트 앤 하우 말쯤에 놈 엔버 the root no 왜 못 i want now change to top 5 3 we 태아 proportion to wax 한번 마음 다바오 telex 부터 끼던 왜 보급 not we roll we took we ask you to power of power koi wa 끼던 왜 이렇게 okay we to town s2 codex 웨이브가 이 볼트도 폴더옵션 센슈얼 to see in korean 포함 들이 airport 펜 전 하우머치 own mouth 해몽이 뭐 입어 tokio 3 idiot and so i was too 없나 쯤엔 vertex woo hoo woo hoo 와콤 셀에 대해서 배워보고 이해 못하겠어 못한 worn only lovers diode is no use ing n 버즈 sinawe no outer and 어서 on wood 255 내가 or 그 지도 마켓 side note or do its tax or 칭 my
10025	https://math.stackexchange.com/questions/368801/getting-an-acute-angle-for-an-obtuse-angle-using-law-of-sines	trigonometry - Getting an acute angle for an obtuse angle using law of Sines. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Getting an acute angle for an obtuse angle using law of Sines. Ask Question Asked 12 years, 5 months ago Modified5 years, 4 months ago Viewed 20k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I have done this problem over and over again. I even looked up tutorials on how to properly use law of sines. It's rather embarrassing that I'm struggling so much wish this simple trigonometric stuff. Here's the picture of the triangle. I'm trying to solve for angle ∠C∠C. Angle ∠C∠C is definitely supposed to be obtuse. I keep getting: sin(21.55)7.7=sin(C)16 sin(21.55)7.7=sin(C)16 I simplify and take the arcsin(16⋅sin(21.55)7.7)arcsin(16⋅sin(21.55)7.7) And I can't get an obtuse angle. Anyone know why? trigonometry triangles Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 22, 2013 at 3:26 user17762 asked Apr 22, 2013 at 2:06 ModdedLifeModdedLife 169 1 1 gold badge 1 1 silver badge 8 8 bronze badges Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. Typically, the range of arcsin(x)arcsin(x) is [−π/2,π/2][−π/2,π/2]. This thereby eliminates the obtuse angle you want. To get the obtuse angle you want, all you need to do is to realize that sin(π−α)=sin(α) sin(π−α)=sin(α) Hence, 180∘−arcsin(16 sin(21.55∘)/7.7)180∘−arcsin(16 sin(21.55∘)/7.7) should give you the answer you need. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 22, 2013 at 2:11 user17762 user17762 0 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Because of the range of arcsin you answer is the supplement to the correct answer. In order to avoid this difficulty always look for angles based on the side lengths in increasing order. In other words if you have a choice as to which angle to solve for first always choose the angle opposite the shortest of the two sides you have. In this problem you are forced to find the angle C first so you must recognize that it is obtuse and therefore the answer you get is the supplement. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 10, 2014 at 17:23 Jayme CronquistJayme Cronquist 1 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. after you simplify you must subtract the value from 180 degrees since you know that angle C must be a obtuse angle which would be approximately 180 - 49.76.the value of C is equal to 130.24 degrees. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 22, 2016 at 2:27 user381281user381281 1 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. In a triangle, the angle opposite the longest side is always the maximum. So, solve the angle opposite the smaller two sides first. Then, subtract the sum of these angles from 180 to get the angle opposite the longest side. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jun 27, 2019 at 5:15 RamrajRamraj 1 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The reason why you can't get an obtuse angle is that there is another triangle that satisfies the law of sines equation that you wrote down, namely triangle △A B C 1△A B C 1: Notice that triangle △A B C 1△A B C 1 satisfies the same equation: sin(21.55)7.7=sin(C 1)16. sin(21.55)7.7=sin(C 1)16. When you solve your equation using arcsin, you're getting the acute angle at vertex C 1 C 1, the one labeled as θ θ in the diagram. The relationship between θ θ and the angle at vertex C C is θ+C=180. θ+C=180. The reason: triangle △A C C 1△A C C 1 is isosceles so ∠A C C 1=∠C C 1 A=θ∠A C C 1=∠C C 1 A=θ. Therefore θ θ and C C are supplementary angles. So to get the obtuse C C you should subtract the acute θ θ from 180 180. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered May 7, 2020 at 17:08 grand_chatgrand_chat 41k 1 1 gold badge 46 46 silver badges 86 86 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions trigonometry triangles See similar questions with these tags. 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10026	https://uspas.fnal.gov/materials/09UNM/Unit_10_Lecture_14_Cyclotron_basics.pdf	1 antaya@psfc.mit.edu / (617) 253-8155 Unit 10 - Lectures 14 Cyclotron Basics MIT 8.277/6.808 Intro to Particle Accelerators Timothy A. Antaya Principal Investigator MIT Plasma Science and Fusion Center 2 antaya@psfc.mit.edu / (617) 253-8155 Outline  Introduce an important class of circular particle accelerators: Cyclotrons and Synchrocyclotrons  Identify the key characteristics and performance of each type of cyclotron and discuss their primary applications  Discuss the current status of an advance in both the science and engineering of these accelerators, including operation at high magnetic ﬁeld Overall aim: reach a point where it will be possible for to work a practical exercise in which you will determine the properties of a prototype high ﬁeld cyclotron design (next lecture) 3 antaya@psfc.mit.edu / (617) 253-8155 Motion in a magnetic ﬁeld 4 antaya@psfc.mit.edu / (617) 253-8155 Magnetic forces are perpendicular to the B ﬁeld and the motion 5 antaya@psfc.mit.edu / (617) 253-8155 Sideways force must also be Centripedal 6 antaya@psfc.mit.edu / (617) 253-8155 Governing Relation in Cyclotrons  A charge q, in a uniform magnetic ﬁeld B at radius r, and having tangential velocity v, sees a centripetal force at right angles to the direction of motion: B v q r r mv r r ! = ˆ 2  The angular frequency of rotation seems to be independent of velocity: m qB / = ! 7 antaya@psfc.mit.edu / (617) 253-8155 Building an accelerator using cyclotron resonance condition  A ﬂat pole H-magnet electromagnet is sufﬁcient to generate require magnetic ﬁeld  Synchronized electric ﬁelds can be used to raise the ion energies as ions rotate in the magnetic ﬁeld  Higher energy ions naturally move out in radius  Highest possible closed ion orbit in the magnet sets the highest possible ion energy 8 antaya@psfc.mit.edu / (617) 253-8155 There is a difﬁculty- we can’t ignore relativity  A charge q, in a uniform magnetic ﬁeld B at radius r, and having tangential velocity v, sees a centripetal force at right angles to the direction of motion: B v q r r mv r r ! = ˆ 2  However:  Picking an axial magnetic ﬁeld B and azimuthal velocity v allows us to solve this relation: qvB r mv = / 2 2 2 1 / 1 c v ! = " ! " = v /r = qB/m ! m = "m0 9 antaya@psfc.mit.edu / (617) 253-8155 Relativistic Limit on Cyclotron Acceleration  The mass in ω= qB/m is the relativistic mass m=γm0  ω≈constant only for very low energy cyclotrons ~52% 1.0 GeV ~21% 250 MeV ~1% 10 MeV % Frequency decrease Proton Energy 10 antaya@psfc.mit.edu / (617) 253-8155 There are 3 kinds of Cyclotrons:  CLASSICAL: (original)  Operate at ﬁxed frequency (ω= qB/m) and ignore the mass increase  Works to about 25 MeV for protons (γ≅1.03)  Uses slowly decreasing magnetic ﬁeld ‘weak focusing’  SYNCHROCYCLOTRON: let the RF frequency ω decreases as the energy increases  ω=ω0/γ to match the increase in mass (m= γm0)  Uses same decreasing ﬁeld with radius as classical cyclotron  ISOCHRONOUS: raise the magnetic ﬁeld with radius such that the relativistic mass increase is just cancelled  Pick B=γB0 {this also means that B increases with radius}  Then ω= qB/m = qB0/m0 is constant.  Field increases with radius- magnet structure must be different How to manage the relativistic change in mass? 11 antaya@psfc.mit.edu / (617) 253-8155 Some Examples of Cyclotrons 12 antaya@psfc.mit.edu / (617) 253-8155 1932 Cyclotron 180˚ ‘Dee’ Internal Energy Analyzer Vacuum Port Ion Source is a gas feed and a wire spark gap Evacuated Beam Chamber sits between magnet poles: 13 antaya@psfc.mit.edu / (617) 253-8155 The Largest…  Gatchina Synchrocyclotron at Petersburg Nuclear Physics… 1000 MeV protons and 10,000 tons 14 antaya@psfc.mit.edu / (617) 253-8155 Superconducting Isochronous Cyclotron 15 antaya@psfc.mit.edu / (617) 253-8155 The Highest Magnetic Field…  Still River Systems 9 Tesla, 250 MeV, synchrocyclotron for Clinical Proton Beam Radiotherapy 16 antaya@psfc.mit.edu / (617) 253-8155 The Newest…  Nanotron: superconducting, cold iron, cryogen free ‘portable’ deuterium cyclotron 17 antaya@psfc.mit.edu / (617) 253-8155 New Cyclotrons and Synchrocyclotrons are coming.. Also:  Gigatron: 1 GeV, 10 mA protons for airborne active interrogation  Megatron: 600 MeV muon cyclotron (requires a gigatron to produce muons and a reverse cyclotron muon cooler for capture for accel.) Isotron -for short lived PET isotope production:  Protons or heavy ions  30-100 MeV  Synchrocyclotron or isochronous cyclotron is possible 18 antaya@psfc.mit.edu / (617) 253-8155 Key Characteristics of the Cyclotron ‘Class’ Cyclotron utility is due to:  Ion capture and Beam formation at low velocity, followed by acceleration to relativistic speeds in a single device  Efﬁcient use of low acceleration voltage makes them robust and uncritical; pulsed or CW operation allowed  Beam characteristics are wrapped up in the design of the static magnetic guide ﬁeld; ions have high orbital stability  Ion species: H+ --> U; neg. ions (e.g. H-), molecular ions (e.g. HeH+)  Intensities; picoamps (one ion per rf bucket) to milliamps  γ: 0.01 --> 2.3 Have resulted in:  2nd largest application base historically and currently (electron linacs used in radiotherapy are 1st)  Science (Nuclear, Atomic, Plasma, Archeology, Atmospheric, Space), Medicine, Industry, Security  Highest energy CW accelerator in the world: K1200 heavy ion at MSU- 19.04 GeV 238U 19 antaya@psfc.mit.edu / (617) 253-8155 Key Characteristics- prob. most important: Cyclotron utility is due to:  Ion capture and Beam formation at low velocity, followed by acceleration to relativistic speeds in a single device  Efﬁcient use of low acceleration voltage makes them robust and uncritical; pulsed or CW operation allowed  Beam characteristics are wrapped up in the design of the static magnetic guide ﬁeld; ions have high orbital stability  Ion species: H+ --> U; neg. ions (e.g. H-), molecular ions (e.g. HeH+)  Intensities; picoamps (one ion per rf bucket) to milliamps  γ: 0.01 --> 2.3 Have resulted in:  2nd largest application base historically and currently (electron linacs used in radiotherapy are 1st)  Science (Nuclear, Atomic, Plasma, Archeology, Atmospheric, Space), Medicine, Industry, Security  Highest energy CW accelerator in the world: K1200 heavy ion at MSU- 19.04 GeV 238U 20 antaya@psfc.mit.edu / (617) 253-8155 Classical Cyclotrons Weak focusing Phase stability Limited by Relativistic Mass Increase 21 antaya@psfc.mit.edu / (617) 253-8155 There are 3 kinds of Cyclotrons:  CLASSICAL: (original)  Operate at ﬁxed frequency (ω= qB/m) and ignore the mass increase  Works to about 25 MeV for protons (γ≅1.03)  Uses slowly decreasing magnetic ﬁeld ‘weak focusing’  SYNCHROCYCLOTRON: let the RF frequency ω decreases as the energy increases  ω=ω0/γ to match the increase in mass (m= γm0)  Uses same decreasing ﬁeld with radius as classical cyclotron  ISOCHRONOUS: raise the magnetic ﬁeld with radius such that the relativistic mass increase is just cancelled  Pick B=γB0 {this also means that B increases with radius}  Then ω= qB/m = qB0/m0 is constant.  Field increases with radius- magnet structure must be different How to manage the relativistic change in mass? 22 antaya@psfc.mit.edu / (617) 253-8155 The 1931 Cyclotron… 23 antaya@psfc.mit.edu / (617) 253-8155 Cyclotron Schematic Diagram (via Lawrence Patent)  A ﬂat pole electromagnet (3) generates a vertical magnetic ﬁeld (m)  Ions (P) rotate in the mid-plane of an evacuated split hollow conductor (1-2)  Time varying electric ﬁelds (4) applied to the outside of this conductor raise the ion energies as ions rotate in the magnetic ﬁeld and cross the split line gap- the only place where electric ﬁelds (e) appear  Higher energy ions naturally move out in radius  Highest allowed closed ion orbit in magnet sets the highest possible ion energy 24 antaya@psfc.mit.edu / (617) 253-8155 Let’s break down the key phenomena that make cyclotrons work…  We’ll do this in a very ‘raw’ manner- using elementary properties of ions, conductors and electromagnetic ﬁelds  Why choose this approach?  To demonstrate just how utterly simple cyclotrons are  To get to better appreciate the key challenges in making cyclotrons work  To understand how the advance machines just shown are possible 25 antaya@psfc.mit.edu / (617) 253-8155 Magnetic Field Generation  A ﬂat pole electromagnet (3) generates a vertical magnetic ﬁeld (m)  Ions (P) rotate in the mid-plane of an evacuated split hollow conductor (1-2)  Time varying electric ﬁelds (4) applied to the outside of this conductor raise the ion energies as ions rotate in the magnetic ﬁeld and cross the split line gap- the only place where electric ﬁelds (e) appear  Higher energy ions naturally move out in radius  Highest allowed closed ion orbit in magnet sets the highest possible ion energy 26 antaya@psfc.mit.edu / (617) 253-8155 Typical large H Magnet 27 antaya@psfc.mit.edu / (617) 253-8155 Magnetic ﬁeld of a H Magnet 28 antaya@psfc.mit.edu / (617) 253-8155 Ion Acceleration-- requires a bit more work…  A ﬂat pole electromagnet (3) generates a vertical magnetic ﬁeld (m)  Ions (P) rotate in the mid-plane of an evacuated split hollow conductor (1-2)  Time varying electric ﬁelds (4) applied to the outside of this conductor raise the ion energies as ions rotate in the magnetic ﬁeld and cross the split line gap- the only place where electric ﬁelds (e) appear  Higher energy ions naturally move out in radius  Highest allowed closed ion orbit in magnet sets the highest possible ion energy 29 antaya@psfc.mit.edu / (617) 253-8155 Acceleration really looks something like this… 30 antaya@psfc.mit.edu / (617) 253-8155 Why not magnetic ﬁeld only acceleration? 31 antaya@psfc.mit.edu / (617) 253-8155 Ion Orbital Rotation Frequency- numerically  Consider an arbitrary positive ion of atomic species (A,Z) with Q orbital electrons removed. The ion cyclotron frequency would be:  Some examples:  Low energy proton in 1 T ﬁeld: 15.23 MHz  250 MeV proton in 8.2T ﬁeld: 98 MHz  3.2GeV 40Ar16+ ion in 5.5T ﬁeld: 30.8 MHz  Where m0 is the rest mass of a nucleon (~940 MeV). Evaluating the constants: ! f = " 2# = qB 2#m = Q A $ % & ' ( ) e 2#m0 B ! f = Q A " # $ % & ' 15.23MHz B ( 32 antaya@psfc.mit.edu / (617) 253-8155 Ion Motion in a cyclotron  A ﬂat pole electromagnet (3) generates a vertical magnetic ﬁeld (m)  Ions (P) rotate in the mid-plane of an evacuated split hollow conductor (1-2)  Time varying electric ﬁelds (4) applied to the outside of this conductor raise the ion energies as ions rotate in the magnetic ﬁeld and cross the split line gap- the only place where electric ﬁelds (e) appear  Higher energy ions naturally move out in radius  Highest allowed closed ion orbit in magnet sets the highest possible ion energy 33 antaya@psfc.mit.edu / (617) 253-8155 Alternative Expression in Momentum  Again we equate the two expressions for the same force:  The momentum at any radius is completely deﬁned by the magnetic ﬁeld there!  Also, at the same ﬁeld B,  If p3>p2>p1  Then r3>r2>r1  Since ω=dθ/dt=qB/m, even though the three orbits are different in size, the ions will make 1 complete revolution at the same angular rate (unless m=γm0 is very different for the three momenta) qvB r mv = / 2 ! p = mv = qBr p3 p2 p1 34 antaya@psfc.mit.edu / (617) 253-8155 Special Challenges in Cyclotrons  Orbit Stability  Initial beam Formation  RF Acceleration  Getting the beam out of the machine!  p=erB --> p/e =rB  we call ℜ≡rB the magnetic rigidity or magnetic stiffness  We will see that ℜ shows up in the Cyclotron ﬁnal energy formula- it’s in KB=e2r2B2/2m0-In cyclotrons, the ﬁnal energy is essentially set by the radius and B ﬁeld at the point of beam extraction 35 antaya@psfc.mit.edu / (617) 253-8155 Built In Orbit Stability- Weak Focusing 36 antaya@psfc.mit.edu / (617) 253-8155 The Field Index and Axial Stability An restoring force is required to keep ions axially centered in the gap We deﬁne the ﬁeld index as: One can show that an axial restoring for exists when n>0 (off median plane Br has right sign) Hence dB/dr<0 is required since B and r enter in ratios This condition can be met with a ﬂat pole H-Magnet dr dB B r n ! = 37 antaya@psfc.mit.edu / (617) 253-8155 Field Index n shows up in Equations of Motions  Small oscillations of ions in r and z about equilibrium orbits:  Have solutions :  Where ω is the cyclotron frequency  Betatron Frequencies (Tunes):  Have real sinusoidal solutions for 0<n<1; this condition is true in a classical cyclotron  It’s also referred to as a weak focusing accelerator ! ˙ ˙ x + (1" n)# 2x = 0 ˙ ˙ z + n# 2z = 0 ! x = xm sin(1" n)1 2#t z = zm sinn1 2#t ! " r = #r /# = 1$ n vz = #z /# = n 38 antaya@psfc.mit.edu / (617) 253-8155 Initial Beam Challenge 39 antaya@psfc.mit.edu / (617) 253-8155 For Example: Initial Proton trajectories at 9T 40 antaya@psfc.mit.edu / (617) 253-8155 Positive Ion Source must be compact  Straight-forward ﬁeld scaling of original 5.5 T ion source of K500 cyclotron  Chimney diameter 3 mm  Test ion source has extra support across median plane  allows separated cathode geometry of Antaya thesis or Harper cyclotron  Pulsed cathode lifetime expected to be months 41 antaya@psfc.mit.edu / (617) 253-8155 RF Acceleration Challenge 42 antaya@psfc.mit.edu / (617) 253-8155 Beam Extraction Challenge 43 antaya@psfc.mit.edu / (617) 253-8155 Orbit Separation impacts Extraction Turn Number Let E1 be the energy gain per revolution Then the total number of revolutions required to reach a ﬁnal kinetic energy T: Let the average ion phase when crossing the acceleration gap phase be φ; V0 is the peak voltage on the dee Energy gain per gap crossing: T1=V0sin φ Gaps per revolution: n Turn number: N=T/nT1=T/(nV0sin φ) 250 MeV protons; 17 KeV/turn: N~15,000 Turn Spacing: dr/dN~r(T1/T) 250 MeV protons r=0.3m: dr/dN≅20 microns! 44 antaya@psfc.mit.edu / (617) 253-8155 Beam Extraction: 5 micron orbit turn spacing to 1 cm in 20 orbit revolutions induced by ﬁeld perturbation 45 antaya@psfc.mit.edu / (617) 253-8155 Phase Stable Acceleration aka Phase Stability 3 General Requirements: required instantaneous acceleration voltage is less than the maximum available voltage a change in ion momentum results in a change in ion orbit rotation period rate of change of the frequency is less than a limiting critical value Second Condition is the most easily accessible: ! d" " = ( 1 # $ 1 % 2 ) dp p 46 antaya@psfc.mit.edu / (617) 253-8155 Acceleration in a 9T Guide Field 47 antaya@psfc.mit.edu / (617) 253-8155 Cyclotrons- Final Energy Scaling with Field and Radius (The origin of Superconducting Cyclotrons and Synchrocyclotrons) 48 antaya@psfc.mit.edu / (617) 253-8155 Cyclotron Energy Scales inversely with Field  The ﬁnal energy can be written as a power series expansion in the relativistic factor γ,  The ﬁrst term in this expansion is : Tﬁnal≅KBQ2/A, for an ion of charge Qe and ion mass Am0  KB represents the equivalent proton ﬁnal energy for the machine, and is related to the ion momentum a.k.a. the particle rigidity (Bρ): KB=(eBρ)2/2m0 49 antaya@psfc.mit.edu / (617) 253-8155 This inverse size scaling is approx. spherical Almost (but not quite) spherical: Efficient cyclotron magnetic circuits include more iron laterally than axially 50 antaya@psfc.mit.edu / (617) 253-8155 Radius and Field Scaling for Fixed Energy 1/729 0.25 9 1/343 0.33 7 1/125 0.46 5 1/27 0.76 3 1 2.28 1 (r1/r)3 rextraction (m) B (T) 51 antaya@psfc.mit.edu / (617) 253-8155 Classical Cyclotrons- Energy Limit  Historically- E<25 MeV, and high acceleration voltages were required  WHY?  Relativistic mass increase lowers the ion orbital frequency: ω=qB/γm0  Ion frequency relative to the ﬁxed RF frequency decreases (rotation time τ increases)  Ions arrive increasing late with respect to the RF voltage on the dee  Eventually crossing the gaps at wrong phase and decelerates  21 MeV proton : γﬁnal=1.022 seems small, but…  Angular rotation slip near full energy dφ/dn=360°Δω/ω=360° [mB0/m0B - 1]≈360[γ-1]-->8°  An ion on peak phase is lost in 11 revolutions  Only solution- very high energy gain per turn - 360kV was required to reach 21 MeV in the LBL 60” Cyclotron!
10027	https://chemistrytalk.org/ice-table-chemistry/	Using an ICE Table \| ChemTalk Skip to content ChemTalk Articles General Chemistry Organic Chemistry Biochemistry Elements Interactive Periodic Table (New) Learn About The Elements Periodic Table with Names (JPG Image) Experiments History of Chemistry Resources Teacher Resources – General Chemistry Teacher Resources – O Chem General Chemistry Worksheets Chemtalk Infographics Podcast Episodes Spotlight on Chemists Academic Success Live Learning Office Hours About Us Our GoFundMe Our Mission Our Team Advisory Board Sponsors & Donors Support Us Join Our Team Donate Join Newsletter Search for: Search Button Using an ICE Table Core Concepts In this tutorial, we will learn about the ICE table for chemistry: a method of completing calculations in equilibrium reactions, either to find the concentrations of reactants and products, or to find the value of the equilibrium constant. Topics Covered in Other Articles How to calculate molarity Le Chatelier’s principle What Are Significant Figures Conditions for Equilibrium Measuring Concentration in Normality Review: Equilibrium Constant & Le Chatelier’s Principle In order to use an ICE table correctly, you need to know how to calculate the equilibrium constant, . is a measure of the extent of the reaction, whether the position of equilibrium favors the reactants or products. For a reaction , then we calculate the equilibrium constant using the equation: . Le Chatelier’s principle states that if a system in a state of dynamic equilibrium is disturbed by a change to its conditions, then the position of equilibrium will shift to counteract the change.For example, if more of one reactant is added to the reaction, then the position of equilibrium will shift towards the products. What is an ICE table? ICE stands for Initial, Change, Equilibrium. An ICE table is a tool used to calculate the changing concentrations of reactants and products in (dynamic) equilibrium reactions. This method first lists the concentrations of both reactants and products, before any changes occur. This is the initialstage. Then, the change is listed, in the form of addition or subtraction of a specific concentration. Alternately, the addition or subtraction of an unknown amount is listed (in the form of + or – M), and the value of is solved for. Finally, the equilibriumconcentration is listed, which is the initial concentration after it has undergone the change described. What is the change in an ICE table? If the described change is unspecified, then the equilibrium concentration is listed in terms of , and the equilibrium constant is used to solve for the variable. If the described change is specific, then the equilibrium concentration is listed as a concrete number, and it is used to solve for the equilibrium constant. We will explore both examples below. Le Chatelier’s principle is especially relevant to ICE tables because the change in the ICE table represents the shift in the position of equilibrium: that means that if the change shows an increase in concentration of reactants, then there will be a subsequent decrease in the concentration of products. Vice versa as well: if the change shows a decrease in concentration of reactants, then there will be a subsequent increase in the concentration of products. This subsequent increase/decrease shows the position of equilibrium shifting to counteract the change to the reactants. Additionally, Le Chatelier’s principle is relevant because the coefficient of the reactant or product affects the change to its concentration. For example, if a reactant has a coefficient of 1, then it changes by , but if a reactant has a coefficient of 2 or more, then it changes by . Examples of an ICE table + how to use them ICE table #1: Find the equilibrium concentrations from the K value. Initial Change Equilibrium Now, we solve for : Now, we recalculate the equilibrium concentrations in the ICE table, using the newly found value: Initial Change Equilibrium ICE table #2: Find the K value from known equilibrium concentrations. Initial Change Equilibrium NOTE: these are purely theoretical examples, neither the concentrations nor values are taken from real-life applications ICE Table Practice Problems Problem 1 Nitrogen monoxide forms through the following reaction: With initial concentrations and , what is the equilibrium concentration of nitrogen monoxide? Assume small . Problem 2 Consider the following reaction and initial concentrations: At equilibrium, A has the following concentration: What is the equilibrium concentration? ICE Table Practice Problem Solutions 1: 2: Further Reading Balancing chemical reactions Carbocation stability Formal charge Our Services Lessons Elements Experiments Resources Contact & About Us admin@chemistrytalk.org Mission Programs Team Advisory Board Get Involved Donate GoFundMe Page Newsletter Follow Us Facebook Instagram Twitter Youtube Tiktok
10028	https://www.themathdoctors.org/degenerate-conics-ii-are-their-parameters-meaningful/	Skip to content Degenerate Conics II: Are Their Parameters Meaningful? February 25, 2022 February 24, 2022 / Algebra, Geometry / Definitions, Degeneracy, Formulas / By Dave Peterson Last time we looked at what a degenerate conic section is, and how it relates on one hand to actual cones, and on the other to the general equation of the conic. Here we’ll look at the parameters of conic sections (focus, directrix, axes, and especially eccentricity) and how they apply to degenerate cases. Does a degenerate conic have an eccentricity? A 2008 question digs into the intersecting lines case more deeply, asking an intriguing question: Straight Lines and Conic Sections Is a pair of intersecting straight lines a conic section? If so, what is its eccentricity value? I heard some say that its eccentricity value tends to infinity, but my math professor disagrees. Please help me. We saw last time that a pair of intersecting lines is one form of degenerate conic (specifically, a degenerate hyperbola). We’ll see what eccentricity means in a moment. Doctor Vogler answered: Hi Nony, Thanks for writing to Dr. Math. The eccentricity of a conic section is usually defined for each conic section, and particularly only for the nondegenerate conics. So the simplistic answer is that the eccentricity of a pair of intersecting lines is not defined. It is defined for the ellipse and for the hyperbola ((e=\frac{c}{a})), where c is the distance from the center to the focus), but not for the parabola, which has no center. For the ellipse, (c=\sqrt{a^2-b^2}), while for the hyperbola, (c=\sqrt{a^2+b^2}), where a and b are the semiaxes. It is not defined directly in terms of the coefficients of the equation. Is it infinite for a pair of lines? Of course, that leaves unanswered the question "Is there a reasonable way to define it?" For example, "infinity" is often a reasonable answer for "undefined" things; sometimes a nonzero quantity divided by zero is taken to be infinity, which makes a certain sense in terms of limits, and in certain other special cases, although this is not always meaningful. Certainly, if someone gives you a (positive) finite eccentricity, you could say what type of nondegenerate conic has that eccentricity, so the natural guess at the eccentricity of a degenerate conic would be infinity, the one positive "number" that was left out of the list. That also jibes with statements you'll find on places like Wikipedia: Eccentricity that eccentricity is a measurement of distance from circular, and eccentricity increases as curvature decreases, since lines have zero curvature. But actually that's an over-simplification, since all circles have zero eccentricity, but larger circles have smaller curvature. Wikipedia explicitly says that the eccentricity of a pair of lines is infinite. Here is a quote (as it is now, 14 years later): In mathematics, the eccentricity of a conic section is a non-negative real number that uniquely characterizes its shape. More formally two conic sections are similar if and only if they have the same eccentricity. One can think of the eccentricity as a measure of how much a conic section deviates from being circular. In particular: The eccentricity of a circleis zero. The eccentricity of an ellipsewhich is not a circle is greater than zero but less than 1. The eccentricity of a parabolais 1. The eccentricity of a hyperbolais greater than 1. The eccentricity of a pair of lines is (\infty). Can that be wrong? Or is Nony’s professor right? Is there a formula that can be generalized? For another thing, the similarity of the definitions of eccentricity for ellipses and hyperbolas lead one to think that maybe there's a general formula for a general conic section in quadratic form Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 which would give the correct values for circles, ellipses, parabolas, and hyperbolas, and then we need only evaluate this formula for the degenerate cases to find out what their eccentricity is. Note that the a and c in the definition of eccentricity are not the A and C in this equation. Doctor Vogler used lower case letters in the equation; I’ve changed them to upper case for clarity. Also, in the following, I have corrected the conditions for the formulas for eccentricity, which were a little off. Unfortunately, things aren't quite so nice. But I'll demonstrate how close it gets, and how things break down for degenerate conics. After a rotation and a translation, you can put most conics (but not parabolas, for example) into the simplified form Ax^2 + By^2 = C. Applying to our formula for eccentricity of an ellipse or a hyperbola, we find that 1) if 0 < A < B and 0 < C, then e = sqrt((B-A)/B) 2) if B < 0 < A and 0 < C, then e = sqrt((B-A)/B) 3) if B < A < 0 and C < 0, then e = sqrt((B-A)/B) 4) if A < 0 < B and C < 0, then e = sqrt((B-A)/B) which looks promising, but 5) if 0 < B < A and 0 < C, then e = sqrt((A-B)/A) 6) if A < 0 < B and 0 < C, then e = sqrt((A-B)/A) 7) if A < B < 0 and C < 0, then e = sqrt((A-B)/A) 8) if B < 0 < A and C < 0, then e = sqrt((A-B)/A). Well, okay, so there's not one formula; there's two. That's a little annoying, and one might wonder which to use in unknown cases (and, of course, I have left out parabolas), but let's work with what we have. Note that A, B, and C in this simplified equation are different from the general equation. In terms of the standard equation for the ellipse (\frac{x^2}{a^2}+\frac{y^2}{b^2}=1), a and b are (\sqrt{\frac{C}{A}}) and (\sqrt{\frac{C}{B}}), a being the larger of the two (the major semiaxis), and b the smaller. For the hyperbola (\frac{x^2}{a^2}-\frac{y^2}{b^2}=1) or (\frac{y^2}{a^2}-\frac{x^2}{b^2}=1), a and b are (\sqrt{\left\|\frac{C}{A}\right\|}) and (\sqrt{\left\|\frac{C}{B}\right\|}), a being the one corresponding to a coefficient with the same sign as C, and b the other. The result is two different formulas, which swap the roles of A and B, because the roles of the axes depend on magnitude (for ellipses) or sign (for hyperbolas) of the coefficients. Cases #3 and #4 are equivalent to #1 and #2 (multiplying through by (-1)), and similarly #7 and #8 are equivalent to #5 and #6. Let’s look at a couple examples. Here are ellipses, cases #1 ((4x^2+9y^2=36\Leftrightarrow\frac{x^2}{9}+\frac{y^2}{4}=1)) and #5 ((9x^2 + 4y^2 = 36\Leftrightarrow\frac{x^2}{4}+\frac{y^2}{9}=1)): For the horizontal ellipse, (a=3) (horizontal), (b=2) (vertical), and (c=\sqrt{a^2-b^2}=\sqrt{3^2-2^2}=\sqrt{5}). So (e=\frac{c}{a}=\frac{\sqrt{5}}{3}). This agrees with the formula $$e=\sqrt{\frac{A-B}{A}}=\sqrt{\frac{9-4}{9}}=\frac{\sqrt{5}}{3}$$ But for the vertical ellipse, (a=3) (vertical), (b=2) (horizontal), so (e=\frac{c}{a}=\frac{\sqrt{5}}{3}). This agrees with the formula $$e=\sqrt{\frac{B-A}{B}}=\sqrt{\frac{9-4}{9}}=\frac{\sqrt{5}}{3}$$ And it’s obvious that these have the same shape, so if eccentricity is to be meaningful, they must be the same. Here are hyperbolas, cases #2 ((9x^2 – 4y^2 = 36\Leftrightarrow\frac{x^2}{4}-\frac{y^2}{9}=1)) and #6 ((-4x^2+9y^2=36\Leftrightarrow\frac{y^2}{4}-\frac{x^2}{9}=1)): For the horizontal hyperbola, (a=2), (b=3), and (c=\sqrt{a^2+b^2}=\sqrt{2^2+3^2}=\sqrt{13}). So (e=\frac{c}{a}=\frac{\sqrt{13}}{2}). This agrees with the formula $$e=\sqrt{\frac{B-A}{B}}=\sqrt{\frac{(-4)-9}{-4}}=\frac{\sqrt{13}}{2}$$ For the vertical hyperbola, (a=2) (vertical), (b=3) (horizontal), so (c=\sqrt{a^2+b^2}=\sqrt{2^2+3^2}=\sqrt{13}). So (e=\frac{c}{a}=\frac{\sqrt{13}}{2}). This agrees with the formula $$e=\sqrt{\frac{A-B}{A}}=\sqrt{\frac{-4-9}{-4}}=\frac{\sqrt{13}}{2}$$ Again, the eccentricities are obviously the same, but come from different formulas. Now, I should add that it is possible to write a “single formula” for eccentricity, but it takes a bit of stretching to call it that. Here is what Wikipedia suggests for a formula: When the conic section is given in the general quadratic form $$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0,$$ the following formula gives the eccentricity e if the conic section is not a parabola (which has eccentricity equal to 1), not a degenerate hyperbola or degenerate ellipse, and not an imaginary ellipse: $$e=\sqrt{\frac{2{\sqrt{(A-C)^{2}+B^{2}}}}{\eta (A+C)+{\sqrt {(A-C)^{2}+B^{2}}}}}$$ where (\eta =1) if the determinant of the 3×3 matrix $${\begin{bmatrix}A&B/2&D/2\B/2&C&E/2\D/2&E/2&F\end{bmatrix}}$$ is negative or (\eta =-1) if that determinant is positive. They’ve really combined two formulas into one by using the determinant as a “switch”; and if it is zero, the eccentricity is undefined. Guess when that happens? Applying it to a pair of lines Now, what happens for degenerate hyperbolas? Now suppose that b = 1 is positive, and a = -m^2 is negative, so that our equation is y^2 - (mx)^2 = c or (y - mx)(y + mx) = c. If c > 0, then we have a hyperbola with eccentricity sqrt(1 + m^(-2)), independent of c. If c < 0, then we have a hyperbola with eccentricity sqrt(1 + m^2), independent of c. If c = 0, then we have a pair of intersecting lines. So what should we consider its eccentricity to be? Here we have (A<00) and case #4 when (C<0). In case #6, $$e=\sqrt{\frac{A-B}{A}}=\sqrt{\frac{-m^2-1}{-m^2}}=\sqrt{\frac{m^2+1}{m^2}}=\sqrt{1+\frac{1}{m^2}}$$ In case #4, $$e=\sqrt{\frac{B-A}{B}}=\sqrt{\frac{1-(-m^2)}{1}}=\sqrt{1+m^2}$$ Here is the family of hyperbolas (y^2-4x^2), with (m=2), whose eccentricities are (\frac{\sqrt{5}}{2}) or (\sqrt{5}): If we graph the eccentricity as a function of the constant c, it looks like this: What should we say the eccentricity is when (c=0)? Well, it's right there on the boundary where the eccentricity changes from one constant to another. The only reasonable choice seems to be undefined, but not because it's infinity. At least, it usually changes from one constant to another. But what if m = 1? If we consider the conic section y^2 - x^2 = c or (y - x)(y + x) = c then we find that we have a hyperbola with eccentricity sqrt(2) for every value of c except zero. When c = 0, we have a pair of intersecting lines. Naturally, we should expect this pair of intersecting lines to have eccentricity sqrt(2), right? Not infinity. Here is this family of (“rectangular”) hyperbolas, all with eccentricity (\sqrt{2}): Here is a graph of the eccentricity vs. c this time: In this special case, an argument could be made that the eccentricity is always (\sqrt{2}); but is that appropriate? Other degenerate conics Anyway, in summary I would conclude that eccentricity is only defined for nondegenerate conic sections, and since it depends on which parameters are larger (where is the "major" axis and the "minor" axis), it doesn't generalize nicely to the degenerate conic sections, though if it did, it would seem that the degenerate conic sections would have eccentricity values in common with some of the nondegenerate ones. For example, x^2 + y^2 = r^2 has eccentricity 0, so the single point (when r = 0) should have eccentricity 0. And y = mx^2 + c has eccentricity 1, so the single line (when m = 0) should have eccentricity 1. And I already argued about the case of two intersecting lines, which are (usually) right on the border between two different eccentricity values. Here is the family of parabolas, with m varying between -1 and 1, with zero producing the line; all have eccentricity 1: Here is the graph of eccentricity as a function of m: Again, should we say the eccentricity of any line is 1? We’ll soon see a reason not to! As for the family of circles approaching a single point, we’ll see that below, too. A circle as a degenerate ellipse We’ll close with several questions about degenerate ellipses. Here is a simple little question from 1998: Graph the Ellipse... [((x-1)^2)/2^2] + [((y-2)^2)/2^2]=1 Graph the ellipse and find the major and minor axes, the foci, the eccentricity, and the center. The graph is actually not hard, but the rest are interesting! Doctor Jaffee answered: Hi Callie, I think your teacher might be trying to trick you with this one (assuming that this is a question you brought from your class). If you multiply both sides of the equation by 4 (that's the 2^2 in your equation), the denominators will cancel and the resulting equation will be (x-1)^2 + (y-2)^2 = 4. Any equation in the form (x-h)^2 + (y-k)^2 = r^2 is a circle whose center is at the point (h,k) and whose radius is r. The general equation of a circle is (x^2+y^2=r^2); written in the standard form for an ellipse, this is (\frac{x^2}{r^2}+\frac{y^2}{r^2}=1), so that semiaxes a and b are both equal to r. So, what you really have is a degenerate ellipse ("degenerate" is really the technical term for a geometric figure that has changed in a particular way to another, simpler form. I'm not trying to insult the ellipse). At any rate, you have a circle whose center is at the point (1,2) and whose radius is 2. The two foci have become the same point, the center, and neither axis through the center is major or minor. In fact, any diameter can be considered one of the axes. So, you could consider the two axes to be the one from (-1,2) to (3,2) and the other from (1,0) to (1,4). Here is a graph of this ellipse that is a circle: At first, you might not think of this as degenerate; it definitely is not a degenerate conic. But it has lost some of the properties of a general ellipse, hasn’t it? Rather than two foci, it has only one; rather than two axes, major and minor, it has infinitely many diameters, any of which might be called an axis, but none of which is “major”. And finally the question about the eccentricity. One definition of eccentricity is the ratio of the distance from the center to a focus and the distance from the center to an endpoint of the major axis. Since the focus is at the center, the distance from the center to the focus is zero, so the ratio is zero. Thus, the eccentricity is 0. This is what Doctor Vogler mentioned above. We have (a=b=2), so the focal distance is (c=\sqrt{a^2-b^2}=\sqrt{2^2-2^2}=0), and the eccentricity is (e=\frac{c}{a}=\frac{0}{2}=0). A degenerate circle can be a point or a line This is from 2001: Degenerate/Nondegenerate FigureWe need to know what a nondegenerate circle is. (We're trying to decide whether or not this is a model of incidence geometry, but we don't know what the definition is.) An exercise (at a somewhat higher level than we are at here) evidently mentions a “non-degenerate” circle, assuming students will know what that means. In order to answer, it’s necessary to explain what a degenerate circle is! Doctor Floor answered: Dear Jan, A circle is called degenerate if: - the length of the radius is zero, so the circle is a single point, - the length of the radius is infinite, so the circle is a straight line. In other cases the circle is nondegenerate. Other geometrical figures can also be degenerate. The degenerate figures always are very special cases, in which the figure is not what one would expect. For instance, a triangle is degenerate if the three vertices lie on a line (or even coincide). (We’ll look at triangles next week!) Here we can see a circle, on one hand, shrinking to a point, and on the other, expanding to a line: The point has equation (x^2+y^2=0), and the line has equation (y+2=0), both of which fit the general form for a conic. Incidentally, observe that these circles all have eccentricity 0; but can we say the line also has eccentricity 0, when we saw (above) reason to call it 1, when we made it as part of a family of parabolas? This is yet more evidence that degenerate conics don’t deserve to be assigned an eccentricity! It would be appropriate to call it indeterminate. Is a point a circle? One last question, from 2004: Circle With Radius of ZeroIs it possible for a circle to have a radius that equals zero? Is it possible for a set of points (e.g., multiple points) to occupy the same location? One textbook defines a circle as the set of points that is equidistant from a center point and that the radius is greater than or equal to zero. I dispute this. Am I wrong? There are two issues here: how a point can be considered to actually be a circle, and how a set of points can consist of only one point. I answered: Hi, Jack. If the radius is zero, then it isn't really a circle, but might be called a degenerate circle--that is, what you get if you slightly stretch the definition of a circle by using the same equation but taking it to extremes by making the radius zero. The point is (no pun intended!) that many things you can say about a circle will still be true if the radius is zero (making a single point), and they have for some reason chosen to allow that. I wouldn't do so, because there are too many other things that would no longer work in that case. I hope, for example, that in theorems about tangents to a circle they specify that the radius has to be greater than zero. If not, then they are inconsistent in their use of the definition, which is not uncommon in textbooks. This is why we generally don’t include degenerate cases within a category, but treat them separately, just as the boundary of a circle is distinct from its interior. A proper circle has one tangent at any point on it; a single point doesn’t have tangents at all, unless you stretch that definition, too! By the way, there is nothing wrong in talking about a set of points that consists only of one point; nothing in that wording should be taken to imply that there are multiple points. The problem with this definition lies in the difficulty of writing theorems based on it, not on how many points there are. It is quite common in mathematics to take a plural to include the possibility of only one. For example, a “polynomial”, whose name comes from Greek for “many names”, is defined as a “sum of terms consisting of a number times a positive integer power of the variables”; but that “sum” may consist of only one term, with no actual addition needed! This is akin to the discussion of inclusive definitions in Is a Square a Rectangle? Classifying Shapes, and the follow-up post, What Is a Trapezoid? More on Inclusive Definitions. Jack replied: Dear Dr. Peterson: Thank you very much for your thoughtful reply. Unfortunately, I lost an argument, but I appreciate knowing the truth even more than being "right." :-) I’m not sure which argument he lost, since I said the book was not really correct on one count, but he was wrong about the word “set”. The important thing, as he says, is to learn! Next week: Degenerate polygons. Leave a Comment Cancel Reply This site uses Akismet to reduce spam. Learn how your comment data is processed.
10029	https://en.m.wikipedia.org/wiki/Spherical_law_of_sines	Law of sines Article Talk (Redirected from Spherical law of sines) This article is about the law of sines in trigonometry. For the law of sines in physics, see Snell's law. In trigonometry, the law of sines (sometimes called the sine formula or sine rule) is a mathematical equation relating the lengths of the sides of any triangle to the sines of its angles. According to the law, where a, b, and c are the lengths of the sides of a triangle, and α, β, and γ are the opposite angles (see figure 2), while R is the radius of the triangle's circumcircle. When the last part of the equation is not used, the law is sometimes stated using the reciprocals; The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the triangle is not uniquely determined by this data (called the ambiguous case) and the technique gives two possible values for the enclosed angle. Law of sines Figure 1, with circumcircle Figure 2, without circumcircle Two triangles labelled with components of the law of sines. The angles α, β and γ are associated with the respective vertices A, B, and C; the respective sides of lengths a, b, and c are opposite these (e.g., side a is opposite vertex A with angle α). The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in scalene triangles, with the other being the law of cosines. The law of sines can be generalized to higher dimensions on surfaces with constant curvature. Contents 1 Proof 2 The ambiguous case of triangle solution 3 Examples 3.1 Example 1 3.2 Example 2 4 Relation to the circumcircle 4.1 Proof 4.2 Relationship to the area of the triangle 5 Spherical law of sines 5.1 Vector proof 5.2 Geometric proof 5.3 Other proofs 6 Hyperbolic case 7 The case of surfaces of constant curvature 8 Higher dimensions 9 History 10 See also 11 References 12 External links Proof edit With the side of length a as the base, the triangle's altitude can be computed as b sin γ or as c sin β. Equating these two expressions gives and similar equations arise by choosing the side of length b or the side of length c as the base of the triangle. The ambiguous case of triangle solution edit When using the law of sines to find a side of a triangle, an ambiguous case occurs when two separate triangles can be constructed from the data provided (i.e., there are two different possible solutions to the triangle). In the case shown below they are triangles ABC and ABC′. Given a general triangle, the following conditions would need to be fulfilled for the case to be ambiguous: The only information known about the triangle is the angle α and the sides a and c. The angle α is acute (i.e., α < 90°). The side a is shorter than the side c (i.e., a < c). The side a is longer than the altitude h from angle β, where h = c sin α (i.e., a > h). If all the above conditions are true, then each of angles β and β′ produces a valid triangle, meaning that both of the following are true: From there we can find the corresponding β and b or β′ and b′ if required, where b is the side bounded by vertices A and C and b′ is bounded by A and C′. Examples edit The following are examples of how to solve a problem using the law of sines. Example 1 edit Given: side a = 20, side c = 24, and angle γ = 40°. Angle α is desired. Using the law of sines, we conclude that Note that the potential solution α = 147.61° is excluded because that would necessarily give α + β + γ > 180°. Example 2 edit If the lengths of two sides of the triangle a and b are equal to x, the third side has length c, and the angles opposite the sides of lengths a, b, and c are α, β, and γ respectively then Relation to the circumcircle edit In the identity the common value of the three fractions is actually the diameter of the triangle's circumcircle. This result dates back to Ptolemy. Proof edit As shown in the figure, let there be a circle with inscribed and another inscribed that passes through the circle's center O. The has a central angle of and thus , by Thales's theorem. Since is a right triangle, where is the radius of the circumscribing circle of the triangle. Angles and lie on the same circle and subtend the same chord c; thus, by the inscribed angle theorem, . Therefore, Rearranging yields Repeating the process of creating with other points gives Relationship to the area of the triangle edit The area of a triangle is given by , where is the angle enclosed by the sides of lengths a and b. Substituting the sine law into this equation gives Taking as the circumscribing radius, It can also be shown that this equality implies where T is the area of the triangle and s is the semiperimeter The second equality above readily simplifies to Heron's formula for the area. The sine rule can also be used in deriving the following formula for the triangle's area: denoting the semi-sum of the angles' sines as , we have where is the radius of the circumcircle: . Spherical law of sines edit The spherical law of sines deals with triangles on a sphere, whose sides are arcs of great circles. Suppose the radius of the sphere is 1. Let a, b, and c be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere, a, b, and c are the angles at the center of the sphere subtended by those arcs, in radians. Let A, B, and C be the angles opposite those respective sides. These are dihedral angles between the planes of the three great circles. Then the spherical law of sines says: Vector proof edit Consider a unit sphere with three unit vectors OA, OB and OC drawn from the origin to the vertices of the triangle. Thus the angles α, β, and γ are the angles a, b, and c, respectively. The arc BC subtends an angle of magnitude a at the centre. Introduce a Cartesian basis with OA along the z-axis and OB in the xz-plane making an angle c with the z-axis. The vector OC projects to ON in the xy-plane and the angle between ON and the x-axis is A. Therefore, the three vectors have components: The scalar triple product, OA ⋅ (OB × OC) is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle OA, OB and OC. This volume is invariant to the specific coordinate system used to represent OA, OB and OC. The value of the scalar triple product OA ⋅ (OB × OC) is the 3 × 3 determinant with OA, OB and OC as its rows. With the z-axis along OA the square of this determinant is Repeating this calculation with the z-axis along OB gives (sin c sin a sin B)2, while with the z-axis along OC it is (sin a sin b sin C)2. Equating these expressions and dividing throughout by (sin a sin b sin c)2 gives where V is the volume of the parallelepiped formed by the position vector of the vertices of the spherical triangle. Consequently, the result follows. It is easy to see how for small spherical triangles, when the radius of the sphere is much greater than the sides of the triangle, this formula becomes the planar formula at the limit, since and the same for sin b and sin c. Geometric proof edit Consider a unit sphere with: Construct point and point such that Construct point such that It can therefore be seen that and Notice that is the projection of on plane . Therefore By basic trigonometry, we have: But Combining them we have: By applying similar reasoning, we obtain the spherical law of sines: See also: Spherical trigonometry, Spherical law of cosines, and Half-side formula Other proofs edit A purely algebraic proof can be constructed from the spherical law of cosines. From the identity and the explicit expression for from the spherical law of cosines Since the right hand side is invariant under a cyclic permutation of the spherical sine rule follows immediately. The figure used in the Geometric proof above is used by and also provided in Banerjee (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices. Hyperbolic case edit In hyperbolic geometry when the curvature is −1, the law of sines becomes In the special case when B is a right angle, one gets which is the analog of the formula in Euclidean geometry expressing the sine of an angle as the opposite side divided by the hypotenuse. See also: Hyperbolic triangle The case of surfaces of constant curvature edit Define a generalized sine function, depending also on a real parameter : The law of sines in constant curvature reads as By substituting , , and , one obtains respectively , , and , that is, the Euclidean, spherical, and hyperbolic cases of the law of sines described above. Let indicate the circumference of a circle of radius in a space of constant curvature . Then . Therefore, the law of sines can also be expressed as: This formulation was discovered by János Bolyai. Higher dimensions edit A tetrahedron has four triangular facets. The absolute value of the polar sine (psin) of the normal vectors to the three facets that share a vertex of the tetrahedron, divided by the area of the fourth facet will not depend upon the choice of the vertex: More generally, for an n-dimensional simplex (i.e., triangle (n = 2), tetrahedron (n = 3), pentatope (n = 4), etc.) in n-dimensional Euclidean space, the absolute value of the polar sine of the normal vectors of the facets that meet at a vertex, divided by the hyperarea of the facet opposite the vertex is independent of the choice of the vertex. Writing V for the hypervolume of the n-dimensional simplex and P for the product of the hyperareas of its (n − 1)-dimensional facets, the common ratio is Note that when the vectors v1, ..., vn, from a selected vertex to each of the other vertices, are the columns of a matrix V then the columns of the matrix are outward-facing normal vectors of those facets that meet at the selected vertex. This formula also works when the vectors are in a m-dimensional space having m > n. In the m = n case that V is square, the formula simplifies to History edit An equivalent of the law of sines, that the sides of a triangle are proportional to the chords of double the opposite angles, was known to the 2nd century Hellenistic astronomer Ptolemy and used occasionally in his Almagest. Statements related to the law of sines appear in the astronomical and trigonometric work of 7th century Indian mathematician Brahmagupta. In his Brāhmasphuṭasiddhānta, Brahmagupta expresses the circumradius of a triangle as the product of two sides divided by twice the altitude; the law of sines can be derived by alternately expressing the altitude as the sine of one or the other base angle times its opposite side, then equating the two resulting variants. An equation even closer to the modern law of sines appears in Brahmagupta's Khaṇḍakhādyaka, in a method for finding the distance between the Earth and a planet following an epicycle; however, Brahmagupta never treated the law of sines as an independent subject or used it systematically for solving triangles. The spherical law of sines is sometimes credited to 10th century scholars Abu-Mahmud Khujandi or Abū al-Wafāʾ (it appears in his Almagest), but it is given prominence in Abū Naṣr Manṣūr's Treatise on the Determination of Spherical Arcs, and was credited to Abū Naṣr Manṣūr by his student al-Bīrūnī in his Keys to Astronomy. Ibn Muʿādh al-Jayyānī's 11th-century Book of Unknown Arcs of a Sphere also contains the spherical law of sines. The 13th-century Persian mathematician Naṣīr al-Dīn al-Ṭūsī stated and proved the planar law of sines: In any plane triangle, the ratio of the sides is equal to the ratio of the sines of the angles opposite to those sides. That is, in triangle ABC, we have AB : AC = Sin(∠ACB) : Sin(∠ABC) By employing the law of sines, al-Tusi could solve triangles where either two angles and a side were known or two sides and an angle opposite one of them were given. For triangles with two sides and the included angle, he divided them into right triangles that he could then solve. When three sides were given, he dropped a perpendicular line and then used Proposition II-13 of Euclid's Elements (a geometric version of the law of cosines). Al-Tusi established the important result that if the sum or difference of two arcs is provided along with the ratio of their sines, then the arcs can be calculated. According to Glen Van Brummelen, "The Law of Sines is really Regiomontanus's foundation for his solutions of right-angled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles." Regiomontanus was a 15th-century German mathematician. See also edit Gersonides – Medieval Jewish philosopher Half-side formula – for solving spherical triangles Law of cosines Law of tangents Law of cotangents Mollweide's formula – for checking solutions of triangles Solution of triangles Surveying References edit ^ a b c "Generalized law of sines". mathworld. ^ Coxeter, H. S. M. and Greitzer, S. L. Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 1–3, 1967 ^ a b "Law of Sines". www.pballew.net. Archived from the original on December 29, 2002. Retrieved 2018-09-18. ^ Mr. T's Math Videos (2015-06-10), Area of a Triangle and Radius of its Circumscribed Circle, archived from the original on 2021-12-11, retrieved 2018-09-18 ^ Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109. ^ Banerjee, Sudipto (2004), "Revisiting Spherical Trigonometry with Orthogonal Projectors" (PDF), The College Mathematics Journal, 35 (5), Mathematical Association of America: 375–381, doi:10.1080/07468342.2004.11922099, archived from the original (PDF) on 2004-10-29, retrieved 2024-02-08 ^ Katok, Svetlana (1992). Fuchsian groups. Chicago: University of Chicago Press. p. 22. ISBN 0-226-42583-5. ^ Eriksson, Folke (1978). "The law of sines for tetrahedra and n-simplices". Geometriae Dedicata. 7 (1): 71–80. doi:10.1007/bf00181352. ^ Toomer, Gerald J., ed. (1998). Ptolemy's Almagest. Princeton University Press. pp. 7, fn. 10, 462, fn. 96. ^ Winter, Henry James Jacques (1952). Eastern Science. John Murray. p. 46. Colebrooke, Henry Thomas (1817). Algebra, with Arithmetic and Mensuration from the Sanscrit of Brahmegupta and Bhascara. London: John Murray. pp. 299–300. ^ Van Brummelen, Glen (2009). The Mathematics of the Heavens and the Earth. Princeton University Press. pp. 109–111. ISBN 978-0-691-12973-0. Brahmagupta (1934). The Khandakhadyaka: An Astronomical Treatise of Brahmagupta. Translated by Sengupta, Prabodh Chandra. University of Calcutta. ^ Sesiano, Jacques (2000). "Islamic mathematics". In Selin, Helaine; D'Ambrosio, Ubiratan (eds.). Mathematics Across Cultures: The History of Non-western Mathematics. Springer. pp. 137–157. ISBN 1-4020-0260-2. Van Brummelen, Glen (2009). The Mathematics of the Heavens and the Earth. Princeton University Press. pp. 183–185. ISBN 978-0-691-12973-0. ^ O'Connor, John J.; Robertson, Edmund F., "Abu Abd Allah Muhammad ibn Muadh Al-Jayyani", MacTutor History of Mathematics Archive, University of St Andrews ^ "Nasir al-Din al-Tusi - Biography". Maths History. Retrieved 2025-03-10. ^ Katz, Victor J. (2017-03-21). A History of Mathematics: An Introduction. Pearson. p. 315. ISBN 978-0-13-468952-4. ^ Van Brummelen, Glen (2009). The Mathematics of the Heavens and the Earth: The Early History of Trigonometry. Princeton University Press. p. 259. ISBN 978-0-691-12973-0. External links edit Wikimedia Commons has media related to Law of sines. "Sine theorem", Encyclopedia of Mathematics, EMS Press, 2001 The Law of Sines at cut-the-knot Degree of Curvature Finding the Sine of 1 Degree Generalized law of sines to higher dimensions Retrieved from "
10030	https://math.stackexchange.com/questions/4106783/angle-chasing-finding-the-anglenrq-in-a-quarter-circle	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Angle chasing: Finding the $\angle{NRQ}$ in a quarter circle Ask Question Asked Modified 4 years, 5 months ago Viewed 244 times 1 $\begingroup$ Let $MON$ be a quarter circle. $P$ is the midpoint of $OM$ and $PQ$ is the angle bisector of $\angle{OPN}$. If $QR\parallel{OM}$, find $\angle{QRN}$. Here is an angle-chasing problem which I am trying to solve. I couldn't make a good progress solving the problem. Though here is my approach: If we assume $\angle{OPQ}=\theta$ and $NP$ intersects $QR$ at $A$, then we have: $$\angle{OPQ}=\angle{NPQ}=\theta,$$ $$\angle{ONP}=\frac{\pi}{2}-\theta,$$ $$\angle{NAQ}=2\theta,$$ $$\angle{PQR}=\theta.$$So, $\Delta{PAQ}$ is isosceles. This doesn't help very much. How to solve the problem? contest-math euclidean-geometry circles recreational-mathematics angle Share asked Apr 18, 2021 at 9:29 user915158user915158 $\endgroup$ 3 1 $\begingroup$ Wow, nice problem. $\endgroup$ F Nishat – F Nishat 2021-04-18 09:32:27 +00:00 Commented Apr 18, 2021 at 9:32 3 $\begingroup$ c.f. Richmond's Method of regular pentagon construction: en.wikipedia.org/wiki/… $\endgroup$ player3236 – player3236 2021-04-18 12:35:49 +00:00 Commented Apr 18, 2021 at 12:35 $\begingroup$ @player3236 Good catch ! $\endgroup$ Jean Marie – Jean Marie 2021-04-18 13:07:53 +00:00 Commented Apr 18, 2021 at 13:07 Add a comment \| 2 Answers 2 Reset to default 1 $\begingroup$ Here is a solution which tries to not use the algebraic knowledge of some trigonometric function (sine, cosine, tangent) of the angle of $36^\circ$ or some related angle (half or double of it). However, since there are "only a few" angles in the picture which are in measure a rational multiple of $\pi$, mainly those in $R$, we still need to involve some algebraic part. The idea is to show that the points $N,R,M$ from the OP are among the vertices of the $20$-gon. I will also use for them alternative notations $X_0$, $X_3$, and respectively $X_4$. To start with, let $S$ be the reflection of $O$ w.r.t. $Q$, so $OQ=QS$. I will work with a figure where the radius of the circle is $$ 4 = OM=ON\ . $$ Then $PN=\sqrt{2^2+4^2}=2\sqrt 5$, and in $\Delta NOP$ the angle bisector theorem gives $OQ:QN=2:2\sqrt 5$, so $\displaystyle\frac{OQ}4=\frac{OQ}{OQ+QN}=\frac2{2\sqrt 5+2}=\frac1{\sqrt 5+1}$, so $\displaystyle OQ=\frac 4{\sqrt 5+1}=\sqrt 5-1$. Some further computations are: $$ \begin{aligned} QS &= OQ=\sqrt 5-1\ ,\ SN &= ON-2OQ=6-\sqrt 5=(\sqrt 5-1)^2\ ,\ QR^2 &=OR^2-OQ^2=16-(5-2\sqrt 5+1)=10+2\sqrt 5=2\sqrt 5(\sqrt 5+1)\ ,\ NR^2 &= QR^2+QN^2=(10+2\sqrt 5) + (5-\sqrt 5)^2=40-8\sqrt 5=8\sqrt 5(\sqrt 5-1)\ ,\text{ so}\ \frac{NR^2}{QR^2} & = \frac{8\sqrt 5(\sqrt 5-1)}{2\sqrt 5(\sqrt 5+1)} = \frac{4(\sqrt 5-1)}{(\sqrt 5+1)} =(\sqrt 5-1)^2 =\left(\frac{NS}{SQ}\right)^2\ . \end{aligned} $$ The reciprocal of the angle bisector theorem (its usage here being the idea of this proof), insures now that the line $RS$ bisects the angle in $R$ in $\Delta QRN$. Let $x$ be the measure of $\widehat{ROM}$. Then we have the following situation with further angles computed in terms of $x$, using parallelism and an isosceles triangle: We obtain an equation involving $x$ from $$ 3x +\frac 12(180^\circ-x)=\widehat{NRM}=\frac 12\cdot (360^\circ-\overset\frown {MN})=\frac 12\cdot 270^\circ\ . $$ We double, and solving for $x$ in $6x+(180^\circ-x)=270^\circ$, we get $$ \color{blue}{\boxed{ \qquad \begin{aligned} x &=\frac 15\cdot90^\circ=18^\circ\ ,\ \widehat {QRN}&=2x =36^\circ \ . \end{aligned} \qquad}} $$ $\square$ We are done, but it may be a good impression to also see the $20$-gon in the picture... Share answered Apr 28, 2021 at 11:27 dan_fuleadan_fulea 38.2k11 gold badge2626 silver badges6464 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ WLOG, we can consider the circle as the unit circle. Preliminary result: $PN=\dfrac{\sqrt{5}}{2}$ by Pythagoras in $\Delta OPN$. Using the bisector angle theorem, we have, for a certain $k$: $$\dfrac{QN}{QO}=\dfrac{PN}{PO}=\sqrt{5} \implies OQ=\dfrac{QN}{\sqrt{5}}$$ As $OQ+QN=1$ we must have $$OQ=\dfrac{1}{2 \Phi}$$ where $\Phi=\dfrac{1+\sqrt{5}}{2}$ is the Golden Ratio. As a consequence, using Pythagoras in triangle $\Delta OQR$: $$OQ^2+QR^2=OR^2 \ \implies QR = \sqrt{1-\dfrac{1}{4 \Phi^2}}=\dfrac{1}{2 \Phi}a$$ where $a:=\sqrt{4 \Phi^2-1}=\sqrt{2 \sqrt{5}+5}=\dfrac{\sqrt{5}}{\sqrt{-2 \sqrt{5}+5}}$ $$\tan \angle QRN =\dfrac{QN}{QR}=\dfrac{\sqrt{5}}{2 \Phi \dfrac{1}{2 \Phi}a}=\dfrac{\sqrt{5}}{a}=\sqrt{5-2 \sqrt{5}}=\tan \dfrac{\pi}{5}$$ (see here); therefore: $$\color{red}{\angle QRN = \dfrac{\pi}{5}}$$ Remark: I wouldn't be surprized that a simpler solution is found for example through a connection with the pentagram. Edit: In fact (see my remark) my intuition is verified. Thanks to @user3236 this issue is like proving the validity of Richmond's construction. Here is another site with complementary informations: Share edited Apr 18, 2021 at 13:32 answered Apr 18, 2021 at 10:56 Jean MarieJean Marie 90.3k77 gold badges5959 silver badges132132 bronze badges $\endgroup$ 0 Add a comment \| You must log in to answer this question. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 4 Trigonometry with multiple angle and exact value of $\tan\pi/5$ Related 1 Problem regarding the intersection of a circumscribed circle and an exterior angle bisector and the midpoint of an arc 4 In $\Delta ABC$ , $\angle A = 20^\circ, \angle C = 90^\circ$. $O$ is a point on $AB$ and $D$ is the midpoint of $OB$ . Find $\angle BCD$ . 4 Find the length of the segment $QA$ in acute-angled $\triangle ABC$ Prove that $\angle IPA=\angle IQB$ in $\triangle ABC$ with incenter $I$. 3 $ABCD$ is a convex quadrilateral. If $\angle BAC=10°$, $\angle CAD=40°$, $\angle ADB=50°$, $\angle BDC=20°$, then find $\angle CBD$. 3 Find the missing angles in the picture containing squares and circles 5 Show that $3$ lines must concur What's the measure of the angle OCB in the triangle below? 0 Find the $\angle FPA$ in the figure below Angle Bisectors and a Parallelogram Problem Hot Network Questions Why, really, do some reject infinite regresses? "Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf Verify a Chinese ID Number how do I remove a item from the applications menu Data lost/Corrupted on iCloud Can a Box have a lifetime less than 'static? Space Princess Space Tours: Black Holes merging - what would you visually see? How to use cursed items without upsetting the player? Can a cleric gain the intended benefit from the Extra Spell feat? Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? 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10031	https://proofwiki.org/wiki/Median_Formula	Median Formula From ProofWiki Jump to navigation Jump to search Contents 1 Theorem 2 Proof 1 3 Proof 2 4 Proof 3 5 Examples 5.1 Triangle $\tuple {1, -2}, \tuple {-3, 4}, \tuple {2, 2}$ 6 Sources Theorem Let $\triangle ABC$ be a triangle. Let $CD$ be the median of $\triangle ABC$ which bisects $AB$. The length $m_c$ of $CD$ is given by: : ${m_c}^2 = \dfrac {a^2 + b^2} 2 - \dfrac {c^2} 4$ Proof 1 \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| \| \| \| \| (\ds a^2 \cdot AD + b^2 \cdot DB) \| (=) \| \| \| \| (\ds CD^2 \cdot c + AD \cdot DB \cdot c) \| \| \| Stewart's Theorem \| \| \| \| \| (\ds \leadsto \ \ ) \| \| \| (\ds a^2 \frac c 2 + b^2 \frac c 2) \| (=) \| \| \| \| (\ds {m_c}^2 \cdot c + \paren {\frac c 2}^2 c) \| \| \| substituting $AD = DB = \dfrac c 2$ and $CD = m_c$ \| \| \| \| \| (\ds \leadsto \ \ ) \| \| \| (\ds \frac c 2 \paren {a^2 + b^2}) \| (=) \| \| \| \| (\ds m_c^2 \cdot c + \frac {c^2} 4 \cdot c) \| \| \| \| \| \| \| \| (\ds \leadsto \ \ ) \| \| \| (\ds \frac {a^2 + b^2} 2) \| (=) \| \| \| \| (\ds m_c^2 + \frac {c^2} 4) \| \| \| \| \| \| \| \| (\ds \leadsto \ \ ) \| \| \| (\ds {m_c}^2) \| (=) \| \| \| \| (\ds \frac {a^2 + b^2} 2 - \frac {c^2} 4) \| \| \| after algebra \| \| $\blacksquare$ Proof 2 Let $\triangle ABC$ be embedded in the complex plane. Let $\mathbf a = \overrightarrow {AC}$ and $\mathbf b = \overrightarrow {BC}$. Then: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| \| \| \| \| (\ds \overrightarrow {AB}) \| (=) \| \| \| \| (\ds \mathbf a - \mathbf b) \| \| \| \| \| \| \| \| \| \| \| (\ds \overrightarrow {AD}) \| (=) \| \| \| \| (\ds \dfrac {\overrightarrow {AB} } 2) \| \| \| \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \dfrac {\mathbf a - \mathbf b} 2) \| \| \| \| \| Then: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| \| \| \| \| (\ds \overrightarrow {AC} + \overrightarrow {CD}) \| (=) \| \| \| \| (\ds \overrightarrow {AD}) \| \| \| \| \| \| \| \| (\ds \leadsto \ \ ) \| \| \| (\ds \overrightarrow {CD}) \| (=) \| \| \| \| (\ds \overrightarrow {AD} - \overrightarrow {AC}) \| \| \| \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac {\mathbf a - \mathbf b} 2 - \mathbf a) \| \| \| \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds -\frac {\mathbf a + \mathbf b} 2) \| \| \| \| \| \| \| \| (\ds \leadsto \ \ ) \| \| \| (\ds {m_c}^2) \| (=) \| \| \| \| (\ds \size {\overrightarrow {CD} }^2) \| \| \| \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \size {-\frac {\mathbf a + \mathbf b} 2}^2) \| \| \| \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac 1 4 \paren {\paren {\mathbf a + \mathbf b} \cdot \paren {\mathbf a + \mathbf b} }) \| \| \| Dot Product of Vector with Itself \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac 1 4 \paren {\mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b + 2 \mathbf a \cdot \mathbf b}) \| \| \| Square of Sum of Vectors \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac 1 4 \paren {\mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b - \paren {\mathbf a - \mathbf b} \cdot \paren {\mathbf a - \mathbf b} + \mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b}) \| \| \| Square of Sum of Vectors \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac 1 4 \paren {2 \size {\mathbf a}^2 + 2 \size {\mathbf b}^2 - \size {\mathbf a - \mathbf b}^2}) \| \| \| Dot Product of Vector with Itself \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac 1 4 \paren {2 a^2 + 2 b^2 - c^2}) \| \| \| \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac {a^2 + b^2} 2 - \frac {c^2} 4) \| \| \| \| \| $\blacksquare$ Proof 3 \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| \| \| \| \| (\ds {m_c}^2) \| (=) \| \| \| \| (\ds b^2 + \paren {\frac c 2}^2 - 2 b \paren {\frac c 2} \cos A) \| \| \| Law of Cosines \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds b^2 + \frac {c^2} 4 - 2 b \paren {\frac c 2} \paren {\frac {b^2 + c^2 - a^2} {2 b c} }) \| \| \| Law of Cosines \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds b^2 + \frac {c^2} 4 - \frac {b^2 + c^2 - a^2} 2) \| \| \| \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac {a^2} 2 + \frac {b^2} 2 - \frac {c^2} 4) \| \| \| \| \| $\blacksquare$ Examples Triangle $\tuple {1, -2}, \tuple {-3, 4}, \tuple {2, 2}$ Consider the triangle $\triangle ABC$ whose vertices are: : $A = \tuple {1, -2}, B = \tuple {-3, 4}, C = \tuple {2, 2}$ The length of the median of $\triangle ABC$ which which bisects $AB$ is $\sqrt {10}$. Sources 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): median formula Retrieved from " Categories: Proven Results Median Formula Medians of Triangles Named Theorems Navigation menu Search
10032	https://www.niddk.nih.gov/health-information/endocrine-diseases/primary-hyperparathyroidism	Published Time: 2025-08-11 22:59:28 -04:00 Primary Hyperparathyroidism - NIDDK Skip to main content Due to current HHS and NIH restructuring, the information provided on niddk.nih.gov is not being updated. Please refer to nih.gov. An official website of the United States government Here’s how you know Here’s how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Menu Search for Information from NIDDK Entire Site Research & Funding Health Information News About NIDDK Search Home [x] Research & FundingExpand Research & Funding Current Funding Opportunities Research Programs & Contacts Human Subjects Research Funding Process Research Training & Career Development Funded Grants & Grant History Research Resources Research at NIDDK Technology Advancement & Transfer Meetings & Workshops [x] Health InformationExpand Health Information Health Topics Diabetes Digestive Diseases Kidney Disease Weight Management Liver Disease Urologic Diseases Endocrine Diseases Diet & Nutrition Blood Diseases Diagnostic Tests La información de la salud en español Health Statistics Clinical Trials For Health Professionals Diabetes Discoveries & Practice Blog Community Health & Outreach [x] NewsExpand News News Archive Meetings & Workshops Media Library Follow Us [x] About NIDDKExpand About NIDDK Meet the Director Offices & Divisions Staff Directory Budget & Legislative Information Advisory & Coordinating Committees Strategic Plans & Reports Research Areas FAQs Jobs at NIDDK Visit Us Contact Us NIDDK 75th Anniversary Home Health Information Endocrine Diseases Primary Hyperparathyroidism Related Topics Endocrine Diseases Acromegaly Adrenal Insufficiency & Addison's DiseaseShow child pages Definition & Facts Symptoms & Causes Diagnosis Treatment Eating, Diet, & Nutrition Clinical Trials Cushing's Syndrome Graves’ Disease Hashimoto's Disease Hyperthyroidism (Overactive Thyroid) Hypothyroidism (Underactive Thyroid) Multiple Endocrine Neoplasia Type 1 Thyroid Disease & Pregnancy Primary Hyperparathyroidism Prolactinoma National Hormone & Pituitary Program (NHPP): Information for People Treated with Pituitary Human Growth HormoneShow child pages Human Growth Hormone & Creutzfeldt-Jakob Disease Resource List Health Alert: Adrenal Crisis Causes Death in Some People Who Were Treated with hGH Primary Hyperparathyroidism On this page: What is primary hyperparathyroidism? What do the parathyroid glands do? How common is primary hyperparathyroidism? Who is more likely to develop primary hyperparathyroidism? What are the complications of primary hyperparathyroidism? What are the symptoms of primary hyperparathyroidism? What causes primary hyperparathyroidism? How do doctors diagnose primary hyperparathyroidism? What tests do doctors use to look for complications of primary hyperparathyroidism? How do doctors treat primary hyperparathyroidism? Should I change my diet if I have primary hyperparathyroidism? What is primary hyperparathyroidism? Primary hyperparathyroidism is a disorder of the parathyroid glands, four pea-sized glands located on or near the thyroid gland in the neck. “Primary” means this disorder begins in the parathyroid glands, rather than resulting from another health problem such as kidney failure. In primary hyperparathyroidism, one or more of the parathyroid glands is overactive. As a result, the gland makes too much parathyroid hormone (PTH). Too much PTH causes calcium levels in your blood to rise too high, which can lead to health problems such as bone thinning and kidney stones. Doctors usually catch primary hyperparathyroidism early through routine blood tests, before serious problems occur. View full-sized imageThe parathyroid glands are located on or near the thyroid gland in the neck. What do the parathyroid glands do? The parathyroid glands’ only purpose is to make PTH, which helps maintain the right balance of calcium in your body. PTH raises blood calcium levels by causing bone, where most of your body’s calcium is stored, to release calcium into the blood helping your intestines absorb calcium from food helping your kidneys hold on to calcium and return it to your blood instead of flushing it out in urine When the level of calcium in your blood falls too low, the parathyroid glands release just enough PTH to bring your blood calcium levels back to normal. You need calcium for good health. This mineral helps build bones and teeth and keep them strong. Calcium also helps your heart, muscles, and nerves work normally. Although their names are similar, the parathyroid glands and the thyroid gland are not related. How common is primary hyperparathyroidism? In the United States, about 100,000 people develop primary hyperparathyroidism each year.1 Primary hyperparathyroidism is one of the most common hormonal disorders. Who is more likely to develop primary hyperparathyroidism? Primary hyperparathyroidism most often affects people between age 50 and 60. Women are affected 3 to 4 times more often than men.1 The disorder was more common in African Americans, followed by Caucasians, in one large study performed in North America.1 What are the complications of primary hyperparathyroidism? Primary hyperparathyroidism most often affects the bones and kidneys, although it also may play a part in other health problems. Weakened bones High PTH levels trigger the bones to release more calcium than normal into the blood. The loss of calcium from the bones may weaken them. Kidney stones The small intestine may absorb more calcium from food, adding to high levels of calcium in your blood. Extra calcium that isn’t used by your bones and muscles goes to your kidneys and is flushed out in urine. Too much calcium in your urine can cause kidney stones. Other complications High blood calcium levels might play a part in other problems, such as heart disease, high blood pressure, and trouble concentrating. However, more research is needed to better understand how primary hyperparathyroidism affects the heart, blood vessels, and brain. What are the symptoms of primary hyperparathyroidism? Most people with primary hyperparathyroidism have no symptoms. When symptoms appear, they’re often mild and similar to those of many other disorders. Symptoms include muscle weakness fatigue depressionNIH external link aches and pains in bones and joints View full-sized imageSymptoms of primary hyperparathyroidism People with more severe disease may have loss of appetite nausea vomiting constipation confusion increased thirst and urination What causes primary hyperparathyroidism? In about 8 out of 10 people with primary hyperparathyroidism, a benign, or noncancerous, tumor called an adenoma has formed in one of the parathyroid glands.2 The tumor causes the gland to become overactive. In most other cases, extra PTH comes from two or more adenomas or from hyperplasia, a condition in which all four parathyroid glands are enlarged. People with rare inherited conditions that affect the parathyroid glands, such as multiple endocrine neoplasia type 1 or familial hypocalciuric hypercalcemiaNIH external link, are more likely to have more than one gland affected. Rarely, primary hyperparathyroidism is caused by cancer of a parathyroid gland. How do doctors diagnose primary hyperparathyroidism? Doctors diagnose primary hyperparathyroidism when a blood test shows high blood calcium and PTH levels. Sometimes PTH levels are in the upper portion of the normal range, when they should drop to low-normal or below normal in response to high calcium levels. Other conditions can cause high calcium, but elevated PTH is the only source in primary hyperparathyroidism. Routine blood tests can detect high blood calcium levels. High blood calcium may cause health care professionals to suspect hyperparathyroidism, even before symptoms appear. Sometimes PTH levels are high but calcium levels are not. Doctors don’t routinely test for PTH but may do so if you have osteoporosisNIH external link or another disorder that affects bone strength. In some cases, this may be the first phase of primary hyperparathyroidism, before calcium levels start to rise. Once doctors diagnose hyperparathyroidism, a 24-hour urine collection can help find the cause. This test measures certain chemicals, such as calcium and creatinine, a waste product that healthy kidneys remove. You will collect your urine over a 24-hour period and your health care professional will send it to a lab for analysis. Results of the test may help tell primary hyperparathyroidism from hyperparathyroidism caused by a kidney disorder. The test can also rule out familial hypocalciuric hypercalcemia, a rare genetic disorder, as a cause. What tests do doctors use to look for complications of primary hyperparathyroidism? Once doctors diagnose primary hyperparathyroidism, they may use other tests to look for bone weakness, kidney problems, and low levels of vitamin D. Bone mineral density test Dual energy x-ray absorptiometry, also called a DXA or DEXA scan, uses low-dose x-rays to measure bone densityNIH external link. During the test, you will lie on a padded table while a technician moves the scanner over your body. A bone expert or radiologist will read the scan. During a DXA scan, you will lie on a padded table while a technician moves the scanner over your body. Kidney imaging tests Doctors may use one of the following imaging tests to look for kidney stones. Ultrasound.Ultrasound uses a device called a transducer that bounces safe, painless sound waves off organs to create an image of their structure. A specially trained technician does the procedure. A radiologist reads the images, which can show kidney stones. Abdominal x-ray. An abdominal x-ray is a picture of the abdomen that uses low levels of radiation and is recorded on film or on a computer. During an abdominal x-ray, you lie on a table or stand up. A technician positions the x-ray machine close to your abdomen and asks you to hold your breath so the picture won’t be blurry. A radiologist reads the x-ray, which can show the location of kidney stones in the urinary tract. Not all stones are visible on an abdominal x-ray. Computed tomography (CT) scans.CT scans use a combination of x-rays and computer technology to create images of your urinary tract. CT scans sometimes use a contrast medium—a dye or other substance that makes structures inside your body easier to see. Contrast medium isn’t usually needed to see kidney stones. For the scan, you’ll lie on a table that slides into a tunnel-shaped machine that takes the x-rays. A radiologist reads the images, which can show the size and location of a kidney stone. Vitamin D blood test Health care professionals test for vitamin D levels because low levels are common in people with primary hyperparathyroidism. In patients with primary hyperparathyroidism, the low vitamin D level can further stimulate the parathyroid glands to make even more parathyroid hormone. Also, a very low vitamin D level may cause a secondary form of hyperparathyroidism, which resolves when vitamin D levels are returned to normal. How do doctors treat primary hyperparathyroidism? Guidelines help doctors to decide whether or not parathyroid surgery should be recommended. You might be a candidate for surgery if you meet any of these guidelines Blood calcium > 1 mg/dL above normal Bone density by DXA < -2.5 at any site (lumbar spine, hip, or forearm) History of kidney stones or evidence of kidney stones or calcifications in the kidney by imaging (e.g., X-ray, ultrasound, CT scan). Evidence for stone risk by 24-hour urine with excessive calcium and other stone risk factors. A fracture resulting from relatively little force, such as a fall from a standing or sitting position (a fragility fracture) Age < 50 Doctors most often recommend parathyroid surgery, particularly if the patient meets one or more of the guidelines noted above. It is also not inappropriate to recommend surgery in those who do not meet guidelines as long as there are no medical contraindications to surgery. In those who do not meet guidelines or do not choose surgery, the doctor will monitor the patient’s condition. If there is evidence for progressive disease (e.g., higher calcium level, lower bone density, a fracture, kidney stone), surgery would be advised. For patients who are not going to have parathyroid surgery, even though guidelines are met, doctors can prescribe medicines to control the high blood calcium or improve the bone density. Surgery Surgery to remove the overactive parathyroid gland or glands is the only sure way to cure primary hyperparathyroidism. Doctors recommend surgery for people with clear symptoms or complications of the disease. In people without symptoms, doctors follow the above guidelines to identify who might benefit from parathyroid surgery.2 Surgery can lead to improved bone density and can lower the chance of forming kidney stones. When performed by experienced surgeons, surgery almost always cures primary hyperparathyroidism. Surgeons often use imaging tests before surgery to locate the overactive gland or glands to be removed. The tests used most often are sestamibi, ultrasound, and CT scans. In a sestamibi scan, you will get an injection, or shot, of a small amount of radioactive dye in your vein. The overactive parathyroid gland or glands then absorb the dye. The surgeon can see where the dye has been absorbed by using a special camera. Surgeons use two main types of operations to remove the overactive gland or glands. Minimally invasive parathyroidectomy.Also called focused parathyroidectomy, surgeons use this type of surgery when they think only one of the parathyroid glands is overactive. Guided by a tumor-imaging test, your surgeon will make a small incision, or cut, in your neck to remove the gland. The small incision means you will probably have less pain and a faster recovery than people who have more invasive surgery. You can go home the same day. Your doctor may use regional or general anesthesiaNIH external link during the surgery. Bilateral neck exploration.This type of surgery uses a larger incision that lets the surgeon find and look at all four parathyroid glands and remove the overactive ones. If you have bilateral neck exploration, you will probably have general anesthesia and may need to stay in the hospital overnight. When performed by experienced surgeons, surgery almost always cures primary hyperparathyroidism. Parathyroid surgery is safe. Rarely, problems can occur after surgery. In about 1 out of every 100 people, the nerves controlling the vocal cords are damaged during surgery, which most often results in hoarseness.3 This condition usually gets better on its own. Low calcium levels in the blood may occur after surgery but usually return to normal in a few days or weeks. On rare occasions, not enough parathyroid tissue is left to make PTH, which can result in hypoparathyroidismNIH external link. Monitoring Some people who have mild primary hyperparathyroidism may not need surgery right away, or even any surgery, and can be safely monitored. You may want to talk with your doctor about long-term monitoring if you don’t have symptoms have only slightly high blood calcium levels have normal kidneys and bone density Long-term monitoring should include regular doctor visits, a yearly blood test to measure calcium levels and check your kidney function, and a bone density test every 1 to 2 years. If you and your doctor choose long-term monitoring, you should drink plenty of water so you don’t get dehydrated get regular physical activityExternal link to help keep your bones strong avoid certain diuretics, such as thiazides Medicines Cinacalcet is a medicine that decreases the amount of PTH the parathyroid glands make and lowers calcium levels in the blood. Doctors may prescribe cinacalcet to treat very high calcium levels in people with primary hyperparathyroidism who can’t have surgery. Cinacalcet does not improve bone density. If you have bone loss, your doctor may prescribe alendronateNIH external link or other medications to help increase bone density. Should I change my diet if I have primary hyperparathyroidism? You don’t need to change your diet or limit the amount of calcium you get from food and beverages. You will need to take a vitamin D supplement if your vitamin D levels are low. Talk with your health care professional about how much vitamin D you should take. If you lose all your healthy parathyroid tissue and develop lasting low-calcium levels, you’ll need to take both calcium and vitamin D for life. References Bilezikian JP. Primary hyperparathyroidism. In: De Groot LJ, Chrousos G, Dungan K, et al., eds. Diseases of Bone and Mineral Metabolism. South Dartmouth, MA: MDTEXT.COM, Inc. 2000-2018. external link. Updated January 15, 2017. Accessed June 6, 2018. Bilezikian JP, Brandi ML, Eastell R, et al. Guidelines for the management of asymptomatic primary hyperparathyroidism: summary statement from the Fourth International Workshop. The Journal of Clinical Endocrinology and Metabolism. 2014;99(10):3561–3569. Callender GG, Udelsman R. Surgery for primary hyperparathyroidism. Cancer. 2014;120(23):3602–3616. Last Reviewed March 2019 Share this page PrintFacebookXEmailMore OptionsWhatsAppLinkedInRedditPinterestCopy Link This content is provided as a service of the National Institute of Diabetes and Digestive and Kidney Diseases (NIDDK), part of the National Institutes of Health. NIDDK translates and disseminates research findings to increase knowledge and understanding about health and disease among patients, health professionals, and the public. Content produced by NIDDK is carefully reviewed by NIDDK scientists and other experts. The NIDDK would like to thank: John P. 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10033	https://www.lehman.edu/faculty/anchordoqui/chapter21.pdf	Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis 21.1 Introduction ........................................................................................................... 1 21.2 Translational Equation of Motion ....................................................................... 1 21.3 Translational and Rotational Equations of Motion ........................................... 1 21.3.1 Summary ......................................................................................................... 6 21.4 Translation and Rotation of a Rigid Body Undergoing Fixed Axis Rotation . 6 21.5 Work-Energy Theorem ........................................................................................ 7 21.6 Worked Examples ................................................................................................. 8 Example 21.1 Angular Impulse ............................................................................... 8 Example 21.2 Person on a railroad car moving in a circle .................................... 9 Example 21.3 Torque, Rotation and Translation: Yo-Yo ................................... 13 Example 21.4 Cylinder Rolling Down Inclined Plane ......................................... 16 Example 21.5 Bowling Ball ..................................................................................... 21 Example 21.6 Rotation and Translation Object and Stick Collision ................. 25 21-1 Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis Accordingly, we find Euler and D'Alembert devoting their talent and their patience to the establishment of the laws of rotation of the solid bodies. Lagrange has incorporated his own analysis of the problem with his general treatment of mechanics, and since his time M. Poinsôt has brought the subject under the power of a more searching analysis than that of the calculus, in which ideas take the place of symbols, and intelligent propositions supersede equations. 1 James Clerk Maxwell 21.1 Introduction We shall analyze the motion of systems of particles and rigid bodies that are undergoing translational and rotational motion about a fixed direction. Because the body is translating, the axis of rotation is no longer fixed in space. We shall describe the motion by a translation of the center of mass and a rotation about the center of mass. By choosing a reference frame moving with the center of mass, we can analyze the rotational motion separately and discover that the torque about the center of mass is equal to the change in the angular momentum about the center of mass. For a rigid body undergoing fixed axis rotation about the center of mass, our rotational equation of motion is similar to one we have already encountered for fixed axis rotation,  τcm ext = d  Lcm spin / dt . 21.2 Translational Equation of Motion We shall think about the system of particles as follows. We treat the whole system as a single point-like particle of mass mT located at the center of mass moving with the velocity of the center of mass  Vcm . The external force acting on the system acts at the center of mass and from our earlier result (Eq. 10.4.9) we have that  Fext = d psys dt = d dt (mT  Vcm). (21.2.1) 21.3 Translational and Rotational Equations of Motion For a system of particles, the torque about a point S can be written as 1 J. C. Maxwell on Louis Poinsôt (1777-1859) in 'On a Dynamical Top' (1857). In W. D. Niven (ed.), The Scientific Papers of James Clerk Maxwell (1890), Vol. 1, 248. 21-2  τS ext = ( ri ×  Fi) i=1 N ∑ . (21.3.1) where we have assumed that all internal torques cancel in pairs. Let choose the point S to be the origin of the reference frame O, then  rS,cm =  Rcm (Figure 21.1). (You may want to recall the main properties of the center of mass reference frame by reviewing Chapter 15.2.1.) Figure 21.1 Torque diagram for center of mass reference frame We can now apply  ri =  Rcm +  r cm,i to Eq. (21.3.1) yielding  τS ext = (( rS,cm +  r cm,i) ×  Fi) i=1 N ∑ = ( rS,cm ×  Fi) i=1 N ∑ + ( r cm,i ×  Fi) i=1 N ∑ . (21.3.2) The term  τS,cm ext =  rS,cm ×  Fext (21.3.3) in Eq. (21.3.2) corresponds to the external torque about the point S where all the external forces act at the center of mass (Figure 21.2). Figure 21.2 Torque diagram for “point-like” particle located at center of mass The term, 21-3  τcm ext = ( r cm,i ×  Fi) i=1 N ∑ . (21.3.4) is the sum of the torques on the individual particles in the center of mass reference frame. If we assume that all internal torques cancel in pairs, then  τcm ext = ( r cm,i ×  Fi ext) i=1 N ∑ . (21.3.5) We conclude that the external torque about the point S can be decomposed into two pieces,  τS ext =  τS,cm ext +  τcm ext . (21.3.6) We showed in Chapter 20.3 that  LS sys =  rS,cm ×  psys + ( rcm,i × mi  vcm,i) i=1 N ∑ , (21.3.7) where the first term in Eq. (21.3.7) is the orbital angular momentum of the center of mass about the point S  LS orbital =  rS,cm ×  psys , (21.3.8) and the second term in Eq. (21.3.7) is the spin angular momentum about the center of mass (independent of the point S )  LS spin = ( rcm,i × mi  vcm,i) i=1 N ∑ . (21.3.9) The angular momentum about the point S can therefore be decomposed into two terms  LS sys =  LS orbital +  LS spin . (21.3.10) Recall that that we have previously shown that it is always true that  τS ext = d  LS sys dt . (21.3.11) Therefore we can therefore substitute Eq. (21.3.6) on the LHS of Eq. (21.3.11) and substitute Eq. (21.3.10) on the RHS of Eq. (21.3.11) yielding as 21-4  τS,cm ext +  τcm ext = d  LS orbital dt + d  LS spin dt . (21.3.12) We shall now show that Eq. (21.3.12) can also be decomposed in two separate conditions. We begin by analyzing the first term on the RHS of Eq. (21.3.12). We differentiate Eq. (21.3.8) and find that d  LS orbital dt = d dt ( rS,cm ×  psys). (21.3.13) We apply the vector identity d dt (  A ×  B) = d  A dt ×  B +  A × d  B dt , (21.3.14) to Eq. (21.3.13) yielding d  LS orbital dt = d rS,cm dt ×  psys +  rS,cm × d psys dt . (21.3.15) The first term in Eq. (21.3.21) is zero because d rS,cm dt ×  psys =  Vcm × mtotal  Vcm =  0. (21.3.16) Therefore the time derivative of the orbital angular momentum about a point S , Eq. (21.3.15), becomes d  LS orbital dt =  rS,cm × d psys dt . (21.3.17) In Eq. (21.3.17), the time derivative of the momentum of the system is the external force,  Fext = d psys dt . (21.3.18) The expression in Eq. (21.3.17) then becomes the first of our relations d  LS orbital dt =  rS,cm ×  Fext =  τS,cm ext . (21.3.19) Thus the time derivative of the orbital angular momentum about the point S is equal to the external torque about the point S where all the external forces act at the center of mass, (we treat the system as a point-like particle located at the center of mass). 21-5 We now consider the second term on the RHS of Eq. (21.3.12), the time derivative of the spin angular momentum about the center of mass. We differentiate Eq. (21.3.9), d  LS spin dt = d dt ( rcm,i × mi  vcm,i) i=1 N ∑ . (21.3.20) We again use the product rule for taking the time derivatives of a vector product (Eq. (21.3.14)). Then Eq. (21.3.20) the becomes d  LS spin dt = d rcm,i dt × mi  vcm,i ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ i=1 N ∑ +  rcm,i × d dt (mi  vcm,i) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ i=1 N ∑ . (21.3.21) The first term in Eq. (21.3.21) is zero because d rcm,i dt × mi  vcm,i ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ i=1 N ∑ = ( vcm,i × mi  vcm,i) i=1 N ∑ =  0 . (21.3.22) Therefore the time derivative of the spin angular momentum about the center of mass, Eq. (21.3.21), becomes d  LS spin dt =  rcm,i × d dt (mi  vcm,i) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ i=1 N ∑ . (21.3.23) The force, acting on an element of mass mi , is  Fi = d dt (mi  vcm,i) . (21.3.24) The expression in Eq. (21.3.23) then becomes d  LS spin dt = ( rcm,i ×  Fi) i=1 N ∑ . (21.3.25) The term, ( r cm,i ×  Fi) i=1 N ∑ , is the sum of the torques on the individual particles in the center of mass reference frame. If we again assume that all internal torques cancel in pairs, Eq. (21.3.25) may be expressed as d  LS spin dt = ( rcm,i ×  Fi ext) i=1 N ∑ =  τcm,i ext i=1 N ∑ =  τcm ext , (21.3.26) which is the second of our two relations. 21-6 21.3.1 Summary For a system of particles, there are two conditions that always hold (Eqs. (21.3.19) and (21.3.26)) when we calculate the torque about a point S ; we treat the system as a point-like particle located at the center of mass of the system. All the external forces  Fext act at the center of mass. We calculate the orbital angular momentum of the center of mass and determine its time derivative and then apply  τS,cm ext =  rS,cm ×  Fext = d  LS orbital dt . (21.3.27) In addition, we calculate the torque about the center of mass due to all the forces acting on the particles in the center of mass reference frame. We calculate the time derivative of the angular momentum of the system with respect to the center of mass in the center of mass reference frame and then apply  τcm ext = ( rcm,i ×  Fi ext) i=1 N ∑ = d  Lcm spin dt . (21.3.28) 21.4 Translation and Rotation of a Rigid Body Undergoing Fixed Axis Rotation For the special case of rigid body of mass m, we showed that with respect to a reference frame in which the center of mass of the rigid body is moving with velocity  Vcm , all elements of the rigid body are rotating about the center of mass with the same angular velocity  ωcm . For the rigid body of mass m and momentum  p = m  Vcm , the translational equation of motion is still given by Eq. (21.2.1), which we repeat in the form  Fext = m  Acm . (21.4.1) For fixed axis rotation, choose the z -axis as the axis of rotation that passes through the center of mass of the rigid body. We have already seen in our discussion of angular momentum of a rigid body that the angular momentum does not necessary point in the same direction as the angular velocity. However we can take the z -component of Eq. (21.3.28) τ cm,z ext = dLcm,z spin dt . (21.4.2) For a rigid body rotating about the center of mass with  ωcm = ωcm,z ˆ k , the z -component of angular momentum about the center of mass is 21-7 Lcm,z spin = Icmωcm,z . (21.4.3) The z -component of the rotational equation of motion about the center of mass is τ cm,z ext = Icm dωcm,z dt = Icmαcm,z . (21.4.4) 21.5 Work-Energy Theorem For a rigid body, we can also consider the work-energy theorem separately for the translational motion and the rotational motion. Once again treat the rigid body as a point-like particle moving with velocity  Vcm in reference frame O . We can use the same technique that we used when treating point particles to show that the work done by the external forces is equal to the change in kinetic energy Wtrans ext =  Fext ⋅d r i f ∫ = d(m  Vcm) dt ⋅d  Rcm i f ∫ = m d(  Vcm) dt ⋅  Vcm dt i f ∫ = 1 2 m d(  Vcm ⋅  Vcm) i f ∫ = 1 2 mVcm, f 2 −1 2 mVcm, i 2 = ΔKtrans. (21.5.1) For the rotational motion we go to the center of mass reference frame and we determine the rotational work done i.e. the integral of the z -component of the torque about the center of mass with respect to dθ as we did for fixed axis rotational work. Then ( τcm ext)z dθ i f ∫ = Icm dωcm,z dt dθ i f ∫ = Icm dωcm,z dθ dt i f ∫ = Icm dωcm,zωcm,z i f ∫ = 1 2 Icmωcm, f 2 −1 2 Icmωcm, i 2 = ΔKrot . (21.5.2) In Eq. (21.5.2) we expressed our result in terms of the angular speed ωcm because it appears as a square. Therefore we can combine these two separate results, Eqs. (21.5.1) and (21.5.2), and determine the work-energy theorem for a rotating and translating rigid body that undergoes fixed axis rotation about the center of mass. W = 1 2 mVcm,f 2 + 1 2 Icmωcm,f 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟−1 2 mVcm,f 2 + 1 2 Icmωcm,i 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ΔKtrans + ΔKrot = ΔK. (21.5.3) Equations (21.4.1), (21.4.4), and (21.5.3) are principles that we shall employ to analyze the motion of a rigid bodies undergoing translation and fixed axis rotation about the center of mass. 21-8 21.6 Worked Examples Example 21.1 Angular Impulse Two point-like objects are located at the points A and B, of respective masses M A = 2M , and M B = M , as shown in the figure below. The two objects are initially oriented along the y-axis and connected by a rod of negligible mass of length D , forming a rigid body. A force of magnitude F = F  along the x direction is applied to the object at B at t = 0 for a short time interval Δt , (Figure 21.3). Neglect gravity. Give all your answers in terms of M and D as needed. What is the magnitude of the angular velocity of the system after the collision? Figure 21.3 Example 21.1 Solutions: An impulse of magnitude F t Δ is applied in the x + direction, and the center of mass of the system will move in this direction. The two masses will rotate about the center of mass, counterclockwise in the figure. Before the force is applied we can calculate the position of the center of mass (Figure 21.4a),  Rcm = M A  rA + M B  rB M A + M B = 2M(D / 2)ˆ j+ M(D / 2)(−ˆ j) 3M = (D / 6)ˆ j. (21.6.1) The center of mass is a distance (2/3)D from the object at B and is a distance (1/3)D from the object at A. 21-9 (a) (b) Figure 21.4 (a) Center of mass of system, (b) Angular momentum about point B Because FΔtˆ i = 3M  Vcm , the magnitude of the velocity of the center of mass is then FΔt / 3M and the direction is in the positive ˆ i -direction. Because the force is applied at the point B, there is no torque about the point B, hence the angular momentum is constant about the point B. The initial angular momentum about the point B is zero. The angular momentum about the point B (Figure 21.4b) after the impulse is applied is the sum of two terms,  0 =  LB, f =  rB, f × 3M  Vcm +  Lcm = (2D / 3)ˆ j× FΔt ˆ i +  Lcm  0 = (2DFΔt / 3)(−ˆ k) +  Lcm. (21.6.2) The angular momentum about the center of mass is given by .  Lcm = Icmω ˆ k = (2M(D / 3)2 + M(2D / 3)2)ω ˆ k = (2 / 3)MD2ω ˆ k . (21.6.3) Thus the angular about the point B after the impulse is applied is  0 = (2DFΔt / 3)(−ˆ k) + (2 / 3)MD2ω ˆ k (21.6.4) We can solve this Eq. (21.6.4) for the angular speed ω = FΔt MD . (21.6.5) Example 21.2 Person on a railroad car moving in a circle A person of mass M is standing on a railroad car, which is rounding an unbanked turn of radius R at a speed v. His center of mass is at a height of L above the car midway 21-10 between his feet, which are separated by a distance of d . The man is facing the direction of motion (Figure 21.5). What is the magnitude of the normal force on each foot? Figure 21.5 Example 21.2 Solution: We begin by choosing a cylindrical coordinate system and drawing a free-body force diagram, shown in Figure 21.6. Figure 21.6 Coordinate system for Example 21.2 We decompose the contact force between the inner foot closer to the center of the circular motion and the ground into a tangential component corresponding to static friction  f1 and a perpendicular component,  N1. In a similar fashion we decompose the contact force between the outer foot further from the center of the circular motion and the ground into a tangential component corresponding to static friction  f2 and a perpendicular component,  N2 . We do not assume that the static friction has its maximum magnitude nor do we assume that  f1 =  f2 or  N1 =  N2 . The gravitational force acts at the center of mass. We shall use our two dynamical equations of motion, Eq. (21.4.1) for translational motion and Eq. (21.4.4) for rotational motion about the center of mass noting that we are considering the special case that  αcm = 0 because the object is not rotating about the center of mass. In order to apply Eq. (21.4.1), we treat the person as a point-like particle located at the center of mass and all the external forces act at this point. The radial component of Newton’s Second Law (Eq. (21.4.1) is given by ˆ r :−f1 −f2 = −m v2 R . (21.6.6) 21-11 The vertical component of Newton’s Second Law is given by ˆ k : N1 + N2 −mg = 0. (21.6.7) The rotational equation of motion (Eq. (21.4.4) is  τcm total = 0. (21.6.8) We begin our calculation of the torques about the center of mass by noting that the gravitational force does not contribute to the torque because it is acting at the center of mass. We draw a torque diagram in Figure 21.7a showing the location of the point of application of the forces, the point we are computing the torque about (which in this case is the center of mass), and the vector  r cm,1 from the point we are computing the torque about to the point of application of the forces. (a) (b) Figure 21.7 Torque diagram for (a) inner foot, (b) outer foot The torque on the inner foot is given by  τcm,1 =  r cm,1 × (  f1 +  N1) = −d 2 ˆ r −Lˆ k ⎛ ⎝ ⎜ ⎞ ⎠ ⎟× (−f1ˆ r + N1ˆ k) = d 2 N1 + Lf1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ˆ θ . (21.6.9) We draw a similar torque diagram (Figure 21.7b) for the forces applied to the outer foot. The torque on the outer foot is given by  τcm,2 =  r cm,2 × (  f2 +  N2) = + d 2 ˆ r −Lˆ k ⎛ ⎝ ⎜ ⎞ ⎠ ⎟× (−f2ˆ r + N2 ˆ k) = −d 2 N2 + Lf2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ˆ θ . (21.6.10) Notice that the forces  f1 ,  N1 , and  f2 all contribute torques about the center of mass in the positive ˆ θ -direction while  N2 contributes a torque about the center of mass in the negative ˆ θ -direction. According to Eq. (21.6.8) the sum of these torques about the center of mass must be zero. Therefore 21-12  τcm ext =  τcm,1 +  τcm,2 = d 2 N1 + Lf1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ˆ θ + −d 2 N2 + Lf2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ˆ θ = d 2 (N1 −N2) + L( f1 + f2) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ˆ θ =  0. (21.6.11) Notice that the magnitudes of the two frictional forces appear together as a sum in Eqs. (21.6.11) and (21.6.6). We now can solve Eq. (21.6.6) for f1 + f2 and substitute the result into Eq. (21.6.11) yielding the condition that d 2 (N1 −N2) + Lm v2 R = 0 . (21.6.12) We can rewrite this Eq. as N2 −N1 = 2Lmv2 Rd . (21.6.13) We also rewrite Eq. (21.6.7) in the form N2 + N1 = mg . (21.6.14) We now can solve for N2 by adding together Eqs. (21.6.13) and (21.6.14), and then divide by two, N2 = 1 2 Mg + 2Lmv2 Rd ⎛ ⎝ ⎜ ⎞ ⎠ ⎟. (21.6.15) We now can solve for N1 by subtracting Eqs. (21.6.13) from (21.6.14), and then divide by two, N1 = 1 2 mg −2Lmv2 Rd ⎛ ⎝ ⎜ ⎞ ⎠ ⎟. (21.6.16) Check the result: we see that the normal force acting on the outer foot is greater in magnitude than the normal force acting on the inner foot. We expect this result because as we increase the speed v , we find that at a maximum speed vmax , the normal force on the inner foot goes to zero and we start to rotate in the positive ˆ θ -direction, tipping outward. We can find this maximum speed by setting N1 = 0 in Eq. (21.6.16) resulting in vmax = gRd 2L . (21.6.17) 21-13 Example 21.3 Torque, Rotation and Translation: Yo-Yo A Yo-Yo of mass m has an axle of radius b and a spool of radius R . Its moment of inertia about the center can be taken to be Icm = (1/ 2)mR2 and the thickness of the string can be neglected (Figure 21.8). The Yo-Yo is released from rest. You will need to assume that the center of mass of the Yo-Yo descends vertically, and that the string is vertical as it unwinds. (a) What is the tension in the cord as the Yo-Yo descends? (b) What is the magnitude of the angular acceleration as the yo-yo descends and the magnitude of the linear acceleration? (c) Find the magnitude of the angular velocity of the Yo-Yo when it reaches the bottom of the string, when a length l of the string has unwound. Figure 21.8 Example 21.3 Figure 21.9 Torque diagram for Yo-Yo Solutions: a) as the Yo-Yo descends it rotates clockwise in Figure 21.9. The torque about the center of mass of the Yo-Yo is due to the tension and increases the magnitude of the angular velocity. The direction of the torque is into the page in Figure 21.9 (positive z -direction). Use the right-hand rule to check this, or use the vector product definition of torque, cm cm,T τ = × r T    . (21.6.18) About the center of mass, cm, ˆ T b = − r i  and ˆ T = − T j  , so the torque is  τcm =  rcm,T ×  T = (−b ˆ i) × (−T ˆ j) = bT ˆ k . (21.6.19) Apply Newton’s Second Law in the ˆ j -direction, mg −T = may . (21.6.20) Apply the rotational equation of motion for the Yo-Yo, 21-14 bT = Icmαz , (21.6.21) where αz is the z -component of the angular acceleration. The z -component of the angular acceleration and the y -component of the linear acceleration are related by the constraint condition ay = bαz , (21.6.22) where b is the axle radius of the Yo-Yo. Substitute Eq. (21.6.22) into (21.6.20) yielding mg −T = mbαz . (21.6.23) Now solve Eq. (21.6.21) for αz and substitute the result into Eq.(21.6.23), mg −T = mb2T Icm . (21.6.24) Solve Eq. (21.6.24) for the tension T , T = mg 1+ mb2 Icm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = mg 1+ mb2 (1/ 2)mR2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = mg 1+ 2b2 R2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . (21.6.25) b) Substitute Eq. (21.6.25) into Eq. (21.6.21) to determine the z -component of the angular acceleration, αz = bT Icm = 2bg (R2 + 2b2) . (21.6.26) Using the constraint condition Eq. (21.6.22), we determine the y -component of linear acceleration ay = bα z = 2b2g (R2 + 2b2) = g 1+ R2 / 2b2 . (21.6.27) Note that both quantities az > 0 and αz > 0 , so Eqs. (21.6.26) and (21.6.27) are the magnitudes of the respective quantities. For a typical Yo-Yo, the acceleration is much less than that of an object in free fall. c) Use conservation of energy to determine the magnitude of the angular velocity of the Yo-Yo when it reaches the bottom of the string. As in Figure 21.9, choose the downward vertical direction as the positive ˆ j-direction and let y = 0 designate the location of the 21-15 center of mass of the Yo-Yo when the string is completely wound. Choose U(y = 0) = 0 for the zero reference potential energy. This choice of direction and reference means that the gravitational potential energy will be negative and decreasing while the Yo-Yo descends. For this case, the gravitational potential energy is U = −mg y . (21.6.28) The Yo-Yo is not yet moving downward or rotating, and the center of mass is located at y = 0 so the mechanical energy in the initial state, when the Yo-Yo is completely wound, is zero Ei = U(y = 0) = 0 . (21.6.29) Denote the linear speed of the Yo-Yo as v f and its angular speed as ω f (at the point y = l ). The constraint condition between v f and ω f is given by v f = bω f , (21.6.30) consistent with Eq. (21.6.22). The kinetic energy is the sum of translational and rotational kinetic energy, where we have used Icm = (1/ 2)mR2 , and so mechanical energy in the final state, when the Yo-Yo is completely unwound, is E f = K f +U f = 1 2 mv f 2 + 1 2 Icmω f 2 −mgl = 1 2 mb2ω f 2 + 1 4 mR2ω f 2 −mgl. (21.6.31) There are no external forces doing work on the system (neglect air resistance), so 0 = E f = Ei . (21.6.32) Thus 1 2 mb2 + 1 4 mR2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ω f 2 = mgl . (21.6.33) Solving for ω f , ω f = 4gl (2b2 + R2) . (21.6.34) We may also use kinematics to determine the final angular velocity by solving for the time interval Δt that it takes for the Yo-Yo to travel a distance l at the constant acceleration found in Eq. (21.6.27)), 21-16 Δt = 2l / ay = l(R2 + 2b2) b2g (21.6.35) The final angular velocity of the Yo-Yo is then (using Eq. (21.6.26) for the z -component of the angular acceleration), ω f = αz Δt = 4gl (R2 + 2b2) , (21.6.36) in agreement with Eq. (21.6.34). Example 21.4 Cylinder Rolling Down Inclined Plane A uniform cylinder of outer radius R and mass M with moment of inertia about the center of mass, 2 cm (1/ 2) I M R = , starts from rest and rolls without slipping down an incline tilted at an angle β from the horizontal. The center of mass of the cylinder has dropped a vertical distance h when it reaches the bottom of the incline Figure 21.10. Let g denote the gravitational constant. The coefficient of static friction between the cylinder and the surface is µs . What is the magnitude of the velocity of the center of mass of the cylinder when it reaches the bottom of the incline? Figure 21.10 Example 21.4 Solution: We shall solve this problem three different ways. 1. Apply the torque condition about the center of mass and the force law for the center of mass motion. 2. Apply the energy methods. 3. Use torque about a fixed point that lies along the line of contact between the cylinder and the surface, First Approach: Rotation about center of mass and translation of center of mass 21-17 We shall apply the torque condition (Eq. (21.4.4)) about the center of mass and the force law (Eq. (21.4.1)) for the center of mass motion. We will first find the acceleration and hence the speed at the bottom of the incline using kinematics. The forces are shown in Figure 21.11. Figure 21.11 Torque diagram about center of mass Choose 0 x = at the point where the cylinder just starts to roll. Newton’s Second Law, applied in the x - and y -directions in turn, yields Mgsinβ −fs = Max , (21.6.37) −N + Mgcosβ = 0. (21.6.38) Choose the center of the cylinder to compute the torque about (Figure 21.10). Then, the only force exerting a torque about the center of mass is the friction force, therefore the rotational equation of motion is cm s z f R I α = . (21.6.39) Use 2 cm (1/ 2) I M R = and the kinematic constraint for the no-slipping condition / z x a R α = in Eq. (21.6.39) to solve for the magnitude of the static friction force yielding (1/ 2) s x f Ma = . (21.6.40) Substituting Eq. (21.6.40) into Eq. (21.6.37) yields Mgsinθ −(1/ 2)Max = Max , (21.6.41) which we can solve for the acceleration 2 sin 3 x a g β = . (21.6.42) 21-18 In the time f t it takes to reach the bottom, the displacement of the cylinder is /sin f x h β = . The x -component of the velocity x v at the bottom is , x f x f v a t = . Thus 2 (1/ 2) f x f x a t = . After eliminating f t , we have 2 , / 2 f x f x x v a = , so the x -component of the velocity when the cylinder reaches the bottom of the inclined plane is vx, f = 2axx f = 2((2 / 3)gsinβ)(h / sinβ) = (4 / 3)gh . (21.6.43) Note that if we substitute Eq. (21.6.42) into Eq. (21.6.40) the magnitude of the frictional force is fs = (1/ 3)Mgsinβ . (21.6.44) In order for the cylinder to roll without slipping fs ≤µsMgcosβ . (21.6.45) Combining Eq. (21.6.44) and Eq. (21.6.45) we have the condition that (1/ 3)Mgsinβ ≤µsMgcosβ (21.6.46) Thus in order to roll without slipping, the coefficient of static friction must satisfy µs ≥1 3tanβ . (21.6.47) Second Approach: Energy Methods We shall use the fact that the energy of the cylinder-earth system is constant since the static friction force does no work. Figure 21.12 Energy diagram for cylinder 21-19 Choose a zero reference point for potential energy at the center of mass when the cylinder reaches the bottom of the incline plane (Figure 21.12). Then the initial potential energy is Ui = Mgh. (21.6.48) For the given moment of inertia, the final kinetic energy is Kf = 1 2 M vx, f 2 + 1 2 Icmω z, f 2 = 1 2 M vx, f 2 + 1 2 (1/ 2)MR2(vx, f / R)2 = 3 4 M vx, f 2. (21.6.49) Setting the final kinetic energy equal to the initial gravitational potential energy leads to 2 , 3 4 x f Mgh M v = . (21.6.50) The magnitude of the velocity of the center of mass of the cylinder when it reaches the bottom of the incline is vx, f = (4 / 3)gh , (21.6.51) in agreement with Eq. (21.6.43). Third Approach: Torque about a fixed point that lies along the line of contact between the cylinder and the surface Choose a fixed point P that lies along the line of contact between the cylinder and the surface. Then the torque diagram is shown in Figure 21.13. Figure 21.13 Torque about a point along the line of contact 21-20 The gravitational force M g = Mgsinβ ˆ i + Mgcosβ ˆ j acts at the center of mass. The vector from the point P to the center of mass is given by  rP,mg = dP ˆ i −R ˆ j, so the torque due to the gravitational force about the point P is given by  τ P,Mg =  rP,Mg × M g = (dP ˆ i −R ˆ j) × (Mgsinβ ˆ i + Mgcosβ ˆ j) = (dPMgcosβ + RMgsinβ) ˆ k. (21.6.52) The normal force acts at the point of contact between the cylinder and the surface and is given by  N = −N ˆ j . The vector from the point P to the point of contact between the cylinder and the surface is  rP,N = dP ˆ i . Therefore the torque due to the normal force about the point P is given by  τ P,N =  rP,N ×  N = (dP ˆ i) × (−N ˆ j) = −dPN ˆ k . (21.6.53) Substituting Eq. (21.6.38) for the normal force into Eq. (21.6.53) yields  τ P,N = −dPMgcosβ ˆ k . (21.6.54) Therefore the sum of the torques about the point P is  τ P =  τ P, Mg +  τ P,N = (dPMgcosβ + RMgsinβ) ˆ k −dPMgcosβ ˆ k = Rmgsinβ ˆ k . (21.6.55) The angular momentum about the point P is given by  LP =  Lcm +  rP,cm × M  Vcm = Icmω z ˆ k + (dP ˆ i −R ˆ j) × (Mvx) ˆ i = (Icmω z + RMvx) ˆ k . (21.6.56) The time derivative of the angular momentum about the point P is then d  LP dt = (Icmα z + RMax) ˆ k . (21.6.57) Therefore the torque law about the point P , becomes RMgsinβ ˆ k = (Icmα z + RMax)ˆ k . (21.6.58) Using the fact that Icm = (1/ 2)MR2 and / x x a R α = , the z -component of Eq. (21.6.58) is then 21-21 sin (1/ 2) (3/ 2) x x x RMg MRa Rma MRa β = + = . (21.6.59) We can now solve Eq. (21.6.59) for the x -component of the acceleration (2/3) sin x a g β = , (21.6.60) in agreement with Eq. (21.6.42). Example 21.5 Bowling Ball A bowling ball of mass m and radius R is initially thrown down an alley with an initial speed vi , and it slides without rolling but due to friction it begins to roll (Figure 21.14). The moment of inertia of the ball about its center of mass is Icm = (2 5)mR2 . Using conservation of angular momentum about a point (you need to find that point), find the speed v f and the angular speed ω f of the bowling ball when it just starts to roll without slipping? Figure 21.14 Example 21.5 Solution: We begin introducing coordinates for the angular and linear motion. Choose an angular coordinate θ increasing in the clockwise direction. Choose the positive ˆ k unit vector pointing into the page in Figure 21.15. Figure 21.15 Coordinate system for ball Then the angular velocity vector is  ω = ω z ˆ k = dθ / dt ˆ k , and the angular acceleration vector is  α = α z ˆ k = d 2θ / dt2 ˆ k . Choose the positive ˆ i unit vector pointing to the right in Figure 21.15. Then the velocity of the center of mass is given by  vcm = vcm,xˆ i = dxcm / dt ˆ i , 21-22 and the acceleration of the center of mass is given by  acm = acm,xˆ i = d 2xcm / dt2 ˆ i . The free-body force diagram is shown in Figure 21.16. Figure 21.16 Free-body force diagram for ball At t = 0 , when the ball is released,  vcm,0 = v0ˆ i and  ω0 =  0 , so the ball is skidding and hence the frictional force on the ball due to the sliding of the ball on the surface is kinetic friction, hence acts in the negative ˆ i -direction. Because there is kinetic friction and non-conservative work, mechanical energy is not constant. The rotational equation of motion is  τS = d  LS / dt . In order for angular momentum about some point to remain constant throughout the motion, the torque about that point must also be zero throughout the motion. As the ball moves down the alley, the contact point will move, but the frictional force will always be directed along the line of contact between the bowling bowl and the surface. Choose any fixed point S along the line of contact then  τS, fk =  rS, fk ×  fk =  0 (21.6.61) because  rS, fk and  fk are anti-parallel. The gravitation force acts at the center of mass hence the torque due to gravity about S is  τS,mg =  rS,mg × m g = dmg ˆ k , (21.6.62) where d is the distance from S to the contact point between the ball and the ground. The torque due to the normal force about S is  τS,N =  rS,N × m g = −dN ˆ k , (21.6.63) with the same moment arm d . Because the ball is not accelerating in the ˆ j-direction, from Newton’s Second Law, we note that mg −N = 0. Therefore  τS,N +  τS,mg = d(mg −N) ˆ k =  0 . (21.6.64) 21-23 There is no torque about any fixed point S along the line of contact between the bowling bowl and the surface; therefore the angular momentum about that point S is constant,  LS,i =  LS, f . (21.6.65) Choose one fixed point S along the line of contact (Figure 21.17). (a) (b) Figure 21.17 Angular momentum about S : (a) initial, (b) final The initial angular momentum about S is only due to the translation of the center of mass (Figure 21.17a),  LS,i =  rS,cm,i × m vcm,i = mRvcm,i ˆ k . (21.6.66) In Figure 21.17b, the ball is rolling without slipping. The final angular momentum about S has both a translational and rotational contribution  LS, f =  rS,cm, f × m vcm, f + Icm  ω f = mRvcm, f ˆ k + Icmω z, f ˆ k . (21.6.67) When the ball is rolling without slipping, vcm, f = Rω z, f and also Icm = (2 / 5)m R2 . Therefore the final angular momentum about S is  LS, f = (mR + (2 / 5)mR)vcm, f ˆ k = (7 / 5)mRvcm, f ˆ k . (21.6.68) Equating the z -components in Eqs. (21.6.66) and (21.6.68) yields mRvcm,i = (7 / 5)mRvcm, f , (21.6.69) which we can solve for vcm, f = (5/ 7)vcm,i . (21.6.70) The final angular velocity vector is 21-24  ω = ω z, f ˆ k = vcm, f R ˆ k = 5vcm,i 7R ˆ k . (21.6.71) We could also solve this problem by analyzing the translational motion and the rotational motion about the center of mass. Gravity exerts no torque about the center of mass, and the normal component of the contact force has a zero moment arm; the only force that exerts a torque is the frictional force, with a moment arm of R (the force vector and the radius vector are perpendicular). The frictional force should be in the negative x - direction. From the right-hand rule, the direction of the torque is into the page, and hence in the positive z -direction. Equating the z -component of the torque to the rate of change of angular momentum about the center of mass yields τ cm = R fk = I cmα z , (21.6.72) where k f is the magnitude of the kinetic frictional force and α z is the z -component of the angular acceleration of the bowling ball. Note that Eq. (21.6.72) results in a positive z -component of the angular acceleration, which is consistent with the ball tending to rotate as indicated Figure 21.15. The frictional force is also the only force in the horizontal direction, and will cause an acceleration of the center of mass, acm,x = −fk / m. (21.6.73) Note that the x -component of the acceleration will be negative, as expected. Now we need to consider the kinematics. The bowling ball will increase its z -component of the angular velocity as given in Eq. (21.6.72) and decrease its x -component of the velocity as given in Eq. (21.6.73), ω z(t) = α z t = Rfk Icm t vcm,x(t) = vcm,i −fk m t. (21.6.74) As soon as the ball stops slipping, the kinetic friction no longer acts, static friction is zero, and the ball moves with constant angular and linear velocity. Denote the time when this happens as t f . At this time the rolling without slipping condition, ω z(t f ) = vcm,x(t f ) / R , holds and the relations in Eq. (21.6.74) become R2 fk Icm t f =vcm,x, f vcm,x,i −fk m t f = vcm,x, f . (21.6.75) 21-25 We can now solve the first equation in Eq. (21.6.75) for f t and find that t f = Icm fkR2 vcm,x, f . (21.6.76) We now substitute Eq. (21.6.76) into the second equation in Eq. (21.6.75) and find that vcm,x, f = vcm,x,i −fk m Icm fkR2 vcm,x, f vcm,x, f = vcm,x,i −Icm mR2 vcm,x, f . (21.6.77) The second equation in (21.6.77) is easily solved for vcm,x, f = v0 1+ Icm / mR2 = 5 7 vcm,x,i , (21.6.78) agreeing with Eq. (21.6.70) where we have used Icm = (2 / 5)mR2 for a uniform sphere. Example 21.6 Rotation and Translation Object and Stick Collision A long narrow uniform stick of length l and mass m lies motionless on ice (assume the ice provides a frictionless surface). The center of mass of the stick is the same as the geometric center (at the midpoint of the stick). The moment of inertia of the stick about its center of mass is Icm . A puck (with putty on one side) has the same mass m as the stick. The puck slides without spinning on the ice with a velocity of  vi toward the stick, hits one end of the stick, and attaches to it (Figure 21.18). You may assume that the radius of the puck is much less than the length of the stick so that the moment of inertia of the puck about its center of mass is negligible compared to Icm . (a) How far from the midpoint of the stick is the center of mass of the stick-puck combination after the collision? (b) What is the linear velocity  vcm, f of the stick plus puck after the collision? (c) Is mechanical energy conserved during the collision? Explain your reasoning. (d) What is the angular velocity  ωcm, f of the stick plus puck after the collision? (e) How far does the stick's center of mass move during one rotation of the stick? 21-26 Figure 21.18 Example 21.6 Solution: In this problem we will calculate the center of mass of the puck-stick system after the collision. There are no external forces or torques acting on this system so the momentum of the center of mass is constant before and after the collision and the angular momentum about the center of mass of the puck-stick system is constant before and after the collision. We shall use these relations to compute the final angular velocity of the puck-stick about the center of mass. Figure 21.19 Center of mass of the system a) With respect to the center of the stick, the center of mass of the stick-puck combination is dcm = mstickdstick + mpuckdpuck mstick + mpuck = m(l / 2) m+ m = l 4 . (21.6.79) where we are neglecting the radius of the puck (Figure 21.19). b) During the collision, the only net forces on the system (the stick-puck combination) are the internal forces between the stick and the puck (transmitted through the putty). Hence, the linear momentum is constant. Initially only the puck had linear momentum  pi = m vi = mviˆ i . After the collision, the center of mass of the system is moving with velocity  vcm, f = vcm, f ˆ i. Equating initial and final linear momenta, 21-27 mvi = (2m)vcm, f ⇒vcm, f = vi 2 . (21.6.80) The direction of the velocity is the same as the initial direction of the puck’s velocity. c) The forces that deform the putty do negative work (the putty is compressed somewhat), and so mechanical energy is not conserved; the collision is totally inelastic. d) Choose the center of mass of the stick-puck combination, as found in part a), as the point S about which to find angular momentum. This choice means that after the collision there is no angular momentum due to the translation of the center of mass. Before the collision, the angular momentum was entirely due to the motion of the puck,  LS,i =  rpuck ×  pi = (l / 4)(mvi) ˆ k , (21.6.81) where ˆ k is directed out of the page in Figure 21.19. After the collision, the angular momentum is  LS, f = Icm, fω cm, f ˆ k , (21.6.82) where Icm, f is the moment of inertia about the center of mass of the stick-puck combination. This moment of inertia of the stick about the new center of mass is found from the parallel axis theorem and the moment of inertia of the puck, which is m(l / 4)2 . Therefore Icm, f = Icm, stick + Icm, puck = (Icm + m(l / 4)2) + m(l / 4)2 = Icm + ml2 8 . (21.6.83) Inserting this expression into Eq. (21.6.82), equating the expressions for  LS, i and  LS, f and solving for ωcm, f yields ωcm, f = m(l / 4) Icm + ml2 / 8 vi . (21.6.84) If the stick is uniform, Icm = ml2 / 12 and Eq. (21.6.84) reduces to ωcm, f = 6 5 vi l . (21.6.85) It may be tempting to try to calculate angular momentum about the contact point C , where the putty hits the stick. If this is done, there is no initial angular momentum, and after the collision the angular momentum will be the sum of two parts, the angular 21-28 momentum of the center of mass of the stick and the angular moment about the center of the stick,  LC, f =  r cm ×  pcm + Icm  ωcm, f . (21.6.86) There are two crucial things to note: First, the speed of the center of mass is not the speed found in part b); the rotation must be included, so that vcm = vi / 2 −ωcm, f (l / 4). Second, the direction of cm cm × r p   with respect to the contact point C is, from the right-hand rule, into the page, or the ˆ −k -direction, opposite the direction of  ωcm, f . This is to be expected, as the sum in Eq. (21.6.86) must be zero. Adding the ˆ k -components (the only components) in Eq. (21.6.86), −(l / 2)m(vi / 2 −ωcm, f (l / 4)) + Icmωcm, f = 0 . (21.6.87) Solving Eq. (21.6.87) for ωcm, f yields Eq. (21.6.84). This alternative derivation should serve two purposes. One is that it doesn’t matter which point we use to find angular momentum. The second is that use of foresight, in this case choosing the center of mass of the system so that the final velocity does not contribute to the angular momentum, can prevent extra calculation. It’s often a matter of trial and error (“learning by misadventure”) to find the “best” way to solve a problem. e) The time of one rotation will be the same for all observers, independent of choice of origin. This fact is crucial in solving problems, in that the angular velocity will be the same (this was used in the alternate derivation for part d) above). The time for one rotation is the period T = 2π /ω f and the distance the center of mass moves is xcm = vcmT = 2π vcm ωcm, f = 2π vi / 2 m(l / 4) Icm + ml2 / 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟vi = 2π Icm + ml2 / 8 m(l / 2) . (21.6.88) Using Icm = ml2 / 12 for a uniform stick gives xcm = 5 6π l . (21.6.89)
10034	https://www.sciencedirect.com/science/article/abs/pii/S1878875011004323	Hyponatremia in Patients with Traumatic Brain Injury: Etiology, Incidence, and Severity Correlation - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract Introduction Section snippets References (22) Cited by (56) World Neurosurgery Volume 76, Issues 3–4, September–October 2011, Pages 355-360 Peer-Review Report Hyponatremia in Patients with Traumatic Brain Injury: Etiology, Incidence, and Severity Correlation Author links open overlay panel Subash Lohani, Upendra Prasad Devkota Show more Add to Mendeley Share Cite rights and content Background Hyponatremia is common in patients with traumatic brain injury (TBI). This study aims at ascertaining the incidence of hyponatremia in TBI with its etiologic diagnosis, biochemical confirmation, and severity correlation with initial Glasgow Coma Scale (GCS) score and computed tomography (CT) abnormality. Methods All patients more than 20 years old with severe and moderate TBI and the mild ones with CT abnormality were included. Daily sodium level was monitored for 14 days. Central venous pressure (CVP) was measured for assessment of volume status. Fractional excretion of uric acid (FEUA) was measured in all patients with hyponatremia, both before and after its correction. Results Of 40 consecutive patients, 33 remained for analysis. Hyponatremia was seen in 9 (27.2%) patients, of whom 6 developed it within the first week. Mean duration of hyponatremia was 1.78 days. We found that 5 patients had an elevated CVP consistent with the syndrome of inappropriate antidiuretic hormone (SIADH), whereas 3 had low CVP consistent with cerebral salt wasting syndrome (CSWS) and 1 had an equivocal reading. Measurement of FEUA did not show consistent pattern to suggest a biochemical distinction. There were 33.3% each of mild and moderate, and 16.6% of severe TBI among hyponatremic patients. Hyponatremia was seen in Rotterdam CT scores I to IV in increasing incidence (r = 0.983, P = 0.017), whereas it had no significant correlation with initial GCS (r = 0.756, P = 0.455). Conclusions Hyponatremia due to SIADH is more common in TBI. FEUA measurement is not consistent enough to make a distinction between SIADH and CSWS. CT scoring of severity is more predictive of hyponatremia than initial GCS. Introduction Traumatic brain injury (TBI) results in a range of neurological and cognitive impairment to personality, behavioral changes, and lifestyle consequences (12). Posttraumatic endocrine complications also pose a great challenge in the management of TBI. Early recognition of these subtle problems can produce a significant impact on outcome (3, 4). Hyponatremia is a common electrolyte disorder in the setting of central nervous system disease including TBI (9). Two common etiologies, namely the syndrome of inappropriate antidiuretic hormone (SIADH) (10, 13) and cerebral salt wasting syndrome (CSWS) (5, 14, 19) are primarily responsible. Differentiation among the two conditions is based on the volume status of the patient (20). Given a typical case scenario, distinction might not appear difficult. Yet, an exceeding number of cases have a borderline picture with diagnostic confusion and impediment in therapeutic intervention. Researchers are thus making an attempt in differentiating the two conditions based on biochemical alterations, especially that of the fractional excretion of uric acid (FEUA) (15). This study primarily aims at identifying the etiology, incidence, and timing of hyponatremia in patients with TBI, including its correlation with initial severity and final outcome. Severity correlation is done with both the clinical and radiological severity status. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Study Design This is a prospective, descriptive, and analytical study conducted at the National Institute of Neurological and Allied Sciences, Bansbari, Kathmandu. Patients admitted with the diagnosis of TBI were the target population. All patients presenting from April to September 2008 who fulfilled the inclusion criteria were enrolled in the study. Study variables were age distribution, sex distribution, Glasgow Coma Scale (GCS) at the time of presentation, Rotterdam computed tomography (CT) score (Table 1 Tools and Techniques Pertinent information was prospectively gathered during admission. Consent was obtained from patients at the time of admission as far as possible. Preformed structured formats were used to make pertinent clinical, radiological, and laboratory documentation. Severity of TBI was classified as mild, moderate, or severe based on GCS 13 to 15, 9 to 11, and 3 to 8, respectively (22). CT scan was performed in all patients, and CT grading was noted. Baseline investigations were done in all patients. Results A total of 40 patients were enrolled in the study, of whom 7 were excluded because they had a hospital stay of less than 1 week for various reasons such as early mortality, transfer to other centers, or associated medical ailments that could produce hyponatremia. Of 33 patients who remained for analysis, hyponatremia was seen in 9 patients within the period of 2 weeks. Discussion To begin with demographics, mean age and sex ratio of overall and hyponatremic patients are comparable. The incidence of hyponatremia was 27.27%, which is comparable to that described by Moro et al. (18) among the subgroup of head injury patients who had intracranial bleeding. Hyponatremia has been shown to have an association with the severity of brain injury (9). However, both mild and moderate head injuries had 33.33% hyponatremia, whereas with increasing severity, incidence decreased to Conclusion and Recommendations Hyponatremia is common in TBI, with an incidence of 27.27% among high-risk patients. Most of them can be attributed to SIADH, although CSWS also occurs in a few. CT scoring of injury has a better correlation with its occurrence than does initial GCS. With prompt identification and treatment, hyponatremia does not result in prolonged hospital stay or any undue morbidity and mortality. Measurement of FEUA does not seem consistent enough for the differentiation of SIADH or CSWS. Because early Recommended articles References (22) J.D. Born et al. Syndrome of inappropriate secretion of antidiuretic hormone after severe head injury Surg Neurol (1985) N. Moro et al. Hyponatremia in patients with traumatic brain injury: incidence, mechanism, and response to sodium supplementation or retention therapy with hydrocortisone Surg Neurol (2007) G. Teasdale et al. Assessment of coma and impaired consciousness: a practical scale Lancet (1974) M.E.A. Abdel-Latif et al. Cerebral salt wasting syndrome following atlantoaxial fracture dislocation in Down syndrome BMJ Case Reports (2009) H.J. Adrogué et al. Hyponatremia N Engl J Med (2000) A. Agha et al. Posterior pituitary dysfunction after traumatic brain injury J Clin Endocrinol Metab (2004) M. Bondanelli et al. Hypopituitarism after traumatic brain injury Eur J Endocrinol (2005) C. Davison et al. Uric acid and kidney T. Doczi et al. Syndrome of inappropriate secretion of antidiuretic hormone (SIADH) after head injury J Neurosurg (1982) M. Fukagawa et al. Fluid and electrolyte disorders M.R. Harrigan Cerebral salt wasting syndrome: a review Neurosurgery (1996) View more references Cited by (56) Hyponatremia in Traumatic Brain Injury: A Practical Management Protocol 2017, World Neurosurgery Citation Excerpt : The 2 common causes of hyponatremia with natriuresis in TBI are SIADH and CSW. The exact incidence, pathophysiology, and accurate clinical distinction between these 2 entities remain unclear.2,5,17,20 SIADH is often overdiagnosed because fluid restriction does raise the serum sodium even in patients with hyponatremia due to other reasons, and this is taken as proof of the diagnosis. Show abstract Hyponatremia (defined as serum sodium <135 mEq/L) is the most common electrolyte abnormality in traumatic brain injury (TBI) and is also an independent predictor of poor neurologic outcome. The reported incidence of hyponatremia varies widely in literature reports, and there is continuing difficulty in clearly differentiating between the 2 common causes of hyponatremia with natriuresis: the syndrome of inappropriate antidiuretic hormone secretion (SIADH) and cerebral salt wasting (CSW). We encounter hyponatremia frequently in our practice, and we therefore decided to review data from our center to estimate the incidence of hyponatremia and the results of our management strategies, and attempt to formulate simple guidelines for the correction of hyponatremia in TBI. A retrospective analysis of 1500 consecutively admitted patients with TBI was performed by the use of electronic records and radiographic review. Hyponatremia was defined as serum sodium <135 mEq/L, and natriuresis as a urine spot sodium of more than >40 mEq/L. The incidence of TBI, its management, and the effect of fludrocortisone were evaluated. The incidence of hyponatremia was 13.2%. Early therapy with fludrocortisone significantly reduced the duration of hospital stay (P< 0.05). Traumatic subarachnoid hemorrhage was the most common abnormality on the admission computed tomographic scan in patients who experienced hyponatremia. Early initiation of fludrocortisone in the setting of hyponatremia with natriuresis decreases the hospital stay. This protocol is probably safer in a tropical country where fluid restriction might be harmful. It also eliminates the need to differentiate between SIADH and CSW. ### The Economic Burden of Hyponatremia: Systematic Review and Meta-Analysis 2016, American Journal of Medicine Show abstract Hyponatremia is the most common electrolyte abnormality observed in clinical practice. Several studies have demonstrated that hyponatremia is associated with an increased length of hospital stay and of hospital resource utilization. To clarify the impact of hyponatremia on the length of hospitalization and costs, we performed a meta-analysis based on published studies that compared hospital length of stay and cost between patients with and without hyponatremia. An extensive Medline, Embase, and Cochrane search was performed to retrieve all studies published up to April 1, 2015 using the following words: “hyponatremia” or “hyponatraemia” AND “hospitalization” or “hospitalisation.” A meta-analysis was performed including all studies comparing duration of hospitalization and hospital readmission rate in subjects with and without hyponatremia. Of 444 retrieved articles, 46 studies satisfied the inclusion criteria, encompassing a total of 3,940,042 patients; among these, 757,763 (19.2%) were hyponatremic. Across all studies, hyponatremia was associated with a significantly longer duration of hospitalization (3.30 [2.90-3.71; 95% CIs] mean days; P< .000). Similar results were obtained when patients with associated morbidities were analyzed separately. Furthermore, hyponatremic patients had a higher risk of readmission after the first hospitalization (odds ratio 1.32 [1.18-1.48; 95% CIs]; P< .000). A meta-regression analysis showed that the hyponatremia-related length of hospital stay was higher in males (Slope= 0.09 [0.05-0.12; 95% CIs]; P= .000 and Intercept=−1.36 [−3.03-0.32; 95% CIs]; P= .11) and in elderly patients (Slope= 0.002 [0.001-0.003; 95% CIs]; P< .000 and Intercept= 0.89 [0.83-0.97; 95% CIs]; P< .001). A negative association between serum [Na+] cutoff and duration of hospitalization was detected. No association between duration of hospitalization, serum [Na+], and associated morbidities was observed. Finally, when only US studies (n= 8) were considered, hyponatremia was associated with up to around $3000 higher hospital costs/patient when compared with the cost of normonatremic subjects. This meta-analysis confirms that hyponatremia is associated with a prolonged hospital length of stay and higher risk of readmission. These observations suggest that hyponatremia may represent one important determinant of the hospitalization costs. ### Dual diagnosis: Traumatic brain injury with spinal cord injury 2014, Physical Medicine and Rehabilitation Clinics of North America Citation Excerpt : Metabolic problems, including disorders of serum sodium regulation and/or neuroendocrine dysfunction, may occur in dual diagnosis patients having moderate–severe TBI, particularly when a lesion involves the hypothalamic-pituitary brain pathways.39–46 Hyponatremia may occur from the syndrome of inappropriate antidiuretic hormone secretion (SIADH) or from cerebral salt-wasting syndrome (CSWS).40–43 In SIADH, there is normovolemic to hypervolemic hyponatremia due to the dilution of serum sodium concentration from renal water retention and sodium excretion caused by excessive release of antidiuretic hormone (ADH) by the hypothalamus.40–43 ### Cerebral salt wasting after traumatic brain injury: A review of the literature 2015, Scandinavian Journal of Trauma Resuscitation and Emergency Medicine Show abstract Electrolyte imbalances are common among patients with traumatic brain injury (TBI). Cerebral salt wasting (CSW) is an electrolyte imbalance characterized by hyponatremia and hypovolemia. Differentiating the syndrome of inappropriate antidiuretic hormone and CSW remains difficult and the pathophysiological mechanisms underlying CSW are unclear. Our intent was to review the literature on CSW within the TBI population, in order to report the incidence and timing of CSW after TBI, examine outcomes, and summarize the biochemical changes in patients who developed CSW. We searched MEDLINE through 2014, hand-reviewed citations, and searched abstracts from the American Association for the Surgery of Trauma (2003–2014). Publications were included if they were conducted within a TBI population, presented original data, and diagnosed CSW. Publications were excluded if they were review articles, discussed hyponatremia but did not differentiate the etiology causing hyponatremia, or presented cases with chronic disease. Fifteen of the 47 publications reviewed met the selection criteria; nine (60%) were case reports, five (33%) were prospective and 1 (7%) was a retrospective study. Incidence of CSW varied between 0.8 - 34.6%. The populations studied were heterogeneous and the criteria used to define hyponatremia and CSW varied. Though believed to play a role in the development of CSW, increased levels of natriuretic peptides in patients diagnosed with CSW were not consistently reported. These findings reinforce the elusiveness of the CSW diagnosis and the need for strict and consistent diagnostic criteria. The online version of this article (doi:10.1186/s13049-015-0180-5) contains supplementary material, which is available to authorized users. ### Inappropriate Antidiuretic Hormone Secretion and Cerebral Salt-Wasting Syndromes in Neurological Patients 2019, Frontiers in Neuroscience ### The screening and management of pituitary dysfunction following traumatic brain injury in adults: British Neurotrauma Group guidance 2017, Journal of Neurology Neurosurgery and Psychiatry View all citing articles on Scopus Conflict of interest statement: National Health Research Council (NHRC) Nepal provided support for laboratory investigations. View full text Copyright © 2011 Elsevier Inc. All rights reserved. Substances (3) Generated by , an expert-curated chemistry database. 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10035	https://www.reddit.com/r/askmath/comments/1kb4nmz/set_theory_and_rational_solutions_finding_a_b/	Set Theory and Rational Solutions – Finding A ∩ B When A ∪ B Is Singleton : r/askmath Skip to main contentSet Theory and Rational Solutions – Finding A ∩ B When A ∪ B Is Singleton : r/askmath Open menu Open navigationGo to Reddit Home r/askmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to askmath r/askmath•5 mo. ago AgileEvening5622 Set Theory and Rational Solutions – Finding A ∩ B When A ∪ B Is Singleton I’m working on a problem involving set operations with rational variables. Let: A = {x²+ 2y, y² + 1} AUB= {x² + 4y, y + 1 - 3x} Ginevn that B≠∅ and x;y∈Q AUB is a singleton. I want to find A∩B What I’ve considered so far: Since has only one element, and both A and B contribute to it, I assumed the two expressions in the union must be equal: x²+4y=y²+1 y+1-3x=x²-2y I tried solving this system under the condition that , but I couldn't find rational solutions that satisfy both equations simultaneously. I'm wondering: Is there a contradiction that makes necessary? Or can we determine rational values such that is non-empty? Read more Share Related Answers Section Related Answers Understanding limits using intuitive examples How to approach word problems in algebra Exploring the Fibonacci sequence in nature Strategies for mastering calculus derivatives Using matrices to solve systems of equations New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
10036	http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/star.html	\| \| \| --- \| \| Double Star Images Suppose we take the common-sense position that if a star is moving toward us, then the light from that star must have a higher velocity toward us. Since light does not depend upon the medium (space) then the light speed surely must add to the velocity of the source? From this point of view, consider the observation of a binary star system. As a youthful clerk in a Swiss patent office, Einstein may have had time to reflect upon the contradictions in star images that arise if the light velocity depends upon the velocity of the star toward us. \| Index Relativity concepts \| \| \| \| \| --- \| \| HyperPhysics Relativity \| R Nave \| \| Go Back \| \| \| \| --- \| \| Aberration of Starlight You have to tilt your telescope to catch the star light just as you have to tilt your umbrella to keep off wind-driven rain. It is a matter of relative velocity. Common experience leads you to expect that the velocity of the light relative to the Earth would be greater than the velocity of the light relative to the star. That is, you expect the velocity of the light with respect to the Earth to be the hypotenuse of the triangle and therefore greater than c. But observations are consistent with the hypotenuse of the triangle being just equal to c. Actual experiments involve measurements of the change in the apparent positions of stars at different times of the year. \| Index Relativity concepts \| \| \| \| \| --- \| \| HyperPhysics Relativity \| R Nave \| \| Go Back \| \| \| \| \| \| --- --- \| \| Apparent Star Positions You have to tilt your telescope to catch the star light just as you have to tilt your umbrella to keep off wind-driven rain. It is a matter of relative velocity. As the Earth moves around the Sun, the star will appear to trace out a small ellipse. The shape of the ellipse depends on the orientation of the star to the ecliptic. \| \| \| --- \| \| \| When a distant star is viewed from the Earth at different times of the year, its apparent position will change as a result of the Earth's velocity through space. This apparent ellipse of motion of the star is called the aberration of starlight. Measurements of the size of this ellipse offer one piece of evidence that the velocity of light is independent of observer motion. This effect is independent of the parallax which permits the determination of the distance of the closest stars by triangulation. \| \| Index Relativity concepts \| \| \| \| \| --- \| \| HyperPhysics Relativity \| R Nave \| \| Go Back \|
10037	https://www.khanacademy.org/math/mr-class-7/x5270c9989b1e59e6:angles-and-pairs-of-angles	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
10038	https://pharmacy.ufl.edu/files/2013/01/5127-28-equations.pdf	Equations/Useful_pharmacokinetic_equ_5127 1 Useful Pharmacokinetic Equations Symbols D = dose  = dosing interval CL = clearance Vd = volume of distribution ke = elimination rate constant ka = absorption rate constant F = fraction absorbed (bioavailability) K0 = infusion rate T = duration of infusion C = plasma concentration General Elimination rate constant     k CL Vd C C t t C C t t e             ln ln ln 1 2 2 1 1 2 2 1 Half-life t Vd CL k k e e 1 2 0693 2 0693 / . ln( ) .     Intravenous bolus Initial concentration C D Vd 0  Plasma concentration (single dose) C C e k t e     0 Plasma concentration (multiple dose)   C C e e k t k e e        0 1  Peak (multiple dose)   C C e k e max     0 1  Trough (multiple dose)   C C e e k k e e min        0 1   Average concentration (steady state) Cp D CL ss   Oral administration Plasma concentration (single dose)    C F D k Vd k k e e a a e k t k t e a           Time of maximum concentration (single dose)   t k k k k a e a e max ln         Plasma concentration (multiple dose)       C F D k Vd k k e e e e a a e k t k k t k e e a a                         1 1   Time of maximum concentration (multiple dose)       t k e k e k k a k e k a e e a max ln                   1 1   Average concentration (steady state) C F D CL    Clearance AUC F Dose Cl   d e V k Cl   Equations/Useful_pharmacokinetic_equ_5127 2 Constant rate infusion Plasma concentration (during infusion)   C k CL e k t e      0 1 Plasma concentration (steady state) C k CL  0 Calculated clearance (Chiou equation)        CL k C C Vd C C C C t t           2 2 0 1 2 1 2 1 2 2 1 Short-term infusion Peak (single dose)   C D CL T e k T e max( ) 1 1       Trough (single dose)   C C e k T e min( ) max( ) 1 1       Peak (multiple dose)     C D CL T e e k T k e e max          1 1  Trough (multiple dose)   C C e k T e min max       Calculated elimination rate constant k C C t e          ln max min  with Cmax = measured peak and Cmin = measured trough, measured over the time interval t Calculated peak C C e k t e max max     with Cmax = measured peak, measured at time t after the end of the infusion Calculated trough C C e k t e min min      with Cmin = measured trough, measured at time t before the start of the next infusion Calculated volume of distribution   )] ( [ 1 min max T e k T e k e e C C e T k D Vd           Calculated recommended dosing interval         ln max( ) min( ) C C k T desired desired e Calculated recommended dose     D C k V T e e desired e k k T e e            max( ) 1 1  Two-Compartment-Body Model C a e b e t t         AUC a b   / /   Vd Vd Vc area ss   Creatinine Clearance CL male age weight Cp creat creat ( ) ( )     140 72 CL female age weight Cp creat creat ( ) ( )     140 85 With weight in kg, age in years, creatinine plasma conc. in mg/dl and CLcreat in ml/min Equations/Useful_pharmacokinetic_equ_5127 3 Ke for aminoglycosides Ke = 0.00293(CrCL) + 0.014 Metabolic and Renal Clearance EH = Cl fu Q Cl fu b H b int int    ClH = E Q H H  = Q Cl fu Q Cl fu H b H b     int int FH = b H H fu Cl Q Q   int Clren = RBFE = in out in C C C GFR   Clren = ion concentrat plasma excretion of rate Clren =         ion concentrat Plasma on reabsorpti of Rate -secretion of Rate GFR fu Clren = ion concentrat Plasma ion concentrat urine flow Urine  Ideal Body Weight Male IBW = 50 kg + 2.3 kg for each inch over 5ft in height Female IBW = 45.5 kg + 2.3 kg for each inch over 5ft in height Obese ABW = IBW + 0.4(TBW-IBW) Volume of Distribution P T P K V V V    T T P fu fu V V V    Clearance AUC Dose Cl  d e V k Cl   Last modified 2010 C:\Current Data\pha5127_Dose_Opt_I\equations\5127-28-equations.doc For One Compartment Body Model If the dosing involves the use of I.V. bolus administration: For a single I.V. bolus administration: V D C  0 t ke e C C    0 For multiple I.V. bolus administration:     t k k nk e e e e e e V D t Cn           1 1 ) ( at peak: t = 0; at steady state n at trough: t =  ) e ( V D C e k ss      1 1 max  e k ss ss e C C    max min If the dosing involves the use of I.V. infusion: For a single short-term I.V. infusion: Since  = t for Cmax   T k e e e T Vk D C     1 max ) ( max min T ke e C C      For multiple short-term I.V. infusion at steady state:      e e k T k e e e T Vk D C       1 1 max ) ( max min T ke e C C      Last modified 2010 C:\Current Data\pha5127_Dose_Opt_I\equations\5127-28-equations.doc If the dosing involves a I.V. infusion (more equations):   t k T k e t e e e e T Vk D C      1 (most general eq.) during infusion t = T so,   t k e t e e T Vk D C     1 (during infusion) at steady state t  , e-ket, t  0 so, CL k Vk k T Vk D Cpss e e 0 0    (steady state) remembering T D k  0 and e k V CL   If the dosing involves oral administration: For a single oral dose:    t k t k e a a a e e e k k V k D F C           e a e a k k k k t          1 ln max For multiple oral doses:                     a a e e k t k k t k e a a e e e e k k V k D F C 1 1       e a k e k a k k e k e k t a e                  1 1 1 ln max  
10039	https://thirdspacelearning.com/us/math-resources/topic-guides/ratio-and-proportion/pressure-force-area/	High Impact Tutoring Built By Math Experts Personalized standards-aligned one-on-one math tutoring for schools and districts Request a demo In order to access this I need to be confident with: Multiplication and division Addition and subtraction Area Metric units of measurement Converting metric units What are pressure, force, and area? Area Force Pressure Pressure triangle Common Core State Standards How to calculate pressure, force, or area Pressure force area examples ↓ Example 1: calculating pressure (Pa) Example 2: calculating pressure (Bar) Example 3: calculating area (m2) Example 4: calculating area (converting units of pressure) Example 5: calculating force (N) Example 6: calculating force (converting units) Example 7: calculating pressure given the mass Teaching tips for pressure force area Easy mistakes to make Related compound measure lessons Practice pressure force area questions Pressure force area FAQs Next lessons Still stuck? Pressure force area Here you will learn about pressure force and area, including what they are and how they are related to one another. Students will first learn about pressure force and area as part of algebra in high school. What are pressure, force, and area? Pressure, force, and area are physical properties. Area Area is a measure of the size of space a flat shape takes up. The SI unit for area is the square meter (m2) although you can use other units including square centimeters (cm2), square millimeters (mm2), square inches (in2), etc. Area is inversely proportional to the pressure exerted; for a constant force, if the area in which the force is being applied increases, the pressure exerted across that area decreases. You can therefore state that A∝P1 for the area, A and the pressure, P. The constant of proportionality is the force, F. Note: When you consider the area of an object, you are referring to the area of contact of the object with the floor, table, ground, wall or shelf, etc. [FREE] Ratio Worksheet (Grade 6 to 7) Use this quiz to check your grade 6 to 7 students’ understanding of ratios. 10+ questions with answers covering a range of 6th and 7th grade ratio topics to identify areas of strength and support! DOWNLOAD FREE x [FREE] Ratio Worksheet (Grade 6 to 7) Use this quiz to check your grade 6 to 7 students’ understanding of ratios. 10+ questions with answers covering a range of 6th and 7th grade ratio topics to identify areas of strength and support! DOWNLOAD FREE Force Force is the energy attributed to a movement or physical action. Force is measured in the standard units of force, Newtons (N). Newton’s Second Law of Motion states that force is equal to the mass of an object, multiplied by the acceleration, or F=MA. This is why weight and mass are not the same. Mass is a measure of how much matter there is in an object (usually measured in kilograms), whereas weight is the size of the pull of gravity on the object (measured in Newtons). The acceleration due to gravity on the surface of the Earth is 9.807 m/s2 and so an average human with a mass of 62 kg would have a weight of 62×9.807=608 N. The force acting on an object is directly proportional to the pressure applied over a constant area. The greater the pressure, the greater the force for a constant area and vice versa. You can therefore state that F∝P for the force F and the pressure P. The constant of proportionality is the area, A. See also: Directly proportional The forces used for this topic act on an object that is in contact with the floor, table, wall, shelf, etc. This means that the force will be perpendicular to the surface for all questions. Pressure Pressure is a compound measure. Pressure is defined as the force per unit area. The standard unit of pressure is Pascals (Pa) where 1 Pa=1 N/m2. You can also measure pressure using a Bar, where 1 Bar =100,000 Pa =100,000N/m2. Imagine an elephant standing on 4 legs. The pressure applied to the ground is distributed over a wider area than if the elephant were standing on one leg. This is because the force remains constant (the weight of the elephant), but the area in which the force is applied has decreased, so the pressure increases. This is an example of inverse proportion. As pressure is inversely proportional to the area, you can state that P∝A1 for the pressure, P, and the area, A. The constant of proportionality is the force, F. See also: Inversely proportional To calculate either the pressure, force, or area of an object, you use the pressure formula (or pressure equation): Pressure=AreaForce Pressure triangle This can be written as a formula triangle (sometimes called the pressure triangle). You can circle what you are trying to find and the formula triangle tells how to calculate the unknown property. Step-by-step guide: Pressure formula What are pressure, force, and area? Common Core State Standards How does this relate to high school math? High School – Algebra – Creating Equations (HS.A.CED.A.1) Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. High School – Algebra – Creating Equations (HS.A.CED.A.2)Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. High School – Algebra – Creating Equations (HS.A.CED.A.4) Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law V=IR to highlight resistance R. How to calculate pressure, force, or area In order to calculate pressure, force, or area using the formula triangle: Draw the triangle and circle the required property. Substitute values for the other two properties and complete the calculation. Write down the solution, including the units. Pressure force area examples Example 1: calculating pressure (Pa) A force of 800 N acts on an area of 200 m2. Calculate the pressure in Pascals. Draw the triangle and circle the required property. To calculate the pressure, you would need to divide the force by the area. 2Substitute values for the other two properties and complete the calculation. Substituting F=800 N and A=200 m2, you have: 800÷200=4 3Write down the solution, including the units. Pressure =4 N/m2=4 Pa. Example 2: calculating pressure (Bar) A force of 6.5×106N acts on an area of 0.2 m2. Calculate the pressure in Bar. Draw the triangle and circle the required property. Show step To calculate the pressure, you would need to divide the force by the area. Substitute values for the other two properties and complete the calculation. Show step Converting 6.5×106 to an ordinary number, you get F=6500000 N. P=6500000÷0.2=32500000 Write down the solution, including the units. Show step You currently have 32,500,000 Pa. As you want the solution in Bar, you need to divide the value by 100,000: 32500000÷100000=325 Bar. Example 3: calculating area (m²) A force of 200 N exerts a pressure of 40 N/m2. Calculate the area of the surface that the force is being applied to in square meters. Draw the triangle and circle the required property. Show step To calculate the area, you would need to divide the force by the pressure. Substitute values for the other two properties and complete the calculation. Show step Substituting F=200 N and P=40 N/m2, you have: A=200÷40=5 Write down the solution, including the units. Show step Area=5m2 Example 4: calculating area (converting units of pressure) A force of 950 N exerts a pressure of 53 Pa. Calculate the area in square meters to the nearest hundredth. Draw the triangle and circle the required property. Show step To calculate the area, you would need to divide the force by the pressure. Substitute values for the other two properties and complete the calculation. Show step As you need the units for the area to be in square meters, you need to convert the pressure from Pascals into N/m2. As 1Pa=1N/m2, you can simply state that our pressure P=53N/m2. Substituting F=950N and P=53N/m2 into the pressure triangle, you have A=950÷53=17.9245283… Write down the solution, including the units. Show step Area =17.92 m2 (nearest hundredth). Example 5: calculating force (N) A box has a cross-sectional area of 1.2 m2 and exerts a pressure of 60 N/m2 on the ground. Calculate the force that the box exerts on the ground, in Newtons. Draw the triangle and circle the required property. Show step To calculate the force, you would need to multiply the pressure by the area. Substitute values for the other two properties and complete the calculation. Show step As P=60N/m2 and A=1.2m2, substituting these into the formula triangle you get: F=60×1.2=72 Write down the solution, including the units. Show step Force =72 N Example 6: calculating force (converting units) A pressure of 35 N/m2 is exerted on an area of 40,000 cm2. Calculate the force in Newtons. Draw the triangle and circle the required property. Show step To calculate the force, you would need to multiply the pressure by the area. Substitute values for the other two properties and complete the calculation. Show step The area is in square centimeters, so you need to divide by 1002 to find the area in square meters. 40000÷1002=4 Substituting P=35 N/m2 and A=4 m2 into the pressure triangle, you have F=35×4=140 Write down the solution, including the units. Show step Force =140 N Example 7: calculating pressure given the mass An object has a mass of 2.4 kg and makes contact with the surface of the Earth with an area of 14.9 m2. Given that the acceleration due to gravity g=9.807 m/s2, calculate the pressure exerted on the Earth by the object to the nearest hundredth. State the units in your answer. Draw the triangle and circle the required property. Show step Before you use the formula triangle, you need to calculate the force exerted on the Earth by the object. This can be found by multiplying the mass by the acceleration due to gravity (F=MA). F=2.4×9.807=23.5368 N As you want to determine the pressure, you need to divide the force by the area: Substitute values for the other two properties and complete the calculation. Show step Now you have F=23.5368 N and A=14.9 m2. Substituting these values into the pressure triangle, you have: P=23.5368÷14.9=1.579651007… Write down the solution, including the units. Show step Pressure =1.58 N/m2 (nearest hundredth). Teaching tips for pressure force area Encourage students to rearrange the formula to solve for force or area, so they can see how each variable impacts the others directly. Explain that pressure applies differently in solids and fluids. In a solid, force is typically distributed over the contact area in one direction. In contrast, fluids (liquids and gases) apply pressure evenly in all directions, which is why pressure gauges can measure uniform fluid pressure. Introduce force as a vector quantity, with both magnitude and direction (example, a push of 10 Newtons applied upward or sideways). In contrast, explain that pressure is a scalar quantity, having only magnitude—when force is spread over an area, it becomes uniformly distributed as pressure, with no specific direction. Show real-world examples like hydrostatic pressure in dams, where water pressure against the dam wall increases with depth. Use this to explain the need for dam walls to be thicker at the base, highlighting the interaction between total pressure, force, and area. Easy mistakes to make Thinking that area is the total surface area of the objectThe area is the amount of surface with which the object is in contact. This will usually be one face of an object, or given as the cross-sectional area. Mixing up the placement of the letters in the triangleFormula triangles have limited use and if the letters are incorrectly written in the wrong place, the subsequent calculation will be incorrect. To overcome this, the standard units of pressure are N/m2 which is a force (N), divided by an area (m2) and so P=F÷A. Related compound measure lessons Speed distance time Speed formula Average speed formula Density formula Density mass volume triangle Mass from density and volume Flow rate Population density formula Practice pressure force area questions A force of 800 N acts on an area of 20 m2. Calculate the pressure: 0.4 N/m2 40 N/m2 4 N/m2 400 N/m2 Using the pressure triangle: and so P=800÷20=40 N/m2 A force of 970 N acts on an area of 350 m2. Calculate the pressure. Give your answer in Pascals to 3 significant figures. 2.78 Pa 0.361 Pa 0.36 Pa 2.77 Pa and so P=970÷350=2.771428571 N/m2=2.77 Pa (3sf) A cabinet has a mass m=61.1 kg and exerts a pressure of 380N/m2 on the surface of the Earth. Given that w=mg where g=9.807m/s2 and w is the weight of an object in Newtons, calculate the area of the base of the cabinet correct to the nearest hundredth. 0.63 m2 0.16 m2 1.58 m2 227698.93 m2 w=mg=61.1×9.807=599.2077 N and so A=599.2077÷380=1.576862368… A=1.58m2 (nearest hundredth) A force of 240 N exerts a pressure of 7.3 Pa. Calculate the area of the surface that the force is being applied to. 1752.0 m2 0.03 m2 30.4 m2 32.9 m2 1Pa=1N/m2 and so 7.3Pa=7.3N/m2 Using the pressure triangle, you have and so A=240÷7.3=32.87671233…A=32.9m2(1dp) A pressure of 40 N/m2 is exerted on an area of 7 m2. Calculate the force. 350 N 280 N 320 N 380 N and so F=40×7=280 N A pressure of 74 N/m2 is exerted on an area of 32000 cm2. Calculate the force. Give the answer correct to 3 significant figures. 236 N 23.1 N 23.2 N 237 N A=32,000 cm2=3.2 m2 and so F=74×3.2=236.8 N Did you know? Air pressure (atmospheric pressure) can change depending on where you are in the world. If you are up a mountain, the air pressure is less than when you are at sea level. Each time a particle comes into contact with a wall, it exerts a force on the wall. If you increase the number of particles within the container (such as an inflating balloon), the frequency of collisions increases, and so does the pressure of the gas inside. At high pressure, the balloon may not be able to withstand the increased amount of force on its surface, and so it bursts. Pressure force area FAQs What is the basic relationship between pressure, force, and area? The relationship is defined by the formula P=AF, where P is pressure, F is force, and A is area. This formula shows that pressure increases when force increases or when the area decreases. What units are used to measure pressure, force, and area? Pressure: Pascals (Pa), where 1 Pa=1 Newton per square meter Force: Newtons (N), the standard SI unit for force Area: Square meters (m2) How is pressure applied in thermodynamics? In thermodynamics, pressure in gases can change due to temperature or volume changes. For a fixed volume, increasing temperature raises gas pressure, which relates to particle collisions and energy transfer. What are some real-world applications of the pressure-force-area relationship? Applications include hydraulic systems, dams, tire design, and pressure-based sensors. In hydraulics, a small force over a small area can generate a larger force over a larger area, demonstrating Pascal’s principle. The next lessons are Ratio Proportion Converting fractions decimals and percentages Still stuck? 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Introduction What are pressure, force, and area? ↓ Area Force Pressure Pressure triangle Common Core State Standards How to calculate pressure, force, or area Pressure force area examples ↓ Example 1: calculating pressure (Pa) Example 2: calculating pressure (Bar) Example 3: calculating area (m2) Example 4: calculating area (converting units of pressure) Example 5: calculating force (N) Example 6: calculating force (converting units) Example 7: calculating pressure given the mass Teaching tips for pressure force area Easy mistakes to make Related compound measure lessons Practice pressure force area questions Pressure force area FAQs Next lessons Still stuck? x [FREE] Common Core Practice Tests (3rd to 8th Grade) Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents. Get your 6 multiple choice practice tests with detailed answers to support test prep, created by US math teachers for US math teachers! 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10040	https://courses.cs.washington.edu/courses/cse311/17sp/lectures/modular-arithmetic/proofs.pdf	CSE 311: Foundations of Computing I Modular Arithmetic Annotated Proofs Relevant Deﬁnitions a \| b (“a divides b”) Definition For a, b ∈Z, where a ̸= 0: a \| b iﬀ∃(k ∈Z) b = ka a ≡b (mod m) (“a is congruent to b modulo m) Definition For a, b ∈Z, m ∈Z+: a ≡b (mod m) iﬀm \| (a −b) Division Theorem Theorem For a ∈Z, d ∈Z+: There exist unique q, r ∈Z, where 0 ≤r < d such that a = dq + r A Modular Arithmetic Property Prove for all integers a, b and positive integers m, a ≡b (mod m) ↔a mod m = b mod m. Proof Commentary & Scratch Work Let a, b ∈Z and m ∈Z+. Prove the ∀’s. . . We want to prove a bi-implication; so, we will have two sub-proofs. First, we’ll assume the left and prove the right. Then, we’ll assume the right and prove the left. Suppose a ≡b (mod m). Begin with assuming the left and proving the right. At this point in the proof, we will be manipulating relevant deﬁnitions until the end. By deﬁnition of congruence, we have m \| (a −b). We can’t work with ≡’s. So, use the deﬁnition to remove the notation. By deﬁnition of divides, we have a −b = km for some integer k. Divides isn’t much better; apply deﬁnitions. Adding b to both sides, we have a = b + km. Taking both sides mod m, we have a mod m = (b + km) mod m = b mod m. So, a mod m = b mod m. Now, re-arrange the equations to get it to mods. Manipulate until we have what we wanted. Now, suppose a mod m = b mod m. Now, we prove the other implication. It’s the same “unroll the deﬁnitions” idea. By the division theorem, we have a = mka + (a mod m) for some ka ∈Z and b = mkb + (b mod m) for some kb ∈Z We need to get to equivalences, which we can do via divides, which we can get via equations. The division theorem seems like the right approach. Re-arranging both equations, we have: a mod m = a −mka and b mod m = b −mkb. We want the equations in terms of mod, because we can set them equal. Since these are equal, we have a−mka = b−mkb. Re-arranging, we have a −b = (ka −kb)m. So, by deﬁnition of divides, m \| (a −b). So, by deﬁnition of mod, we have a ≡b (mod m). Re-rolling the deﬁnitions in reverse. It’s worth not-ing that this feels a lot like the ﬁrst half of the proof in reverse. The only diﬀerence is that it uses diﬀerent variables. Another Modular Arithmetic Property Prove for all integers m ∈Z+, a, b ∈Z, if a ≡b (mod m) and c ≡d (mod m), then a + c ≡b + d (mod m). Proof Commentary & Scratch Work Let a, b ∈Z and m ∈Z+. Prove the ∀’s. . . Suppose a ≡b (mod m) and c ≡d (mod m). Prove the implication. . . Then, by deﬁnition of modular equivalences, we have m \| (a −b) and m \| (c −d). Apply a deﬁnition Furthermore, by deﬁnition of divides, we have k, l ∈ Z such that a −b = km and c −d = lm. Apply a deﬁnition Now, we actually have to think about what to do. In particular, we’re going to “re-roll” deﬁnitions. But how? Working backwards, we want a + c ≡b + d (mod m) ↔m \| ((a + c) −(b + d)) So, we put our pieces together to get there. Adding the equations together and re-arranging, we have (a + c) −(b + d) = (a −b) + (c −d) = km + lm = (k + l)m By deﬁnition of divides, we have m \| (a+c)−(b+ d). Apply a deﬁnition By deﬁnition of congruences, we have a + c ≡b + d (mod m). Apply a deﬁnition Another-nother Modular Arithmetic Property Prove for all integers m ∈Z+, a, b ∈Z, if a ≡b (mod m) and c ≡d (mod m), then ac ≡bd (mod m). Proof Commentary & Scratch Work Let a, b ∈Z and m ∈Z+. Prove the ∀’s. . . Suppose a ≡b (mod m) and c ≡d (mod m). Then, by deﬁnition of modular equivalences, we have m \| (a −b) and m \| (c −d). Apply a deﬁnition Furthermore, by deﬁnition of divides, we have k, l ∈ Z such that a −b = km and c −d = lm. Apply a deﬁnition Solving for a and c, and multiplying the results, we get ac = (km + b)(lm + d) = (klm)m + (dk)m + (bl)m + bd We want equations in terms of ac and bd; so, we solve for a and c. Taking both sides mod m, we get ac mod m = bd mod m By the ﬁrst theorem we proved, it follows that ac ≡bd (mod m) Always use theorems that have already been proven whenever possible! A Modular Arithmetic Proof Prove for all integers n ∈Z, n2 ≡0 (mod 4) or n2 ≡1 (mod 4). Proof Commentary & Scratch Work Let n ∈Z be arbitrary. We go by cases. After trying small examples, it looks like mod 2 is a good way to go! We split up our eﬀorts into the two cases mod 2. Case 1 (n is even): Suppose n is even. Then, there is some k ∈Z such that n = 2k. Multiplying both sides by n, we have n2 = (2k)2 = 4k2. So, by deﬁnition of divides and congruences, we have n2 ≡0 (mod 4). We want to prove something about n2; so, we get an equation for n2 and start manipulating and ap-plying theorems. . . Case 2 (n is odd): Suppose n is odd. Then, there is some k ∈Z such that n = 2k + 1. Multiplying both sides by n, we have n2 = (2k + 1)2 = 4k2 + 4k + 1. Taking both sides mod 4, we get n2 mod 4 = 1 mod 4. By the ﬁrst theorem we proved, it follows that n2 ≡1 (mod 4). We want to prove something about n2; so, we get an equation for n2 and start manipulating and ap-plying theorems. . . Since the claim is true for both cases, it’s true in general.
10041	https://thirdspacelearning.com/us/blog/teaching-addition-subtraction-elementary/	NEW LOWER-COST TUTORING Introducing Skye, your students’ AI voice tutor Adaptive, dialogue-driven one-on-one tutoring built by math teachers Unlimited sessions for as many grade 3-8 students as need it One fixed low yearly cost no matter how many sessions you schedule Meet Skye Operations & Algebraic Thinking Teaching Addition and Subtraction: A Guide For Elementary School Teachers From 2nd To 5th Grade March 31, 2025 \| 9 min read Neil Almond Addition and subtraction in upper elementary school math builds on the foundational skills students acquired in lower elementary school, helping them perform more sophisticated mathematics, and solve more complex problems. This post will show you what that progression looks like from 2nd grade to 5th grade, what your students should be able to do before moving on, and finally offer some practical suggestions as to how some objectives could be taught in the classroom. Addition and subtraction Teaching addition and subtraction from 2nd grade: before you begin… Teaching addition and subtraction: The theory Addition and subtraction 2nd Grade Addition and subtraction 3rd grade Addition and subtraction 4th grade Addition and subtraction 5th grade What is addition and subtraction? Addition and subtraction are two of the ‘four operations’–the four core math concepts children need a strong understanding of in order to tackle the rest of the subject. Addition is the act of putting two or more numbers together to obtain a larger result, and subtraction is the reverse – removing one or more numbers from another to obtain a smaller result. Addition and subtraction are among the first math skills children are taught, and are key in developing their number sense. Addition and subtraction Earlier in elementary school, students learn basic addition and subtraction. Kindergarten students should be able to solve addition and subtraction problems within 10 and develop methods to compose and decompose numbers within 19. 1st grade students should be able to solve addition and subtraction problems within 20 (i.e. one and two-digit addition) and develop methods to add within 100 and subtract multiples of 10. Students will only encounter relatively basic number sentences, are not required to write their own number sentences (however, this should be encouraged) and most teaching will incorporate manipulatives and simple visual representations, such as flashcards and number lines. Addition And Subtraction Check For Understanding Use this quiz to check your grade 2, 3, 4 and 7 students’ understanding of addition and subtraction. 15 questions with answers to identify areas of strength and support! Download Free Now! Teaching addition and subtraction from 2nd grade: before you begin… It is an understatement to say that a secure understanding of how place value works in base 10 is a key component to success in addition and subtraction in math. Along with students developing their mental models of number and understanding the ‘numberness’ of numbers between 1-20 e.g. four can be made the following ways: 4+0 1+3 2+2 2+1+1 1+1+1+1 Number bonds within 20 are also a key element that should be near to a high degree of fluency – meaning that students should be able to solve these problems mentally – by this stage. Students that have this conceptual understanding of numbers and the number system are far more likely to be successful when it comes to manipulating numbers in elementary math and beyond, especially when tasked with doing so mentally. Therefore, before you begin teaching this unit, it is worth knowing that when we teach for mastery, our first step is to ensure the prerequisite knowledge needed to be successful – it is not enough to rely on the fact that students have progressed into 2nd grade that they fully grasped the fundamentals and are ready for 2nd grade content. It is recommended that regardless of the curriculum your school follows, you ensure that you’ve finished teaching place value before you begin addition and subtraction in 2nd grade. If you are comfortable that your students are secure, then carry on reading. Meet Skye, the voice-based AI tutor making math success possible for every student. Built by teachers and math experts, Skye uses the same pedagogy, curriculum and lesson structure as our traditional tutoring. But, with more flexibility and a low cost, schools can scale online math tutoring to support every student who needs it. Find out more Teaching addition and subtraction: The theory It’s important to remember that students should still be using manipulatives at this point to help them with their conceptual understanding of the mathematical knowledge they are gaining. A possible error that new teachers may fall into is that because curriculums, including the Common Core State Standards, mentions that calculations should be done ‘mentally’, they may take that to mean that manipulatives cannot be used. Here it is worth reminding teachers that these objectives are the outcomes; it is totally appropriate (and I would argue necessary) to use manipulatives as part of a concrete, representational, abstract (CRA) approach when beginning to teach this unit. Of course, the purpose of any manipulative is to show the underlying mathematical structure so that it is understood; then gradually reduce the need for its requirement. Math manipulatives teachers may want to consider to help teach this unit include: Place value counters Base 10 blocks Visual representations a teacher should use, but eventually discourage use of include: Bar models Part-whole models Place value charts (Number lines are, by this point, too simple a representation to use with most students.) If sourcing these manipulatives is hard, the Didax website has some virtual ones that can be shown on a large interactive whiteboard. Addition and subtraction 2nd Grade For states following the Common Core State Standards in 2nd grade, building fluency with addition and subtraction is one of the main focuses of instructional time. Below are some key skills expected of 2nd grade students. Add and subtract numbers, including: add up to four two-digit numbers Fluently add and subtract within 100 add and subtract within 1000, using concrete models or drawings Mentally add 10 or 100 to a given number 100–900, and mentally subtract 10 or 100 from a given number 100–900 Use addition and subtraction within 100 to solve one- and two-step word problems For schools following the Texas state standards, or the TEKS, students will be expected to solve one-step and multi-step word problems, involving addition and subtraction, within 1,000 by the end of 2nd grade. Addition and subtraction activity ideas 2nd grade A good easy way into the objective ‘add and subtract numbers mentally…’ is to look at the addition and subtraction of multiples of one hundred. Base ten blocks are useful here as the students should be familiar with them from previous years. Furthermore, you can demonstrate the relationship between ones, tens and hundreds by counting to 10 in ones, 100 in tens and 1,000 in hundreds. By asking students ‘what do you notice?’, they will soon see the relationship between the amount of physical pieces you have and the quantity they represent. Once students have seen this pattern and are familiar, teachers can encourage whole-class skip counting, both forwards and backwards, playing games like showing a set amount of the 100 block, getting the students to close their eyes and hiding a set amount. The students will then have to tell you how many there were originally, how many were taken and how many are left. This can then be modeled to show a formal calculation. This will help improve the students’ understanding of place value and mental calculations. Furthermore, it will allow the students to feel successful, which will inspire them to try some of the more difficult objectives. The next step would be to take multiples of 100 away from numbers that have a value in the hundred, tens and ones. Addition and subtraction word problems 2nd grade A typical example of a word problem that students may be expected to solve by the of this period of teaching would look like this: There are 26 ducks in the pond. 45 more ducks arrive. How many birds are there? For this addition word problem, students may use many different strategies to solve it. One student may take 4 from 45 and add it to 26 to make a more friendly number of 30. And then add 41+30 = 71. Another student may add the 10s first (20+40 = 60) and then add the 1s (6+5 = 11). They would then add 60+11 = 71. It’s important to note that students using the formal algorithm method here is not the most important. Allowing students the ability to reason and think about how numbers are created can only help them in the long run. Mr. Almond has 425 marbles in a box. He loses 39. How many are left over? Similarly, the end goal of the unit would be for students to solve this subtraction word problem with the formal algorithm subtraction, but once again discussing possible mental subtraction strategies is highly recommended. Addition and subtraction: reasoning and problem solving 2nd grade There is, of course, more to the learning of math than just learning these objectives, and reasoning and problem solving should not just be limited to word problems. These questions will help develop the reasoning and problem-solving questions from this unit. Is this the most effective method? Discuss. Once the formal algorithm method has been learned, there is a tendency to overuse it. By giving students questions like this, it reinforces that we are merely providing new mathematical tools for the learner to use as they wish when they deem it appropriate. We need to be reminding students that there is always more than one method they can draw on. Creating problems that vary slightly to what a student has typically seen or experienced, is a good way to see if a student understands the underlying math or is merely able to parrot a method back at you. This bar model question may cause some issues at first as there are three numbers that have been added together to make the whole and the missing part comes between the two other parts and not at the end, as is typically seen in a classroom. This problem allows greater discussion of the underlying mathematics particularly the commutative property of addition which would allow the students to rearrange the bars as below. Read more: What is a bar model It is these types of questions that will push students’ addition and subtraction skills and set them on the path to being true mathematicians. These skills can then be used in other topics – for example, as a key part of teaching statistics and data handling. Addition and subtraction 3rd grade Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction. Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Addition and subtraction lesson ideas 3rd grade As well as revisiting the objectives from 2nd grade (remember those prerequisites), in 3rd grade students move towards fluently working with numbers within one thousand. Because of the hierarchical nature of math (there is a certain order that knowledge of the domain needs to be taught in for the rest of it to make sense and stick), it is so crucial that students are comfortable with counting in 100s. Building on the advice given in 2nd grade, considering using place value counters could be one such way into this unit – though it is hoped that by this point, students are secure in their understanding of place value. For this section, I want to focus on the objective, ‘Assess the reasonableness of answers using mental computation and estimation strategies including rounding.’ Depending on the prior experiences of the students, using Cuisenaire rods can be helpful in showing this. With plenty of practice of counting from the smaller number up to the larger number, coupled with the physical taking away using place value counters, students should quickly grasp this idea. If students are already familiar with the idea they can move to the next part. The standard algorithm: addition and subtraction In 3rd grade, students encounter the formal written method of standard algorithm addition and subtraction. It is common practice for the standard algorithm in addition to be taught first, swiftly followed by the standard algorithm in subtraction. This is an area that is generally taught well by teachers, as they would likely have been taught this method themselves when at school. Assuming automaticity within both methods, when students reach 4th grade, it is possible to combine using the standard algorithm in both addition and subtraction in order to ensure that students can use the inverse to check their answers. As students are now becoming familiar and increasingly familiar with the algorithm method for addition and subtraction, now is an excellent time to add in one more step to the method, which is to perform the inverse calculation as part of the process. This would be what the students are familiar with already: ``` 4,532 +3,653 7,185 ``` What I would propose is the following: ``` 4,532 +3,653 7,185 3,653 - 4,532 ``` Here the students have taken the sum and an addend from the addition part of the calculation and used them to form a minuend and a subtrahend of a subtraction calculation. The difference between the minuend and the subtrahend was also the first addend of the addition question. If this is the case, then the original question has been answered correctly. Addition and subtraction word problems 3rd grade A typical problem you would expect students to answer would be the following: Mr. Almond buys a laptop for $482 and a tablet for $239. How much did he spend altogether? Students should use a bar model to represent this addition word problem. And then use a formal written method to solve – including the use of the inverse calculation at the end. 482 +239 721 239 - 482 Addition and subtraction: Reasoning and problem-solving 3rd grade When looking at creating reasoning and problem-solving activities, it is highly appropriate to look back on the objectives from previous years and create a reasoning or problem-solving activity based on them. In mathematics, maturation matters. If some ideas are still too novel for students, despite them showing some success with them, then solving problems with them can be overwhelming. As we want students to attend to the mathematics, creating difficult problems with numbers they are more comfortable with frees up the students thinking to consider the structures and not worry about the numbers. A good problem to look at would be missing numbers in a calculation. Plenty of reasoning is available in these questions. Students can look at the addend and sum and reason the missing number must be an even number as adding an even number to an odd number produces another odd number. These questions also allow students to practice their fluency of number bonds. These questions can vary in difficulty by bridging to the next place value or not – something that students often find difficult during these tasks. Addition and subtraction 4th grade Fluently add and subtract multi-digit whole numbers using the standard algorithm. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Solve addition and subtraction multi-step problems in contexts, deciding which operations and methods to use and why Addition and subtraction activities 4th grade As students continue to develop their math skills in this area, it is hoped that they have developed a strong understanding of the place value system. In 4th grade place value, students learn numbers to at least 1,000,000. It is therefore likely that the questions you use will include 4-digit numbers, and possibly up to 6-digit numbers. To add an element of challenge into the teaching at this step, teachers could try some of the following: Provide equations that require balancing on both the left and the right-hand side of the equal sign. E.g. 142,530 + 432,943 = 354,954 + ________ Ask students round the numbers in the question to a nearest whole place value to estimate the answer first. E.g. 142,530 + 432,943 = 354,954 + 220,519 142,500 + 432,900 = 355,950 + 219,450 Doing this further enhances the students’ understanding of the equals sign while practicing rounding skills. You could challenge students to round the same numbers to the nearest hundred thousand, ten thousand, thousand, hundred and tens to investigate which rounding will give an estimate nearer the final answer. Addition and subtraction word problems 4th grade At this stage, in accordance with the Common Core State Standards, students should be solving multi-step problems in context. A typical problem could be something similar to the following: Journey Distance (miles) London to Paris 580 miles London to Rome 908 miles Paris to Rome 737 miles A plane flies from London to Rome and then on to Paris. How much further is this than flying direct to Paris from London? 908 miles + 737 miles = 1645 miles 1645 miles – 580 miles = 1065 miles This question relies on students having an understanding of measurement and uses numbers that they are familiar with. Again, with two-step problems, using slightly easier numbers is actually beneficial, as it allows students to concentrate on understanding the language and structure of the question as to why it is a multiple step problem. Students should encounter plenty of worked examples of these before attempting their own. Addition and subtraction: Reasoning and problem-solving 4th grade Reasoning and problem-solving in 4th grade gets more sophisticated. Numbers are exchanged for symbols and a greater use of unknown quantities is used to get students to reason about their knowledge of numbers, laying the foundations for algebra in high school. A typical problem may take the form of the following: The way to tackle a problem such as this is to ask students what is the same and what is different about the first two calculations. Teasing out that the difference between the two equations is one square and that the difference between the two answers is 700. From this, students can deduce that a square is equal to 700 and therefore a triangle is 500. When adding two triangles and a square together, you get the answer 1,700. Read more: Teaching Addition In Elementary School: How To Identify and Fix The Most Common Gaps Addition and subtraction 5th grade Perform mental calculations, including with mixed operations and large numbers Use their knowledge of the order of operations to carry out calculations involving the 4 operations Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Addition and subtraction word problems 5th grade: When looking at 5th grade word problems, pupils should be able to solve them in a range of real-world contexts (money and measurement, for example) as well as with up to 2 decimal places. A common subtraction problem for this age group could be something on the lines of the following: The Children and Green School are raising money for a charity. Their target is to collect $380. So far they have collected $77.73. How much more money do they need to reach their target? The difficulty in this subtraction word problem is the requirement that when performing the subtraction, students must remember to include two-zero place holders from the target number to ensure the place value is correct. 380.00 77.73 – Students will then have to cross three place value holders to the tens column in order to perform all the necessary exchanges, which adds an additional level of difficulty. Once the calculation has been performed, students should get the answer $302.27. Addition and subtraction: Reasoning and problem-solving 5th grade The following question provides an insight into the type of reasoning and problem solving that is expected by the end of 5th grade. Students are expected to take the information from the table and perform a 2-step calculation – find the combined height of Kilimanjaro and Ben Nevis and then the differences between this total height of both mountains and Everest. This could be represented in the following way using bar modelling to make the two calculations clearer to see. Adding both heights of Kilimanjaro and Everest gives a height of 7,239 meters. When this is subtracted from the height of Everest (8,848m), you are left with the answer 1,609 meters. Read more: Teaching Subtraction In Elementary School: How To Identify and Fix The Most Common Gaps Addition and subtraction are the first mathematical skills that students really get to grips with, and having a strong foundation in them is key to becoming better mathematicians. Hopefully this post has given you some good ideas to achieve exactly that for your class, no matter which grade they are in! READ MORE: 4th grade math problems Subtraction worksheets Addition and subtraction worksheets Do you have students who need extra support in math? Skye—our AI math tutor built by experienced teachers—provides students with personalized one-on-one, spoken instruction that helps them master concepts, close skill gaps, and gain confidence. Since 2013, we’ve delivered over 2 million hours of math lessons to more than 170,000 students, guiding them toward higher math achievement. Discover how our AI math tutoring can boost student success, or see how our math programs can support your school’s goals: – 3rd grade tutoring – 4th grade tutoring – 5th grade tutoring – 6th grade tutoring – 7th grade tutoring – 8th grade tutoring The content in this article was originally written by primary school lead teacher Neil Almond and has since been revised and adapted for US schools by elementary math teacher Christi Kulesza. Share: Related articles What is the sum? A Guide For Teachers And Students 6 min read What Is The Quotient? A Step-by-Step Guide For Students And Teachers 5 min read What Is Box Method Multiplication? Explained For Elementary School Teachers, Parents And Pupils 3 min read What Is Order Of Operations: Explained For Elementary School 3 min read x [FREE] Ultimate Math Vocabulary Lists (K-5) An essential guide for your Kindergarten to Grade 5 students to develop their knowledge of important terminology in math. Use as a prompt to get students started with new concepts, or hand it out in full and encourage use throughout the year. Download free
10042	https://www.bbc.co.uk/bitesize/guides/zqhk7ty/revision/4	One number as a fraction of another - Fractions - National 5 Application of Maths Revision - BBC Bitesize BBC Homepage Skip to content Accessibility Help Sign in Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds More menu More menu Search Bitesize Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Close menu Bitesize Menu Home Learn Study support Careers Teachers Parents Trending My Bitesize More England Early years KS1 KS2 KS3 GCSE Functional Skills Northern Ireland Foundation Stage KS1 KS2 KS3 GCSE Scotland Early Level 1st Level 2nd Level 3rd Level 4th Level National 4 National 5 Higher Core Skills An Tràth Ìre A' Chiad Ìre An Dàrna Ìre 3mh ìre 4mh ìre Nàiseanta 4 Nàiseanta 5 Àrd Ìre Wales Foundation Phase KS2 KS3 GCSE WBQ Essential Skills Cyfnod Sylfaen CA2 CA3 CBC TGAU International KS3 IGCSE More from Bitesize About us All subjects All levels Primary games Secondary games National 5 Fractions One number as a fraction of another Calculations can be carried out using fractions of shapes and quantities. Mixed fractions can be added or subtracted to find the number of fractional parts in a mixed number. Part ofApplication of MathsNumeracy skills Save to My Bitesize Save to My Bitesize Saving Saved Removing Remove from My Bitesize Save to My Bitesize close panel In this guide Revise Test Pages Fractions of a quantity Equivalent fractions Cancelling fractions One number as a fraction of another Adding and subtracting fractions Fractions and whole numbers Fractions - video summaries One number as a fraction of another Imagine that there are 10 questions in a test and you get 7 of them correct. You would say that you got 7 10, because 7 as a fraction of 10 is 7 10. In the same way, 4 as a fraction of 12 is 4 12 or 1 3 Similarly 20 as a fraction of 48 is 20 48 or 5 12. Seems easy, but just be careful with the units. For example, 20 p as a fraction of £20 is not: 20 20 but 20 2000 because £20 is 2000 p. Similarly, 30 c m as a fraction of 5 m is 30 500 because 5 m=500 c m. Next page Adding and subtracting fractions Previous page Cancelling fractions More guides on this topic Order of operations Multiplication Division Negative numbers Distance, speed and time Rounding Rounding and estimating Percentages Ratio Related links BBC Podcasts: Maths BBC Radio 4: Maths collection BBC Skillswise SQA Lifeskills Maths GOV.UK Maths is Fun NRICH Maths Club Language: Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Terms of Use About the BBC Privacy Policy Cookies Accessibility Help Parental Guidance Contact the BBC BBC emails for you Advertise with us Do not share or sell my info Copyright © 2025 BBC. The BBC is not responsible for the content of external sites. Read about our approach to external linking.
10043	https://www.numbersaplenty.com/set/lucky_number/	lucky numbers Lucky numbers are those numbers which survive a sieving process which is similar to the Eratosthenes sieve that can be used to compute the prime numbers. Let us declare that 1 is a lucky number and let us start with a sieve containing only the odd numbers: The first number greater than 1 is 3, so we declare 3 lucky and we delete from the sieve all the numbers in a position which is a multiple of 3, i.e., 5, 11, 17, 23, and so on. We are left with The first new survivor is 7, so we declare 7 lucky and we delete from the sieve all the numbers in a position which is a multiple of 7, i.e., 19, 39, and so on. The next lucky number is thus 9 and continuing this ideal process we could identify all the lucky numbers. The lucky numbers have been studied because they have some properties (like density) which resemble the properties of primes. The first lucky numbers are 1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 49, 51, 63, 67, 69, 73, 75, 79, 87, 93, 99, 105 more terms Below, the spiral pattern of lucky numbers up to 3600. See the page on prime numbers for an explanation and links to similar pictures. Pictorial representation of remainders (mod 2, 3, ...,11) frequency. For a table of values and more details click here A graph displaying how many lucky numbers are multiples of the primes p from 2 to 71. In black the ideal line 1/p. Useful links Mathworld, Lucky Number Wikipedia, Lucky number OEIS, Sequence A000959 Lucky numbers can also be... (you may click on names or numbers and on + to get more values) a-pointer 13 631 1801 + 9923227 ABA 1029 2187 10125 + 7381125 aban 13 15 21 + 9000985 abundant 1575 2835 3465 + 9999825 Achilles 1323 7803 8575 + 9948123 admirable 4095 6435 9555 + 6220665 alt.fact. 619 alternating 21 25 43 + 9898927 amenable 13 21 25 + 9999997 amicable 12285 14595 69615 + 4482765 anti-perfect 285 133857 apocalyptic 285 361 529 + 29995 arithmetic 13 15 21 + 9999997 astonishing 15 429 133533 1301613 automorphic 25 90625 balanced p. 211 1123 3307 + 9983977 Bell 15 21147 bemirp 1606081 betrothed 75 195 1575 + 3676491 binomial 15 21 105 + 9988215 brilliant 15 21 25 + 9999811 c.decagonal 31 151 211 + 9779011 c.heptagonal 43 463 841 + 9731947 c.nonagonal 1711 3403 3655 + 9916831 c.octagonal 25 49 169 + 9941409 c.pentagonal 31 51 141 + 9726891 c.square 13 25 421 + 9994921 c.triangular 31 235 361 + 9957529 cake 15 93 23479 + 9437505 canyon 105 201 205 + 9876589 Carmichael 1105 2821 6601 + 9890881 Carol 223 Catalan 429 Chen 13 31 37 + 9999397 congruent 13 15 21 + 9999997 constructible 15 51 1285 983055 cube 729 29791 50653 + 9393931 Cullen 25 385 897 4609 Cunningham 15 31 33 + 9884737 Curzon 21 33 69 + 9998841 cyclic 13 15 31 + 9999997 D-number 15 21 33 + 7043109 d-powerful 43 63 135 + 9996789 de Polignac 127 331 997 + 9999997 decagonal 297 451 855 + 9942985 deceptive 259 451 2821 + 9890881 deficient 13 15 21 + 9999997 dig.balanced 15 21 37 + 9999963 double fact. 15 105 135135 Duffinian 21 25 49 + 9999997 economical 13 15 21 + 9999823 emirp 13 31 37 + 9999739 emirpimes 15 49 51 + 9999721 enlightened 119911 3720087 equidigital 13 15 21 + 9999823 eRAP 2079 14905 35167 + 9472585 esthetic 21 43 67 + 9876787 Eulerian 152637 evil 15 33 43 + 9999963 fibodiv 75 427 6505 + 3769767 Fibonacci 13 21 1597 + 75025 Friedman 25 127 289 + 995347 frugal 729 1029 1369 + 9985759 gapful 105 135 195 + 9997135 Gilda 49 997 good prime 37 67 127 + 9993223 happy 13 31 49 + 9999553 Harshad 21 63 111 + 9999825 heptagonal 189 235 2059 + 9967027 hex 37 127 169 + 9975457 hexagonal 15 231 2415 + 9988215 hoax 319 361 391 + 9998311 Hogben 13 21 31 + 9976123 Honaker 1039 1933 2221 + 9985231 house 933 1285 23101 + 9566817 hyperperfect 21 2133 19521 + 3420301 iban 21 43 73 + 777777 iccanobiF 13 1053 104937 idoneal 13 15 21 + 1365 inconsummate 63 75 195 + 999951 insolite 111 interprime 15 21 69 + 9999957 Jacobsthal 21 43 171 + 699051 junction 105 111 115 + 9999961 Kaprekar 99 297 9999 + 9372385 katadrome 21 31 43 + 9876541 Kynea 79 1087 16639 Lehmer 15 51 133 + 9973693 Leyland 131361 268705 lonely 211 31429 203713 + 2010807 Lucas 3571 9349 710647 Lynch-Bell 15 135 735 + 9315 m-pointer 1123 1231 15121 4121113 magic 15 111 1105 + 9951391 magnanimous 21 25 43 + 6640063 metadrome 13 15 25 + 2345689 modest 13 49 69 + 9992727 Moran 21 63 111 + 9999283 Motzkin 21 51 127 + 2356779 mountain 141 151 163 + 8986531 narcissistic 92727 nialpdrome 21 31 33 + 9999997 nonagonal 75 111 261 + 9980301 nude 15 33 99 + 9991395 oban 13 15 25 + 997 octagonal 21 133 645 + 9857281 odious 13 21 25 + 9999997 Ormiston 18397 36997 37813 + 9953071 palindromic 33 99 111 + 9986899 palprime 151 727 787 + 9514159 pancake 37 67 79 + 9965881 panconsummate 15 21 31 + 3097 pandigital 15 21 75 + 9998163 partition 15 135 231 + 5392783 pentagonal 51 477 651 + 9867555 pernicious 13 21 25 + 9999997 Perrin 51 367 1497 + 727653 Pierpont 13 37 73 + 5038849 plaindrome 13 15 25 + 8899999 Poulet 645 1105 1387 + 9920401 power 25 49 169 + 9941409 powerful 25 49 169 + 9948123 prim.abundant 1575 3465 4095 + 9999825 prime 13 31 37 + 9999823 primeval 13 37 10279 12379 Proth 13 25 33 + 9990145 pseudoperfect 1575 2835 3465 + 999075 repdigit 33 99 111 + 7777777 repfigit 75 7909 355419 1084051 repunit 13 15 21 + 9976123 Rhonda 2835 19435 33475 + 9843255 Ruth-Aaron 15 25 49 + 9997417 Sastry 106755 self 31 75 211 + 9999807 semiprime 15 21 25 + 9999997 sliding 25 133 205 + 7073125 Smith 319 391 483 + 9998613 sphenic 105 195 231 + 9999961 square 25 49 169 + 9941409 star 13 37 73 + 9746101 straight-line 111 135 159 + 7777777 strobogrammatic 69 111 619 + 6911169 strong prime 37 67 79 + 9999823 super-d 31 69 105 + 9999807 tau 4761 12321 15129 + 9941409 taxicab 46683 149389 216027 + 7882245 tetrahedral 1771 4495 9139 + 9886435 tetranacci 15 10671 triangular 15 21 105 + 9988215 tribonacci 13 927 1705 10609 trimorphic 25 49 51 + 9218751 truncatable prime 13 31 37 + 9981373 twin 13 31 43 + 9999049 uban 13 15 21 + 9000079 Ulam 13 69 87 + 9999823 undulating 141 151 171 + 9696969 unprimeable 535 895 1645 + 9999855 upside-down 37 73 159 + 997311 vampire 1395 1435 1827 + 792585 wasteful 33 51 63 + 9999997 weak prime 13 31 43 + 9999049 weakly prime 604171 2474431 6463267 Wieferich 1093 10533 14209 + 3697083 Woodall 63 159 895 + 524287 Zeisel 105 1419 5719 + 8712985 Zuckerman 15 111 115 + 7133175 Zumkeller 1575 2835 3465 + 97335 zygodrome 33 99 111 + 9999555 Giovanni Resta, 2013-2025 • e-mail: info -at- numbersaplenty.com • search limits
10044	https://www.youtube.com/watch?v=0rPMBtY4zwM	Fluid Mechanics Lesson 08C: Fully Developed Pipe Flow John Cimbala 14400 subscribers 106 likes Description 9361 views Posted: 17 Sep 2022 Fluid Mechanics Lesson Series - Lesson 08C: Fully Developed Pipe Flow In this 12.5-minute video, Professor Cimbala examines more closely the region beyond the entrance of a pipe - the fully developed region, where shear stress and velocity profile do not change, but pressure drops linearly due to irreversible losses. He also compares laminar and turbulent flow and works through an example problem. This video incorporates material from Section 8-4 of the Fluid Mechanics textbook by Cengel and Cimbala. An Excel file listing of all the videos in this series can be found at . If you liked this video, please subscribe to Dr. Cimbala's YouTube channel at to be informed when new videos are posted. You can also watch all related short videos with one click by going to one of Dr. Cimbala's playlists: Fluid Mechanics Lesson Series: Two-Minute Bible Videos: Two-Minute Excel Tutorials: Two-Minute Fluid Mechanics: Two-Minute Math and Statistics Videos: Two-Minute Science Videos: Short Marshmallow Peep Videos: Thirty-Second Engineering: Dr. John M. Cimbala is Professor of Mechanical Engineering at Penn State. He is an educator, textbook author, Christian author, husband, father, and grandfather. He also created and maintains a website for helping people grow in their faith called Christian Faith Grower at His YouTube channel is at 4 comments Transcript: welcome to lesson 8c fully developed pipe flow in this lesson we discuss what happens beyond the entrance region what we call the fully developed pipe flow region we'll look at differences between laminar and turbulent fully developed pipe flow such as velocity profile wall shear stress and pressure drop and we'll do one example problem first let's compare laminar and turbulent fully developed pipe flow first we look at velocity profiles i'll split the page in half between laminar and turbulent flow turns out that we can solve laminar fully developed pipe flow exactly we'll do that in a later lesson but there is no analytical solution for turbulent fully developed pipe flow for laminar flow the flow is steady turbulent flow is always unsteady recall we mentioned the 3d swirling eddies that you get in a turbulent flow but a turbulent pipe flow can be steady in the mean the laminar flow velocity profile is parabolic the mean velocity profile is all that we talk about in the turbulent case and it's fuller than that of the laminar case i'll illustrate with some sketches this is the velocity profile for fully developed laminar pipe flow where lowercase r is the radius d is pipe diameter we have a maximum speed at the center and u is a function of r we generally use x as the downstream coordinate it turns out that v average equals one half u max which i sketch here as i mentioned we do have an analytical solution which is given by this equation and we usually just use v instead of v average we'll be able to derive this equation later when we talk about differential analysis of fluid flows now let's draw the turbulent case we use the same coordinate system and u is still a function of r in the mean but the instantaneous velocity profile can look something like this since there are all these turbulent eddies a short time later it might look like that and if we keep taking lots of different profile tests we see that there's lots of scatter because of all the turbulent eddies so what we generally do is average these over time and generate a mean velocity profile which looks like this thick red line which is now u of r or u bar of r since it's an average u max still occurs at the center line and we can define v or v average as the average speed through the pipe when i say the velocity profile is fuller what that means is it's more like a top hat or one-dimensional profile compared to the laminar case which is parabolic in other words it's pretty flat in the middle of the pipe and then very rapidly decays to zero near the wall whereas v average was a half umax for laminar flow the average is about 85 percent of umax for turbulent flow and this is a ballpark number it does depend on reynolds number there's no analytical solution for turbulent pipe flow like we had for the laminar case but you can see the textbook for some empirical equations such as a power law and the log law i won't get into any detail about these at this point instead i want to look now at the differences with respect to the wall shear stress recall that shear stress tau is mu d u d y for simple shear flows again i sketched the laminar profile u of r we're interested in the wall shear stress this is the shear stress right at the wall acting on the fluid if we let y be the distance from the wall you see that y and r are in opposite directions in fact y equal capital r minus little r where capital r is the radius so d u d y is minus d u d r you can show this by using the chain rule taking d u d y in terms of r since y is a function of r to avoid problems with negative signs i'm going to just use the absolute value so tau w is mu d u d r absolute value at the wall and we can figure out the sign it depends if we're talking about tau wall acting on the fluid or the fluid acting on the pipe here the pipe is trying to slow down the fluid due to viscosity so tau w is in this direction tau w is proportional to the slope d u d y at the wall we do the same with turbulent flow drawing the mean velocity profile and notice that the slope d u d y is much bigger at the wall in other words speed u increases rapidly with distance y from the wall compared to the more slow increase for the laminar case so tau w is much bigger tau w turbulent is greater than tau w laminar because of the steeper slope at the wall how much bigger in some cases about a factor of 3 if you have the same conditions except laminar versus turbulent flow the practical implication of this is that turbulent pipe flow has a larger pressure drop for the same flow rate physically since we have a larger shear stress we have more friction and you have to push the fluid harder to get it to have the same flow rate compared to the laminar case so because of this the pressure drop for turbulent flow is greater and also the irreversible head loss is greater for turbulent flow than for laminar flow again if you have the same flow rate at the same v-dot let's analyze the pressure drop for fully developed pipe flow this applies to either the laminar or the turbulent case we've done problems like this before using the conservation laws we consider a horizontal pipe which is round you can use hydraulic diameter if it's not round and we're far enough downstream from the entrance that this section of pipe is fully developed so the velocity profile at that point is identical downstream call these points one and two tau w at the wall is also constant since it's fully developed and the slope doesn't change note that i draw it here but it's actually all around the pipe the inside wall of the pipe there's some volume flow rate and the flow is fully developed incompressible and steady if it's laminar flow or steady in the mean if it's turbulent flow the length of this pipe section is capital l the pressure at point one is p1 acting to the right pressure at 2 is p2 acting to the left i'll draw p1 longer because p1 has to be greater than p2 in order to overcome friction to push this fluid through the pipe let's pick a control volume inside the pipe crossing one and two along the walls of the pipe now we apply our conservation laws conservation of mass is m dot one equal m.2 equal m-dot or rho v dot one equal row v dot two for incompressible flow the densities cancel so this equation becomes b 1 i d squared over 4 equal v 2 pi d squared over 4 since volume flow rate is average speed times cross sectional area well these terms cancel and so v1 equal v2 we already knew that since the profile is not changing in a fully developed pipe flow the average speed does not change now let's apply conservation of momentum in the x direction this is a good review of previous lessons by the way recall our workhorse formula sigma fx equals sigma fx gravity plus sigma fx pressure plus sigma fx viscous plus sigma fx other equal the sum of all the outlets of beta m dot u minus sum over all the inlets of beta m dot u well let's work on these terms there's no gravity in the x direction the pressure term is p1 pi d squared over 4 minus p2 id squared over four the viscous force is in the negative x direction negative tau wall times pi d l where pi d l is the total wall surface area in this control volume so tau wall acts on the entire surface area which is pi d the circumference times length l there are no other forces either in this control volume for the momentum flow rates everything coming out is the same as everything coming in since it's fully developed so these terms cancel betas are the same m dot is the same and u is the same u here being the average speed in the x direction so x momentum reduces to p 1 minus p 2 pi d squared over 4 since these are the only two terms left i'll put this term on the right side so we have tau w pi dl on the right the pi's cancel one of the d's cancel and we can rewrite this as p1 minus p2 equal 4 tau w l over d this is the result of the x momentum equation again this is valid for laminar or turbulent flow i haven't said which we're talking about at this point now let's apply the head form of the energy equation for the same control volume i'll rewrite that equation again as good review at the inlet we have p1 over rho g plus alpha 1 v1 squared over 2g plus z1 plus h pump u plus the same first three terms at the outlet with subscripts 2 and then the extracted turbine head and the irreversible head loss hl well some of these terms cancel again since it's fully developed alphas are the same and speeds are the same so these two terms cancel it's a horizontal pipe so the z's cancel we have no pump we have no turbine so the energy equation reduces to p1 minus p2 equal rho ghl physically this is the pressure drop expressed as a head and the reason there's a pressure drop is because we have irreversible head losses due to friction in the pipe i'll call this equation two again this holds for either laminar or turbulent flow now let's equate one and two since both of them have p1 minus p2 on the left hand side as you can see so 4 tau wl over d from equation 1 must equal rho ghl from equation 2 where hl is 4 tau w over rho g l over d i'll call this equation 3. we still can't use this equation because we don't know what tau w is what is tau w for laminar flow we can solve for it exactly for turbulent flow we'll need some empirical equations for either case we previously performed a dimensional analysis and we found that darcy friction factor f which is 8 tau w over rho v squared is a function of reynolds number and non-dimensional roughness parameter in terms of f the darcy friction factor equation 3 becomes hl equal 4 f rho v squared over 8 rho g l over d equal f b squared over 2g l over d so we have another expression for h l the irreversible head loss namely f l over d v squared over 2 g i'll call this equation 4 and this will be our workhorse equation for fully developed pipe flow now let's do an example problem water at 20 degrees c flows through a long horizontal straight section of round pipe the section under consideration is fully developed and at a set of operating conditions these are the known values pipe diameter pipe length reynolds number this is definitely turbulent flow and darcy friction factor we want to calculate the irreversible head losses through the pipe hl i'll label these parameters d l reynolds number and darcy friction factor f let's use our workhorse equation h l is f l over d v squared over 2 g we're not given v but we are given reynolds number so we'll have to calculate v from that first of all for water at 20 degrees c i look up the properties kinematic viscosity is 1.004 times 10 to the minus 6 meters squared per second and since re is vd over nu i can solve for v v is re times nu over d at this point most students would plug in their calculators and get v but i like to stay in variable form as far as possible so hl becomes fl over d our expression for v squared and 1 over 2g this is our answer in variable form we now know everything in this equation so we plug in the numbers f l d reynolds number nu d again that whole quantity squared and 1 over 2g you can verify that all the units cancel except meters and our final answer is hl is 17.5 meters what does hl mean it's the irreversible head loss due to irreversible losses in the pipe thank you for watching this video please subscribe to my youtube channel for more videos [Music]
10045	https://linbo.one/files/courses/2018Fall-M325K/M325K%20Lecture%205%20-%20Section%203.1&3.2.pdf	Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quantiﬁed statements Bo Lin September 13th, 2018 Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Outline Predicates and quantiﬁers. Universal and existential quantiﬁed statements. Other forms of quantiﬁed statements. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Motivation Now we can do reasoning by valid argument forms. But there are a lot more complicated arguments. For example: all humans are mortal; Socrates is a human; ∴Socrates is mortal. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Motivation Now we can do reasoning by valid argument forms. But there are a lot more complicated arguments. For example: all humans are mortal; Socrates is a human; ∴Socrates is mortal. This should be a valid argument form, but it does not ﬁt any patterns we introduced last time. In fact, our tools to deal with statements are not enough. And as a result, we need to introduce some new concepts and tools to study these arguments, which is called predicate logic. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Predicates In grammar, the word predicate refers to the part of a sentence that gives information about the subject. Deﬁnition A predicate is a sentence that contains a ﬁnite number of variables and becomes a statement when speciﬁc values are substituted for the variables. The domain of a predicate variable is the set of all values that may be substituted in place of the variable. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Predicates In grammar, the word predicate refers to the part of a sentence that gives information about the subject. Deﬁnition A predicate is a sentence that contains a ﬁnite number of variables and becomes a statement when speciﬁc values are substituted for the variables. The domain of a predicate variable is the set of all values that may be substituted in place of the variable. Remark Usually we can interpret a predicate as a function whose co-domain consists of statements. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Examples of predicates as functions Example Let S(x) be the predicate x2 > x with domain R. Rewrite the following statements in sentences: (a) S(2); (b) ~S(1) ∧S(0). Are they true or false? Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Examples of predicates as functions Example Let S(x) be the predicate x2 > x with domain R. Rewrite the following statements in sentences: (a) S(2); (b) ~S(1) ∧S(0). Are they true or false? Solution (a) is simply 22 > 2 and it is true. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Examples of predicates as functions Example Let S(x) be the predicate x2 > x with domain R. Rewrite the following statements in sentences: (a) S(2); (b) ~S(1) ∧S(0). Are they true or false? Solution (a) is simply 22 > 2 and it is true. (b) is ’(not 12 > 1) and 02 > 0’, which is equivalently ’12 ≤1 and 02 > 0’, so it is false. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Truth set Suppose we have a predicate P(x) with domain D, then for each y ∈D, P(y) is a speciﬁc statement and it is either true or false. Then we would like to know that for what elements y ∈D, P(y) is true? Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Truth set Suppose we have a predicate P(x) with domain D, then for each y ∈D, P(y) is a speciﬁc statement and it is either true or false. Then we would like to know that for what elements y ∈D, P(y) is true? Deﬁnition If P(x) is a predicate with domain D, the truth set of P(x) is the set of all elements of D that make P(x) true when they are substituted for x. The truth set of P(x) is denoted {x ∈D \| P(x) is true }. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Quantiﬁers In a sentence, even if we ﬁx the subject and the predicate, there is still a twist: the number of subjects referred to? For example, the following sentences have very diﬀerent meanings. As a result, we must consider the quantiﬁers too. all humans are mortal; some humans are mortal; one human is mortal; no human is mortal. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Quantiﬁers In a sentence, even if we ﬁx the subject and the predicate, there is still a twist: the number of subjects referred to? For example, the following sentences have very diﬀerent meanings. As a result, we must consider the quantiﬁers too. all humans are mortal; some humans are mortal; one human is mortal; no human is mortal. Deﬁnition Quantiﬁers are words that refer to quantities such as ’some’ or ’all’ and tell for how many elements a given predicate is true. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements The quantiﬁers ∀and ∃ In predicate logic, there are two quantiﬁers that we use all the time: ∀and ∃. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements The quantiﬁers ∀and ∃ In predicate logic, there are two quantiﬁers that we use all the time: ∀and ∃. Deﬁnition The universal quantiﬁer, written as ∀and read ’for all’, refers to all elements in the domain. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements The quantiﬁers ∀and ∃ In predicate logic, there are two quantiﬁers that we use all the time: ∀and ∃. Deﬁnition The universal quantiﬁer, written as ∀and read ’for all’, refers to all elements in the domain. Deﬁnition The existential quantiﬁer, written as ∃and read ’there exists/there exist’, refers to at least one element in the domain. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Universal statements Deﬁnition Let Q(x) be a predicate and D the domain of x. A universal statement is a statement of the form ’∀x ∈D, Q(x)’. It is deﬁned to be true if and only if Q(x) is true for every x in D. It is deﬁned to be false if and only if Q(x) is false for at least one x in D. A value for x for which Q(x) is false is called a counterexample to the universal statement. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Universal statements Deﬁnition Let Q(x) be a predicate and D the domain of x. A universal statement is a statement of the form ’∀x ∈D, Q(x)’. It is deﬁned to be true if and only if Q(x) is true for every x in D. It is deﬁned to be false if and only if Q(x) is false for at least one x in D. A value for x for which Q(x) is false is called a counterexample to the universal statement. Remark One needs to specify the domain of universal statements. Sometimes this set may be implicitly given. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Examples of universal statements Example Let R(x) be the predicate ’√x is a rational number’ with domain N (why not Z?), and L(x) be the predicate ’x < 2x’ with domain N. Rewrite the following universal statements in sentences and ﬁgure out whether they are true or false. (a) ∀x ∈N, R(x); (b) ∀x ∈N, L(x). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Examples of universal statements Example Let R(x) be the predicate ’√x is a rational number’ with domain N (why not Z?), and L(x) be the predicate ’x < 2x’ with domain N. Rewrite the following universal statements in sentences and ﬁgure out whether they are true or false. (a) ∀x ∈N, R(x); (b) ∀x ∈N, L(x). Solution (a) is ’the square root of all positive integers are rational numbers’. It is false as 2 is a counterexample. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Examples of universal statements Example Let R(x) be the predicate ’√x is a rational number’ with domain N (why not Z?), and L(x) be the predicate ’x < 2x’ with domain N. Rewrite the following universal statements in sentences and ﬁgure out whether they are true or false. (a) ∀x ∈N, R(x); (b) ∀x ∈N, L(x). Solution (a) is ’the square root of all positive integers are rational numbers’. It is false as 2 is a counterexample. (b) is ’for all positive integers x, x is less than 2x’, which is true. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Existential statements Deﬁnition Let Q(x) be a predicate and D the domain of x. An existential statement is a statement of the form ’∃x ∈D such that Q(x)’. It is deﬁned to be true if and only if Q(x) is true for at least one x in D. It is deﬁned to be false if and only if Q(x) is false for all x in D. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Existential statements Deﬁnition Let Q(x) be a predicate and D the domain of x. An existential statement is a statement of the form ’∃x ∈D such that Q(x)’. It is deﬁned to be true if and only if Q(x) is true for at least one x in D. It is deﬁned to be false if and only if Q(x) is false for all x in D. Remark Existential statements are somehow dual to universal statements, and we will introduce their connections in a while. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Examples of existential statements Example Rewrite the following existential statement in symbols and ﬁgure out whether it is true or false: there exists an integer that is prime and greater than 10. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Examples of existential statements Example Rewrite the following existential statement in symbols and ﬁgure out whether it is true or false: there exists an integer that is prime and greater than 10. Solution The answer is not unique: ∃n ∈Z such that n is prime and n > 10; ∃n ∈{x ∈Z \| x > 10} such that n is prime; ∃n ∈the set of prime numbers such that n > 10. Since 11 is prime and greater than 10, this existential statement is true. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Universal conditional statements Deﬁnition A universal conditional statement is of the form ∀x ∈D, if P(x) then Q(x), where P(x), Q(x) are predicates with domain D. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Universal conditional statements Deﬁnition A universal conditional statement is of the form ∀x ∈D, if P(x) then Q(x), where P(x), Q(x) are predicates with domain D. Remark Let U be the truth set of P(x), then the above universal conditional statement is logically equivalent to ∀x ∈U, Q(x). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Implicit quantiﬁcation As we have already seen, some statements are written in a way without quantiﬁers. However, we can equivalently rewrite them with ∀or ∃. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Implicit quantiﬁcation As we have already seen, some statements are written in a way without quantiﬁers. However, we can equivalently rewrite them with ∀or ∃. Example Write the following statement using symbols: if a person is a member of UT, then he/she has a UT EID. You may need to deﬁne predicates by yourselves. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Implicit quantiﬁcation As we have already seen, some statements are written in a way without quantiﬁers. However, we can equivalently rewrite them with ∀or ∃. Example Write the following statement using symbols: if a person is a member of UT, then he/she has a UT EID. You may need to deﬁne predicates by yourselves. Solution Let H be the set of humans, M(x) be the predicate ’x is a member of UT’ with domain H and E(x) be the predicate ’x has a UT EID’ with domain H. The statement becomes ∀x ∈H, M(x) →E(x). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Equivalent notations Deﬁnition Let P(x) and Q(x) be predicates with domain D. For convenience we write P(x) ⇒Q(x) for ∀x ∈D, P(x) →Q(x). And we write P(x) ⇔Q(x) for ∀x ∈D, P(x) ↔Q(x). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Finding the truth value of quantiﬁed statements Given a universal or existential statement, how to ﬁnd its truth value? Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Finding the truth value of quantiﬁed statements Given a universal or existential statement, how to ﬁnd its truth value? If the domain is ﬁnite, we can simply substitute the variable by each element in the domain. This approach is called exhaustion and it is guaranteed to work in this case. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Finding the truth value of quantiﬁed statements Given a universal or existential statement, how to ﬁnd its truth value? If the domain is ﬁnite, we can simply substitute the variable by each element in the domain. This approach is called exhaustion and it is guaranteed to work in this case. But in mathematics, many sets are inﬁnite, which means we need to consider more eﬃcient way. There is one shortcut: one counterexample is enough to show a universal statement being false and one example is enough to show a existential statement being true. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Negation of universal statements Proposition The negation of a statement of the form ∀x ∈D, Q(x) is logically equivalent to a statement of the form ∃x ∈D such that ~Q(x). Symbolically, ~(∀x ∈D, Q(x)) ≡(∃x ∈D, such that ~Q(x)). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Negation of universal statements Proposition The negation of a statement of the form ∀x ∈D, Q(x) is logically equivalent to a statement of the form ∃x ∈D such that ~Q(x). Symbolically, ~(∀x ∈D, Q(x)) ≡(∃x ∈D, such that ~Q(x)). Remark As a corollary, universal statements and existential statements can be deﬁned by each other plus negation. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Negation of existential statements Similarly we have Proposition The negation of a statement of the form ∃x ∈D such that Q(x) is logically equivalent to a statement of the form ∀x ∈D, ~Q(x). Symbolically, ~(∃x ∈D such that Q(x)) ≡(∀x ∈D, ~Q(x)). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Negation of universal conditional statements Proposition The negation of a statement of the form ∀x ∈D, P(x) →Q(x) is logically equivalent to a statement of the form ∃x ∈D such that ~(P(x) →Q(x)). Symbolically, ~(∀x ∈D, P(x) →Q(x)) ≡∃x ∈D such that ~(P(x) →Q(x)). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Negation of universal conditional statements Proposition The negation of a statement of the form ∀x ∈D, P(x) →Q(x) is logically equivalent to a statement of the form ∃x ∈D such that ~(P(x) →Q(x)). Symbolically, ~(∀x ∈D, P(x) →Q(x)) ≡∃x ∈D such that ~(P(x) →Q(x)). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Example: ﬁnd the negations Example Write the formal (with symbols) negations of the following quantiﬁed statements: (a) All computer programs are ﬁnite. (b) There is a computer program in the programming language Lisp. (c) If a computer program has more than 100,000 lines, then it contains a bug. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Example: ﬁnd the negations Solution Let P be the set of all computer programs. F(x) be the predicate ’x is ﬁnite’; L(x) be the predicate ’x is in Lisp’; T(x) be the predicate ’x has more than 100, 000 lines’; and B(x) be be the predicate ’x contains a bug’, all with domain P. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Example: ﬁnd the negations Solution Let P be the set of all computer programs. F(x) be the predicate ’x is ﬁnite’; L(x) be the predicate ’x is in Lisp’; T(x) be the predicate ’x has more than 100, 000 lines’; and B(x) be be the predicate ’x contains a bug’, all with domain P. (a) is ∀x ∈P, F(x). So the negation is ∃x ∈P such that ~F(x). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Example: ﬁnd the negations Solution Let P be the set of all computer programs. F(x) be the predicate ’x is ﬁnite’; L(x) be the predicate ’x is in Lisp’; T(x) be the predicate ’x has more than 100, 000 lines’; and B(x) be be the predicate ’x contains a bug’, all with domain P. (a) is ∀x ∈P, F(x). So the negation is ∃x ∈P such that ~F(x). (b) is ∃x ∈P such that L(x). So its negation is ∀x ∈P, ~L(x). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Example: ﬁnd the negations Solution Let P be the set of all computer programs. F(x) be the predicate ’x is ﬁnite’; L(x) be the predicate ’x is in Lisp’; T(x) be the predicate ’x has more than 100, 000 lines’; and B(x) be be the predicate ’x contains a bug’, all with domain P. (a) is ∀x ∈P, F(x). So the negation is ∃x ∈P such that ~F(x). (b) is ∃x ∈P such that L(x). So its negation is ∀x ∈P, ~L(x). (c) is ∀x ∈P, T(x) →B(x). So its negation is ∃x ∈P such that ~(T(x) →B(x)), equivalently ∃x ∈P such that (T(x) ∧~B(x)). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Connection with ∧and ∨ Let P(x) be a predicate with a ﬁnite domain D = {x1, x2, . . . , xn}. Then we have Proposition The universal statement ∀x ∈D, P(x) is logically equivalent to P(x1) ∧P(x2) ∧· · · ∧P(xn). And the existential statement ∃x ∈D such that P(x) is logically equivalent to P(x1) ∨P(x2) ∨· · · ∨P(xn). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Connection with ∧and ∨ Let P(x) be a predicate with a ﬁnite domain D = {x1, x2, . . . , xn}. Then we have Proposition The universal statement ∀x ∈D, P(x) is logically equivalent to P(x1) ∧P(x2) ∧· · · ∧P(xn). And the existential statement ∃x ∈D such that P(x) is logically equivalent to P(x1) ∨P(x2) ∨· · · ∨P(xn). Remark We don’t have a similar result when D is inﬁnite. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Vacuous truth of universal statements Example Is the following universal conditional statement true? For all M325K lectures in Fall 2018, if it is on Saturday, then all students are required to bring textbooks. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Vacuous truth of universal statements Example Is the following universal conditional statement true? For all M325K lectures in Fall 2018, if it is on Saturday, then all students are required to bring textbooks. Solution It is of the form ∀x ∈D, P(x) →Q(x) while for all x ∈D, P(x) is false. Then for each x ∈D, the conditional statement P(x) →Q(x) is by default true, hence the universal conditional statement is also true. And this is called the vacuous truth of them. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Other forms of universal conditional statements Deﬁnition Let P(x), Q(x) be predicates with domain D and s be the statement ∀x ∈D, P(x) →Q(x) The contrapositive of s is ∀x ∈D, ~Q(x) →~P(x). The converse of s is ∀x ∈D, Q(x) →P(x). And the inverse of s is ∀x ∈D, ~P(x) →~Q(x). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Other forms of universal conditional statements Deﬁnition Let P(x), Q(x) be predicates with domain D and s be the statement ∀x ∈D, P(x) →Q(x) The contrapositive of s is ∀x ∈D, ~Q(x) →~P(x). The converse of s is ∀x ∈D, Q(x) →P(x). And the inverse of s is ∀x ∈D, ~P(x) →~Q(x). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Warning: ∀and ∃are not commutative As we will discuss in detail next time, the quantiﬁers are not commutative in general. Example Let G(x, y) be the binary predicate x < y with domain N × N. Are the following statements true or false? (a) ∀x ∈N, ∃y ∈N such that G(x, y). (b) ∃y ∈N such that ∀x ∈N, G(x, y). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Warning: ∀and ∃are not commutative As we will discuss in detail next time, the quantiﬁers are not commutative in general. Example Let G(x, y) be the binary predicate x < y with domain N × N. Are the following statements true or false? (a) ∀x ∈N, ∃y ∈N such that G(x, y). (b) ∃y ∈N such that ∀x ∈N, G(x, y). Solution (a) is true, as given x ∈N, we can always choose y = x + 1 to justify the existential statement. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements Warning: ∀and ∃are not commutative As we will discuss in detail next time, the quantiﬁers are not commutative in general. Example Let G(x, y) be the binary predicate x < y with domain N × N. Are the following statements true or false? (a) ∀x ∈N, ∃y ∈N such that G(x, y). (b) ∃y ∈N such that ∀x ∈N, G(x, y). Solution (a) is true, as given x ∈N, we can always choose y = x + 1 to justify the existential statement. (b) is false, as given y ∈N, x = y + 1 is a counterexample to the universal statement, so for all y it is false. So is (b). Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant Predicates and quantiﬁers Universal and existential quantiﬁed statements Other forms of quantiﬁed statements HW #2 - these sections Section 3.1 Exercise 6, 17(b), 23(b), 28(a)(c). Section 3.2 Exercise 2, 14, 23, 46. Bo Lin Math 325K - Lecture 5 Section 3.1 & 3.2 Predicates and quant
10046	https://pdfs.semanticscholar.org/4298/961a7b50fab41691cb1442f124159effd2fb.pdf	Over the past quarter of a century, several scientific develop-ments have challenged traditional concepts in ovarian cancer. First, it was recognized that ovarian cancer is not a homo-geneous disease, but rather a group of diseases-each with different morphology and biological behavior. Approximately 90% of ovarian cancers are carcinomas (malignant epithelial tumors) and, based on histopathology, immunohistochemis-try, and molecular genetic analysis, at least five main types are currently distinguished: high-grade serous carcinoma (HGSC, 70%); endometrioid carcinoma (EC,10%); clear-cell carcinoma (CCC,10%); mucinous carcinoma (MC, 3%); and low-grade serous carcinoma (LGSC, <5%) [1,2]. These tumor types (which account for 98% of ovarian carcinomas) can be reproducibly diagnosed by light microscopy and are inherently different diseases, as indicated by differences in epidemiologic and genetic risk factors; precursor lesions; patterns of spread; and molecular events during oncogenesis, response to chemotherapy, and prognosis [2,3]. Much less common are malignant germ cell tumors and potentially malignant sex cord-stromal tumors. The biomarker expression profile within a given histotype is consistent across stages. Ovarian cancers differ primarily based on histologic type. In the era of personalized cancer medicine, reproducible histopathologic diagnosis of tumor cell type is a sine qua non for successful treatment. Different tumor histotypes respond differently to chemotherapy. The International Federation of Gynecology and Obstetrics (FIGO) Committee on Gynecologic Oncology unanimously agreed that histologic type should be designated at staging. The finding of high-grade serous tubal intraepithelial car ci noma (STIC), in patients with BRCA mutation undergo-ing risk-reducing salpingo-oophorectomy (RRSO) also influenced the new FIGO staging. Although STIC is capable of metastasizing and, therefore, cannot be considered a true carcinoma in situ, compelling evidence for a tubal origin of BRCA-positive HGSC has accumulated over the past decade [5,6]. The relative proportion of HGSCs of ovarian and tubal derivation is unknown, mainly because tumor growth in advanced-stage cancers conceals the primary site. Even in cases involving BRCA mutation, evidence of a tubal origin of HGSCs is incomplete and a multicentric origin of these tumors cannot be excluded. The process of the proposed changes to the staging of ovarian, fallopian tube, and primary peritoneal cancer started three years ago under the leadership of the Chair of the FIGO Committee on Gynecologic Oncology, Professor Lynette Denny. The proposal was sent to all relevant gynecologic oncology organizations and societies worldwide. The new staging was reached by consensus of those participating in the FIGO meeting held in Rome, Italy, on October 7, 2012 and approved two weeks later. The following is the consensus agreement that resulted from these efforts and represents new criteria for staging of these gynecologic cancer (Table 1). FIGO’s staging classification for cancer of the ovary, fallopian tube, and peritoneum: abridged republication Jaime Prat; for the FIGO Committee on Gynecologic Oncology 1 Department of Pathology, Hospital de la Santa Creu i Sant Pau, Autonomous University of Barcelona, Barcelona, Spain Copyright © 2015. Asian Society of Gynecologic Oncology, Korean Society of Gynecologic Oncology This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. www.ejgo.org FIGO Guideline J Gynecol Oncol Vol. 26, No. 2:87-89 pISSN 2005-0380 · eISSN 2005-0399 This article is an abridged version of the staging classification published in the International Journal of Gynecology and Obstetrics in January 2014. Please cite the original article as: Prat J; FIGO Committee on Gynecologic Oncology. Staging classification for cancer of the ovary, fallopian tube, and peritoneum. Int J Gynecol Obstet 2014;124(1):1-5. Copyright Elsevier (2013) Correspondence to Jaime Prat Department of Pathology, Hospital de la Santa Creu I Sant Pau, Autonomous University of Barcelona, Building C, Floor 2, SantQuintí, 87-89, Barcelona 08041, Spain. E-mail: jprat@santpau.cat Jaime Prat, et al. 88 www.ejgo.org MAIN CHANGES The primary site (i.e., ovary, fallopian tube, or peritoneum) should be designated where possible. In some cases, it might not be possible to delineate the primary site clearly; such cases should be listed as “undesignated”. The histologic type should be recorded. Stage I ovarian or fallopian tube cancer is confined to the ovaries or the fallopian tubes and peritoneal fluid/washings. Tumor rupture, surface involvement by tumor cells or pres-ence of malignant cells in the ascites or peritoneal washings warrants a stage of IC. It is not possible to have stage I perito-neal cancer. Stage II ovarian cancer comprises a small and heteroge-neous group making up less than 10% of ovarian cancers. It is defined as extension or metastasis to extraovarian/ extratubal pelvic organs and may include curable tumors that have directly extended to adjacent organs but have Table 1. 2014 FIGO ovarian, fallopian tube, and peritoneal cancer staging system and corresponding TNM Stage I. Tumor confined to ovaries or fallopian tube(s) T1-N0-M0 IA: tumor limited to one ovary (capsule intact) or fallopian tube; no tumor on ovarian or fallopian tube surface; no malignant cells in the ascites or peritoneal washings T1a-N0-M0 IB: tumor limited to both ovaries (capsules intact) or fallopian tubes; no tumor on ovarian or fallopian tube surface; no malignant cells in the ascites or peritoneal washings T1b-N0-M0 IC: tumor limited to one or both ovaries or fallopian tubes, with any of the following: IC1: surgical spill T1c1-N0-M0 IC2: capsule ruptured before surgery or tumor on ovarian or fallopian tube surface T1c2-N0-M0 IC3: malignant cells in the ascites or peritoneal washings T1c3-N0-M0 Stage II. Tumor involves one or both ovaries or fallopian tubes with pelvic extension (below pelvic brim) or primary peritoneal cancer T2-N0-M0 IIA: extension and/or implants on uterus and/or fallopian tubes and/or ovaries T2a-N0-M0 IIB: extension to other pelvic intraperitoneal tissues T2b-N0-M0 Stage III. Tumor involves one or both ovaries or fallopian tubes, or primary peritoneal cancer, with cytologically or histologically confirmed spread to the peritoneum outside the pelvis and/or metastasis to the retroperitoneal lymph nodes T1/T2-N1-M0 IIIA1: positive retroperitoneal lymph nodes only (cytologically or histologically proven): IIIA1(i) Metastasis up to 10 mm in greatest dimension IIIA1(ii) Metastasis more than 10 mm in greatest dimension IIIA2: microscopic extrapelvic (above the pelvic brim) peritoneal involvement with or without positive retroperitoneal lymph nodes T3a2-N0/N1-M0 IIIB: macroscopic peritoneal metastasis beyond the pelvis up to 2 cm in greatest dimension, with or without metastasis to the retroperitoneal lymph nodes T3b-N0/N1-M0 IIIC: macroscopic peritoneal metastasis beyond the pelvis more than 2 cm in greatest dimension, with or without metastasis to the retroperitoneal lymph nodes (includes extension of tumor to capsule of liver and spleen without parenchymal involvement of either organ) T3c-N0/N1-M0 Stage IV. Distant metastasis excluding peritoneal metastases Stage IVA: pleural effusion with positive cytology Stage IVB: parenchymal metastases and metastases to extra-abdominal organs (including inguinal lymph nodes and lymph nodes outside of the abdominal cavity) Any T, any N, M1 FIGO’s staging classification for cancer of the ovary J Gynecol Oncol Vol. 26, No. 2:87-89 www.ejgo.org 89 not yet metastasized, as well as tumors that have seeded the pelvic peritoneum by metastasis and, therefore, have a poor prognosis. The Committee felt that subdividing this small category further into IIB1 and IIB2 (i.e., microscopic and macroscopic pelvic peritoneal metastases) was not based on evidence/biology. All stage II disease is treated with adjuvant chemotherapy, so subclassification is not essential. Also, the old substage IIC (i.e., IIA or IIB but with tumor on surface, capsule ruptured, or ascites or positive peritoneal washing) was considered redundant and eliminated. Most ovarian cancers are HGSCs that usually present in stage III, with the vast majority (84%) stage IIIC . These tumors characteristically spread along peritoneal surfaces involving both pelvic and abdominal peritoneum. Less than 10% of ovarian carcinomas extend beyond the pelvis with exclusively retroperitoneal lymph node involvement. Evidence in the literature indicates that these cases have a better prognosis than that of tumors with abdominal peritoneal involvement [8-14]. The new staging includes a revision of stage III patients and assignment to stage IIIA1 based on spread to the retro-peritoneal lymph nodes without intraperitoneal dissemina-tion. Stage IIIA1 is further subdivided into IIIA1(i) (metastasis ≤10 mm in greatest dimension) and IIIA1(ii) (metastasis >10 mm in greatest dimension), even if there are no retrospective data supporting quantification of the size of metastasis in IIIA1. Involvement of retroperitoneal lymph nodes must be proven cytologically or histologically. Stage IV is defined as distant metastasis and includes patients with parenchymal liver/splenic metastases and extra-abdominal metastases; 12% to 21% of patients present with stage IV disease . Extension of tumor from omentum to spleen or liver (stage IIIC) should be differentiated from isolated parenchymal metastases (stage IVB). ACKNOWLEDGMENTS 1Committee members: H. Belhadj (Switzerland), J. Berek (USA), A. Bermudez (Argentina), N. Bhatla (India), J. Cain (USA), L. Denny (Chair; South Africa), K. Fujiwara (Japan), N. Hacker (Australia), E. Åvall-Lundqvist (Sweden), D. Mutch (USA), F. Odi-cino (Italy), S. Pecorelli (Italy), J. Prat (Spain), M. Quinn (Co-chair; Australia), M.A-F. Seoud (Lebanon), S.K. Shrivastava (India). REFERENCES 1. Kurman RJ, Carcangiu ML, Herrington CS, Young RH. WHO classification of tumours of female reproductive organs. 4th ed. Lyon: International Agency for Research on Cancer; 2014. 2. Prat J. Ovarian carcinomas: five distinct diseases with different origins, genetic alterations, and clinicopathological features. Virchows Arch 2012;460:237-49. 3. Gilks CB, Prat J. Ovarian carcinoma pathology and genetics: recent advances. Hum Pathol 2009;40:1213-23. 4. Piek JM, van Diest PJ, Zweemer RP, Jansen JW, Poort-Keesom RJ, Menko FH, et al. Dysplastic changes in prophylactically removed Fallopian tubes of women predisposed to developing ovarian cancer. J Pathol 2001;195:451-6. 5. Callahan MJ, Crum CP, Medeiros F, Kindelberger DW, Elvin JA, Garber JE, et al. Primary fallopian tube malignancies in BRCA-positive women undergoing surgery for ovarian cancer risk reduction. J Clin Oncol 2007;25:3985-90. 6. Kindelberger DW, Lee Y, Miron A, Hirsch MS, Feltmate C, Medeiros F, et al. Intraepithelial carcinoma of the fimbria and pelvic serous carcinoma: evidence for a causal relationship. Am J Surg Pathol 2007;31:161-9. 7. Heintz AP, Odicino F, Maisonneuve P, Quinn MA, Benedet JL, Creasman WT, et al. Carcinoma of the ovary. FIGO 26th Annual Report on the Results of Treatment in Gynecological Cancer. Int J Gynaecol Obstet 2006;95 Suppl 1:S161-92. 8. Onda T, Yoshikawa H, Yasugi T, Mishima M, Nakagawa S, Yamada M, et al. Patients with ovarian carcinoma upstaged to stage III after systematic lymphadenctomy have similar survival to Stage I/ II patients and superior survival to other Stage III patients. Cancer 1998;83:1555-60. 9. Kanazawa K, Suzuki T, Tokashiki M. The validity and significance of substage IIIC by node involvement in epithelial ovarian cancer: impact of nodal metastasis on patient survival. Gynecol Oncol 1999;73:237-41. 10. Panici PB, Maggioni A, Hacker N, Landoni F, Ackermann S, Campagnutta E, et al. Systematic aortic and pelvic lymphadenec tomy versus resection of bulky nodes only in optimally debulked advanced ovarian cancer: a randomized clinical trial. J Natl Cancer Inst 2005;97:560-6. 11. Cliby WA, Aletti GD, Wilson TO, Podratz KC. Is it justified to classify patients to Stage IIIC epithelial ovarian cancer based on nodal involvement only? Gynecol Oncol 2006;103:797-801. 12. Ferrandina G, Scambia G, Legge F, Petrillo M, Salutari V. Ovarian cancer patients with "node-positive-only" Stage IIIC disease have a more favorable outcome than Stage IIIA/B. Gynecol Oncol 2007;107:154-6. 13. Baek SJ, Park JY, Kim DY, Kim JH, Kim YM, Kim YT, et al. Stage IIIC epithelial ovarian cancer classified solely by lymph node metastasis has a more favorable prognosis than other types of stage IIIC epithelial ovarian cancer. J Gynecol Oncol 2008;19:223-8. 14. Bakkar R, Gershenson D, Fox P, Vu K, Zenali M, Silva E. Stage IIIC ovarian/peritoneal serous carcinoma: a heterogeneous group of patients with different prognoses. Int J Gynecol Pathol 2014;33:302-8. █ █ █
10047	https://www.merckmanuals.com/professional/injuries-poisoning/how-to-do-skin-soft-tissue-and-minor-surgical-procedures/how-to-debride-and-dress-a-burn	How To Debride and Dress a Burn - Injuries; Poisoning - Merck Manual Professional Edition honeypot link skip to main content Professional Consumer Professional edition active ENGLISH Merck Manual Professional Version MEDICAL TOPICSRESOURCESDRUG INFOCOMMENTARYPROCEDURESQUIZZESABOUT US MEDICAL TOPICSRESOURCESDRUG INFOCOMMENTARYPROCEDURESQUIZZES Professional/ Injuries; Poisoning/ How To Do Skin, Soft Tissue, and Minor Surgical Procedures/ How To Debride and Dress a Burn/ IN THIS TOPIC Indications Contraindications Complications Equipment Relevant Anatomy Positioning Step-by-Step Description of Procedure Aftercare Warnings and Common Errors Tips and Tricks OTHER TOPICS IN THIS CHAPTER How To Debride and Dress a Burn How To Do Burn Escharotomy How To Do Nail Trephination How To Incise and Drain an Abscess How To Remove a Shallow Fishhook How To Remove a Deep Fishhook How To Remove a Tick How To Treat an Ingrown Toenail How To Debride and Dress a Burn ByMatthew J. Streitz, MD, San Antonio Uniformed Services Health Education Consortium Reviewed ByDiane M. Birnbaumer, MD, David Geffen School of Medicine at UCLA Reviewed/Revised Modified May 2025 v43612882 View Patient Education Burns are injuries of skin or other tissue caused by thermal, radiation, chemical, or electrical contact. Damage to the epidermal barrier allows for bacterial invasion, external fluid loss and impaired thermoregulation. Burn wounds typically need debridement and/or dressing. Indications\| Contraindications\| Complications\| Equipment\| Relevant Anatomy\| Positioning\| Step-by-Step Description of Procedure\| Aftercare\| Warnings and Common Errors\| Tips and Tricks\| Debridement (removal of nonviable tissue) and wound dressings are used to decrease the risk of infection and decrease pain in superficial and partial thickness burns. (See also Burns.) Indications for Debriding and Dressing a Burn Superficial and partial thickness burn wounds Contraindications to Debriding and Dressing a Burn Absolute contraindications None Relative contraindications Wounds or other burn-related injuries that require transfer to a specialized burn unit (see treatment of burns) In these patients, consult with the receiving burn center to decide whether to initiate certain aspects of burn care before transfer. Complications of Debriding and Dressing a Burn Allergic reactions to topical antibiotics Equipment for Debriding and Dressing a Burn Nonsterile gloves Cleansing solution (eg, 2% chlorhexidine)Cleansing solution (eg, 2% chlorhexidine) 25- and 21-gauge needles 10-mL syringe Local injectable anesthetic (eg, 1%lidocaine)Local injectable anesthetic (eg, 1%lidocaine) Sterile scissors, forceps Nonadherent dressing Absorptive bulk dressing (such as 4 × 4 gauze dressings and tape, flexible rolled gauze wrap for extremity burns) Relevant Anatomy for Debriding and Dressing a Burn Burns involving the hands, feet, face, genitals, perineum, or involving major joints or burns that are circumferential or extensive often require transfer to a burn center. Depth of skin injury: Superficial (formerly called 1st-degree burns): Involving the epidermis only Partial-thickness (formerly called 2nd-degree burns): Extending into the dermis Full-thickness (formerly called 3rd-degree burns): Destroying the entire skin For partial-thickness and full-thickness burns, estimate and document the size of burn, expressed as percentage of total body surface area (see figure[A] Rule of nines [for adults] and [B] Lund-Browder chart [for children]). Positioning for Debriding and Dressing a Burn Position to provide excellent exposure of burn wound Step-by-Step Description of Debriding and Dressing a Burn Initial care of all burn wounds Diagnose and treat serious associated injuries. Any patient with other traumatic injuries should be fully evaluated to identify and treat life threatening injuries first. Remove all clothing and gross debris from the burned area. Remove all jewelry from the burn and also any that is distal to the burn to prevent entrapment from potential edema. For chemical burns, irrigate with tap water for at least 20 minutes to remove any residual chemicals. For thermal burns, irrigation will cool the injured area to prevent additional thermal damage, but optimal duration of irrigation has not been established (1). During the first 30 minutes after injury, use room temperature (20 to 25° C) or cold tap-water irrigation, immersion, or compresses to limit the extent of the burn and provide significant pain relief (2). Do not immerse burned tissue in ice or ice water because ice immersion increases pain and burn depth and increases the risk of frostbite and, if the burn surface is large, systemic hypothermia. Treat pain as soon as possible. Analgesics can be given concurrently with initiation of irrigation. For severe pain, IV opioids (eg, fentanyl 1 mcg/kg, or morphine 0.1 mg/kg) can be administered and titrated as needed. For mild to moderate pain, nonsteroidal anti-inflammatory drugs (NSAIDs) and acetaminophen may be sufficient (Treat pain as soon as possible. Analgesics can be given concurrently with initiation of irrigation. For severe pain, IV opioids (eg, fentanyl 1 mcg/kg, or morphine 0.1 mg/kg) can be administered and titrated as needed. For mild to moderate pain, nonsteroidal anti-inflammatory drugs (NSAIDs) and acetaminophen may be sufficient (3). Ensure that irrigation has removed all clothing and debris from the burned area. Cover the burn with a moist, sterile dressing soaked in room temperature water or saline. The dressing should be kept cool and moist to provide continued pain relief. Give tetanus toxoid-containing vaccine (eg, Td, Tdap) depending on patient's vaccination history (see tableTetanus Prophylaxis in Routine Wound Management). Incompletely immunized patients should also receive tetanus immune globulin 250 units IM.). Incompletely immunized patients should also receive tetanus immune globulin 250 units IM. Transfer stable patients with major burns (full-thickness burns>1% TBSA, partial-thickness burns>5% TBSA, burns of the hands, face, feet, or perineum (partial-thickness or deeper) to a burn center. If patients do not require transfer to a burn center or other appropriate facility, definitive burn care may be provided. Definitive burn wound care Cleanse the burned area gently with a clean cloth or gauze and soap and water or a mild antibacterial wound cleanser such as chlorhexidine.Cleanse the burned area gently with a clean cloth or gauze and soap and water or a mild antibacterial wound cleanser such as chlorhexidine. Irrigate the wound with saline or water. Some clinicians recommend leaving unruptured blisters intact, and others recommend opening them with sterile scissors and forceps. Regardless, desquamated skin and broken blisters are devitalized tissue that should be debrided by peeling from the wound and excising with scissors close to the border with viable, attached epidermis (4). Apply a sterile burn dressing, with or without a topical antimicrobial cream/ointment. There are several options for burn dressings (5). Some are impregnated with antimicrobials (eg, silver). Most are a form of gauze, but there are biosynthetic dressings with some of the characteristics of skinX that adhere to the wound and can be left in place for extended periods of time (6). Some contain an antimicrobial, those that do not are typically applied over a layer of antimicrobial cream or ointment. In all cases, dressings should be sterile and have an absorptive layer sufficient for the amount of exudate expected. Consider applying a layer of antibiotic cream or ointment such as bacitracin or mupirocin directly to all wounds except superficial burns. Silver sulfadiazine, once a mainstay of topical burn treatment, is Consider applying a layer of antibiotic cream or ointment such as bacitracin or mupirocin directly to all wounds except superficial burns. Silver sulfadiazine, once a mainstay of topical burn treatment, is no longer recommended because it is not more effective than other topical antibiotic preparations and may impair wound healing. However, it is sometimes still used for partial thickness burns (7). Cover the wound surface. There are many commercial dressings available but a fine-mesh gauze or commercial nonadherent gauze is appropriate. Cover and pad the wound with additional layers of loose gauze fluffs which can absorb exudate from the burn wounds more effectively than gauze pads. If fingers and toes are involved, pad the web spaces and the digits individually and separate them with strips of gauze. Wrap the entire dressing with an absorbent, slightly elastic material. Aftercare for Debriding and Dressing a Burn Give instructions about taking analgesics at home, and provide or prescribe if appropriate. Instruct the patient to elevate an affected limb to prevent swelling, which may cause delayed healing or infection. Instruct the patient to return for follow-up visit about 24 hours after initial burn care. At this visit, remove the dressing and reassess the burn for depth of injury and need for further debridement, then redress. Frequency of additional visits depend on several factors, discussed below. The timing and location (eg, clinic, home) of subsequent dressing changes depend on The type of dressing used: The frequency of dressing changes vary, depending on the dressing material and the wound. Most should be changed daily. Patient and family ability to perform wound care: Large burns, wound locations requiring awkward or complicated dressings, and/or patients who are not able to manage wound care (with or without help of others at home), may need more frequent professional care and/or less frequent dressing changes. The amount of exudate produced by the wound: Drier burns need less frequent dressing changes. For wound self-care, patients should wash hands with soap and water, gently remove the old dressing, rinse the wound with lukewarm tap water, and apply a similar dressing material as first used. Warnings and Common Errors When Debriding and Dressing a Burn Do not underestimate the need for procedural analgesia and sometimes sedation, particularly for complicated debridement or dressing changes. Inadequate analgesia deters thorough wound care. Tips and Tricks for Debriding and Dressing a Burn For burns on the face and neck: Cleanse the wound with chlorhexidine and debride blisters and any loose skin, then apply a topical antibiotic such as bacitracin but leave the wound uncovered. The wound can be washed 2 or 3 times per day, followed by reapplication of the topical agent. Encourage patients to sleep with their head elevated to help minimize or decrease swelling.For burns on the face and neck: Cleanse the wound with chlorhexidine and debride blisters and any loose skin, then apply a topical antibiotic such as bacitracin but leave the wound uncovered. The wound can be washed 2 or 3 times per day, followed by reapplication of the topical agent. Encourage patients to sleep with their head elevated to help minimize or decrease swelling. Alternatives to IV opioids as analgesia for initial management include regional or nerve block anesthesia; inhaled nitrous oxide; or IV ketamine. Alternatives to IV opioids as analgesia for initial management include regional or nerve block anesthesia; inhaled nitrous oxide; or IV ketamine. For debridement of small burns, local anesthetic injection may provide adequate analgesia. Home burn care and dressing changes may be quite painful. For severe pain, an adequate supply of an oral opioid analgesic should be provided, and responsible analgesic use should be encouraged. References Djärv T, Douma M, Palmieri T, et al. Duration of cooling with water for thermal burns as a first aid intervention: A systematic review.Burns. 2022;48(2):251-262. doi:10.1016/j.burns.2021.10.007 International Society for Burn Injury (ISBI) Practice Guidelines Committee, Steering Committee, Advisory Committee.ISBI practice guidelines for burn care.Burns. 2016;42(5):953–1021. doi: 10.1016/j.burns.2016.05.013 Romanowski KS, Carson J, Pape K, et al. American Burn Association Guidelines on the Management of Acute Pain in the Adult Burn Patient: A Review of the Literature, a Compilation of Expert Opinion, and Next Steps.J Burn Care Res. 2020;41(6):1129-1151. doi:10.1093/jbcr/iraa119 Greenhalgh DG. Management of Burns.N Engl J Med. 2019;380(24):2349-2359. doi:10.1056/NEJMra1807442 Żwierełło W, Piorun K, Skórka-Majewicz M, et al. Burns: Classification, Pathophysiology, and Treatment: A Review.Int J Mol Sci. 2023;24(4):3749. Published 2023 Feb 13. doi:10.3390/ijms24043749 Aggarwala S, Harish V, Roberts S, et al. Treatment of Partial Thickness Burns: A Prospective, Randomized Controlled Trial Comparing Four Routinely Used Burns Dressings in an Ambulatory Care Setting.J Burn Care Res. 2021;42(5):934-943. doi:10.1093/jbcr/iraa158 7.Heyneman A, Hoeksema H, Vandekerckhove D, et al. The role of silver sulphadiazine in the conservative treatment of partial thickness burn wounds: A systematic review. Burns. 2016;42(7):1377–1386. doi:10.1016/j.burns.2016.03.029 Drugs Mentioned In This Article Test your KnowledgeTake a Quiz! Copyright © 2025 Merck & Co., Inc., Rahway, NJ, USA and its affiliates. All rights reserved. About Disclaimer Cookie Preferences Copyright© 2025 Merck & Co., Inc., Rahway, NJ, USA and its affiliates. All rights reserved. Find In Topic This Site Uses Cookies and Your Privacy Choice Is Important to Us We suggest you choose Customize my Settings to make your individualized choices. Accept Cookies means that you are choosing to accept third-party Cookies and that you understand this choice. See our Privacy Policy Customize my Settings Reject Cookies Accept Cookies
10048	https://byjus.com/maths/rouches-theorem/	In complex analysis, we may come across various formulas, theorems, such as Rouche’s theorem, and functions such as harmonic function, entire function, analytic function, etc. Analytic functions play an important role in defining many related functions and theorems. Let’s recall what an analytic function is, along with some terms related to Rouche’s theorem such that we can quickly understand the use of Rouche’s theorem. Analytic function Consider a complex function f(z), it is said to be an analytic function at a certain point z0 if f(z) is differentiable not only at z0 but also at each point in some neighbourhood of z0. It is also known as a regular, monogenic or holomorphic function. Besides, a value of z is called the zero of an analytic function f(z) if f(z) = 0. Click here to get more information about analytic functions. Meromorphic function A complex function f(z) is meromorphic on a domain D if and only if f(z) is an analytic function in D except at finitely many poles. Argument principle In complex analysis, the argument principle describes the difference between the number of zeros and poles of a meromorphic function to a contour integral of the logarithmic derivative of that particular function. This is also called Cauchy’s argument principle. What is Rouche’s Theorem in Complex Analysis? Rouche’s Theorem Statement: If f(z) and g(z) are two analytic functions within and on a simple closed curve C such that \|f(z)\| > \|g(z)\| at each point on C, then both f(z) and f(z) + g(z) have the same number of zeros inside C. Rouche’s Theorem Proof Two analytic functions f(z) and g(z) inside and on a closed contour C. Also, \|g(z)\| < \|f(z)\| or \|f(z)\| > \|g(z) at each point on C. To prove: f(z) and f(z) + g(z) have the same number of zeros inside C. First, let us prove neither f(z) nor f(z) + g(z) has a zero on C. Let us assume that f(z) has a zero at a, i.e., z = a. So, f(a) = 0 Thus, \|g(z)\| < \|f(z)\| can be written as: \|g(a)\| < \|f(a)\| ⇒ g(a) = 0 {since f(a) = 0} ⇒ \|f(a)\| = \|g(a)\| That means \|f(z)\| = \|g(z)\| at z = a. This is the contradiction to the initial statement that \|g(z)\| < \|f(z)\| on C. Now, let f(z) + g(z) has a zero at z = b on C. So, f(b) \| g(b) = 0 ⇒ f(b) = -g(b) ⇒ \|f(b)\| = \|g(b)\| Or \|f(z)\| = \|g(z)\| at z = b. Therefore, neither f(z) nor f(z) + g(z) has a zero on C. Let N1 and N2 be the number of zeros of f(z) and f(z) + g(z), respectively inside C. Now, we need to prove that N1 = N2. As we know, f(z) and f(z) + g(z) are analytic functions within and on C. Also, they have no pole inside C. Thus, from the argument principle, we can write as: (\begin{array}{l}\frac{1}{2\pi i}\int_{c}\frac{f'(z)}{f(z)}dz=N_{1}-P\end{array} ) When P = 0, (\begin{array}{l}\frac{1}{2\pi i}\int_{c}\frac{f'(z)}{f(z)}dz=N_{1}\end{array} ) Similarly, (\begin{array}{l}\frac{1}{2\pi i}\int_{c}\frac{f'(z) + g'(z)}{f(z) + g(z)}dz=N_{2}\end{array} ) The difference of the above two equations is: (\begin{array}{l}\frac{1}{2\pi i}\int_{c}\left [ \frac{f'(z)+g'(z)}{f(z)+g(z)}-\frac{f'(z)}{g'(z)} \right ]dz=N_{2}-N_{1}…(1)\end{array} ) Let Φ(z) = g(z)/f(z) From this we can write g(z) as: g(z) = f(z) Φ(z), i.e., g = fΦ Consider \|g(z)\| < \|f(z)\| That means, \|g(z)/f(z)\| < 1 \|Φ(z)\| < 1 Also, [f’(z) + g’(z)]/ [f(z) + g(z)] = [f’(z) + f’(z) Φ(z) + f(z) Φ’(z)]/ [f(z) + f(z) Φ(z)] {since g(z) = f(z) Φ(z)} = [(1 + Φ(z))f’(z) + f(z) Φ’(z)]/ [f(z) (1 + Φ(z))] = [f’(z)/f(z)] + [Φ’(z)/ (1 + Φ(z))]……………(2) From (1) and (2), we have; (\begin{array}{l}N_{2}-N_{1}=\frac{1}{2\pi i}\int_{c}\frac{\Phi'(z)}{1+\Phi(z)} dz=\frac{1}{2\pi i}\int_{c}\Phi'(z)(1+\Phi(z))^{-1} dz…(3)\end{array} ) As we already got \|Φ(z)\| < 1 that means the binomial expansion of (1 + Φ(z))-1 is possible. However, the powers of Φ(z) in the binomial expansion of (1 + Φ(z))-1 will be uniformly convergent. Therefore, ∫c Φ’(z) (1 + Φ(z))-1 dz = ∫c Φ’(z) [1 – Φ(z) + Φ2(z) – Φ3(z) + ……] dz………….(4) From the given, f(z) and g(z) are analytic functions within and on C such that f(z) ≠ 0 and g(z) ≠ 0 for any point on C (which is proved above). Therefore, Φ(z) is an analytic function since Φ(z) = g(z)/f(z), and is non-zero for any point on C. Consequently, we can say that Φ(z) and all of its derivatives are analytic functions. Thus, by Cauchy’s theorem, the RHS of equation (4) vanishes. That means, ∫c Φ’(z) (1 + Φ(z))-1 dz = 0. Then from equation (3) and the above equation, we can say that N2 – N1 = 0 Therefore, N1 = N2 i.e., f(z) and f(z) + g(z) have the same number of zeros inside C. Hence proved. \| \| \| Read more: Real functions Complex analysis Uniform convergence \| Applications of Rouche’s Theorem Let’s have a look at some important applications of Rouche’s theorem. We can use Rouche’s theorem to simplify an analytic function for finding the zeros. Rouche’s theorem helps us to prove a short type proof for the fundamental theorem of algebra. It also helps in proving the open mapping theorem for analytic functions in complex analysis. Rouche’s Theorem Solved Example Question: Find the number of zeros of p(z) = z6 + 9z4 + z3 + 2z + 4 in the unit disk. Solution: Given, p(z) = z6 + 9z4 + z3 + 2z + 4 Let U = ∆ that means the boundary of U is the unit circle. Let us write p(z) as the sum of two functions such that one of them is greater than the other on the unit circle. Also, on the unit circle, \|zk\| = 1, regardless of k. Consequently, the power of z is invisible. Now, f(z) = 9z4 (greatest term) Therefore, g(z) = z6 + z3 + 2z + 4. Let us find \|f(z)\| and \|g(z)\|. \|f(z)\| = \|9z4\| = 9(1) = 9 \|g(z)\| = \|z6 + z3 + 2z + 4\| = \|z6\| + \|z3\| + \|2z\| + \|4\| = 1 + 1 + 2 + 4 = 8 ≤ 9 = \|f(z)\| Thus , we can say that p(z) = f(z) + h(z) has the same number of zeroes as f(z) on the unit disk. However, f(z) = 9z4 has four zeroes. Therefore, p(z) has four zeroes on the unit disk. Problems on Rouche’s Theorem Consider f(z) = 1 + 2z + 7z2 + 3z5, show that f has exactly two roots inside the unit disc. Show all 5 zeros of z5 + 3z + 1 are inside the curve C2 : \|z\| = 2. Find the number of zeros of p(z) = 2z4 – 2z3 + 2z2 = 2z + 9 in \|z\| < 1. Frequently Asked Questions on Rouche’s Theorem Q1 State Rouche’s theorem in complex analysis. Suppose f(z) and g(z) are two analytic functions within and on a closed contour C such that \|f(z)\| > \|g(z)\| at each point on C, then both f(z) and f(z) + g(z) have the same number of zeros inside C. Q2 Is it possible to find the number of zeros for an analytic function using Rouche’s theorem? Yes, it is possible to find the number of zeros for an analytic function using Rouche’s theorem. Q3 What do you mean by argument principle? The argument principle shows a special relationship between the number of zeros and the number of poles of an analytic function f(z) in a domain D and how the function f(z) maps the boundary of the derivative of the domain. Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
10049	https://www.researchgate.net/publication/331000440_Solving_the_Latin_Square_Completion_Problem_by_Memetic_Graph_Coloring	Published Time: 2019-02-10 (PDF) Solving the Latin Square Completion Problem by Memetic Graph Coloring Article PDF Available Solving the Latin Square Completion Problem by Memetic Graph Coloring February 2019 IEEE Transactions on Evolutionary Computation PP(99) DOI:10.1109/TEVC.2019.2899053 Authors: Yan Jin Huazhong University of Science and Technology Jin-Kao Hao University of Angers Download full-text PDFDownload full-text PDFRead full-text Download full-text PDFDownload full-text PDF Read full-text Download citation Copy link Link copied Read full-textDownload citation Copy link Link copied Citations (21)References (40)Figures (2) Abstract and Figures The Latin square completion problem (LSC) involves completing a partially filled Latin square of order n by assigning numbers from 1 to n to the empty grids such that each number occurs exactly once in each row and each column. LSC has numerous applications and is however NP-complete. In this paper, we investigate an approach for solving LSC by converting a LSC instance to a domain-constrained Latin square graph and then solving the associated list coloring problem. To be effective, we first employ a constraint propagation based kernelization technique to reduce the graph model and then call for a dedicated memetic algorithm to find a legal list coloring. The population-based memetic algorithm combines a problem-specific crossover operator to generate meaningful offspring solutions, an iterated tabu search procedure to improve the offspring solutions and a distance-quality-based pool updating strategy to maintain a healthy diversity of the population. Extensive experiments on more than 1800 LSC benchmark instances in the literature show that the proposed approach can successfully solve all the instances, surpassing the state-of-the-art methods. To our knowledge, this is the first approach achieving such a performance for the considered problem. We also report computational results for the related partial Latin square extension problem. An illustrative example of converting a partial Latin square (a) to a domain-constrained Latin square graph (b). … The relaxation-based perturbation. … Figures - uploaded by Jin-Kao Hao Author content All figure content in this area was uploaded by Jin-Kao Hao Content may be subject to copyright. Discover the world's research 25+ million members 160+ million publication pages 2.3+ billion citations Join for free Public Full-texts 2 JinHaoIEEE20 19.pdf Content available from Jin-Kao Hao: JinHaoIEEE2019.pdf JinHaoIEEE2019.pdf Content uploaded by Jin-Kao Hao Author content All content in this area was uploaded by Jin-Kao Hao on Mar 05, 2019 Content may be subject to copyright. JinHaoIEEE20 19.pdf Content available from Jin-Kao Hao: JinHaoIEEE2019.pdf JinHaoIEEE2019.pdf Content uploaded by Jin-Kao Hao Author content All content in this area was uploaded by Jin-Kao Hao on Feb 10, 2019 Content may be subject to copyright. 1 Solving the Latin square completion problem by memetic graph coloring Y an Jin and Jin-Kao Hao IEEE Tr ansactions on Evolutionary Computation. Accepted on 8 February 2019 Abstract—The Latin square completion problem (LSC) in- volves completing a partially ﬁlled Latin square of order n by assigning numbers from 1 to n to the empty grids such that each number occurs exactly once in each row and each column. LSC has numerous applications and is however NP-complete. In this paper, we investigate an approach for solving LSC by con verting a LSC instance to a domain-constrained Latin square graph and then solving the associated list coloring problem. T o be effective, we ﬁrst employ a constraint propagation based kernelization technique to reduce the graph model and then call for a dedicated memetic algorithm to ﬁnd a legal list coloring. The population- based memetic algorithm combines a problem-speciﬁc crossover operator to generate meaningful offspring solutions, an iterated tabu search procedure to impr ove the offspring solutions and a distance-quality-based pool updating strategy to maintain a healthy diversity of the population. Extensive experiments on more than 1800 LSC benchmark instances in the literature show that the proposed approach can successfully solve all the instances, surpassing the state-of-the-art methods. T o our knowl- edge, this is the ﬁrst approach achieving such a performance f or the considered problem. W e also report computational results for the related partial Latin square extension problem. Index T erms—Latin square completion, Graph coloring, List coloring, Memetic search, T abu search. I. I NTRODUCTION A Latin square L of order n is composed of n×n grids (or cells) such that each grid is ﬁlled with a number in {1, . . . , n} (n∈N+) and each number occurs in each row and each column exactly once. If some grids of L remain unﬁlled (or empty), L is a partial Latin square. The Latin square completion problem (LSC) of order n involves completing the empty grids of a partial Latin square with numbers in {1, . . . , n}to obtain a legal Latin square. LSC was ﬁrst studied by Hall and Ryser , and was known to be NP-complete in the general case , , . LSC can be considered as a special case of the partial Latin square extension problem (PLSE), which is to assign numbers in {1, . . . , n}to as many empty grids as possible under the condition that each number has to occur at most once in each row and each column. Both LSC and PLSE arise naturally in a variety of practical applications, such as scheduling, optical routing, error correcting codes as well as combinatorial design , , , , . Y. Jin is with the School of Computer Science and Technology, Huazhong University of Science &T echnology, 1037 Luoyu Road, W uhan 430074, China. J.K. Hao (corresponding author) is with the Department of Computer Science, LERIA, Universit´ e d’Angers, 2 Boulevard Lavoisier, 49045 Angers 01, France, and is also afﬁliated with the Institut Universitaire de France. (E-mail: yanjin.china@hotmail.com and jin-hao.hao@univ-angers.fr). Given their theoretical and practical importance, a number of studies on LSC and PLSE have been reported in the literature. For instance, in 1999, Kumar et al. proposed two approximation algorithms for PLSE with nontrivial worst-case performance guarantees . In 2002, Gomes and Shmoys studied three complete solution methods for solving LSC: a Constraint Satisfaction based approach (CSP), a hybrid 0/1 Linear Programming/CSP based strategy (LP/CSP), and a Boolean Satisﬁability based approach (SA T) . In 2004, Ans´ otegui et al. focused on a systematic comparison of SA T and CSP models for the Latin square (quasigroup) completion problem . The same year, Gomes et al. presented a natural randomized rounding algorithm based on a packing linear pro- gramming relaxation, which yields an e/(e−1)-approximation algorithm . These algorithms are able to solve small LSC instances within a reasonable time, but fail to solve most of the large and hard traditional benchmark instances. Recently in 2016, Haraguchi introduced several powerful iterated local search algorithms with multiple neighborhoods to solve PLSE as well as LSC . Assessed on a large set of 1800 new instances of various sizes and characteristics, these local search algorithms showed state-of-the-art performances. In particular, the Trellis-neighborhood search algorithm (Tr-ILS∗) proves to outperform other tested ILS variants and two general optimization solvers (IBM-ILOG CPLEX and LocalSolver). The instances and the associated results presented in will be used as the main references for our computational studies. Despite much research effort dedicated to LSC and the resulting advances, there are still very few methods that are able to solve the problem effectiv ely. For instance, no existing algorithm can ﬁnd a solution for some traditional instances tested in and many new instances introduced by Haraguchi in . It is thus quite useful and challenging to devise a method able to solve large and dif ﬁcult instances. In this work, we investigate for the ﬁrst time a solution method for solving LSC by converting the problem to a particular graph coloring problem (i.e., precoloring extension , then list coloring , ). With reference to the particular features of the resulting coloring model, we propose a memetic coloring algorithm (MMCOL) to solve it. Note that as a heuristic, if MMCOL ﬁnds a legal coloring for a given LSC instance, then the problem is solved. Otherwise, it says nothing about whether the LSC instance is solvable or not. W e summarize the contributions of this work as follows. First, from a perspective of solution method, the proposed approach considers the Latin square completion problem as a particular graph coloring problem. In this approach, we 2 start by converting a LSC instance to a domain-constrained Latin square graph (Section II-A). Then we reduce the graph model by applying a constraint propagation based kerneliza- tion technique (Section II-B), leading to an instance of the list coloring problem. Finally, we seek a le gal list coloring of the graph by running a dedicated memetic algorithm (Section III). The kernelization technique recursively uses constraint propagation to remove the vertices with a ﬁxed color (cor- responding to ﬁlled grids). The memetic algorithm adapts ideas from graph coloring algorithms to effectively solv e the underlying list coloring problem. In particular, the algorithm integrates a problem-speciﬁc crossover to generate promising offspring solutions, an effectiv e iterated tabu search procedure to improve each offspring solution, and a distance-and-quality based pool updating strategy to ensure a healthy diversity of the population. Second, from a perspective of computational performance, we provide experimental results on a large number of LSC benchmark instances available in the literature (over 1800 in total including 19 traditional benchmark instances from and 1800 new instances from ) and show comparisons with various state-of-the-art approaches including four recent iterated local search algorithms, general IP and exact CP solvers, and a general heuristic solver. While the reference approaches can only solve a subset of the tested instances, our approach is able to solve all the instances consistently. Such a performance has never been reported in the literature, demonstrating the high effectiveness of considering LSC as a graph coloring problem and using the proposed population- based memetic algorithm to color Latin square graphs. W e also adapt the method to the general partial Latin square extension problem and report computational results on additional 1800 PLSE benchmark instances from . Third, and more generally, the proposed method can be used to solve the list coloring and precoloring extension problems, which are relevant graph models both in theory and in practice . Indeed, for these two important coloring problems, al- though the literature offers many theoretical studies on speciﬁc graphs, we are not aware of any dedicated and effecti ve algorithm able to handle large graphs. Our work thus ﬁlls in this gap. Moreover, since precoloring extension and list coloring are useful models to formulate various applications, our method can be applied in these practical settings as well. The rest of the paper is organized as follows. Section II describes the converted graph coloring model. Section III presents the proposed MMCOL algorithm. Section IV reports computational results obtained with the proposed method and provides comparisons with state-of-the-art algorithms. Section V shows an analysis of two key components of the method, followed by concluding comments in the last section. The Appendix reports computational results of the proposed method on the related partial Latin square extension problem. II. L ATIN SQU ARE COMPLETION AND GRAPH COLORING A.Partial Latin square and Latin squar e graph Let P be a partial Latin square with n×n grids, an associated graph G= (V, E ), called Latin square graph, can be conveniently deﬁned with the verte x set V={{1, . . . , n}× {1, . . . , n}} (\|V\|=n 2) representing the grids and edge set E(\|E\|=n 2(n−1)) where {u, v} ∈ E if and only if u and v represent two grids of the same row or column . Then the Latin square completion problem is equivalent to ﬁnding a legal n-coloring of the associated Latin square graph G by using the colors {1, . . . , n}as follows. Let D(v) denote the color domain of vertex v of the graph. Obviously, if v corresponds to a grid already ﬁlled with number k (k∈ {1, . . . , n}), D(v)is a singleton domain {k}; otherwise, D(v)is initially set to {1, . . . , n}. The above coloring problem is the so-called precoloring extension problem , where some vertices have a ﬁxed color and the remaining v ertices are to be assigned a color in {1, . . . , n}. Note that a legal n-coloring of G can also be deﬁned as a partition of V into n color classes V 1, . . . , V n such that ∀u, v∈ V i(i= 1, . . . , n),{u, v}/∈E holds. Basically, in order to legally complete a partial Latin square, each color class must contain exactly n vertices when all the grids are ﬁlled. Let \|V i\|be the cardinality of color class V i(i= 1, . . . , n), we use \|R i\|=n− \|V i\|to denote the residual capacity of color class V i. Fig. 1 shows a partial Latin square of order 3, with 2 ﬁlled grids and 7 empty grids (Fig. 1(a)) and the corresponding domain-constrained graph G with 9 vertices and 18 edges (Fig. 1(b)). Let L xy represent the grid with x th row and y th column, then the connection between L xy and its corresponding vertex v i is given by i= (x−1) ×n+y. The objectiv e is to ﬁnd a legal 3-coloring of the associated G by using the colors {1,2,3}. The vertices with the blue and red colors (indicated by colors 1 and 2 respectively) represent the ﬁlled grids in Fig. 1(a) while the black vertices represent the empty grids. In this example, D(v 3) = {1}and D(v 6) = {2}while the color domain of each uncolored vertex is D(v i) = {1,2,3}(i= 1,2,4,5,7,8,9). The residual capacities of V 1,V 2 and V 3 are 2, 2 and 3 respectively (\|R 1\|= 2,\|R 2\|= 2,\|R 3\|= 3). Now, completing the partial Latin square is equivalent to ﬁnding a legal coloring of the graph by assigning a color in {1,2,3}to each uncolored vertex of G. One notices that unlike the general graph coloring problem, the precoloring extension problem associated to a Latin square graph has a speciﬁc feature. That is, if a vertex v of the graph represents a grid already ﬁlled with k∈ {1, . . . , n}, v has a singleton color domain D(v) = {k}and thus receives deﬁnitiv ely the unique color k. Moreover, this color is forbidden for any vertex u adjacent to v and should be excluded from the color domain D(u). From a perspective of graph coloring, we can beneﬁcially use this property to perform a preprocessing of the graph to obtain a reduced graph and then color the reduced graph instead of the initial Latin square graph. B.Preprocessing to simplify the Latin square graph The preprocessing procedure (Algorithm 1) aims to reduce the given Latin square graph by using the colored vertices (i.e., those with a singleton color domain). For this pur- pose, we apply a kernelization technique based on constraint 3 31 32 11 12 1 2 1 2 3 13 21 23 22 33 1 2 3 (a) (b) v 1( v 2(v 3( v 4( v 5( v 6( v 7( v 8( v 9( ) )) ) ) ) ) ) ) {1} {1,2,3} {2} {1,2,3} {1,2,3} {1,2,3} {1,2,3} {1,2,3} {1,2,3} Fig. 1.An illustrative example of con verting a partial Latin square (a) to a domain-constrained Latin square graph (b). propagation as follows. W e ﬁrst remove the pre-colored vertices (corresponding to the ﬁlled grids) as well as the edges connected to a colored vertex. Moreover, considering the coloring constraint stating that two adjacent vertices cannot receive the same color, once a vertex v receives color k,k is forbidden for any adjacent vertex u and can be safely removed from its color domain D(u). If the color domain of a vertex u becomes a singleton, vertex u deﬁnitively recei ves the unique color. Since verte x u is now a colored vertex, it can be used to further reduce the graph. This process is repeated until no color domain can be reduced. Notice that if the color domain of a vertex is reduced to the empty set during the preprocessing procedure, then the given LSC instance has no solution, i.e., it cannot be fully completed. Algorithm 1 Preprocessing procedure for graph reduction Require: A Latin square graph G= (V, E )with some vertices already colored, each vertex v’s color domain D(v) (v∈V) Ensure: A reduced graph 1: while ∃a vertex v∈V with singleton color domain D(v) = {k}do 2: Pick such a vertex v∈V with D(v) = {k}// v is colored by k 3: V←V\ {v}// Remove this colored vertex v from the graph 4: E←E\ {{u, v} ∈ E}// Remove the edges linked to v 5: for each uncolored u∈V adjacent to v do 6: D(u)←D(u)\ {k}// Remove color k from the color domain D(u) 7: end for 8: end while 9: return G= (V, E) Consider again the example of Fig. 1. After applying the preprocessing to the Latin square graph in Fig. 2(a), we obtain the reduced graph shown in Fig. 2(b). In this particular case, since v 9 is connected to the two colored vertices v 3 and v 6, the colors 1 and 2 are removed from the color domain of v 9, causing D(v 9)to become a singleton {3}. As a result, v 9 receives the unique color 3. The color domains of other vertices adjacent to v 3,v 6 or v 9 are also reduced, leading to the graph of Fig. 2(b) with D(v 1) = D(v 2) = {2,3}, D(v 4) = D(v 5) = {1,3}, and D(v 7) = D(v 8) = {1,2}. In terms of graph coloring, a reduced Latin square graph like Fig. 2(b) is a domain-constrained graph because the permissible colors of a vertex are limited to a list of colors in {1, . . . , n}(instead of the whole set {1, . . . , n}). In fact, the underlying coloring problem is the so-called list coloring problem , , which, like the classic vertex coloring problem, is NP-complete in the general case. Our literature review on list coloring indicates that no practical algorithm is currently able to color large graphs. Meanwhile, it is known that the list coloring problem can be transformed to the vertex coloring problem . However, this transformation needs to create an auxiliary graph which is larger than the input graph by adding k≥n vertices and k 2edges. Note that in the case of LSC, the Latin square graphs include already 2500- 4900 vertices for n= 50,60,70 for the main benchmark instances tested in this work. T o our knowledge, few vertex coloring algorithms are able to effectively color graphs of these sizes given that the benchmark graphs from the well-known DIMACS Challenge ( are limited to 1000 vertices. For these reasons, we introduce below a dedicated algorithm speciﬁcally designed to solve the list coloring problem of Latin square graphs. III. M EMETIC ALGORITHM FOR COLORING L ATIN S QUA RE GR AP HS W e describe in this section the population-based memetic al- gorithm for coloring domain-constrained Latin square graphs, i.e., solving the associated list coloring problem where each vertex v can only take a color from its giv en color domain D(v). A.General procedure The proposed algorithm (called MMCOL, shown in Al- gorithm 2) follows the general memetic framework which combines population-based evolutionary search and local op- timization , , , . One notices that memetic 4 (a) (b) Fig. 2.The Latin square graph of Fig 1(b) and the residual Latin square graph obtained by the preprocessing procedure. approaches have proved to be highly successful to solv e graph coloring and partition problems , , , , , . Algorithm 2 Graph coloring algorithm for Latin square com- pletion (MMCOL) Require: A reduced Latin square graph G= (V, E ), the number of colors n, population size p, color domain D(v) of each vertex v∈V Ensure: The best n-coloring c∗and f∗found so far 1: Population Initialization(P,p); // Generate p initial solu- tions of G, Sect. III-C 2: c∗←c; // c∗records the best coloring found so far 3: f∗←f(c∗); // f∗records the smallest number of conﬂicting edges 4: repeat 5: (P 1, P 2)←Selection(P) // Select two parents at random for crossover 6: o←MAGX(P 1, P 2) // Crossover to get an of fspring coloring, Sect. III-D 7: o←ITS(o) // Improve o with an iterated local search procedure based on T abucol and a relaxation-based perturbation, Sect. III-E 8: if f(o)< f∗then 9: c∗←o;f∗←f(o); 10: end if 11: Population Updating(P, o) // Use the improved off- spring o to update the population, Sect. III-F 12: until a stopping condition is met 13: return f∗,c∗ The algorithm takes a reduced Latin square graph G as its input and tries to ﬁnd a legal list coloring in the search space deﬁned in Section III-B. For this purpose, the algorithm starts with an initial population (Algorithm 2 line 1, Sect. III-C). Then, to ﬁnd a legal n-coloring, MMCOL repeats a number of generations to improve the population until a stopping condition (limited to maxGenerations) is met. At each generation, MMCOL randomly selects two parent colorings from the population (Algorithm 2 line 5, Sect. III-C) and recombines them to generate an offspring coloring by a dedicated crossover operator (Algorithm 2 line 6, Sect. III-D). This offspring coloring is then improved by an iterated tabu search procedure ITS (Algorithm 2 line 7, Sect. III-E). Finally, the impro ved offspring is used to update the population according to an updating strategy based on a distance-quality criterion (Algorithm 2 line 11, Sect. III-F). During the memetic search process, if a legal coloring is found, MMCOL stops and returns the legal coloring found. B.Search space and evaluation function Let G= (V, E )be a Latin square graph with L ver- tices {v 1, . . . , v L}and color domains D(v i)⊆ {1, . . . , n} (i= 1, . . . , L). Our MMCOL algorithm explores the following space C of candidate list colorings. C={c:V→ {1, . . . , n}:c(v i)∈D(v i), i= 1, . . . , L} Given a candidate coloring c in C, if c(u) = c(v)and {u, v} ∈ E(i.e., two adjacent vertices u and v receive the same color), then {u, v}is a conﬂicting edge in c while u and v are called conﬂicting vertices. T o assess the quality of the coloring c, we use the evaluation or ﬁtness function f gi ven in Eq. (1), which counts the number of conﬂicting edges in c. f(c) =X {i,j}∈E max{0,1− \|c(v i)−c(v j)\|} (1) Consequently, if f(c)=0,c is conﬂict-free and identiﬁes a legal list coloring. Otherwise (f(c)>0), c is an illegal coloring with conﬂicting edges. For two candidate solutions c 1 and c 2,c 1 is considered to be better than c 2 if f(c 1)< f(c 2) (c 1 contains fewer conﬂicting edges). Given the above e valuation function, the objective of our memetic algorithm is to ﬁnd a legal (conﬂict-free) list coloring in the search space C by minimizing f. C.Population initialization The MMCOL algorithm applies a randomized coloring strategy to create the initial population P that is composed 5 of p colorings sampled in C(p is the population size and set to 20 in this work). Let G= (V, E )be the gi ven graph with V={v 1, . . . , v L}and D(v i)⊆ {1, . . . , n}(i= 1, . . . , L). T o build an initial coloring of G, we iteratively select an uncolored vertex v at random and then assign it a random color k from its color domain D(v). Such an initial solution can be obtained very quickly in O(L), but may in volve a high number of conﬂicting edges. T o obtain an initial coloring of reasonable quality, we impro ve this solution by the local optimization procedure (see Sect. III-E) and then insert the improved solution into the population if the solution does not exist in P. Otherwise, the solution is discarded and a new solution is generated. This initialization process is repeated until the population is ﬁlled up with p different colorings. D.Crossover operator Recombination is an important component of our MMCOL algorithm that aims to transmit meaningful features from parents to offspring solutions . For the con ventional graph coloring problem, the Greedy Partition Crossover (GPX) is known to be highly effectiv e. However, given that list coloring graphs have restricted color domains (instead of the set {1, . . . , n}), GPX cannot be applied directly in the context of the list coloring problem. On the other hand, the key idea of GPX, i.e., inheriting large color classes, is of interest even in the case of list coloring. As a result, we propose an adaptation of GPX by taking into account the constrained color domains of our graphs. This leads to our maximum approximate group based crossover (MAGX) for Latin square graph coloring. Algorithm 3 Pseudo-code of the MAGX crossover operator Require: Parent solutions P 1={V 1 1, . . . , V 1 n},P 2= {V 2 1, . . . , V 2 n}, and color domain D(v)of each vertex v∈V Ensure: An offspring solution o={V o 1, . . . , V o n} 1: g←0// Count the number of color classes already built in o 2: while g < n do 3: Identify from parents (P 1 and P 2) the largest color class V b i(b= 1 or 2) satisfying \|V b i\| ≤ \|R i\|and color class V o i is empty 4: V o i←V b i// Color class V b i is transmitted to the offspring 5: Remove the vertices of V b i from P 1 and P 2 6: g←g+ 1 7: end while 8: for each empty color class V o i in o do 9: V o i←V 1 i T V 2 i// For the residual vertices, transmit the vertices that share the same color in both parents 10: end for 11: for each uncolored v∈V in o do 12: v is randomly assigned a color from its color domain D(v) 13: end for 14: return o The proposed MAGX crossover operator generates one offspring solution from two randomly selected parent solutions (see Algorithm 3). Let P 1={V 1 1, . . . , V 1 n}and P 2= {V 2 1, . . . , V 2 n}be the parent solutions, MAGX generates, in three phases, the offspring solution o={V o 1, . . . , V o n}where each V o i(i= 1, . . . , n) is initially set to empty. First, MAGX builds a number of color classes of o by inheriting color classes from the parent solutions. T o build a new color class, MAGX selects, among the color classes of both P 1 and P 2, one largest class (call it V b i) such that its cardinality does not exceed the residual capacity R i of the corresponding color class (Algorithm 3 line 3). MAGX then uses V b i to form the new color class V o i and removes the vertices of V b i from both parent solutions (Algorithm 3 lines 4- 6). One notices that the color class whose cardinality is larger than its residual capacity must contain conﬂicting vertices. So, a color class whose cardinality is equal to (or slightly smaller than) the residual capacity is preferred in order to obtain a offspring class without conﬂicts. Moreover, unlike the general coloring problem where the colors are interchangeable during the recombination operation (like GPX of does), for our list coloring problem of Latin square graphs, each color class of the offspring must inherit the color of the selected parent due to the constrained color domains of the vertices. Second, for each color j such that V o j=∅in o, if V 1 j and V 2 j share common vertices, these vertices are used to form the color class V o j of the offspring. Third, for each vertex v missing in o,v is assigned a random color class in its color domain D(v). At this stage, a complete offspring solution o is obtained. In case that the offspring is the same as one of the parent solutions (this rarely happens), MAGX applies a slightly different strategy for the ﬁrst phase such that the largest color class is selected by considering alternatively P 1 and P 2(instead of considering simultaneously P 1 and P 2). Since the three phases have a time complexity of O(n 2),O(n) and O(n 2)respectively, the time complexity of the MAGX crossover is bounded by O(n 2). Fig. 3 shows an illustration example of the MA GX crossover. This example involves a Latin square graph of order 3 (n= 3) with 9 vertices a, b, c, d, e, f, g, h, i to be assigned to 3 color classes V 1, V 2, V 3. Suppose that the color domains D(a) = D(c) = D(g) = {1,3},D(h) = {1,2} and D(x) = {1,2,3}for x∈ {b, d, e, f, i}. At the beginning, no color class exists in o, so \|R i\|= 3 (i= 1,2,3). In the ﬁrst step, V 2={d, e, f }of P 1 is identiﬁed as the lar gest color class whose \|V 2\| ≤ \|R 2\|and V 2 of the offspring o is empty. Thus, this color class {d, e, f}becomes the color class V 2 of the offspring and the vertices d, e, f are removed from both P 1 and P 2. Notice that due to the fact that vertices may have different color domains, the vertices of the inherited color class {d, e, f }of o recei ves the same color as the donor parent (here color 2). Similarly, V 3={b, c, i}and V 1={a, g}of P 2 are inherited as color classes V 3 and V 1 of o. After these operations, vertex h is still missing in o. Since this vertex belongs to different classes in P 1 and P 2,h is assigned a random color class from its color domain D(h) = {1,2}. 6 a c d i f b e h g P 1 c a d i f b e h g a c d i f b e h g P 2 c a d i f b e h g P P 1 2 o P P 1 2 o step 1 step 2 step 3 d e f V V V 1 2 3 d e f d e f P P 1 2 o d e f d e f P P 1 2 o d e f d e f V V V 1 2 3 V V V 1 2 3 V V V 1 2 3 a b h i c g a g f e h b c i d e h a g f b c i d a b h i c g a b h i c g b c i d e h a g f b c i a b h i c g a g f e h b c i d b c i a g Fig. 3.Illustration of the ﬁrst phase of the MAGX crossover operator (Algorithm 3 lines 2–7). E.Iterated tabu search procedur e The iterated tabu search procedure (ITS) (Algorithm 4) takes an offspring solution c generated by the MAGX crosso ver operator as its input and tries to improve its quality in terms of ﬁtness function f(Eq. (1), Sect. III-B). For this purpose, ITS iterates between a tabu search procedure followed by a relaxation-based perturbation procedure to try to attain a legal coloring by resolving the conﬂicts (Algorithm 4 lines 3–13). TS iteratively improves c by recoloring conﬂicting v ertices (see Sect. III-B). At the end of each tabu search run, if the conﬂicts are resolved, then a legal list coloring c∗is found, and the whole search terminates immediately. If conﬂicts remain in the solution, ITS triggers a perturbation procedure to modify the solution and then uses the modiﬁed solution as its starting solution for the next tabu search run (Algorithm 4 line 9). ITS repeats the above process until a pre-ﬁxed maximum number of iterations maxLSI ters is reached or a legal coloring is obtained. 1)Tabu searc h based coloring procedure:As its key optimization procedure, ITS uses the tabu search method to improve a given ille gal list coloring. Speciﬁcally, the tabu search (TS) procedure used in this work is based on the implementations presented in , of the pop- ular T abuCol algorithm for the conventional graph coloring 7 Algorithm 4 Pseudo-code of iterated tabu search (ITS) Require: A n-coloring c, depth of tabu search α, color domain D(v)of each vertex v∈V Ensure: A legal coloring c∗ 1: c∗←c; // c∗records the best solution found so far 2: f∗←f(c∗); // f∗records the smallest number of conﬂicting edges 3: repeat 4: (c, f )←TS(c, α); // Apply the tab u search procedure with search depth α to improve the input coloring c, see Sect. III-E1 5: if f< f∗then 6: c∗←c;f∗←f(c); 7: end if // c is not legal coloring, trigger perturbation 8: if f>0 then 9: (c, f )←Perturbation Procedure(c); // Apply the perturbation procedure to locate at a promising re- gion, see Sect. III-E2 10: else 11: return the legal coloring c∗; 12: end if 13: until maxLSI ters is reached problem . Suppose that the solution c is composed of L vertices {v 1, v 2, . . . , v L}and each vertex v i receiv es a permissible color in its constrained color domain D(v i) (i∈ {1,2, . . . , L}). The tabu search procedure explores the space C composed of all possible colorings (see Sect. III-B) to seek a legal list coloring. T o improve the solution, TS iteratively makes transitions from the current solution c to one neighboring solution. T o obtain a neighboring solution c 0 from solution c, TS displaces a conﬂicting vertex v from its current color class V i to another eligible color class V j such that j∈D(v)(i.e., the current color i of vertex v is changed to a new permissible color j in v’s color domain D(v)). Thus, c and c 0 dif fer only by the color of a conﬂicting vertex v. Since the color domains are bounded by n, the size of this neighborhood is bounded by O(n c×n)where n c is the number of conﬂicting vertices in coloring c. At each iteration, TS selects among the eligible neighboring solutions the best neighbor c 0 b according to the evaluation function f(Eq (1), Sect. III-B) and uses c 0 b to replace c. A neighboring solution is eligible if it is not forbidden by the tabu list (see explanation below) or if it is better than the best recorded solution. Suppose that the selected neighboring solution is obtained by changing the color i of conﬂicting vertex v,(v, i)is recorded in the tabu list, indicating that vertex v is forbidden to receive the color i again for the next β consecutive iterations (β is called the tabu tenure). Following , , β is dynamically tuned by β= 0.6∗f(c) + random(10) where random(10) is a random number in {1,...,10}. The TS procedure stops when its iteration counter reaches the given limit α(α is called the tabu search depth). The best solution c and the number of conﬂicting edges in c recorded during the search are returned as its output when the procedure terminates. 2) Relaxation-based perturbation:It is possible that no legal list coloring is found at the end of a tabu search run (see Algorithm 4 line 8). In this case, the search is considered to be trapped in a local optimum and we trigger a relaxation- based perturbation procedure to escape from the trap. X Im pr ov e d b y TS o n t h e s ub gr ap h …... c P …... c P c + c Fig. 4.The relaxation-based perturbation. The overall procedure of the relaxation-based perturbation is illustrated in Fig. 4. Let X⊂V be the set of conﬂicting vertices (i.e., each vertex of X is in volved in at least one conﬂicting edge in c). The perturbation procedure is performed in three steps: 1) Extract a subgraph G 0 by randomly removing d\|X\|/2 e conﬂicting vertices along with the incident edges; 2) Improve the coloring on G 0 using TS; and 3) Construct a new coloring on G by getting it back to G. The perturbation procedure is based on the consideration that the conﬂicting vertices of the local optimum are critical vertices for obtaining a legal list coloring. Meanwhile, these are also difﬁcult vertices for conﬂict resolution. By ignoring some of these difﬁcult vertices, TS has a higher chance to resolve the conﬂicts of the relaxed subproblem, thus provid- ing new search opportunity when the improved solution of the relaxed suproblem is added back to the ignored partial solution. Notice that, in case that the improved c∗ p is not a legal coloring after the second step of the perturbation procedure, c∗ p is still a high quality partial coloring which could be close to a complete solution. Using c∗ p as its starting point to be extended, TS will explore a new search trajectory and hopefully encounters a legal coloring. F.P opulation updating In order to avoid premature con vergence of our MMCOL algorithm, we apply a population updating strategy similar to those used in , , , . The adopted strategy simultaneously considers solution quality and diversity when using an offspring solution to update the population. Given two list colorings c i and c j, we use the so-called set-theoretic partition or transfer distance D i,j , to measure the dissimilarity of c i and c j, which is deﬁned as the minimum number of vertices that need to be moved between color classes of c i to transform c i to c j. The diversity between one solution and the entire population P is given by D i,P = min j 6=i{D ij }. Furthermore, we deﬁne the goodness score of one n-colorings c i of P in terms of both solution quality and diversity by s(c i) = f(c i)+e 0.08 n 2/D i,P ,∀c i∈P where f(c i) 8 is the number of conﬂicting edges of c i. A small (large) s(c i)value indicates a good (bad) solution with respect to the individuals of P. Given of fspring o, the population P is updated with o according to the following procedure. Step 1 Insert the offspring solution o into P and compute the score s(c i)of each individual c i of P. Step 2 Identify the worst individual c w(i.e., with the largest value of the scoring function s) and second worst individual c sw (with the second largest s value). Step 3 If c w is different from o, remove c w from P. Step 4 If c w is o, remove c w with probability 0.8 and remov e c sw with probability 0.2. This updating strategy ensures that the individuals of the population are not only of high quality, b ut also sufﬁciently distanced. This property provides a basis for the random strategy used in our algorithm to select the parents for the crossover. IV. E XP ER IM EN TAL R E SU LTS In this section, we assess the proposed approach by report- ing computational results on the Latin square completion prob- lem and showing comparisons with state-of-the-art methods. W e show in the Appendix additional results on the related partial Latin square extension problem. A.Benchmark instances and experimental protocol T o evaluate the performance of the proposed approach for solving LSC, we carry out extensive experiments on the set of 1800 random LSC benchmark instances recently introduced in 1. These LSC benchmark instances (named as LSC-n- r) are evenly divided into 18 types (n∈ {50,60,70},r∈ {0.3,0.4,0.5,0.6,0.7,0.8}) where n is the order of the partial Latin square and r(r∈[0,1]) denotes the ratio of ﬁlled grids over the n×n grids. So each type (n, r)has 100 instances. These instances were generated by randomly removing (1 − r)n 2 grids from an arbitrary legal Latin square. As a result, these LSC instances always admit a complete Latin square. By converting these instances to Latin square graphs (see Sect. II-A), we obtain graphs with 2500 to 4900 vertices and 122,500 to 338,100 edges 2. T o solve each instance, we ﬁrst apply the preprocessing procedure of Section II-B to obtain a reduced list coloring graph which is then colored with the MMCOL algorithm. The preprocessing step takes typically from several seconds to dozens of seconds. In addition to these 1800 random instances, we also assess our approach on the set of 19 traditional benchmark instances from the COLOR03 competition 3 that were tested, for in- stance, in , . MMCOL was coded in C++4 and compiled using g++ with the ‘-O3’ option on a computer running Linux equipped with 2.83 GHz and 8 GB RAM. When running the DIMACS machine benchmark procedure ‘dfmax.c’5 on our machine, we 1 Av ailable at 2 Av ailable at anJINFR/Latin-Square-Completion.git 3 Av ailable at 4 Av ailable at 5 Av ailable at ftp://dimacs.rutgers.edu/pub/dsj/clique/ obtain the following results: 0.20, 1.23 and 4.68 seconds for graphs r300.5, r400.5 and r500.5 respectively. The computa- tional results reported in this section were obtained with the parameter setting shown in T able I. In the following subsections, we ﬁrst show the results on the 19 traditional instances, and then present a comparative analysis of our computational results on the large set of 1800 benchmark instances with respect to the state-of-the-art results in the literature. Given the stochastic nature of MMCOL, each instance was independently solved 30 times with different random seeds. TABLE I P ARA MET ER S ETT IN G Parameter Sect. Description V alue maxLSI ters III-A Maximum iterations of ITS procedure 100 maxGenerations III-A Maximum number of generations 100 p III-C population size 20 α III-E Depth of tabu search 10 5 B.Results on 19 traditional benchmark instances The computational results of MMCOL on the 19 traditional Latin square graphs are summarized in T able II. Columns 1–3 of T able II indicate the characteristics of each instance: the name, the Latin square order n and the ratio r. Columns 4–5 present the success rate over 30 trials (SR) and the computation time over the successful runs t(s)in seconds (a successful run means that a legal Latin square is attained for this run). From T able II, one observes that MMCOL can complete the partial Latin square for all the 19 instances. Besides, our MMCOL requires a very short computation time even for the large instances with n≥50. Moreov er, the last ﬁve instances are critically constrained and fully “balanced”, where the number of empty grids is approximately the same over rows and columns. These instances are kno wn to be particularly difﬁcult and only the two smallest ones of these ﬁve instances (qwhdec.order33.holes381.bal.1 and qwhdec.order50.holes825.bal.1) can be solved by very few approaches presented in including CSP, hybrid strategy mixing LP/CSP and SA T-based method. The difﬁculty of these instances are further conﬁrmed by the most recent study reported in , where even the best performing heuristic T r- ILS cannot solve any of these “balanced” instances. W e also ran the source code of Tr-ILS on our computer for a long computation time of 3600 seconds and still failed to solve any of these ﬁve balanced instances. It is thus remarkable that our MMCOL approach solves these instances consistently, even though MMCOL has a low success rate for 2 instances. W e conclude that MMCOL performs very competitively with respect to all of the existing approaches for solving these traditional instances. C. Comparative r esults on the set of 1800 benchmark in- stances T able III summarizes the computational statistics of our MMCOL algorithm on the set of 1800 benchmark instances, 9 TABLE II C OM PUTATI ONA L RES ULTS O N TH E SET O F 19 TRADITIONAL BENCHMARK INSTANCES Instance MMCOL Name n r SR t(s) qwhdec.order5.holes10.1 5 0.60 30/30 0.00 qwhdec.order18.holes120.1 18 0.63 30/30 0.00 qg.order30 30 0.00 30/30 0.23 qwhdec.order30.holes316.1 30 0.65 30/30 0.19 qwhdec.order30.holes320.1 30 0.64 30/30 0.65 qg.order40 40 0.00 30/30 1.25 qg.order60 60 0.00 30/30 1.93 qg.order100 100 0.00 30/30 18.49 qwhdec.order35.holes405.1 35 0.67 30/30 24.80 qwhdec.order40.holes528.1 40 0.67 30/30 19.18 qwhdec.order60.holes1440.1 60 0.60 30/30 2.84 qwhdec.order60.holes1620.1 60 0.55 30/30 0.77 qwhdec.order70.holes2940.1 70 0.40 30/30 0.74 qwhdec.order70.holes2450.1 70 0.50 30/30 0.80 qwhdec.order33.holes381.bal.1 33 0.65 30/30 238.23 qwhdec.order50.holes825.bal.1 50 0.67 30/30 133.03 qwhdec.order50.holes750.bal.1 50 0.70 3/30 207.91 qwhdec.order60.holes1080.bal.1 60 0.70 5/30 397.94 qwhdec.order60.holes1152.bal.1 60 0.68 30/30 430.11 together with the results of seven most recent methods in the literature reported in . The reference methods in- clude CPX-IP, CPX-CP, LSSOL, 1-ILS, 2-ILS, 3-ILS and Tr-ILS, where CPX-IP and CPX-CP are IP (Integer Pro- gramming) and CP (Constraint Programming) solvers from IBM/ILOG CPLEX, LSSOL denotes the LocalSolver 6, and 1- ILS, 2-ILS, 3-ILS and Tr-ILS are four iterated local search algorithms with (1,∞)-neighborhood, (2,∞)-neighborhood, (3,∞)-neighborhood and Trellis-neighborhood search respec- tively. All the reference algorithms are performed on an Intel core i7-4770 Processor with 3.90 GHz and 8 GB RAM (which is faster than our computer), with a time limit of 30 seconds for CPX-IP, CPX-CP and LSSCOL, and 10 seconds for 1-ILS, 2-ILS, 3-ILS and Tr-ILS. T able IV additionally presents the detailed results of our approach on a subset of 600 difﬁcult benchmark instances. Columns 1–3 of T able III show the characteristics of the tested instances: the order n of each Latin square, the ratio r and the number of instances I nst#for each type (n, r). Following , columns 4–10 present the results of the sev en reference algorithms (CPX-IP, CPX-CP, LSSOL, 1-ILS, 2- ILS, Tr-ILS and 3-ILS), “Suc#” shows for each type of 100 instances the number of instances for which an algorithm can obtain a legal solution. Columns 11–12 give the results of our MMCOL algorithm in terms of “Suc#” and the average time t avg(s)in seconds is deﬁned by t avg(s) =P i=100 i=1 t i(s) 100 , where t i(s)is the average time over the successful runs for the i th instance. The last row (Perfect success times) shows the number of instance types with Suc# = 100, i.e., the number of instance types among the 18 types (n, r)for which all the 100 instances are solved by an algorithm. From T able III, one observes that both exact methods CPX- IP and CPX-CP solve all 100 instances for only 3 out of the 18 types. The ﬁve ILS heuristics (LSSOL, 1-ILS, 2-ILS, 3- 6 ILS and Tr-ILS) solve all 100 instances for 1, 4, 5, 1 and 6 of 18 types (in bold) respectively. In contrast, our MMCOL algorithm can solve all the instances for all 18 types. The average time to ﬁnd a solution is less than 11 seconds except for the 300 instances with r= 0.7 for which MMCOL needs less than 300 seconds to attain a solution while all reference algorithms fail to solve any of these instances. Besides, we observe that the seven reference algorithms hav e a worse performance for the types (n∈ {50,60,70}, r∈ {0.6,0.7}) which are known to be more difﬁcult . On the other hand, MMCOL has no particular difﬁculty to solve these hard instances. In order to verify if the ILS algorithms can solve more instances by using more computation time, we ran the source code of the best performing algorithm Tr-ILS under a much relaxed time limit of 3600 seconds on the instances of types (n∈ {50,60,70}, r∈ {0.6,0.7}). One observes that the 100 instances of LSC-50-60 (n= 50, r= 0.6) can be fully completed, 98 and 95 instances of LSC-60-60 (n= 60, r= 0.6) and LSC-70-60 (n= 70, r= 0.6) can be fully completed respectively. Nevertheless, no instance of the types (n∈ {50,60,70}, r= 0.7) can be fully completed even if these instances have more ﬁlled grids for the LSC. Furthermore, we show in T able IV the detailed results of MMCOL on the 600 difﬁcult instances (n∈ {50,60,70}, r∈ {0.6,0.7}) for which most reference algorithms perform badly. For each instance, we present the success rate over 30 trials SR and the average computation time over the successful runs t(s)in seconds. From T able IV, we observe that MMCOL achieves the perfect success rate 30/30 on three types (n∈ {50,60,70},r= 0.6). MMCOL has a lower success rate only for 28 instances of the type LSC-50-70 (n= 50, r= 0.7), 3 instances of the type LSC-60-70 (n= 60, r= 0.7) and 3 instances of the type LSC-70-70 (n= 70, r= 0.7) (in italic). In summary, MMCOL competes v ery favorably with seven most recent methods in the literature and proves to be highly effective in solving the set of 1800 benchmark instances with no exception. V. A NALYS IS O F TW O KE Y CO MP ON EN TS In this section, we present an analysis of two key compo- nents of the proposed method: the constraint propagation based kernelization and the role of the MAGX crossov er operator. A.Impact of constraint propagation based kernelization The constraint propagation based kernelization of Section II-B is used to preprocess an initial Latin square graph and thus reduces the search space of the subsequent list coloring task. In order to investigate the impact of this kernelization technique, we evaluate the reduced vertices and the color domains for the 1800 benchmark instances. T able V summarizes the statistics of the reduced graphs for each type of instances. Columns 1–3 recall the instance characteristics: the order n, the ratio r and the number of instances I nst#. For each type (n, r)of 100 instances, column 4 “#V avg” indicates the average number of the pre- colored vertices (i.e., the ﬁlled grids) in the initial Latin square graph. Since the color domain of some vertices becomes a 10 TABLE III C OMPARA TIVE RESULTS OF MMCOL WITH BEST-PERF OR MIN G ALG OR ITH MS O N THE S ET O F 1800 LSC BEN CH MAR K IN STAN CES Instance CPX-IP CPX-CP LSSOL 1-ILS∗2-ILS 3-ILS Tr-ILS∗MMCOL n r Inst#Suc#S uc#Suc#S uc#Suc#Suc#S uc#Suc#t av g (s) 0.3 100 9 94 10 100 100 95 100 100 0.22 0.4 100 3 71 8 99 99 92 100 100 0.18 50 0.5 100 0 12 6 96 96 83 100 100 0.22 0.6 100 0 0 0 30 23 5 36 100 1.99 0.7 100 0 0 0 0 0 0 0 100 299.45 0.8 100 100 100 100 100 100 100 100 100 0.00 0.3 100 0 71 1 100 100 51 100 100 0.55 0.4 100 0 22 0 96 99 52 100 100 0.41 60 0.5 100 0 1 0 89 95 17 95 100 0.51 0.6 100 0 0 0 16 12 0 23 100 4.77 0.7 100 0 0 0 0 0 0 0 100 209.65 0.8 100 100 100 99 98 100 99 99 100 0.00 0.3 100 0 34 0 100 100 19 99 100 1.27 0.4 100 0 8 0 95 97 8 98 100 0.90 70 0.5 100 0 0 0 82 87 0 84 100 1.03 0.6 100 0 0 0 5 2 0 10 100 10.02 0.7 100 0 0 0 0 0 0 0 100 272.11 0.8 100 100 100 46 93 97 95 98 100 0.06 Perfect success times over 18 types (n, r)3 3 1 4 5 1 6 18 singleton during the kernelization process, the graph can be further reduced. Hence, column 5 “#V 0 avg” shows the average number of further reduced vertices. Column 6 “#D avg” presents for each type of 100 instances the average cardinality of the color domains after kernelization, i.e., #D avg = P j=Inst# j=1 (P i=n×n i=1 \|D(v i)\|) Inst#. And column 7 “S olved#” gives the number of instances that are solved during the kernelization process (i.e., all color domains are reduced to a singleton). From T able V, one observes that the vertices of the con- verted graphs are dramatically reduced by the kernelization technique. For the types of instances (n∈ {50,60,70}, r= 0.3,0.4,0.5 and (n= 60, r= 0.6)), only the pre-colored ver- tices (the ﬁlled grids) are removed from the graphs, meaning that removing the pre-colored vertices does not lead to any new singleton domain. Howev er, for the types of instances (n∈ {50,60,70}, r= 0.8), many vertices can be further removed by the kernelization technique. Furthermore, the color domains of the uncolored vertices are obviously reduced from their initial sizes n. In particular, for 32 instances of the type (n= 50, r= 0.8) and 9 instances of the type (n= 60, r= 0.8), a solution is found during kernelization (i.e., all color domains are reduced to a singleton), without needing to run the subsequent coloring algorithm. W e conclude that the kernelization technique plays an important role in reducing the initial Latin square graphs and helps to ease the subsequent list coloring task. B.Impact of the MAGX crossover operator Within the proposed memetic algorithm, the MA GX crossover described in Section III-D is another key component. T o assess its impact, we present an experiment to compare MMCOL (with the crossover MAGX), MMCOL’ (with an uniform assignment crossover) and ITS (without MAGX). The uniform assignment crossover builds an offspring solution by inheriting, for each vertex, the color either from parent P 1 or P 2 with an equal probability of 0.5. For a fair comparison, we use the same parameter setting for MMCOL and MMCOL’ and set maxLSI ters = 10 4 for each ITS run, which corresponds roughly to the same search effort of MMCOL/MMCOL’ with 100 generations (see T able 1). The initial solutions of all the algorithms are generated by the initialization procedure given in Section III-C. W e ran MMCOL, MMCOL’ and ITS 30 times to solve each of the 600 difﬁcult benchmark instances (n∈ {50,60,70}, r∈ {0.6,0.7}) listed in Table IV. The comparative results are shown in Fig. 5 where we indicate for each type of 100 instances, the number of instances for which an algorithm can ﬁnd a solution with a perfect success rate 30/30 (i.e., each of its 30 trials ﬁnds a legal list coloring). From Fig. 5, one notices that both MMCOL, MMCOL’ and ITS have a perfect success rate 30/30 on the 300 instances of types (n∈ {50,60,70}, r= 0.6). On the other hand, for the other 300 instances of types (n∈ {50,60,70}, r= 0.7), MMCOL achieves a perfect success rate 30/30 on 72, 97 and 97 instances respectively, against 48, 83 and 90 instances for ITS and 71, 72 and 28 instances for MMCOL’. Additionally, T able VI presents the detailed results on the 22 most difﬁcult instances of LSC-50-70 where we show for each instance and each algorithm (MMCOL, MMCOL’ and ITS), the success rate SR over 30 trials and the average computation time t(s)in seconds over the successful runs. This table indicates that MMCOL obtains a higher success rate SR than MMCOL’ and ITS for 17 and 22 instances re- spectively (in bold) with shorter computation times t(s)except for 2 cases. This experiment conﬁrms that the population- based evolutionary framew ork implemented in our memetic algorithm and its crossover operator contribute positi vely to the performance of the proposed algorithm. 11 TABLE IV D ETAI LE D COM PUTATI ONA L RE SULT S OF MMCOL ON A S UBS ET O F 600 INSTANCES ID LSC-50-60 LSC-50-70 LSC-60-60 LSC-60-70 LSC-70-60 LSC-70-70 SR t(s)S R t(s)SR t(s)SR t(s)S R t(s)SR t(s) 1 30/30 1.80 30/30 229.52 30/30 6.14 30/30 237.78 30/30 7.50 30/30 211.15 2 30/30 1.67 29/30 353.16 30/30 4.59 30/30 262.11 30/30 9.23 30/30 320.59 3 30/30 1.88 30/30 77.12 30/30 3.71 30/30 143.20 30/30 10.59 30/30 192.00 4 30/30 1.40 27/30 302.56 30/30 4.09 30/30 135.90 30/30 12.11 30/30 262.11 5 30/30 1.91 30/30 295.33 30/30 3.83 30/30 143.76 30/30 9.89 30/30 213.62 6 30/30 1.91 29/30 511.98 30/30 3.62 30/30 58.73 30/30 11.77 30/30 354.76 7 30/30 1.82 30/30 219.66 30/30 6.81 30/30 246.25 30/30 12.01 30/30 278.17 8 30/30 1.63 24/30 994.24 30/30 4.44 30/30 101.65 30/30 9.25 30/30 108.13 9 30/30 1.34 29/30 219.90 30/30 3.95 30/30 432.35 30/30 11.35 30/30 129.48 10 30/30 2.31 30/30 156.48 30/30 5.35 30/30 262.10 30/30 10.51 30/30 268.28 11 30/30 2.18 30/30 245.21 30/30 5.28 30/30 174.23 30/30 10.79 30/30 238.76 12 30/30 2.52 29/30 279.76 30/30 5.21 30/30 232.61 30/30 9.44 30/30 239.79 13 30/30 3.20 17/30 1154.08 30/30 4.31 30/30 178.88 30/30 9.18 30/30 310.61 14 30/30 2.47 30/30 78.89 30/30 6.62 30/30 220.65 30/30 12.60 30/30 315.94 15 30/30 1.51 30/30 55.16 30/30 4.89 30/30 277.19 30/30 8.36 30/30 392.82 16 30/30 2.28 30/30 58.69 30/30 5.02 30/30 172.64 30/30 10.69 30/30 217.64 17 30/30 2.02 30/30 315.53 30/30 6.35 30/30 171.60 30/30 10.49 30/30 280.76 18 30/30 2.17 30/30 63.03 30/30 3.54 30/30 109.55 30/30 10.52 30/30 391.01 19 30/30 2.42 30/30 140.20 30/30 4.55 30/30 212.71 30/30 10.37 30/30 146.66 20 30/30 3.23 10/30 912.57 30/30 4.74 30/30 197.22 30/30 12.84 30/30 260.88 21 30/30 2.46 27/30 516.05 30/30 4.36 30/30 124.49 30/30 10.92 30/30 185.08 22 30/30 2.26 30/30 69.85 30/30 5.08 30/30 192.56 30/30 13.49 30/30 169.89 23 30/30 2.89 30/30 170.39 30/30 4.22 30/30 189.37 30/30 10.05 30/30 411.34 24 30/30 2.45 30/30 304.91 30/30 4.82 30/30 228.04 30/30 7.06 29/30 472.18 25 30/30 3.16 28/30 598.79 30/30 5.60 30/30 396.87 30/30 9.24 30/30 711.93 26 30/30 2.35 13/30 1006.82 30/30 5.17 30/30 113.45 30/30 8.60 30/30 383.70 27 30/30 2.91 30/30 275.97 30/30 4.11 30/30 265.82 30/30 13.47 30/30 198.42 28 30/30 2.28 29/30 634.01 30/30 4.37 30/30 90.16 30/30 9.69 30/30 314.49 29 30/30 1.89 30/30 381.79 30/30 4.48 30/30 122.30 30/30 8.21 30/30 387.83 30 30/30 2.67 30/30 249.37 30/30 5.91 30/30 530.25 30/30 11.60 30/30 459.71 31 30/30 2.11 29/30 362.59 30/30 6.41 30/30 266.20 30/30 14.32 30/30 210.70 32 30/30 1.97 30/30 287.39 30/30 6.04 28/30 687.59 30/30 10.82 30/30 463.97 33 30/30 2.03 30/30 56.08 30/30 4.56 30/30 96.57 30/30 8.69 30/30 250.81 34 30/30 1.56 30/30 153.74 30/30 3.88 30/30 235.41 30/30 14.16 30/30 253.28 35 30/30 2.14 29/30 583.45 30/30 6.77 30/30 142.38 30/30 7.78 30/30 211.31 36 30/30 2.14 30/30 373.41 30/30 6.65 30/30 136.03 30/30 9.71 30/30 433.29 37 30/30 1.41 30/30 69.45 30/30 6.45 30/30 175.92 30/30 9.54 30/30 234.66 38 30/30 2.57 28/30 373.18 30/30 5.12 30/30 89.58 30/30 14.84 30/30 182.65 39 30/30 2.65 30/30 226.60 30/30 4.25 30/30 304.11 30/30 16.04 30/30 168.72 40 30/30 2.09 29/30 460.62 30/30 3.62 30/30 468.96 30/30 10.13 30/30 170.03 41 30/30 1.82 30/30 95.53 30/30 8.16 30/30 108.12 30/30 10.50 30/30 390.81 42 30/30 1.98 30/30 93.70 30/30 3.27 30/30 185.71 30/30 6.91 30/30 134.69 43 30/30 3.06 30/30 85.74 30/30 4.00 30/30 163.11 30/30 10.70 30/30 267.40 44 30/30 1.74 30/30 46.91 30/30 4.40 30/30 93.55 30/30 10.14 30/30 190.74 45 30/30 1.71 30/30 300.44 30/30 5.12 30/30 157.89 30/30 7.92 30/30 511.40 46 30/30 2.20 30/30 268.83 30/30 5.54 30/30 276.91 30/30 11.06 30/30 232.72 47 30/30 1.80 30/30 180.58 30/30 2.70 30/30 129.40 30/30 10.39 30/30 224.85 48 30/30 2.04 30/30 52.25 30/30 4.32 30/30 219.03 30/30 9.19 30/30 259.98 49 30/30 1.41 29/30 117.13 30/30 5.33 30/30 210.02 30/30 8.10 30/30 154.96 50 30/30 1.18 30/30 389.08 30/30 5.92 27/30 1104.26 30/30 7.59 30/30 470.39 51 30/30 2.95 29/30 293.51 30/30 5.87 30/30 238.48 30/30 11.52 30/30 295.06 52 30/30 1.84 30/30 71.20 30/30 5.50 30/30 261.65 30/30 10.73 30/30 167.39 53 30/30 1.50 30/30 89.39 30/30 5.09 30/30 73.60 30/30 11.03 30/30 177.52 54 30/30 1.60 30/30 137.29 30/30 4.67 30/30 108.94 30/30 10.21 30/30 145.50 55 30/30 1.29 30/30 327.82 30/30 6.11 30/30 260.51 30/30 8.25 30/30 199.61 56 30/30 1.63 30/30 137.14 30/30 4.08 30/30 141.16 30/30 9.52 30/30 383.23 57 30/30 3.63 4/30 1724.49 30/30 3.48 30/30 95.86 30/30 9.87 30/30 197.62 58 30/30 2.60 23/30 1160.19 30/30 3.39 30/30 167.31 30/30 11.81 30/30 456.38 59 30/30 2.28 30/30 276.06 30/30 7.84 30/30 213.65 30/30 7.95 30/30 677.79 60 30/30 1.43 30/30 143.88 30/30 4.50 30/30 131.84 30/30 7.89 30/30 226.17 61 30/30 1.46 30/30 27.79 30/30 5.37 30/30 144.70 30/30 6.78 30/30 259.97 62 30/30 2.52 30/30 55.34 30/30 6.00 30/30 70.70 30/30 7.85 30/30 281.82 63 30/30 1.65 30/30 193.20 30/30 4.16 30/30 219.72 30/30 10.08 30/30 298.14 64 30/30 1.61 30/30 141.12 30/30 4.02 30/30 349.99 30/30 10.41 30/30 546.86 65 30/30 1.66 30/30 293.92 30/30 4.85 30/30 130.68 30/30 6.80 30/30 156.37 66 30/30 1.63 30/30 240.08 30/30 2.92 30/30 74.45 30/30 7.06 30/30 262.80 67 30/30 2.39 30/30 95.70 30/30 6.48 30/30 157.17 30/30 8.60 30/30 198.07 68 30/30 1.28 30/30 168.85 30/30 4.84 27/30 781.71 30/30 6.12 30/30 474.08 69 30/30 1.23 30/30 52.04 30/30 3.77 30/30 123.35 30/30 11.13 30/30 163.11 70 30/30 2.01 27/30 561.12 30/30 5.19 30/30 151.39 30/30 7.00 30/30 297.63 71 30/30 1.08 30/30 106.96 30/30 4.25 30/30 168.97 30/30 10.12 30/30 194.81 72 30/30 2.91 29/30 423.29 30/30 3.76 30/30 106.15 30/30 11.49 30/30 456.06 73 30/30 2.58 30/30 71.39 30/30 5.78 30/30 221.44 30/30 10.78 30/30 279.10 74 30/30 2.15 17/30 918.90 30/30 4.84 30/30 448.23 30/30 7.58 30/30 275.67 75 30/30 1.57 30/30 202.87 30/30 4.27 30/30 189.70 30/30 5.78 30/30 192.49 76 30/30 2.18 30/30 67.06 30/30 5.68 30/30 267.95 30/30 11.53 30/30 165.14 77 30/30 1.45 30/30 55.37 30/30 6.55 30/30 159.72 30/30 11.54 30/30 189.93 78 30/30 2.30 29/30 265.87 30/30 5.76 30/30 197.63 30/30 12.40 30/30 162.90 79 30/30 2.45 30/30 58.50 30/30 3.56 30/30 119.92 30/30 10.48 29/30 478.63 80 30/30 1.33 30/30 112.48 30/30 5.11 30/30 159.74 30/30 9.22 30/30 324.70 81 30/30 1.12 30/30 124.85 30/30 4.18 30/30 233.84 30/30 11.83 30/30 365.06 82 30/30 1.56 30/30 240.88 30/30 5.28 30/30 109.26 30/30 10.49 29/30 318.54 83 30/30 1.26 25/30 438.82 30/30 3.64 30/30 61.54 30/30 14.78 30/30 305.79 84 30/30 1.28 30/30 163.42 30/30 3.98 30/30 309.11 30/30 11.62 30/30 130.06 85 30/30 1.67 30/30 44.76 30/30 4.61 30/30 120.79 30/30 8.78 30/30 122.38 86 30/30 1.91 30/30 324.36 30/30 2.99 30/30 93.56 30/30 11.39 30/30 98.84 87 30/30 1.56 30/30 518.07 30/30 3.66 30/30 184.49 30/30 9.06 30/30 140.46 88 30/30 1.15 30/30 130.92 30/30 4.47 30/30 162.81 30/30 8.56 30/30 338.45 89 30/30 1.88 30/30 497.59 30/30 4.60 30/30 176.23 30/30 8.86 30/30 233.54 90 30/30 1.65 28/30 968.10 30/30 3.12 30/30 171.07 30/30 10.33 30/30 261.34 91 30/30 2.94 30/30 401.46 30/30 4.36 30/30 100.40 30/30 9.65 30/30 113.15 92 30/30 2.36 30/30 118.06 30/30 4.95 30/30 366.14 30/30 8.94 30/30 127.49 93 30/30 1.22 30/30 161.35 30/30 2.66 30/30 141.83 30/30 11.57 30/30 121.07 94 30/30 1.37 30/30 297.39 30/30 3.75 30/30 95.65 30/30 10.69 30/30 218.80 95 30/30 1.93 30/30 95.87 30/30 3.21 30/30 115.31 30/30 10.56 30/30 129.73 96 30/30 1.62 30/30 175.05 30/30 4.97 30/30 386.67 30/30 6.86 30/30 251.44 97 30/30 1.93 30/30 213.69 30/30 3.53 30/30 165.69 30/30 11.01 30/30 288.06 98 30/30 1.71 29/30 354.62 30/30 3.89 30/30 130.48 30/30 9.58 30/30 190.03 99 30/30 1.55 30/30 204.81 30/30 4.25 30/30 136.39 30/30 6.25 30/30 222.49 100 30/30 2.73 17/30 550.96 30/30 5.60 30/30 292.85 30/30 9.19 30/30 404.90 12 L S C-50-6 0 L S C-5 0-70 L S C-6 0-60 L SC-6 0-7 0 L SC-7 0-60 L S C-7 0-70 ITS 1 0 0 4 8 1 0 0 8 3 1 0 0 9 0 MMC OL' 1 0 0 7 1 1 0 0 7 2 1 0 0 2 8 0 2 0 4 0 6 0 8 0 1 0 0 1 2 0 Nu mb e r of ins ta nc e s (SR=30/3 0) Fig. 5.Comparison of MMCOL, MMCOL’ and ITS on 600 dif ﬁcult instances. TABLE V S TATIST IC S OF TH E RE DUC ED L ATI N SQUA RE G RAP HS A FTE R KERNELIZATION Instance #V avg #V 0 avg #D avg S olved# n r I nst# 0.3 100 750.0 0.00 25.01 0 0.4 100 1000.0 0.00 18.64 0 50 0.5 100 1250.0 0.00 13.24 0 0.6 100 1500.0 0.19 8.83 0 0.7 100 1750.0 9.19 5.37 0 0.8 100 2000.0 495.24 2.01 32 0.3 100 1080.0 0.00 29.91 0 0.4 100 1440.0 0.00 22.23 0 60 0.5 100 1800.0 0.00 15.74 0 0.6 100 2160.0 0.00 10.43 0 0.7 100 2520.0 4.63 6.29 0 0.8 100 2880.0 516.63 2.43 9 0.3 100 1470.0 0.00 34.79 0 0.4 100 1960.0 0.00 25.83 0 70 0.5 100 2450.0 0.00 18.24 0 0.6 100 2940.0 0.03 12.03 0 0.7 100 3430.0 2.25 7.19 0 0.8 100 3920.0 162.24 3.47 0 VI. C ONCLUSION W e have proposed an approach to solve the Latin square completion problem (LSC) by converting LSC to a domain- constrained graph coloring problem. By taking advantage of the particular features of Latin square graphs, we have developed a constraint propagation based kernelization tech- nique to preprocess the given Latin square graph to obtain a reduced graph, for which an associated list coloring problem is deﬁned. T o effectively solve the list coloring problem, we have devised a dedicated memetic algorithm MMCOL which integrates a tailored crossover operator to generate new solutions, an iterated tabu search procedure to improve each offspring solution and a distance-quality based pool updating strategy to ensure a healthy diversity of the population. Extensive ev aluations on a large number of benchmark instances in the literature (19 traditional instances and 1800 random instances) have showed that the proposed approach performs very well with respect to the state-of-the-art methods including those introduced very recently in 2016. In particular, our approach is able to ﬁnd a solution for all the benchmark instances consistently and effectively, a performance never attained by any existing approach. W e have also used a slightly modiﬁed version of the method to solve the general partial Latin square extension problem and reported computational results on the set of 2018 PLSE instances in the Appendix. Given that LSC and PLSE hav e a number of applications, the proposed approach can help to solve these applications. More generally, the method proposed in this w ork can be used to approximate the important list coloring and precoloring extension problems, for which few practical algorithms exist in the literature. The method will be particularly useful if large problem instances are considered. For future work, it would be interesting to identify addi- tional features of Latin square graphs and use them to design effective search strate gies and operators. Approaches based on vertex coloring algorithms are also worthy of in vestigation. A CKNOWLEDGMENT W e would like to thank Dr. K. Haraguchi for making the benchmark instances tested in available and sharing the codes of his ILS algorithm with us and Dr. U. Benlic for her comments on the paper. This work w as partially supported by the Franco-Chinese PHC Cooperation Program (No. 41342NC) and National Natural Science Foundation of China (Grants No. 61602196, 61472147). R EFERENCES L. Anderson, 1985. Completing partial latin squares. Mathematisk Fysiske Meddelelser, 41, 23–69. C. Ans´ otegui, A. del V al, I. Dot´ u, C. Fern´ andez, F. Man y` a, 2004. Mod- eling choices in quasigroup completion: SA T vs. CSP. In McGuinness, D. L., & Ferguson, G. (Eds.), AAAI, pp. 137–142, San Jose, California, USA. AAAI Press / The MIT Press. R.A. Barry, P.A. Humblet, 1993. Latin routers, design and implementa- tion. Journal of Lightwave T echnology, 11(56): 891–899. 13 TABLE VI C OMPARA TIVE RESULTS ON 22 D IFFIC ULT CA SE S OF LSC-50-70 Instance ITS MMCOL’MMCOL Instance ITS MMCOL’MMCOL SR t(s) S R t(s) SR t(s) SR t(s) S R t(s) SR t(s) LSC-50-70-4 24/30 988.17 30/30 394.21 27/30 302.56 LSC-50-70-51 26/30 643.12 27/30 627.41 29/30 293.51 LSC-50-70-6 24/30 585.44 28/30 722.32 29/30 511.98 LSC-50-70-55 26/30 495.45 30/30 587.49 30/30 327.82 LSC-50-70-8 16/30 1232.98 17/30 1664.40 24/30 994.23 LSC-50-70-57 1/30 1942.50 0/30 0.00 4/30 1724.49 LSC-50-70-13 10/30 1002.09 16/30 1375.53 17/30 1154.08 LSC-50-70-58 13/30 664.00 11/30 899.54 23/30 1160.19 LSC-50-70-20 9/30 1127.22 4/30 2086.53 10/30 912.57 LSC-50-70-70 21/30 716.79 25/30 1155.18 27/30 561.12 LSC-50-70-21 22/30 1131.82 23/30 1182.76 27/30 516.05 LSC-50-70-72 26/30 618.60 28/30 883.57 29/30 423.29 LSC-50-70-25 22/30 1029.20 30/30 776.40 28/30 598.79 LSC-50-70-74 8/30 1297.03 9/30 1530.22 17/30 918.89 LSC-50-70-26 4/30 1151.33 4/30 1166.57 13/30 1006.81 LSC-50-70-83 20/30 994.51 18/30 1734.15 25/30 438.82 LSC-50-70-28 20/30 888.14 23/30 902.15 29/30 634.01 LSC-50-70-87 22/30 562.83 26/30 1592.43 30/30 518.07 LSC-50-70-35 26/30 803.14 30/30 11069.10 29/30 583.45 LSC-50-70-90 20/30 1002.29 21/30 1529.61 28/30 968.10 LSC-50-70-40 24/30 764.63 29/30 893.85 29/30 460.62 LSC-50-70-100 14/30 1304.27 14/30 1013.30 17/30 550.96 U. 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T o apply the proposed method to solve PLSE, we make the following two adjustments. First, unlike LSC, the color domains of some vertices in the case of PLSE may become empty during the constraint propagation based kernalization process, implying that the corresponding grids cannot be legally ﬁlled by any giv en number. In terms of graph coloring, if the color domain of a vertex v becomes empty when applying the preprocessing 14 procedure, any color for vertex v is deﬁnitiv ely conﬂicting with at least one of its adjacent vertices. If this happens, we randomly assign a color {1, . . . , n}to vertex v and keep this color unchanged during the search process. Second, at the end of the MMCOL algorithm, there are two possibilities. If the returned ﬁnal coloring c∗is conﬂict- free (i.e., f(c∗)=0), the given partial Latin square is fully completed and an optimal solution is found. Otherwise, some vertices are assigned conﬂicting colors in c∗(i.e., f(c∗)>0). In this case, we obtain a legal partial solution by dropping from c∗some conﬂicting vertices. The dropped vertices correspond to the grids that cannot be legally ﬁlled while the remaining vertices in the legal partial coloring deﬁne a solution for the given PLSE instance. T o remove conﬂicting vertices, we ﬁrst drop any vertex v with empty color domain caused by the preprocessing procedure. Then, if conﬂicts remain, we repetitively remove the v ertex u which is conﬂicting with the largest number of other vertices in the coloring until we obtain a partial conﬂict-free coloring. T able VII summarizes the results of our MMCOL al- gorithm and seven reference methods on the set of 1800 PLSE benchmark instances introduced in . Like the LSC instances of Section IV-A, these 1800 PLSE instances are evenly divided into 18 types (n∈ {50,60,70}, r∈ {0.3,0.4,0.5,0.6,0.7,0.8})(so 100 instances per type (n, r)). For these instances, n 2 is a trivial upper bound of their optimal solutions. In this experiment, we used the same experimental condition as for solving LSC (Section IV-C). Like , we solved each instance once. Columns 1–3 indicate the char- acteristics of the instances with the same information as in T able III. Columns 4–17 present, for each reference algorithm and for each type (n, r)of 100 instances, the number of fully completed Latin squares “suc#”, and the average completed grids over 100 instances “Avg”. Columns 18–20 show the results of our MMCOL algorithm in terms of “suc#”,“Avg” and the average computation time “t avg(s)” in seconds. Notice that if a partial Latin square is fully completed, the optimum is attained (so 1186 instances out of 1800 are solved to optimality). Otherwise, the reported result in terms of the ﬁlled grids gives a lower bound of the gi ven PLSE instance. T able VII shows that MMCOL obtains improved or equal average results for 15 out of 18 types (in bold) except the types (n∈ {50,60,70}, r= 0.8). In particular, for the 1000 instances of types (n∈ {50,60,70}, r∈ {0.3,0.4,0.5}), (n= 70, r= 0.6), and 186 instances of types (n∈ {50,60}, r= 0.6),(n= 70, r= 0.7), MMCOL attains an optimal solution. Meanwhile, for the instances of types (n∈ {50,60,70}, r= 0.8), MMCOL performs worse than the reference algorithms. W e mention that since the instances of types (n∈ {50,60,70}, r= 0.8) are strongly constrained, the color domains of some uncolored vertices are reduced to the empty set during the constraint propagation based preprocessing of Section II-B, indicating that these instances cannot be fully completed. Let α >0 be the number of vertices with an empty color domain identiﬁed during the preprocessing, n 2−α deﬁnes an upper bound of the given instance, which is strictly tighter than the trivial n 2 bound. TABLE VII C OMPARA TIVE RESULTS OF MMCOL ON T HE SE T OF 1800 PLSE B ENC HM ARK I NS TANC ES Instance CPX-IP CPX-CP LSSOL 1-ILS∗2-ILS Tr-ILS∗3-ILS MMCOL n r Inst#Suc#Av g. Suc#Av g. Suc#Av g. Suc#Avg. Suc#Avg. Suc#Avg. S uc#Avg. S uc#Avg. t avg (s) 0.3 100 10 2496.03 98 2499.87 13 2496.35 100 2500 99 2499.98 100 2500 98 2499.96 100 2500 0.22 0.4 100 1 2493.78 66 2498.02 4 2494.65 99 2499.98 100 2500 100 2500 93 2499.86 100 2500 0.16 50 0.5 100 0 2488.52 4 2489.92 1 2492.96 95 2499.89 98 2499.95 100 2500 67 2499.25 100 2500 0.31 0.6 100 0 2476.21 0 2478.87 0 2489.21 7 2496.23 7 2496.3 20 2497.18 0 2494.67 85 2499.64 17.55 0.7 100 0 2446.4 0 2451.04 0 2463.45 0 2469.47 0 2469.78 0 2470.07 0 2467.77 0 2478.94 3513.73 0.8 100 0 2394.58 0 2388.1 0 2393.67 0 2394.14 0 2394.11 0 2394.14 0 2394.09 0 2364.61 0.10 0.3 100 0 3593.07 77 3598.29 0 3593.2 99 3599.98 100 3600 100 3600 64 3599.28 100 3600 0.69 0.4 100 0 3590.68 19 3592.55 0 3591.17 99 3599.97 98 3599.96 100 3600 43 3598.58 100 3600 0.52 60 0.5 100 0 3585.29 1 3585.83 0 3587.5 83 3599.65 81 3599.58 97 3599.94 21 3597.53 100 3600 0.67 0.6 100 0 3572.61 0 3573.7 0 3585.52 5 3595.82 2 3595.85 13 3596.67 1 3592.77 97 3599.94 13.41 0.7 100 0 3534.71 0 3540.45 0 3561.05 0 3571.47 0 3570.58 0 3572.12 0 3566.51 0 3589.82 4297.18 0.8 100 0 3478.58 0 3464.14 0 3476.44 0 3478.59 0 3478.37 0 3478.49 0 3478.05 0 3431.85 0.10 0.3 100 0 4890.2 38 4893.75 0 4890.25 99 4899.98 99 4899.98 100 4900 13 4897.32 100 4900 0.90 0.4 100 0 4887.73 5 4888.36 1 4887.98 98 4899.96 99 4899.98 99 4899.98 4 4896.4 100 4900 0.65 70 0.5 100 0 4881.09 0 4881.17 0 4882.9 76 4899.41 78 4899.44 81 4899.57 0 4893.97 100 4900 1.51 0.6 100 0 4868.21 0 4868.74 0 4877.77 2 4895.3 0 4894.93 9 4896.19 0 4888.52 100 4900 19.82 0.7 100 0 4829.65 0 4831.94 0 4859.71 0 4872.41 0 4870.97 0 4872.95 0 4864.38 4 4894.58 4617.20 0.8 100 0 4761.44 0 4737.73 0 4761.17 0 4766.67 0 4765.81 0 4765.91 0 4763.93 0 4698.78 0.10 15 Dr. Yan Jin received the B.Sc (2009) and M.Sc (2012) degrees from the Huazhong University of Science and Technology (China). She obtained the Ph.D. degree in Computer Science from the Univer- sity of Angers (France) in 2015. She is currently an Associate Professor in the School of Computer Science and Technology, Huazhong University of Science and Technology. Her research interests in- clude reinforcement learning based intelligence com- puting and metaheuristics for solving large-scale combinatorial optimization problems and transporta- tion, including graph coloring problem, maximum independent set problem, packing, scheduling and routing problem. Dr. Jin-Kao Hao holds the title of Distinguished Professor (Professeur de classe exceptionnelle) with the Computer Science Department, University of Angers (France) and is Senior Fellow of the Institut Universit´ e de France. His research lies in the design of effective algorithms and intelligent computational methods for solving large-scale combinatorial search problems. He is interested in various application areas including data science, complex networks, and transportation. He has authored or co-authored more than 250 peer-reviewed publications and co-edited 9 books in Springers LNCS series. He has served as an Invited Member of more than 200 Program Committees of International Conferences and is on the Editorial Board of 7 International Journals. Dr. Hao graduated in 1982 from the National University of Defense T echnology (School of Computer Science) (China). He received the Master degree (Oct. 1987) from the National Institute of Applied Sciences (INSA Lyon, France), the Ph.D. in Constraint Programming (Feb. 1991) from the University of Franche-Comt´ e (France), and the Professorship Diploma HDR (Habilitation ` a Diriger des Recherches) (Jan. 1998) from the University of Science and T echnology of Montpellier (France). Citations (21) References (40) ... Besides, the author also proposed a novel swap operation called Trellisswap, resulting in a novel Trellis-neighborhood search algorithm named Tr-ILS. According to the literature, the current best heuristic algorithm for the LSC problem is called MMCOL (Jin and Hao 2019), which is mainly based on a constraint propagation technique, a problem-specific crossover operator, an iterated tabu search procedure and a distance-quality-based pool updating strategy. Besides, a transformation method is also proposed to convert an LSC instance to a domain-constrained Latin square graph (Jin and Hao 2019). ... ... According to the literature, the current best heuristic algorithm for the LSC problem is called MMCOL (Jin and Hao 2019), which is mainly based on a constraint propagation technique, a problem-specific crossover operator, an iterated tabu search procedure and a distance-quality-based pool updating strategy. Besides, a transformation method is also proposed to convert an LSC instance to a domain-constrained Latin square graph (Jin and Hao 2019). Although the MMCOL algorithm performs very well on some hard graphs, it has to cost lots of computation time for obtaining an arbitrary legal solution. ... ... To improve the performance of local search for the LSC problem on the hard instances with large sizes, three reduction rules are introduced in which the first reduction rule has been already used into reducing the scale of LSC instances (Jin and Hao 2019). Reduction Rule 1: If a vertex v has only one optional color class (i.e., The correctness of the second and third reduction rules is very obvious. ... A Fast Local Search Algorithm for the Latin Square Completion Problem Article Jun 2022 Shiwei Pan Yiyuan Wang Minghao Yin The Latin square completion (LSC) problem is an important NP-complete problem with numerous applications. Given its theoretical and practical importance, several algorithms are designed for solving the LSC problem. In this work, to further improve the performance, a fast local search algorithm is developed based on three main ideas. Firstly, a reduction reasoning technique is used to reduce the scale of search space. Secondly, we propose a novel conflict value selection heuristic, which considers the history conflicting information of vertices as a selection criterion when more than one vertex have equal values on the primary scoring function. Thirdly, during the search phase, we record previous history search information and then make use of these information to restart the candidate solution. Experimental results show that our proposed algorithm significantly outperforms the state-of-the-art heuristic algorithms on almost all instances in terms of success rate and run time. View Show abstract ... Starting from an initial solution typically constructed using a greedy heuristic, a local search algorithm improves the current solution by considering the best moves in a given neighborhood. To escape local optima traps, local search algorithms usually incorporate dedicated mechanisms such as tabu lists [2,17] or perturbation strategies [22,33]. However for very difficult instances of graph coloring, the single trajectory local search approach is not powerful enough to locate very high quality solutions mainly due to its limited diversification capacity. ... A deep learning guided memetic framework for graph coloring problems Article Full-text available Oct 2022 KNOWL-BASED SYST Olivier Goudet Cyril Grelier Jin-Kao Hao Given an undirected graph G=(V,E) with a set of vertices V and a set of edges E, a graph coloring problem involves finding a partition of the vertices into different independent sets. In this paper we present a new framework that combines a deep neural network with the best tools of classical heuristics for graph coloring. The proposed method is evaluated on two popular graph coloring problems (vertex coloring and weighted coloring). Computational experiments on well-known benchmark graphs show that the proposed approach is able to obtain highly competitive results for both problems. A study of the contribution of deep learning in the method highlights that it is possible to learn relevant patterns useful to obtain better solutions to graph coloring problems. View Show abstract ... e hybridization between global and local search algorithms has been experimentally proven to provide better search performance . Many examples have proved the effectiveness of this strategy, including many difficult problems, such as multimodal optimization [33,34], large-scale global optimization , combinatorial optimization [32,35], single-objective optimization [36,37], and multiobjective optimization [38,39]. For large-scale global optimization problems, most of the best-performing algorithms were hybrid algorithms combining global and local search (e.g., SHADE-ILS, MLSHADE-SPA, and MOS). ... A Novel Memetic Algorithm Based on Multiparent Evolution and Adaptive Local Search for Large-Scale Global Optimization Article Full-text available Mar 2022 Comput Intell Neurosci Wenfen Zhang Yulin Lan In many fields, including management, computer, and communication, Large-Scale Global Optimization (LSGO) plays a critical role. It has been applied to various applications and domains. At the same time, it is one of the most challenging optimization problems. This paper proposes a novel memetic algorithm (called MPCE & SSALS) based on multiparent evolution and adaptive local search to address the LSGO problems. In MPCE & SSALS, a multiparent crossover operation is used for global exploration, while a step-size adaptive local search is utilized for local exploitation. A new offspring is generated by recombining four parents. In the early stage of the algorithm execution, global search and local search are performed alternately, and the population size gradually decreases to 1. In the later stage, only local searches are performed for the last individual. Experiments were conducted on 15 benchmark functions of the CEC′2013 benchmark suite for LSGO. The results were compared with four state-of-the-art algorithms, demonstrating that the proposed MPCE & SSALS algorithm is more effective. View Show abstract Learning a Prior for Monte Carlo Search by Replaying Solutions to Combinatorial Problems Chapter Sep 2024 Tristan Cazenave View A massively parallel evolutionary algorithm for the partial Latin square extension problem Article Full-text available May 2023 COMPUT OPER RES Olivier Goudet Jin-Kao Hao View An area optimization approach taking into account polarity conversion sequence Article May 2023 APPL SOFT COMPUT Yuhao Zhou Zhenxue He Chen Chen Xiang Wang View Fast Area Optimization Approach for XNOR/OR-based Fixed Polarity Reed-Muller Logic Circuits Based on Multi-strategy Wolf Pack Algorithm Article Mar 2023 Yuhao Zhou Zhenxue He Jianhui Jiang Xiang Wang Area optimization is one of the most important contents of circuits logic synthesis. The smaller area has stronger testability and lower cost. However, searching for a circuit with the smallest area in a large-scale space of polarity is a combinatorial optimization problem. The existing optimization approaches are inefficient and do not consider the time cost. In this paper, we propose a multi-strategy wolf pack algorithm (MWPA) to solve high-dimension combinatorial optimization problems. MWPA performs global search based on the proposed global exploration strategy, extends the search area based on the Levy flight strategy, and performs local search based on the proposed deep exploitation strategy. In addition, we propose a fast area optimization approach (FAOA) for fixed polarity Reed-Muller (FPRM) logic circuits based on MWPA, which searches the best polarity corresponding to a FPRM circuit. The experimental results confirm that FAOA is highly effective and can be used as a promising EDA tool. View Show abstract Collaborative Neurodynamic Algorithms for Solving Sudoku Puzzles Conference Paper Oct 2022 Hongzong Li Jun Wang View Solving the clustered traveling salesman problem via traveling salesman problem methods Article Full-text available Jun 2022 Yongliang Lu Jin-Kao Hao Qinghua Wu The Clustered Traveling Salesman Problem (CTSP) is a variant of the popular Traveling Salesman Problem (TSP) arising from a number of real-life applications. In this work, we explore a transformation approach that solves the CTSP by converting it to the well-studied TSP. For this purpose, we first investigate a technique to convert a CTSP instance to a TSP and then apply powerful TSP solvers (including exact and heuristic solvers) to solve the resulting TSP instance. We want to answer the following questions: How do state-of-the-art TSP solvers perform on clustered instances converted from the CTSP? Do state-of-the-art TSP solvers compete well with the best performing methods specifically designed for the CTSP? For this purpose, we present intensive computational experiments on various benchmark instances to draw conclusions. View Show abstract An Efficient Power Optimization Approach for Fixed Polarity Reed–Muller Logic Circuits Based on Metaheuristic Optimization Algorithm Article Dec 2022 Yuhao Zhou Zhenxue He Chen Chen Xiang Wang With the emergence of multi-core architecture and the increase of chip operating frequency, power optimization has become a key step of circuit logic synthesis. Aiming at the XNOR/OR circuits, with the goal of minimizing power, construct the optimal polarity FPRM circuits power optimization scheme. However, the power optimization for FPRM circuits is a multipeak combinatorial optimization problem, we first propose a meta-heuristic optimization algorithm (MOA), which includes global exploration optimizer, local deep exploitation optimizer and initial population and uses the proposed differential evolution optimization, fierce wolf siege algorithm based tabu search, and improved Skew Tent map to make the population evolve. Based on the proposed Huffman tree construction algorithm and MOA, we propose an efficient power optimization approach (EPOA) to find the minimum power FPRM circuit. Experimental results on the benchmark circuits confirm the effectiveness of EPOA. View Show abstract Show more Opposition-Based Memetic Search for the Maximum Diversity Problem Article Full-text available Feb 2017 IEEE T EVOLUT COMPUT Yangming Zhou Jin-Kao Hao Beatrice Duval As a usual model for a variety of practical applications, the maximum diversity problem (MDP) is computational challenging. In this paper, we present an opposition-based memetic algorithm (OBMA) for solving MDP, which integrates the concept of opposition-based learning (OBL) into the wellknown memetic search framework. OBMA explores both candidate solutions and their opposite solutions during its initialization and evolution processes. Combined with a powerful local optimization procedure and a rank-based quality-and-distance pool updating strategy, OBMA establishes a suitable balance between exploration and exploitation of its search process. Computational results on 80 popular MDP benchmark instances show that the proposed algorithm matches the best-known solutions for most of instances, and finds improved best solutions (new lower bounds) for 22 instances. We provide experimental evidences to highlight the beneficial effect of opposition-based learning for solving MDP. View Show abstract A Gentle Introduction to Memetic Algorithms Chapter Full-text available Jan 2003 Pablo Moscato Carlos Cotta View Iterated local search with Trellis-neighborhood for the partial Latin square extension problem Article Full-text available Oct 2016 J HEURISTICS Kazuya Haraguchi A partial Latin square (PLS) is a partial assignment of n symbols to an n×n n\times n grid such that, in each row and in each column, each symbol appears at most once. The PLS extension problem is an NP-hard problem that asks for a largest extension of a given PLS. We consider the local search such that the neighborhood is defined by (p, q)-swap , i.e., the operation of dropping exactly p symbols and then assigning symbols to at most q empty cells. As a fundamental result, we provide an efficient (p,∞)(p,\infty )-neighborhood search algorithm that finds an improved solution or concludes that no such solution exists for p∈{1,2,3}p\in {1,2,3}. The running time of the algorithm is O(n p+1)O(n^{p+1}). We then propose a novel swap operation, Trellis-swap, which is a generalization of (p, q)-swap with p≤2 p\le 2. The proposed Trellis-neighborhood search algorithm runs in O(n 3.5)O(n^{3.5}) time. The iterated local search (ILS) algorithm with Trellis-neighborhood is more likely to deliver a high-quality solution than not only ILSs with (p,∞)(p,\infty )-neighborhood but also state-of-the-art optimization solvers such as IBM ILOG CPLEX and LocalSolver. View Show abstract Hybrid evolutionary search for the minimum sum coloring problem of graphs Article Full-text available Mar 2016 INFORM SCIENCES Yan Jin Jin-Kao Hao Given a graph G, a proper k-coloring of G is an assignment of k colors to the vertices of G such that two adjacent vertices receive two different colors. The minimum sum coloring problem (MSCP) is to find a proper k-coloring while minimizing the sum of the colors assigned to the vertices. This paper presents a stochastic hybrid evolutionary search algorithm for computing upper and lower bounds of this NP-hard problem. The proposed algorithm relies on a joint use of two dedicated crossover operators to generate offspring solutions and an iterated double-phase tabu search procedure to improve offspring solutions. A distance-and-quality updating rule is used to maintain a healthy diversity of the population. We show extensive experimental results to demonstrate the effectiveness of the proposed algorithm and provide the first landscape analysis of MSCP to shed light on the behavior of the algorithm. View Show abstract Handbook of combinatorial designs Book Jan 2006 Charles J. Colbourn Jeff Dinitz Continuing in the bestselling, informative tradition of the first edition, the Handbook of Combinatorial Designs, Second Edition remains the only resource to contain all of the most important results and tables in the field of combinatorial design. This handbook covers the constructions, properties, and applications of designs as well as existence results. Over 30% longer than the first edition, the book builds upon the groundwork of its predecessor while retaining the original contributors' expertise. The first part contains a brief introduction and history of the subject. The following parts focus on four main classes of combinatorial designs: balanced incomplete block designs, orthogonal arrays and Latin squares, pairwise balanced designs, and Hadamard and orthogonal designs. Closely connected to the preceding sections, the next part surveys 65 additional classes of designs, such as balanced ternary, factorial, graphical, Howell, quasi-symmetric, and spherical. The final part presents mathematical and computational background related to design theory. New to the Second Edition • An introductory part that provides a general overview and a historical perspective of the area • New chapters on the history of design theory, various codes, bent functions, and numerous types of designs • Fully updated tables, including BIBDs, MOLS, PBDs, and Hadamard matrices • Nearly 2,200 references in a single bibliographic section Meeting the need for up-to-date and accessible tabular and reference information, this handbook provides the tools to understand combinatorial design theory and applications that span the entire discipline. The author maintains a website with more information. View Show abstract Handbook of Memetic Algorithms Book Jan 2012 Ferrante Neri Carlos Cotta Pablo Moscato Memetic Algorithms (MAs) are computational intelligence structures combining multiple and various operators in order to address optimization problems. The combination and interaction amongst operators evolves and promotes the diffusion of the most successful units and generates an algorithmic behavior which can handle complex objective functions and hard fitness landscapes. “Handbook of Memetic Algorithms” organizes, in a structured way, all the the most important results in the field of MAs since their earliest definition until now. A broad review including various algorithmic solutions as well as successful applications is included in this book. Each class of optimization problems, such as constrained optimization, multi-objective optimization, continuous vs combinatorial problems, uncertainties, are analysed separately and, for each problem, memetic recipes for tackling the difficulties are given with some successful examples. Although this book contains chapters written by multiple authors, a great attention has been given by the editors to make it a compact and smooth work which covers all the main areas of computational intelligence optimization. It is not only a necessary read for researchers working in the research area, but also a useful handbook for practitioners and engineers who need to address real-world optimization problems. In addition, the book structure makes it an interesting work also for graduate students and researchers is related fields of mathematics and computer science. View Show abstract A Guide to Graph Colouring Article Jan 2016 R.M.R. Lewis This book treats graph colouring as an algorithmic problem, with a strong emphasis on practical applications. The author describes and analyses some of the best-known algorithms for colouring arbitrary graphs, focusing on whether these heuristics can provide optimal solutions in some cases; how they perform on graphs where the chromatic number is unknown; and whether they can produce better solutions than other algorithms for certain types of graphs, and why. The introductory chapters explain graph colouring, and bounds and constructive algorithms. The author then shows how advanced, modern techniques can be applied to classic real-world operational research problems such as seating plans, sports scheduling, and university timetabling. He includes many examples, suggestions for further reading, and historical notes, and the book is supplemented by a website with an online suite of downloadable code. The book will be of value to researchers, graduate students, and practitioners in the areas of operations research, theoretical computer science, optimization, and computational intelligence. The reader should have elementary knowledge of sets, matrices, and enumerative combinatorics. View Show abstract Completing partial latin squares Article Jan 1985 L. Anderson View Memetic search with interdomain learning: A realization between CVRP and CARP Article Jan 2014 IEEE T EVOLUT COMPUT Liang Feng Yew Soon Ong Meng-Hiot Lim Ivor W Tsang In recent decades, a plethora of dedicated evolutionary algorithms (EAs) have been crafted to solve domain specific complex problems more efficiently. Many advanced EAs have relied on the incorporation of domain specific knowledge as inductive biases that is deemed to fit the problem of interest well. As such, the embedment of domain knowledge about the underlying problem within the search algorithms is becoming an established mode of enhancing evolutionary search performance. In this paper, we present a study on evolutionary memetic computing paradigm that is capable of learning and evolving knowledge meme that traverses different but related problem domains, for greater search efficiency. Focusing on combinatorial optimization as the area of study, a realization of the proposed approach is investigated on two NP-hard problem domains (i.e., capacitated vehicle routing problem (CVRP) and capacitated arc routing problem (CARP)). Empirical studies on well established routing problems and their respective state-of-the-art optimization solvers are presented to study the potential benefits of leveraging knowledge memes that are learned from different but related problem domains on future evolutionary search. View Show abstract Sur quelques aspects mathématiques des problèmes de classification automatique. I, II Article Jan 1983 S. Regnier View Show more Recommended publications Discover more Article Full-text available Memetic search for the max-bisection problem January 2013 · Computers & Operations Research Qinghua Wu Jin-Kao Hao Given an undirected graph G=(V,E)G=(V,E) with weights on the edges, the max-bisection problem (MBP) is to find a partition of the vertex set VV into two subsets V1 and V2 of equal cardinality such that the sum of the weights of the edges crossing V1 and V2 is maximized. Relaxing the equal cardinality, constraint leads to the max-cut problem (MCP). In this work, we present a memetic algorithm for ... [Show full abstract] MBP which integrates a grouping crossover operator and a tabu search optimization procedure. The proposed crossover operator preserves the largest common vertex groupings with respect to the parent solutions while controlling the distance between the offspring solution and its parents. Extensive experimental studies on 71 well-known G-set benchmark instances demonstrate that our memetic algorithm improves, in many cases, the current best known solutions for both MBP and MCP. View full-text Article New neighborhood and memetic heuristics for the Maximum Parsimony problem November 2006 Adrien Goëffon Phylogenetic reconstruction aims at reconstructing the evolutionary history of a set of species, represented by a tree. Among the reconstruction methods, the Maximum Parsimony (MP) problem consists, given a set of aligned sequences to find a binary tree, whose leaves are associated to the sequences and which minimizes the parsimony score. Traditionally, existing resolution approaches of this ... [Show full abstract] NP-complete problem apply basic heuristic methods, like greedy algorithms and local search. One of the difficulties concerns the handling of binary trees and the definition of tree neighborhoods. In this thesis, we first focus on an improvement of descent algorithms. We empirically show the limits of the existing tree neighborhoods, and introduce a progressive neighborhood which evolves during the search to limit the evaluation of inappropriate neighbors. This algorithm is combined with a genetic algorithm which uses a specific tree crossover based on topological distances between each pair of leaves. This memetic algorithm shows very competitive results, both on real benchmarks taken from the literature as well as with randomly generated instances. Read more Article Full-text available Multi-objectives memetic discrete differential evolution algorithm for solving the container Pre-mar... January 2019 · Journal of Information and Communication Technology Hossam Mustafa Masri Ayob Mohd Zakree Ahmad Nazri Sawsan Abu-Taleb The Container Pre-marshalling Problem (CPMP) has the significant effect of reducing ship berthing time, and can help in increasing terminal turnover rate. In order to solve the CPMP, this research proposes a Multi-objectives Memetic Discrete Differential Evolution algorithm (MODDE). To date, existing research in CPMP only focuses on single-objective approaches. However, this is not a suitable ... [Show full abstract] approach due to the considerable effort required to validate the hard constraints of CPMP. Therefore, this work aims at addressing the effect of minimizing the number of miss-overlaid containers on the total number of movements in building the final feasible bay layout by embedding it in the multi-objectives evaluation function. The proposed algorithm combines the Discrete Differential Evolution mutation with the Memetic Algorithm evolutionary steps in order to find high quality CPMP solutions, achieve high convergence rate and avoid premature convergence and local optima problems. In addition, it improves the exploration and exploitation capabilities of the algorithm. The standard pre-marshalling benchmark dataset (i.e., Bortfeldt-Forster) is used to evaluate the effectiveness of the proposed algorithm. The experimental results reveal that the proposed MODDE algorithm can find good solutions on instances of the standard pre-marshalling benchmarks. This demonstrates that using the multi-objectives approach with a combination of the Discrete Differential Evolution mutation and the Memetic Algorithm evolutionary is a suitable approach for solving multi-objectives CPMP. View full-text Article Full-text available A Memetic Approach to the Nurse Rostering Problem November 2001 · Applied Intelligence Edmund K. Burke Peter I. Cowling Patrick De Causmaecker Greet Vanden Berghe Constructing timetables of work for personnel in healthcare institutions is known to be a highly constrained and difficult problem to solve. In this paper, we discuss a commercial system, together with the model it uses, for this rostering problem. We show that tabu search heuristics can be made effective, particularly for obtaining reasonably good solutions quickly for smaller rostering ... [Show full abstract] problems. We discuss the robustness issues, which arise in practice, for tabu search heuristics. This paper introduces a range of new memetic approaches for the problem, which use a steepest descent improvement heuristic within a genetic algorithm framework. We provide empirical evidence to demonstrate the best features of a memetic algorithm for the rostering problem, particularly the nature of an effective recombination operator, and show that these memetic approaches can handle initialisation parameters and a range of instances more robustly than tabu search algorithms, at the expense of longer solution times. Having presented tabu search and memetic approaches (both with benefits and drawbacks) we finally present an algorithm that is a hybrid of both approaches. This technique produces better solutions than either of the earlier approaches and it is relatively unaffected by initialisation and parameter changes, combining some of the best features of each approach to create a hybrid which is greater than the sum of its component algorithms. View full-text Discover the world's research Join ResearchGate to find the people and research you need to help your work. Join for free ResearchGate iOS App Get it from the App Store now. Install Keep up with your stats and more Access scientific knowledge from anywhere or Discover by subject area Recruit researchers Join for free LoginEmail Tip: Most researchers use their institutional email address as their ResearchGate login Password Forgot password? - [x] Keep me logged in Log in or Continue with Google Welcome back! Please log in. 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10050	https://www.mdapp.co/mca-calculator-for-fetal-anemia-534/	MCA Calculator for Fetal Anemia MDApp © 2020 MDApp. All Rights Reserved. Home Specialties Blog Contact Specialties Allergy And Immunology Anesthesiology Cardiology Deficiency View All Home Specialties Allergy And Immunology Anesthesiology Cardiology Deficiency View All Blog Contact Search Email Password [x] Show Login Password - [x] Remember Me Forgot your password?Not a Member yet? Please Sign Up! Email MCA Calculator for Fetal Anemia Screens for risk of fetal anemia based on middle cerebral artery measurement during doppler scan in pregnancy. Refer to the text below the tool for more information about the MCA velocity by gestational age in weeks and the multiple of median (MoM value). Purpose Key Facts Contents The middle cerebral artery is one of the three largest vessels that supply the brain with blood and is very visible on Doppler ultrasound during fetal wellbeing checks. Middle Cerebral Artery Peak Systolic Velocity (MCA-PSV) = e (2.31 + 0.046 GA) Multiple of the Median (MoM) = Doppler MCA peak systolic velocity / MoM 1 (Median) With the following properties: R2=0.78, P<0.001. The risk of fetal anemia is highest at observed peak systolic velocities of 1.5 times the median or higher. Jump to: MCA Calculator Multiple of Median and MCA Velocity References Gestational age (weeks) Doppler MCA peak systolic velocity EmbedPrintShare × To embed this calculator, please copy this code and insert it into your desired page: Close Click to copy × To Save This Calculator As A Favourite You Must Be Logged In... Creating an account is free and takes less than 1 minute. Log InSign Up Close Other Tools Feedback How to Print Pregnancy Weight Calculator Pregnancy Week Calculator Pregnancy Iron Deficiency Calculator hCG Levels Calculator Send Us Your Feedback [x] By using this form you agree with the storage and handling of your data by this website. Read our Privacy Policy Steps on how to print your input & results: Fill in the calculator/tool with your values and/or your answer choices and press Calculate. Then you can click on the Print button to open a PDF in a separate window with the inputs and results. You can further save the PDF or print it. Please note that once you have closed the PDF you need to click on the Calculate button before you try opening it again, otherwise the input and/or results may not appear in the pdf. Multiple of Median and MCA Velocity The middle cerebral artery is one of the three largest vessels that supply the brain with blood and is very visible on Doppler ultrasound during fetal wellbeing checks. This can replace serial amniocenteses for fetal anemia screening. Evaluation of the MCA is employed in: Diagnosis of growth restriction; Screening and assessment risk of pre-eclampsia; Suspected serological Rh+/Rh- conflict; Complications in multiple pregnancy. There are established values of the median peak systolic velocity, based on gestational age (week 18 to 40) that can be retrieved from published tables or estimated through: Middle Cerebral Artery Peak Systolic Velocity (MCA-PSV) = e (2.31 + 0.046 GA) With the following properties: R2=0.78, P<0.001. The median MCA-PSV is then compared with the observed MCA during Doppler ultrasound, to determine the Multiple of the Mean (MoM) which s a measure of how the observed MCA differs from the median population MCA. Multiple of the Median (MoM) = Doppler MCA peak systolic velocity / MoM 1 (Median) The MCA peak systolic velocity increases by 0.2 cm/s with every 100g (3.5 ounces) of fetal weight gain. An increased velocity of blood flow through the middle cerebral artery (and/or dilatation) may be indicative of hypoxia, with a redirection of blood flow to vital organa, such as brain, heart and lungs. The risk of fetal anemia is highest at observed peak systolic velocities of 1.5 times the median or higher. There are several indications to be respected, to obtain an accurate measurement during the Doppler: The MCA must occupy more than 50% of the enlarged image; The MCA must be sampled 2mm after internal carotid artery origin; The angle between the ultrasound beam and the direction of blood flow should be 0 degrees. References Original reference Mari G, Deter RL, Carpenter RL, et al. Noninvasive diagnosis by Doppler ultrasonography of fetal anemia due to maternal red-cell alloimmunization. Collaborative Group for Doppler Assessment of the Blood Velocity in Anemic Fetuses. N Engl J Med. 2000; 342(1):9-14. Other references Zimmerman R, Carpenter RJ Jr, Durig P, Mari G. Longitudinal measurement of peak systolic velocity in the fetal middle cerebral artery for monitoring pregnancies complicated by red cell alloimmunisation: a prospective multicentre trial with intention-to-treat. BJOG. 2002;109(7):746-752. Mari G. Middle cerebral artery peak systolic velocity: is it the standard of care for the diagnosis of fetal anemia?. J Ultrasound Med. 2005; 24(5):697-702. Delle Chiaie L, Buck G, Grab D, Terinde R. Prediction of fetal anemia with Doppler measurement of the middle cerebral artery peak systolic velocity in pregnancies complicated by maternal blood group alloimmunization or parvovirus B19 infection. Ultrasound Obstet Gynecol. 2001; 18(3):232-236. Specialty:Obstetrics Gynecology System:Reproductive Objective:Screening Year Of Study:2000 Abbreviation:MCA Article By:Denise Nedea Published On:July 30, 2020 Last Checked:July 30, 2020 Next Review:July 30, 2025 Browse specialties Allergy And Immunology Anesthesiology Cardiology Deficiency Dermatology Disability Emergency View All Become A Member Save your favourites! Get latest medical calculators! 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10051	https://mathoverflow.net/questions/97307/polynomials-all-of-whose-roots-are-rational	solvable groups - Polynomials all of whose roots are rational - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Polynomials all of whose roots are rational Ask Question Asked 13 years, 4 months ago Modified13 years, 4 months ago Viewed 2k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. I have two questions about the class of integer-coefficient polynomials all of whose roots are rational. I asked this at MSE, but it attracted little interest (perhaps because it is not interesting!) Q1. Is there some way to recognize such a polynomial from its coefficients a 0,a 1,…,a n a 0,a 1,…,a n? I am aware of the rational-root theorem, which says that each rational root is of the form ±p/q±p/q, where p p is a factor of a 0 a 0 and q q a factor of a n a n. Example. The roots of 12544 x 5+24976 x 4−23994 x 3−51721 x 2−17080 x+1275 12544 x 5+24976 x 4−23994 x 3−51721 x 2−17080 x+1275 are {3 2,−5 7,−5 7,1 16,−17 8}.{3 2,−5 7,−5 7,1 16,−17 8}. Here a 0=1275=3⋅5⋅17 a 0=1275=3⋅5⋅17 and a 5=12544=2 8⋅7 2 a 5=12544=2 8⋅7 2. As Mark Bennet commented at MSE, perhaps an analog of Sturm's theorem would serve. Q2. Has this class of polynomials been studied in its own right? In other words, is this class interesting? I can see it has at least a monoid structure, as the product of two such polynomials also has all rational roots. These are naive questions, well out of my expertise. Thanks in advance for educating me! polynomials solvable-groups factorization reference-request Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Apr 13, 2017 at 12:19 CommunityBot 1 2 2 silver badges 3 3 bronze badges asked May 18, 2012 at 13:17 Joseph O'RourkeJoseph O'Rourke 153k 36 36 gold badges 371 371 silver badges 994 994 bronze badges 15 3 @Joseph: Why is the rational-root theorem not enough? It gives a relatively fast algorithm. What kind of characterization do you want?user6976 –user6976 2012-05-18 13:34:07 +00:00 Commented May 18, 2012 at 13:34 1 Your questions are about rational fully-factorizable polynoms, and I must admit that I don't know more than what you said about them. For question Q2, but for real polynoms, the derivative of a fully-factorizable polynom is also fully-factorizable. Is it the kind of results your after?Julien Puydt –Julien Puydt 2012-05-18 13:54:55 +00:00 Commented May 18, 2012 at 13:54 @Mark: My understanding is that that theorem gives the form of possible rational roots, but does not tell you that all the roots will be rational...Joseph O'Rourke –Joseph O'Rourke 2012-05-18 13:55:52 +00:00 Commented May 18, 2012 at 13:55 @Julien: Yes, exactly, structural theorems giving properties of the whole class.Joseph O'Rourke –Joseph O'Rourke 2012-05-18 13:57:18 +00:00 Commented May 18, 2012 at 13:57 1 This is a naive question: is this equivalent to the condition that the Galois group is trivial?Bruce Westbury –Bruce Westbury 2012-05-18 15:16:58 +00:00 Commented May 18, 2012 at 15:16 \|Show 10 more comments 5 Answers 5 Sorted by: Reset to default This answer is useful 10 Save this answer. Show activity on this post. It seems to me that the obvious algorithm via the rational root theorem is somewhat inefficient in at least two cases: a 0 a 0 or a n a n is BIG (so that we might not even be able to factor it), or they have A LOT of prime factors. Instead, I believe the following algorithm based on Hensel's lifting lemma is more suited here. Let F=∑i=0 n a i X i∈Z[X]F=∑i=0 n a i X i∈Z[X] be our polynomial, which we may assume to have no multiple root. Now pick a prime p p which does not divide a n a n and pass to Z/p Z.Z/p Z. If F F has no root over Z/p Z Z/p Z (this requires p p checks), then F F has no rational root. (The fact we assumed F F has no multiple root over the integers does not necessarily mean it still has no multiple root over Z/p Z,Z/p Z, but this can easily be circumvented by a suitable choice of p.p.) Otherwise, use Hensel's lemma to lift the roots r k r k from Z/p k Z Z/p k Z to Z/p 2 k Z,Z/p 2 k Z, where k k is to be chosen later. (this works fine since p∤F′(r k)p∤F′(r k)) Finally, we need to get back to the integers, from a root r k∈Z/p k Z r k∈Z/p k Z (where we may choose k k). To an element from Z/p k Z Z/p k Z, we associate the unique integer from its congruence class mod p p which is between −p k/2−p k/2 and p k/2.p k/2. If we choose k k so large that p k p k is greater than 2\|a n a 0\|,2\|a n a 0\|, then a n X−a n/b a a n X−a n/b a (which is a factor of F F if b X−a∈Z[X]b X−a∈Z[X] is) remains unchanged by the above association, but a n X−a n/b a=a n(X−r k)a n X−a n/b a=a n(X−r k) in (Z/p k Z)X[X] and a n(X−r k)=a n X−ρ a n(X−r k)=a n X−ρ where ρ ρ is obtained by the above association. We are done now: divide a n X−ρ a n X−ρ (which is an integer multiple of b X−a b X−a) by gcd(a n,ρ)gcd(a n,ρ) and check the divisibility of F F by this reduced factor. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited May 18, 2012 at 18:13 Joseph O'Rourke 153k 36 36 gold badges 371 371 silver badges 994 994 bronze badges answered May 18, 2012 at 17:52 Daniel m3Daniel m3 276 2 2 silver badges 6 6 bronze badges 6 (I fixed the broken link.)Joseph O'Rourke –Joseph O'Rourke 2012-05-18 18:13:47 +00:00 Commented May 18, 2012 at 18:13 I am confused. In your first step you either have to factor a n a n (to see that p p does not divide it), or take p p so large that it cannot divide a n,a n, in which case your p p checks will kill you, not to mention finding the requisite large prime. Admittedly, doing something more sensible, like Berlekamp's factoring algorithm will get you back to (probabilistic) polynomial time, but I am guessing this will be be pretty bad in practice.Igor Rivin –Igor Rivin 2012-05-18 18:18:09 +00:00 Commented May 18, 2012 at 18:18 2 I think this will lead to a polynomial time algorithm because that is what LLL do in the paper Lenstra, A. K.; Lenstra, H. W., Jr.; Lovász, L. Factoring polynomials with rational coefficients. Math. Ann. 261 (1982), no. 4, 515–534. user6976 –user6976 2012-05-18 19:13:03 +00:00 Commented May 18, 2012 at 19:13 1 @Igor, if you need a prime that doesn't divide an N N-digit number, it suffices (for N>10 N>10) to look among the first N N primes.Barry Cipra –Barry Cipra 2012-05-18 22:38:44 +00:00 Commented May 18, 2012 at 22:38 1 It does not take p p checks to determine if f f has roots in F p F p. Compute G C D(f,t p−t)G C D(f,t p−t) using the Euclidean algorithm, with the first computation of t p−t mod f t p−t mod f done by repeated squaring. This takes something like (deg f)2 log p(deg⁡f)2 log⁡p steps, so it beats (deg f)p(deg⁡f)p unless p p is very small.David E Speyer –David E Speyer 2016-12-16 13:36:37 +00:00 Commented Dec 16, 2016 at 13:36 \|Show 1 more comment This answer is useful 9 Save this answer. Show activity on this post. Given a bound for the possible denominators, small enough intervals containing each root will each contain at most one candidate for a rational root, and it is easy to find that candidate (e.g. using continued fractions) and check whether it really is a root. It seems to me that this should all be possible to do in polynomial time. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered May 18, 2012 at 17:33 Robert IsraelRobert Israel 54.7k 1 1 gold badge 79 79 silver badges 156 156 bronze badges 5 @Robert: I already posted a new question mathoverflow.net/questions/97329/… . Your answer seems promising but how are you going to avoid factorization of a n a n and a 0 a 0? user6976 –user6976 2012-05-18 17:48:08 +00:00 Commented May 18, 2012 at 17:48 The denominator is a factor of a n a n, so it is at most a n a n. For any two distinct rationals p/q p/q, p′/q′p′/q′ with \|q\|,\|q′\|≤a n\|q\|,\|q′\|≤a n, \|p/q−p′/q′\|≥1/(q q′)≥1/a 2 n\|p/q−p′/q′\|≥1/(q q′)≥1/a n 2. So you just need an interval of length <1/a 2 n<1/a n 2 around each root.Robert Israel –Robert Israel 2012-05-18 19:22:20 +00:00 Commented May 18, 2012 at 19:22 Suppose a n a n is 1, so we are looking at the integer roots (in fact we can always assume that) how can you avoid factoring a 0 a 0? user6976 –user6976 2012-05-18 19:40:10 +00:00 Commented May 18, 2012 at 19:40 I guess you are then localize the roots enough so that there are at most one integer in each interval? Can it be done in polynomial time?user6976 –user6976 2012-05-18 19:41:33 +00:00 Commented May 18, 2012 at 19:41 1 Any integer root is at most \|a 0\|\|a 0\|. WLOG we may assume the roots are distinct (otherwise divide by the gcd of the polynomial and its derivative). Construct the Sturm sequence of the polynomial, and bisect the interval (O(n log\|a 0\|)O(n log⁡\|a 0\|) times until you isolate all the roots in intervals of length 1 1.Robert Israel –Robert Israel 2012-05-18 21:01:25 +00:00 Commented May 18, 2012 at 21:01 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Maybe this book has pertinent information: Alas, I have no access to a copy now. ALSO, MathSciNet has this paper: MR1342405 Luo, Yong Chao Some criteria for polynomials with integer coefficients to have rational roots, and their applications. (Chinese. English, Chinese summary) Guizhou Shifan Daxue Xuebao Ziran Kexue Ban 12 (1994), no. 4, 21–30. 12D10 Alas, the paper was not reviewed, so I have no idea what it contained and if it was valid. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited May 18, 2012 at 14:42 answered May 18, 2012 at 13:48 Felix GoldbergFelix Goldberg 7,072 4 4 gold badges 32 32 silver badges 57 57 bronze badges 2 @Felix: That paper title seems quite relevant, although I don't yet see how to access the English summary... Joseph O'Rourke –Joseph O'Rourke 2012-05-18 14:04:17 +00:00 Commented May 18, 2012 at 14:04 2 @Joseph and Felix: This book does not seem to have any relevant information. The book is quite good, though!user6976 –user6976 2012-05-18 15:49:32 +00:00 Commented May 18, 2012 at 15:49 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Random comments: The rational root test might be good for finding all rational roots but less so if one is happy to abort as soon as an irrational root is found (i.e one not of the form t a 0 t a 0. If a n=1 a n=1 then check if ±1±1 are roots. If so great! if not then you can factor a 0 a 0 searching for a factor less than a 1/n 0 a 0 1/n. There is some gain from looking further for very small integer divisors, but perhaps not much. If a 0=1 a 0=1 then factor u n f(1/u)u n f(1/u) This may not be so great if a 0 a 0 is huge. For example if we replace f f by a n−1 n f(u/a n)a n n−1 f(u/a n) to get a monic polynomial with constant term a 0 a n−1 n a 0 a n n−1 Repeated roots can be tricky for some methods so one might wish to compute f′f′ and find the polynomial gcd since any repeated roots will be roots of that. In your case the gcd of 7 x+5 7 x+5 reveals a double root of −5 7−5 7 leaving 256 x 3+144 x 2−826 x+51 256 x 3+144 x 2−826 x+51 Given f′f′, even without bothering with the gcd, one is set to use Newton's method (or some other) to quickly find approximate real roots. Then given a somewhat accurate real root r,r, one can see if it is close to a rational root. The continued fraction should have a convergent which is remarkably good. Seeing a root near −0.7−0.7 gives −.7138457729−.7138457729 after 5 5 iterations. The convergents are −2/3,−5/7,−227/318,−232/325,−5563/7793−2/3,−5/7,−227/318,−232/325,−5563/7793 which gives two reasonable candidates. A couple more iterations would leave no doubt. Your example is not great for illustrating that because the "round off error" quickly gives the exact rational root (as a decimal) if it is of the form t 10 k t 10 k for k k small. I was excited that Newton's method (although others might be better) returns rationals given rationals, however the denominators grow very quickly. However, the previous observation gives the idea of using Newton's method plus rounding to always get approximants of the form t a 0.t a 0. That will quickly get to a root if there is one (I'd think.) If f is a product of n n linear factors the same is true mod m m for any m m. Famously, the converse is not true. However there are algorithms to factor mod p p and one failure tells you to stop. I recall methods to lift to factorizations mod p k p k but that is back to general integer factorization. Maybe that is easier if you already have linear factors though. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered May 18, 2012 at 20:45 Aaron MeyerowitzAaron Meyerowitz 30.3k 2 2 gold badges 50 50 silver badges 105 105 bronze badges 2 @Aaron: Thanks for your illuminating thoughts. I have to admit that my example was chosen quite randomly, rather than to illustrate a particular point, so I am not surprised that it "is not great for illustrating" X X. :-)Joseph O'Rourke –Joseph O'Rourke 2012-05-19 01:58:58 +00:00 Commented May 19, 2012 at 1:58 1 I just got a kick out of the fact that exact rational numbers explode: 1/10↦133933/2049550↦203135852565605401626332803/3249200102395371629529717250 1/10↦133933/2049550↦203135852565605401626332803/3249200102395371629529717250 so I went to middling accuracy reals and got 0.1↦0.065347515↦0.062518727↦0.06250000 0.1↦0.065347515↦0.062518727↦0.06250000 which is a root.Aaron Meyerowitz –Aaron Meyerowitz 2012-05-19 04:54:57 +00:00 Commented May 19, 2012 at 4:54 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Robert Israel's comment about using Sturm sequences got me thinking about how it could be done using only the original polynomial f(x)f(x) (assumed to have distinct rational zeros). If the degree is odd, it is easy to identify an interval in which the sign changes. If the degree is even, it is easy to find three points so that f(x)f(x) is least at the middle point of the three, then golden section search will find a point where the polynomial is negative. (Golden section search finds a local minimum, and all local minima have negative f(x)f(x) in this case.) Once an interval with a sign change is found, use binary search to find a root. Divide it out and repeat. I think that only a polynomial number of evaluations of f(x)f(x) are required altogether, since all of the searches only need to continue to precision 1/a 0 1/a 0. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered May 19, 2012 at 7:55 Brendan McKayBrendan McKay 38.3k 3 3 gold badges 85 85 silver badges 153 153 bronze badges Add a comment\| You must log in to answer this question. 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10052	https://www.youtube.com/watch?v=mcyXUM5aorg	Imre Barany: The Steinitz lemma, its matrix version, and balancing vectors II Hausdorff Center for Mathematics 9540 subscribers Description 152 views Posted: 12 Apr 2024 The Steinitz lemma, a classic from 1913, states that a sequence a_1,...,a_n of (at most) unit vectors in R^d whose sum is the origin, can be rearranged so that every partial sum of the rearranged sequence has norm at most 2d. It is an important result with several applications. I plan to mention a few. I also explain its connection to vector balancing. In the matrix version of the Steinitz lemma A is a k by n matrix whose entries unit vectors in R^d and their sum is the origin. Oertel, Paat, Weismantel have proved recently that there is a rearrangement of row j of A (for every j) such that the sum of the entries in the first m columns of the rearranged matrix has norm at most 40d^5 (for every m). We improve this bound to 4d-2. Transcript: it's topic is chines theorem and those who missed yester last week's spook I I tell them what it is but first want to let everybody know that tomorrow at 11:30 we have a guided tour at the Arium arum is a is a mathematical Museum not far from here and part of a building is also nice and it's part of the operations discret mathematics research group as well so so we meet there at 11:30 we meet there at 11:30 and the guy did to last till 1:00 and uh and we the entrance fee was waved because we are colleagues this is my little Victory sotin is the from 1913 says the following Stein is B is a subset of the unit B of A Unit B in Rd unit Bard B this is finite and there is one more condition the sum of the vectors in V is zero so this is a set of vectors and you start from the origin and it zigzags out and comes back and then it says that there is a there is an ordering V = V1 VN such that the maximum over or k of su one to k b i partial sums along this ordering are all smaller than a constant that only depends on on diens does not depend on the vector set here and the result I proved last time was that it is always smaller than equal to D which is a which is an excellent result very strong result I I gave you a wrong proof of the same thing for the ukan or and that was a nice proof as well but it gave two to the D basically so this is much better and uh and the the best constant that you can right here is uh is a certain number s of B which is thein is constant of the norm or the unit and the I proved last time is that it is smaller than Dimension and it's a beautiful proof but it does not it does not see the norm at all the norm comes to play in the proof in the last moment and you only use it that Alpha time x Norm is smaller than Alpha absolute value times x norm and cannot use that it is ukian or maximum or L1 Norm ET the question was motivated or was by another question of rean that says that he proved that a conditional convergence series of real numbers uh can be made to converge to either nothing or a single point or every point in the line by a suitable re Arrangement and here that question was what is the case in higher Dimensions if you take a conditionally convergent sequ series of vectors what are the set of points that it can be made to converge to and the answer is ACC using this theorem this is the key step in the proof is that it is an fi Subspace so I a single point or the empty set or a line or a SP or a plane Etc um I gave you several applications one of them I did not explain this application uh uh the Hilbert's question the positive answer to Ran's question but I give you an example in in scheduling right I give you another example in uh [Music] um Vic I me balancing balancing vectors yeah uh yeah and I want to give you another application which is there are several applications and is another application which is I of what this un some some Norm which is written here so this is a this is typically a central symmetric convex body containing the origin it's interior but for this proof you don't have to it does not have to be [Music] cetric I want to give you another another application but first I want to mention a similar result so uh you have a the same system finite uh no not the same system vectors set of vectors in Min -1 one to the D and uh you know that finite finite and the sum of the vectors in V is zero so same same condition but for this Norm and what you want to have is an ordering find an ordering ordering V1 VN such that sum from 1 to K VI I J's component J's component of this Vector is smaller than a function that only depends on J so this is a this is an infinite versional infinite version of the previous theorem because it tells you that you if it's true then you can no matter what you have in what dimension you can you can find an ordering so that that the first component is less than 10 the second component is less than 10 squ the third components less than 10 to the three Etc so just I give you an example for F of G and so this was a question and of course it has some applic implications for the conditional convergence series as well which I don't explain in detail I Prov this one with a very we B here long ago with about two to the 3 to the J and uh and I won't tell you that proof but but I tell you that there was a much stronger approv by y back which shows that this is less than this can be made less than J to 4 plus Epsilon for any Epsilon and here is a universal constant so this is a much stronger result this is us back and his proof is wrong and highly nonrival and I won't be able to explain the anything about it but the basic idea later uh and so another application is is this one you have a set of vectors in -1 one to the D and finite and the sum of the vectors in V is again in -11 to the D and the of the set size V equals n uh if this is the so this is a theorem byin and someone else whose name and and then there is a there is a subset no n is larger than 4 2D to the D n is large enough then there is a subset U of V proper subset uh at least two elements such that the sum of the vectors in U also lies in minus1 one to d d so if if a certain set of vectors in this box sum to a vector in this in this box as well that there is a non-rival subset of this set V that also some the elements of which some to in this box as well this was needed for some computer science problem but I I don't want to go into the details um I can tell you the SS of the proof of this application so the cality of you is bigger than 4 2 to 2D to the D this is V but but u u u is larger than two because one element satisfy this property automatically right so that's why I insist on this being larger than two it says that every Vector here lies in this box and I want to have a subset whose sum lies also in this box other than the one element set and the full set and I tell you what the proof is sorry so it may happen that that the cality of you has to be really bigger than two yeah yeah you can't guarantee that uh just for two [Music] yeah so the proof is ask about so you say that this is WR proof so B is not what was the question no I he wrong Pro you are interested in this group not interesting I I'm interested what you're saying I also understood the the proof is wrong you said but you you said long you said long it's a it's a correct Pro it's a long and and complicated proof with construction which guarantees a certain family of family of subsets with seven properties and I don't have time for that today uh so the proof goes this way I this some of these vectors is a vector that I call minus V node if you wish then the SU some of the vectors in V plus v note is zero right so by stain theorem I can order them so that every partial sum is small within the bow of radius of D so every partial sum of that guide is in the Box D minus D to the D right and and I may assume that V note is on the upper half of this sequence so I have V1 V2 Etc V and the last one and V note is somewhere lat and the middle because it doesn't if I change the order this the same remains true so I go the opposite order and now I choose this as the four first parti this as the second parti this as the third par sub this is s not this is S2 this is S4 Etc this the last one is going to be S almost n half are you with me they all lie in the D by D box that if there are enough many of them and two of them lie in the same box are you with me so far so good so the difference is that at least two element set which which has some in this box here in this box so this is a fairly direct application of of S is theorem but you see it's how useful it is you just you just take it out from the black box and apply it and you get it so a related topic is balancing balancing vectors and this is my next topic and uh and uh um I gave you an example last time which is a which is a game and the game consists of a pusher and a Chooser and it is being played on the pl and The Pusher offers a unit Vector in ukan Norm for every step so push them and Chooser offers vector v n in Step n and the Chooser chooses an Epsilon and the unit Vector in in the plane and this chooses an plus or minus one uh and after end steps the position is some one to n Epsilon KV K so this a vector in the pl and Pusher wants this this Vector to be as big as possible wants to push away from the origin this position and Chooser wants to keep it close to the origin and then it's a very simple game because at every moment Pusher chooses a vector orthogonal to the present position right then the length is the square of the length is increased by at least one no matter what choice Chooser makes and Chooser always chooses the two vectors for which the distance from the origin is smaller so that means that the value of the game is square root of n exactly right so this a very simple game because The Pusher get reach that the square of the distance is increases by one at least and uh and the Chooser can reach that increase in the square is at most one each step uh and the variant I offered is the following instead of offering the whole unit wall Pusher can only offer this vectors here this vectors of course then it can offer these vectors as well because V and minus V is just the same thing right so what what what is the value of the game in that case and I have some partial result that in that case the value of the game is larger than root 2 - 1un n so the value is smaller and I don't have anything better than square root of n so not even .99 square root of n so this is a this is a question that I could not solve um and any any successes in a week yeah so this is just a side cck and and let's see Vector question goes this way V is a sequence of vectors this time V1 V1 V2 Etc V and can be actually infinite that I in the unit be of Rd and find Epsilon one Etc Epsilon 2 in -1 1 - 1 1 so that the sum of Epsilon i v i from 1 to K [Music] maximum is smaller than a constant that similar to this one does not depend on anything but the norm right so this question was was around for quite some time vetki considered it in the plane uh in 1963 and in 1981 I proved that together with Greenberg that this is always smaller than 2dus one so then there exist a little theorem and this green B Green uh and I want to give you the proof of this theorem okay uh um um um so uh I to is is K is going to be a mapping from me1 v k + D to 0 no to - one 1 interal so this is a coefficient real number between minus one and one associated with every Vector here right so that it is a linear dependence that me that Al k v i i = to k + D is equal to Z and I want to make sure that very few of the alpha K are different from plus and minus one I want to make sure very few of the coefficients here are different from minus one and one so more precisely number of I such that RHA k v i is larger than and smaller than plus one is at most D so this is my main requirement but I also need one one more thing that if Alpha K of v i is equal to 1 or -1 then Al k + 1al to Al K Al v i sorry Alp Al so if somebody reaches one or minus one it remains minus one or one no change in that one from that one so is this clear I make a picture of this one as well so this is Alpha K is and here I say that this is on the or the vectors V1 V K plus d but I could say is it depends on 1 2 3 and k + D and then the picture is this one here is one 2 3 ET k+ D and the other numbers here is plus one here isus one this is plus one this isus one and my function looks like that color choke did uh so there are several values that it is one and then several values that it is minus one and then again several values that it is plus one and then there are atos D that are between right this is what I want to achieve and and these I call them floating variables the floating variable the number is at most D and how how shall we do it so let me say alpha 1 Alpha One is should be a linear dependence of the first D+ one vectors right there are D plus Factor there is a linear dependence I multiply it by a factor so that the maximum absolute value coefficient is one right then I kick out one value to here and here and at most def floating variables right now assume that we are in this position Alpha K is constructed alpha 1 is done now I step k 2 K2 k + 1 comes and then I have an alpha K linear dependen here satisfying everything and I choose k + V + 1 with zero level I have another linear dependence the coefficients are all same as in Alpha K but this Vector comes with coefficient zero I have D+ one at D plus one floating variables they have a linear dependence I add this linear dependence to my present linear dependence so that one gu kicked out to plus or minus one is this clear so I get a a new picture here uh somebody's kick out to one and then I get another floting variable here at was D and I kept every zero and every plus and minus one the same because they did not touch them right so this is almost the end of the proof what happens is that Alpha Cas stabilizes one more and more Alpha case alpha alpha VI stabilizes it becomes plus or minus one it remains plus or minus one and so in the end I will choose Epsilon Epsilon VI I equal Alpha K VI I if it is one or 1 once so then Epsilon is defined for every Vector here except at D and for those one I choose whatever I want I don't care okay and then let's check what's happens with with this sum here so let me write it out sum from 1 to k + Z Epsilon i v i I may assume that K is K is smaller than if I take all epon V High and if I take only I wrote here K plus d but if the the upper index is smaller than D then we are done because the number of vectors is at most B no matter what vectors I choose the sum is smaller than the so in this case uh I I consider I can take away from this one one to k + D Alp k v i v i so this is nothing but epon IUS Al k v i times V sub I now this number here is zero except at most D places right this Su is zero except for at most D places and then I have these summons each has value one here in that two here so then I get that this is smaller than equal to 2D I promise 2D minus one and I leave it to you to check how to reach that one now this is a this is a uh let me write here s s of b as the S sequence constant of the norm B the smallest number that you can write over here the smallest number that you can write over here and this proof tells you that this number here is always smaller than oral to 2us one so uh yeah so are you happy so far sure um [Music] I'm so um so there are various to this question um so here what you are given is a on step I choice between VI and minus V right the vector and its opposite what happens if I give instead a set of vectors here and the set of vectors here and the set of vectors here you at most UN vectors and you have to choose one from each so that these partial sums are small well if if the if this let me call this subset instead V sub One V sub 2 if V sub I don't contain the origin in the convex then say all of them can be separated from the origin by line then you just push it out to Infinity quickly so natural assumption is that the convex of H VI contains the origin just like in the Plus minus V Cas right so what happens if you are given sequence of subsets of the unit B with with the the convex are containing the origin can you pick one element from me so that this Su is small small essentially almost identical Pro Works in that Cas us is linear algebra so this is one variant of the which I don't go into thisa and another one is is so what you what you want to do in this original set is you want to split this set V into two sets the set v+ and the set V minus so that so the v+ is where Epsilon is + one and the V minus is where Epsilon I is minus one and then you want to see that if you go up to level K in each set then the sum of the first one up to K and the sum of the second up to K is close to each other and you can check that it means that so the sum of the sum of the vectors in the first of V plus set up to K minus the minus the sum of the v+ 2 K is in the B of radius 2 D V right this is what thisal say right are you with me this is an unusual notation this means that I add the vectors in this set up to index K and I add the vectors in this set this is minus up to K and of course the sum of the K up in v plus and plus this one sum of the vectors in V minus is equal to the sum of the vectors in V up to K so if you add these two together and then then this cancels and you see that this this sum here is not is in the half of the sum up to B up to K plus d B so it tells you that that this sum is in uh some v v v in up to K 12 plus d V and the same applies to V minus v+ V minus and then there is a question can you split it into five parts so that each part this sum up to K is close to the one over 5 this so this is yeah this one is in in yeah are you with me can I repeat okay so the question is whether this can be done and the answer is yes you can do it and you can split it into r r sets so that uh each partial sum uh of any set is in the ball of 1 / R this sum plus 2.5 D B right so this you can you can um uh yeah so this is this is another result that I I have have no time to prove but this is a yeah um uh more interesting is the next theorem one of my favorites in this topic uh which is transference theor this say the following for a symmetric Norm for a symmetric Norm s of B is smaller than oral to s s of T So this this makes a connection between between the Steines problem and the S sequence problem right so so this [Music] is UN B symmetric C symmetri Central symmetric this the works for non symmetric BS as well but that one does not so this is a this is a a very powerful and interesting theorem and actually the proof of this this result is based on a balancing Vector result which tells you that given a sequence of vectors in in in in in plus minus yeah in the maximum Norm then there are signs so that the case the J's component the J component of every partial sum is smaller than J to the 4 plus Epsilon uh and instead of giving you Chan's beautiful proof I tell you how y back proof proceeds for for this one with the signs there and and and he starts with so what what is so here he proves it first that every coordinate is plus one minus one or zero right a very special case and so here is a big Matrix Infinity if you wish and here are in the Cs there are there are uh Zer plus ones minus ones Etc and you want to give a sign to every column so that so that the partial sums are small the the following is the starting idea if I see a zero here I don't care what sign it has I just fix it if I have a one here and a minus one here or a one and a one then I to this one I will give sign plus one and to this one I will sign minus one and then I glue together this color completely and then the the new in the new Matrix this is going to be a zero right and here I get numbers which are not too big so each time I I either glue together two columns or just leave the columns alone and what I have here is 0 0 0 0 Z right and now this is a new Vector it's a new vector slightly larger than my original Vector but I can give you any sign I want later because this this will just Flip Flip the signs original signs here so this is easy but once in the second step I have uh say I have this these two vectors here and I know that the coefficients are - 2 - one 0 one 2 and you fixed that if you have 21 of them then you can find five among them that sum to zero with you plusus ones so if out of every 21 you can find five that gives you a zero value in both places and those five will be glued together with a zero coefficient and then so out of the 21 use that five and then you you add five more and find another five that sums to zero and fix the so this is this is going to be another block here five block and later another five block but no matter how far you go that as this is zero this is zero this is doesn't matter what side they get dat and only the sum of these ones matters but there are only L than 21 of them each at each stage right so then after this one or two first these two columns are fixed and then I have another set of larger vectors and I know that all of them are bonded right bonded components and then this is a pitching principle that you find a zero s because this is his proof for plus minus one Z vectors and then there is another idea highly not Reve that carries over to any anything so this this is how you get to to the this quantity here and ch's transference theorem I want to mention about ch's transference theorem that that I I think it was 19 83 or four that I was I was delegated toi from my Institute in Budapest to the m Institute in b e Georgia the Caucasus Mountains is the probabil me is the probabilist yes yes he's a probabilist uh a probabilist Armenian origin living in the Georgia capital but the Georgia capital has a large Armenian population and once you go there you don't expect anything to happen I mean I'm introduced to the director of The Institute shake hands with everybody and then give a to and go back home but CH was different he was interested in sign sequences and we talked about side sequences and I keep still in touch with him so that that's a good thing I mean it's good thing that a mathematician goes anywhere and can find somebody who can who he can talk to Easy and meaningfully anyway so this is since then he moved to the he he had a job in the states and now he's back in Georgia yeah anyway so um so I promised you Matrix version of shin and that will happen on the [Music] on so so this time Matrix ver Matrix verion uh a is a k by n Matrix and AI J is an element of a unit bow so in Rd so this is a this is a matrix whose entries are vectors not numbers and each entry is in the unit B of some norm and uh the assumption is again that the sum of the entries is zero total sum of the entri is zero and we are interested in this quantity Sigma M of a is equal to so here is my Matrix this is I have it has K rows and N columns and this is M and the sum of the vectors in this block is Sigma F so this is equal to some k j = I = 1 to K and J = 1 to m a i j uh this is the the sum of the first the the the first M columns right and you want them you want this one to be small in order to make this one small for every M you have to have Sigma n a small right if Sigma and the total sum is Big then you can't make everybody this small so we assume that Sigma n of a is zero and sigma and uh and C is r permuted copy of a if this is the J the [Music] iow and in the I I can rearrange the vectors any way I like right in dependently of in every row I can choose another ordering another ordering another ordering so this the sum of these K n vectors is zero and there is a a good order according to Shin this theorem that all partial sums along that ordering are small but here I restrict the possibilities of rearrangement I can rearrange the vectors only rowwise so that's why I call it rer copy of a and the theorem and and that is a theem i i or part and viice Manel um there is saying that under these conditions there is a r permuted copy of a the copy C of a with uh Sigma m a c Norm Max M smaller than a number that only depends on Dimension and this is um this is uh 40 D to the 5 uh so this is the result it's interesting because it does not depend on the size of the Matrix only the dimension of the space right and there is another condition that is Trivial because the yeah I can you can do it different way but you see The Columns of this Matrix are unit vectors in in B B B product and the sum of the columns is zero right so you can rearrange them by by sign theem so that every partial sum is smaller than dimension of the space which is C V so this is always smaller than a d uh so whichever is better is abound yeah and so what the names I'm sorry this is I can write it out it's o l advice and the paper appeared in the Met programming um maybe p a vice no no e a a t and o e r t e l uh and here is a new result of mine saying that here that this can be smaller than uh four 4 d 4 D- 2 so this is an improvement over this part this part I I don't have an improvement so this is what I want to explain okay and uh and uh actually it is a we will see that it is a stronger theorem in in a certain sense but but uh uh um something first before that um the is so let me uh for these together form a set of matrices this is a k b k n and B is a set of matrices K by and matrices each entry equals an element of the unit B and the sum is zero okay so if B is a is in and I drop a at that b is in a actually in this case you don't have to have it zero some and and Epsilon is a Epsilon i j k byn k byn Matrix Matrix of plus and minus on then the of Epsilon is um Epsilon i j a i j Matrix this is again a matrix a again I don't know if the sum is zero but doesn't matter okay and then there is a and give a LMA [Music] LMA there is always a an Epsilon Matrix with uh six Sigma M uh the Epsilon smaller than uh- one for every M so this is a very simple number and you will see in a moment that I can list the vectors in this order and go up and come back here and then go up and come down here then go up and come down here it's an ordering of the vectors of d right then there is a according to that theorem there is a choice of the epsilons so that every sum is smaller than 2D minus one and of course every block here is such a small such a s so this is to even more is to up to this level it is more than 2 but I only use it this way and uh and then the theorem is the theorem is that that s of the matric M this is the best constant that you can write over here here over here is smaller than twice SS Matrix B so here another transference theorem it says that the that the Matrix constant the shinest constant of Matrix is smaller than twice the S sequence constants of a matrix of the same nor B now Cally symmetric me B is now centrally symmetric again yes yes yes I will not have time to finish this proof but I uh well I try my best um so for the proof you choose a matrix a in a which reaches this constant here right which means that uh and I assume that this Matrix a is in this form a A1 a11 a12 a13 Etc okay so that means that that Sigma Ma the maximum of Sigma m a b isal to S of B and for any row permuted copy of a c of a sigma M of C is larger than S of B for some M right so this at this this is the the best one that reaches this level it means that that this is equal to SB but no matter how I reorganize the rearrange the rows of the Matrix I get a partial sum which is larger than S so and then and then so here is my Matrix and uh um um assume that any n is even n is = 2 and then I create a new Matrix D which is equal to 12 a a it is I it is this 2 J [Music] -1 minus a i to J so what happens is that in this Matrix I choose this is the I throw and this is this element the next element I take this one minus this one you have it so this is the Matrix is the set a but this time a n half and in the same me because in the same me right because the average of two unit vectors is a unit vector and then for this one there is a b of EP is there is an Epsilon matx that Sigma M of D Epsilon is smaller than 2dus one no smaller than SSB of matrices for every M right so this is what we have and now I create a new R permuted copy of C so I instead of writing out this whole Matrix I already write out the I throw which is uh a one a a i 2 a i1 a i 2 a i 3 a i 4 Etc a i n a i nus one and then I will I will there will be a new matx here instead reorganized Matrix and what happens is that these two vectors go to this spot and this spot and these two vectors here go to either this spot or that spot Etc the rule is that that if so for this difference I have an Epsilon Epsilon and if it's is positive this goes here and that goes over there if this is negative then this goes over there this comes over here right and the same applies to everybody here I think I bet I will lose the ODS very soon so so so this way I get a new Matrix and uh and uh after some manipulations you understand that that this Matrix satisfies s of B for some so there are some inequalities that come into play I prove it but so actually this is Chan's idea original idea but it in his case it was different he did not have to create a new Matrix out of it but I better stop here because I'm afraid who go of you soon you won't I will not you will not because you stay patiently uh shall [Music] I so so so what happens is that uh I get a sequence here and um so if if this one was positive then this goes the first this becomes the first element and this becomes the last element which if I open up this Epsilon here then this goes with positive sign and this goes with negative sign from over there so this goes negative sign over there same applies here if if it Epsilon was negative this would go to the end and this would come here so what happens is that in this sequence the ones that come with positive coefficient go to the first half and in this order and one comes with negatives and comes in this order uh now if you want to um if you want to compute the partial sum up to this point right then what you add up [Music] is I have to write it as c i j is equal to 12 a r I 2 J - 1 + a i 2 j epon i j a i 2 J - 1 minus a [Music] i2j so this is this is this is how the rule applies when G is up to here if Epsilon is positive then this one cancels and you get a i this one exactly if Epsilon is negative then this cancels and you get this VOR here right and uh so if you ENT the C JS uh for fixed I and J = 1 2 okay uh no for fixed for fixed J one to n then here you get uh uh uh so in this case f is positive so you get these two oh yeah no no no sorry if you sum all of these then you get sum a i j j = 1 to 2 N sorry this is 1 12 s right if I sum all of these I get the first two two M elements of the vectors and in this second half I get uh this is D this is d i j so I get the sum of the D J is up to M so this is sum d i j Epsilon i j and this is what you have to sum and now you sum everything from a = 1 to uh no I = 1 to k i to K and then what you have here is a 12 Sigma M of Sigma 2 m of [Music] a are you with me the D JS are this D is this difference and times this one is written here so here you get the some that I'm interested in and here I get the sigma M uh the Epsilon and and this one this one uh yeah this one the two is disappeared because this is this is divided by two is the J right so this is Sigma Sigma M of C and this is uh in the in the ball with r with multiplier Sigma 2 m a and sigma M DC this is smaller than S of B time 12 and this is smaller than um SS of B uh for so so this one here is smaller and the sum of these two right and and you you so this is the sum up to here and for the sum up to here you do the same thing I mean sum up to here is the same as the some from here to here and this is an identical argument and tells you that Sigma m c is smaller than equal to that quantity okay but I know from this condition that Sigma is lar than S of B for some so what I get is that Sigma of B is is is smaller than 12 s of is smaller than S Plus s SB which is exactly what I want it to do right so I say I jumped over some f computations uh but but I hope it you can understand what's happening you can construct a a new ordering which which gives you what you want today that's it that's all for today thank you
10053	https://control.asu.edu/Classes/MAE318/318Lecture09.pdf	Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 9: Dynamics of Response: Complex Poles Overview In this Lecture, you will learn: Characteristics of the Response Complex Poles • Rise Time • Settling Time • Percent Overshoot Performance Speciﬁcations • Geometric Pole Restrictions M. Peet Lecture 9: Control Systems 2 / 31 Complex Poles Recall: Damping and Frequency 3 Diﬀerent Forms: Each with yss = 1 ˆ G(s) = b s2 + as + b ˆ G(s) = ω2 n s2 + 2ζωns + ω2 n ˆ G(s) = σ2 + ω2 d s2 −2σs + (σ2 + ω2 d) = σ2 + ω2 d (s −σ + ıωd)(s −σ −ıωd) = σ2 + ω2 d (s −σ)2 + ω2 d Two Complex Poles at s = σ ± ıωd. • Damped Frequency: ωd ▶ωd = q b −a2 4 = ωn p 1 −ζ2 • Decay Rate: σ ▶σ = −a 2 = −ζωn • Natural Frequency: ωn ▶ωn = √ b = p σ2 + ω2 d • Damping Ratio: ζ ▶ζ = a 2ωn = \|σ\| ωn M. Peet Lecture 9: Control Systems 3 / 31 Step Response for Complex Poles ˆ y(s) = ω2 d + σ2 s2 + 2σs + ω2 d + σ2 1 s = ω2 n s2 + 2ζωns + ω2 n 1 s = k1s + k2 s2 + 2ζωns + ω2 n + r2 s The poles of ˆ y are at s = σ ± ωdı. Using PFE, the solution is: y(t) = 1 −eσt cos(ωdt) + σ ωd sin(ωdt) = 1 −eσt ωn ωd sin (ωdt + φ) Where σ = ζωn, ωd = ωn p 1 −ζ2 and φ = tan−1 ωd ζωn . The result is oscillation with an Exponential Envelope. • Envelope decays at rate σ • Speed of oscillation is ωd, the Damped Frequency M. Peet Lecture 9: Control Systems 4 / 31 Complex Poles How NOT to calculate Rise Time (Tr) Recall: • Tr is the time to go from .1 to .9 of the ﬁnal value. Suppose there were no damping (σ = 0). Then the normalized solution is y(t) = 1 −cos(ωdt) The points t1 and t2 occur at ωt1 = cos−1(.9) = .45, ωdt2 = cos−1(.1) = 1.47 So that Tr = t2 −t1 = 1.02 ωd WRONG!!!! M. Peet Lecture 9: Control Systems 5 / 31 Complex Poles An Approximation for Rise Time (Tr) However, with Damping, the situation changes. • Damping slows the response. • No easy formula for rise time of a complex pole! • When ζ = .5, Tr ∼ = 1.8 ωn (We will use this approximation) M. Peet Lecture 9: Control Systems 6 / 31 Complex Poles How to Calculate Settling Time (Ts) Recall the step response y(t) = 1 −eσt cos(ωdt) + σ ωd sin(ωdt) Oscillations are conﬁned within an Exponential Envelope • The exponential envelop decays at rate σ. • Settling Time for a complex pole is given by contraction of the envelope ∥1 −y(t)∥≤eσt ≤.01 • Thus we use the same formula as for a real pole - i.e. Ts = 4.6 −σ M. Peet Lecture 9: Control Systems 7 / 31 Complex Poles What is Time to Peak (Peak Time) The time of MAXIMUM deﬂection. • Only Complex poles have peaks. Deﬁnition 1. The Peak Time, Tp is time at which the step response obtains its maximum value. To calculate Tp, we must ﬁnd when y(t) = 1 −eσt cos(ωdt) + σ ωd sin(ωdt) = 1 −eσt ωn ωd sin (ωdt + φ) Achieves its maximum. M. Peet Lecture 9: Control Systems 8 / 31 Complex Poles How to Calculate Peak Time (Tp) To ﬁnd the extrema, we set ˙ y(t) = 0, where it can be shown that ˙ y(t) = ωn p 1 −ζ2 eσt sin ωdt = ω2 n ωd eσt sin ωdt So ˙ y(t) = 0 when t = nπ ωd . Because of the exponential envelope, the ﬁrst peak will always be largest (n = 1). Tp = π ωd = π ωn p 1 −ζ2 M. Peet Lecture 9: Control Systems 9 / 31 Complex Poles What is Percent Overshoot? Unique to complex poles is the concept of overshoot: Deﬁnition 2. The Percent Overshoot, Mp is the peak value of the signal, as a percentage of steady-state ( cmax−cfinal cfinal · 100%). To ﬁnd Mp, we need the max of y(t) (We need (cmax)). • cmax = y(Tp) where Tp = π ωd . Systems with high overshoot may move violently before settling. • May diverge from acceptable path • Can cause crashes, un-modeled dynamics, etc. M. Peet Lecture 9: Control Systems 10 / 31 Complex Poles How to Calculate Percent Overshoot (Mp) To calculate Mp, we need the maximum value of y(t). • Occurs at time Tp = π ωd . Since cfinal = 1 (yss = 1): Mp = y(Tp) −1 = −eσTp cos(ωdTp) + σ ωd sin(ωdTp) = −eσTp cos(π) + σ ωd sin(π) = eσTp = e πσ ωd = e − πζ √ 1−ζ2 Mp = e πσ ωd = e − πζ √ 1−ζ2 Mp depends only on ζ. M. Peet Lecture 9: Control Systems 11 / 31 Complex Poles Lab Example Estimate: • Rise/Peak Time • Percent Overshoot M. Peet Lecture 9: Control Systems 12 / 31 Complex Poles Numerical Example Lets look at the suspension problem Open Loop: ˆ G(s) = s2 + s + 1 s4 + 2s3 + 3s2 + s + 1 The poles are: • p1,2 = −.9567 ± 1.2272ı • p3,4 = −.0433 ± .6412ı x1 x2 mc mw u Because there are two sets of poles, we should consider both. σ1 = −.9567 ωd,1 = 1.2272 σ2 = −.0433 ωd,2 = .6412 ωn,1 = q σ2 1 + ω2 d,1 = 1.5561 ζ1 = \|σ\| ωn,1 = .6148 ωn,2 = .6427 ζ2 = .0674 M. Peet Lecture 9: Control Systems 13 / 31 Complex Poles Numerical Example Closed Loop (Upper Feedback): Let k = 1 ˆ G(s) = s2 + s + 1 s4 + 2s3 + (3 + k)s2 + (1 + k)s + (1 + k) The poles are: • p1,2 = −.8624 ± 1.4391ı • p3,4 = −.1376 ± .8316ı x1 x2 mc mw u Consider both sets of poles. σ1 = −.8624 ωd,1 = 1.4391 σ2 = −.1376 ωd,2 = .8316 The natural frequency and damping ratios are ωn,1 = 1.6777 ζ1 = .5140 ωn,2 = .8429 ζ2 = .1632 M. Peet Lecture 9: Control Systems 14 / 31 Complex Poles Numerical Example: Percent Overshoot 5 10 15 20 25 30 t 0.4 0.5 0.6 0.7 yHtL Figure: Closed Loop 5 10 15 20 25 30 t 0.5 1.0 1.5 yHtL Figure: Open Loop Overshoot: Open Loop (easiest to use σ and ω directly) Mp,1 = e πσ1 ω1 = .0864 Mp,2 = .8088 Overshoot: Closed Loop (Upper Feedback) Mp,1 = .152 Mp,2 = .5946 A substantial improvement in performance. M. Peet Lecture 9: Control Systems 15 / 31 Complex Poles Numerical Example: Settling Time 5 10 15 20 25 30 t 0.4 0.5 0.6 0.7 yHtL Figure: Closed Loop 5 10 15 20 25 30 t 0.5 1.0 1.5 yHtL Figure: Open Loop Settling Time: Open Loop Ts,1 = 4.6 −σ = 4.81 Ts,2 = 106.23 Settling Time: Closed Loop (Upper Feedback) Ts,1 = 5.33 Ts,2 = 33.43 M. Peet Lecture 9: Control Systems 16 / 31 Complex Poles Numerical Example: Peak Time 5 10 15 20 25 30 t 0.4 0.5 0.6 0.7 yHtL Figure: Closed Loop 5 10 15 20 25 30 t 0.5 1.0 1.5 yHtL Figure: Open Loop Peak Time: Open Loop Tp,1 = π ωd = 2.56 Tp,2 = 4.90 Peak Time: Closed Loop Tp,1 = 2.18 Tp,2 = 3.78 M. Peet Lecture 9: Control Systems 17 / 31 Performance Speciﬁcations Dynamic response is determined by pole locations. • Except Steady-State Error Usually, dynamic response improves with feedback. • Recall the numerical examples. • Pole locations change under feedback. • The choice of k = 1 was just a guess. Im(s) Re(s) Consider: The goal of a controller is to change the location of the poles. • But where do we want them? Performance Speciﬁcations create Geometric Constraints in the Complex Plane. M. Peet Lecture 9: Control Systems 18 / 31 Pole Locations Constraint on Peak Time Suppose we have performance specs for • Tp, Tr, Mp etc. We can translate this to regions of the complex plane. Maximum Peak Time: Tp,desired. Im(s) Re(s) We usually require Tp < Tp,desired. π ωd = Tp < Tp,desired Which means ωd > π Tp,desired The geometric interpretation is that the imaginary part be suﬃciently large. M. Peet Lecture 9: Control Systems 19 / 31 Pole Locations Constraint on Settling Time Maximum Settling Time: Ts,desired. We want quick convergence. • So we require Ts < Ts,desired. Hence, 4.6 −σ = Ts < Ts,desired Im(s) Re(s) Which translates to σ < − 4.6 Ts,desired The geometric interpretation is that the real part be suﬃciently negative. M. Peet Lecture 9: Control Systems 20 / 31 Pole Locations Constraint on Percent Overshoot Maximum Overshot: Mp,desired. We don’t like hitting things, • So we need Mp < Mp,desired. 2 roots (σ ± ωdı) give two constraints: e πσ ±ωd = Mp < Mp,desired πσ ±ωd < ln(Mp,desired) or since ln(Mp,desired) < 0, ωd < π ln(Mp,desired)σ, ωd > − π ln(Mp,desired)σ Im(s) Re(s) A sector constraint on σ and ω? M. Peet Lecture 9: Control Systems 21 / 31 Complex Poles Percent Overshoot Alternatively, Mp,desired is determined by damping ratio alone: Invert: Mp = e − πζ √ 1−ζ2 , ζ = \|σ\| ωn M. Peet Lecture 9: Control Systems 22 / 31 Complex Poles Percent Overshoot A ﬁxed ζdesired deﬁnes an angle in the complex plane. θ = π 2 −sin−1(ζdesired) M. Peet Lecture 9: Control Systems 23 / 31 Pole Locations Constraint on Rise Time The expression for rise time is complicated. We use ζ = .5, to get Tr ∼ = 1.8 ωn Maximum Rise Time: Tr,desired. • We want quick response. • We require Tr < Tr,desired. 1.8 ωn = Tr < Tr,desired Im(s) Re(s) Thus we require ωn > 1.8 Tr,desired Recall that ωn = p σ2 + ω2 d, so the geometric interpretation is a circle: ∥s∥> 1.8 Tr,desired . M. Peet Lecture 9: Control Systems 24 / 31 Complex Poles M. Peet Lecture 9: Control Systems 25 / 31 Pole Locations Multiple Constraints Mostly, we have several constraints ωd > π Tp,desired σ < − 4.6 Ts,desired ωd < π ln(Mp,desired)σ ωn > 1.8 Tr,desired Im(s) Re(s) Any pole locations not prohibited are allowed. M. Peet Lecture 9: Control Systems 26 / 31 Pole Locations Multiple Constraints: Example High Performance Aircraft: • Overshoot: Reduce overshoot to less than 5%. Mp,desired = .05 ωd < π ln(Mp,desired)σ = −1.05σ • A diﬃcult requirement to meet? • Rise Time: Quick response is critical. Limit Rise Time to 1s or less Tr,desired = 1 ωn > 1.8 Tr,desired = 1.8 • Settling Time: Limit settling time to Ts,desired = 3.5s. Ts,desired = 3.5s σ < − 4.6 Ts,desired = −1.333 M. Peet Lecture 9: Control Systems 27 / 31 Pole Locations Multiple Constraints: Example We have the required Overshoot: Along a line of about θ = atan ωd σ = atan 1 −.9535 = 46◦ Which means a damping ratio of ζ = sin(90 −46◦) = .69. • Roughly ωd = σ Im(s) Re(s) To satisfy Ts, σ < −1.333, so lets try σ = −1.5. • Then ωd < 1.5 • Choose ωd = 1.4 M. Peet Lecture 9: Control Systems 28 / 31 Pole Locations Multiple Constraints: Example For Tr, need ωn > 1.8. However, ωn = q ω2 d + σ2 = 2.05 So rise time is already satisﬁed. Im(s) Re(s) If we need to decrease rise time, increase omega, while staying on lines of constant overshoot M. Peet Lecture 9: Control Systems 29 / 31 Pole Locations Missile Video Estimate Performance Specs: M. Peet Lecture 9: Control Systems 30 / 31 Summary What have we learned today? In this Lecture, you will learn: Characteristics of the Response Complex Poles • Rise Time • Settling Time • Percent Overshoot Performance Speciﬁcations • Geometric Pole Restrictions Next Lecture: Designing Controllers M. Peet Lecture 9: Control Systems 31 / 31
10054	https://fraze.it/sentences_using_celibate	347 English sentences using 'celibate' Toggle navigation Login Sign Up API Blog Discover more writing Writing write Filter by (Press Ctr Key for multiple selections): Source Hist. ALL Daily Use Famous ....Proverbs ....Poets ....Philosophers [View All] My Content Form Interrogative Negative Exclamative Tense Active Form: Simple Present Present Continuous Simple Past Past Continuous Present Perfect Present Perf. Cont. Past Perfect Past Perf. Cont. Simple Future Future Continuous Future Perfect Future Perf. Cont. Past Future Perf. Past Future Cont. Past Future Perf. Cont. Passive Form: Simple Present(p) Present Continuous(p) Simple Past(p) Past Continuous(p) Present Perfect(p) Present Perf. Cont.(p) Past Perfect(p) Past Perf. Cont.(p) Simple Future(p) Future Continuous(p) Future Perfect(p) Future Perf. Cont.(p) Rule Starts With Ends With Zone US UK Context Art Business Classifieds Education Entertainment Environment Food Fun Health Legal Military Politic Relationship Science Showbiz Sport Style Technology Zen FRAZE·IT in English ▶■French■Spanish■German■Italian■Portuguese Search ■Usages■Definitions■Synonyms■Translations■Pronunciations■Images Toggle filters About 347 results found using 'CELIBATE'. Source: 'Daily Use'. I did not always want to be celibate but was anyway-eventually for such reasons.(open, save, copy) allconsidering.com After she persisted with her vocation, entering a convent, he remained celibate.(open, save, copy) en.wikipedia.org A year ago, after losing the love of my life I fled to live with celibate monks.(open, save, copy) guardian.co.uk This is why, the Church is convinced, God chooses certain people to be celibate.(open, save, copy) cnn.com They must be celibate, foregoing even hugs, handshakes and extended eye contact.(open, save, copy) bostonherald.com The Chinese government prefers the celibate model of Buddhism for Taoist clergy.(open, save, copy) en.wikipedia.org I have heard it said that Caesar Augustus was celibate prior to meeting Cornelia.(open, save, copy) economist.com A celibate priesthood, he says, is a product of church law, which can be changed.(open, save, copy) washingtonpost.com Morrissey claimed to be celibate, although the British pop press was skeptical.(open, save, copy) npr.org 1 2 3 4 5 Next » Discover more write Writing writing Define 'celibate': Adjective: Abstaining from marriage and sexual relations, typically for religious reasons Having or involving no sexual relations Noun: celibates (plural) A person who abstains from marriage and sexual relations Web Definitions for 'celibate': An unmarried person who has taken a religious vow of chastity [source)] Abstaining from sexual intercourse; "celibate priests" [source)] (celibacy) an unmarried status [source)] Celibacy is defined as the lifestyle of someone who is voluntarily abstaining from all sexual activities, remaining without any sexual or romantic partner, and unmarried, all of his or her life. ... [source)] (celibately) In a celibate way [source)] (Celibacy) The unmarried state of life. Unlike the Roman Church, Orthodoxy permits a clergyman to be married; however, his marriage must occur before the ordination to a deacon or presbyter. ... [source)] Celibacy is the free choice to remain unmarried for the whole of one's life for the sake of the Kingdom of God. ... [source)] (Celibacy) The state of being unmarried; required of western clergy in the major orders (bishop, priest, deacon, subdeacon) since the twelfth century. [source)] (CELIBACY) (Arabic Uzubah), although not absolutely condemned by Muhammad, is held to be a lower form of life to that of marriage. It is related thatUsman ibn Maz`un wished to lead a celebate life and the Prophet forbade him, for, said he, "When a Muslim marries he perfects his religion. ... [source)] Synonyms for 'celibate': Adjective: unmarried , single , celibatarian Noun: bachelor , celibatarian Translate 'celibate' in : No language selected. Images for 'celibate': Native speakers pronounce 'celibate': Visit YouGlish.com ■English■French■Italian■Spanish■Portuguese■German HOMEABOUTCONTACTPRIVACY & TERMSAPIBLOG × Confirmation Do you really want to delete this phrase? Close Delete × Notice Login is required to perform this operation. Close Login Sign up × Notice Copy to clipboard: Ctrl+C, Enter: Close CLOSE
10055	https://physics.stackexchange.com/questions/413604/newtons-law-requires-two-initial-conditions-while-the-taylor-series-requires-in	newtonian mechanics - Newton's law requires two initial conditions while the Taylor series requires infinite! - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Physics helpchat Physics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Newton's law requires two initial conditions while the Taylor series requires infinite! Ask Question Asked 7 years, 3 months ago Modified7 years, 3 months ago Viewed 6k times This question shows research effort; it is useful and clear 27 Save this question. Show activity on this post. From Taylor's theorem, we know that a function of time x(t)x(t) can be constructed at any time t>0 t>0 as x(t)=x(0)+x˙(0)t+x¨(0)t 2 2!+x...(0)t 3 3!+...(1)(1)x(t)=x(0)+x˙(0)t+x¨(0)t 2 2!+x⃛(0)t 3 3!+... by knowing an infinite number of initial conditions x(0),x˙(0),x¨(0),x...(0),...x(0),x˙(0),x¨(0),x⃛(0),... at t=0 t=0. On the other hand, it requires only two initial conditions x(0)x(0) and x˙(0)x˙(0), to obtain the function x(t)x(t) by solving Newton's equation m d 2 d t 2 x(t)=F(x,x˙,t).(2)(2)m d 2 d t 2 x(t)=F(x,x˙,t). I understand that (2) is a second order ordinary differential equation and hence, to solve it we need two initial conditions x(0)x(0) and x˙(0)x˙(0). But how do we reconcile (2) which requires only two initial conditions with (1) which requires us to know an infinite number of initial informations to construct x(t)x(t)? How is it that the information from higher order derivatives at t=0 t=0 become redundant? My guess is that due to the existence of the differential equation (2), all the initial conditions in (1) do not remain independent but I'm not sure. newtonian-mechanics classical-mechanics mathematical-physics differential-equations Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Jun 25, 2018 at 13:10 SRSSRS asked Jun 25, 2018 at 12:48 SRSSRS 27.9k 13 13 gold badges 120 120 silver badges 376 376 bronze badges 3 4 As a rule, I do not comment about anonymous down-votes I receive because honestly, I don't see the need to give them a second thought. But I just noticed that this question has received 3 anonymous down-votes and each answer has at least one anonymous down-vote. That seems a bit odd to me.Alfred Centauri –Alfred Centauri 2018-06-25 22:00:26 +00:00 Commented Jun 25, 2018 at 22:00 1 @AlfredCentauri Best to cast a flag for that sort of thing. There's no guarantee the moderators will be able to figure out whether anything nefarious is going on, but we can at least try to look into it.David Z –David Z 2018-06-26 01:51:25 +00:00 Commented Jun 26, 2018 at 1:51 see physics.stackexchange.com/q/399647/45664 for the same question applied to the wave equation--and a similar accepted answer.user45664 –user45664 2018-06-26 18:26:51 +00:00 Commented Jun 26, 2018 at 18:26 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 39 Save this answer. Show activity on this post. On the other hand, it requires only two initial conditions x(0) and x˙(0), to obtain the function x(t) by solving Newton's equation For notational simplicity, let x 0=x(0)x 0=x(0) v 0=x˙(0)v 0=x˙(0) and then write your equations as x(t)=x 0+v 0 t+x¨(0)t 2 2!+x...(0)t 3 3!+⋯x(t)=x 0+v 0 t+x¨(0)t 2 2!+x⃛(0)t 3 3!+⋯ m x¨(t)=F(x,x˙,t)m x¨(t)=F(x,x˙,t) Now, see that x¨(0)=F(x 0,v 0,0)m x¨(0)=F(x 0,v 0,0)m x...(0)=F˙(x 0,v 0,0)m x⃛(0)=F˙(x 0,v 0,0)m and so on. Thus x(t)=x 0+v 0 t+F(x 0,v 0,0)m t 2 2!+F˙(x 0,v 0,0)m t 3 3!+⋯x(t)=x 0+v 0 t+F(x 0,v 0,0)m t 2 2!+F˙(x 0,v 0,0)m t 3 3!+⋯ In other words, the initial value of the 2nd and higher order time derivatives of x(t)x(t) are determined by F(x,x˙,t)F(x,x˙,t). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 25, 2018 at 14:42 answered Jun 25, 2018 at 14:37 Alfred CentauriAlfred Centauri 60.7k 3 3 gold badges 76 76 silver badges 211 211 bronze badges 0 Add a comment\| This answer is useful 12 Save this answer. Show activity on this post. FGSUZ has given part of the answer in his comment, but he has not given full details. Consider x¨(t)=F(x,x˙,t)x¨(t)=F(x,x˙,t). In this case you have the second derivative in terms of lower order terms. You can therefore use this to remove the second derivative in favor of lower order items. You can then take the time derivative of this equation. This will give you the third order time derivative of x x in terms of lower order derivatives. And you can use the first equation and its derivative to write everything in terms of at most the first derivative. So, order by order, you can construct the Taylor expansion. Now the general case may require you to deal with derivatives of F(x,x˙,t)F(x,x˙,t). That is because you need the following (if I've recalled my calculus correctly). d 3 x d t 3=F˙(x,x˙,t)=∂∂x F(x,x˙,t)d x d t+∂∂x˙F(x,x˙,t)d x˙d t+∂∂t F(x,x˙,t)d 3 x d t 3=F˙(x,x˙,t)=∂∂x F(x,x˙,t)d x d t+∂∂x˙F(x,x˙,t)d x˙d t+∂∂t F(x,x˙,t) This will often not be explicitly solvable. However, it also can be Taylor expanded in a similar fashion. And, at each order you keep only the corresponding order in the expansion of this equation. So, order by order, you can construct the Taylor series. At each step you can use the equation of motion to remove all but the x x, x˙x˙, and t t dependence. And so you will only need two initial conditions. Tedious, but possible. The nice cases are those few where you can derive a simple formula that gives an easy recursion. So you might, for simple forms of F F, get some simple thing that the (n+1)(n+1) derivative is some simple function of the n n derivative. In such cases, it is potentially useful in numerical solutions, since you can write things in terms of the time step and a nice Taylor expansion. Though, even in such cases, there are often more efficient methods. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jun 25, 2018 at 13:42 user93146 user93146 1 1 The nice cases are those few where you can derive a simple formula that gives an easy recursion. Nice insight!SRS –SRS 2018-06-26 07:10:13 +00:00 Commented Jun 26, 2018 at 7:10 Add a comment\| This answer is useful 9 Save this answer. Show activity on this post. Power series expansion does not hold for all functions f(t)f(t) or for all t∈R t∈R, but only for real analytic functions and for t t in the radius of convergence. In particular, it does not hold at any point e.g. for functions C 2(R,R d)∖C 3(R,R d)C 2(R,R d)∖C 3(R,R d). Therefore it is not possible to define any function by giving countably many real numbers (x(n)(0))n∈N(x(n)(0))n∈N. In particular, Newton's equation may have solutions in C 2(R,R d)∖C 3(R,R d)C 2(R,R d)∖C 3(R,R d), that therefore do not admit a power series expansion, or in general solutions that are not real analytic for all times, and therefore that do not always admit a Taylor expansion. Nonetheless, these functions are uniquely defined by two real numbers (x(0)x(0), x˙(0)x˙(0)) and by being solution of Newton's equation (i.e. they are also determined by m m and the functional form F F of the force). In case that a solution of the Newton equation is real analytic, then the value of the higher order derivatives in zero is determined uniquely by the solution itself, and thus they also depend only on x(0)x(0), x˙(0)x˙(0), m m and F F; no further knowledge is required. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jun 25, 2018 at 13:53 yuggibyuggib 12.4k 1 1 gold badge 26 26 silver badges 49 49 bronze badges 8 2 In pure mathematics, this is true. But this is physics. Any realistic physical system can be considered analytic, since not only is it impossible to make measurements to infinite precision, it is impossible to even define the quantities being measured to such. (Consider for example, the volume of an object - beyond a certain precision, you can't even say where the boundary of the space filled by the object is.)Paul Sinclair –Paul Sinclair 2018-06-25 16:34:33 +00:00 Commented Jun 25, 2018 at 16:34 3 @PaulSinclair I really do not see the point of your comment. The OP is asking about the mathematical properties of a function, and how it can be obtain as either the solution of a differential equation, or as the expansion in power series. If one should argue taking into account the limitations of accuracy in measurements, then one should very well deny the concept of derivative as being physical, since it involves a limit procedure with quantities becoming arbitrary small. So even Newton's equation does not make sense as such, since it involves two derivatives.yuggib –yuggib 2018-06-25 16:43:53 +00:00 Commented Jun 25, 2018 at 16:43 You did indeed completely miss the point. However, as I see that in editing I accidently misphrased the second sentence, this is understandable. Mathematics is used in physics for modelling. Anywhere a non-analytic function is used to model a physical phenomenon, an analytic function could also be used with the same or better accuracy. Thus in physics it is entirely reasonable to assume all functions are analytic and not worry about the fiddly details that mathematicians must spend so much attention on.Paul Sinclair –Paul Sinclair 2018-06-25 23:06:01 +00:00 Commented Jun 25, 2018 at 23:06 3 @PaulSinclair It is not true in general that whenever a non-analytic function is used, it could be replaced by an analytic one from the physical point of view. It is often necessary from the physical perspective to use rough functions. An example that comes into my mind is the brownian motion. You cannot explain as accurately the trajectory of a particle in brownian motion using a smooth path as you would using a continuous nowhere differentiable path. Not to mention the necessity of using distributions (that are usually not even functions) in classical electromagnetism and signal theory.yuggib –yuggib 2018-06-26 07:43:48 +00:00 Commented Jun 26, 2018 at 7:43 1 @SRS The smoothness of the solution depends on the forces involved in the process. An harmonic oscillator has smooth solutions, but other potentials (e.g. the Newton/Coulomb potential) may have singularities, and so the solution may fail to be smooth. Nonetheless, once you have the solution, smooth or otherwise, you can compute its derivatives explicitly. Since the solution depends only on the parameters of the system (mass and force field), and on the initial conditions, then also the higher-order derivatives will only depend on these. At least to me, it seems quite natural/straightforward.yuggib –yuggib 2018-06-26 07:52:36 +00:00 Commented Jun 26, 2018 at 7:52 \|Show 3 more comments This answer is useful 4 Save this answer. Show activity on this post. Long story short, to get to the core of your question, I hope First, some functions don't correspond to their Taylor series at 0 0. But let's ignore that for this answer. But, more importantly: The Taylor series representation has more degrees of freedom simply because not all functions are solutions to the equation (2)! This should be rather obvious if you think of it: If I throw a ball, then if you didn't know any physics or not have any experience on real world, its path could be anything, it could fly to Mars and return to me, it could vibrate between two points, it could draw your name on air. If you only use (1), you can't discard these possibilities. But once you realize it's following Newton's equations, the possible paths are very limited. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jun 26, 2018 at 12:04 JiKJiK 906 7 7 silver badges 14 14 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. As an example, suppose we have Hooke's Law, F = -kx. Writing the Taylor (technically, Maclaurin, since it's centered at zero) series as x(t)=∑∞n=0 x(n)(0)t n n!x(t)=∑n=0∞x(n)(0)t n n! Where x(n)x(n) is the nth derivative of x x, then x(2)(t)=∑∞n=2 x(n)(0)t n−2(n−2)!x(2)(t)=∑n=2∞x(n)(0)t n−2(n−2)! Shifting the index, this can be written as x(2)(t)=∑∞n=0 x(n+2)(0)t n n!x(2)(t)=∑n=0∞x(n+2)(0)t n n! We can then write Hooke's Law as m∑∞n=0 x(n+2)(0)t n n!=−k∑∞n=0 x(n)(0)t n n!m∑n=0∞x(n+2)(0)t n n!=−k∑n=0∞x(n)(0)t n n! Setting like terms equal, we have m x(n+2)(0)n!=−k x(n)(0)n!m x(n+2)(0)n!=−k x(n)(0)n! or x(n+2)(0)=−k m x(n)(0)x(n+2)(0)=−k m x(n)(0) So given any n, we can find the (n+2)th coefficient in terms of the nth coefficient. This means that the even coefficients are determined by the 0th coefficient, and the odd coefficients are determined by the 1st coefficient. (The even powers correspond to a solution in terms of cosine, and the odd powers correspond to a solution in terms of sine, and the general solution is a linear combination of the two.) This is known as an analytic solution of the ODE. In general, it won't be as simple as this. However, since the LHS of Newton's Equation has only a second order term, and the RHS is first order, the (n+2)th coefficient will be able to be expressed in terms of the nth and (n+1)th coefficients, giving the 0th and 1st coefficients as initial conditions. So, the key is that for polynomials to be equal to each other, the coefficients of corresponding powers must be equal, and this can be extended to Taylor series. This gives a recurrence relation giving coefficients in terms of lower order coefficients, and the infinite Taylor series collapses down to being determined by two coefficients. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jun 26, 2018 at 15:32 AcccumulationAcccumulation 9,707 19 19 silver badges 33 33 bronze badges Add a comment\| Protected question. To answer this question, you need to have at least 10 reputation on this site (not counting the association bonus). The reputation requirement helps protect this question from spam and non-answer activity. Start asking to get answers Find the answer to your question by asking. 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10056	http://lampx.tugraz.at/~hadley/ss2/problems/hextb/s.pdf	Question: The unit vectors of a two-dimensional hexagonal lattice are: ⃗ a1 = aˆ x ⃗ a2 = a 2 ˆ x + √ 3 2 aˆ y Calculate the dispersion relation for this two-dimensional crystal using tight-binding model. The equation for energy in the tight binding model is: E = ϵ −t P m ei⃗ k⃗ ρm Answer: In the equation for energy ϵ and t are just constance. So all you need to calculate is the sum: P m ei⃗ k⃗ ρm Where m means the sum over the next nearest neighbours. ⃗ ρm are the vectors pointing to the next nearest neighbours. And ⃗ k is: ⃗ k = kx ky ! First you have to ﬁnd the ⃗ ρm: 1 Nearest neighbours and ⃗ ρm As you can see in the ﬁgure the ⃗ ρm are: ⃗ ρ1 = ⃗ a2 = a 2 ∗ 1 √ 3 ! ⃗ ρ2 = ⃗ a1 = a ∗ 1 0 ! ⃗ ρ3 = a 2 ∗ 1 − √ 3 ! ⃗ ρ4 = a 2 ∗ −1 − √ 3 ! = −a 2 ∗ 1 √ 3 ! ⃗ ρ5 = a ∗ −1 0 ! = −a ∗ 1 0 ! ⃗ ρ6 = a 2 ∗ −1 √ 3 ! = −a 2 ∗ 1 − √ 3 ! Now the sum over the next nearest neighbours can simply be calculated: P m ei⃗ k⃗ ρm = ei a 2 (kx+ √ 3ky) + e−i a 2 (kx+ √ 3ky) + ei a 2 (kx− √ 3ky) + e−i a 2 (kx− √ 3ky) + eiakx + e−iakx Since 2 ∗cos(x) = eix + e−ix, the sum can also be written as: 2 P m ei⃗ k⃗ ρm = 2 ∗[cos( a 2(kx + √ 3ky)) + cos( a 2(kx − √ 3ky)) + cos(akx)] If you put that in the equation for the Energy you get the dispersion relation: E(⃗ k) = ϵ −2t ∗[cos( a 2(kx + √ 3ky)) + cos( a 2(kx − √ 3ky)) + cos(akx)] 3
10057	https://pmc.ncbi.nlm.nih.gov/articles/PMC4040134/	Acyl-coenzyme A synthetases in metabolic control - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Advanced Search Journal List User Guide NewTry this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Curr Opin Lipidol . Author manuscript; available in PMC: 2014 May 31. Published in final edited form as: Curr Opin Lipidol. 2010 Jun;21(3):212–217. doi: 10.1097/mol.0b013e32833884bb Search in PMC Search in PubMed View in NLM Catalog Add to search Acyl-coenzyme A synthetases in metabolic control Jessica M Ellis Jessica M Ellis 1Department of Nutrition, University of North Carolina, Chapel Hill, North Carolina, USA Find articles by Jessica M Ellis 1, Jennifer L Frahm Jennifer L Frahm 1Department of Nutrition, University of North Carolina, Chapel Hill, North Carolina, USA Find articles by Jennifer L Frahm 1, Lei O Li Lei O Li 1Department of Nutrition, University of North Carolina, Chapel Hill, North Carolina, USA Find articles by Lei O Li 1, Rosalind A Coleman Rosalind A Coleman 1Department of Nutrition, University of North Carolina, Chapel Hill, North Carolina, USA Find articles by Rosalind A Coleman 1 Author information Copyright and License information 1Department of Nutrition, University of North Carolina, Chapel Hill, North Carolina, USA ✉Correspondence to Rosalind A. Coleman, MD, Department of Nutrition, University of North Carolina, CB#7461 Chapel Hill, NC 27599, USA Tel: +1 919 966 7213; fax: +1 919 843 8555; rcoleman@unc.edu © 2010 Wolters Kluwer Health \| Lippincott Williams & Wilkins PMC Copyright notice PMCID: PMC4040134 NIHMSID: NIHMS586007 PMID: 20480548 The publisher's version of this article is available at Curr Opin Lipidol Abstract Purpose of review The 11 long-chain (ACSL) and very long chain acyl-coenzyme A (acyl-CoA) synthetases [(ACSVL)/fatty acid transport protein] are receiving considerable attention because it has become apparent that their individual functions are not redundant. Recent findings Recent studies have focused on the structure of the acyl-CoA synthetases, their post-translational modification, their ability to activate fatty acids of varying chain lengths, and their role in directing fatty acids into different metabolic pathways. An unsettled controversy focuses on the ACSVL isoforms and whether these have both enzymatic and transport functions. Another issue is whether conversion of a fatty acid to an acyl-CoA produces an increase in the AMP/ATP ratio that is sufficient to activate AMP-activated kinase. Summary FuturestudiesarerequiredtodeterminethesubcellularlocationofeachACSLandACSVL isoform and the functional importance of phosphorylation and acetylation. Purification and crystallization of mammalian ACSL and ACSVL isoforms is needed to confirm the mechanism of action and discover how these enzymesdiffer in their affinity for fatty acids of differentchainlengths.Functionally,itwillbeimportanttolearnhowtheACSLisoformscan direct their acyl-CoA products toward independent downstream pathways. Keywords: β-oxidation, acyl-CoA synthetase, AMP-activated kinase, fatty acid, fatty acid transport protein, glycerolipid synthesis Introduction Of the 26 mammalian acyl-coenzyme A (acyl-CoA) synthetases (ACS), at least 11 can activate the major long-chain fatty acids (FAs) of 16–22 carbons [1,2•]. These include members of the long-chain [long-chain acyl-CoA synthetases (ACSL)] and very long chain [very long chain acyl-CoA synthetases (ACSVL)/fatty acid transport protein (FATP)] subfamilies that catalyze a two-step reaction: This review will focus primarily on very recent advances in the study of these long-chain ACSs in metabolic regulation, highlighting new studies on structure, function, and regulation. Structure and sequence of acyl-coenzyme A synthetase Acyl-CoA synthetases are homologous in their sequences. The ACS catalytic domain shares homology with the ANL (Acyl-CoA synthetases, Nonribosomal peptide synthetase adenylation domains, and Luciferase enzymes) super-family, a family consisting of ACS, nonribosomal peptide synthetase, and luciferase [3••]. The homology between ACS and luciferase is so similar that mutation of a single residue (L345S) allows the Drosophilia melangaster ACS to gain luciferase activity [4••]. For all ANL superfamily enzymes, the C-terminal domain is rotated ~140° after adenylation-formation; in ACS this rotation results in the conformation that forms thioesters. Analysis of the only reported mammalian ACS crystal structure revealed the following: (1) in the adenylate-forming conformation a conserved lysine, Mg2+, and the P-loop holds the pyro-phosphate in the active site; (2) in the thioester conformation the Arg472-Glu365 salt bridge sterically blocks ATP binding; and (3) the switch of the adenylate-forming conformation to the thioesterase conformation requires the presence of pyrophosphate [5•]. Acyl-coenzyme A synthetase and fatty acid uptake FA may enter cells via both concentration-based flipping and protein-mediated transport . Cellular retention may depend on ‘vectorial acylation’, the trapping of exogenous FA as an acyl-CoA within the cell and its subsequent use in downstream metabolism . Although FATP1 was discovered in a functional FA uptake screen , several FATP family members are not present on the plasma membrane , and so they are also called ACSVL to distinguish their ability to activate FA with chain lengths as long as 26 carbons . Although FA uptake increases in NIH fibroblasts and COS (CV-1 (simian) in Origin, and carrying the SV40 genetic material) cells overexpressing ACSL1 , it is unaffected by over-expressing or knocking out ACSL1 in primary hepatocytes [12••] or 3T3-L1 adipocytes [13••]. The conflicting data concerning results of FA uptake by the same ACSL isoforms likely reflect differences in techniques for measuring uptake , the artificial effect of overexpression, tissue-specific subcellular location, and tissue-specific interacting proteins. Fatty acid channeling Although each ACSL and ACSVL isoform activates a broad range of long-chain FA, their respective physiological roles remain unknown. We hypothesized that different ACS isoforms channel FA into specific downstream pathways that could vary in a tissue-specific manner . For example, despite a 50% decrease in total ACSL activity, primary hepatocytes from ACSL1 liver-specific knockout mice had small reductions in FA incorporation into both triacylglycerol (TAG) and β-oxidation products [12••]. The ACSL1 knockout livers had reduced long-chain acyl-carnitine content and reduced rates of FA oxidation. The minimal phenotype suggests that alternative pathways for FA oxidation and TAG synthesis are active in the AcslL−/− liver. In contrast, primary epididymal adipocytes from ACSL1 adipose-specific knockout mice have unchanged FA incorporation into TAG, diacylglycerol (DAG), or phospholipid (PL), but reduced β-oxidation products (J.M. Ellis, R.A. Coleman, unpublished data). However, siRNA knockdown of ACSL1 in 3T3-L1 adipocytes does not affect de-novo lipogenesis or FA oxidation, but increases FA efflux [13••], leading the authors to conclude that ACSL1 activates FA that is re-esterified after lipolysis. Pulse-chase labeling experiments could verify reduced rates of re-esterification and confirm this interpretation. Knocking down ACSL3 in rat primary hepatocytes decreases labeled acetate incorporation into TAG and PL [14••]. In human hepatoma Huh7 cells, siRNA knockdown of ACSL3 decreases oleate incorporation into phosphatidylcholine (PC) and secreted VLDL and increases cellular-free oleate, whereas knockdown of ACSL1 and ACSL4 does not alter oleate incorporation into cholesterol ester (CE), TAG, phospholipids, or free FA [15••]. Differential effects of individual ACSL iso-forms support the notion that ACSL isoforms channel FA to distinct metabolic fates. Several overexpression studies are inconsistent with the knockdown studies and provide strikingly contradictory interpretations of the role of ACSL1. For example, when ACSL1 is overexpressed in rat primary hepatocytes, FAs are incorporated into DAG and phospholipids but not into CEorsecretedTAG.OverexpressedACSL5increases FA incorporation into DAG and TAG but does not affect FA used for β-oxidation . In contrast, however, over-expressing either human ACSL3 or rat ACSL5 in HepG2 cells increases palmitate oxidation . Further, TAG content increases in NIH-3T3 fibroblasts when ACSL1/ FATP1 is overexpressed and in insulin-stimulated 3T3-L1 adipocytes when ACSL1 is knocked down [13••]. We see two potential problems. The first is that over-expression of an enzyme may overwhelm downstream pathways that cannot readily accommodate excess substrate; the second problem is that studies in cell lines are not always translatable to primary cells. Thus, proteomic analysis of mitochondria isolated from white adipose tissue, brown adipose tissue, and 3T3-L1 adipocytes reveals substantial differences [19••]. Compared with mitochondria from adipose tissue, many proteins are downregulated in 3T3-L1 mitochondria. Further, medium-chain ACS (ACSM) isoforms are more abundant than ACSL isoforms in white adipose tissue mitochondria, suggesting greater medium-chain FA oxidative metabolism. ACSL1 and ACSL6 are particularly enriched in white adipose mitochondria, whereas ACSL5 is more abundant in mitochondria from brown adipose. However, despite the high abundance of ACSL5 in brown adipose mitochondria, ACSL1 is required for adaptive thermo-genesis and normal FA oxidation in brown adipose (J.M. Ellis, R.A. Coleman, unpublished data). Thus, the function of ACSL5 in BAT remains unknown. FA channeling could occur via protein–protein interactions. Immunoprecipitation of endogenous FATP1 from 3T3-L1 adipocytes followed by mass spectrometry identified mitochondrial 2-oxoglutarate dehydrogenase (OGDH), a key enzyme in the tricarboxylic acid cycle [20••]. FATP1 enhances OGDH activity in proteoliposomes, whereas FATP1 knockdown in 3T3-L1 adipocytes showed decreased OGDH and TCA cycle activity. Over-expressed FATP1 and carnitine palmitoyltransferase-1 (CPT1) also co-immunoprecipitate from L6E9 myotubes and rat skeletal muscle in vivo [21•]. In L6E9 myotubes, FATP1 overexpression enhances FA oxidation, suggesting that protein–protein interactions of FATP1 facilitate activities of FA oxidation proteins. Overexpressed ACSL1 and FATP1 co-immunoprecipitate in 3T3-L1 adipocytes , but it is not known whether endogenous ACSL1 and FATP1 proteins interact. The subcellular location of ACS enzymes, which was recently reviewed [23•], may influence the downstream metabolism of the acyl-CoA product. However, sub-cellular location studies have yielded conflicting interpretations. Most have examined subcellular fractions that are limited in purity. Data from overexpression studies are also problematic because the location of over-expressed proteins may not mimic the endogenous condition. In one such example, the subcellular locations of overexpressed and endogenous FATP4/ACSVL5 differed . Confocal microscopy with tested isoformspecific antibodies will be required to determine precise locations of the acyl-CoA synthetases. Acyl-coenzyme A synthetase enzyme regulation Broad mass spectrometry surveys reveal a number of post-translational modifications to ACS enzymes, although their functional significance is generally unknown. An exception is the regulation on Lys-609 located on the C-terminus of the Salmonella enterica, ACS, for which lysine acetylation inhibits activity by preventing formation of the thioester . Similar regulation occurs by the sirtuin (SIRT) acetylases on Lys-661 and Lys-635 of acetyl-CoA synthetase-1 and acetyl-CoA synthetase-2, respectively, and human acetyl-CoA synthetase 2 is regulated on Lys-642 . In response to a calorie-restricted diet, acetylation of Lys-534 of ACSM1 and Lys-510 of acyl-CoA synthetase family member-2 (Q8VCW8) increased 2.75-fold and 2.3-fold, respectively [27••]. Role of acyl-coenzyme A synthetase in transcriptional regulation ACSL can theoretically regulate transcription by modulating the intracellular content of FA and acyl-CoA, which are ligands for transcription factors. FA, for example, is a ligand for peroxisome proliferator-activated receptor (PPAR) α, β, or δ and may suppress SREBP1c target genes, and acyl-CoA is a possible ligand for HNF4α . The role of ACS in regulating transcription was addressed by a study that used siRNA to knock down the long-chain ACS isoforms that are expressed in liver, ACSL1, 3, 4, and 5, and FATP2, 4, and 5 [14••]. In primary rat hepatocytes, only an siRNA that targeted ACSL3 had effects on transcription. This siRNA elevated intracellular FA 82% and diminished the expression of PPARγ, as well as reporter gene activity for PPARγ, ChREBP, SREBP1c, and liver X receptor α (LXRα), together with the expression of target genes for these transcription factors. Decreased use of [1-14C]acetate for de-novo lipogenesis and complex lipid synthesis confirmed the effect of diminishing the expression of genes involved in lipid synthesis. Although one might conclude that the knockdown of ACSL3 decreased these transcriptional activities via an increase in a specific pool of FA ligand, Bu et al. argue that ACSL3 must control transcription in a ligand-independent manner because synthetic ligands did not normalize transcriptional activities of LXR and PPARγ. Long-chain acyl-coenzyme A synthetase's role in disease ACSL is required for complex lipid synthesis in rapidly growing cells and the fungal metabolite triacsin C was initially reported to inhibit ACSL, diminish the synthesis of phospholipids, and decrease cell proliferation . Triacsin C caused apoptosis in glioma cells, and, although it inhibits ACSL1, 3, and 4, but not ACSL5, overexpression of ACSL5 rescued the glioma cells [30• ,31•]. In this case, ACSL5 appeared to compensate for the inhibited isoforms. In a related publication, ACSVL3 (FATP3) was elevated in human gliomas [32•]. ACSVL3 siRNA inhibited glioma cell growth and the cells became less tumorigenic. Because these features were reversed when active myr-Akt was expressed, ACSVL3 was thought to maintain oncogenesis by regulating Akt function. Mutations in ACS genes have been identified in large screening studies. An Acsl5 single nucleotide polymorphism (SNP) (rs2419621; C>T) increased Acsl5 mRNA abundance in skeletal muscle and was associated with improved weight loss. This SNP increases the transcriptional rate of Acsl5 by enhancing gene activation by MyoD [33•]. Because this ACSL5 SNP increases the differentiation of skeletal myoblast cells, persons with this mutation may have an advantageous ability to increase muscle mass and caloric expenditure with diet and exercise. High levels of Acsl3 gene methylation in umbilical cord white blood cells were strongly correlated with levels of maternal polycyclic aromatic hydrocarbons and with the mRNA abundance of Acsl3 [34•]. In a separate cohort, Acl3 methylation was associated with polycyclic aromatic hydrocarbon exposure and childhood asthma [34•]. These data suggest that maternal environmental exposures may affect childhood asthma through the methylation of the Acl3 gene. The knockdown of Acsl4 in neurites reduces dendritic spine formation, suggesting a role for ACSL4 in the maturation and remodeling of neurons and providing a link to the X-linked mental retardation associated with human mutations in ACSL4 [35••]. In another study, a SNP (rs1324805) in the Acsl4 gene was more prevalent in patients with the metabolic syndrome [36•]. A third study found an Acsl4 SNP (rs7887981) associated with increased liver fat or with increased circulating insulin and TAG in overweight participants from two different cohorts [37•]. It is unknown how these SNPs affect the function of ACSL4, nor is it clear how the preference of ACSL4 for long-chain polyunsaturated FA may be involved. A mutation within the Fatp4 gene is associated with the ichthyosis prematurity syndrome (IPS) [38•]. Fibroblasts from an IPS patient had reduced ACSVL activity and reduced incorporation rates of very long chain fatty acids into complex lipids. The skin findings are like those of Fatp4 null mice, which die in early life with hyper-proliferative hyperkeratosis , and suggest that FATP4 plays a role in early epidermal development and that other ACS enzymes cannot compensate for a deficit of this enzyme. Effect of fatty acid activation on AMP-activated kinase Because the synthesis of each acyl-CoA uses the energy equivalent of two ATP molecules and produces one AMP, it has been suggested that FA activation might increase cellular AMP/ATP ratios and activate AMP-activated kinase (AMPK) (Fig. 1). AMPK would then increase FA oxidation while blocking TAG synthesis. It would be surprising, however, if the conversion of FA to acyl-CoA triggered the activity of an energy-deficit indicator in settings of energy surplus when TAG synthesis is prominent. Figure 1. Activation of fatty acids by long-chain acyl-coenzyme A synthetase may activate AMP-activated kinase. Open in a new tab When exogenous fatty acids enter the cell, are synthesized de novo, or are hydrolyzed from intracellular triacylglycerol stores, they must first be activated by a long chain acyl-CoA synthetase (ACSL) to form fatty acyl-CoAs. They can then enter downstream metabolic pathways. Because ATP is a substrate for ACSL and AMP is a product, the increased AMP/ATP ratio, a signal of low cellular energy stores, could theoretically activate AMP-activated kinase (AMPK). Activated AMPK inhibits triacylglycerol synthesis and acetyl-CoA carboxylase (ACC). The resulting decrease in malonyl-CoA levels diminishes fatty acid synthesis and also relieves the inhibition of carnitine palmitoyltransferase-1 (CPT1) by malonyl-CoA, allowing fatty acyl-CoA to be converted to acyl-carnitine and enter the mitochondria for β-oxidation. Thus, activated AMPK would generate an increase in ATP production. A prolonged lipogenic stimulus [which activates hormone-sensitive lipase (HSL)], may also inactivate AMPK to modulate these effects. It seems unlikely, however, that ACSL would activate AMPK under physiological conditions that stimulate triacylglycerol synthesis. Studies in animals have not been performed to compare ACSL-mediated activation of AMPK under different physiological conditions, but metabolic adaptations to fasting and chronic caloric restriction in heart, muscle, and liver do not include changes in AMPK activity , and when mice are placed in a cold environment, AMPK is not activated for several days despite increased ACSL-mediated FA activation . Some experiments, however, do support ACSL-mediated activation of AMPK. In 3T3-L1 adipocytes, stimulation of lipolysis resulted in activated AMPK [42••]. The authors suggested that activation of AMPK during lipolysis would ‘restrain the energy depletion and oxidative stress’. It might also permit use of released FA for β-oxidation within the adipocyte and conserve glucose during a ‘fast’. In a situation more akin to ‘feeding’, AMPK activation increases in 3T3-L1 adipocytes during insulin-stimulated palmitate uptake, although the AMP/ ATP ratio changes minimally [43••]. Studies in heart and skeletal muscle [45• ,46,47] suggest that AMPK may be sensitive to the ‘lipid status’ of the cell and that activation may be influenced by intracellular FA availability, independent of cellular AMP levels, with FAs, themselves, activating AMPK [48•]. FA added to L6 myotubes increases AMPK activity by 40–50%, together with phosphorylation of AMPK Thr172 and acetyl-CoA carboxylase-β Ser218, and the rate of palmitate oxidation is almost doubled, without an increase in AMP content or LKB1/AMPKK activation . Dominant-negative AMPK adenovirus reduced basal FA oxidation and inhibited the stimulatory effects of FA pretreatment on FA oxidation, suggesting that FA might improve AMPK as a substrate for LKB1. However, in assays with purified components, 10–50 μmol/l long-chain acyl-CoAs directly inhibited LKB1, suggesting that the rise in long-chain acyl-CoA content that occurs when FA synthesis is enhanced would negate the effect of increased AMP to prevent simultaneous synthesis and FA oxidation . In addition, strong evidence supports the idea that protein kinase A phosphorylates and inactivates AMPK , thereby allowing hormone-sensitive lipase activation to diminish and slowing the rate of lipolysis. Although the cellular concentration of long-chain acyl-CoAs has been estimated at between 5 and 160 μmol/l , acyl-CoAs are amphipathic and are probably bound either to cell membranes or acyl-CoA-binding protein; neither their free concentration nor their availability as substrates or inhibitors can be known with certainty. Conclusion Although much has been accomplished, many important questions remain. Future studies are required to determine the subcellular location of each ACSL and ACSVL isoform. Purification and crystallization of mammalian ACSL and ACSVL isoforms is needed to confirm the mechanism of action and discover how these enzymes differ in their affinity for FAs of different chain lengths. Functionally, it will be important to learn the conditions under which ACS activity activates AMPK and how the ACSL isoforms can direct their acyl-CoA products toward independent downstream pathways. Acknowledgements This work was supported by NIH grants DK056598, R01-DK59935 and DK59935-S1 (ARRA) (R.A.C.), DK40936, U24-DK59635, a postdoctoral fellowship DK082099 (J.L.F.), and the UNC Clinical Nutrition Research Unit (P30 DK56350). Postdoctoral (L.O.L.) and Predoctoral (J.M.E.) fellowships were from the American Heart Association-Mid-Atlantic Region. References and recommended reading Papers of particular interest, published within the annual period of review, have been highlighted as: • of special interest •• of outstanding interest Additional references related to this topic can also be found in the Current World Literature section in this issue (p. 260). 1.Watkins PA, Maiguel D, Jia Z, Pevsner J. Evidence for 26 distinct acylcoenzyme A synthetase genes in the human genome. J Lipid Res. 2007;48:2736–2750. doi: 10.1194/jlr.M700378-JLR200. [DOI] [PubMed] [Google Scholar] 2•.Watkins PA. Very-long-chain acyl-CoA synthetases. 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[DOI] [PMC free article] [PubMed] [Google Scholar] ACTIONS View on publisher site PDF (402.0 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Structure and sequence of acyl-coenzyme A synthetase Acyl-coenzyme A synthetase and fatty acid uptake Effect of fatty acid activation on AMP-activated kinase Conclusion Acknowledgements References and recommended reading Cite Copy Download .nbib .nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter) NCBI on Facebook NCBI on LinkedIn NCBI on GitHub NCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter) NLM on Facebook NLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
10058	https://www.statisticshowto.com/stirling-numbers-second-kind/	Skip to content Stirling Numbers of the Second Kind Combinatorics> In combinatorics, the Stirling numbers of the second kind tell us how many ways there are of dividing up a set of n objects (all different, or at least all labeled) into k nonempty subsets. The k subsets aren’t labeled. We write Stirling numbers of the second kind as S(n,k) or: . Examples of Stirling Numbers of the Second Kind What is S(3,3)? This question asks “How many ways can we partition a set of three elements into three subsets?” A set of three elements — say, the set {0,1,2} — can be partitioned into three subsets exactly one way: {1,2,3}, so S(3,3) is just 1. In fact, since a set of k elements can always be partitioned into a set of k nonempty subsets exactly one way, S(n,n) is always 1. S(3,2) will be the number of ways we can partition our set of three elements into two subsets. There are three possible ways to do this; each splits the set into two pieces made up of one element and two elements. Using our example set of {0,1,2}, we can come up with: {{0,1}{2}} {{0}{1,2}} {{1}{0,2}}. So S(3,2)= 3. S(3,1) is the number of ways we can partition our elements into just one subset; since the only possibility of that is the set including every element, {0,1,2}, S(3,1) is also 1. This can also be generalized; S(n,1) = 1 for every n. The following table (from Quaintance, 2015) shows the first few possibilities for Stirling numbers of the second kind: Calculating Stirling Numbers of the Second Kind There are two ways of calculating Stirling numbers of the second kind. First,they can be calculated recursively; i.e, with reference to lower order Stirling numbers of the second kind. S(m,n) = S(m – 1,n – 1) + nS(m – 1,n). Where: m is the number of elements in the original set, n is the number of subsets. This simple formula makes it easy to make tables of Stirling numbers of the second kind, or to find a number when we know the figures for smaller sets. We saw above that S(3,2) = 3 and S(3,1) = 1. We can use these two facts to enable us to find S(4,2), since with m = 4 and 2 = n, the formula above becomes: S(4,2) = S(3,1) + 2 S(3,2) = 1 + 2 3 = 7. There’s also a formula that can be used to calculate S(n,k) for any n or k. This is given by Here ! means a factorial, and the large Σ means that what comes after it is summed up from i = 0 to n. References Quaintance, J. (2015). Combinatorial Identities for Stirling Numbers: The Unpublished Notes of H W Gould. World Scientific, Oct 27. Comments? Need to post a correction? Please Contact Us.
10059	https://www.media4math.com/library/math-example-place-value-comparing-and-ordering-whole-numbers-using-place-value-example-20	Math Example--Place Value--Comparing and Ordering Whole Numbers Using Place Value--Example 20 \| Media4Math Skip to main content Enable accessibility for low vision Open the accessibility menu Skip to content Home Search Subscribe ContentNewest ResourcesLesson Plan LibraryContent ShowcaseAnimated Math Clip ArtSAT Test PrepMath Fluency CentersMath Visual GlossaryStudent Tutorials Standards AlignmentsStandards Alignments (K-12)Standards Reports Getting StartedEducatorsTutorsParents Partners Shop@Media4Math Persistent Menu Library Classroom Log in Register Register to SaveSubscribe to DownloadPreviewAdd to Slideshow (Subscribers Only) If you are a subscriber, please log in. Arithmetic>>Numbers and Patterns>>Place Value Resource Related Resources Display Title Math Example--Place Value--Comparing and Ordering Whole Numbers Using Place Value--Example 20 Math Example--Place Value--Comparing and Ordering Whole Numbers Using Place Value--Example 20 Topic Comparing and Ordering Whole Numbers Description Two numbers (14,891 and 12,607) are compared, emphasizing thousands place differences in a chart. Example 20. Compare 14,891 and 12,607. Ten thousands digits are equal. Thousands digits differ: 4 > 2. Therefore, 14,891 > 12,607. Understanding place value helps students learn how to compare and order whole numbers accurately. By examining the digits from left to right, students grasp which number is larger based on their place values. These examples offer visual aids and step-by-step comparisons that make these abstract concepts tangible. Seeing multiple worked-out examples is essential for students to grasp comparing and ordering numbers. Each example reinforces the steps and nuances involved in comparing place values, highlighting how slight variations in digits impact overall value. Teacher’s Script:"Look closely at each place value to compare the numbers. For instance, here we compare 12,607 and 14,891. This method helps us determine which number is larger or smaller by examining each digit’s position." For a complete collection of math examples related to Comparing and Ordering Whole Numbers click on this link: Math Examples: Comparing and Ordering Whole Numbers Collection. Loading… An error occurred while loading related resources. No related resources found. \| Common Core Standards \| CCSS.MATH.CONTENT.2.NBT.A.4, CCSS.MATH.CONTENT.4.NBT.A.2 \| \| Grade Range \| 2 - 4 \| \| Curriculum Nodes \| Arithmetic •Numbers and Patterns •Place Value \| \| Copyright Year \| 2021 \| \| Keywords \| place value, comparing and ordering whole numbers \| The Math Examples Library Media4Math has a huge collection of instructional math examples covering a wide range of math topics. This collection of resources is made up images that you can easily incorporate into your lesson plans. To see the complete collection of Math Examples, click on this link. In answer to the question, "Why so many examples?": Math concepts are best learned through multiple examples of increasing complexity. For example, in learning about the slope formula, students need to see examples covering these scenarios: Two points in the first quadrant, positive slope Two points in the first quadrant, negative slope Two points in the first quadrant, zero slope Two points in the first quadrant, no slope But this only covers points in the first quadrant. Students also need to see examples in the other quadrants, to see the impact of negative coordinates on the formula. Plus, students need to see points on two separate coordinates. As you can see, just with this one topic, there are dozens of possible examples to cover in order for students to get a better understanding of the nuances of working with the slope formula. Textbooks and other curriculum solutions provide a handful of examples, but hardly a comprehensive. This is where Media4Math comes in. We provide significantly more examples, each providing a detailed solution. In the case of the slope formula, click on this link to see the full set of math examples on the slope formula. Below is a listing of the categories of math examples covered. To see the complete library of Math Examples, click on this link. 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For example the Practice Guide entitledTeaching Strategies for Improving Algebra Knowledge in Middle and High School Studentshas one recommendation involving showing students solved problems in order for them to better understand abstract algebraic concepts. Watch the video below to learn more about this approach. Having students study solved problems allows them to analyze and internalize the mathematical reasoning involved. This is what the Math Examples Library attempts to do: Solve enough problems for students to gain a firm understanding of the underlying concepts. The other two recommendations in the practice guide involve having students look at the mathematical structure from the standpoint of multiple representations of mathematical quantities, as well as choosing from different strategies for solving problems. The Math Examples Library also has a wealth of examples that show multiple representations and different approaches to solving math problems. 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10060	https://emergencycarebc.ca/clinical_resource/clinical-summary/methanol-ingestion/	Go back INDEX Methanol Ingestion Context Signs and Symptoms General HEENT (Head, Eyes, Ears, Nose, Throat) CVS Respiratory Neurologic Gastrointestinal Genitourinary Metabolic Musculoskeletal Other Late Sequelae Toxic Dose Diagnostic Process Calculate Anion and Osmolar Gaps Clinical Pitfalls Recommended Treatment Criteria for Hospital Admission Criteria for Transfer to Another Facility Consult Criteria For Safe Discharge Home Quality Of Evidence? Related Information References: Relevant Resources Methanol Ingestion Toxicology Last Reviewed on Dec 17, 2018 Read Disclaimer By Roy Purssell Context Ingestion of 30-60 mL is potentially lethal to an adult. The onset of anion gap metabolic acidosis and visual impairment are delayed and death can occur within hours to days, usually due to respiratory failure or cardiovascular collapse. Survivors of severe poisoning may suffer permanent blindness and/or neurologic impairment. Signs and Symptoms General Initial symptoms (onset about 2 hours) may resemble ethanol intoxication. Latent period of 4-12 hours prior to onset of metabolic acidosis and associated symptoms. Massive exposure results in early onset of acidosis; Co-ingestion of ethanol will delay onset of toxicity. Contact the BC Drug and Poison Information Centre 604-682-5050, 1-800-567-8911 early. HEENT (Head, Eyes, Ears, Nose, Throat) Ocular effects begin 8-36 hours after exposure. Early symptoms include blurred vision, decreased visual acuity, sensation of “being in a snowstorm”. Findings range from normal to decreased pupillary light reactions, hyperemia or pallor of optic disc, retinal edema. Bilateral blindness may develop. Retinal and optic nerve damage can be permanent; partially or fully reversible in some cases. CVS Cardiovascular effects may be seen in severely poisoned, moribund patients. Sinus bradycardia indicates a poor prognosis. Respiratory Tachypnea or Kussmaul respirations secondary to metabolic acidosis are common. Respiratory arrest may occur. Neurologic Early symptoms may resemble ethanol intoxication; may be of short duration. Progressively worsening headache, vertigo, weakness, apprehension, confusion develop within hours, and can progress to, seizures and coma. Late symptoms (> 12-24 hours) include cerebral edema, cerebral hemorrhage and/or necrosis; basal ganglia infarcts. Necrosis of the putamen is characteristic of severe methanol poisoning. Gastrointestinal Nausea, vomiting, abdominal pain (may be severe). Genitourinary Renal failure is uncommon; may be secondary to rhabdomyolysis. Metabolic Progressive metabolic acidosis with elevated anion gap is characteristic. Onset usually within 12 hours; may be delayed with ethanol co-ingestion. May be rapid following massive exposure. Musculoskeletal Rhabdomyolysis may occur. Other Elevation of serum amylase with or without pancreatitis has been reported. Late Sequelae Partial or complete visual impairment, Parkinsonian-like symptoms, pseudobulbar palsy, cognitive defects Toxic Dose 30-60 mL is potentially lethal to an adult; 2.5-5 mL could produce serious toxicity in a young child. Diagnostic Process Laboratory testing must be used to make the diagnosis because clinical symptoms are delayed and it is better to provide treatment before symptoms develop. However, only three hospitals, the Royal Columbian Hospital, the Royal Jubilee Hospital and Vancouver General Hospital can perform definitive testing using gas chromatography and the approval of the clinical chemist on call must be obtained. Therefore, surrogate testing must be used. The osmolar gap, the anion gap and the serum bicarbonate from an arterial blood gas are commonly used surrogate tests. In hospitals where surrogate testing is not available, the patient will need to be transferred to complete this testing. Contact the Poison Control Centre. Calculate Anion and Osmolar Gaps Anion Gap (AG): Elevated AG indicates accumulation of toxic acid metabolite. AG may remain normal for several hours after exposure, or with ethanol co-ingestion. AG = Na+ – (Cl- + HCO3-) Reference range ~6-12 Osmole Gap (OG): OG gives an imprecise estimate of serum methanol concentration, has poor sensitivity for detecting low methanol levels and may be within normal limits in late presentation cases. Serum osmolality (Om), sodium, urea, glucose and ethanol must be measured simultaneously. OG = Om–([2 x Na+]+ urea+ glucose+ ethanol); all values in mmol/L Reference range -10 to +10 Clinical Pitfalls Onset of anion gap metabolic acidosis and visual impairment are delayed. Qualitative (colorimetric) methanol assay is prone to false positive results. Recommended Treatment Recommended treatment includes: supportive care, aggressive treatment of acidosis, the use of the antidotes fomepizole and folate and hemodialysis. a) Gastrointestinal decontamination Charcoal, laxatives and enemas are not effective and have no role in pure methanol overdoses. If co-ingestants are possible, consider if a large quantity of toxin consumed and < 1 hour post ingestion. b) Treat metabolic acidosis aggressively. Prompt correction of acidosis reduces tissue penetration of formate and improves outcome. Monitor serum sodium and potassium closely. Maintain fluid and electrolyte balance. Administer IV sodium bicarbonate for arterial pH < 7.25-7. c) Antidote therapy: Begin fomepizole (preferred) or ethanol therapy if: Serum methanol level ³ 6 mmol/L, OR Osmole gap > 10 (corrected for ethanol) with history of methanol exposure, OR History or strong clinical suspicion of methanol poisoning with at least two of the following criteria: Serum bicarbonate < 20 mmol/L AG > 16 arterial pH < 7.3 Osmole gap > 10 d) Monitor vital signs and electrolytes. Monitor blood gases in symptomatic patients. e) Fomepizole Administration: Administration WITHOUT dialysis: Initial loading dose 15 mg/kg to a maximum of 1500 mg (one vial), subsequent doses 10 mg/kg every 12 hours, dosage adjustment may be required after 48 hours of treatment because fomepizole induces its own metabolism. Administration WITH dialysis: Initial loading dose 15 mg/kg to a maximum of 1500 mg (one vial), the second dose 10 mg/kg should be given 4 hours after the start of dialysis, the third dose should be given at the end of dialysis. Treatment Endpoint: Continue treatment until metabolic acidosis has resolve and methanol level is < 6mmol/L.f. f) Folate therapy: Administer either folic acid or leucovorin to enhance detoxification of formic acid. Indicated for all high risk patients. 50 mg of folate should be given every 4 hours. g) Hemodialysis Indications for hemodialysis include: Arterial pH < 7.25 that fails to rapidly normalize with treatment, OR Visual defects, OR Other serious clinical effects (e.g. coma, seizures, severe electrolyte imbalance, renal impairment), OR Clinical condition deteriorates during antidote treatment, OR Initial measured or estimated methanol serum level > 15 mmol/L. Methanol is eliminated slowly during antidote therapy (serum level of 15 mmol/L would require ~ 3 days of treatment with antidote alone to reach sub-toxic levels). Continuous renal replacement therapy such as CVVHD clears the toxins more slowly than conventional hemodialysis (limited data). Conventional hemodialysis is preferred for rapid detoxification, if the patient can tolerate the procedure. Criteria for Hospital Admission Any patient who may have ingested methanol who has: any symptoms acidosis an elevated anion gap or an elevated osmolar gap Criteria for Transfer to Another Facility If patient requires ICU care or dialysis not available locally Consult BC Drug and Poison Information Centre 604-682-5050, 1-800-567-8911 Criteria For Safe Discharge Home Patients may be discharged after observation periods have been met and the patient’s psychiatric condition has been treated if the overdose was intentional. Quality Of Evidence? High We are highly confident that the true effect lies close to that of the estimate of the effect. There is a wide range of studies included in the analyses with no major limitations, there is little variation between studies, and the summary estimate has a narrow confidence interval. Moderate We consider that the true effect is likely to be close to the estimate of the effect, but there is a possibility that it is substantially different. There are only a few studies and some have limitations but not major flaws, there are some variations between studies, or the confidence interval of the summary estimate is wide. Low When the true effect may be substantially different from the estimate of the effect. The studies have major flaws, there is important variations between studies, of the confidence interval of the summary estimate is very wide. Justification Case Reports and Case Series Low Related Information OTHER RELEVANT INFORMATION EM Cases podcast Feb 2018 “Toxic alcohols – minding the gaps.” References: Poison Management Manual (PMM), BC Drug and Poison Information Centre, 2015 Kruse JA. Methanol poisoning. Intensive Care Medicine. 1992;18(7): 391-7. Hovda, K.E., et al., Methanol outbreak in Norway 2002-2005: epidemiology, clinical features and prognostic signs. Journal of Internal Medicine, 2005. 258: p. 181-190. Megarbane B, Borron SW, Baud F. Current recommendations for treatment of severe toxic alcohol poisonings. Intensive Care Medicine. 2005;31:189-195 American Academy of Clinical Toxicology practice guidelines on the treatment of methanol poisoning. Journal of Toxicology – Clinical Toxicology. 2002;40(4):415-446. Relevant Resources RELEVANT CLINICAL RESOURCES Venlafaxine (Effexor XR) and Desvenlafaxine (Pristiq) Overdoses Clinical Summary August 18, 2017 • Posted by Roy Purssell Opioid Overdoses – Management Clinical Summary November 01, 2017 • Posted by Andrew Kestler, Christopher DeWitt, Roy Purssell, Jesse Godwin Tricyclic Antidepressant Overdose Clinical Summary September 18, 2017 • Posted by Roy Purssell View all Resources RELEVANT RESEARCH IN BC System Response to Toxicologic Emergencies Print Summary RESOURCE AUTHOR(S) ### Dr Roy Purssell UBC Department of Emergency Medicine Interim Co-Head DISCLAIMER The purpose of this document is to provide health care professionals with key facts and recommendations for the diagnosis and treatment of patients in the emergency department. This summary was produced by Emergency Care BC (formerly the BC Emergency Medicine Network) and uses the best available knowledge at the time of publication. However, healthcare professionals should continue to use their own judgment and take into consideration context, resources and other relevant factors. Emergency Care BC is not liable for any damages, claims, liabilities, costs or obligations arising from the use of this document including loss or damages arising from any claims made by a third party. Emergency Care BC also assumes no responsibility or liability for changes made to this document without its consent. Last Updated Dec 17, 2018 Visit our website at COMMENTS (0) POST COMMENT We welcome your contribution! If you are a member, log in here. If not, you can still submit a comment but we just need some information. Cancel Add public comment… SHARE POST Thank you for your submission. We will contact you shortly.
10061	https://dictionary.cambridge.org/us/dictionary/english-chinese-traditional/braggadocio	Translation of braggadocio – English–Traditional Chinese dictionary Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio (Translation of braggadocio from the Cambridge English-Chinese (Traditional) Dictionary © Cambridge University Press) Examples of braggadocio Translations of braggadocio Get a quick, free translation! Browse Word of the Day Victoria sponge Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio a soft cake made with eggs, sugar, flour, and a type of fat such as butter. It is made in two layers with jam or cream, or both, between them Blog Calm and collected (The language of staying calm in a crisis) New Words vibe coding © Cambridge University Press & Assessment 2025 © Cambridge University Press & Assessment 2025 Learn more with +Plus Learn more with +Plus To add braggadocio to a word list please sign up or log in. Add braggadocio to one of your lists below, or create a new one. {{message}} {{message}} Something went wrong. {{message}} {{message}} Something went wrong. {{message}} {{message}} There was a problem sending your report. {{message}} {{message}} There was a problem sending your report.
10062	https://askfilo.com/user-question-answers-smart-solutions/differntation-of-sin-ax-b-3338303938363933	World's only instant tutoring platform Instant TutoringPrivate Courses Explore TutorsBecome Tutor StudentTutor Higher Education Smart Solutions Differntation of sin(ax+b) Question asked by Filo student Differntation of sin(ax+b) Views: 5,411students Updated on: Sep 10, 2025 Not the question you're searching for? Ask your question Ask your question Or Upload the image of your question Text solutionVerified Concepts Differentiation, Chain Rule, Derivative of Sine Function Explanation To differentiate a composite function like sin(ax+b), use the chain rule. The chain rule states that the derivative of sin(u) with respect to x is cos(u)⋅dxdu, where u is a function of x (here, u=ax+b). Step-By-Step Solution Step 1 Let y=sin(ax+b). Step 2 Let u=ax+b. So y=sin(u). Step 3 Differentiate y with respect to u: dudy=cos(u) Step 4 Differentiate u with respect to x: dxdu=a Step 5 By Chain Rule: dxdy=dudy⋅dxdu=cos(u)⋅a Step 6 Substitute u=ax+b: dxd[sin(ax+b)]=acos(ax+b) Final Answer dxd[sin(ax+b)]=acos(ax+b) Ask your next question Or Upload the image of your question Found 2tutors discussing this question OliviaDiscussed Differntation of sin(ax+b) 7mins ago Discuss this question LIVE 7mins 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 Trusted by 4 million+ students Students who ask this question also asked Question 1 Views:5,380 Q.5 A circuit consisting of a coil of resistance 12 ohms and inductance 0.15 henry is in series with a capacitor of 12 microfarad. It is connected to a variable frequency supply which has a constant voltage of 24 volt. Calculate: The resonant frequency The current at resonance The voltage across capacitor and the coil at resonance Topic: Smart Solutions View solution Question 2 Views:5,153 A survey of students shows: Total students: 200 Students reading Indian Economic Review (IER): 120 Students reading Harvard Business Review (HBR): 100 Find: Number of students who do not read either IER or HBR. Number of students who read both IER and HBR. Topic: Smart Solutions View solution Question 3 Views:5,836 Which of the following BEST describes when an emergency stop button should be tested on a winch or windlass? Never as it causes a surge in the power system. Once a month with the other stop button tests. During the pre-use checks. During the mooring operation. Topic: Smart Solutions View solution Question 4 Views:5,977 Differentiate between a bill and cheque Topic: Smart Solutions View solution View more Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. 231students are takingLIVE classes \| \| \| --- \| \| Question Text \| Differntation of sin(ax+b) \| \| Updated On \| Sep 10, 2025 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Undergraduate \| \| Answer Type \| Text solution:1 \| Are you ready totake control of your learning? Download Filo and start learning with yourfavoritetutors right away!
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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Algebra and Trigonometry Practice Test Algebra and Trigonometry Practice Test Contents Contents Highlights Table of contents Preface 1 Prerequisites 2 Equations and Inequalities 3 Functions 4 Linear Functions 5 Polynomial and Rational Functions 6 Exponential and Logarithmic Functions Introduction to Exponential and Logarithmic Functions 6.1 Exponential Functions 6.2 Graphs of Exponential Functions 6.3 Logarithmic Functions 6.4 Graphs of Logarithmic Functions 6.5 Logarithmic Properties 6.6 Exponential and Logarithmic Equations 6.7 Exponential and Logarithmic Models 6.8 Fitting Exponential Models to Data Chapter Review Exercises Review Exercises Practice Test 7 The Unit Circle: Sine and Cosine Functions 8 Periodic Functions 9 Trigonometric Identities and Equations 10 Further Applications of Trigonometry 11 Systems of Equations and Inequalities 12 Analytic Geometry 13 Sequences, Probability, and Counting Theory A \| Proofs, Identities, and Toolkit Functions Answer Key Index Search for key terms or text. Close Practice Test 1. The population of a pod of bottlenose dolphins is modeled by the function A(t)=8(1.17)t,A(t)=8(1.17)t,A(t)=8(1.17)t, where t t t is given in years. To the nearest whole number, what will the pod population be after 3 3 3 years? Find an exponential equation that passes through the points (0, 4)(0, 4)(0, 4) and (2, 9).(2, 9).(2, 9). 3. Drew wants to save $2,500 to go to the next World Cup. To the nearest dollar, how much will he need to invest in an account now with 6.25%6.25%6.25% APR, compounding daily, in order to reach his goal in 4 4 4 years? An investment account was opened with an initial deposit of $9,600 and earns 7.4%7.4%7.4% interest, compounded continuously. How much will the account be worth after 15 15 15 years? 5. Graph the function f(x)=5(0.5)−x f(x)=5(0.5)−x f(x)=5(0.5)−x and its reflection across the y-axis on the same axes, and give the y-intercept. The graph shows transformations of the graph of f(x)=(1 2)x.f(x)=(1 2)x.f(x)=(1 2)x. What is the equation for the transformation? 7. Rewrite log 8.5(614.125)=a log 8.5(614.125)=a log 8.5(614.125)=a as an equivalent exponential equation. Rewrite e 1 2=m e 1 2=m e 1 2=m as an equivalent logarithmic equation. 9. Solve for x x x by converting the logarithmic equation l o g 1 7(x)=2 l o g 1 7(x)=2 l o g 1 7(x)=2 to exponential form. Evaluate log(10,000,000)log(10,000,000)log(10,000,000) without using a calculator. 11. Evaluate ln(0.716)ln(0.716)ln(0.716) using a calculator. Round to the nearest thousandth. Graph the function g(x)=log(12−6 x)+3.g(x)=log(12−6 x)+3.g(x)=log(12−6 x)+3. 13. State the domain, vertical asymptote, and end behavior of the function f(x)=log 5(39−13 x)+7.f(x)=log 5(39−13 x)+7.f(x)=log 5(39−13 x)+7. Rewrite log(17 a⋅2 b)log(17 a⋅2 b)log(17 a⋅2 b) as a sum. 15. Rewrite log t(96)−log t(8)log t(96)−log t(8)log t(96)−log t(8) in compact form. Rewrite log 8(a 1 b)log 8(a 1 b)log 8(a 1 b) as a product. 17. Use properties of logarithm to expand ln(y 3 z 2⋅x−4−−−−√3).ln(y 3 z 2⋅x−4 3).ln(y 3 z 2⋅x−4 3). Condense the expression 4 ln(c)+ln(d)+ln(a)3+ln(b+3)3 4 ln(c)+ln(d)+ln(a)3+ln(b+3)3 4 ln(c)+ln(d)+ln(a)3+ln(b+3)3 to a single logarithm. 19. Rewrite 16 3 x−5=1000 16 3 x−5=1000 16 3 x−5=1000 as a logarithm. Then apply the change of base formula to solve for x x x using the natural log. Round to the nearest thousandth. Solve (1 81)x⋅1 243=(1 9)−3 x−1(1 81)x⋅1 243=(1 9)−3 x−1(1 81)x⋅1 243=(1 9)−3 x−1 by rewriting each side with a common base. 21. Use logarithms to find the exact solution for −9 e 10 a−8−5=−41−9 e 10 a−8−5=−41−9 e 10 a−8−5=−41 . If there is no solution, write no solution. Find the exact solution for 10 e 4 x+2+5=56.10 e 4 x+2+5=56.10 e 4 x+2+5=56. If there is no solution, write no solution. 23. Find the exact solution for −5 e−4 x−1−4=64.−5 e−4 x−1−4=64.−5 e−4 x−1−4=64. If there is no solution, write no solution. Find the exact solution for 2 x−3=6 2 x−1.2 x−3=6 2 x−1.2 x−3=6 2 x−1. If there is no solution, write no solution. 25. Find the exact solution for e 2 x−e x−72=0.e 2 x−e x−72=0.e 2 x−e x−72=0. If there is no solution, write no solution. Use the definition of a logarithm to find the exact solution for 4 log(2 n)−7=−11 4 log(2 n)−7=−11 4 log(2 n)−7=−11 27. Use the one-to-one property of logarithms to find an exact solution for log(4 x 2−10)+log(3)=log(51)log(4 x 2−10)+log(3)=log(51)log(4 x 2−10)+log(3)=log(51) If there is no solution, write no solution. The formula for measuring sound intensity in decibels D D D is defined by the equation D=10 log(I I 0),D=10 log(I I 0),D=10 log(I I 0), where I I I is the intensity of the sound in watts per square meter and I 0=10−12 I 0=10−12 I 0=10−12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a rock concert with a sound intensity of 4.7⋅10−1 4.7⋅10−1 4.7⋅10−1 watts per square meter? 29. A radiation safety officer is working with 112 112 112 grams of a radioactive substance. After 17 17 17 days, the sample has decayed to 80 80 80 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest day, what is the half-life of this substance? Write the formula found in the previous exercise as an equivalent equation with base e.e.e. Express the exponent to five significant digits. 31. A bottle of soda with a temperature of 71°71°71° Fahrenheit was taken off a shelf and placed in a refrigerator with an internal temperature of 35° F.35° F.35° F. After ten minutes, the internal temperature of the soda was 63° F.63° F.63° F. Use Newton’s Law of Cooling to write a formula that models this situation. To the nearest degree, what will the temperature of the soda be after one hour? The population of a wildlife habitat is modeled by the equation P(t)=360 1+6.2 e−0.35 t,P(t)=360 1+6.2 e−0.35 t,P(t)=360 1+6.2 e−0.35 t, where t t t is given in years. How many animals were originally transported to the habitat? How many years will it take before the habitat reaches half its capacity? 33. Enter the data from Table 1 into a graphing calculator and graph the resulting scatter plot. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic. xf(x) 1 3 2 8.55 3 11.79 4 14.09 5 15.88 6 17.33 7 18.57 8 19.64 9 20.58 10 21.42 Table 1 The population of a lake of fish is modeled by the logistic equation P(t)=16,120 1+25 e−0.75 t,P(t)=16,120 1+25 e−0.75 t,P(t)=16,120 1+25 e−0.75 t, where t t t is time in years. To the nearest hundredth, how many years will it take the lake to reach 80%80%80% of its carrying capacity? For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places. 35. xf(x) 1 20 2 21.6 3 29.2 4 36.4 5 46.6 6 55.7 7 72.6 8 87.1 9 107.2 10 138.1 xf(x) 3 13.98 4 17.84 5 20.01 6 22.7 7 24.1 8 26.15 9 27.37 10 28.38 11 29.97 12 31.07 13 31.43 37. xf(x) 0 2.2 0.5 2.9 1 3.9 1.5 4.8 2 6.4 3 9.3 4 12.3 5 15 6 16.2 7 17.3 8 17.9 PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? 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10064	https://www.tutorchase.com/answers/ib/physics/how-to-convert-celsius-to-kelvin	How to convert Celsius to Kelvin? \| TutorChase Revision PlatformHire a tutor All Tutors GCSE IGCSE A-Level IB AP Oxbridge / UK Admissions US Admissions Resources More All Tutors GCSE IGCSE A-Level IB AP Oxbridge / UK Admissions US Admissions Resources More Revision PlatformHire a tutor Answers IB Physics Article How to convert Celsius to Kelvin? To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. The Kelvin scale is an absolute temperature scale, which means it starts at absolute zero, the theoretical lowest possible temperature where all molecular motion stops. This is different from the Celsius scale, which starts at the freezing point of water. The Kelvin and Celsius scales are linearly related, meaning they increase at the same rate. However, they have different zero points. The conversion formula from Celsius to Kelvin is K = °C + 273.15. This means that to convert a temperature from Celsius to Kelvin, you simply add 273.15. For example, if you have a temperature of 25 degrees Celsius and you want to convert it to Kelvin, you would calculate 25 + 273.15 = 298.15 Kelvin. It's important to note that while we often use degrees to denote temperatures in Celsius or Fahrenheit, we do not use degrees when talking about temperatures in Kelvin. So, we would say "298.15 Kelvin," not "298.15 degrees Kelvin." This conversion is fundamental in physics, particularly in thermodynamics, where absolute temperatures are often required. The Kelvin scale is used because it allows for the calculation of absolute temperatures, which are necessary for equations that involve temperature, such as the ideal gas law or the Boltzmann constant. Remember, when converting from Celsius to Kelvin, you're not actually changing the temperature, just representing it in a different way. The actual thermal energy doesn't change, just the number we use to describe it. This is similar to converting between different units of length, such as metres and feet. The actual length doesn't change, just the number and unit we use to describe it. In summary, converting from Celsius to Kelvin is a straightforward process of adding 273.15 to the Celsius temperature. This conversion is essential in many areas of physics, particularly those involving thermodynamics. Answered by Aaron - Qualified Tutor \| B.A. in Maths IB Physics tutor Study and Practice for Free Trusted by 100,000+ Students Worldwide Achieve Top Grades in your Exams with our Free Resources. Practice Questions, Study Notes, and Past Exam Papers for all Subjects! IB ResourcesA-Level ResourcesGCSE ResourcesIGCSE Resources Need help from an expert? 4.93/5 based on733 reviews in The world’s top online tutoring provider trusted by students, parents, and schools globally. #### Vera PGCE Qualified Teacher \| IB Examiner \| BA World Literature #### Stefan PGCE Qualified Teacher \| Cambridge University \| BA Mathematics #### Flora PGCE Qualified Teacher \| Cambridge University MSci Natural Sciences Hire a tutor Related Physics ib Answers How to determine the uncertainty of a measurement? What units are used for time in the SI system? What's the appropriate instrument to measure the volume of a liquid? How do you calculate the percentage uncertainty? Read All Answers Earn £ as a Student Ambassador! Loading... This website uses cookies to improve your experience, by continuing to browse we’ll assume you’re okay with this. Accept Loading...
10065	https://study.com/skill/learn/how-to-determine-the-oxidation-state-for-atoms-in-compound-explanation.html	How to Determine the Oxidation State for Atoms in Compound \| Chemistry \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright How to Determine the Oxidation State for Atoms in Compound High School Chemistry Skills Practice Click for sound 5:18 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:04 How to determine the… 03:19 How to determine the… Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Wesley Smith, Kirsten Wordeman Instructors Wesley Smith Wesley D. Smith has a Ph.D. in theoretical chemistry from Brigham Young University. He has taught freshman chemistry a total 30 years at five different universities. View bio Kirsten Wordeman Kirsten has taught high school biology, chemistry, physics, and genetics/biotechnology for three years. She has a Bachelor's in Biochemistry from The University of Mount Union and a Master's in Biochemistry from The Ohio State University. She holds teaching certificates in biology and chemistry. View bio Example SolutionsPractice Questions Steps for Determining the Oxidation State of Atoms in A Compound Step 1: Assign oxidation numbers to every atom using the n X notation, and set up an equation using the Rule. Step 2: Assign as many oxidation numbers as you can according to the Six Cases. Step 3:Solve the equation for the one remaining oxidation number. One Rule and Six Cases for Determining the Oxidation State of Atoms in Compound In order to keep track of the number of electrons being transferred in oxidation-reduction reactions, you need some kind of bookkeeping system. The system of oxidation states or oxidation numbers works for that purpose. An oxidation number, n is assigned to every atom in a compound. The oxidation number is defined as the charge the atom has in an ionic compound, or, in a covalent compound, it is the charge it would have if all shared electrons in any bond are given to the more electronegative atom. For example, in the Lewis structure of ionic sodium chloride, If you do not have the Lewis structure, you can either draw it, or you can use the following procedure, which, with a little practice, is actually less complicated. The first step is to give every atom in the molecule or ion a symbolic oxidation number using the n notation. Then add them all up to make an equation according to the One Rule. Rule:The sum of all the oxidation numbers in any molecule or ion must equal the charge on the species. e.g., in P O 4 3− the sum of n P+4 n O=−3; in C 6 H 12 O 22, the sum of 6 n C+12 n H+22 n O=0 The next step is to give numerical values to as many oxidation numbers as you can according to the following Six Cases: Case 1: Elements, either uncombined or combined only with themselves, have oxidation numbers equal to zero. e.g., in I 2,n I=0; in A l,n A l=0; in P 4,n P=0 Case 2: Monatomic ions have oxidation numbers equal to their charges. e.g., in S 2−,n S=−2; in A l 3+,n A l=+3; in N a 1+,n N a=+1 Case 3: Except for Case 1 above, n F=−1,n a l k a l i m e t a l=+1,n a l k a l i n e e a r t h=+2,n A l=+3 e.g., in N F 3,n F=−1; in A l 2 O 3,n A l=+3; in K 3 P O 4,n K=+1; in M g H 2,n M g=+2 Case 4: Except for Case 1 above, n h a l o g e n=−1,n S=−2, provided they are the most electronegative atom in the compound. e.g., in K C l,n C l=−1, but in K C l O 3,n C l≠−1; in H 2 S,n S=−2, but in H 2 S O 4,n S≠−2 Case 5: Except for Case 1 above, n H=+1 when hydrogen is combined with nonmetals, and n H=−1 when hydrogen is combined with metals. e.g., in N H 3,H 2 S,H 3 P O 4,n H=+1 but in M g H 2,L i H,n H=−1 Case 6: Except for Case 1 above, n O=−2, except in peroxides where n O=−1. e.g., in H 3 P O 4,C r 2 O 7 2−,n O=−2, but in H 2 O 2,H O 2 1−,n O=−1 After taking care of all of the above cases, you will find (in a whole molecule or a recognizable polyatomic ion within the molecule) only one element remains whose oxidation number is yet to be determined. You can deduce its oxidation number by solving the simple equation you made according to the Rule (There are rare instances where this procedure will leave more than one oxidation number unknown, but you will not see such examples). For methyl iodide C H 3 I, then, the equation according to the Rule is n C+3 n H+n I=0. But n H=+1 by Case 5 and n I=−1 by Case 4. Only n C remains, and it must satisfy n C+3(+1)+(−1)=0 Thus, n C=+2. What about permanganate ion, M n O 4 1−? n M n+4 n O=−1; n O=−2 by Case 6. And n M n=+7. Here are 3 example problems illustrating the Six Cases and the One Rule: you see that n N a=+1 and n C l=−1. In the Lewis structure for water, you give the shared electrons to oxygen, and you get so that n H=+1 and n O=−2. Thus, it is easy to assign oxidation numbers if you have the Lewis structure. Example Problem 1 - How to Determine the Oxidation State for Atoms in Compound What is the oxidation number of C r in C r 2 O 7 2−? Step 1: Assign oxidation numbers to every atom using the n X notation, and set up an equation using the Rule. 2 n C r+7 n O=−2 Step 2: Assign as many oxidation numbers as you can according to the Six Cases. 2 n C r+7(−2)=−2 from Case 6 Step 3:Solve the equation for the one remaining oxidation number. n C r=+6 Example Problem 2 - How to Determine the Oxidation State for Atoms in Compound What is the oxidation number of chlorine in K C l O 3? Step 1: Assign oxidation numbers to every atom using the n X notation, and set up an equation using the Rule. n K+n C l+3 n O=0 Step 2: Assign as many oxidation numbers as you can according to the Six Cases. (+1)+n C l+3(−2)=0 from Case 3 and Case 6 Step 3:Solve the equation for the one remaining oxidation number. n C l=+5 Example Problem 3 - How to Determine the Oxidation State for Atoms in Compound What is the oxidation number of nitrogen and arsenic in (N H 4)2 H A s O 3? This is really a two-part problem. In the first part, you must recognize the ammonium ion, N H 4 1+ component, and use its formula to deduce n N. Step 1: Assign oxidation numbers to every atom using the n X notation, and set up an equation using the Rule. n N+4 n H=+1 Step 2: Assign as many oxidation numbers as you can according to the Six Cases. n N+3(+1)=+1 from Case 5 Step 3:Solve the equation for the one remaining oxidation number. n N=−3 Knowing that the charge on each ammonium ion is +1, you can deduce the charge on the rest of the molecule must be -2. Step 1: Assign oxidation numbers to every atom using the n X notation, and set up an equation using the Rule. n H+n A s+3 n O=−2 Step 2: Assign as many oxidation numbers as you can according to the Six Cases. (+1)+n A s+3(−2)=−2 from Case 5 and Case 6 Step 3:Solve the equation for the one remaining oxidation number. n A s=+4 Get access to thousands of practice questions and explanations! Create an account Table of Contents Steps for Determining the Oxidation State of Atoms in A Compound One Rule and Six Cases for Determining the Oxidation State of Atoms in Compound Example Problem 1 Example Problem 2 Example Problem 3 Test your current knowledge Practice Determining Oxidation State for Atoms in Compound Related Courses National Eligibility Test (AIPMT): Study Guide Chemistry 112L: Chemistry II with Lab General Science Lessons Chemistry 101: General Chemistry AP Chemistry Study Guide and Exam Prep Related Lessons Electrochemical Cell \| Definition, Types & Examples Equilibrium Constant Expressions \| Overview, Equation & Examples Electrophoresis Definition, Types & Uses Electrochemical Cells and Electrochemistry Anode vs. Cathode in Electrochemical Cells \| Reaction & Notation Recently updated on Study.com Videos Courses Lessons Articles Quizzes Concepts Teacher Resources The Invention of Writing Ethnic Groups in Indonesia \| Demographics & People Gods of the Winter Solstice Holes by Louis Sachar \| Themes, Quotes & Analysis Aurangzeb \| Empire, Achievements & Failures Libya Ethnic Groups \| Demographics, Population & Cultures Cold War Lesson for Kids: Facts & Timeline The Chronicles of Narnia Series by C.S. 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10066	https://www.youtube.com/watch?v=otxE_VFd3G0	they say we can't factor x^2+y^2 blackpenredpen 1390000 subscribers 5724 likes Description 224179 views Posted: 23 May 2019 Factoring the sum of two squares. Sign up for a free account at and try their daily challenges now. You can also get a 20% discount for their annual premium subscription so you can get access to ALL of their awesome designed courses! Subscribe for more math for fun videos 👉 💪 Support this channel, 🛍 Shop math t-shirt & hoodies: (10% off with the code "WELCOME10") 🛍 I use these markers: 480 comments Transcript: okay as we all know people will tell you that we can affect an expression x squared plus y squared and I greased them but that's only true in the real and polynomial world today if you're willing to get into the complex world or maybe the non polynomial world we can do a lot more let me show you how let's take care of this in the complex world first so x squared plus y squared unfortunately we don't have a nice formula for this yet but we do know the following today if we have the difference of two squares namely if we have a square minus B Square of course we can factor this out nicely and the answer to that is just a minus B times a plus B like this and this is very nice well let's see we have x squared plus y squared okay let me just write down the X square like this and you would have first use that we have to have a minus in between okay let me just put a minus and everybody knows a negative times a negative is equal to a past year by no negative times I Square will also give you a positive and the reason is because we know that I squared is equal to negative one and now with the I think both we are in the complex world so keep this in mind I squared is equal to negative one and as we can see we can still just write down the Y Square and this is still the same as that and the beauty of doing so is we can see the first term is X and put parentheses around it and then square so now this is much fancier and this is minus for the second part we pour indices I can put the I and the Y inside and then I squared out very nice and now this is our a and this is our B and we can just use the formula and to factor this the first factor is just X minus iy huh why cool and then the second one is X plus iy and that's it so this is the answer in the complex world if you want to factor x squared plus y squared now you think give a sugar cane through algebra you know like that see we'll just give the algebra sure look at this thank you so much what design the t-shirt thank you you know who you are thank you okay now let's take a look of this one right here when people say we are going to factor usually we want to just keep in the polynomial form namely the power of the eggs shall be positive whole numbers what maybe just zero so this is not possible if you want to keep it as polynomial but as I said it'll willing to use some kind of square roots it's actually possible let me show you so let me write this down again x squared plus y squared but again we don't know about this too much and I don't want to get into the complex world for this one so let me use another one right here for you guys this is actually how we know much better and we would like to see if we can show any kind of come natural all right now let's see we have x squared plus y squared would it be nice if we have something the middle that will actually help us out what do I mean well let me just put this down here for you guys suppose today we have a square and if we have plus 2 a B haha this right here after we add a B Square after that we know this is just nicely equal to a plus B squared very nice and what started as we can see this right here it's a square and this right here is a piece we're pretty much and now we just have to put the middle term right here so let's go ahead and make that happen now here we have X square and let me just put a plus 2 a B and of course in this case the ASD X and then Y is the B right so we can't just put up 2xy and we have the plus y squared after that that's very nice of course this is no longer the same as that it's okay because we can just go ahead right here and then put on -2 X Y and now you might be wondering why don't I want to use the minus version right here if I have a minus version this will be just a minus B and square that's also possible but I really want to use this one because now I'm adding here and I will have to subtract this you will see that the first three terms here will keep us a pretty square namely we have X plus y square like this and then for this one because we have the minus in between so let's put that down and yes you guessed it we can use the difference of two squares to help us out and were the first to use that be sure you open a parenthesis put a square on outside now what to the second power will give us two and the answer to that square root of two good and what to the second power will give us X and the answer to that is square root of x and of course I can just make the square ruler and put the X inside and that's the of course same speech so we put a white inside like this and from where the clear square root of 2xy unfortunately this is no longer a polynomial so people don't usually do this when they are talking about factoring x squared plus y squared and now of course we can just apply this right here to help us out and we will see this right here is just X plus y and we will just have to minus this part which is square root of 2xy and that will be the first factor for the second factor it's pretty much just X plus y and we add square root of 2xy and guess what which is factor this guy in the real world but fortunately this is not in the polynomial world but still not yet so hopefully you guys all liked this video and let me show you guys okay for the people were like math and science let me tell you this a popular work this right here the interactive learning website that focuses on problem solving and as you guys can see they offered daily challenges for you guys and these questions are really fun and really interesting you guys won't like them too and because I have the annual subscription actually get access to their archived questions so they Michelle gets this question from a few days ago this geometry question and the question is which ones bigger the blue region were the green region well you guys can try this on your own or you can just click on keep reading and it's going to show you guys the strategy so the way to do it is if you to a vertical line right here a common strategy in solving geometry question and now it should be clear that this and they are the same and this in the artists a and of course the blue portion here it's the same sense that green portion so the fact is okay they have the same area and now based on that can you use the similar strategies to solve this challenge and I'm going to leave that you guessed right here which areas bigger the yellow one were the red one and of course you guys can come here and choose the option and after you guys finish it you guys can click on the quiz right here and you just can continue learning at this and if you guys will also like to get their premium subscription you guys can use the link print our work slash black and repent this way you guys can get 20% off discount for the air and 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10067	http://everyspec.com/NIST/NIST-General/SP_811_3831/	SP 811 NIST SPECIAL PUBLICATION GUIDE USE INTERNATIONAL Home Library Popular Specs Hardware Categories Quick Search Search Resources Home>Library>NIST>NIST-General>SP 811 SP 811, NIST SPECIAL PUBLICATION: GUIDE FOR THE USE OF INTERNATIONAL SYSTEM OF UNITS (SI) (1995) Discover more DoD standards access Specification search tools Technical document translation Engineering handbook downloads SP 811, NIST SPECIAL PUBLICATION: GUIDE FOR THE USE OF INTERNATIONAL SYSTEM OF UNITS (SI) (1995)., The International System of Units, universally abbreviated SI is the modern metric system of measurement. Long the dominant measurement system used in science, the SI is becoming the dominant measurement system used in international commerce. The Omnibus Trade and Competitiveness Act of August 1988 [Public Law (PL) 100-418] changed the name of the National Bureau of Standards (NBS) to the National Institute of Standards and Technology (NIST) and gave to NIST the added task of helping United States industry increase its competitiveness in the global marketplace. It also recognized the rapidly expanding use of the SI by amending the Metric Conversion Act of 1975 (PL 94-168). In particular, section 5164 (Metric Usage) of PL 100-418 designates the metric system of measurement as the preferred system of weights and measures for United States trade and commerce and requires that each Federal agency, by a date certain and to the extent economically feasible by the end of fiscal year 1992, use the metric system of measurement in its procurements, grants, and other business-related activities, except to the extent that such use is impractical or is likely to cause significant inefficiencies or loss of markets for United States firms. In January 1991, the Department of Commerce issued an addition to the Code of Federal Regulations entitled "Metric Conversion Policy for Federal Agencies," 15 CFR 1170, which removes the voluntary aspect of the conversion to the SI for Federal agencies and gives in detail the policy for that conversion. Executive Order 12770, issued in July 1991, reinforces that policy by providing Presidential authority and direction for the use of the metric system of measurement by Federal agencies and departments. Because of the importance of the SI to both science and technology, NIST has over the years published documents to assist NIST authors and other users of the SI, especially to inform them of changes in the SI and in SI usage. For example, this second edition of the Guide replaces the first edition prepared by Arthur O. McCoubrey and published in 1991. That edition, in turn, replaced NBS Letter Circular LC 1120 (1979), which was widely distributed in the United States and which was incorporated into the NBS Communications Manual for Scientific, Technical, and Public Information, a manual of instructions issued in 1980 for the preparation of technical publications at NBS. SP-811 Rev. 1995 ---------------------------------------------------------------------------------------------- Download File - 327.92 KB Report Problem (email) Version: 1995 04-1995 327.92 KB SP_811 Discover more purchase military test procedures order DoD standardization documents where to buy MIL-PRF Industry standards Standardization documents Acquire military specifications binder buy aerospace materials specifications Simple Search Number only or Find by FSC code Search Discover more buy specific MIL-STD documents buy military standard publications Standardization documents military parts catalog online buy engineering drawing files purchase military handbook documents ADF AIAA AN SPECS AND SPECS ANSI ARMY ASQC ASTM CCSDS COMML ITEM DESC DATA ITEM DESC DOC. TEMPLATES DoD DODSSP DOE DOT ESA FAA FED SPECS FED-STD JAN SPECS JAXA MIL-HDBK MIL-PRF MIL-SPECS MIL-STD MISC MS Specs NASA NATO NIST NUREG SAE DEF STAN USAB USAF USCG USMC USN About Us\|Terms of Use\|DMCA\|Privacy\| EverySpec LLC © 2009 - 2025 All rights reserved.
10068	https://pmc.ncbi.nlm.nih.gov/articles/PMC8574467/	TP10.2.2: Adherence to the new Royal College of Surgeons (RCS) guidance pertaining to acute appendicitis management during COVID-19 - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Br J Surg . 2021 Oct 28;108(Suppl 7):znab362.139. doi: 10.1093/bjs/znab362.139 Search in PMC Search in PubMed View in NLM Catalog Add to search TP10.2.2: Adherence to the new Royal College of Surgeons (RCS) guidance pertaining to acute appendicitis management during COVID-19 Rajab Khan Rajab Khan 1 Southend University Hospital NHS Foundation Trust Find articles by Rajab Khan 1, Sabina Shamsad Sabina Shamsad 1 Southend University Hospital NHS Foundation Trust Find articles by Sabina Shamsad 1, Umaimah Rahimi Umaimah Rahimi 1 Southend University Hospital NHS Foundation Trust Find articles by Umaimah Rahimi 1, Hamisha Salih Hamisha Salih 1 Southend University Hospital NHS Foundation Trust Find articles by Hamisha Salih 1, Humayun Razzaq Humayun Razzaq 1 Southend University Hospital NHS Foundation Trust Find articles by Humayun Razzaq 1, Janine Adedeji Janine Adedeji 1 Southend University Hospital NHS Foundation Trust Find articles by Janine Adedeji 1 Author information Article notes Copyright and License information 1 Southend University Hospital NHS Foundation Trust Collection date 2021 Oct. © The Author(s) 2021. Published by Oxford University Press on behalf of BJS Society Ltd. All rights reserved. For permissions, please email: journals.permissions@oup.com This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model ( This article is made available via the PMC Open Access Subset for unrestricted re-use and analyses in any form or by any means with acknowledgement of the original source. These permissions are granted for the duration of the COVID-19 pandemic or until permissions are revoked in writing. Upon expiration of these permissions, PMC is granted a perpetual license to make this article available via PMC and Europe PMC, consistent with existing copyright protections. PMC Copyright notice PMCID: PMC8574467 Abstract Aims The mainstay of acute appendicitis treatment is a surgical approach. However, in the tumultuous COVID-19 era, the approach to acute appendicitis management has altered. We sought to assess the applicability of the new RCS COVID-19 guidance after resolution of the global pandemic. Methods A retrospective study was conducted on 244 patients presenting between 01/03/2020 and 17/07/2020. Three sources of data were sought: patients presenting to A&E with signs of appendicitis, operative logs for patients who underwent removal of their appendix and all CT/US scans where the clinician had queried appendicitis in the request. Results 139 patients were treated conservatively with antibiotics (57.0%). 35 (25.2%) represented within 6 months. Conservative treatment was successful in 92.1% of cases. 65 appendectomies were completed during that time. 45 cases presented acutely whereby the surgical management of acute appendicitis was the primary choice. The decision to operate was due to: 13 (20%) because the patient was 16 or younger. 4 cases presenting with signs of severe sepsis. 4 cases had a complicated appendicitis on their imaging results. 22 (33.8%) cases were completed without initial conservative management. 9 (3.7%) cases failed conservative treatment thus requiring surgical management. 11 (4.5%) cases represented within 6 months due to recurrent appendicitis despite successful antibiotic treatment. Therefore, the recurrence rate following conservative treatment was 7.9%. Conclusions There was mixed adherence to the new guidance. Surgical management remains the best approach towards acute appendicitis. However, excellent results can still be achieved with appropriately targeted antibiotic therapy. 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10069	https://www.math.utah.edu/lectures/math2210/17PostNotes.pdf	1 Maxima and Minima 2 Recall from Calculus I: 1) Critical points (where f'(x) = 0 or DNE) are the candidates for where local min and max points can occur. 2) You can use the Second Derivative Test (SDT) to test whether a given critical point is a local min or max. SDT is not always conclusive. 3) Global max and min of a function on an interval can occur at a critical point in the interior of the interval or at the endpoints of the interval. 3 Extreme Values 1) f has a global maximum at a point (a,b) if f(a,b) ≥ f(x,y) for all (x,y) in the domain of f. f has a local maximum at a point (a,b) if f(a,b)) ≥ f(x,y) for all (x,y) near (a,b). 2) f has a global minimum at a point (a,b) if f(a,b) ≤ f(x,y) for all (x,y) in the domain of f. f has a local minimum at a point (a,b) if f(a,b)) ≤ f(x,y) for all (x,y) near (a,b). 4 Theorem (Critical Point) Let f be defined on a set S containing (a,b). If f (a,b) is an extreme value (max or min), then (a,b) must be a critical point, i.e. either (a,b) is a) a boundary point of S b) a stationary point of S (where ∇f (a,b) = 0, i.e. the tangent plane is horizontal) c) a singular point of S (where f is not differentiable). Fact: Critical points are candidate points for both global and local extrema. Theorem (Max-Min Existence) If f is continuous on a closed, bounded set S, then f attains both a global max value and a global min value there. ⇀ 5 Second Partials Test Theorem Suppose f(x,y) has continuous second partial derivatives in a neighborhood of (a,b) and ∇ f(a,b) = 0. Let D = D(a,b) = fxx(a,b)fyy(a,b) - fxy2(a,b) then 1) If D>0 and fxx(a,b)<0, then f(a,b) is a local max. 2) If D>0 and fxx(a,b)>0, then f(a,b) is a local min. 3) If D<0, then f(a,b) is not an extreme value. ((a,b) is a saddle point.) 4) If D = 0 the test is inconclusive. ⇀ 6 EX 1 For f(x,y) = xy2 - 6x2 - 3y2, find all critical points, indicating whether each is a local min, a local max or saddle point. 7 EX 2 Find the global max and min values for f(x,y) = x2 - y2 - 1 on S = {(x,y)\| x2 + y2 ≤ 1} 8 EX 3 Find the points where the global max and min occur for f(x,y) = x2 + y2 on S = {(x,y)\|x∈[-1,3], y∈[-1,4]}. 9 EX 4 Find the 3-D vector of length 9 with the largest possible sum of its components.
10070	https://math.stackexchange.com/questions/2578416/what-is-the-best-way-to-calculate-the-average-of-individual-averages	Skip to main content What is the "best" way to calculate the average of individual averages? Ask Question Asked Modified 7 years, 8 months ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. In preparing for GRE, I see questions looking for the mean of two or more individual means. Here is an example: The average (arithmetic mean) of 100 measurements is 24, and the average of 50 additional measurements is 28 ``` Quantity A Quantity B The average of the 150 measurements 26 ``` A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given. My question regards the best way to determine the mean (average). I have found two methods that always basically solve it correctly, although the results differ by ~ 0.1. One method is adding the weighted means: find sum of each sample, add the sums for a total sum, determine the proportion of each individual sum to the total sum to calculate the "weight", then add the products of each individual mean. In other words, Weighted mean = (proportion)(group A mean) + (proportion)(group B mean) + .... So, to solve the example question: sum of Group A measurements = 100⋅24=2400 sum of Group B measurements = 50⋅28=1400 total sum = 3800 proportion of Group A to total = 2400÷3800=0.631579 weighted mean of Group A = 24⋅0.631579=15.157896 proportion of Group B to total = 1400÷3800=0.368421 weighted mean of Group B to total = 28⋅0.368421=10.315788 The average (mean) of the 150 measurements = 15.157896 + 10.315788 = 25.473684 The other method is to just divide the total sum by the 150 measurements, which is how a normal mean is calculated, to equal 25.333333. Although both methods make B) the answer, they could be significantly different in other contexts. It seems to me that the first, weighted, method is better since it probably takes outliers, repeated measurement, etc. into account. What is the "best" way to calculate the average of individual averages? statistics average means Share CC BY-SA 3.0 Follow this question to receive notifications asked Dec 24, 2017 at 5:43 Matt LondonMatt London 1122 bronze badges Add a comment \| 2 Answers 2 Reset to default This answer is useful 2 Save this answer. Show activity on this post. Your second method is correct and the mean is 2513. The proportion in your first method should be the proportion of the measurements, not the proportion of the sum, so it would be 24⋅100150+28⋅50150=2513, which agrees with the other. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Dec 24, 2017 at 6:01 Ross MillikanRoss Millikan 383k2828 gold badges262262 silver badges472472 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. The best way might be this: with 100 sitting at 24 and 50 sitting at 28, you need the balance point between the two. The balance point is closer to 24 than to 28, because there’s more mass sitting there. Therefore, it’s less than 26. If you want to be more precise, there’s half as much mass at 28, so 28 is twice as far away from the balance. You need the point “one third” of the way from 24 to 28, so that’s 24+43=2513. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Dec 24, 2017 at 7:01 G Tony JacobsG Tony Jacobs 32.1k44 gold badges5757 silver badges115115 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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10071	https://miamioh.ecampus.com/logan-turners-diseases-nose-throat-ear/bk/9780429772450?srsltid=AfmBOop9-ruu7GFGJ39pc7meysECDThWXTouOjXaloPTQQI9K5IMT4cZ	Logan Turner's Diseases of the Nose, Throat and Ear Skip Navigation MIAMI UNIVERSITY OFFICIAL BOOKSTORE Login/Sign Up Home Miami Alumni Miami Athletics MENU Shop Textbooks Campus Locations Login/Sign Up Back\| Campus Locations Home Home Miami Alumni Miami Alumni Miami Athletics Miami Athletics Back\| Shop Clothing Clothing Accessories Accessories Gifts Gifts Graduation Graduation Supplies Supplies Back\| Clothing Kids Men Sweatshirts Women View All Back\| Accessories For You For Your Car For Your Home For Your Pet For Your Tech View All Back\| Gifts Artwork Cooking Essentials Games Gift Wraps Holiday Home Decor Mascot Office Decor Outdoor/Recreation View All Back\| Graduation Graduation Gear Graduation Gifts View All Back\| Supplies Art Supplies For Your Office Medical Supplies Office Supplies School Supplies View All Kids Back\| Kids Bibs Bottoms Dresses Headwear Hoodies Matching Sets Onesies Shirts Sweatshirts Men Back\| Men Bottoms Footwear Hoodies Jerseys Mens Apparel Outerwear Polos Shirts Sweatshirts T-Shirts Sweatshirts Back\| Sweatshirts Women Back\| Women Bottoms Dresses Headwear Hoodies Outerwear Pants Shirts Sweatshirts Undergarments View All Clothing > Miami Merger Myaamia Heritage Collection Alumni Collection Miami Regionals Custom Items Miami Cradle of Coaches collection For You Back\| For You Backpacks Bags Buttons Drinkware Fan Gear Flags Hair Accessories Headwear ID Holders Jewelry Keychains Knitwear Lanyards Lapel Pins Pennants Socks Ties Umbrellas Wallets For Your Car Back\| For Your Car Decals License Plates For Your Home Back\| For Your Home Banners Blankets Cooking Essentials Decals Drinkware Flags Frames Home Decor Magnets Pillows Signs Stickers For Your Pet Back\| For Your Pet Beds Bowls Charms Clothes Toys For Your Tech Back\| For Your Tech Computer Accessories Phone Accessories View All Accessories > Artwork Back\| Artwork Ornaments Wall Art Cooking Essentials Back\| Cooking Essentials Food Games Back\| Games Balls Puzzles Gift Wraps Back\| Gift Wraps Gift Bags Holiday Back\| Holiday Ornaments Stocking Home Decor Back\| Home Decor Candles Mascot Back\| Mascot Plushies Office Decor Back\| Office Decor Desk Accessories Outdoor/Recreation Back\| Outdoor/Recreation Tailgate View All Gifts > Graduation Gear Back\| Graduation Gear Caps and Gowns Hoods Stoles Study Abroad Sashes Tassel Tassels Graduation Gifts Back\| Graduation Gifts Diploma Frames Yard Signs View All Graduation > Art Supplies Back\| Art Supplies Art Supply Products For Your Office Back\| For Your Office Desk Accessories Pens Medical Supplies Back\| Medical Supplies Kits Office Supplies Back\| Office Supplies Padfolios School Supplies Back\| School Supplies Folders Notebooks Planners View All Supplies > Erin Condren Shopping Cart (0) Shop Textbooks Search Shopping Cart (0) Write a Review Logan Turner's Diseases of the Nose, Throat and Ear byMusheer Hussain, S. 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10072	https://www.khanacademy.org/science/ap-biology/cell-structure-and-function/facilitated-diffusion/v/sodium-potassium-pump-video	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
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10074	https://calcworkshop.com/limits/limits-at-infinity/	Limits At Infinity How To Solve Em w/ 9 Examples! // Last Updated: - Watch Video // Is it possible to determine what is happening to a function if we let x get really large or small? Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher) In other words, what’s the limit at infinity? What Is Infinity First, we need to understand that infinity is not a tangible value but an idea. It is used to represent a number that is greater than any real number. Largely (pun intended), infinity is used to explain the end behavior of a function that is either increasing or decreasing without bound. So, if infinity isn’t actually attainable, how can we approach it? Convergence (The Main Ingredient) First, we will need to understand the notion of convergence, which is the idea that the value of a function eventually becomes arbitrarily close to some number. For example, look at the graph below. As the x-value increases, what is happening to the y-value? 1 Divided by X — Graph As x gets bigger and bigger (approaching positive infinity), the y-values are getting smaller and smaller (approaching zero). And that’s the secret to limits at infinity, or as some textbooks say, limits approaching infinity. All you have to do is find the function’s end-behavior. Constant Over Infinity And this brings us to a cool idea — any number divided by a really big number is approximately zero! In fact, we have the following rule that states if n is a positive rational number and c is any real number, then: Theorem For Limits At Infinity Example For instance, let’s apply our new limit rule to evaluate the following limit. How To Solve Limits At Infinity See, not so bad! But here’s the really big question…why does this work? Horizontal Asymptotes Because as x gets larger and larger without bound, f(x) gets closer and closer a tangible number. Look back at the graph above of y = 1/x and notice that as x approaches infinity, f(x) approaches zero. This nicely shows that when x approaches infinity the y-value is approaching a horizontal asymptote! Which means horizontal asymptotes help us to determine the end behavior of a graph and is defined as follows: Horizontal Asymptote — Definition Therefore, if we can determine the horizontal asymptote of a function, then we can find the limit at infinity! Example In this question, we will evaluate the limit as x approaches infinity by finding the horizontal asymptote algebraically. Calculating the Limit at Infinity Now, I know that you’re probably thinking… …that seems like a lot of work. And you’re right. The Shortcut Thankfully there’s a shortcut to finding horizontal asymptotes and limits at infinity, as the table below nicely highlights. How To Find Horizontal Asymptotes Using Limits Sweet! Summary So, all we have to do is look for the degrees of the numerator and denominator, and we can evaluate limits approaching infinity as Khan Academy nicely confirms. Together we will look at nine examples, so you’ll know exactly how to handle these questions. In fact, we’re going to utilize this technique countless times throughout our journey of calculus, so get excited! Let’s get to it. Video Tutorial w/ Full Lesson & Detailed Examples (Video) Get access to all the courses and over 450 HD videos with your subscription Monthly and Yearly Plans Available Get My Subscription Now Still wondering if CalcWorkshop is right for you? Take a Tour and find out how a membership can take the struggle out of learning math. 4.9 / 5 1,194 Reviews slide 5 to 8 of 50 Perry R.04-15-25 - GA, United States Jenn thoroughly explains every move that she makes! She' s easy to understand and responds when emailed with questions! Jenn is saving my Differential Equations grade! William M.04-15-25 - GA, United States Calcworkshop was/is an excellent resource for anyone interested in or currently learning math. I started my Math degree at 26, spending absolutely 0 time doing math between 18 and then. I did not even know how to FOIL. The subscription got me up to speed pretty quick and then I used it for a few other courses. Couldn't recommend it enough Joey M.04-14-25 - AZ, United States They make the complex simple. They provide a systematic approach to ensure your success. Patrick04-09-25 - NJ, United States I went through the discrete math course prior to taking it at a university. Jen made difficult concepts easy to understand, leaving me well prepared for my Discrete college course. NB08-18-25 - ID, United States Very good courses with detailed videos and practice problems. It was helpful to have extra resources in calculus. Sabrina08-18-25 - CA, United States This course help prepare me for Calculus 2! Daily08-18-25 - OR, United States Calcworkshop.com for the win. I used this as a supplement for college level Pre-Calculus II, Calculus I, and Calculus II courses. The Calculus courses were during an accelerated 4-week summer term, which made them that much more difficult. Jenn's teaching ability is amazing! I was regularly able to impress my professors with my grasp of the content. They regularly commented on how much I was improving. For every topic that was covered in each class, Calcworkshop.com has the same content in an easy to digest video. U-Sub, Trig Sub, Diff-Equations, the list goes on. Best money I've spent in a very long time. Three classes supplemented with Calcworkshop.com so far, and three A letter grades! Austin08-08-25 - AZ, United States I was lost in an AP Calculus class and Jenn's explanations of things like limits and dedicates helped me out tremendously! The break downs of every lesson made it seem like I was learning basic algebra again, and allowed me to do so much better in Calculus. A Reviewer08-07-25 - CT, United States I have been so super confident ever since I encountered the Calcworksop. This platform was so convenient for me it enabled me to study effectively. Jenn is an absolute guru, I love her so much. She gets it! I do feel I come to tears with the amount of progress I've made with calculus. If anyone ever has any struggle with calculus, Calcworkshop will rescue you. And in a brief amount of time will you start to make progress. This website is a miracle TRUST ME. Calculus 1,2 and 3 can all be courses that you can get great grades out of from this website. And there are more courses you can do well in for mathematics. Again this website saved my life! Jenn Forever! A Reviewer08-04-25 - GA, United States Great online videos and super helpful prep for an 8 week college Calculus class. I felt well prepared and did very well in my course overall. Clara08-01-25 - United States CalcWorkshop was wonderful! My Honors Calculus IV professor was a wonderful personality to be around but was incredibly smart and only taught about the theory of the math... rather than with example problems. I would end up spending hours upon hours trying to complete three problems and I knew that I needed more examples to show different workflows. With the Jenns help, I ended up pulling through with a high B in this course, which is honestly my most prized grade that I worked so hard for in undergrad studies so far. Rosie T.07-30-25 - United States I was struggling with my online calculus class and was watching YouTube videos when I saw an ad for calcworkshop and was able to watch a couple videos and decided to get the subscription for the remainder of the semester. Jenn explains everything so well and is very thorough, easy to follow and I would remember some of the little phrases she would say that really helped. I highly recommend the site for your math classes. A Reviewer07-28-25 - CA, United States Outstanding... keep up the high quality content. Kerry K.07-22-25 - United States This program and Jenn especially are the ONLY reason I got through calc 1,2,3 and Differential equations. Better than any professor I've ever had Tom07-15-25 - VA, United States Extremely helpful! In the age of online classes this website delivers classic style lectures that really helped me. Zoe R.07-14-25 - NC, United States Calcworkshop really helped me gain a deeper understanding of my calculus curriculum. I kept feeling like I was 'almost there' with every unit we did in class but something was missing. Jenn really breaks down every topic in a way that makes sense and doesn't skip any steps. Calcworkshop definitely played a huge role in improving my calculus scores this year. I would highly recommend it to anyone who feels like they need a little extra help to be successful. Sean P.07-14-25 - SC, United States This is a great service for anyone who struggles with math. Jenn breaks everything down to basic levels so you can understand the whole process. For me I chose to go back to school in my 40's and many of the simple basic fundamentals had been forgotten and with her workshop I was able to recall them without taking a step back and losing traction in the process. Mike07-02-25 - FL, United States It's a great service working through example problems is a great way to learn. Dana06-29-25 - MN, United States Math is hard and math can be boring, so I really appreciate Jenn's enthusiasm in ALL of her videos. She explains concepts so well, points out the common mistakes, and goes through many examples. I am retired and learning math, which I didn't take in my adult years, so it's been really hard to learn on my own. But I want to learn physics and I will need calculus for that. I love how Jenn has organized her courses to focus on what will be needed for calculus and then the calculus courses, as well as linear algebra and differential equations. Julian06-27-25 - NC, United States Jenn and Calcworkshop are one of the best resources outside of school for mastering any math course. I went from failing Pre-Calculus to getting an A in Calculus 2. There's nowhere else you need to go. Nika06-26-25 - CA, United States Excellent explanations on every topic of precalculus and calculus. Great examples. Easy to understand. Thank you! A Reviewer06-23-25 - United States Literally taught me calc because my actual teacher didn't. Jen explains in a way that is simple and just makes sense. My calc teacher in high school taught super vaguely and just expected us to understand how to make connections right away. This website actually saved me from failing and helped a lot with studying for the AP test!! Lottie06-17-25 - AL, United States It was very helpful to have videos where the instructor did not assume that part of process was already known by the person watching. Maureen S.06-13-25 - NE, United States I used CalcWorkshop because I needed help with college calculus I required for entrance to a grad program I am pursuing. The instruction provided through my university was vague at best. As a full-time professional, wife, and mother of 5, it was difficult for me to go in for the tutoring sessions offered by my university. I love how Jenn provided the repetition I needed to solidify the processes in my brain and reviewed the algebra tricks while going through the comprehensive variety of problems in detail. I believe that anything can be learned if enough time is spent trying. The one-stop-shop in CalcWorkshop saved me the frustration of searching YouTube for examples and helped me make more efficient use of my time. In the course evaluation for my university, I mentioned that I used CalcWorkshop for support. I am likely to re-subscribe for myself and for my children. Thanks, again. Kevin M.06-09-25 - FL, United States This has been the best resource I have used in my whole academic career. I was struggling in my courses and this helped me go from barely passing my college classes to getting As. I highly recommend this to everyone. The calc workshop is a game changer and is just absolutely amazing. Thank you so much! Jacqueline06-06-25 - WA, United States My son used the Calculus videos as a supplement to his high school class. The explanations were clear, and helped him get over some of the hurdles he experienced in the class. Lindsay06-03-25 - FL, United States I have only been using this for about a week, and I have already greatly benefited from this resource. The videos are clear and straightforward, with just the right amount of background information and explanation. I love that I am able to access any course that I need in order to fill in any gaps in skill. I am not just granted access to one particular course. I also love that when I contacted Jenn about a question that I had, she reached out with clarity and specific videos that would help my situation. This was an amazing help!! I would not have known where to find them on my own, and they really helped me to understand concepts that I needed.I highly recommend this resource:) Joann L.06-03-25 - TX, United States I am very satisfied with format Calcworkshop has to offer. I highly recommend this site. After six months of reviewing sites and texts I decided to try Calcworkshop.I stopped searching because it met all my needs.I will take my grandchildren from PreAgebra to Calculus 1 through Calcworkshop. Joann from Salado, TX Andrea A.05-30-25 - United States Thanks to all of the support offered on your website, I as able to save a lot of tears. Truly one of the best sources for calculus aid. I ended up finished Calc 1-3 with A's. I thank you and your team for the amazing website you have put together! A Reviewer05-28-25 - United States This website has helped me understand the concepts better of step by step. Grace05-24-25 - MD, United States As a college student and math major, CalcWorkshop was the perfect resource for me as it cleared up what I didn't understand in lectures or even taught me the whole topic! Jenn really does go through every step and understands where students may be confused during lessons. I totally would recommend CalcWorkshop to ANYONE, no matter what math class their taking. Jenna05-24-25 - KS, United States I abolsutely love Jenn! She made calc easier to understand and I genuinely enjoyed her videos. Without Jenn, I never would have made it through calc II. I wish I had her for Calc I! CJ05-21-25 - NC, United States I was way out of my element in Calc II in college, and my professor was no help, and I was unsuccessfully trying to teach myself calculus from a textbook. I had failed my first test and was looking for tutors, but all of the ones I tried just gave me help on homework and did not teach the content to me. I fully credit CalcWorkshop for being able to pass Calc II and not having to retake it. The way things are explained is so clear and simple, and can even help someone who is self-proclaimed "bad at math". I would recommend this site to anyone struggling in calc!! Maya05-19-25 - United States I just passed the Calc 2 course at my local collage with an A as a junior in high school and couldn't have done it without Calcworkshop! Jenn gives tons of great examples and explains things so clearly and calmly without skipping over the "little unimportant steps" that would continuously trip me up while trying to learn the new material. The website is also very well designed and makes it very easy to find exactly what you are looking for. 10/10 RetiredMD05-15-25 - FL, United States My overall experience has been excellent. I am a retired physician now teaching. Mathematics has always been important to me. I have needed review from time to time and to have this as a source was extremely helpful. Jenn shows you the thinking done in solving problems and not just the mechanics. I most likely will renew in the future but I am pleased with the quality. WeezyMathGirl05-15-25 - GA, United States These videos are amazing. She explains everything so well and this was a lifesaver for Calculus BC for my daughter! Brenda05-12-25 - KY, United States absolutely amazing! Jose C.05-12-25 - FL, United States I am very grateful to have found this site and above all I am grateful for Jen's work. It is notorious the passion for teaching and doing good to the community that continues with the hope and dream of dedicating themselves to mathematics despite the fact that the educational system or teachers that we may encounter along the way are not the most cooperative. As a personal experience, I had a terrible year academically in terms of mathematics, actually no one in my class understood the teacher, I thought that this was no longer my thing, but this online course made me regain confidence in me and knowing that from now on I will have a tool that can accompany me if I have any doubt is a great relief. John05-10-25 - NM, United States Being in my thirties and returning to school to pursue a degree in Mechanical Engineering and hadn't taken a math class in over 25 years, Calc Workshop was the only reason I survived pre calc/trig, calc 1, and calc 2. Jenn breaks down every step of every type of problem into a logical and easy to remember process. This subscription has been worth every penny and I have recommended it to everyone I know taking math. I will be re-subscribing for calc 3 and differential equations next year. Thank you Jenn for such an amazing tool and helping me achieve success in these challenging courses, I couldn't have done it without you! Deathbyintegrals05-09-25 - United States Overall it served my intended purpose which was to get me through double, triple, and line integrals. I like how the site is organized and I absolutely appreciate the clear and concise video lessons. Should I need a math refresher or assistance in a future course I will absolutely be back. Dillan05-06-25 - UT, United States Her videos are very well made, you can tell she put a lot of work into how she introduces and walks you through all the material. I used her videos to help me with calculus 3. I just canceled my subscription because I wasn't in need of the information anymore and 30$ is a lot to pay if your not gonna watch the videos. I will probably re subscribe for my future math classes. Her videos are super good!Her website is well organized and easy to get around. I did not use any of her other features on her website I just used her posted videos from the website and I found that plenty sufficient for me to learn the material. Overall, I felt like 30$ a month was a little expensive but this is such a great product that I would 100% subscribe again! Amanda M.05-02-25 - AZ, United States Calcworkshop is an outstanding resource for anyone looking to strengthen their understanding of calculus. Jen's clear and approachable teaching style made even the most complex topics feel manageable. As a math teacher, I found her step-by-step explanations incredibly helpful in breaking down difficult concepts for my own students. The scaffolding she provides is thoughtfully designed, building confidence and deepening comprehension at every stage. Whether you're looking for a refresher or a fresh perspective on solving problems, I highly recommend Calcworkshop as a go-to tool for mastering calculus. Perry R.04-15-25 - GA, United States Jenn thoroughly explains every move that she makes! She' s easy to understand and responds when emailed with questions! Jenn is saving my Differential Equations grade! William M.04-15-25 - GA, United States Calcworkshop was/is an excellent resource for anyone interested in or currently learning math. I started my Math degree at 26, spending absolutely 0 time doing math between 18 and then. I did not even know how to FOIL. The subscription got me up to speed pretty quick and then I used it for a few other courses. Couldn't recommend it enough Joey M.04-14-25 - AZ, United States They make the complex simple. They provide a systematic approach to ensure your success. Patrick04-09-25 - NJ, United States I went through the discrete math course prior to taking it at a university. Jen made difficult concepts easy to understand, leaving me well prepared for my Discrete college course. NB08-18-25 - ID, United States Very good courses with detailed videos and practice problems. It was helpful to have extra resources in calculus. Sabrina08-18-25 - CA, United States This course help prepare me for Calculus 2! Daily08-18-25 - OR, United States Calcworkshop.com for the win. I used this as a supplement for college level Pre-Calculus II, Calculus I, and Calculus II courses. The Calculus courses were during an accelerated 4-week summer term, which made them that much more difficult. Jenn's teaching ability is amazing! I was regularly able to impress my professors with my grasp of the content. They regularly commented on how much I was improving. For every topic that was covered in each class, Calcworkshop.com has the same content in an easy to digest video. U-Sub, Trig Sub, Diff-Equations, the list goes on. Best money I've spent in a very long time. Three classes supplemented with Calcworkshop.com so far, and three A letter grades! Austin08-08-25 - AZ, United States I was lost in an AP Calculus class and Jenn's explanations of things like limits and dedicates helped me out tremendously! The break downs of every lesson made it seem like I was learning basic algebra again, and allowed me to do so much better in Calculus. See more reviews on Shopper Approved
10075	https://edu.wyoming.gov/wp-content/uploads/2024/01/Math-Ch.-10-WYCPS.pdf	2 s \| 1 023 Math W CPS with PLD Y 2023 Mathematics (Emended Feb. 2025) Wyoming Content & Performance Standards (WYCPS) Effective - July 17, 2024 To be Fully Implemented in Districts by the Beginning of School Year 2025-26 Rationale: Mathematics is the language that defines the blueprint of the universe. Mathematics is woven into all parts of our lives and is more than a list of skills to be mastered. The essence of mathematics is the ability to employ critical thinking and reasoning to solve problems. To be successful in mathematics, one must see mathematics as sensible, useful, and worthwhile. Organization of the Standards: Standard Code = Grade.Math Domain.Standard # Key: K.G.4 = Grade K.Domain Geometry (G).Standard 4 Domain: The domains are the core concepts to be studied in math. The Math Standards usually consist of 5-6 domains in each grade level. The math domains are listed below with the grade level(s) in which they are found. • Kindergarten – Counting & Cardinality (CC) • K-5 – Operations & Algebraic Thinking (OA) • K-5 – Number & Operations in Base Ten (NBT) • K-5 – Measurement & Data (MD) • K-HS – Geometry (G) • 3-5 – Number & Operations – Fractions (NF) • 6-7 – Ratios & Proportional Relationships (RP) • 8-12 – Functions (F) • 6-8 – Expressions & Equations (EE) • 6-8 – The Number System (NS) • 6-12 – Statistics & Probability (SP) • 9-12 – Number & Quantity (N) • 9-12 – Algebra (A) Notes for Accessibility: For best results—if using screen reader technology to access this document—adjust punctuation settings/ speech verbosity to read parentheses and other special characters aloud. 2023 M WYCPS ath with PLDs \| 2 Standards for Mathematical Practices: 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Standards for Mathematical Practice “The Standards for Mathematical Practice describe varieties of expertise that mathematics educators at all levels should seek to develop in their students. These practices rest on important “processes and proficiencies” with longstanding importance in mathematics education. Procedural knowledge alone will not prepare our 21st Century students to be globally competitive. Mathematical thinkers also visualize problems and recognize that multiple strategies may lead to a single solution. They realize mathematics is applicable outside of the classroom and are confident in their ability to apply mathematical concepts to all aspects of life. The Standards for Mathematical Practice cultivate mathematically literate and informed citizens. Using mathematics as a means of synthesizing complex concepts and making informed decisions is paramount to college and career success. The Standards for Mathematical Practice develop skills that serve students beyond the math classroom. 1. Make sense of problems and persevere in solving them. Mathematically proficient students start by explaining to themselves the meaning of a problem and looking for entry points to its solution. They analyze givens, constraints, relationships, and goals. They make conjectures about the form and meaning of the solution and plan a solution pathway rather than simply jumping into a solution attempt. They consider analogous problems, and try special cases and simpler forms of the original problem in order to gain insight into its solution. They monitor and evaluate their progress and change course if necessary. Older students might, depending on the context of the problem, transform algebraic expressions or change the viewing window on their graphing calculator to get the information they need. Mathematically proficient students can explain correspondences between equations, verbal descriptions, tables, and graphs, or draw diagrams of important features and relationships, graph data, and search for regularity or trends. Younger students might rely on using concrete objects or pictures to help conceptualize and solve a problem. Mathematically proficient students check their answers to problems using a different method, and they continually ask themselves, “Does this make sense?” They can understand the approaches of others to solving complex problems and identify correspondences between different approaches. 2023 Math YCPS with PLD W s \| 3 2. Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents —and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. 3. Construct viable arguments and critique the reasoning of others. Mathematically proficient students understand and use stated assumptions, definitions, and previously established results in constructing arguments. They make conjectures and build a logical progression of statements to explore the truth of their conjectures. They are able to analyze situations by breaking them into cases and can recognize and use counterexamples. They justify their conclusions, communicate them to others, and respond to the arguments of others. They reason inductively about data, making plausible arguments that take into account the context from which the data arose. Mathematically proficient students are also able to compare the effectiveness of two plausible arguments, distinguish correct logic or reasoning from that which is flawed, and—if there is a flaw in an argument—explain what it is. Elementary students can construct arguments using concrete referents such as objects, drawings, diagrams, and actions. Such arguments can make sense and be correct, even though they are not generalized or made formal until later grades. Later, students learn to determine domains to which an argument applies. Students at all grades can listen or read the arguments of others, decide whether they make sense, and ask useful questions to clarify or improve the arguments. 4. Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of interest depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose. 2023 M 4 ath WYC with PLDs \| PS 5. Use appropriate tools strategically. Mathematically proficient students consider the available tools when solving a mathematical problem. These tools might include pencil and paper, concrete models, a ruler, a protractor, a calculator, a spreadsheet, a computer algebra system, a statistical package, or dynamic geometry software. Proficient students are sufficiently familiar with tools appropriate for their grade or course to make sound decisions about when each of these tools might be helpful, recognizing both the insight to be gained and their limitations. For example, mathematically proficient high school students analyze graphs of functions and solutions generated using a graphing calculator. They detect possible errors by strategically using estimation and other mathematical knowledge. When making mathematical models, they know that technology can enable them to visualize the results of varying assumptions, explore consequences, and compare predictions with data. Mathematically proficient students at various grade levels are able to identify relevant external mathematical resources, such as digital content located on a website, and use them to pose or solve problems. They are able to use technological tools to explore and deepen their understanding of concepts. 6. Attend to precision. Mathematically proficient students try to communicate precisely to others. They try to use clear definitions in discussion with others and in their own reasoning. They state the meaning of the symbols they choose, including using the equal sign consistently and appropriately. They are careful about specifying units of measure, and labeling axes to clarify the correspondence with quantities in a problem. They calculate accurately and efficiently, expressing numerical answers with a degree of precision appropriate for the problem context. In the elementary grades, students give carefully formulated explanations to each other. By the time they reach high school they have learned to examine claims and make explicit use of definitions. 7. Look for and make use of structure. Mathematically proficient students look closely to discern a pattern or structure. Young students, for example, might notice that three and seven more is the same amount as seven and three more, or they may sort a collection of shapes according to how many sides the shapes have. Later, students will see 7 × 8 equals the well-remembered 7 × 5 + 7 × 3, in preparation for learning about the distributive property. In the expression x² + 9x + 14, older students can see the 14 as 2 × 7 and the 9 as 2 + 7. They recognize the significance of an existing line in a geometric figure and can use the strategy of drawing an auxiliary line for solving problems. They also can step back for an overview and shift perspective. They can see complicated things, such as some algebraic expressions, as single objects or as being composed of several objects. For example, they can see 5 − 3(x − y)² as 5 minus a positive number times a square and use that to realize that its value cannot be more than 5 for any real numbers x and y. 8. Look for and express regularity in repeated reasoning. Mathematically proficient students notice if calculations are repeated, and look both for general methods and for shortcuts. Upper elementary students might notice when dividing 25 by 11 that they are repeating the same calculations over and over again, and conclude they have a repeating decimal. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1, 2) with slope 3, middle school students might abstract the equation (y − 2) ⁄ (x − 1) = 3. Noticing the regularity in the way terms cancel when expanding (x − 1)(x + 1), (x − 1)(x² + x + 1), and (x − 1)(x³ + x² + x + 1) might lead them to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of their intermediate results. 2023 Ma s \| 5 th WYCP with PLD S Kindergarten Math Content & Performance Standards Counting and Cardinality Know number names and the count sequence. K.CC.1 K.CC.1a Count to 100 by ones and by tens. K.CC.1b Count backwards by ones from 20. • The Proficient student is able to: A1. Count to 100 by ones, starting at one. A2. Count to 100 by multiples of ten, starting at ten. B. Count backwards by ones from 20. K.CC.3 Write numbers from 0 to 20. Represent a number of objects with a written numeral from 0 to 20 [with a 0 (zero) representing a count of no objects]. • The Proficient student is able to: A. Write numbers from 0 to 20. B. Represent a number of objects with a written numeral from 0 to 20 [with 0 (zero representing a count of no objects]. Count to tell the number of objects. K.CC.4 Understand the relationship between numbers and quantities; connect counting to cardinality. K.CC.4a Use one-to-one correspondence when counting objects. K.CC.4b Understand that the last number name said, tells the number of objects counted regardless of their arrangement. K.CC.4c Understand that each successive number name refers to a quantity that is one more, and each previous number name refers to a quantity that is one less. • The Proficient student is able to count and tell the number of objects in a range from 10 to 39. A. Use one-to-one correspondence when counting objects. B. Understand that the last number name said, tells the number of objects counted regardless of their arrangement. C. Understand that each successive number name refers to a quantity that is one more, and each previous number name refers to a quantity that is one less. Operations and Algebraic Thinking Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from. K.OA.2 Solve word problems using objects and drawings to find sums up to 10 and differences within 10. • The Proficient student is able to solve word problems using objects and drawings to find sums up to 10 and differences within 10. K.OA.3 Decompose numbers less than or equal to 10 in more than one way. • The Proficient student is able to decompose numbers less than or equal to 10 in more than one way. K.OA.4 For any number from 1 to 9, find the number that makes 10 when added to the given number. • The Proficient student is able to for any number from 1 to 9, find the number that makes 10 when added to the given number. 2023 Math WYCPS Ds \| 6 wit PL h K.OA.5 Fluently add and subtract within 5. • The Proficient student is able to fluently add and subtract within 5. Measurement and Data Classify objects and count the number of objects in each category. K.MD.3 Classify objects into given categories; count the numbers of objects in each category and sort the categories by count. (Limit category counts to be less than or equal to 10.) • The Proficient student is able to classify objects into given categories; count the numbers of objects in each category and sort the categories by count. (Limit category counts to be less than or equal to 10.) Geometry Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). K.G.2 Correctly name shapes regardless of their orientations or overall size. • The Proficient student is able to correctly name shapes regardless of their orientations or overall size. Analyze, compare, create, and compose shapes. K.G.4 Analyze and compare two- and three-dimensional shapes, using informal language to describe their similarities, differences, and attributes. • The Proficient student is able to analyze and compare two- and three-dimensional shapes, using informal language to describe their similarities, differences, and attributes. K.G.6 Use simple shapes to compose squares, rectangles, and hexagons. • The Proficient student is able to use simple shapes to compose squares, rectangles, and hexagons. Grade 1 Math Content & Performance Standards Operations & Algebraic Thinking Represent and solve problems involving addition and subtraction. 1.OA.1 Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, by using objects, drawings, or equations with a symbol for the unknown number to represent the problem. • The Proficient student is able to use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, by using objects, drawings, or equations with a symbol for the unknown number to represent the problem. Add and subtract within 20. 1.OA.6 Add and subtract within 20, demonstrating fluency in addition and subtraction within 10. Use strategies such as counting on; making ten using the relationship between addition and subtraction. • The Proficient student is able to add and subtract within 20, demonstrating fluency in addition and subtraction within 10. Use strategies such as counting on; making ten using the relationship between addition and subtraction. 2023 Math WYC s \| 7 PS w PLD ith Work with addition and subtraction equations. 1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. • The Proficient student is able to understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. Number and Operations in Base Ten Extend the counting sequence. 1.NBT.1 Extend the number sequences to 120. In this range: 1.NBT.1a Count forward and backward, starting at any number less than 120. 1.NBT.1b Read numerals. 1.NBT.1c Write numerals. 1.NBT.1d Represent a number of objects with a written numeral. • The Proficient student is able to extend the number sequences to 120. In this range: A. Count forward and backward, starting at any number less than 120. B. Read numerals. C. Write numerals. D. Represent a number of objects with a written numeral. Understand place value. 1.NBT.2 Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: 1.NBT.2a 10 can be thought of as a bundle of ten ones — called a “ten. ” 1.NBT.2b The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. 1.NBT.2c The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). • The Proficient student is able to understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten. ” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). Use place value understanding and properties of operations to add and subtract. 1.NBT.4 Add within 100, using concrete models or drawings and strategies based on place value: 1.NBT.4a Including adding a two-digit number and a one-digit number. 1.NBT.4b Adding a two-digit number and a multiple of 10. 1.NBT.4c Understand that in adding two-digit numbers, adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten. 1.NBT.4d Relate the strategy to a written method and explain the reasoning used. • The Proficient student is able to add within 100, using concrete models or drawings and strategies based on place value: A. Including adding a two-digit number and a one-digit number. B. Adding a two-digit number and a multiple of 10. C. Understand that in adding two-digit numbers, adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten. D. Relate the strategy to a written method and explain the reasoning used. 2023 Math WYCP \| 8 S Ds with PL Measurement and Data Work with time and money. 1.MD.3a Tell and write time in hours and half-hours using analog and digital clocks. • The Proficient student is able to tell and write time in hours and half-hours using analog and digital clocks. 1.MD.3b Identify U.S. coins by value (pennies, nickels, dimes, quarters). • The Proficient student is able to identify U.S. coins by value (pennies, nickels, dimes, quarters). Geometry Reason with shapes and their attributes. 1.G.1 Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size); for a wide variety of shapes; build and draw shapes to possess defining attributes. • The Proficient student is able to distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size); for a wide variety of shapes; build and draw shapes to possess defining attributes. 1.G.3 Partition circles and rectangles into two and four equal shares and: 1.G.3a Describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. 1.G.3b Describe the whole as two of, or four of the shares. 1.G.3c Recognize that decomposing into more equal shares creates smaller shares. • The Proficient student is able to partition circles and rectangles into two and four equal shares and: A. Describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. B. Describe the whole as two of, or four of the shares. C. Recognize that decomposing into more equal shares creates smaller shares. Grade 2 Math Content & Performance Standards Operations & Algebraic Thinking Represent and solve problems involving addition and subtraction. 2.OA.1 Use addition and subtraction within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, by using drawings and equations with a symbol for the unknown number to represent the problem. • The Proficient student is able to use addition and subtraction within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, and taking apart, and comparing with unknowns in all positions, by using drawings and equations with a symbol for the unknown number to represent the problem. Add and subtract within 20. 2.OA.2 Fluently add and subtract within 20 using mental strategies. By end of Grade 2, know automatically all sums of two one-digit numbers based on strategies. • The Proficient student is able to fluently add and subtract within 20 using mental strategies. By end of Grade 2, know automatically all sums of two one-digit numbers based on strategies. 2023 Math W s \| 9 YCP h PLD S wit Number and Operations in Base Ten Understand place value. 2.NBT.1 Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones; and demonstrate that: 2.NBT.1a 100 can be thought of as a bundle of ten tens — called a “hundred. ” 2.NBT.1b The numbers 100, 200, 300, 400, 500, 600, 700, 800, 900 refer to one, two, three, four, five, six, seven, eight, or nine hundreds (and 0 tens and 0 ones). 2.NBT.1c Three-digit numbers can be decomposed in multiple ways (e.g., 524 can be decomposed as 5 hundreds, 2 tens and 4 ones or 4 hundreds, 12 tens, and 4 ones, etc.) • The Proficient student is able to understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones; and demonstrate that: A. 100 can be thought of as a bundle of ten tens — called a “hundred. ” B. The numbers 100, 200, 300, 400, 500, 600, 700, 800, 900 refer to one, two, three, four, five, six, seven, eight, or nine hundreds (and 0 tens and 0 ones). C. Three-digit numbers can be decomposed in multiple ways (e.g., 524 can be decomposed as 5 hundreds, 2 tens and 4 ones or 4 hundreds, 12 tens, and 4 ones, etc.) 2.NBT.4 Compare pairs of three-digit numbers based on meanings of the hundreds, tens, and ones digits, using the words “is greater than, ” “is equal to, ” “is less than, ” and with the symbols >, =, and < to record the results of comparisons. • The Proficient student is able to compare pairs of three-digit numbers based on meanings of the hundreds, tens, and ones digits, using the words “is greater than, ” “is equal to, ” “is less than” and with the symbols >, =, and < to record the results of comparisons. Use place value understanding and properties of operations to add and subtract. 2.NBT.5 Add and subtract within 100 using strategies based on place value, properties of addition, and/or the relationship between addition and subtraction. • The Proficient student is able to add and subtract within 100 using strategies based on place value, properties of addition, and/or the relationship between addition and subtraction. 2.NBT.7 Add and subtract within 1,000, using concrete models or drawings and strategies based on place value, properties of addition, and/or the relationship between addition and subtraction: 2.NBT.7a Relate the strategy to a written method and explain the reasoning used. 2.NBT.7b Understand that in adding or subtracting three-digit numbers, add or subtract hundreds and hundreds, tens and tens, ones and ones. 2.NBT.7c Understand that sometimes it is necessary to compose or decompose tens or hundreds. • The Proficient student is able to add and subtract within 1,000, using concrete models or drawings and strategies based on place value, properties of addition, and/or the relationship between addition and subtraction: A. Relate the strategy to a written method and explain the reasoning used. B. Understand that in adding or subtracting three-digit numbers, add or subtract hundreds and hundreds, tens and tens, ones and ones. C. Understand that sometimes it is necessary to compose or decompose tens or hundreds. 202 ith PLDs \| 10 3 Ma h WYCPS w t Measurement and Data Measure and estimate lengths in standard units. 2.MD.1 Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes. • The Proficient student is able to measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes. Work with time and money. 2.MD.7 Tell and write time from analog and digital clocks in five minute increments using a.m. and p.m. • The Proficient student is able to tell and write time from analog and digital clocks in five minute increments using a.m. and p.m. 2.MD.8 Solve word problems up to $10 involving dollar bills, quarters, dimes, nickels, and pennies, using $ (dollars) and ¢ (cents) symbols appropriately. • The Proficient student is able to solve word problems up to $10 involving dollar bills, quarters, dimes, nickels, and pennies, using $ (dollars) and ¢ (cents) symbols appropriately. Geometry Reason with shapes and their attributes. 2.G.2 Partition a rectangle into rows and columns of same-size squares and count to find the total number of them. • The Proficient student is able to partition a rectangle into rows and columns of same-size squares and count to find the total number of them. 2.G.3 Partition circles and rectangles into two, three, or four equal shares by: 2.G.3a Describing the shares using the words halves, thirds, half of, a third of, etc. 2.G.3b Describing the whole as two halves, three thirds, four fourths. 2.G.3c Recognizing that equal shares of identical wholes need not have the same shape. • The Proficient student is able to partition circles and rectangles into two, three, or four equal shares by: A. Describing the shares using the words halves, thirds, half of, a third of, etc. B. Describing the whole as two halves, three thirds, four fourths. C. Recognizing that equal shares of identical wholes need not have the same shape. 2023 Math W h PLDs \| 11 YCP wit S Grade 3 Math Content & Performance Standards Operations and Algebraic Thinking Multiply and divide within 100. 3.OA.7 Fluently multiply and divide with factors from 1 to 10 using mental strategies. By end of Grade 3, know automatically all products of one-digit factors based on strategies. • The Proficient student is able to fluently multiply and divide with factors from 1 to 10 using mental strategies. By end of Grade 3, know automatically all products of one-digit factors based on strategies. Solve problems involving the four operations, and identify and explain patterns in arithmetic. 3.OA.8 Solve two-step word problems (limited to the whole number system) using the four basic operations. Students should apply the Order of Operations when there are no parentheses to specify a particular order. 3.OA.8a Represent these problems using equations with a symbol standing for the unknown quantity. 3.OA.8b Assess the reasonableness of answers using mental computation and estimation strategies including rounding. • The Proficient student is able to solve two-step word problems (limited to the whole number system) using the four basic operations. Students should apply the Order of Operations when there are no parentheses to specify a particular order. A. Represent these problems using equations with a symbol standing for the unknown quantity. B. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Number and Operations in Base Ten Use place value understanding and properties of operations to perform multi-digit arithmetic (a range of algorithms may be used). 3.NBT.2 Fluently add and subtract within 1,000 using strategies and algorithms based on place value, properties of addition, and/or the relationship between addition and subtraction. • The Proficient student is able to fluently add and subtract within 1,000 using strategies and algorithms based on place value, properties of addition, and/or the relationship between addition and subtraction. Number and Operations - Fractions Develop understanding of fractions as numbers (limited to denominators 2, 3, 4, 6, and 8) (use horizontal fractions). 3.NF.1 Understand a fraction 1 ⁄ b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a ⁄ b as the quantity formed by a parts of size 1 ⁄ b. • The Proficient student is able to understand a fraction 1 ⁄ b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a ⁄ b as the quantity formed by a parts of size 1 ⁄ b. 3.NF.2 Understand and represent fractions on a number line diagram. 3.NF.2a Represent a fraction 1 ⁄ b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1 ⁄ b and that the endpoint of the part based at 0 locates the number 1 ⁄ b on the number line. 2023 Ma with PLDs \| 12 th W CPS Y 3.NF.2b Represent a fraction a ⁄ b on a number line diagram by marking off a lengths 1 ⁄ b from 0. Recognize that the resulting interval has size a ⁄ b and that its endpoint locates the number a ⁄ b on the number line. • The Proficient student is able to understand and represent fractions on a number line diagram. A. Represent a fraction 1 ⁄ b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1 ⁄ b and that the endpoint of the part based at 0 locates the number 1 ⁄ b on the number line. B. Represent a fraction a ⁄ b on a number line diagram by marking off a lengths 1 ⁄ b from 0. Recognize that the resulting interval has size a ⁄ b and that its endpoint locates the number a ⁄ b on the number line. 3.NF.3 Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size. 3.NF.3a Understand two fractions as equivalent if they are the same size, or the same point on a number line. 3.NF.3b Recognize and generate simple equivalent fractions. Explain why the fractions are equivalent. 3.NF.3c Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. 3.NF.3d Compare two fractions with the same numerator or the same denominator, by reasoning about their size. Recognize that valid comparisons rely on the two fractions referring to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions. • The Proficient student is able to explain equivalence of fractions in special cases, and compare fractions by reasoning about their size. A. Understand two fractions as equivalent if they are the same size, or the same point on a number line. B. Recognize and generate simple equivalent fractions. Explain why the fractions are equivalent. C. Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. D. Compare two fractions with the same numerator or the same denominator, by reasoning about their size. Recognize that valid comparisons rely on the two fractions referring to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions. Measurement and Data Represent and interpret data. 3.MD.4 Generate measurement data by measuring lengths using rulers marked with halves and fourths of an inch. Use the data to create a line plot, where the horizontal scale is marked off in appropriate units—whole numbers, halves, or quarters. • The Proficient student is able to generate measurement data by measuring lengths using rulers marked with halves and fourths of an inch. Use the data to create a line plot, where the horizontal scale is marked off in appropriate units—whole numbers, halves, or quarters. 2023 Ma WYCPS wit th h PLDs \| 13 Geometric measurement: understand concepts of area and relate area to multiplication and to addition. 3.MD.7 Relate area to the operations of multiplication and addition. 3.MD.7a Find the area of a rectangle with whole-number side lengths (dimensions) by multiplying them. Show that this area is the same as when counting unit squares. 3.MD.7b Multiply side lengths to find areas of rectangles with whole-number side lengths in the context of solving real-world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning. 3.MD.7c Use area models to represent the Distributive Property in mathematical reasoning. Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. • The Proficient student is able to relate area to the operations of multiplication and addition. A. Find the area of a rectangle with whole-number side lengths (dimensions) by multiplying them. Show that this area is the same as when counting unit squares. B. Multiply side lengths to find areas of rectangles with whole-number side lengths in the context of solving real world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning. C. Use area models to represent the Distributive Property in mathematical reasoning. Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Geometry Reason with shapes and their attributes. 3.G.1 Use attributes of quadrilaterals to classify rhombuses, rectangles, and squares. Understand that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories. • The Proficient student is able to use attributes of quadrilaterals to classify rhombuses, rectangles, and squares. Understand that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories. 2023 Math WYCPS with PLDs \| 14 Grade 4 Math Content & Performance Standards Operations and Algebraic Thinking Use the four operations with whole numbers to solve problems. 4.OA.3 Solve multi-step word problems posed with whole numbers, including problems in which remainders must be interpreted. 4.OA.3a Represent these problems using equations with a letter standing for the unknown quantity. 4.OA.3b Assess the reasonableness of answers using mental computation and estimation strategies including rounding. • The Proficient student is able to solve multi-step word problems posed with whole numbers, including problems in which remainders must be interpreted. A. Represent these problems using equations with a letter standing for the unknown quantity. B. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Number and Operations in Base Ten Generalize place value understanding for multi-digit whole numbers (limited to numbers less than or equal to 1,000,000). 4.NBT.2 Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols. • The Proficient student is able to read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols. 4.NBT.3 Use place value understanding to round multi-digit whole numbers to any place. • The Proficient student is able to use place value understanding to round multi-digit whole numbers up to 1,000,000 to any place. Use place value understanding and properties of operations to perform multi-digit arithmetic (limited to whole numbers less than or equal to 1,000,000). 4.NBT.5 Use strategies based on place value and the properties of multiplication to: 4.NBT.5a Multiply a whole number of up to four digits by a one-digit whole number. 4.NBT.5b Multiply a pair of two-digit numbers. 4.NBT.5c Use appropriate models to explain the calculation, such as by using equations, rectangular arrays, and/or area models. • The Proficient student is able to use strategies based on place value and the properties of multiplication to: A. Multiply a whole number of up to four digits by a one-digit whole number. B. Multiply a pair of two-digit numbers. C. Use appropriate models to explain the calculation, such as by using equations, rectangular arrays, ratio tables, or area models. 2023 Math WYCPS with PLDs \| 15 4.NBT.6 Use strategies based on place value, the properties of multiplication, and/or the relationship between multiplication and division to find quotients and remainders with up to four-digit dividends and one-digit divisors. Use appropriate models to explain the calculation, such as by using equations, rectangular arrays, and/or area models. • The Proficient student is able to use strategies based on place value, the properties of multiplication, and/or the relationship between multiplication and division to find quotients and remainders with up to four-digit dividends and one-digit divisors. Use appropriate models to explain the calculation, such as by using equations, rectangular arrays, ratio tables, or area models. Number and Operations - Fractions Extend understanding of fraction equivalence and ordering (limited to denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100). 4.NF.1 Explain why a fraction a ⁄ b is equivalent to a fraction n×a ⁄n×b by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. • The Proficient student is able to explain why a fraction a ⁄ b is equivalent to a fraction n×a ⁄ n×b by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. 4.NF.2 Compare two fractions with different numerators and different denominators by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1 ⁄ 2. 4.NF.2a Recognize that comparisons are valid only when the two fractions refer to the same whole. 4.NF.2b Record the results of comparisons with symbols >, =, or <. 4.NF.2c Justify the conclusions by using a visual fraction model. • The Proficient student is able to compare two fractions with different numerators and different denominators by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1 ⁄ 2. A. Recognize that comparisons are valid only when the two fractions refer to the same whole. B. Record the results of comparisons with symbols >, =, or <. C. Justify the conclusions by using a visual fraction model. Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers (limited to denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100). 4.NF.3 Understand a fraction a ⁄ b with a > 1 as a sum of unit fractions (1 ⁄ b). 4.NF.3a Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. 4.NF.3b a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions by using a visual fraction model. Decompose 4.NF.3c Add and subtract mixed numbers with like denominators by replacing each mixed number with an equivalent fraction, and/or by using properties of addition and the relationship between addition and subtraction. 2023 Ma with PLDs \| 16 th W CPS Y 4.NF.3d Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators. • The Proficient student is able to understand a fraction a ⁄ b with a > 1 as a sum of unit fractions (1 ⁄ b). A. Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. B. Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions by using a visual fraction model. C. Add and subtract mixed numbers with like denominators by replacing each mixed number with an equivalent fraction, and/or by using properties of addition and the relationship between addition and subtraction. D. Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators. Understand decimal notation for fractions, and compare decimal fractions. 4.NF.7 Compare and order decimal numbers to hundredths and justify by using concrete and visual models. Record the results of comparisons with the words “is greater than, ” “is equal to, ” “is less than, ” and with the symbols >, =, and <. • The Proficient student is able to compare and order decimal numbers to hundredths and justify by using concrete and visual models. Record the results of comparisons with the words “is greater than, ” “is equal to, ” “is less than, ” and with the symbols >, =, and <. Measurement and Data Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit. 4.MD.3 Apply the area and perimeter formulas for rectangles in real-world and mathematical problems. • The Proficient student is able to apply the area and perimeter formulas for rectangles in real-world and mathematical problems. Geometric measurement: understand concepts of angle and measure angles. 4.MD.7 Solve addition and subtraction problems to find unknown angles on a diagram in real-world and mathematical problems. • The Proficient student is able to solve addition and subtraction problems to find unknown angles on a diagram in real-world and mathematical problems. Geometry Draw and identify lines and angles, and classify shapes by properties of their lines and angles. 4.G.2 Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles. • The Proficient student is able to classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category and identify right triangles. 2023 Math WYCPS with PLDs \| 17 Grade 5 Math Content & Performance Standards Operations and Algebraic Thinking Write, interpret, and/or evaluate numerical expressions. 5.OA.1 Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols. • The Proficient student is able to use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols. 5.OA.2 Write simple expressions requiring parentheses that record calculations with numbers, and interpret numerical expressions without evaluating them. • The Proficient student is able to write simple expressions requiring parentheses that record calculations with numbers, and interpret numerical expressions without evaluating them. Number and Operations in Base Ten Understand the place value system. 5.NBT.1 Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1 ⁄ 10 of what it represents in the place to its left. • The Proficient student is able to recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1 ⁄ 10 of what it represents in the place to its left. 5.NBT.2 Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole number exponents to denote powers of 10. • The Proficient student is able to explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole number exponents to denote powers of 10. Perform operations with multi-digit whole numbers and with decimals to hundredths. 5.NBT.7 Add, subtract, multiply, and divide decimals to hundredths using concrete models or drawings, and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. • The Proficient student is able to add, subtract, multiply, and divide decimals to hundredths using concrete models or drawings, and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Number and Operations - Fractions Use equivalent fractions as a strategy to add and subtract fractions. 5.NF.2 Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. • The Proficient student is able to solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. 2023 Math WYCPS with PLDs \| 18 Apply and extend previous understandings of multiplication and division to multiply and divide fractions. 5.NF.3 Interpret a fraction as division of the numerator by the denominator (a ⁄ b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers by using visual fraction models or equations to represent the problem. • The Proficient student is able to interpret a fraction as division of the numerator by the denominator (a ⁄ b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers by using visual fraction models or equations to represent the problem. 5.NF.6 Solve real-world problems involving multiplication of fractions and mixed numbers by using visual fraction models or equations to represent the problem. • The Proficient student is able to solve real-world problems involving multiplication of fractions and mixed numbers by using visual fraction models or equations to represent the problem. 5.NF.7 Extend the concept of division to divide unit fractions and whole numbers by using visual fraction models and equations. 5.NF.7a Interpret division of a unit fraction by a non-zero whole number and compute the quotient. 5.NF.7b Interpret division of a whole number by a unit fraction and compute the quotient. 5.NF.7c Solve real-world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions by using visual fraction models and equations to represent the problem. • The Proficient student is able to extend the concept of division to divide unit fractions and whole numbers by using visual fraction models and equations. A. Interpret division of a unit fraction by a non-zero whole number and compute the quotient. B. Interpret division of a whole number by a unit fraction and compute the quotient. C. Solve real-world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions by using visual fraction models and equations to represent the problem. Measurement and Data Geometric measurement: understand concepts of volume and relate volume to multiplication and to addition. 5.MD.5 Relate volume to the operations of multiplication and solve real-world and mathematical problems involving volume. 5.MD.5a Find the volume of a right rectangular prism with whole number dimensions by multiplying them. Show that this volume is the same as when counting unit cubes. 5.MD.5b Find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real-world and mathematical problems given the formulas V=(l)(w)(h) and V=(B)(h) for rectangular prisms. • The Proficient student is able to relate volume to the operations of multiplication and solve real-world and mathematical problems involving volume. A. Find the volume of a right rectangular prism with whole number dimensions by multiplying them. Show that this volume is the same as when counting unit cubes. B. Find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real-world and mathematical problems given the formulas V=(l)(w)(h) and V=(B)(h) for rectangular prisms. 2023 Math WYCPS with PLDs \| 19 Geometry Graph points on the coordinate plane to solve real-world and mathematical problems. 5.G.2 Plot and interpret points in the first quadrant of the coordinate plane to represent real-world and mathematical situations. • The Proficient student is able to plot and interpret points in the first quadrant of the coordinate plane to represent real-world and mathematical situations. Grade 6 Math Content & Performance Standards Ratios and Proportional Relationships Understand ratio concepts and use ratio reasoning to solve problems. 6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems. 6.RP.3a Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. 6.RP.3b Solve unit rate problems including those involving unit pricing and constant speed. 6.RP.3c Understand that a percentage is a rate per 100 and use this to solve problems involving wholes, parts, and percentages. 6.RP.3d Use ratio reasoning to convert measurement units; convert units appropriately when multiplying or dividing quantities. • The Proficient student is able to: A. Make tables of equivalent ratios relating quantities with whole number measurements and plot the pairs of values on the coordinate plane. Use tables to compare ratios. B. Solve unit rate problems with whole number measurements including those involving unit pricing and constant speed. C. In mathematical and real-world contexts solve one-step problems involving wholes, parts, and percentages. D. Use ratio reasoning to convert measurement units and to transform units appropriately when multiplying or dividing quantities in one-step problems. The Number System Apply and extend previous understandings of multiplication and division to divide fractions by fractions. 6.NS.1 Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions by using visual fraction models and equations to represent the problem. • The Proficient student is able to interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions by using visual fraction models and equations to represent the problem. Compute fluently with multi-digit numbers and find common factors and multiples. 6.NS.3 Add, subtract, multiply, and divide manageable multi-digit decimals using efficient and generalizable procedures including, but not limited to, the standard algorithm for each operation. • The Proficient student is able to add, subtract, multiply, and divide multi-digit decimals using efficient and generalizable procedures including, but not limited to, the standard algorithm for each operation. 2023 Math WYCPS with PLDs \| 20 Apply and extend previous understandings of numbers to the system of rational numbers. 6.NS.7 Understand ordering and absolute value of rational numbers. 6.NS.7a Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. 6.NS.7b Write, interpret, and explain statements of order for rational numbers in real-world contexts. 6.NS.7c Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. 6.NS.7d Distinguish comparisons of absolute value from statements about order. • The Proficient student is able to understand ordering and absolute value of rational numbers. A. Interpret statements of inequality as statements about the relative position of two numbers on a (vertical or horizontal) number line diagram. B. Write, interpret, and explain statements of order for rational numbers in real-world contexts. C. Describe the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. D. Distinguish comparisons of absolute value from statements about order. 6.NS.8 Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Find distances between points with the same first coordinate or the same second coordinate; relate absolute value and distance. • The Proficient student is able to solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Find distances between points with the same first coordinate or the same second coordinate; relate absolute value and distance. Expressions and Equations Apply and extend previous understandings of arithmetic to algebraic expressions. 6.EE.2 Write, read, and evaluate expressions in which letters stand for numbers. 6.EE.2a Write expressions that record operations with numbers and with letters standing for numbers. 6.EE.2b Identify parts of an expression using mathematical terms (sum, difference, term, product, factor, quotient, coefficient, constant). 6.EE.2c Use Order of Operations to evaluate algebraic expressions using positive rational numbers and whole-number exponents. Include expressions that arise from formulas in real-world problems. • The Proficient student is able to write, read, and evaluate expressions in which letters stand for numbers. A. Write two-step algebraic expressions. B. Identify parts of an expression using mathematical terms (sum, difference, term, product, factor, quotient, coefficient, constant). C. Use Order of Operations to evaluate algebraic expressions using positive rational numbers and whole-number exponents. Include expressions that arise from formulas relative to sixth grade standards in real-world problems. 6.EE.3 Apply the properties of operations to generate equivalent expressions. • The Proficient student is able to apply the properties of operations to generate equivalent expressions (Commutative Property, Associative Property, Distributive Property, Additive Identity Property, Multiplicative Identity Property, and Zero Product Property). 2023 Math WYCPS with PLDs \| 21 Reason about and solve one-variable equations and inequalities. 6.EE.6 Use variables to represent unknown numbers and write expressions when solving a real-world or mathematical problem. • The Proficient student is able to use variables to represent unknown numbers and write expressions to represent real-world or mathematical problems. 6.EE.7 Write and solve real-world and mathematical problems in the form of one-step, linear equations involving non negative rational numbers. • The Proficient student is able to solve problems in both real-world and mathematical contexts by writing and solving equations in the form of one-step, linear equations involving non negative rational numbers. 6.EE.8 Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams. • The Proficient student is able to write an inequality of the form x > c or x < c (with the variable on either side) to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities have infinitely many solutions and represent solutions by graphing on a number line. Geometry Solve real-world and mathematical problems involving area, surface area, and volume. 6.G.1 Find area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems. • The Proficient student is able to find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems. 6.G.4 Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures in the context of solving real-world and mathematical problems. • The Proficient student is able to represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of right triangular prisms, right rectangular prisms, and right rectangular pyramids (given lateral height) in the context of solving real-world and mathematical problems. Statistics and Probability Summarize and describe distributions. 6.SP.4 Display numerical data in plots on a number line, including dot plots, stem-and-leaf plots, histograms, and box plots. • The Proficient student is able to display numerical data in plots on a number line, including dot plots, stem-and-leaf plots, histograms, and box plots. 6.SP.5 Summarize numerical data sets in relation to their real-world context. 6.SP.5a Report the sample size. 6.SP.5b Describe the context of the data under investigation, including how it was measured and its units of measurement. 2023 Math WYCPS with PLDs \| 22 6.SP.5c Find quantitative measures of center (median, mode and mean) and variability (range and interquartile range). Describe any overall pattern (including outliers, clusters, and distribution), with reference to the context in which the data was gathered. 6.SP.5d Justify the choice of measures of center (median, mode, or mean) based on the shape of the data distribution and the context in which the data was gathered. • The Proficient student is able to summarize numerical data sets in relation to their real-world context. A. Report the sample size. B. Describe the context of the data under investigation, including how it was measured and its units of measurement. C. Find quantitative measures of center (median, mode, and mean) and variability (range and interquartile range). Describe any overall pattern (including outliers, clusters, and distribution), with reference to the context in which the data was gathered. D. Justify the choice of measures of center (median, mode, or mean) based on the shape of the data distribution and the context in which the data was gathered. Grade 7 Math Content & Performance Standards Ratios and Proportional Relationships Analyze proportional relationships and use them to solve real-world and mathematical problems. 7.RP.2 Recognize and represent proportional relationships between quantities. 7.RP.2a Decide whether two quantities in a table or graph are in a proportional relationship. 7.RP.2b Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. 7.RP.2c Represent proportional relationships with equations. 7.RP.2d Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate. • The Proficient student is able to recognize and represent proportional relationships between quantities. A. Decide whether two quantities in a table or graph are in a proportional relationship. B. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. C. Represent proportional relationships with equations. D. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate. 7.RP.3 Solve multi-step real-world and mathematical problems involving ratios and percentages. • The Proficient student is able to solve multi-step real-world and mathematical problems involving ratios and percentages (e.g., simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error). 2023 Math WYCPS w 23 ith PL \| Ds The Number System Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers. 7.NS.3 Solve real-world and mathematical problems involving the four arithmetic operations with rational numbers. (Computations with rational numbers extend the rules for manipulating fractions to complex fractions.) • The Proficient student is able to solve real-world and mathematical problems involving the four arithmetic operations with rational numbers. (Computations with rational numbers extend the rules for manipulating fractions to complex fractions.) Expressions and Equations Use properties of operations to generate equivalent expressions. 7.EE.1 Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients. • The Proficient student is able to add, subtract, factor, and expand linear expressions with rational coefficients. Solve real-life and mathematical problems using numerical and algebraic expressions and equations. 7.EE.4 Apply the concepts of linear equations and inequalities in one variable to real-world and mathematical situations. 7.EE.4a Write and fluently solve linear equations of the form ax +b = c and a(x + b) = c where a, b, and c are rational numbers. 7.EE.4b Write and solve multi-step linear equations that include the use of the Distributive Property and combining like terms. Exclude equations that contain variables on both sides. 7.EE.4c Write and solve two-step linear inequalities. Graph the solution set on a number line and interpret its meaning. 7.EE.4d Identify and justify the steps for solving multi-step linear equations and two-step linear inequalities. • The Proficient student is able to apply the concepts of linear equations and inequalities in one variable to real-world and mathematical situations. A. Write and fluently solve linear equations of the form ax + b = c and a(x + b) = c where a, b, and c are rational numbers. B. Write and solve multi-step linear equations that include the use of the Distributive Property and combining like terms. Exclude equations that contain variables on both sides. C. Write and solve two-step linear inequalities. Graph the solution set on a number line and interpret its meaning. D. Identify and justify the steps for solving multi-step linear equations and two-step linear inequalities. Geometry Solve real-life and mathematical problems involving angle measure, area, surface area, and volume. 7.G.4 Investigate the concept of circles. 7.G.4a Demonstrate an understanding of the proportional relationships between diameter, radius, and circumference of a circle. 7.G.4b Understand that π is defined by the constant of proportionality between the circumference and diameter. 2023 Math WYCPS with PLDs \| 24 7.G.4c Given the formulas for circumference and area of circles, solve real-world and mathematical problems. • The Proficient student is able to investigate the concept of circles. A. Demonstrate an understanding of the proportional relationships between diameter, radius, and circumference of a circle. B. Understand that π is defined by the constant of proportionality between the circumference and diameter. C. Given the formulas for circumference and area of circles, solve real-world and mathematical problems. 7.G.5 Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure. • The Proficient student is able to use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure. 7.G.6 Solve real-world and mathematical problems involving: 7.G.6a Area and surface area of objects composed of triangles and quadrilaterals; 7.G.6b Volume of objects composed only of right prisms having triangular or quadrilateral bases. • The Proficient student is able to solve real-world and mathematical problems involving: A. Area and surface area of objects composed of triangles and quadrilaterals; B. Volume of objects composed only of right prisms having triangular or quadrilateral bases. Statistics and Probability Use random sampling to draw inferences about a population. 7.SP.1 Solve real-world and mathematical problems involving: 7.SP.1a Understand that a sample is a subset of a population. 7.SP.1b Differentiate between random and non-random sampling. 7.SP.1c Understand that generalizations from a sample are valid only if the sample is representative of the population. 7.SP.1d Understand that random sampling is used to gather a representative sample and tends to support valid inferences about the population. • The Proficient student is able to solve real-world and mathematical problems involving: A. Describing a sample that is a subset of a population. B. Differentiating between random and non-random sampling. C. Determining if a generalization is valid by justifying whether or not the sample is representative of the population. D. Determining if inferences about the population are valid based on how the given sample was collected. Draw informal comparative inferences about two populations. 7.SP.4 Given measures of center and variability (mean, median and/or mode; range, interquartile range, and/or standard deviation), for numerical data from random samples, draw appropriate informal comparative inferences about two populations. • The Proficient student is able to given measures of center and variability (mean, median, and/or mode; range, interquartile range, and/or standard deviation), for numerical data from random samples, draw appropriate informal comparative inferences about two populations. 2023 Math WYCPS with PLDs \| 2 5 Investigate chance processes and develop, use, and evaluate probability models. 7.SP.5 Find and interpret the probability of a random event. Understand that the probability of a random event is a number between, and including, 0 and 1 that expresses the likelihood of the event occurring. • The Proficient student is able to find and interpret the probability of a random event. Understand that the probability of a random event is a number between, and including, 0 and 1 that expresses the likelihood of the event occurring. 7.SP.6 Collect multiple samples to compare the relationship between theoretical and experimental probabilities for simple events. • The Proficient student is able to collect multiple samples to compare the relationship between theoretical and experimental probabilities for simple events. Grade 8 Math Content & Performance Standards The Number System Know that there are numbers that are not rational, and approximate them by rational numbers. 8.NS.1 Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number. Explore the real number system and its appropriate usage in real-world situations. 8.NS.1a Make comparisons between rational and irrational numbers. 8.NS.1b Understand that all real numbers have a decimal expansion. 8.NS.1c Model the hierarchy of the real number system, including natural, whole, integer, rational, and irrational numbers. 8.NS.1d Convert repeating decimals to fractions. • The Proficient student is able to know that numbers that are not rational are called irrational. Show that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number. Explore the real number system and its appropriate usage in real-world situations. A. Make comparisons between rational and irrational numbers. B. Show that real numbers (excluding irrational numbers) have a decimal expansion. C. Model the hierarchy of the real number system, including natural, whole, integer, rational, and irrational numbers. D. Convert repeating decimals to fractions. Expressions and Equations Understand the connections between proportional relationships, lines, and linear equations. 8.EE.5 Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. • The Proficient student is able to graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. 2023 Math WYCPS with PLDs \| 26 Analyze and solve linear equations and pairs of simultaneous linear equations. 8.EE.7 Extend concepts of linear equations and inequalities in one variable to more complex multi-step equations and inequalities in real-world and mathematical situations. 8.EE.7a Solve linear equations and inequalities with rational number coefficients that include the use of the Distributive Property, combining like terms, and variable terms on both sides. 8.EE.7b Recognize the three types of solutions to linear equations: one solution, infinitely many solutions, or no solutions. • The Proficient student is able to extend concepts of linear equations and inequalities in one variable to more complex multi-step equations and inequalities in real-world and mathematical situations. A. Solve linear equations and inequalities with rational number coefficients that include the use of the Distributive Property, combining like terms, and variable terms on both sides. B. Recognize the three types of solutions to linear equations: one solution, infinitely many solutions, or no solutions. 8.EE.8 Analyze and solve pairs of simultaneous linear equations. 8.EE.8a Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously. 8.EE.8b Solve systems of two linear equations in two variables with integer solutions by graphing the equations. 8.EE.8c Solve simple real-world and mathematical problems leading to two linear equations in two variables given y = mx + b form with integer solutions. • The Proficient student is able to analyze and solve a system of linear equations. A. Show that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously, including systems with one, infinitely many, and no solutions. B. Solve systems of two linear equations in two variables with integer solutions by graphing the equations. C. Solve simple real-world and mathematical problems leading to two linear equations in two variables given y = mx + b form with integer solutions. Functions Define, evaluate, and compare functions. 8.F.2 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). • The Proficient student is able to compare properties (intercepts, domain, and range) of two linear functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). Use functions to model relationships between quantities. 8.F.4 Apply the concepts of linear functions to real-world and mathematical situations. 8.F.4a Understand that the slope is the constant rate of change and the y-intercept is the point where x = 0. 8.F.4b Determine the slope and the y-intercept of a linear function given multiple representations, including two points, tables, graphs, equations, and verbal descriptions. 2023 Math WYCPS with PLDs \| 27 8.F.4c Construct a function in slope-intercept form that models a linear relationship between two quantities. 8.F.4d Interpret the meaning of the slope and the y-intercept of a linear function in the context of the situation. • The Proficient student is able to apply the concepts of linear functions to real-world and mathematical situations. A. Recognize that the slope is the constant rate of change and the y-intercept is the point where x = 0 from an equation, graph, table, and verbal description. B. Determine the slope and the y-intercept of a linear function given multiple representations, including two points, tables, graphs, equations, and verbal descriptions. C. Construct a function in slope-intercept form that models a linear relationship between two quantities. D. Interpret the meaning of the slope and the y-intercept of a linear function in the context of the situation. Geometry Understand congruence and similarity using physical models, transparencies, or geometry software. 8.G.2 Recognize through visual comparison that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them. • The Proficient student is able to recognize through visual comparison that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of at most two transformations (rotations, reflections, and translations); given two congruent figures, describe a sequence of at most two transformations that exhibits the congruence between them. 8.G.5 Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. • The Proficient student is able to use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. Understand and apply the Pythagorean Theorem. 8.G.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems. • The Proficient student is able to apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems. 8.G.8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. • The Proficient student is able to apply the Pythagorean Theorem to find the distance between two points in a coordinate system. 2023 Math WYCPS with PLDs \| 28 Statistics and Probability Investigate patterns of association in bivariate data. 8.SP.3 Use an equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. • The Proficient student is able to interpret the slope and y-intercept in the context of the bivariate measurement data when given a scatter plot with a line of best fit and an equation. 8.SP.4 Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. 8.SP.4a Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. 8.SP.4b Use relative frequencies calculated for rows or columns to describe possible association between the two variables. • The Proficient student is able to understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. A. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. B. Use relative frequencies calculated for rows or columns to describe possible association between the two variables. High School Math Content & Performance Standards Number and Quantity The Real Number System Extend the properties of exponents to rational exponents. N.RN.1 Explain how the meaning of the definition of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. • The Proficient student is able to explain and use the meaning of rational exponents in terms of properties of integer exponents and use proper notation for radicals in terms of rational exponents. N.RN.2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. • The Proficient student is able to rewrite expressions involving radicals and rational exponents, using the properties of exponents. Quantities Reason quantitatively and use units to solve problems. N.Q.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; and choose and interpret the scale and the origin in graphs and data displays. • The Proficient student is able to use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; and choose and interpret the scale and the origin in graphs and data displays. 2023 Math WYCPS with PLDs \| 29 The Complex Number System Use complex numbers in polynomial identities and equations. N.CN.7 Solve quadratic equations with real coefficients that have complex solutions. • The Proficient student is able to solve quadratic equations with real coefficients that have complex solutions. Algebra Seeing Structure in Expressions Write expressions in equivalent forms to solve problems. A.SSE.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. A.SSE.3a Factor a quadratic expression to reveal the zeros of the function it defines. • The Proficient student is able to choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. A. Factor a quadratic expression to reveal the zeros of the function it defines. Arithmetic with Polynomials and Rational Expressions Perform arithmetic operations on polynomials. A.APR.1 Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. • The Proficient student is able to understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. Understand the relationship between zeros and factors of polynomials. A.APR.3 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. • The Proficient student is able to identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. Creating Equations Create equations that describe numbers or relationships. A.CED.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. • The Proficient student is able to create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. A.CED.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. • The Proficient student is able to create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. 2023 Math WYCPS with PLDs \| 30 A.CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. • The Proficient student is able to represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. Reasoning with Equations and Inequalities Understand solving equations as a process of reasoning and explain the reasoning. A.REI.2 Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise. • The Proficient student is able to solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise. Solve equations and inequalities in one variable. A.REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. • The Proficient student is able to solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. A.REI.4 Solve quadratic equations in one variable. A.REI.4b Solve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b • The Proficient student is able to solve quadratic equations in one variable. B. Solve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. Solve systems of equations. A.REI.6 Estimate solutions graphically and determine algebraic solutions to linear systems, focusing on pairs of linear equations in two variables. • The Proficient student is able to estimate solutions graphically and determine algebraic solutions to linear systems, focusing on pairs of linear equations in two variables. A.REI.7 Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. • The Proficient student is able to solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. Functions Interpreting Functions Understand the concept of a function and use function notation. F.IF.1 Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). • The Proficient student is able to demonstrate that a function’s domain is assigned to exactly one element of the range in equations, tables, graphs, and context. 2023 Math WYCPS with PLDs \| 3 1 F.IF.2 Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. • The Proficient student is able to use function notation, evaluate functions for inputs in their domain, and interpret statements that use function notation in terms of a context. Interpret functions that arise in applications in terms of the context. F.IF.4 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. • The Proficient student is able to, for a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. Analyze functions using different representations. F.IF.7 Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. F.IF.7a Graph linear and quadratic functions and show intercepts, maxima, and minima. F.IF.7b Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. F.IF.7c Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. F.IF.7e Graph exponential and logarithmic functions, showing intercepts and end behavior. • The Proficient student is able to graph linear, quadratic, and exponential functions expressed symbolically and show appropriate key features of the graph showing intercepts, maxima, and minima, and end behavior. A. Graph linear and quadratic functions and show intercepts, maxima, and minima. B. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. C. Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. E. Graph exponential and logarithmic functions, showing intercepts and end behavior. F.IF.9 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). • The Proficient student is able to compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). Building Functions Build a function that models a relationship between two quantities. F.BF.1 Write a function that describes a relationship between two quantities. F.BF.1a Determine an explicit expression, a recursive process, or steps for calculation from a context. • The Proficient student is able to write a function that describes a relationship between two quantities. A. Determine an explicit expression, a recursive process, or steps for calculation from a context. 2023 Math WYCPS with PLDs \| 32 Build new functions from existing functions. F.BF.3 Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them. • The Proficient student is able to identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them. F.BF.4 Find inverse functions. F.BF.4a Write an expression for the inverse of a simple, invertible function f(x). Understand that an inverse function can be obtained by expressing the dependent variable of one function as the independent variable of another, as f and g are inverse functions, if and only if, f(x) = y and g(y) = x, for all values of x in the domain of f and all values of y in the domain of g. • The Proficient student is able to find inverse functions. A. Write an expression for the inverse of a simple, invertible function f(x). Understand that an inverse function can be obtained by expressing the dependent variable of one function as the independent variable of another, as f and g are inverse functions, if and only if, f(x) = y and g(y) = x, for all values of x in the domain of f and all values of y in the domain of g. Linear, Quadratic, and Exponential Models Construct and compare linear, quadratic, and exponential models and solve problems. F.LE.1 Distinguish between situations that can be modeled with linear functions and with exponential functions. F.LE.1a Verify that linear functions grow by equal differences over equal intervals and that exponential functions grow by equal factors over equal intervals. F.LE.1b Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. F.LE.1c Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another. • The Proficient student is able to distinguish between situations that can be modeled with linear functions and with exponential functions. A. Verify that linear functions grow by equal differences over equal intervals and that exponential functions grow by equal factors over equal intervals. B. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. C. Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another. F.LE.2 Construct linear and exponential functions using a graph, a description of a relationship, or two input-output pairs (include reading these from a table). • The Proficient student is able to construct linear and exponential functions using a graph, a description of a relationship, or two input-output pairs (include reading these from a table). 2023 Math WYCPS with PLDs \| 33 Geometry Congruence Experiment with transformations in the plane. G.CO.3 Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. • The Proficient student is able to, given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. Understand congruence in terms of rigid motions. G.CO.8 Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions. • The Proficient student is able to explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions. Prove geometric theorems. G.CO.9 Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. • The Proficient student is able to prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. G.CO.10 Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180 degrees; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. • The Proficient student is able to prove theorems about triangles. Theorems include: measure of interior angles of a triangle sum to 180 degrees; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. Similarity, Right Triangles, and Trigonometry Prove theorems involving similarity. G.SRT.5 Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. • The Proficient student is able to use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. Define trigonometric ratios and solve problems involving right triangles. G.SRT.8 Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. • The Proficient student is able to use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. 2023 Math WYCPS with PLDs \| 34 Circles Find arc lengths and areas of sectors of circles. G.C.5 Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector. • The Proficient student is able to derive, using similarity, the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector. Expressing Geometric Properties with Equations Use coordinates to prove simple geometric theorems algebraically. G.GPE.5 Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point). • The Proficient student is able to prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point). Geometric Measurement and Dimension Explain volume formulas and use them to solve problems. G.GMD.3 Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. • The Proficient student is able to use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. Statistics and Probability Interpreting Categorical and Quantitative Data Summarize, represent, and interpret data on a single count or measurement variable. S.ID.2 Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. • The Proficient student is able to use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. Summarize, represent, and interpret data on two categorical and quantitative variables. S.ID.6 Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. S.ID.6a Use a function to describe data trends to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear, quadratic, and exponential models. • The Proficient student is able to represent data on two quantitative variables on a scatter plot, and describe how the variables are related. - Use a function to describe data trends to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear, quadratic, and exponential models. 2023 Math WYCPS with PLDs \| 35 Interpret linear models. S.ID.7 Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. • The Proficient student is able to interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. S.ID.9 Distinguish between correlation and causation. • The Proficient student is able to distinguish between correlation and causation. Conditional Probability and the Rules of Probability Understand independence and conditional probability and use them to interpret data. S.CP.1 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or, ” “and, ” “not”). • The Proficient student is able to describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or, ” “and, ” “not”).
10076	https://stattrek.com/random-variable/independence	Stat Trek Teach yourself statistics Stat Trek Teach yourself statistics Stat Trek Independent Random Variables When a study involves pairs of random variables, it is often useful to know whether or not the random variables are independent. This lesson explains how to assess the independence of random variables. Independence of Random Variables If two random variables, X and Y, are independent, they satisfy the following conditions. P(x\|y) = P(x), for all values of X and Y. P(x ∩ y) = P(x) P(y), for all values of X and Y. The above conditions are equivalent. If either one is met, the other condition is also met; and X and Y are independent. If either condition is not met, X and Y are dependent. Note: If X and Y are independent, then the correlation between X and Y is equal to zero. Joint Probability Distributions The table below shows the joint probability distribution between two discrete random variables - X and Y. \| X \| \| 0 \| 1 \| 2 \| \| Y \| 3 \| 0.1 \| 0.2 \| 0.2 \| \| 4 \| 0.1 \| 0.2 \| 0.2 \| In a joint probability distribution table, numbers in the cells of the table represent the probability that particular values of X and Y occur together. From this table, you can see that the probability that X=0 and Y=3 is 0.1; the probability that X=1 and Y=3 is 0.2; and so on. You can use tables like this to figure out whether two discrete random variables are independent or dependent. Problem 1 below shows how. Test Your Understanding Problem 1 The table below shows the joint probability distribution between two random variables - X and Y. \| X \| \| 0 \| 1 \| 2 \| \| Y \| 3 \| 0.1 \| 0.2 \| 0.2 \| \| 4 \| 0.1 \| 0.2 \| 0.2 \| And the next table shows the joint probability distribution between two random variables - A and B. \| A \| \| 0 \| 1 \| 2 \| \| B \| 3 \| 0.1 \| 0.2 \| 0.2 \| \| 4 \| 0.2 \| 0.2 \| 0.1 \| Which of the following statements are true? I. X and Y are independent random variables. II. A and B are independent random variables. (A) I only (B) II only (C) I and II (D) Neither statement is true. (E) It is not possible to answer this question, based on the information given. Solution The correct answer is A. The solution requires several computations to test the independence of random variables. Those computations are shown below. X and Y are independent if P(x\|y) = P(x), for all values of X and Y. From the probability distribution table, we know the following: P(x=0) = 0.2; P(x=0 \| y=3) = 0.2; P(x=0 \| y = 4) = 0.2 P(x=1) = 0.4; P(x=1 \| y=3) = 0.4; P(x=1 \| y = 4) = 0.4 P(x=2) = 0.4; P(x=2 \| y=3) = 0.4; P(x=2 \| y = 4) = 0.4 Thus, P(x\|y) = P(x), for all values of X and Y, which means that X and Y are independent. We repeat the same analysis to test the independence of A and B. P(a=0) = 0.3; P(a=0 \| b=3) = 0.2; P(a=0 \| b = 4) = 0.4 P(a=1) = 0.4; P(a=1 \| b=3) = 0.4; P(a=1 \| b = 4) = 0.4 P(a=2) = 0.3; P(a=2 \| b=3) = 0.4; P(a=2 \| b = 4) = 0.2 Thus, P(a\|b) is not equal to P(a), for all values of A and B. For example, P(a=0) = 0.3; but P(a=0 \| b=3) = 0.2. This means that A and B are not independent. Last lesson Next lesson
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[x] Non vendere né condividere le mie informazioni personali Rifiuta Accetta selezionati Personalizza Accetta tutti +39 0331 968130/1info@brancaidealair.it it inglese francese spagnolo LOGIN HOME PRODOTTI CONDIZIONAMENTO DI PRECISIONE STRUMENTI TESSILI SATURAZIONE ADIABATICA IDRONICO IMPEDENZA ACUSTICA HVAC DATALOGGING PER LABORATORI SOLUZIONI PER LABORATORI Laboratori TESSILE Laboratori PELLAME CUOIO CALZATURE Laboratori CLASSIFICAZIONE COTONE Laboratori METROLOGIA Laboratori CARTA, CARTONE Laboratori CEMENTI, COLLE, ADESIVI Laboratori prove INCHIOSTRI Laboratori per le POLVERI SOTTILI ASSISTENZA TECNICA SUPPORTO TECNICO SUPPORTO REMOTO BLOG Contatti Questionario Laboratorio Cart 0 Home>blog>Necessità di avere un controllo temperatura costante in Metrologia? Contattaci! Menu blog Categoria Analisi Tessile Condizionamento di precisione Archivio giugno 2024 ottobre 2022 luglio 2022 aprile 2022 aprile 2021 luglio 2020 maggio 2020 aprile 2020 agosto 2017 gennaio 2016 ottobre 2013 settembre 2013 febbraio 2013 Necessità di avere un controllo temperatura costante in Metrologia? Contattaci! 29 08 2017 scritto da STA branca idealair in Condizionamento di precisione Necessità di ottenere un controllo di temperatura costante in Metrologia? Contattaci subito! Il controllo costante e preciso della temperatura è l'elemento fondamentale per il settore della Metrologia dimensionale come dichiarato dallo Standard Internazionale ISO1 Branca Idealair offre soluzioni di qualità per il controllo di precisione della temperatura e umidità relativa nei laboratori e sale metrologiche Un controllo climatico ambientale non è assolutamente un lavoro complesso se eseguito con criteri di progettazione e installazione giusti, ma al contrario può diventare un'incubo molto difficile da risolvere se progettato male e si tenta di metterci una pezza per riparare errori ai quali non si può tornare indietro. NESSUNA PAURA! CI SIAMO NOI. S.T.A. BRANCA IDEALAIR può fornire tutta l'esperienza e know-how di progettazione e consulenza tecnica su come muoversi in fase di progetto considerando tutti gli aspetti di fondamentale importanza e soprattutto senza incorrere in rischi irreparabili per il bene delle attività svolte all'interno del laboratorio di metrologia e misurazioni di precisione. Nella maggior parte dei laboratori è necessario ottenere un controllo di temperatura di: Laboratori di metrologia dimensionale hanno operato con temperature di (20± 2)°C e in casi particolari con mantenimento più stretto delle condizioni di controllo temperatura di (20± 1)°C. Sarebbe meglio avere una miglior stabilizzazione della temperature rispetto al raggiungimento del valore di temperatura ideale. Spieghiamoci meglio.. Ciò che è di eguale importanza è il profilo delle temperature in tutto il laboratorio L'impiego corretto di diffusori d'aria diffonderà solamente l'aria trattata attorno al laboratorio per non avere getti d'aria su uno spazio di lavoro specifico. Il controllo di umidità relativa è di minor importanza, ma deve essere mantenuto costantemente all'interno delle soglie di tolleranza di (45 ± 10)%, ma soprattutto non deve superare la soglia superiore di tolleranza del +5%. Di buon costume, gli ambienti controllati in ambito metrologica necessitano di due stadi di accesso al laboratorio attraverso una bussola con sistema di interblocco porte che permetta di ridurre al minimo le variazioni termiche che potrebbe causare un ambiente esterno al laboratorio e magari non condizionato. Il primo fattore favorevole è che un impianto di condizionamento sia in grado di fornire Comfort all'interno dell'ambiente di lavoro, e in secondo luogo che quest'ultimo in base ad una corretta progettazione dell'impianto riesca a tenere sotto controllo i gradienti termici spaziali e temporali. Questo significa che il solo impianto di condizionamento di precisione del laboratorio si occupi esclusivamente di garantire lievi variazioni termiche all'interno della struttura assicurando contenute fluttuazioni nonostante condizioni climatiche critiche all'esterno. In alcuni casi non è necessariamente richiesta una precamera di accesso al laboratorio, ma il laboratorio deve garantire un valore positivo di sovrapressione e tramite anche l'impiego di un filtro assoluto deve garantire che polvere e batteri vengano intercettati all'esterno e all'interno del laboratorio. L'alimentazione elettrica e l'illuminazione richiedono un studio di progetto elettrico ed illuminotecnico necessario alla fase di sviluppo di un laboratorio metrologico. L'illuminazione di un laboratorio richiede in vari casi 500Lux sulle postazioni di lavoro e in altri 300 Lux nelle aree di stoccaggio pezzi e strumentazione. Alcuni tra i laboratori in Metrologia adottano una combinazione di luci fluorescenti e incandescenti per far sì che se si sta effettuando una misurazione sensibile al rumore elettrico, possono spegnere le luci fluorescenti. Quest'ultime generano una grande quantità di rumore elettrico. QUALI SONO I PRINCIPALI CASI DELLE APPLICAZIONI METROLOGICHE DOVE IL CONTROLLO DEI GRADIENTI DI TEMPERATURA RISULTA INEVITABILE? METROLOGIA DELLE MASSE: Tipo 1 Temperatura ambiente:(23 ± 2)°C per ambienti di lavoro di medio/alto livello o (23 ± 5)°C per attività generiche Temperatura:(20-25)°C +/-1°C Velocità massima del gradiente di temperatura:0,5 °C/ora Umidità relativa (%):(40-60)% +/-5% ogni 4 ore Sovrapressione:10 Pa Classe di filtrazione:E1 e E2 METROLOGIA DELLE MASSE: Tipo 2 Temperatura:(20-25) +/-2 °C Velocità massima del gradiente di temperatura:1.0 °C/ora Umidità relativa (%):(40-60) +/- 15% ogni 10 ore Sovrapressione:10 Pa Classe di filtrazione:F1 e F2 METROLOGIA DELLE MASSE: Tipo 3 Temperatura:(18-27)°C +/-2°C Velocità massima del gradiente di temperatura:1,0 °C/ora Umidità relativa (%):(40-60) +/- 15% ogni 10 ore Sovrapressione:Pressione atmosferica normale Classe di filtrazione:M1, M2, M3 La strumentazione richiesta per misurare la pressione barometrica dell'aria e una sonda di controllo termoigrometrico. Livello di calibrazione Pressione barometrica Temperatura Umidità relativa Classe E1 e E2+/- 65 Pa (+/-0.5 mmHg)+/-0.1°C+/- 5% Classe F1 e F2+/- 135 Pa (+/-1.0 mmHg)+/-0.5 °C+/10 % Classe M1, M2, M3 Il laboratorio metrologico deve avere documentate le condizioni di temperatura tramite datalogger nel lungo periodo. METROLOGIA DEI VOLUMI Le condizioni ambientali richieste nei laboratori di Metrologia per l'analisi dei volumi fanno riferimento a quelli indicati per i laboratori di Metrologia delle Masse. METROLOGIA DELLE PRESSIONI Le condizioni di temperatura e umidità relativa nei laboratori di calibrazione metrologica dove essere mantenuta alle seguenti condizioni seguenti: Temperatura:23 +/- 2 °C Umidità relativa:50% +/- 10% Usando condizionatori d'aria a parete e umidificatori. La pressione barometrica all'interno del laboratorio dev'essere registrata costantemente. Il valore dell'accelerazione locale dovuta alla gravità (g) locale deve essere misurato e conosciuto con la migliore incertezza di misura possibile. S.T.A. Branca Idealair offre tre diverse soluzioni per il controllo di temperatura per SALE METROLOGICHE: SCOPRI DI PIU' Tags: Laboratori metrologici,Laboratori di taratura,ISO 1,Elementi Climatici Branca Idealair,Laboratorio Metrologico,Controllo temperatura,Misure dimensionali,Gradiente termico,Metrology,BRANCA IDEALAIR,laboratory for metrology,metrological room,need for constant temperature,laboratori metrologia,metrologia masse,massima gradiente temperatura,temperatura costante metrologia,controllo temperatura costante,metrologia dimensionale,metrology masses type,rate temperature gradient,temperature gradient,air conditioning,constant temperature control,temperature close control,metrology temperature control,metrological room project design,environmentally controlled metrological room STA Branca Idealair Mission Pubblicazioni ISO 9001:2015 Referenze Clienti Modulo intervento tecnico Parti di ricambio Politica della qualità Network Utility Privacy Sitemap Social LinkedIn © 2025 - STA BRANCA IDEALAIR di Branca Barbara & C. s.a.s. 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10078	https://en.wikipedia.org/wiki/Dividing_a_square_into_similar_rectangles	Jump to content Dividing a square into similar rectangles Español Edit links From Wikipedia, the free encyclopedia Mathematical problem Dividing a square into similar rectangles (or, equivalently, tiling a square with similar rectangles) is a problem in mathematics. Three rectangles [edit] See also: Plastic ratio There is only one way (up to rotation and reflection) to divide a square into two similar rectangles. However, there are three distinct ways of partitioning a square into three similar rectangles: The trivial solution given by three congruent rectangles with aspect ratio 3:1. The solution in which two of the three rectangles are congruent and the third one has twice the side length of the other two, where the rectangles have aspect ratio 3:2. The solution in which the three rectangles are all of different sizes and where they have aspect ratio ρ2, where ρ is the plastic ratio. The fact that a rectangle of aspect ratio ρ2 can be used for dissections of a square into similar rectangles is equivalent to an algebraic property of the number ρ2 related to the Routh–Hurwitz theorem: all of its conjugates have positive real part. Generalization to n rectangles [edit] In 2022, the mathematician John Baez brought the problem of generalizing this problem to n rectangles to the attention of the Mathstodon online mathematics community. The problem has two parts: what aspect ratios are possible, and how many different solutions are there for a given n. Frieling and Rinne had previously published a result in 1994 that states that the aspect ratio of rectangles in these dissections must be an algebraic number and that each of its conjugates must have a positive real part. However, their proof was not a constructive proof. Numerous participants have attacked the problem of finding individual dissections using exhaustive computer search of possible solutions. One approach is to exhaustively enumerate possible coarse-grained placements of rectangles, then convert these to candidate topologies of connected rectangles. Given the topology of a potential solution, the determination of the rectangle's aspect ratio can then trivially be expressed as a set of simultaneous equations, thus either determining the solution exactly, or eliminating it from possibility. The numbers of distinct valid dissections for different values of n, for n = 1, 2, 3, ..., are: 1, 1, 3, 11, 51, 245, 1372, 8522, ... (sequence A359146 in the OEIS). See also [edit] Squaring the square References [edit] ^ Ian Stewart, A Guide to Computer Dating (Feedback), Scientific American, Vol. 275, No. 5, November 1996, p. 118 ^ Spinadel, Vera W. de; Redondo Buitrago, Antonia (2009), "Towards Van der Laan's Plastic Number in the Plane" (PDF), Journal for Geometry and Graphics, 13 (2): 163–175. ^ Jump up to: a b Freiling, C.; Rinne, D. (1994), "Tiling a square with similar rectangles", Mathematical Research Letters, 1 (5): 547–558, doi:10.4310/MRL.1994.v1.n5.a3, MR 1295549 ^ Laczkovich, M.; Szekeres, G. (1995), "Tilings of the square with similar rectangles", Discrete & Computational Geometry, 13 (3–4): 569–572, doi:10.1007/BF02574063, MR 1318796 ^ Baez, John (2022-12-22). "Dividing a Square into Similar Rectangles". golem.ph.utexas.edu. Retrieved 2023-03-09. ^ "John Carlos Baez (@johncarlosbaez@mathstodon.xyz)". Mathstodon. 2022-12-15. Retrieved 2023-03-09. ^ Jump up to: a b Roberts, Siobhan (2023-02-07). "The Quest to Find Rectangles in a Square". The New York Times. ISSN 0362-4331. Retrieved 2023-03-09. ^ "cutting squares into similar rectangles using a computer program". ianhenderson.org. Retrieved 2023-03-09. ^ Baez, John Carlos (2023-03-06). "Dividing a Square into 7 Similar Rectangles". Azimuth. Retrieved 2023-03-09. External links [edit] Python code for dissection of a square into n similar rectangles via "guillotine cuts" by Rahul Narain \| v t e Tessellation \| \| \| --- \| \| Periodic \| \| --- \| \| Pythagorean Rhombille Schwarz triangle Rectangle + Domino Uniform tiling and honeycomb + Coloring + Convex + Kisrhombille Wallpaper group Wythoff \| \| \| \| \| \| \| Aperiodic \| \| --- \| \| Ammann–Beenker Aperiodic set of prototiles + List Einstein problem + Socolar–Taylor Gilbert Penrose Pinwheel Quaquaversal Rep-tile and Self-tiling + Sphinx Socolar Truchet \| \| \| \| \| \| \| \| --- \| \| Anisohedral and Isohedral Architectonic and catoptric Circle Limit III Computer graphics Honeycomb Isotoxal List Packing Pentagonal Problems + Domino - Wang + Heesch's + Squaring + Dividing a square into similar rectangles Prototile + Conway criterion + Girih Regular Division of the Plane Regular grid Substitution Voronoi Voderberg \| \| \| \| \| \| By vertex type \| \| --- \| \| \| \| \| --- \| \| Spherical \| 2n 33.n V33.n 42.n V42.n \| \| Regular \| 2∞ 36 44 63 \| \| Semi- regular \| 32.4.3.4 V32.4.3.4 33.42 33.∞ 34.6 V34.6 3.4.6.4 (3.6)2 3.122 42.∞ 4.6.12 4.82 \| \| Hyper- bolic \| 32.4.3.5 32.4.3.6 32.4.3.7 32.4.3.8 32.4.3.∞ 32.5.3.5 32.5.3.6 32.6.3.6 32.6.3.8 32.7.3.7 32.8.3.8 33.4.3.4 32.∞.3.∞ 34.7 34.8 34.∞ 35.4 37 38 3∞ (3.4)3 (3.4)4 3.4.62.4 3.4.7.4 3.4.8.4 3.4.∞.4 3.6.4.6 (3.7)2 (3.8)2 3.142 3.162 3.∞2 42.5.4 42.6.4 42.7.4 42.8.4 42.∞.4 45 46 47 48 4∞ (4.5)2 (4.6)2 4.6.12 4.6.14 V4.6.14 4.6.16 V4.6.16 4.6.∞ (4.7)2 (4.8)2 4.8.10 V4.8.10 4.8.12 4.8.14 4.8.16 4.8.∞ 4.102 4.10.12 4.122 4.12.16 4.142 4.162 4.∞2 54 55 56 5∞ 5.4.6.4 (5.6)2 5.82 5.102 5.122 64 65 66 68 6.4.8.4 (6.8)2 6.82 6.102 6.122 6.162 73 74 77 7.62 7.82 7.142 83 84 86 88 8.62 8.122 8.162 ∞3 ∞4 ∞5 ∞∞ ∞.62 ∞.82 \| \| \| \| \| Retrieved from " Categories: Rectangular subdivisions Mathematical problems Recreational mathematics Hidden categories: Articles with short description Short description is different from Wikidata
10079	https://www.youtube.com/watch?v=q3sJr4xJm74	Volume of a Frustum - Geometry Mr. Allen Math 53900 subscribers 8 likes Description 700 views Posted: 9 May 2022 Finding the volume of a frustum by subtracting the small cones volume from the large cones volume. Similarity is involved. 2 comments Transcript: what's popping dogs mr allen here about to find the volume of a frustum what the frustum is that it's that lampshade looking thing down there at the bottom and basically what it is is it's we're taking a big old cone and we're chopping the top off which is why i have that all dashed in there all right so our overall formula for the volume of a frustum is i'm going to find the volume of the big cone and i'm going to subtract the volume of that small cone okay so we're going to keep that in mind that means i need to find my volume of the big cone and the volume of the small cone well what kind of information do we have here i've got a radius of the big cone is 10 meters the radius of the small cone is 6 meters and i know that the distance between the two centers is eight meters okay well i need the height of both the bitcoin and the small cone in order to figure this thing out i don't have either how long am i going to get that well it looks to me that i actually have a triangle situation going on here as well as going right across here i can get myself similar triangles due to the fact that i have a right angle here i got another right angle here right those things are going to have to be that way if it's going to be cone okay and then i've got this angle is congruent to itself up here so angle angle similarity tells me these two triangles are similar so i have 6 here i've got 8 here i've got 10 here and if i were to find this x here that's the height of my small cone it's also the height of the big cone when i add 8 to it so i'm in the money if i can get that that dimension right there so if i set up x divided by 6 that's my height of that small triangle over its base is equal to the height of the big triangle which is x plus 8 over its base and when i cross multiply i have 10x equals 6x plus 48 i'm going to subtract 6x and get 4x equals 48 so x is equal to 12. and i'm gonna go ahead and erase that pop a 12 right here oh baby we good we are all sorts of good all right now it's time to find the volume of our big cone and subtract our small cone from it so the volume of my big cone that's going to be let's see here we've got our base big b right that's going to be 10 squared pi that's because it's a circle multiplied by my height which is 12 plus 8 which is 20 divided by 3. so i've got 100 times 20 which is 2 000 pi divided by 3. okay i'm going to leave that divided by 3 for now i'm not going to try and simplify things you're getting that's more like that i'm just going to leave that now i'm going to find the volume my small cone which i change to a green color right so the volume of that small cone is going to be let's see here pi r squared my big b my area of my base is going to be 6 squared pi times my height which is 12 divided by 3. again i'm going to leave this over 3 for now so that's going to end up being 36 36 times 12 that's going to give me 432 pi divided by 3. so the reason why i'm leaving them over 3 even though the one on the right can be simplified but the one on the left can't is because if i'm going to subtract these two if i'm going to subtract these two to get my frustum volume i'm going to need a common denominator anyways so i'm going to have the 2000 pi over 3 minus my 432 pi divided by 3 and i'm going to end up with a final answer of 1568 1568 pi over three and we got meters cubed what if we can divide that by three let's see let's see if we can divide that by three nope it's gonna be repeating decimal so we dunseys we dunzies that's the frustum that's what i'm talking about hope you guys enjoyed it american freedom rock roll costco rubber duck jenny on the gram yo-ro you only round once if you are plugging into that calculator you guys have an awesome day deuces
10080	https://www.govinfo.gov/content/pkg/GOVPUB-HE20-PURL-gpo64719/pdf/GOVPUB-HE20-PURL-gpo64719.pdf	Diet for Kidney Stone Prevention National Kidney and Urologic Diseases Information Clearinghouse How does diet affect the risk of developing kidney stones? Kidney stones can form when substances in the urine—such as calcium, oxalate, and phosphorus—become highly concentrated. The body uses food for energy and tissue repair. After the body uses what it needs, waste products in the bloodstream are carried to the kidneys and excreted as urine. Diet is one of several factors that can promote or inhibit kidney stone formation. Certain foods may promote stone formation in people who are susceptible, but scientists do not believe that eating any specific food causes stones to form in people who are not susceptible. Other factors that affect kidney stone formation include genes, environment, body weight, and fluid intake. More information about kidney stones can be found in the National Kidney and Urologic Diseases Information Clearinghouse fact sheets Kidney Stones in Adults and Kidney Stones in Children at www.kidney.niddk.nih.gov. What are the types of kidney stones? Four major types of kidney stones can form: • Calcium stones are the most common type of kidney stone and occur in two major forms: calcium oxalate and calcium phosphate. Calcium oxalate stones are more common. Calcium oxalate stone formation may be caused by high calcium and high oxalate excretion. Calcium phosphate stones are caused by the combination of high urine calcium and alkaline urine, meaning the urine has a high pH. • Uric acid stones form when the urine is persistently acidic. A diet rich in purines—substances found in animal protein such as meats, fish, and shellfish—may increase uric acid in urine. If uric acid becomes concentrated in the urine, it can settle and form a stone by itself or along with calcium. • Struvite stones result from kidney infections. Eliminating infected stones from the urinary tract and staying infection-free can prevent more struvite stones. • Cystine stones result from a genetic disorder that causes cystine to leak through the kidneys and into the urine, forming crystals that tend to accumulate into stones. Why is knowing which type of kidney stone a person has important? The first step in preventing kidney stones is to understand what is causing the stones to form. This information helps the health care provider suggest diet changes to prevent future kidney stones. For example, limiting oxalate in the diet may help prevent calcium oxalate stones but will do nothing to prevent uric acid stones. Some dietary recommendations may apply to more than one type of stone. Most notably, drinking enough fluids helps prevent all kinds of kidney stones by keeping urine diluted and flushing away materials that might form stones. How does a health care provider determine the type of kidney stone? If a person can catch a kidney stone as it passes, it can be sent to a lab for analysis. Stones that are causing symptoms can be retrieved surgically or with a scope inserted through the urethra into the bladder or ureter, then sent to a lab for analysis. Blood and urine can also be tested for unusual levels of chemicals such as calcium, oxalate, and sodium to help determine what type of kidney stone a person may have had. Dietary Changes to Help Prevent Kidney Stones People can help prevent kidney stones by making changes in fluid intake and, depending on the type of kidney stone, changes in consumption of sodium, animal protein, calcium, and oxalate. Drinking enough fluids each day is the best way to help prevent most types of kidney stones. Health care providers recommend that a person drink 2 to 3 liters of fluid a day. People with cystine stones may need to drink even more. Though water is best, other fluids may also help prevent kidney stones, such as citrus drinks. Recommendations based on the specific type of kidney stone include the following: Calcium Oxalate Stones • reducing sodium • reducing animal protein, such as meat, eggs, and fish • getting enough calcium from food or taking calcium supplements with food • avoiding foods high in oxalate, such as spinach, rhubarb, nuts, and wheat bran Calcium Phosphate Stones • reducing sodium • reducing animal protein • getting enough calcium from food or taking calcium supplements with food Uric Acid Stones • limiting animal protein 2 Diet for Kidney Stone Prevention How much fluid should a person drink to prevent kidney stone formation? People who have had a kidney stone should drink enough water and other fluids to produce at least 2 liters of urine a day. People who have had cystine stones may need to drink even more. The amount of fluid each person needs to drink depends on the weather and the person’s activity level—people who work or exercise in hot weather need more fluid to replace the fluid they lose through sweat. A 24-hour urine collection may be used to determine the volume of urine produced during a day. If the volume of urine produced is too low, the person can be advised to increase fluid intake. Drinking enough fluid is the most important thing a person can do to prevent kidney stones. Some studies suggest citrus drinks like lemonade and orange juice protect against kidney stones because they contain citrate, which stops crystals from growing into stones. How does sodium in the diet affect kidney stone formation? Sodium, often from salt, causes the kidneys to excrete more calcium into the urine. High concentrations of calcium in the urine combine with oxalate and phosphorus to form stones. Reducing sodium intake is preferred to reducing calcium intake. The U.S. recommended dietary allowance (RDA) of sodium is 2,300 milligrams (mg), but Americans’ intake averages 3,400 mg, according to the U.S. Department of Agriculture.1 The risk of kidney stones increases with increased daily sodium consumption. People who form calcium oxalate or calcium phosphate stones should limit their intake to the U.S. RDA level, even if they take medications to prevent kidney stones. 1U.S. Department of Agriculture and U.S. Department of Health and Human Services. Dietary Guidelines for Americans, 2010. 7th ed. Washington, D.C.: U.S. Government Printing Office; December 2010. How can a person limit sodium intake? Learning the sodium content of foods can help people control their sodium intake. Food labels provide information about sodium and other nutrients. Keeping a sodium diary can help a person limit sodium intake to 2,300 mg. When eating out, people should ask about the sodium content of the foods they order. Some foods have such large amounts of sodium that a single serving provides a major portion of the RDA. Foods that contain high levels of sodium include • hot dogs • canned soups and vegetables • processed frozen foods • luncheon meats • fast food 3 Diet for Kidney Stone Prevention People who are trying to limit their sodium intake should check labels for ingredients and hidden sodium, such as • monosodium glutamate, or MSG • sodium bicarbonate, the chemical name for baking soda • baking powder, which contains sodium bicarbonate and other chemicals • disodium phosphate • sodium alginate • sodium nitrate or nitrite How does animal protein in the diet affect kidney stone formation? Meats and other animal protein—such as eggs and fish—contain purines, which break down into uric acid in the urine. Foods especially rich in purines include organ meats, such as liver. People who form uric acid stones should limit their meat consumption to 6 ounces each day. Animal protein may also raise the risk of calcium stones by increasing the excretion of calcium and reducing the excretion of citrate into the urine. Citrate prevents kidney stones, but the acid in animal protein reduces the citrate in urine. How does calcium in the diet affect kidney stone formation? Calcium from food does not increase the risk of calcium oxalate stones. Calcium in the digestive tract binds to oxalate from food and keeps it from entering the blood, and then the urinary tract, where it can form stones. People who form calcium oxalate stones should include 800 mg of calcium in their diet every day, not only for kidney stone prevention but also to maintain bone density. A cup of low-fat milk contains 300 mg of calcium. Other dairy products such as yogurt are also high in calcium. For people who have lactose intolerance and must avoid dairy products, orange juice fortified with calcium or dairy with reduced lactose content may be alternatives. Calcium supplements may increase the risk of calcium oxalate stones if they are not taken with food. How does oxalate in the diet affect kidney stone formation? Some of the oxalate in urine is made by the body. However, eating certain foods with high levels of oxalate can increase the amount of oxalate in the urine, where it combines with calcium to form calcium oxalate stones. Foods that have been shown to increase the amount of oxalate in urine include • spinach • rhubarb • nuts • wheat bran Avoiding these foods may help reduce the amount of oxalate in the urine. 4 Diet for Kidney Stone Prevention What diet plan should a person follow to prevent future kidney stones? A dietitian can help a person plan meals that lower the risk of forming stones based on the type of stone the person formed in the past. A person with a history of kidney stones may want to talk to a dietitian who specializes in kidney stone prevention or nutrition for people with kidney problems. A dietitian can also help overweight people plan meals to help them lose weight. Studies have shown that being overweight increases the risk of kidney stones, particularly uric acid stones. Diets that are low in carbohydrates have been shown to further increase the risk of uric acid stones and should be avoided. Studies have shown the Dietary Approaches to Stop Hypertension (DASH) diet can reduce the risk of kidney stones. The DASH diet is high in fruits and vegetables, moderate in low-fat dairy products, and low in animal protein. More information about the DASH diet can be found on the National Heart, Lung, and Blood Institute’s website at www.nhlbi.nih.gov/health/ health-topics/topics/dash. Points to Remember • Kidney stones can form when substances in the urine—such as calcium, oxalate, and phosphorus— become highly concentrated. Diet is one of several factors that can promote or inhibit kidney stone formation. • Four major types of kidney stones can form: calcium stones, uric acid stones, struvite stones, and cystine stones. • Drinking enough fluid is the most important thing a person can do to prevent kidney stones. • People who have had a kidney stone should drink enough water and other fluids to make at least 2 liters of urine a day. • Sodium, often from salt, causes the kidneys to excrete more calcium into the urine. High concentrations of calcium in the urine combine with oxalate and phosphorus to form stones. Reducing sodium intake is preferred to reducing calcium intake. • Meats and other animal protein—such as eggs and fish—contain purines, which break down into uric acid in the urine. • Calcium from food does not increase the risk of calcium oxalate stones. Calcium in the digestive tract binds to oxalate from food and keeps it from entering the blood, and then the urinary tract, where it can form stones. • A dietitian can help a person plan meals that lower the risk of forming stones based on the type of stone the person formed in the past. 5 Diet for Kidney Stone Prevention Hope through Research The National Institute of Diabetes and Digestive and Kidney Diseases (NIDDK) funds research on the causes, treatments, and prevention of kidney stones. The International Registry for Hereditary Kidney Stone Diseases, funded under National Institutes of Health (NIH) clinical trial number NCT00588562, collects medical information from a large number of patients with kidney stones to create a registry that will help researchers compare similarities and differences in patients and their symptoms. The Study of the Biological and Physical Manifestations of Spontaneous Uric Acid Kidney Stone Disease, funded under NIH clinical trial number NCT00904046, aims to determine how much fat accumulates within cells and how it affects the kidneys by correlating kidney fat content with urine test results. A second aim is to evaluate the effect of the medication thiazolidinedione on excess fatty acid accumulation in kidney tissue and its correlation with uric acid stone formation. Tamsulosin for Urolithiasis in the Emergency Department, funded under NIH clinical trial number NCT00382265, studies the effectiveness and safety of tamsulosin in treatment of kidney stones. Other areas of focus include reduction in time to pain-free recovery, decrease in narcotic medication for pain, less need for follow-up, decrease in the need for surgery, and cost savings. Clinical trials are research studies involving people. Clinical trials look at safe and effective new ways to prevent, detect, or treat disease. Researchers also use clinical trials to look at other aspects of care, such as improving the quality of life for people with chronic illnesses. To learn more about clinical trials, why they matter, and how to participate, visit the NIH Clinical Research Trials and You website at www.nih.gov/ health/clinicaltrials. For information about current studies, visit www.ClinicalTrials.gov. 6 Diet for Kidney Stone Prevention For More Information National Kidney Foundation 30 East 33rd Street New York, NY 10016 Phone: 1–800–622–9010 or 212–889–2210 Fax: 212–689–9261 Internet: www.kidney.org Oxalosis and Hyperoxaluria Foundation 201 East 19th Street, Suite 12E New York, NY 10003 Phone: 1–800–OHF–8699 (1–800–643–8699) or 212–777–0470 Fax: 212–777–0471 Email: kimh@ohf.org Internet: www.ohf.org Urology Care Foundation 1000 Corporate Boulevard Linthicum, MD 21090 Phone: 1–800–828–7866 or 410–689–3700 Fax: 410–689–3998 Email: info@urologycarefoundation.org Internet: www.UrologyHealth.org Acknowledgments Publications produced by the Clearinghouse are carefully reviewed by both NIDDK scientists and outside experts. The original version of this fact sheet was reviewed by Gary Curhan, M.D., Harvard Medical School, and David Goldfarb, M.D., New York University School of Medicine. You may also find additional information about this topic by visiting MedlinePlus at www.medlineplus.gov. This publication may contain information about medications. When prepared, this publication included the most current information available. For updates or for questions about any medications, contact the U.S. Food and Drug Administration toll-free at 1–888–INFO–FDA (1–888–463–6332) or visit www.fda.gov. Consult your health care provider for more information. 7 Diet for Kidney Stone Prevention National Kidney and Urologic Diseases Information Clearinghouse 3 Information Way Bethesda, MD 20892–3580 Phone: 1–800–891–5390 TTY: 1–866–569–1162 Fax: 703–738–4929 Email: nkudic@info.niddk.nih.gov Internet: www.kidney.niddk.nih.gov The National Kidney and Urologic Diseases Information Clearinghouse (NKUDIC) is a service of the National Institute of Diabetes and Digestive and Kidney Diseases (NIDDK). The NIDDK is part of the National Institutes of Health of the U.S. Department of Health and Human Services. Established in 1987, the Clearinghouse provides information about diseases of the kidneys and urologic system to people with kidney and urologic disorders and to their families, health care professionals, and the public. The NKUDIC answers inquiries, develops and distributes publications, and works closely with professional and patient organizations and Government agencies to coordinate resources about kidney and urologic diseases. This publication is not copyrighted. The Clearinghouse encourages users of this publication to duplicate and distribute as many copies as desired. This publication is available at www.kidney.niddk.nih.gov. NIH Publication No.13–6425 February 2013 The NIDDK prints on recycled paper with bio-based ink.
10081	https://en.wikipedia.org/wiki/Nuclear_densitometry	Nuclear densitometry - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Sources 2 Modes of use 3 UsesToggle Uses subsection 3.1 Ground compaction 3.2 Density of liquids in pipes 3.3 Locating underground water 3.4 Testing separators 4 See also 5 References Nuclear densitometry [x] 2 languages Bahasa Indonesia Norsk bokmål Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Density metering technique A density gauge being used to ensure proper compaction for the foundation of a school construction project. Nuclear densitometry is a technique used in civil construction and the petroleum industry, as well as for mining and archaeology purposes, to measure the density and inner structure of the test material. The processes uses a nuclear density gauge, which consists of a radiation source that emits particles and a sensor that counts the received particles that are either reflected by the test material or pass through it. By calculating the percentage of particles that return to the sensor, the gauge can be calibrated to measure the density. In geotechnical engineering, a nuclear densometer or soil density gauge is a field instrument used to determine the density of a compacted material. The device uses the interaction of gamma radiation with matter to measure density, either through direct transmission or the "backscatter" method. The device determines the density of material by counting the number of photons emitted by a radioactive source (cesium-137) that are read by the detector tubes in the gauge base. A 60-second time interval is typically used for the counting period. Sources [edit] Different variants are used for different purposes. For density analysis of very shallow objects such as roads or walls, a gamma source emitter such as 137Cesium is used to produce gamma radiation. These isotopes are effective in analyzing the top 10 inches (25 centimeters) with high accuracy. 226Radium is used for depths of 328 yards (300 meters). Such instruments can help find caves or identify locations with lower density that would make tunnel construction hazardous. Modes of use [edit] Nuclear density gauges are typically operated in one of two modes: Direct transmission: The retractable rod is lowered into the mat through a pre-drilled hole. The source emits radiation, which then interact with electrons in the material and lose energy and/or are redirected (scattered). Radiation that loses sufficient energy or is scattered away from the detector is not counted. The denser the material, the higher the probability of interaction and the lower the detector count. Therefore, the detector count is inversely proportional to material density. A calibration factor is used to relate the count to the actual density. Backscatter: The retractable rod is lowered so that it is even with the detector but still within the instrument. The source emits radiation, which then interact with electrons in the material and lose energy and/or are redirected (scattered). Radiation that is scattered towards the detector is counted. The denser the material, the higher the probability that radiation will be redirected towards the detector. Therefore, the detector count is proportional to the density. A calibration factor is used to correlate the count to the actual density. Many devices are built to measure both the density and moisture content of material. This is important to the civil construction industry specifically as both are essential to verifying suitable soil conditions to support structures, streets, highways, and airport runways. Uses [edit] Ground compaction [edit] Asphalt Density Gauge A nuclear densometer is used on a compacted base to establish its percentage of compaction. Before field tests are performed, the technician performs a calibration on the gauge which records the 'standard count' of the machine. Standard counts are the amount of radiation released by the two nuclear sources inside the machine, with no loss or leakage. This allows the machine to compare the amount of radiation released to the amount of radiation received. With the use of a 3/4" diameter rod a hole is created in the compacted base by hammering the rod into the base to produce a hole that the densometer's probe can be inserted into. The densometer is placed on top of the hole, and then the probe is inserted into the hole by unlocking the handle at the top of the probe. One source produces radiation that interacts with the atoms in the soil, and is then compared to the standard count, to calculate the density. The other source interacts with hydrogen atoms to calculate the percentage of water in the soil. In direct transmission mode, the source extends through the base of the gauge into a predrilled hole, positioning the source at the desired depth. The testing procedure is analogous to burying a known quantity of radioactive material at a specific depth, and then using a Geiger counter at the ground surface to measure how effectively the soil's density blocks the penetration of gamma radiation through the soil. As the soil's density increases, less radiation can pass through it, owing to dispersion from collisions with electrons in the soil being tested. Since the soil's moisture level is partly responsible for its in-place density, the gauge also contains a neutron moisture gauge consisting of an americium/beryllium high-energy neutron source and a thermal neutron detector. The high-energy neutrons are slowed when they collide with hydrogen atoms, and the detector then counts the "slowed" neutrons. This count is proportional to the soil's water content, since the hydrogen in this water (H2 O) is responsible for almost all the hydrogen found in most soils. The gauge calculates the moisture content, subtracts it from the soil's in-place (wet) density, and reports the soil's dry density. Density of liquids in pipes [edit] Nuclear density gauges can also be used to measure the density of a liquid in a pipe. If a source is mounted on one side of a pipe and a detector on the other, the amount of radiation seen at the detector is dependent upon the shielding provided by the liquid in the pipe. Tracerco pioneered the use of radiation to measure density in the 1950s and determined that the Beer–Lambert law also applied to radiation as well as optics. Gauges are normally calibrated using gas and a liquid of known density to find the unknowns in the equation. Once it has been calibrated and as long as the source detector alignment remains constant, it is possible to calculate the density of the liquid in the pipe. One factor is the half-life of the radioactive source (30 years for 137 Cs), which means that the system needs to be recalibrated at regular intervals. Modern systems incorporate correction for source decay. Locating underground water [edit] Another variant is to use a strong neutron source like 241 americium/beryllium to produce neutron radiation and then measure the energy of returning neutron scattering. As hydrogen characteristically slows down neutrons, the sensor can calculate the density of hydrogen - and find pockets of underground water, humidity up to a depth of several meters, moisture content, or asphalt content. Testing separators [edit] Neutron sources can also be used to assess the performance of a separator (oil production) in the same way. Gas, oil, water and sand all have different concentrations of hydrogen atoms which reflect different amounts of slow neutrons. Using a head which contains an 241 AmBe neutron source and a slow neutron detector, by scanning it up and down a separator it is possible to determine the interface levels within the separator. See also [edit] Nuclear power References [edit] ^M. Falahati; et al. (2018). "Design, modelling and construction of a continuous nuclear gauge for measuring the fluid levels". Journal of Instrumentation. 13 (2): P02028. doi:10.1088/1748-0221/13/02/P02028. S2CID125779702.{{cite journal}}: CS1 maint: article number as page number (link) ^ASTM D2922-05 Standard Test Methods for Density of Soil and Soil-Aggregate in Place by Nuclear Methods (Shallow Depth), doi:10.1520/D2922-05 (Standard withdrawn 2007) ^ASTM D6938 - 08a Standard Test Method for In-Place Density and Water Content of Soil and Soil-Aggregate by Nuclear Methods (Shallow Depth), doi:10.1520/D6938-08A ^Jackson, Peter (2004). Radioisotope Gauges for Industrial. Chichester: John Wiley & Sons Ltd. doi:10.1002/0470021098.fmatter. ISBN0-471-48999-9. \| hide v t e Geotechnical engineering \| \| Offshore geotechnical engineering \| \| Investigation and instrumentation \| \| Field (in situ) \| Core drill Cone penetration test Geo-electrical sounding Permeability test Load test Static Dynamic Statnamic Pore pressure measurement Piezometer Well Ram sounding Rock control drilling Rotary-pressure sounding Rotary weight sounding Sample series Screw plate test Deformation monitoring Inclinometer Settlement recordings Shear vane test Simple sounding Standard penetration test Total sounding Trial pit Visible bedrock Nuclear densometer test Exploration geophysics Crosshole sonic logging Pile integrity test Wave equation analysis \| \| Laboratory testing \| Soil classification Atterberg limits California bearing ratio Direct shear test Hydrometer Proctor compaction test R-value Sieve analysis Triaxial shear test Oedometer test Hydraulic conductivity tests Water content tests \| \| \| Soil \| \| Types \| Clay Silt Sand Gravel Peat Loam Loess \| \| Properties \| Hydraulic conductivity Water content Void ratio Bulk density Thixotropy Reynolds' dilatancy Angle of repose Friction angle Cohesion Porosity Permeability Specific storage Shear strength Sensitivity \| \| \| Structures (Interaction) \| \| Natural features \| Topography Vegetation Terrain Topsoil Water table Bedrock Subgrade Subsoil \| \| Earthworks \| Shoring structures Retaining walls Gabion Ground freezing Mechanically stabilized earth Pressure grouting Slurry wall Soil nailing Tieback Land development Landfill Excavation Trench Embankment Cut Causeway Terracing Cut-and-cover Cut and fill Fill dirt Grading Land reclamation Track bed Erosion control Earth structure Expanded clay aggregate Crushed stone Geosynthetics Geotextile Geomembrane Geosynthetic clay liner Cellular confinement Infiltration \| \| Foundations \| Shallow Deep \| \| \| Mechanics \| \| Forces \| Effective stress Pore water pressure Lateral earth pressure Overburden pressure Preconsolidation pressure \| \| Phenomena/ problems \| Permafrost Frost heaving Consolidation Compaction Earthquake Response spectrum Seismic hazard Shear wave Landslide analysis Stability analysis Mitigation Classification Sliding criterion Slab stabilisation Bearing capacity Stress distribution in soil \| \| \| Numerical analysis software \| SEEP2D STABL SVFlux SVSlope UTEXAS Plaxis \| \| Related fields \| Geology Geochemistry Petrology Earthquake engineering Geomorphology Soil science Hydrology Hydrogeology Biogeography Earth materials Archaeology Agricultural science Agrology \| Retrieved from " Categories: Radiography Density meters In situ geotechnical investigations Hidden categories: CS1 maint: article number as page number Articles with short description Short description matches Wikidata This page was last edited on 17 February 2023, at 15:46(UTC). 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10083	https://www.hmdb.ca/metabolites/HMDB32024	Your source for quantitative metabolomics technologies and bioinformatics. Showing metabocard for 4-Ethylbenzaldehyde (HMDB0032024) Jump To Section: Identification Taxonomy Ontology Physical properties Spectra Biological properties Concentrations Links References XML \| Record Information \| \| --- \| \| Version \| 5.0 \| \| Status \| Expected but not Quantified \| \| Creation Date \| 2012-09-11 17:47:19 UTC \| \| Update Date \| 2023-02-21 17:21:29 UTC \| \| HMDB ID \| HMDB0032024 \| \| Secondary Accession Numbers \| HMDB32024 \| \| Metabolite Identification \| \| \| Common Name \| 4-Ethylbenzaldehyde \| \| Description \| 4-Ethylbenzaldehyde belongs to the class of organic compounds known as benzoyl derivatives. These are organic compounds containing an acyl moiety of benzoic acid with the formula (C6H5CO-). 4-Ethylbenzaldehyde is a sweet, almond, and bitter tasting compound. 4-Ethylbenzaldehyde has been detected, but not quantified in, several different foods, such as green tea, teas (Camellia sinensis), nuts, black tea, and herbal tea. This could make 4-ethylbenzaldehyde a potential biomarker for the consumption of these foods. Based on a literature review a small amount of articles have been published on 4-Ethylbenzaldehyde. \| \| Structure \| MOL 3D MOL SDF 3D SDF PDB 3D PDB SMILES InChI MOL for HMDB0032024 (4-Ethylbenzaldehyde) Mrv1652303202019012D 10 10 0 0 0 0 999 V2000 0.0000 -1.6500 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.7145 -1.2375 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 1.4289 0.0000 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0000 0.0000 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 1.4289 0.8250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0000 0.8250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.7145 2.0625 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.7145 -0.4125 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.7145 1.2375 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0000 2.4750 0.0000 O 0 0 0 0 0 0 0 0 0 0 0 0 2 1 1 0 0 0 0 5 3 1 0 0 0 0 6 4 2 0 0 0 0 8 2 1 0 0 0 0 8 3 2 0 0 0 0 8 4 1 0 0 0 0 9 5 2 0 0 0 0 9 6 1 0 0 0 0 9 7 1 0 0 0 0 10 7 2 0 0 0 0 M END Download 3D MOL for HMDB0032024 (4-Ethylbenzaldehyde) HMDB0032024 RDKit 3D 4-Ethylbenzaldehyde 20 20 0 0 0 0 0 0 0 0999 V2000 -2.8518 0.1296 0.7739 C 0 0 0 0 0 0 0 0 0 0 0 0 -2.1280 -0.1948 -0.5115 C 0 0 0 0 0 0 0 0 0 0 0 0 -0.6668 -0.1172 -0.3266 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0057 1.0577 -0.5386 C 0 0 0 0 0 0 0 0 0 0 0 0 1.3708 1.1366 -0.3676 C 0 0 0 0 0 0 0 0 0 0 0 0 2.1070 0.0460 0.0202 C 0 0 0 0 0 0 0 0 0 0 0 0 3.5636 0.1412 0.1999 C 0 0 0 0 0 0 0 0 0 0 0 0 4.1239 1.2455 -0.0095 O 0 0 0 0 0 0 0 0 0 0 0 0 1.4522 -1.1484 0.2398 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0792 -1.2218 0.0662 C 0 0 0 0 0 0 0 0 0 0 0 0 -2.1976 -0.0518 1.6440 H 0 0 0 0 0 0 0 0 0 0 0 0 -3.7417 -0.5307 0.8894 H 0 0 0 0 0 0 0 0 0 0 0 0 -3.2626 1.1697 0.7477 H 0 0 0 0 0 0 0 0 0 0 0 0 -2.3675 -1.2580 -0.7532 H 0 0 0 0 0 0 0 0 0 0 0 0 -2.5072 0.4833 -1.2863 H 0 0 0 0 0 0 0 0 0 0 0 0 -0.5894 1.9277 -0.8490 H 0 0 0 0 0 0 0 0 0 0 0 0 1.8697 2.0731 -0.5417 H 0 0 0 0 0 0 0 0 0 0 0 0 4.1091 -0.7378 0.5056 H 0 0 0 0 0 0 0 0 0 0 0 0 2.0585 -1.9894 0.5442 H 0 0 0 0 0 0 0 0 0 0 0 0 -0.4273 -2.1607 0.2399 H 0 0 0 0 0 0 0 0 0 0 0 0 1 2 1 0 2 3 1 0 3 4 2 0 4 5 1 0 5 6 2 0 6 7 1 0 7 8 2 0 6 9 1 0 9 10 2 0 10 3 1 0 1 11 1 0 1 12 1 0 1 13 1 0 2 14 1 0 2 15 1 0 4 16 1 0 5 17 1 0 7 18 1 0 9 19 1 0 10 20 1 0 M END Download 3D SDF for HMDB0032024 (4-Ethylbenzaldehyde) Mrv1652303202019012D 10 10 0 0 0 0 999 V2000 0.0000 -1.6500 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.7145 -1.2375 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 1.4289 0.0000 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0000 0.0000 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 1.4289 0.8250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0000 0.8250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.7145 2.0625 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.7145 -0.4125 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.7145 1.2375 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0000 2.4750 0.0000 O 0 0 0 0 0 0 0 0 0 0 0 0 2 1 1 0 0 0 0 5 3 1 0 0 0 0 6 4 2 0 0 0 0 8 2 1 0 0 0 0 8 3 2 0 0 0 0 8 4 1 0 0 0 0 9 5 2 0 0 0 0 9 6 1 0 0 0 0 9 7 1 0 0 0 0 10 7 2 0 0 0 0 M END > <DATABASE_ID> HMDB0032024 > <DATABASE_NAME> hmdb > <SMILES> CCC1=CC=C(C=O)C=C1 > <INCHI_IDENTIFIER> InChI=1S/C9H10O/c1-2-8-3-5-9(7-10)6-4-8/h3-7H,2H2,1H3 > <INCHI_KEY> QNGNSVIICDLXHT-UHFFFAOYSA-N > <FORMULA> C9H10O > <MOLECULAR_WEIGHT> 134.1751 > <EXACT_MASS> 134.073164942 > <JCHEM_ACCEPTOR_COUNT> 1 > <JCHEM_ATOM_COUNT> 20 > <JCHEM_AVERAGE_POLARIZABILITY> 15.122325447636655 > <JCHEM_BIOAVAILABILITY> 1 > <JCHEM_DONOR_COUNT> 0 > <JCHEM_FORMAL_CHARGE> 0 > <JCHEM_GHOSE_FILTER> 0 > <JCHEM_IUPAC> 4-ethylbenzaldehyde > <ALOGPS_LOGP> 2.71 > <JCHEM_LOGP> 2.643738215 > <ALOGPS_LOGS> -2.57 > <JCHEM_MDDR_LIKE_RULE> 0 > <JCHEM_NUMBER_OF_RINGS> 1 > <JCHEM_PHYSIOLOGICAL_CHARGE> 0 > <JCHEM_PKA_STRONGEST_BASIC> -7.102752048197052 > <JCHEM_POLAR_SURFACE_AREA> 17.07 > <JCHEM_REFRACTIVITY> 42.284200000000006 > <JCHEM_ROTATABLE_BOND_COUNT> 2 > <JCHEM_RULE_OF_FIVE> 1 > <ALOGPS_SOLUBILITY> 3.60e-01 g/l > <JCHEM_TRADITIONAL_IUPAC> 4-ethylbenzaldehyde > <JCHEM_VEBER_RULE> 1 $$$$ Download 3D-SDF for HMDB0032024 (4-Ethylbenzaldehyde) HMDB0032024 RDKit 3D 4-Ethylbenzaldehyde 20 20 0 0 0 0 0 0 0 0999 V2000 -2.8518 0.1296 0.7739 C 0 0 0 0 0 0 0 0 0 0 0 0 -2.1280 -0.1948 -0.5115 C 0 0 0 0 0 0 0 0 0 0 0 0 -0.6668 -0.1172 -0.3266 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0057 1.0577 -0.5386 C 0 0 0 0 0 0 0 0 0 0 0 0 1.3708 1.1366 -0.3676 C 0 0 0 0 0 0 0 0 0 0 0 0 2.1070 0.0460 0.0202 C 0 0 0 0 0 0 0 0 0 0 0 0 3.5636 0.1412 0.1999 C 0 0 0 0 0 0 0 0 0 0 0 0 4.1239 1.2455 -0.0095 O 0 0 0 0 0 0 0 0 0 0 0 0 1.4522 -1.1484 0.2398 C 0 0 0 0 0 0 0 0 0 0 0 0 0.0792 -1.2218 0.0662 C 0 0 0 0 0 0 0 0 0 0 0 0 -2.1976 -0.0518 1.6440 H 0 0 0 0 0 0 0 0 0 0 0 0 -3.7417 -0.5307 0.8894 H 0 0 0 0 0 0 0 0 0 0 0 0 -3.2626 1.1697 0.7477 H 0 0 0 0 0 0 0 0 0 0 0 0 -2.3675 -1.2580 -0.7532 H 0 0 0 0 0 0 0 0 0 0 0 0 -2.5072 0.4833 -1.2863 H 0 0 0 0 0 0 0 0 0 0 0 0 -0.5894 1.9277 -0.8490 H 0 0 0 0 0 0 0 0 0 0 0 0 1.8697 2.0731 -0.5417 H 0 0 0 0 0 0 0 0 0 0 0 0 4.1091 -0.7378 0.5056 H 0 0 0 0 0 0 0 0 0 0 0 0 2.0585 -1.9894 0.5442 H 0 0 0 0 0 0 0 0 0 0 0 0 -0.4273 -2.1607 0.2399 H 0 0 0 0 0 0 0 0 0 0 0 0 1 2 1 0 2 3 1 0 3 4 2 0 4 5 1 0 5 6 2 0 6 7 1 0 7 8 2 0 6 9 1 0 9 10 2 0 10 3 1 0 1 11 1 0 1 12 1 0 1 13 1 0 2 14 1 0 2 15 1 0 4 16 1 0 5 17 1 0 7 18 1 0 9 19 1 0 10 20 1 0 M END Download PDB for HMDB0032024 (4-Ethylbenzaldehyde) HEADER PROTEIN 20-MAR-20 NONE TITLE NULL COMPND NULL SOURCE NULL KEYWDS NULL EXPDTA NULL AUTHOR Marvin REVDAT 1 20-MAR-20 0 HETATM 1 C UNK 0 0.000 -3.080 0.000 0.00 0.00 C+0 HETATM 2 C UNK 0 1.334 -2.310 0.000 0.00 0.00 C+0 HETATM 3 C UNK 0 2.667 0.000 0.000 0.00 0.00 C+0 HETATM 4 C UNK 0 0.000 0.000 0.000 0.00 0.00 C+0 HETATM 5 C UNK 0 2.667 1.540 0.000 0.00 0.00 C+0 HETATM 6 C UNK 0 0.000 1.540 0.000 0.00 0.00 C+0 HETATM 7 C UNK 0 1.334 3.850 0.000 0.00 0.00 C+0 HETATM 8 C UNK 0 1.334 -0.770 0.000 0.00 0.00 C+0 HETATM 9 C UNK 0 1.334 2.310 0.000 0.00 0.00 C+0 HETATM 10 O UNK 0 0.000 4.620 0.000 0.00 0.00 O+0 CONECT 1 2 CONECT 2 1 8 CONECT 3 5 8 CONECT 4 6 8 CONECT 5 3 9 CONECT 6 4 9 CONECT 7 9 10 CONECT 8 2 3 4 CONECT 9 5 6 7 CONECT 10 7 MASTER 0 0 0 0 0 0 0 0 10 0 20 0 END Download 3D PDB for HMDB0032024 (4-Ethylbenzaldehyde) COMPND HMDB0032024 HETATM 1 C1 UNL 1 -2.852 0.130 0.774 1.00 0.00 C HETATM 2 C2 UNL 1 -2.128 -0.195 -0.511 1.00 0.00 C HETATM 3 C3 UNL 1 -0.667 -0.117 -0.327 1.00 0.00 C HETATM 4 C4 UNL 1 0.006 1.058 -0.539 1.00 0.00 C HETATM 5 C5 UNL 1 1.371 1.137 -0.368 1.00 0.00 C HETATM 6 C6 UNL 1 2.107 0.046 0.020 1.00 0.00 C HETATM 7 C7 UNL 1 3.564 0.141 0.200 1.00 0.00 C HETATM 8 O1 UNL 1 4.124 1.246 -0.009 1.00 0.00 O HETATM 9 C8 UNL 1 1.452 -1.148 0.240 1.00 0.00 C HETATM 10 C9 UNL 1 0.079 -1.222 0.066 1.00 0.00 C HETATM 11 H1 UNL 1 -2.198 -0.052 1.644 1.00 0.00 H HETATM 12 H2 UNL 1 -3.742 -0.531 0.889 1.00 0.00 H HETATM 13 H3 UNL 1 -3.263 1.170 0.748 1.00 0.00 H HETATM 14 H4 UNL 1 -2.367 -1.258 -0.753 1.00 0.00 H HETATM 15 H5 UNL 1 -2.507 0.483 -1.286 1.00 0.00 H HETATM 16 H6 UNL 1 -0.589 1.928 -0.849 1.00 0.00 H HETATM 17 H7 UNL 1 1.870 2.073 -0.542 1.00 0.00 H HETATM 18 H8 UNL 1 4.109 -0.738 0.506 1.00 0.00 H HETATM 19 H9 UNL 1 2.059 -1.989 0.544 1.00 0.00 H HETATM 20 H10 UNL 1 -0.427 -2.161 0.240 1.00 0.00 H CONECT 1 2 11 12 13 CONECT 2 3 14 15 CONECT 3 4 4 10 CONECT 4 5 16 CONECT 5 6 6 17 CONECT 6 7 9 CONECT 7 8 8 18 CONECT 9 10 10 19 CONECT 10 20 END Download SMILES for HMDB0032024 (4-Ethylbenzaldehyde) CCC1=CC=C(C=O)C=C1 Download INCHI for HMDB0032024 (4-Ethylbenzaldehyde) InChI=1S/C9H10O/c1-2-8-3-5-9(7-10)6-4-8/h3-7H,2H2,1H3 Download Structure for HMDB0032024 (4-Ethylbenzaldehyde) Download 3D Structure for HMDB0032024 (4-Ethylbenzaldehyde) View in JSmol View Stereo Labels \| \| Synonyms \| \| Value \| Source \| --- \| \| 4-Ethyl-benzaldehyde \| HMDB \| \| BENZALDEHYDE,4-ethyl \| HMDB \| \| EBAL \| HMDB \| \| Ethyl benzaldehyde \| HMDB \| \| Ethyl-benzaldehyde \| HMDB \| \| FEMA 3756 \| HMDB \| \| P-Ethyl-benzaldehyde \| HMDB \| \| P-Ethylbenzaldehyde \| HMDB \| \| \| Chemical Formula \| C9H10O \| \| Average Molecular Weight \| 134.1751 \| \| Monoisotopic Molecular Weight \| 134.073164942 \| \| IUPAC Name \| 4-ethylbenzaldehyde \| \| Traditional Name \| 4-ethylbenzaldehyde \| \| CAS Registry Number \| 4748-78-1 \| \| SMILES \| CCC1=CC=C(C=O)C=C1 \| \| InChI Identifier \| InChI=1S/C9H10O/c1-2-8-3-5-9(7-10)6-4-8/h3-7H,2H2,1H3 \| \| InChI Key \| QNGNSVIICDLXHT-UHFFFAOYSA-N \| \| Chemical Taxonomy \| \| \| Description \| Belongs to the class of organic compounds known as benzoyl derivatives. These are organic compounds containing an acyl moiety of benzoic acid with the formula (C6H5CO-). \| \| Kingdom \| Organic compounds \| \| Super Class \| Benzenoids \| \| Class \| Benzene and substituted derivatives \| \| Sub Class \| Benzoyl derivatives \| \| Direct Parent \| Benzoyl derivatives \| \| Alternative Parents \| Benzaldehydes Organic oxides Hydrocarbon derivatives \| \| Substituents \| Benzoyl Benzaldehyde Aryl-aldehyde Organic oxygen compound Organic oxide Hydrocarbon derivative Organooxygen compound Aldehyde Aromatic homomonocyclic compound \| \| Molecular Framework \| Aromatic homomonocyclic compounds \| \| External Descriptors \| Not Available \| \| Ontology \| \| \| Physiological effect \| Organoleptic effect Taste + Sweet + Bitter Odor + Almond \| \| Disposition \| Biological location Cellular substructure + Membrane (HMDB: HMDB0032024) + Cell membrane (HMDB: HMDB0032024) Route of exposure Enteral + Ingestion (HMDB: HMDB0032024) Source Endogenous + Endogenous (HMDB: HMDB0032024) Plant - Plant (HMDB: HMDB0032024) Exogenous Food - Food (HMDB: HMDB0032024) Tea Tea (FooDB: FOOD00038) Black tea (FooDB: FOOD00907) Green tea (FooDB: FOOD00908) Herbal tea (FooDB: FOOD00989) Red tea (FooDB: FOOD00925) Nut Nuts (FooDB: FOOD00869) Beverage Alcoholic beverage + Alcoholic beverages (FooDB: FOOD00854) + Exogenous (HMDB: HMDB0032024) \| \| Process \| Not Available \| \| Role \| Industrial application Food and nutrition Food additive Flavoring agent Flavoring Agent (HMDB: HMDB0032024) \| \| Physical Properties \| \| \| State \| Not Available \| \| Experimental Molecular Properties \| \| Property \| Value \| Reference \| \| Melting Point \| Not Available \| Not Available \| \| Boiling Point \| 220.00 to 222.00 °C. @ 760.00 mm Hg \| The Good Scents Company Information System \| \| Water Solubility \| 397.7 mg/L @ 25 °C (est) \| The Good Scents Company Information System \| \| LogP \| 2.408 (est) \| The Good Scents Company Information System \| \| \| Experimental Chromatographic Properties \| Not Available \| \| Predicted Molecular Properties \| \| Property \| Value \| Source \| \| Water Solubility \| 0.36 g/L \| ALOGPS \| \| logP \| 2.71 \| ALOGPS \| \| logP \| 2.64 \| ChemAxon \| \| logS \| -2.6 \| ALOGPS \| \| pKa (Strongest Basic) \| -7.1 \| ChemAxon \| \| Physiological Charge \| 0 \| ChemAxon \| \| Hydrogen Acceptor Count \| 1 \| ChemAxon \| \| Hydrogen Donor Count \| 0 \| ChemAxon \| \| Polar Surface Area \| 17.07 Å² \| ChemAxon \| \| Rotatable Bond Count \| 2 \| ChemAxon \| \| Refractivity \| 42.28 m³·mol⁻¹ \| ChemAxon \| \| Polarizability \| 15.12 Å³ \| ChemAxon \| \| Number of Rings \| 1 \| ChemAxon \| \| Bioavailability \| Yes \| ChemAxon \| \| Rule of Five \| Yes \| ChemAxon \| \| Ghose Filter \| No \| ChemAxon \| \| Veber's Rule \| Yes \| ChemAxon \| \| MDDR-like Rule \| No \| ChemAxon \| \|Show more... \| Predicted Chromatographic Properties \| Predicted Collision Cross Sections \| Predictor \| Adduct Type \| CCS Value (Å2) \| Reference \| --- --- \| \| DarkChem \| [M+H]+ \| 129.842 \| 31661259 \| \| DarkChem \| [M-H]- \| 128.603 \| 31661259 \| \| DeepCCS \| [M+H]+ \| 130.133 \| 30932474 \| \| DeepCCS \| [M-H]- \| 126.303 \| 30932474 \| \| DeepCCS \| [M-2H]- \| 163.738 \| 30932474 \| \| DeepCCS \| [M+Na]+ \| 139.277 \| 30932474 \| \| AllCCS \| [M+H]+ \| 127.0 \| 32859911 \| \| AllCCS \| [M+H-H2O]+ \| 122.3 \| 32859911 \| \| AllCCS \| [M+NH4]+ \| 131.5 \| 32859911 \| \| AllCCS \| [M+Na]+ \| 132.8 \| 32859911 \| \| AllCCS \| [M-H]- \| 128.8 \| 32859911 \| \| AllCCS \| [M+Na-2H]- \| 130.5 \| 32859911 \| \| AllCCS \| [M+HCOO]- \| 132.3 \| 32859911 \| Predicted Retention TimesUnderivatized \| Chromatographic Method \| Retention Time \| Reference \| --- \| Measured using a Waters Acquity ultraperformance liquid chromatography (UPLC) ethylene-bridged hybrid (BEH) C18 column (100 mm × 2.1 mm; 1.7 μmparticle diameter). Predicted by Afia on May 17, 2022. Predicted by Afia on May 17, 2022. \| 5.56 minutes \| 32390414 \| \| Predicted by Siyang on May 30, 2022 \| 15.7462 minutes \| 33406817 \| \| Predicted by Siyang using ReTip algorithm on June 8, 2022 \| 4.02 minutes \| 32390414 \| \| Fem_Long = Waters ACQUITY UPLC HSS T3 C18 with Water:MeOH and 0.1% Formic Acid \| 1902.4 seconds \| 40023050 \| \| Fem_Lipids = Ascentis Express C18 with (60:40 water:ACN):(90:10 IPA:ACN) and 10mM NH4COOH + 0.1% Formic Acid \| 584.1 seconds \| 40023050 \| \| Life_Old = Waters ACQUITY UPLC BEH C18 with Water:(20:80 acetone:ACN) and 0.1% Formic Acid \| 218.9 seconds \| 40023050 \| \| Life_New = RP Waters ACQUITY UPLC HSS T3 C18 with Water:(30:70 MeOH:ACN) and 0.1% Formic Acid \| 387.3 seconds \| 40023050 \| \| RIKEN = Waters ACQUITY UPLC BEH C18 with Water:ACN and 0.1% Formic Acid \| 250.1 seconds \| 40023050 \| \| Eawag_XBridgeC18 = XBridge C18 3.5u 2.1x50 mm with Water:MeOH and 0.1% Formic Acid \| 595.6 seconds \| 40023050 \| \| BfG_NTS_RP1 =Agilent Zorbax Eclipse Plus C18 (2.1 mm x 150 mm, 3.5 um) with Water:ACN and 0.1% Formic Acid \| 753.2 seconds \| 40023050 \| \| HILIC_BDD_2 = Merck SeQuant ZIC-HILIC with ACN(0.1% formic acid):water(16 mM ammonium formate) \| 153.0 seconds \| 40023050 \| \| UniToyama_Atlantis = RP Waters Atlantis T3 (2.1 x 150 mm, 5 um) with ACN:Water and 0.1% Formic Acid \| 1305.2 seconds \| 40023050 \| \| BDD_C18 = Hypersil Gold 1.9µm C18 with Water:ACN and 0.1% Formic Acid \| 507.5 seconds \| 40023050 \| \| UFZ_Phenomenex = Kinetex Core-Shell C18 2.6 um, 3.0 x 100 mm, Phenomenex with Water:MeOH and 0.1% Formic Acid \| 1333.1 seconds \| 40023050 \| \| SNU_RIKEN_POS = Waters ACQUITY UPLC BEH C18 with Water:ACN and 0.1% Formic Acid \| 428.1 seconds \| 40023050 \| \| RPMMFDA = Waters ACQUITY UPLC BEH C18 with Water:ACN and 0.1% Formic Acid \| 502.0 seconds \| 40023050 \| \| MTBLS87 = Merck SeQuant ZIC-pHILIC column with ACN:Water and :ammonium carbonate \| 527.3 seconds \| 40023050 \| \| KI_GIAR_zic_HILIC_pH2_7 = Merck SeQuant ZIC-HILIC with ACN:Water and 0.1% FA \| 517.2 seconds \| 40023050 \| \| Meister zic-pHILIC pH9.3 = Merck SeQuant ZIC-pHILIC column with ACN:Water 5mM NH4Ac pH9.3 and 5mM ammonium acetate in water \| 89.3 seconds \| 40023050 \| Predicted Kovats Retention IndicesUnderivatized \| Metabolite \| SMILES \| Kovats RI Value \| Column Type \| Reference \| --- --- \| 4-Ethylbenzaldehyde \| CCC1=CC=C(C=O)C=C1 \| 1743.7 \| Standard polar \| 33892256 \| \| 4-Ethylbenzaldehyde \| CCC1=CC=C(C=O)C=C1 \| 1156.8 \| Standard non polar \| 33892256 \| \| 4-Ethylbenzaldehyde \| CCC1=CC=C(C=O)C=C1 \| 1183.3 \| Semi standard non polar \| 33892256 \| \|Show more... \| Spectra \| \| \| \| GC-MS Spectra \| Spectrum Type \| Description \| Splash Key \| Deposition Date \| Source \| View \| --- --- --- \| \| Experimental GC-MS \| GC-MS Spectrum - 4-Ethylbenzaldehyde EI-B (Non-derivatized) \| splash10-001i-7900000000-6a3ae9843f99712bf916 \| 2017-09-12 \| HMDB team, MONA, MassBank \| View Spectrum \| \| Experimental GC-MS \| GC-MS Spectrum - 4-Ethylbenzaldehyde EI-B (Non-derivatized) \| splash10-001i-3900000000-d03f4f4d16760a07a4e0 \| 2017-09-12 \| HMDB team, MONA, MassBank \| View Spectrum \| \| Experimental GC-MS \| GC-MS Spectrum - 4-Ethylbenzaldehyde EI-B (Non-derivatized) \| splash10-001i-8900000000-82ca7060f7613dd16c1d \| 2017-09-12 \| HMDB team, MONA, MassBank \| View Spectrum \| \| Experimental GC-MS \| GC-MS Spectrum - 4-Ethylbenzaldehyde EI-B (Non-derivatized) \| splash10-001i-7900000000-6a3ae9843f99712bf916 \| 2018-05-18 \| HMDB team, MONA, MassBank \| View Spectrum \| \| Experimental GC-MS \| GC-MS Spectrum - 4-Ethylbenzaldehyde EI-B (Non-derivatized) \| splash10-001i-3900000000-d03f4f4d16760a07a4e0 \| 2018-05-18 \| HMDB team, MONA, MassBank \| View Spectrum \| \| Experimental GC-MS \| GC-MS Spectrum - 4-Ethylbenzaldehyde EI-B (Non-derivatized) \| splash10-001i-8900000000-82ca7060f7613dd16c1d \| 2018-05-18 \| HMDB team, MONA, MassBank \| View Spectrum \| \| Predicted GC-MS \| Predicted GC-MS Spectrum - 4-Ethylbenzaldehyde GC-MS (Non-derivatized) - 70eV, Positive \| splash10-067i-2900000000-d010447bf25b4ddfa431 \| 2017-09-01 \| Wishart Lab \| View Spectrum \| \| Predicted GC-MS \| Predicted GC-MS Spectrum - 4-Ethylbenzaldehyde GC-MS (Non-derivatized) - 70eV, Positive \| Not Available \| 2021-10-12 \| Wishart Lab \| View Spectrum \| MS/MS Spectra \| Spectrum Type \| Description \| Splash Key \| Deposition Date \| Source \| View \| --- --- --- \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 10V, Positive-QTOF \| splash10-000i-0900000000-0ae14cf704183806f2c6 \| 2017-09-01 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 20V, Positive-QTOF \| splash10-000i-0900000000-09a3700ff74df97af5da \| 2017-09-01 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 40V, Positive-QTOF \| splash10-0i2l-9300000000-2edeec0cc3b65da2078c \| 2017-09-01 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 10V, Negative-QTOF \| splash10-001i-0900000000-869ebcc7c9552522b81e \| 2017-09-01 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 20V, Negative-QTOF \| splash10-001i-0900000000-b4c298347aafc757d258 \| 2017-09-01 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 40V, Negative-QTOF \| splash10-0a4i-3900000000-9bd655656f5f06a1f101 \| 2017-09-01 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 10V, Negative-QTOF \| splash10-001i-0900000000-b646282963ddd573b4f4 \| 2021-09-21 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 20V, Negative-QTOF \| splash10-053r-1900000000-491e7fa9956be0e702f8 \| 2021-09-21 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 40V, Negative-QTOF \| splash10-00or-9400000000-0b78a8bd01423970c638 \| 2021-09-21 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 10V, Positive-QTOF \| splash10-000i-0900000000-bb9d5b1dcb5a8f18b314 \| 2021-09-25 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 20V, Positive-QTOF \| splash10-0a4i-7900000000-f76d045c51970e2bcb0e \| 2021-09-25 \| Wishart Lab \| View Spectrum \| \| Predicted LC-MS/MS \| Predicted LC-MS/MS Spectrum - 4-Ethylbenzaldehyde 40V, Positive-QTOF \| splash10-0fb9-9000000000-75a5666babf76ef6791e \| 2021-09-25 \| Wishart Lab \| View Spectrum \| IR Spectra \| Spectrum Type \| Description \| Deposition Date \| Source \| View \| --- --- \| Predicted IR Spectrum \| IR Ion Spectrum (Predicted IRIS Spectrum, Adduct: [M+H]+) \| 2023-02-04 \| FELIX lab \| View Spectrum \| \|Show more... \| Biological Properties \| \| \| Cellular Locations \| Membrane \| \| Biospecimen Locations \| Not Available \| \| Tissue Locations \| Not Available \| \| Pathways \| Not Available \| Name \| SMPDB/PathBank \| KEGG \| --- \| \| Normal Concentrations \| \| \| \| Not Available \| \| Abnormal Concentrations \| \| \| \| Not Available \| \| Associated Disorders and Diseases \| \| \| Disease References \| None \| \| Associated OMIM IDs \| None \| \| External Links \| \| \| DrugBank ID \| Not Available \| \| Phenol Explorer Compound ID \| Not Available \| \| FooDB ID \| FDB008722 \| \| KNApSAcK ID \| Not Available \| \| Chemspider ID \| 21105903 \| \| KEGG Compound ID \| Not Available \| \| BioCyc ID \| Not Available \| \| BiGG ID \| Not Available \| \| Wikipedia Link \| 4-Ethylbenzaldehyde \| \| METLIN ID \| Not Available \| \| PubChem Compound \| 20861 \| \| PDB ID \| Not Available \| \| ChEBI ID \| Not Available \| \| Food Biomarker Ontology \| Not Available \| \| VMH ID \| Not Available \| \| MarkerDB ID \| Not Available \| \| Good Scents ID \| rw1038111 \| \| References \| \| \| Synthesis Reference \| Not Available \| \| Material Safety Data Sheet (MSDS) \| Not Available \| \| General References \| 1. Caviglioli G, Valeria P, Brunella P, Sergio C, Attilia A, Gaetano B: Identification of degradation products of ibuprofen arising from oxidative and thermal treatments. J Pharm Biomed Anal. 2002 Oct 15;30(3):499-509. [PubMed:12367674 ] 2. Chen QX, Song KK, Wang Q, Huang H: Inhibitory effects on mushroom tyrosinase by some alkylbenzaldehydes. J Enzyme Inhib Med Chem. 2003 Dec;18(6):491-6. [PubMed:15008513 ] 3. Pennarun AL, Prost C, Haure J, Demaimay M: Comparison of two microalgal diets. 2. Influence on odorant composition and organoleptic qualities of raw oysters (Crassostrea gigas). J Agric Food Chem. 2003 Mar 26;51(7):2011-8. [PubMed:12643667 ] 4. Racz G, Csenki Z, Kovacs R, Hegyi A, Baska F, Sujbert L, Zsakovics I, Kis R, Gustafson R, Urbanyi B, Szende B: Subacute toxicity assessment of water disinfection byproducts on zebrafish. Pathol Oncol Res. 2012 Jul;18(3):579-84. doi: 10.1007/s12253-011-9479-3. Epub 2011 Dec 11. [PubMed:22161134 ] 5. (). Yannai, Shmuel. (2004) Dictionary of food compounds with CD-ROM: Additives, flavors, and ingredients. Boca Raton: Chapman & Hall/CRC.. . \|
10084	https://arxiv.org/abs/2404.05565	[2404.05565] Extremal problems in BMO and VMO involving the Garsia norm The Scheduled Database Maintenance 2025-09-17 11am-1pm UTC has been completed The scheduled database maintenance has been completed. We recommend that all users logout and login again.. Blog post Skip to main content We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors.Donate >math> arXiv:2404.05565 Help \| Advanced Search Search GO quick links Login Help Pages About Mathematics > Complex Variables arXiv:2404.05565 (math) [Submitted on 8 Apr 2024 (v1), last revised 11 Feb 2025 (this version, v2)] Title:Extremal problems in BMO and VMO involving the Garsia norm Authors:Konstantin M. Dyakonov View a PDF of the paper titled Extremal problems in BMO and VMO involving the Garsia norm, by Konstantin M. Dyakonov View PDFHTML (experimental) Abstract:Given an L^2function fon the unit circle \mathbb T, we put \Phi_f(z):=\mathcal P(\|f\|^2)(z)-\|\mathcal Pf(z)\|^2,\qquad z\in\mathbb D,where \mathbb Dis the open unit disk and \mathcal Pis the Poisson integral operator. The Garsia norm \\|f\\|Gis then defined as \sup{z\in\mathbb D}\Phi_f(z)^{1/2}, and the space {\rm BMO}is formed by the functions f\in L^2with \\|f\\|G<\infty. If \\|f\\|^2_G=\Phi_f(z_0)for some point z_0\in\mathbb D, then fis said to be a norm-attaining {\rm BMO}function, written as f\in{\rm BMO}{\rm na}. Note that {\rm BMO}{\rm na}contains {\rm VMO}, the space of functions with vanishing mean oscillation. We study, first, the functions fin L^\infty(as well as in L^\infty\cap{\rm BMO}{\rm na}) with the property that \\|f\\|G=\\|f\\|\infty. The analytic case, where L^\inftygets replaced by H^\infty, is discussed in more detail. Secondly, we prove that every function f\in{\rm BMO}_{\rm na}with \\|f\\|_G=1is an extreme point of {\rm ball}\,({\rm BMO}), the unit ball of {\rm BMO}with respect to the Garsia norm. This implies that the extreme points of {\rm ball}\,({\rm VMO})are precisely the unit-norm {\rm VMO}functions. As another consequence, we arrive at an amusing "geometric" characterization of inner functions. Comments:19 pages Subjects:Complex Variables (math.CV); Classical Analysis and ODEs (math.CA); Functional Analysis (math.FA) MSC classes:30H10, 30H35, 30H50, 30J05, 30J10, 46A55, 46J15 Cite as:arXiv:2404.05565 [math.CV] (or arXiv:2404.05565v2 [math.CV] for this version) Focus to learn more arXiv-issued DOI via DataCite Journal reference:J. Funct. Anal. 288 (2025), no. 7, Paper No. 110833 Related DOI: Focus to learn more DOI(s) linking to related resources Submission history From: Konstantin Dyakonov [view email] [v1] Mon, 8 Apr 2024 14:37:35 UTC (17 KB) [v2] Tue, 11 Feb 2025 17:27:28 UTC (17 KB) Full-text links: Access Paper: View a PDF of the paper titled Extremal problems in BMO and VMO involving the Garsia norm, by Konstantin M. Dyakonov View PDF HTML (experimental) TeX Source Other Formats view license Current browse context: math.CV <prev \| next> new \| recent \| 2024-04 Change to browse by: math math.CA math.FA References & Citations NASA ADS Google Scholar Semantic Scholar export BibTeX citation Loading... BibTeX formatted citation × Data provided by: Bookmark Bibliographic Tools Bibliographic and Citation Tools [x] Bibliographic Explorer Toggle Bibliographic Explorer (What is the Explorer?) [x] Connected Papers Toggle Connected Papers (What is Connected Papers?) [x] Litmaps Toggle Litmaps (What is Litmaps?) 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Author Venue Institution Topic About arXivLabs arXivLabs: experimental projects with community collaborators arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website. Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them. Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs. Which authors of this paper are endorsers? \| Disable MathJax (What is MathJax?) About Help Contact Subscribe Copyright Privacy Policy Web Accessibility Assistance arXiv Operational Status Get status notifications via email or slack
10085	https://mathbitsnotebook.com/Algebra1/RationalEquations/REundefined.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| Rational Expressions MathBitsNotebook.com Topical Outline \| Algebra 1 Outline \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| --- \| Rational numbers are denoted by a script Q. \| \| \| \| Q stands for "quotient" \| \| \| \| \| --- \| \| \| A rational number is a number that can be expressed as a fraction p/q where p and q are integers and q ≠ 0. It is the ratio of two integers. \| You are familiar with rational numbers from your work with fractions. \| \| \| --- \| \| \| A rational expression is an expression that is the ratio of two polynomials. (where P(x) and Q(x) are polynomials) \| Rational expressions are algebraic fractions in which the numerator is a polynomial and the denominator is also a polynomial (usually different from the numerator). The polynomials used in creating a rational expression may contain one term (monomial), two terms (binomial), three terms (trinomial), and so on. \| \| \| \| --- \| Rational Expressions (monomial/monomial) \| Rational Expression (binomial/binomial) \| Rational Expression (binomial/trinomial) \| \| \| \| \| Expressions that are not polynomials cannot be used in the creation of rational expressions. \| For example: is not a rational expression, since is not a polynomial. Since rational expressions represent division, we must be careful to avoid division by zero. If a rational expression has a variable in its denominator, we must ensure that any value (or values) substituted for that variable will not create a zero denominator. If it is not obvious which values will cause a division by zero error in a rational expression,set the denominator equal to zero and solve for the variable. \| \| \| --- \| \| Rational expression: Could it possiblybe undefined? When? \| Rational expression: Could it possiblybe undefined? When? \| \| Obviously, when x = 1, the denominator will be zero, making the expression undefined. Domain: All Real numbers but not x = 1. \| Set the denominator = 0 and solve. a2 - 4 = 0 a2 = 4 \| \| For this rational expression, we must limit the x's which may be used, to avoid a division by zero error, and leaving the expression undefined. Notation: read "all x's such that x ≠ 1." \| For this rational expression, we must prevent two x-values from being used in the expression. Domain: All Real numbers but not a = 2 nor a = -2. \| \| Rational expression: Could it possiblybe undefined? When? \| Rational expression: Could it possiblybe undefined? When? \| \| Set: 8 - y = 0 8 = y Domain: All Real numbers, except y = 8. \| Set: x2 + x - 12 = 0 (x - 3)(x + 4) = 0 x - 3 = 0; x = 3 x + 4 = 0; x = -4 Domain: All Real numbers, but not x = 3 and not x = -4. \| \| \| \| \| \| When rational expressions are "improper", division can be used to simplify the expression into a proper form. For this course, the simplification process will be limited to determining common factors between the numerator and the denominator, and reducing. See Simplifying Rational Expressions. \| \| \| NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". \| Topical Outline \| Algebra 1 Outline \| MathBitsNotebook.com \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts \| \|
10086	https://www.raymondcamden.com/2025/09/11/recognizing-abundant-deficient-and-perfect-numbers	Recognizing Abundant, Deficient, and Perfect Numbers Raymond Camden Code and Cats Home About Speaking Stuff 🔍 Recognizing Abundant, Deficient, and Perfect Numbers September 11, 2025developmentboxlang Ok, this post falls into the "I'll never actually use this again" category, which frankly, my normal readers know happens all the time, but it was a fun little diversion and a reminder of why I used to love math so much. Yesterday I found out that one of my kids' homework was to look at a set of numbers and determine if they were abundant, deficient, or perfect. Right now you are probably (at least I know I was) asking, "what in the actual heck is that???" A quick bit of Googling turned up this explainer that basically boils it down to a simple principal. Given a number, find all the divisors of that number, excluding the number itself, and add them up. If the result is less than the original number, it is deficient. If the number is equal, it's perfect. Finally, if the sum is over the input, it's abundant. So for example, 6 is considered perfect because the divisors, 1+2+3, equal 6 when added together. 5, which only has a divisor of 1 (remember, you exclude the number itself) is deficient. 12 is abundant (1+2+3+4+6 == 16). You can read more, and see more variations, on the Encyclopedia.com article if you would like. Honestly, I see absolutely see use for this, but I thought - why not whip up a quick demo of this in BoxLang Version One The first version I built uses one main method to get divisors, minus the input itself, with a special case for 1: js function getDivisors(n) { local.divisors = []; if(n === 1) return divisors; for (local.i = 1; i <= ceiling(abs(n)/2); i++) { if (n % i == 0) { divisors.push(i); } } return divisors; } Copy The result of this is an array of divisors. I then built one function each for the three types of numbers: ```js function isPerfectNumber(required int x) { local.divisors = getDivisors(x); local.sum = divisors.reduce((prev, cur) => return cur + prev, 0); return sum == x; } function isAbundantNumber(required int x) { local.divisors = getDivisors(x); local.sum = divisors.reduce((prev, cur) => return cur + prev, 0); return sum > x; } function isDeficientNumber(required int x) { local.divisors = getDivisors(x); local.sum = divisors.reduce((prev, cur) => return cur + prev, 0); return sum < x; } ``` Copy Fairly simple, right? I whipped up a quick test: js for(i=1;i<30;i++) { println("is #i# a perfect number? #isPerfectNumber(i)#"); println("is #i# an abundant number? #isAbundantNumber(i)#"); println("is #i# a deficient number? #isDeficientNumber(i)#"); println(""); } Copy And you can try this yourself below. Finally, given that a number must be one of three types, it's probably easier to just have one function: js function getNumberType(required int x) { local.divisors = getDivisors(x); local.sum = divisors.reduce((prev, cur) => return cur + prev, 0); if(sum < x) return 'deficient'; if(sum === x) return 'perfect'; return 'abundant'; } Copy You can try this below: Image from the National Gallery of Art, Cats and Kittens, by an unknown artist. Hire Me! I'm currently looking for my next role in developer evangelism and advocacy. I have a long history of helping companies work with developers and love to write, create demos, and present at conferences. You can find my resume to learn more and drop me an email (raymondcamden@gmail.com) to reach out. Support this Content! If you like this content, please consider supporting me. You can become a Patron, visit my Amazon wishlist, or buy me a coffee! Any support helps! Want to get a copy of every new post? Use the form below to sign up for my newsletter. Subscribe © 2025 Raymond Camden. Powered by Eleventy 3.0.0. NowGitHubYouTubeMastodonLinkedIn amazon.com Sign up to Amazon Prime for unlimited free delivery amazon.com >
10087	https://www2.seas.gwu.edu/~ayoussef/cs1311/BooleanAlgebra.pdf	1 Boolean Algebra Definition: A Boolean Algebra is a math construct (B,+, . , ‘, 0,1) where B is a non-empty set, + and . are binary operations in B, ‘ is a unary operation in B, 0 and 1 are special elements of B, such that: a) + and . are communative: for all x and y in B, x+y=y+x, and x.y=y.x b) + and . are associative: for all x, y and z in B, x+(y+z)=(x+y)+z, and x.(y.z)=(x.y).z c) + and . are distributive over one another: x.(y+z)=xy+xz, and x+(y.z)=(x+y).(x+z) d) Identity laws: 1.x=x.1=x and 0+x=x+0=x for all x in B e) Complementation laws: x+x’=1 and x.x’=0 for all x in B Examples:  (B=set of all propositions, ∨, ∧, ¬, T, F)  (B=2A, U, ∩, c, Φ,A) Theorem 1: Let (B,+, . , ‘, 0,1) be a Boolean Algebra. Then the following hold: a) x+x=x and x.x=x for all x in B b) x+1=1 and 0.x=0 for all x in B c) x+(xy)=x and x.(x+y)=x for all x and y in B Proof: a) x = x+0 Identity laws = x+xx’ Complementation laws = (x+x).(x+x’) because + is distributive over . = (x+x).1 Complementation laws = x+x Identity laws x = x.1 Identity laws = x.(x+x’) Complementation laws = x.x +x.x’ because + is distributive over . = x.x+0 Identity laws = x.x b) x+1 =x+(x+x’) Complementation laws = (x+x)+x’ + is associative = x+x’ using (a) = 1 Complementation laws 0.x =(x’.x).x Complementation laws = x’.(x.x) . is associative = x’.x using (a) =0 Complementation laws c) x+(xy) = x.1+x.y Identity laws =x.(1+y) because + is distributive over . 2 =x.1 using (b) =x Identity laws x.(x+y) = x.x+x.y Distributivity laws =x+x.y by (a) =x Just shown above. Q.E.D. Definition: An element y in B is called a complement of an element x in B if x+y=1 and xy=0 Theorem 2: For every element x in B, the complement of x exists and is unique. Proof:  Existence. Let x be in B. x’ exists because ‘ is a unary operation. X’ is a complement of x because it satisfies the definition of a complement (x+x’=1 and xx’=0).  Uniqueness. Let y be a complement of x. We will show that y=x’. Since y is a complement of x, we have x+y=1 and xy=yx=0. y=y.1=y.(x+x’)=yx+yx’=0+yx’=xx’+yx’=(x+y)x’=1.x’=x’ => y=x’. QED Corollary 1: (x’)’=x. Proof, since x’+x=1 and x’x=0, it follows that x is a complement of x’. Since the complement of x’ is unique, it follows then that (x’)’ , which is a complement of x’, and x, which is also a complement of x’, must be equal. Thus, (x’)’=x. QED Theorem 3 (De Morgan’s Laws): a) (x+y)’=x’y’ b) (xy)’=x’+y’ Proof: a) Show that x’y’+(x+y)=1 and (x’y’)(x+y)=0. x’y’+(x+y)=(x’y’+x)+y=(x’+x)(y’+x)+y=1.(y’+x)+y=(y’+x)+y=(x+y’)+y=x+(y’+y)=x+1=1 (x’y’)(x+y)=(x’y’)x+(x’y’)y=(y’x’)x+x’(y’y)=y’(x’x)+x’0=y’0+0=0+0=0 b) The proof is similar and left as an exercise. QED. Definition: Let (B,+, . , ‘, 0,1) be a Boolean Algebra. Define the following ≤ relation in B: x ≤ y if xy=x Theorem 4: The relation ≤ is a partial order relation. Proof: We need to prove that ≤ is reflexive, antisymmetric and transitive  Reflexivity: since xx=x (by Theorem 1-a), it follows that x ≤ x  Antisymmetry: need to show that x≤y and y≤x => x=y. x≤y and y≤x => xy=x and yx=y => 3 x =xy because x≤y =yx because . is commutative =y because y≤x Therefore, x=y.  Transitivity: x≤y and y≤z => x≤z ? xz =(xy)z because xy=x since x≤y =x(yz) because . is associative =xy because yz=y since y≤z =x because xy=x since x≤y Therefore, xz=x and hence x≤z. We conclude that ≤ is a partial order relation. Theorem 5 (without proof): If B is a finite Boolean Algebra, then \|B\| is a power of 2 and the Hasse Diagram of B with respect to ≤ is a hypercube. Definition: A Boolean variable x is a variable (placeholder) where the set from which it takes on its values is a Boolean algebra. Definition: A Boolean expression is any string that can be derived from the following rules and no other rules: a) 0 and 1 are Boolean expressions b) Any Boolean variable is a Boolean expression c) If E and F are Boolean expressions, then (E), (E+F), (E.F), and E’ are Boolean expressions. Note that we can omit the parentheses when no ambiguity arises. Examples:  x+y, x’+y, x.y, and x.(y+z’) are all Boolean expressions  xyz+x’yz’+xyz’+(x+y)(x’+z) is a Boolean expression  x/y is not a Boolean expression  xy is not a Boolean expression. Definition: Let B be a Boolean Algebra. A Boolean function of n variables is a function f: Bn B where f(x1,x2,…,xn) is a Boolean expression in x1,x2,…,xn. Examples: f(x,y,z)=xy+x’z is a 3-variable Boolean function. The function g(x,y,z,w)=(x+y+z’)(x’+y’+w)+xyw’ is also a Boolean function. Definition: Two Boolean expressions are said to be equivalent if their corresponding Boolean functions are the same. 4 Definition: A literal is any Boolean variable x or its complement x’. Truth Tables of Boolean functions:  Much like the truth tables for logical propositions  If f(x,y,z, …) is an n-variable Boolean function, a truth table for f is a table of n+1 columns (one column per variable, and one column for f itself), where the rows represent all the 2n combinations of 0-1 values of the n variables, and the corresponding value of f for each combination.  Examples: f(x,y)=xy+x’y’; x y f 1 1 1 1 0 0 0 1 0 0 0 1 g(x,y,z) = xy’z’+x’y’z+x’yz’; x y z g 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 0 0 0 1 h(x,y,z,w) = x’y’w’+xyzw+xz’; x y z w h 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 1 0 1 1 1 0 0 1 1 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 u(x,y,z,w)= 1 if the string xyzw has an odd number of 1’s; otherwise, it is 0. x y z w u 1 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 0 0 1 0 0 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 0 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 5 Definitions of Minterms and Maxterms:  Suppose we’re dealing with n Boolean variables. A minterm is any product of n literals where each of the n variable appears once in the product. o Example, where n=3 and the variables are x, y and z:  Then, xyz, xy’z, xy’z’ are all miterms.  xy is not a minterm because z is missing.  Also, xyzy’ is not a minterm because y appears multiple times (once as y, and another time as y’). o For n=2 where the variables are x and y, there are 4 minterms in total: xy, xy’, x’y, x’y’.  A maxterm is any sum of n literals where each of the n variable appears once in the sum. o Example, where n=3 and the variables are x, y and z:  x+y+z, x+y’+z’ are both maxterms (of 3 variables).  x+y’ is not a maxterm because z is missing. Definition (Disjunctive Normal Form): A Boolean function/expression is in Disjunctive Normal Form (DNF), also called minterm canonical form, if the function/expression is a sum of minterms. Examples:  f(x,y,z)= xyz+xy’z+x’yz’+x’y’z is in DNF  g(x,y)=xy+x’y’ is in DNF  But h(x,y,z)=xy+x’y’z is not in DNF because xy is not a minterm of size 3. Definition (Conjunctive Normal Form): A Boolean function/expression is in Conjunctive Normal Form (CNF), also called maxterm canonical form, if the function/expression is a product of maxterms. Examples:  f(x,y,z)= (x+y+z)(x+y+z’)(x’+y+z’)(x’+y’+z) is in CNF  g(x,y)=(x+y)(x’+y’) is in CNF  But h(x,y,z)=(x+y)(x’+y’+z) is not in CNF because x+y is not a maxterm of size 3. Observation: Thanks to De Morgan’s Laws, if f is in DNF, then f’ derived from the DNF using De Morgan’s Laws (that is, changing every literal to its complement, and every “.” to “+”, and every “+” to “.”) is in CNF, and vice versa. Method of Putting a Function in DNF, using Truth Tables: 1. Create the truth table of the given Boolean function f 6 2. For each row where the value of f is 1, create a minterm as follows: put in the position of every variable x in the minterm either x or x’ according to whether the corresponding value in that combination is 1 or 0.For example:  For combination 111, the midterm is xyz.  For combination 010, the minterm is x’yz’. 3. The DNF of f is the sum of all the minterms created in step 2. Examples: For the function f(x,y,z) = xy’z’+y’z+xz’; x y z f 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 1 0 1 1 0 0 1 0 0 0 0 1 1 0 0 0 0 The minterms of f (where f is 1) are: xyz’, xy’z, xy’z’, xyz’. Therefore, the DNF of f is: xyz’+ xy’z+xy’z’+xyz’. Method of Putting a Function in CNF, using Truth Tables: 1. Create the truth table of the given Boolean function f 2. Add a column for f’ to the right of the column of f, and fill it with the complements of the column of f (that is, wherever f is 1, put 0 under f’, and wherever f is 0, put 1 under f’) 3. Create the DNF of f’ by applying steps 2 and 3 of the DNF method. 4. Apply De Morgan’s laws on the DNF of f’, we get the CNF of f. For example, for the same function f(x,y,z) = xy’z’+y’z+xz’, we take its truth from the previous example, and the we add the column of f’: x y z f f’ 1 1 1 0 1 1 1 0 1 0 1 0 1 1 0 1 0 0 1 0 0 1 1 0 1 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 Then, we obtain the DNF of f’: f’=xyz+x’yz+x’yz’+x’y’z’ 7 Finally, applying De Morgan’s, we get the CNF of f: f=(f’)’= (x’+y’+z’)(x+y’+z’)(x+y’+z)(x+y+z). Optimization of Boolean functions using Karnaugh Maps: x’y’z + xz + xy’z’ y y’ x 1 1 1 x’ 1 z z’ z Minimized form: xz + xy’ + y’z xz + yz’ + y’z’ y y’ x 1 1 1 1 x’ 1 1 z z’ z Minimized form: x+z’ xyz’ + xy’z’w’ + x’y’zw + x’yw + y’z’w y y’ x 1 1 w 1 1 w’ x’ 1 1 1 1 w z z’ z Minimized form: xz’+x’w x’zw’ + yz’w’ + x’y’z’ + y’z’w’ + x’yz’ y y’ x w 1 1 w’ x’ 1 1 1 1 1 1 w z z’ z Minimized form: x’w’+z’w’+x’z’ General procedure for Karnaugh-map-based minimization of Boolean functions: 1. The Karnaugh map is a table of squares (2n squares when you have n variables) 2. Divide the map into regions so that each variable “owns” half of the squares, and its complement owns the other half. Each square will end up being owned by n literals (making up a minterm). 8 For example, for 3 variables, the map (of 8 sqaures) is y y’ x x’ z z’ z The variable x owns the top 4 yellow squares. Its complement x’ owns the bottom yellow row of squares. The variable y owns the left half of the squares, and y’ owns the right half. The variavle z owns the left and right column of yellow squares, and its complement z’owns the two middle columns of yellow squares. For 4 variables, the map (of 16 squares) is: y y’ x w w’ x’ w z z’ z o The variable x owns the top two rows of yellow squares. Its complement x’ owns the bottom two yellow rows of squares. o The variable y owns the left half of the squares, and y’ owns the right half. o The variavle z owns the left and right column of yellow squares, and its complement z’owns the two middle columns of yellow squares. o Finally, the variable w owns the top and bottom row of yellow squares, and its complement owns the two middle rows of the yellow squares. 3. Fill the map for a given function f: for each minterm in f, put “1” inside the square corresponding to (or owned by) that minterm. 4. Grouping the filled squares: Group the 1’s into rectangles of totally filled squares such that o the length and width of each rectangle are powers of 2 o no filled square remains ungrouped o rectangles can overlap o each rectangle must exclusively own at least one filled square. 5. Convert each rectangle to a product of literals: For each rectangle, identify the literals where each literal owns the entire rectangle, then multiply those literals. 6. The minimized form is the sum of the products derived in the previous steps.
10088	https://pmc.ncbi.nlm.nih.gov/articles/PMC11349631/	Fundamental equations and hypotheses governing glomerular hemodynamics - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Front Physiol . 2024 Aug 14;15:1440627. doi: 10.3389/fphys.2024.1440627 Search in PMC Search in PubMed View in NLM Catalog Add to search Fundamental equations and hypotheses governing glomerular hemodynamics Serena Y Kuang Serena Y Kuang 1 Department of Foundational Medical Studies, Oakland University William Beaumont School of Medicine, Rochester, MI, United States Find articles by Serena Y Kuang 1,,†, Besjana Ahmetaj Besjana Ahmetaj 1 Department of Foundational Medical Studies, Oakland University William Beaumont School of Medicine, Rochester, MI, United States Find articles by Besjana Ahmetaj 1, Xianggui Qu Xianggui Qu 2 Department of Mathematics and Statistics, Oakland University, Rochester, MI, United States Find articles by Xianggui Qu 2 Author information Article notes Copyright and License information 1 Department of Foundational Medical Studies, Oakland University William Beaumont School of Medicine, Rochester, MI, United States 2 Department of Mathematics and Statistics, Oakland University, Rochester, MI, United States Edited by:Irena Levitan, University of Illinois Chicago, United States Reviewed by:Praghalathan Kanthakumar, University of Missouri, United States Elizabeth LeMaster, University of Illinois Chicago, United States ✉ Correspondence: Serena Y. Kuang, kuang@oakland.edu † ORCID: Serena Y. Kuang, orcid.org/0000-0001-6492-6839; Xianggui Qu, orcid.org/0000-0003-2448-4890 Received 2024 May 29; Accepted 2024 Jul 19; Collection date 2024. Copyright © 2024 Kuang, Ahmetaj and Qu. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. PMC Copyright notice PMCID: PMC11349631 PMID: 39206387 Abstract The glomerular filtration rate (GFR) is the outcome of glomerular hemodynamics, influenced by a series of parameters: renal plasma flow, resistances of afferent arterioles and efferent arterioles (EAs), hydrostatic pressures in the glomerular capillary and Bowman’s capsule, and plasma colloid osmotic pressure in the glomerular capillary. Although mathematical models have been proposed to predict the GFR at both the single-nephron level and the two-kidney system level using these parameters, mathematical equations governing glomerular filtration have not been well-established because of two major problems. First, the two-kidney system-level models are simply extended from the equations at the single-nephron level, which is inappropriate in epistemology and methodology. Second, the role of EAs in maintaining the normal GFR is underappreciated. In this article, these two problems are concretely elaborated, which collectively shows the need for a shift in epistemology toward a more holistic and evolving way of thinking, as reflected in the concept of the complex adaptive system (CAS). Then, we illustrate eight fundamental mathematical equations and four hypotheses governing glomerular hemodynamics at both the single-nephron and two-kidney levels as the theoretical foundation of glomerular hemodynamics. This illustration takes two steps. The first step is to modify the existing equations in the literature and establish a new equation within the conventional paradigm of epistemology. The second step is to formulate four hypotheses through logical reasoning from the perspective of the CAS (beyond the conventional paradigm). Finally, we apply the new equation and hypotheses to comprehensively analyze glomerular hemodynamics under different conditions and predict the GFR. By doing so, some concrete issues are eliminated. Unresolved issues are discussed from the perspective of the CAS and a desinger’s view. In summary, this article advances the theoretical study of glomerular dynamics by 1) clarifying the necessity of shifting to the CAS paradigm; 2) adding new knowledge/insights into the significant role of EAs in maintaining the normal GFR; 3) bridging the significant gap between research findings and physiology education; and 4) establishing a new and advanced foundation for physiology education. Keywords: glomerular hemodynamics, renal plasma flow, glomerular filtration rate, efferent arteriole, mathematical model, colloid osmotic pressure, net filtration pressure, complex adaptive system 1 Introduction The mammalian kidney is a vital organ responsible for maintaining homeostasis by regulating fluid balance, electrolytes, and waste removal through urine production. It plays a crucial role in the overall health and functionality of the body. The kidney is unique in the body as it is the only organ that has two arterioles and two capillary beds aligned in a series. The renal autoregulation mechanisms (myogenic response and tubuloglomerular feedback [TGF]) keep the renal plasma flow (RPF) and glomerular filtration rate (GFR) stable when arterial blood pressure fluctuates within a broad range. This indicates the importance of maintaining a normal GFR that is critical to the homeostasis of the internal environment. Precise regulation of the GFR depends on the balance between the resistances of afferent arterioles (AAs) and efferent arterioles (EAs), which together determine the net filtration pressure (NetP) that drives glomerular filtration. NetP is the sum of the four Starling forces across a glomerular capillary wall. The two Starling forces that favor glomerular filtration are the hydrostatic pressure in the glomerular capillary (P GC) and the colloid osmotic pressure in Bowman’s capsule (π BC), where π BC is zero or negligible in the normal situation and becomes significant in patients with various renal diseases. The two Starling forces that oppose glomerular filtration are the plasma colloid osmotic pressure in the glomerular capillary (π GC) and the hydrostatic pressure in Bowman’s capsule (P BC). The filtration fraction (FF) is the ratio of the GFR/RPF and is about 20% in the normal situation. All of these parameters together characterize glomerular hemodynamics and determine the GFR. The parameters of glomerular hemodynamics are well-known and have been used to establish mathematical models to predict the GFR at both the single-nephron (SN) level and the two-kidney system level (Deen et al., 1974; Huss et al., 1975; Brenner et al., 1976; Navar et al., 1977; Chang, 1978; Papenfuss and Gross, 1978; Tucker and Blantz, 1981; Sgouralis and Layton, 2015). Nevertheless, two major problems exist: first, the two-kidney system-level equations are simply extended from the equations at the SN level, which is inappropriate in epistemology and methodology. Second, EAs play an important role in glomerular hemodynamics and, thus, the GFR, but the role of EAs in the maintenance of the normal GFR is underappreciated. These two problems are elaborated in terms of a total of six concrete issues in the next section and collectively show the need for a shift in epistemology toward a more holistic and evolved way of thinking reflected in the concept of the complex adaptive system (CAS; Holland, 2006; Carmichael and Hadžikadić, 2019). After this elaboration, we illustrate eight fundamental equations and four hypotheses that govern glomerular hemodynamics at both the SN and two-kidney system levels as the theoretical foundation of glomerular hemodynamics. This illustration modifies some equations in the literature, establishes a new equation in the conventional paradigm of epistemology, and formulates four new hypotheses through logical reasoning from the perspective of the CAS (beyond the conventional paradigm). Finally, we apply the new equation and hypotheses we established to comprehensively analyze glomerular hemodynamics under different conditions and predict the GFR. By doing so, some concrete issues are eliminated. Unresolved issues are discussed from the perspective of the CAS and a desinger’s view. The methodology in this article is logical, largely quantitative, and systematic. The significance of this article is as follows: 1) it makes clear the necessity of shifting the epistemology that guides research from a conventional paradigm toward a CAS paradigm; 2) it adds new knowledge/insights to understand the significant potential role of EAs in maintaining the normal GFR, which has been underappreciated; 3) it bridges the significant gap between research findings and physiology education; and 4) it establishes a new and advanced foundation for physiology education in which glomerular hemodynamics should be illustrated at the SN and two-kidney system levels. 2 How are system-level equations extended from the SN level, and why is the role of EAs in the GFR underappreciated? Different from the pressure profile in the peripheral capillaries, the decrease in P GC during glomerular filtration is insignificant, so P GC is constant throughout glomerular filtration (Figure 1). Meanwhile, π GC is a variable that increases linearly (Figure 1A) or nonlinearly (Figures 1B,C) during glomerular filtration (Brenner et al., 1976; Giebisch et al., 2017; Hall and Hall, 2021). The rising orange line labeled Q in Figure 1 shows the sum of the two opposing pressures to glomerular filtration: π GC + P BC. P BC is constant under normal conditions (Giebisch et al., 2017) and in Figure 1, and π BC is ignored under normal situations and, thus, not drawn. If the orange line rises, it means that π GC increases because P BC remains constant. FIGURE 1. Open in a new tab Starling forces across the wall of a glomerular capillary (GC) in three different situations. (A) Normal stable glomerular hydrostatic pressure (P GC) and increasing plasma colloid osmotic pressure (π GC); (B) initial moderate elevations in P GC and π GC due to moderate efferent arteriole (EA) constriction; and (C) initial significant elevations in P GC and π GC due to severe EA constriction. P BC: hydrostatic pressure in Bowman’s capsule, x 1 and x 2: distances in the capillary from the afferent arteriole end. Based on Figure 1, multiple issues can be addressed as follows. In the literature, different symbols are used to refer to the parameters of glomerular hemodynamics. In this article, we deal with the equations that model glomerular hemodynamics at both the SN and system levels; hence, the parameters at these levels are clearly differentiated to avoid confusion, and new terms are defined when necessary. 2.1 Issue 1 The widely used equation, NetP = P GC – π GC – P BC, is qualitative and vague. π GC increases during glomerular filtration, so it is unclear which value of π GC, such as an instantaneous π GC or the mean π GC throughout glomerular filtration (), should be used in the equation. It is also unclear how should be quantified when it increases nonlinearly (Figures 1B,C). Subsequently, whether NetP is instantaneous or the total NetP SN or the mean NetP throughout glomerular filtration () is not addressed. These questions are sometimes addressed in research (Brenner et al., 1976; Navar et al., 1977; Papenfuss and Gross, 1978) but do not appear in physiology education, leaving the equation quantitatively inappropriate and useless. 2.2 Issue 2 Another widely used equation, GFR = K f(NetP) = K f(P GC – π GC – P BC), where K f refers to the filtration coefficient (Drumond and Deen, 1994; Leatherby et al., 2021), is more vague or ambiguous. It has the same problem addressed above. Moreover, • P GC, π GC, and P BC are pressures across a glomerular capillary wall, whereas the GFR is the filtration achieved by the two kidneys (with numerous nephrons) per unit time. It is inappropriate to calculate the GFR at the system level using these single-capillary-level pressures. • The filtration coefficient is the product of the hydraulic permeability of the filtration membrane (K or L P A) and the filtration area (Brenner et al., 1976; Tucker and Blantz, 1977; Hall and Hall, 2021). The total filtration area of the two kidneys is remarkably different from the filtration area of an SN. However, the symbols that refer to them are inconsistent: either SNK f or K f is used to refer to the filtration coefficient at the SN level (Brenner et al., 1976; Ott et al., 1976; Navar et al., 1977; Marchand and Mohrman, 1980; Savin and Terreros, 1981; Arendshorst and Gottschalk, 1985), whereas K f also refers to the filtration coefficient at the two-kidney level (Giebisch et al., 2017; Hall and Hall, 2021). It should be noted that the unit of the filtration coefficient for the SN is nl/min/mmHg (Brenner et al., 1976; Navar et al., 1977; Marchand and Mohrman, 1980; Savin and Terreros, 1981) or nl/sec/mmHg (Tucker and Blantz, 1977; Chang, 1978; Arendshorst and Gottschalk, 1985), whereas the unit of the coefficient for the two-kidney system is ml/min/mmHg (Costanzo, 2018; Hall and Hall, 2021). • It is unclear what NetP refers to in the equation. In other words, it remains unclear whether it should be averaged at the SN level or the mean NetP averaged from numerous nephrons at the system level (). • If NetP in this equation applies to an SN, then it should be total NetP SN or , the GFR should be SNGFR, and the filtration coefficient should be SNK f. If NetP refers to , then the GFR should remain the GFR; the filtration coefficient should be K f; and P GC, π GC, and P BC should be averaged at the system levels , , and , respectively. To the best of our knowledge, this is the first time that two sets of parameters have been defined and differentiated systematically to avoid confusion. Specifically, P GC, P BC, dπ GC/dx, , SNK f, d(NetP)/dx, total NetP SN, SNRPF, SNGFR, and SNFF apply to the SN level, and RPF, GFR, FF, K f, , , , and apply to the system level. 2.3 Issue 3 Whether filtration equilibrium occurs in mammalian kidneys remains a continuous debate (Osgood et al., 1982; Arendshorst and Gottschalk, 1985). Filtration equilibrium refers to the phenomenon when NetP decreases to zero at a point before the blood reaches the EA so that no filtration occurs after this point (Figure 1C). Filtration equilibrium has been reported in some experimental studies on Munich–Wistar rats (Marchand and Mohrman, 1980) and squirrel monkeys (Maddox et al., 1974) but has not been observed in studies on dogs (Ott et al., 1976), Wistar rats (Seiller and Gertz, 1977), and rabbits (Denton and Anderson, 1991). In other words, glomerular filtration in the kidneys of these animals is characterized by filtration disequilibrium. 2.4 Issue 4 A critical gap in current research is the notable lack of research questions and efforts to investigate whether there is a direct and/or indirect communication between an upstream AA and its downstream EA. This situation may lead to missing crucial insights into glomerular hemodynamics. The advantages of these types of communication are obvious. For example, in the design of artificial nephrons or kidneys, enabling these communications could potentially improve the coordination between AAs and EAs and lead to more efficient function. On the other hand, Davis (1991) reported that under certain circumstances, TGF may involve EA vasomotion either in the same or opposite direction of AA vasomotion. If there is no communication between EAs and AAs, necessary vasomotion of AAs and/or EAs to maintain the normal GFR may be mediated through the TGF. If so, it is neither efficient nor economical. 2.5 Issue 5 Physiologists often note that FF = GFR/RPF ≈ 20%. Obviously, this means that a much larger fraction of RPF (80%) is not filtered but exits through the EAs under normal conditions. This 80% fraction is apparently ignored because its implications for glomerular hemodynamics and maintaining the normal GFR are not mentioned, appreciated, or discussed in the literature. The importance of having 80% RPF exiting through EAs becomes clear gradually in this article, and its implications are addressed in Conclusion. 2.6 Issue 6 In terms of how AA resistance influences glomerular hemodynamics, there is no disagreement among physiologists in general. However, the explanations of how EA resistance influences glomerular hemodynamics are inconsistent, incomplete, and inappropriate: • Some literature only introduce the effect of AA resistance on the GFR but not the effect of EA constriction (Pal et al., 2017; Kibble, 2020; Loscalzo et al., 2022; Eaton and Pooler, 2023). • EA constriction may increase both P GC and π GC but may or may not reduce RPF. However, textbooks often merely mention that EA constriction increases P GC and/or GFR and do not address how it influences RPF and/or π GC (Bijlani and Manjunatha, 2011; Krishna, 2015; Koeppen and Stanton, 2018; Barrett et al., 2019). • Some literature briefly note that EA constriction has a biphasic effect on the GFR depending on whether EA constriction reduces RPF and how significantly it increases π GC. Specifically, if the EAs constrict slightly, which reduces RPF insignificantly or not at all, then the GFR increases slightly. However, if the EAs constrict severely (causing a threefold or more increase in the EA resistance), the RPF and GFR are both reduced because under this circumstance, the increase in π GC (∆π GC) becomes greater than the increase in P GC (∆P GC), i.e., ∆π GC> ∆P GC (Giebisch et al., 2017; Hall and Hall, 2021). This means that the role of each phase of the biphasic effect of EA constriction on the RPF and, thus, GFR is conditional. Omitting the analysis of RPF, π GC, and whether ∆π GC> ∆P GC but stating that EA constriction increases or decreases the GFR, is logically flawed. The comparison between ∆π GC and ∆P GC is an indispensable step that determines whether the GFR increases or remains unchanged or decreases in response to a change in EA resistance. However, comparing the two is not appropriate because of the lack of logical rigor, as shown in issue 1 above. Theoretically, ∆π GC> ∆P GC needs to be replaced by > ∆P GC or a comparison of the total NetP SN in a situation with the normal total NetP SN illustrated in Section 4 (see Hypothesis 1). 2.7 Summary These issues, together, indicate the following: • The qualitative, vague equations cause insufficient and confusing definitions of the parameters at both the SN and system levels. • A system-level understanding of glomerular hemodynamics is mechanically extended from the SN level due to the lack of an appropriate epistemology. • A comprehensive understanding of the role of the EAs as a type of resistance vessel on glomerular hemodynamics and, thus, the GFR has not been well-established. 3 Eight fundamental mathematical equations within the conventional paradigm of epistemology The mathematical equations illustrated in this section can be reasoned out by anyone who understands the fundamentals of calculus or can be modified from the literature Eq. 1, Eq. 2, Eq. 6, and Eq. 7. SN-level parameters (SNRPF, d(NetP)/dx, total NetP SN, , , SNK f, SNGFR, and SNFF) and system-level parameters (RPF, , , , , K f, GFR, and FF) are easy to differentiate. Following common practice in the literature, all capillaries in a glomerulus are considered one tube with the same filtration area as all the capillaries together (Brenner et al., 1976; Chang, 1978; Drumond and Deen, 1994). Since π GC is the function of the distance (x) from a point of glomerular filtration to the beginning of the filtration (the AA end of the capillary), the vague expression NetP = P GC – π GC – P BC needs to be derived to quantify a derivative NetP [d(NetP)/dx)] and the total NetP SN [the integration of d(NetP)/dx)]: (1) and (2) where = , m = represents the slope of the orange line in Figure 1A, and D refers to the distance from the AA end to the end of glomerular filtration. The gray area in Figure 1A represents the total NetP SN determined by Eq. 2. If the orange line is a curve (Figures 1B,C), integrating the total NetP SN becomes complex. It requires conducting experiments, setting points to collect data, and then performing mathematical modeling, which is beyond the scope of these fundamental equations. Nevertheless, regardless of whether the total NetP SN can be easily integrated using Eq. 2 or needs a complex model, the gray area in Figures 1A–C represents the total NetP SN determined by the line of P GC and the orange line. Furthermore, an instantaneous SNGFR can be reasoned out or modified from the equation provided by Brenner et al. (1976) or Navar et al. (1977) using the symbols at the SN level defined in this article: (3) The total SNGFR can be reasoned out as follows or modified from the equation provided by Deen et al. (1974) or Sgouralis and Layton (2015) using the symbols at the SN level defined in this article: (4) Similarly, the vague expression GFR = K f(NetP) = K f(P GC – π GC – P BC) needs to be derived to estimate the GFR using the symbols at the two-kidney system level defined in this article: (5) Obviously, the many mean values in the equation can only be estimated for the millions of nephrons at the two-kidney level. This equation makes better sense than its original form [GFR = K f(NetP) = K f(P GC – π GC – P BC)]. It is theoretically meaningful but is still of no practical use. Practically and clinically, the GFR can be calculated using inulin clearance or estimated using creatinine clearance. The following equation estimates K f: (6) Hladunewich et al. (2004) estimated K f for pregnant women using this equation, where the estimated mean total number of nephrons for healthy women between the ages of 20 and 50 years is 1.4 × 10 6 (Nyengaard and Bendtsen, 1992). The relationship between the GFR and SNGFR can be expressed as (7) Since RPF is the plasma flow that enters AAs and FF = GRF/RPF ≈ 20%, we establish the following equation to describe the distribution of RPF after entering AAs at the two-kidney system level: (8) Eq. 8 leads to the formulation of the last hypothesis in the next section and is critical to resolve issue 6 and understand the significant potential role of EAs in maintaining the normal GFR when renal autoregulation fails to maintain the normal RPF. The two kidneys as a whole have numerous nephrons (about 30,000 in a rat kidney and 10 6 in a human kidney; Sgouralis and Layton, 2015). These nephrons not only have similarities in their structures and functions but also exhibit heterogeneity in their structural and functional aspects from the molecular level to the cellular, nephron, and regional levels. All of these contribute to the complexity of the system (the two kidneys). Moreover, a system with numerous components exhibits emergent properties that its components or agents (in this context, single nephrons) do not possess, such as a great capacity of resilience and adaptability to internal and external perturbations, as well as nonlinearity. Nonlinearity means that the response of such a system toward a perturbation is often unproportional to the strength of the perturbation (Janson, 2012), and a perturbation to the system may cause a large nonproportional response, a proportional response, or no response at all. For instance, Denton et al. (2000) reported that administering intrarenal angiotensin II caused a decrease in RPF with a concomitant increase in FF in a dose-dependent manner so that the GFR does not decrease but is maintained with no change. Their research also showed the following observations: • From the outer cortex to the juxtamedullary cortex, the diameters of the EAs show a gradient: those with the smallest diameters are in the outer cortex, whereas those with the largest diameters lie in the juxtamedullary cortex. • The diameters of the EA in the outer and mid cortexes are smaller than those of the AA, but the diameters of the EA in the juxtamedullary cortex are similar to those of the AA. • Such heterogeneity in the diameters of the EA seems to be one of the reasons that angiotensin II has differential degrees of vasoconstrictive effects on the AAs and EAs. According to Poiseuille’s equation, which states that resistance is inversely proportional to the fourth power of the radius, it can be predicted that the EAs in the outer cortex with the smallest diameters can impact glomerular hemodynamics most substantially. • Since outer and midcortical glomeruli account for about 90% of all glomeruli, in general, angiotensin II tends to cause a higher increase in EA resistance than in AA resistance. Therefore, the sensitivity of glomerular hemodynamics to a minor change in EA resistance should not be ignored. In other words, the EAs possess great potential to regulate glomerular hemodynamics in various ways to maintain a normal GFR. If we consider the two kidneys a renal CAS (Holland, 2006; Carmichael and Hadžikadić, 2019) with both similarities and heterogeneities in its agents and emergent properties at the system level, it becomes clear why extending the SN-level equations to describe the system-level equations is inappropriate in epistemology and methodology. However, so far, mathematical modeling of a CAS is difficult and needs further advancement. From the perspective of the CAS and a designer’s view, the debate about whether filtration equilibrium does or does not occur in mammalian nephrons may be reconsidered to avoid a mechanical, mutually exclusive approach and facilitate a holistic and dynamic approach: • “Fitness functions that are inherent in nature are always pushing the system, any system, toward more efficient use of resources” (Carmichael and Hadžikadić, 2019). Filtration equilibrium makes a fraction of the capillary useless for filtration. Therefore, it is worth considering that filtration equilibrium might not be the normal condition and may occur only under specific circumstances. Theoretically, if SNRPF is low and/or SNK f is high and/or the constriction of an EA is severe, filtration equilibrium may occur (Arendshorst and Gottschalk, 1985) in some glomeruli. Practically, multiple factors may encourage or prevent it. It is crucial to identify these factors and determine whether filtration equilibrium is more or less likely to occur in specific regions of the kidney. • If we design an artificial kidney, it is important to determine whether it is beneficial for filtration equilibrium and disequilibrium to be mutually transformable under some conditions for the sole purpose of increasing the capacity of both resilience and adaptability of the kidneys. Alternatively, it should be assessed whether filtration equilibrium should be more likely to appear in some nephrons and less likely to occur in others for the same purpose. • Renal heterogeneity could be a consequence of the past adaptive processes of the renal CAS toward internal and external perturbations for the purpose of maintaining a normal GFR. It is essential to explore whether renal heterogeneity should exhibit different patterns at various levels, from molecular to cellular, nephron, and system, in response to different perturbations. • Developing methods to study and recognize different patterns of renal heterogeneity is critical for advancing our understanding of kidney function. The eight equations given above are generally linear or simple models; thus, we consider them fundamental in the study of glomerular hemodynamics, or more specifically, glomerular filtration. To model other aspects of glomerular hemodynamics or, more broadly, renal hemodynamics, such as renal autoregulation (myogenic response and TGF) and coupled nephrons, much more complex mathematical models are needed, and readers may refer to the review article by Sgouralis and Layton (2015). Like the eight fundamental mathematical models mentioned above, complex models have the same problem, i.e., how the SN and two-kidney levels of models can be well-integrated by taking both the similarity and heterogeneity of nephrons into consideration in the direction of the CAS. 4 Four hypotheses from the perspective of the complex adaptive system Obviously, the gray area in Figure 1C due to severe EA constriction is smaller than the normal gray area in Figure 1A. If the SNGFR were to be calculated for the condition in Figure 1C, it would be less than the normal SNGFR in Figure 1A. Depending on the concrete value of SNRPF, the exact degree of EA constriction, and the resulting P GC and π GC, the gray area due to moderate EA constriction (Figure 1B) may be greater than, equal to, or smaller than the normal gray area in Figure 1A. Note that in Figure 1B, before x 1, NetP (x < x 1) is greater than that in the normal situation (Figure 1A); after x 1, NetP (x > x 1) becomes smaller than normal; and at x 1, NetP (x 1, Figure 1B) = the normal (x 1 in Figure 1A)1 (this analysis also applies to Figure 1C). The closer x 1 is to the AA end in Figure 1B, the more likely it is that the resulting gray area is smaller than the normal gray area in Figure 1A; on the contrary, the farther x 1 is from the AA end in Figure 1B, the more likely the gray area is to be equal to or greater than the normal gray area in Figure 1A. This analysis makes it clearer that a comparison of whether ∆π GC> ∆P GC is not practical, but a comparison of a gray area with the normal gray area is doable. Hence, logically, the comparison of ∆π GC and ∆P GC should be replaced by Hypothesis 1: Hypothesis 1 If the total NetP NS (the gray area) < or = or > the normal, then SNGFR < or = or > the normal. In other words, the SNGFR decreases or remains unchanged or increases. Hypothesis 1 is qualitative and specifically useful for physiology education, which is so far not math-heavy. From now on, if the total NetP SN is reduced compared to the normal, it means that the mean increase in π GC is greater than the increase in P GC, i.e., > ∆P GC. Subsequently, Hypothesis 2 If the total NetP SN in a significant number of nephrons increases/decreases, increases/decreases so that the GFR increases/decreases. Note that from the total NetP SN at the SN level to at the system level, the description “significant number of nephrons increases/decreases” reflects the CAS. Since renal autoregulation mechanisms conventionally involve the AA and macula densa, not the EA, from a designer’s view, we hypothesize that there are unknown direct or indirect communications (mechanical, electrical, chemical, or biological) between an upstream AA and its downstream EA, or between a glomerulus and its downstream EA, or between the macula densa and its adjacent EA, so that these parties work efficiently and in coordination to precisely maintain the normal GFR as far as possible, especially when renal autoregulation mechanisms fail to maintain stable RPF: Hypothesis 3 If the GFR is greater than the normal upper range, generally speaking, the AAs should constrict, or the EAs should dilate, or both; on the contrary, if the GFR is less than the normal lower range, the AAs should dilate or the EAs should constrict or both. To date, the occasional involvement of the EA in TGF has been reported under some circumstances (Davis, 1991; Ren et al., 2001); some autocoids (e.g., nitric oxide and prostgalndins etc.) produced in the glomerulus may diffuse to influence EA vasomotion (Ito and Abe, 1997; Leipziger and Praetorius, 2020). Chilton et al. (2008) speculated a direct communication between an upstream AA and its downstream EA via electrical potential change in the smooth muscles of the EA. Denton et al. (2000) reported that angiotensin II changed the geometry of the glomerular pole including the extraglomerular mesangium, whereas Elger et al. (1998) suggested a direct functional influence of an AA on an EA via the extraglomerular mesangium and the presence of a specific sheer stress receptor located in the intraglomerular portion of the EA. Further research to explore the hypothesized communications between an upstream AA and its downstream EA will be of great value. In Section 5, we show the pivotal role of this hypothesis in guiding our analyses of glomerular hemodynamics and GFR under various conditions and eventually resolve issue 6. Next, we reason out Hypothesis 4 using the following data. If the RPF of a healthy man is approximately 600 mL/min, the GFR is approximately 125 mL/min, and FF is approximately 20%, then his RPF EA (renal plasma flow that exits through the EAs) should be approximately 475 mL/min. Below, the unit ml/min is omitted for RPF, GFR, and RPF EA. Assume that his blood pressure decreases too much for some reason so that renal autoreglation can no longer maintain stable RPF, e.g., RPF decreases to 300. In order to maintain the GFR at approximately 125, according to Eq. 8, the EAs should constrict to cause RPF EA to be approximately 175. If RPF EA> or <175, then GFR < or > normal, indicating that the total NetP SN in a significant number of nephrons in the two kidneys < or > their normal values. Hence, 175 is the critical point of RPF EA in response to the primary change of RPF = 300. If RPF is approximately 400, then the critical point of RPF EA should be approximately 275. Due to the heterogeneity of the nephrons, it is not easy to obtain or estimate the total NetP SN for the majority of the nephrons. The purpose of introducing the concept of the critical point of RPF EA at the two-kidney level is to use it to estimate what is more likely to happen in the majority of the nephrons in terms of their total NetP SN when RPF is below normal: Hypothesis 4a If RPF EA> or = or < a critical point in response to a particular value of RPF, the total NetP SN in a significant number of nephrons is < or = or > their normal values. On the contrary, if EA vasomotion is the primary change, then RPF has a critical point in response to a particular EA resistance, which also predicts the total NetP SN: Hypothesis 4b If RPF > or = or < a critical point in response to a particular value of RPF EA, the total NetP SN in a significant number of nephrons is > or = or < their normal value. Hypothesis 4 is inferred from Eq. 8. This is the first significance of Eq. 8. The illustration of the equations and hypotheses leads to the criteria to define the following terms in this article: • Slight EA constriction means that the EAs constrict slightly, which does not reduce RPF but redistributes it to the GFR and RPF EA. • Severe EA constriction means that the EAs constrict significantly, causing a threefold or more increase in EA resistance and reducing RPF, which results in the total NetP SN in the majority of the nephrons becoming smaller than normal (Figure 1C) and, thus, a decrease in and the GFR. • Moderate EA constriction is between slight and severe constriction, which may lead to an increase or decrease or no change in the GFR depending on the resulting RPF, P GC, π GC, and total NetP SN in the majority of the nephrons, as shown in Figure 1B. Figure 2 depicts a flowchart of how the parameters of the glomerular hemodynamics of a single nephron should be analyzed after a change in either AA or EA resistance or both without missing links. This flowchart will help eliminate the logical flaws addressed in issue 6. Figure 3 depicts a flowchart of how to view glomerular hemodynamics at the two-kidney level by considering the two kidneys as a renal CAS. FIGURE 2. Open in a new tab Logical steps to analyze the parameters of the glomerular hemodynamics of a single nephron after a change in the resistance of either an afferent arteriole (AA) or an efferent arteriole (EA) or both. SNGFR: single-nephron GFR; SNFF: single-nephron FF; orange lines: single-nephron autoregulation mechanisms; dashed orange lines: some supportive findings in the literature; dashed black line: authors’ hypothesis. FIGURE 3. Open in a new tab Renal complex adaptive system (CAS) for glomerular filtration with emergent, macroscopic properties (in red). 5 A comprehensive analysis to understand the impact of EA constriction on glomerular hemodynamics Guided by the flowcharts given in Figures 2, 3, issue 6 is resolved in this section by applying Eq. 8 and Hypothesis 3 and Hypothesis 4b. 5.1 The essential roles of the EAs and EA baseline resistance When renal autoregulation maintains stable RPF and GFR, why is the P GC higher than that in the peripheral tissue and relatively stable throughout the glomerular filtration (Figure 1A)? A normal EA tone (a certain level of constant constriction) maintains the P GC high and relatively constant (Savedchuk et al., 2023). This is the essential role of the EAs. This essential role is conditional upon the renal autoregulation of RPF functioning normally. A normal EA tone results in normal EA resistance (level-0 resistance). This normal EA resistance is the baseline when renal autoregulation mechanisms function normally. EA constriction means that the EAs constrict more than the level of the normal EA tone; thus, EA resistance becomes higher than the baseline. EA dilation means that the EAs constrict less than the normal EA tone; thus, EA resistance becomes smaller than the baseline. Throughout our analysis, each level of EA constriction sets a new baseline for EA resistance. EA constriction/dilation means that the EAs constrict more/less than a corresponding level of EA constriction. 5.2 Comprehensive analysis of the effect of EA constriction on glomerular hemodynamics The data on RPF, GFR, and FF of the healthy man mentioned above when we addressed Hypothesis 4 are used to facilitate the analysis. Due to renal heterogeneity, glomerular hemodynamics is analyzed at the system level not the SN level. 5.2.1 Slight EA constriction, which sets a level-1 baseline EA resistance This situation does not reduce RPF significantly or at all but redistributes RPF (∼600) to RPF EA and GFR (Eq. 8). The more the EAs constrict, the more , GFR, and FF increase. According to Hypothesis 3, the EA should stop constricting and dilate (with reference to level 1-EA baseline resistance) to reduce the GFR to normal. 5.2.2 Moderate EA constriction, which sets a level-2 baseline EA resistance, e.g., EA resistance ≈1.5-fold of level-0 EA baseline resistance or RPF EA ≈ 316.7, 2/3 of the original (475) This situation causes a reduction in RPF and an increase in both and , in general. How it influences , GFR, and FF depends on the exact degree of EA constriction and the concrete RPF. If the resulting is around normal, then the GFR is around normal, but FF increases because of a reduction in RPF. If RPF EA remains approximately 316.7, RPF needs to be adjusted to 441.7 (441.7 = 125 + 316.7, Eq. 8) to maintain the GFR at approximately 125. According to Hypothesis 4b, RPF of approximately 441.7 is the critical point for RPF EA of approximately 316.7. If RPF > or <441.7, then GFR > or < normal, > or < normal, and FF should decrease or increase. According to Hypothesis 3, the EA should dilate or constrict with reference to the level-2 baseline EA resistance. Therefore, level-2 EA constriction is certain to cause a reduction in RPF and an increase in FF but embraces multiple possibilities for the values of and GFR. 5.2.3 Severe EA constriction, which sets a level-3 baseline EA resistance, e.g., EA resistance = or > 3-fold of level-0 EA baseline resistance or RPF EA ≈ 158.3 (1/3 of 475) This situation causes significant reduction in RPF and , which leads to a decrease in the GFR and an increase in FF. Under such circumstances (most likely, renal autoregulation mechanisms have failed to maintain the normal, stable RPF), the critical point of RPF with regard to RPF EA of approximately 158.3 is 278.3 (278.3 = 125 + 158.3, Eq. 8), if possible. If RPF > or <278.3, then the EA should dilate or constrict with reference to the level-3 baseline EA resistance to adjust the GFR to be not too far from normal. This means, logically, that even though RPF is reduced significantly, as long as the EA can constrict more, it still has the potential to maintain the normal GFR. FF should always increase at level-3 EA constriction. Practically, the EAs may not be able to constrict more, especially if they are smaller in diameter. In addition, it should be noted that if the reduction of RPF is not due to EA constriction, but due to other reasons, the analysis should still revolve around how to maintain the normal GFR by applying Eq. 8, Hypothesis 3 and Hypothesis 4a. 5.2.4 More severe EA constriction that causes RPF (e.g., 100) < normal GFR (e.g., ∼120) This situation can occur under some pathological conditions such as acute renal failure, when both the AAs and EAs constrict significantly as part of systemic vascular constriction. Regardless of how severely the EAs constrict, a normal GFR cannot be maintained. Therefore, administering vasodilator(s) to the AA and EA is a preventive action if acute renal failure is likely to occur and life-saving if acute renal failure is already occurring. This is an exception to Hypothesis 3. 5.2.5 Summary The analyses in this section are guided by the logical relationships shown in Figures 2, 3, Eq. 8, and Hypothesis 3 and Hypothesis 4b. They are much more comprehensive than those in the literature and in standard physiology textbooks, without logical errors, an address the following two important points: • The impact of EA constriction on glomerular hemodynamics is quite conditional, depending on concrete situations. • The parameters that should and should not be analyzed must be differentiated in this context in consideration of renal heterogeneity. 6 Conclusion This article presents the following outcomes. First, it makes clear the need for a shift in epistemology to adopt the concept of the CAS and a designer’s view. Second, some fundamental equations are modified/improved, and one new equation is established in the conventional paradigm. Four new hypotheses are formulated from the perspective of the CAS with guiding significance for future research. Third, new insights to understand the role of EAs as resistance arterioles are developed, specifically: • RPF EA (80% RPF) serves as an adequate reserve of the normal GFR. This reserve becomes significant when renal autoregulation fails to maintain normal RPF, and RPF is significantly low. Theoretically, as long as RPF > normal GFR, EA constriction has the potential to adjust EA resistance to maintain the GFR at a normal level. Therefore, the distribution of 80% RPF to EAs, in particular, plays a protective role in the maintenance of the normal GFR. If the fractions of a normal GFR and RPF EA in RPF are reversed, i.e., normal GFR ≈80% RPF and RPF EA ≈ 20% RPF, the EA will not be able to effectively protect the normal GFR in any scenario. This is the second significance of Eq. 8. • Having an EA aligned in series with an AA and a glomerulus for each nephron is the necessary condition to maintain the normal GFR, whereas having 80% RPF entering the EAs as a significant reserve to maintain the normal GFR when renal autoregulation fails to maintain the normal stable RPF is the sufficient condition to maintain the normal GFR. Without the sufficient condition, the kidney will lack resilience and adaptability and will be unable to cope with various internal and external perturbations. This analysis of necessary and sufficient conditions is borrowed from cybernetics. It theorizes our understanding of renal autoregulation of many of its functions at the philosophical level. If biomedical research studies adopt this perspective, more insights into the biomedical disciplines will emerge. • It is possible that if the pre-glomerular resistance increases or decreases inappropriately, the EAs have the sensitivity and potential to constrict or dilate to a certain degree to correct the error. This is because the heterogeneity in the diameters of the EAs and the distribution of EAs with different diameters in the renal cortex in contrast to AAs suggest the substantial power and potential of the EAs in the regulation of glomerular hemodynamics in various ways to maintain the normal GFR. Future research on glomerular hemodynamics should focus on recognizing patterns of renal heterogeneity in response to various perturbations, dynamic interactions among nephrons, and the emergent properties of the renal CAS. Acknowledgments The authors thank Cody Bailey-Crow for modifying Figure 1 and Savanna Lavander for helping with the literature. Funding Statement The author(s) declare that no financial support was received for the research, authorship, and/or publication of this article. Footnotes 1 ΔP GC = P GC (Figure 1B) – P GC (Figure 1A) = 60–53 = 7 (mmHg) and ∆Q(x 1) = Q(x 1) (Figure 1B) – Q(x 1;) (Figure 1A) = 42–35 = 7 (mmHg), meaning at x 1, ∆Q(x 1) = ΔP GC, instantaneous NetP(x 1) (Figure 1B) = instantaneous NetP(x 1) (Figure 1A). Data availability statement The original contributions presented in the study are included in the article further inquiries can be directed to the corresponding authors. Author contributions SK: conceptualization, formal analysis, investigation, methodology, project administration, resources, supervision, validation, visualization, writing–original draft, and writing–review and editing. BA: investigation, resources, validation, and writing–review and editing. XQ: conceptualization, investigation, methodology, resources, supervision, validation, visualization, and writing–review and editing. Conflict of interest The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. Publisher’s note All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors, and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher. References Arendshorst W. J., Gottschalk C. W. (1985). Glomerular ultrafiltration dynamics: historical perspective. Am. J. Physiol. 248 (2 Pt 2), F163–F174. 10.1152/ajprenal.1985.248.2.F163 [DOI] [PubMed] [Google Scholar] Barrett K. E., Barman S. M., Brooks H. L., Yuan J. X.-J. 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Articles from Frontiers in Physiology are provided here courtesy of Frontiers Media SA ACTIONS View on publisher site PDF (1.3 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1 Introduction 2 How are system-level equations extended from the SN level, and why is the role of EAs in the GFR underappreciated? 3 Eight fundamental mathematical equations within the conventional paradigm of epistemology 4 Four hypotheses from the perspective of the complex adaptive system 5 A comprehensive analysis to understand the impact of EA constriction on glomerular hemodynamics 6 Conclusion Acknowledgments Funding Statement Footnotes Data availability statement Author contributions Conflict of interest Publisher’s note References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
10089	https://www.youtube.com/watch?v=5zIK-D1DVgo	Symmetric and skew symmetric matrices \| Matrices \| Grade 12 \| Math \| Khan Academy Khan Academy India - English 540000 subscribers 15 likes Description 902 views Posted: 10 Apr 2025 In this video, we talk about symmetric and skew symmetric matrices. We first look at examples of each and then understand the constraints on the elements. We then verify 2 results that work for sum and difference of a matrix and its transpose. We finally derive these results using properties of transpose of a matrix. Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now! ( Timestamps: 0:00 What are symmetric matrices? 1:17 What are skew symmetric matrices? 4:10 Verifying results on sum and difference (with transpose) 6:15 Proving the results using properties Khan Academy India is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We have videos and exercises that have been translated into multiple Indian languages, and 15 million people around the globe learn on Khan Academy every month. Support Us: Created by Ashish Gupta Transcript: What are symmetric matrices? in this video we're going to talk about symmetric and skew symmetric matrices let's start with an example a is 321 2 4 -1 1 -1 7 it's a 3X3 and I'm saying that this is a symmetric Matrix pause the video think about where the Symmetry is in this symmetric Matrix okay if you look at it closely you'll see some elements repeating two here two here one here one here Min - one here- one here here not only are these elements matching some rows and columns are also matching look at the first row 3 2 1 look at the First Column 3 2 1 second Row 2 4 -1 second column 2 4 -1 third row 1 - 1 7 third column 1 - 1 7 you get the idea a is its own transpose if you take the transpose of a you'll get the same Matrix and this is what makes these Matrix is symmetric we can also write this as a i j is equal to a ji this is the condition that we need for a matrix to be symmetric if we switch the ith row and jth column we'll get the same element a i j is same as a ji now let's talk about SK symmetric What are skew symmetric matrices? matrices the condition is almost identical but with a negative sign B J is equal to negative of bji or we can write this as B is equal to negative of B transpose so in this case when you take the transpose you don't get the same Matrix you get the negative of it pause the video try writing a negative Matrix try writing an example okay let's do this together if you take some of these elements if you borrow them 3 4 7 2 1 -1 2 1 - 1 but this time let's not have 2 1 -1 let's have negatives of them Min -2 -1 and 1 let's see if this condition works if we switch the row and column we should get a negative element first row second column is minus 2 so second row First Column that should be negative of it that's two so this works for minus1 we have 1 for one we have minus one so this should work but is it really a q symmetric not really we didn't check the diagonals so let's check the diagonal as well for first row First Column we have three if we switch one and one we'll get the first row and First Column again but but that's not the negative of three that's still three so the diagonals are not working let's write this down if you write 1 one here if you take I and J as 1 b11 the first row First Column element that should be equal to the negative of itself well how can that be 3 is not negative of three so this does not work but if you have to make it work let's solve this equation we'll have twice of this element as zero which means the element itself is zero and Z really helps us out here because negative of 0 is also 0 so this element is 0 this element is 0o and this element is zero all elements on the diagonal are actually zero so let's correct this this is our SK symmetric Matrix 0 0 0 for the diagonals and on one side we have some elements on the other side we have the negatives of them - 2 -1 1 then 2 1 - 1 let's check for its transpose B transpose is 0 - 2 - 1 so 0 - 2 - 1 2 0 1 2 0 1 1 - 1 0 1 - 1 0 now let's zoom out let's see if B transpose is really the negative of B let's have a check if we put minus sign on all of these elements do we get B transpose yes we do negatives of 0 is still 0er minus of-2 is 2 minus of -1 is 1 minus of one is minus one and we get the same thing here as well so yes this works this is a skew symmetric Matrix we have to remember that all diagonal elements of all skew symmetric matrices are zero let's move on so now we know what symmetric and S Verifying results on sum and difference (with transpose) symmetric matrices are let's verify some results we have this Matrix a 1 2 3 2 4 6 5 2 1 Let's verify these two results if you add its transpose to itself you'll get a symmetric Matrix if you subtract its transpose from itself you'll get a ske symmetric Matrix pause the video try this on your own okay let's do this together let's first find its transpose we have a we need a dash 1 2 3 1 2 3 2 4 6 2 4 6 5 2 1 5 2 1 now let's add these two a + a dash that's going to be 1 + 1 that's 2 2 + 2 4 3 + 5 8 2 + 2 4 4 + 4 8 6 + 2 8 5 + 3 8 2 + 6 8 and 1 + 1 2 this is a + a Das now let's check whether this is symmetric or not so 2 48 first row is same as 2 48 First Column 488 second row 488 second column 882 third row and 8 82 third column so yes this is in fact a symmetric Matrix let's check for a minus a Das so if we subtract this we'll get 1 - 1 0 2 - 2 0 3 - - 5 is - 2 2 - 2 is 0 4 - 4 is 0 6 - 2 is 4 5 - 3 is 2 2 - 6 is -4 and then 1 - 1 is 0 Let's see if this is C symmetric we have all the zeros and for the other elements do we have negatives for 0 we have 0er that's the negative of itself Min - 2 2 this works 4 and minus 4 this also works so if we take its transpose we will get the negative of this metric so this means this is a skus metric so yes we have verified that this actually works now it's a good thing that we verified but pause the video try using the properties of transpose of a matrix to prove these two results as Proving the results using properties well okay let's do this together if you want to prove this first thing let's assume this new Matrix B is actually a plus a Das now let's take transpose on both sides B DH that's a + a d whole dash sum of two matrices taking transpose so we can take the transpose first and then add so B Das is equal to a d + a d DH and transpose of a transpose is the same Matrix so B Das is equal to a d + a now look at this B is A+ a D and B Das is a d plus a these two are the same thing which means B is same as B Dash which means B is actually a symmetric Matrix so whenever you add a matrix to its transpose you will always get a symmetric Matrix let's prove this other result as well if we take c as a minus a d and take the transpose on both sides what do we get C- is a minus a d-h this is A- minus A- D and transpose of transpose is the same thing so C- is A- minus a now let's compare C is a minus a d and c d is A- minus a C- is actually the negative of C that's the exact definition for skew symmetric which means C- is negative of C which means C is a skew symmetric so whenever you subtract a transpose from a matrix you will get skew symmetric when you add you'll get a symmetric
10090	https://www.quora.com/How-does-this-math-trick-work-See-link-below	How does this math trick work? See link below - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Multiplication Tricks Math Hacks Mathematics Trick Math Short Tricks Tricks of Maths Mathematics , Math Tricks 4 How does this math trick work? See link below All related (43) Sort Recommended Raman Sood I like maths, sport, science and the world. ·9y The steps are: Think of a number between 1 and 1000 multiply it by 20 (later when testing again i found that they change it every time but it is meant to be a factor of 100) add 35 to the answer (a random number that changes every time, it doesn't matter as long as it is below 100 and a multiple of 5) multiply it by 5 (the matching factor of 100 from step 2) subtract 85 from the answer (same as 3) add your age (1-99) to the answer type in the answer Then it gives you the result. From what i can see, all the steps above make it such that every output on step five (before you add your age) is one hund Continue Reading The steps are: Think of a number between 1 and 1000 multiply it by 20 (later when testing again i found that they change it every time but it is meant to be a factor of 100) add 35 to the answer (a random number that changes every time, it doesn't matter as long as it is below 100 and a multiple of 5) multiply it by 5 (the matching factor of 100 from step 2) subtract 85 from the answer (same as 3) add your age (1-99) to the answer type in the answer Then it gives you the result. From what i can see, all the steps above make it such that every output on step five (before you add your age) is one hundred numbers apart. What this means is that it knows the output that will come out after step five, and after adding your age it finds the closest number that is not larger than the number and subtracts it from the closest number that is smaller. (If it doesn't make sense now it will in the example) For example. When thinking of the number 1 at step five I get 530. When thinking of 2, 630 3,730 4, 830 and the list goes on. after adding your age to your number (for example i will take 1 as my age and will think of the number 1) it will take the number you give it (in my case 531), find the closest output from step five possible (530), subtract it, (1) and take that as well as the out put from step five and present it to you saying that it is a genius (1,1). Note that in my example i got 530 from thinking of 1, as every time it changes step 3 and 5 i won't all ways get 530. but it will always be 100 apart. Hope it helped and if you have any questions than please comment. Upvote · 99 15 9 2 Related questions More answers below What is the secret thing behind this math trick? How can the following maths trick's working be explained? What's the math behind the "boobless" (55378008) calculator trick? How do this math trick work (Guess a number and I know the answer)? Why does this math trick always yield 5? David Rutter B.S. Discrete Mathematics, Georgia Tech · Upvoted by Michael Jørgensen , PhD in mathematics · Author has 9.2K answers and 11.8M answer views ·9y The number you get after the first 6 steps is equal to 100 x+a+r 100 x+a+r, where x is the number you originally thought of, a is your age (which must be less than 100 if it is to work... notice it says "1-99", though I can't imagine a 1-year-old using this app), and r is an arbitrary number which depends on exactly which numbers they told you to add and subtract in the previous steps, but not on the number you are thinking of. Importantly, the app can readily calculate r and subtract it out of your answer. After it does so, suppose getting y as a result, computing your numbers is easy. The number you we Continue Reading The number you get after the first 6 steps is equal to 100 x+a+r 100 x+a+r, where x is the number you originally thought of, a is your age (which must be less than 100 if it is to work... notice it says "1-99", though I can't imagine a 1-year-old using this app), and r is an arbitrary number which depends on exactly which numbers they told you to add and subtract in the previous steps, but not on the number you are thinking of. Importantly, the app can readily calculate r and subtract it out of your answer. After it does so, suppose getting y as a result, computing your numbers is easy. The number you were thinking of is ⌊y/100⌋⌊y/100⌋ and your age is y mod 100 y mod 100. Upvote · 9 8 9 1 Assistant Bot · 1y It looks like you mentioned a link, but I can't access external content. However, if you describe the math trick or provide the details about it, I'd be happy to explain how it works! Upvote · Augustin Wenger 9y Maybe include the trick in the question ? Upvote · 9 4 9 2 Related questions More answers below What are some cool calculator math tricks for fun? Which is the best math tricks site? What are some easy maths tricks that can "trick" anyone? Which is the most unique maths trick? What are some tricks to solve math quickly? Balaji Kumar 9y I think you can get the solution by backtracking all the operations. (Go from the number you entered and go back to step 1). Let's say you followed these steps to get the result and x be the number you thought of. (((((x 20)+85)5) - 100)+ 30(age)) and you got the result 1355. Since the app asks you to enter the age it is easy to tell what your age is. Now we need to find the value of x. 1355 - 30 =1325 1325 + 100 = 1425 1425 ÷ 5 = 285 285 - 85 = 200 200 ÷ 20 = 10 10 is the number you thought. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 1 Kevin Lei Studied at University of Oxford · Upvoted by Emerson Studt , B.S. Mathematics, Massachusetts Institute of Technology (2019) and Vladimir Novakovski , silver medals, IOI 2001 and IPhO 2001 ·11y Related What's a math trick that is not very well-known? Wu's Squaring Trick, named after the famous Scott Wu, is a technique used to quickly square numbers over 25 in your head. It uses the identity n 2=(n−25)⋅100+(n−50)2 n 2=(n−25)⋅100+(n−50)2 For example, 62 2=37⋅100+12 2=3844 62 2=37⋅100+12 2=3844 44 2=19⋅100+6 2=1936 44 2=19⋅100+6 2=1936 77 2=52⋅100+27 2=5929 77 2=52⋅100+27 2=5929 These mental computations can be done very quickly with just a little practice, so amaze your friends with how fast you can square two digit numbers! Upvote · 9.5K 9.5K 999 128 99 10 David Prifti High School student. I love maths, programming and movies. · Upvoted by Ondrej Sluciak , PhD in signal processing, Math enthusiast, Languages and Peter Ferguson , MSci Theoretical Physics & Mathematics, Lancaster University (2018) · Author has 55 answers and 2.1M answer views ·8y Related What's a math trick that is not very well-known? x x % of y=y y=y % of x x y∗x/100=x∗y/100 y∗x/100=x∗y/100 y x/100=x y/100 y x/100=x y/100 So, in order to calculate a percentage in your head it might be easier to turn it around. Example What is 2% of 50? It's the same thing as 50% of 2." I know some of you already know this, but for those of you who don't hopefully this made percentages easier. Upvote · 17.3K 17.3K 99 91 99 10 Jinang Shah Worked at Pradeep Singhi & Associates (2013–2016) ·10y Related What are some useful mental math tricks? Solving a cube-root of a digit in less than a 3 second : 1) First, Go through the following instructions : Cube of 1 to 10 1=1 2=8 3=2 7 4=6 4 5=12 5 6=21 6 7=34 3 8=51 2 9=72 9 10=100 0 2) Now, note the last digit of the above answers. For 1,4,5,6,9 the last digit of them will always be the same number For 2 and 8 it's vice versa similarly for 3 and 7. 3) Now, take cube of any number. For Example, Cube of 73 is 389,017. 4) Steps to find the cube root of the 389,017. The last digit of the above number is 7. Hence on the one's place it would be 3. 5) Now, to find out tenth's place, Avoid the last 3 digit Continue Reading Solving a cube-root of a digit in less than a 3 second : 1) First, Go through the following instructions : Cube of 1 to 10 1=1 2=8 3=2 7 4=6 4 5=12 5 6=21 6 7=34 3 8=51 2 9=72 9 10=100 0 2) Now, note the last digit of the above answers. For 1,4,5,6,9 the last digit of them will always be the same number For 2 and 8 it's vice versa similarly for 3 and 7. 3) Now, take cube of any number. For Example, Cube of 73 is 389,017. 4) Steps to find the cube root of the 389,017. The last digit of the above number is 7. Hence on the one's place it would be 3. 5) Now, to find out tenth's place, Avoid the last 3 digit of the number and note down the rest digit. For the above case, it would be 389. Now, find the cube of number which is lower than 389 and place it on tenth place. For the above case, the cube of 7 is 343 which is lower than 389. Hence take 7 at tenth place. Thus, the final answer would be 73. Example, 328,509 one's place is 9 and on tenth place it would be 6 as cube of 6=216 being lower than 328. Therefore, it would be 69. Now, this would be only applicable for the perfect cube roots. Thanks, Upvote · 999 128 9 1 9 2 Tim Farage Professor, Mathematics and Computer Science, Retired · Upvoted by Paul LeFevre , Masters Mathematics, California State University, Fullerton (1987) and Michael Jørgensen , PhD in mathematics · Author has 4.9K answers and 17.7M answer views ·Updated 9mo Related What are very useful mathematics short tricks? This isn’t the most useful math trick in the world, but here it comes. What is √1.124 1.124? I just made that up. And my immediate estimate is that it is 1.062 1.062. My calculator gives 1.060 1.060 rounded to three significant digits. Not a bad guess. Here’s one more. What is √1.0062 1.0062? My immediate estimate is 1.0031 1.0031. My calculator gives 1.0031 1.0031. See? Here’s the trick. If you take √1+x 1+x where x is a very small positive value, then you can estimate this root as 1+x 2 1+x 2. So in my last example, I just took 1.0062 1.0062 and added 1 1 to half of 0.0062 0.0062 to get 1.0031 1.0031. Why does this work? The Taylor series for \s\s Continue Reading This isn’t the most useful math trick in the world, but here it comes. What is √1.124 1.124? I just made that up. And my immediate estimate is that it is 1.062 1.062. My calculator gives 1.060 1.060 rounded to three significant digits. Not a bad guess. Here’s one more. What is √1.0062 1.0062? My immediate estimate is 1.0031 1.0031. My calculator gives 1.0031 1.0031. See? Here’s the trick. If you take √1+x 1+x where x is a very small positive value, then you can estimate this root as 1+x 2 1+x 2. So in my last example, I just took 1.0062 1.0062 and added 1 1 to half of 0.0062 0.0062 to get 1.0031 1.0031. Why does this work? The Taylor series for √1+x 1+x is: 1+x 2−1 8 x 2+1 16 x 3+….1+x 2−1 8 x 2+1 16 x 3+…. If x is very small, the terms after the first two become insignificant, giving us our trick. Not too useful, but you can impress your friends with it. Maybe. Upvote · 999 127 99 14 9 3 Balaji Viswanathan Using Android apps since 2011 · Author has 1.5K answers and 5.4M answer views ·9y Originally Answered: How does this math trick work? · It's simple. There is no magic here. If you pay attention then you can figure it out. The first set of cards or things shown when you pick aren't shown after you pick. For example if first cards of 1, 2, 3 and 4 of spades is shown for you to pick one then after few seconds none of these cards are shown. So whatever you picked out of the initial four doesn't appear. Since you are only thinking about one card you don't notice that the entire set of cards is changed.. Upvote · 9 2 Mohit Mohapatra I love maths. ·9y Originally Answered: How do this math trick work (Guess a number and I know the answer)? · Actually it's simple algebra... Let the no. be x. According to the question, [(2x+8)/2]. (Without including the substraction) =x+4 We find that any number would come 4 more than the original and when it is subtracted from the original we get no number other than 4. The place where you are deceived... When you multiply 2 and later divide by 2 they are cancelled out. BUT, in the middle of the process, 8 is added. So upon dividing, you get the number, but not the original because of the addition. So, it gets included in the equation i.e (8/2)=4 gets added and number increases by 4. This seems si Continue Reading Actually it's simple algebra... Let the no. be x. According to the question, [(2x+8)/2]. (Without including the substraction) =x+4 We find that any number would come 4 more than the original and when it is subtracted from the original we get no number other than 4. The place where you are deceived... When you multiply 2 and later divide by 2 they are cancelled out. BUT, in the middle of the process, 8 is added. So upon dividing, you get the number, but not the original because of the addition. So, it gets included in the equation i.e (8/2)=4 gets added and number increases by 4. This seems simple but you won't be able to get it when the calculations are somewhat complex. If you want the solution of a trick , try(if possible) taking a variable. That way, you can interpret the trickery. But there are cases when you can't do this, like when you have to add the digits separately. In these cases, simply take simple but unusual numbers (that are not usually picked), like 0 or 1. You can easily spot out the twist. Cheers. Upvote · 9 1 Sanjay Mali Have published a translatio of a book by Dr. APJ Abdul Kalam · Author has 547 answers and 767.6K answer views ·9y Let the number be x 20 x 20 x+35 100 x+185 100 x+100 100 x+100+47 ' 47 is age 100 x + 147 ' Now the last two digits of 100 x are 0 & 0 for any x So the last two digits of the sum 100 x+147 are 4 & 7 which when concatenated show the age Look at step, To get x you should subtract 147 Then we get 100 x Just divide by 100 u get x Upvote · Alex Mazanek Studied at Pima Community College · Upvoted by Federico Riveroll , M.S. Data Science & Mathematics, Mexico Autonomous Institute of Technology and Samuel Gomes da Silva , Ph.D. Mathematics & Set Theory, University of São Paulo (2004) ·Updated 9y Related What are some useful basic math tricks that most people don't know about? Very simple one I learned in 5th grade that I still use to this day that most people don't know: what's 4x18? This isn't the most complex problem but it would take people at least a couple seconds to solve and some may have to write it down. Instead what is 8x9? Just double the first number and halve the second, will give you the same answer in half the time. Upvote · 7.6K 7.6K 999 519 99 35 Related questions What is the secret thing behind this math trick? How can the following maths trick's working be explained? What's the math behind the "boobless" (55378008) calculator trick? How do this math trick work (Guess a number and I know the answer)? Why does this math trick always yield 5? What are some cool calculator math tricks for fun? Which is the best math tricks site? What are some easy maths tricks that can "trick" anyone? Which is the most unique maths trick? What are some tricks to solve math quickly? How can I improve my mathematics trick? What are easy math calculation tricks? How would I solve this math problem in the link below? What can I find some mathematics tricks? How does one solve this math problem? Related questions What is the secret thing behind this math trick? How can the following maths trick's working be explained? What's the math behind the "boobless" (55378008) calculator trick? How do this math trick work (Guess a number and I know the answer)? Why does this math trick always yield 5? What are some cool calculator math tricks for fun? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10091	https://www.uptodate.com/contents/poliomyelitis-and-post-polio-syndrome	Your Privacy To give you the best possible experience we use cookies and similar technologies. We use data collected through these technologies for various purposes, including to enhance website functionality, remember your preferences, and show the most relevant content. You can select your preferences by clicking the link. For more information, please review our Privacy & Cookie Notice Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device. Because we respect your right to privacy, you can choose not to allow certain types of cookies on our website. Click on the different category headings to find out more and manage your cookie preferences. However, blocking some types of cookies may impact your experience on the site and the services we are able to offer. Privacy & Cookie Notice Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function. They are usually set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, this may have an effect on the proper functioning of (parts of) the site. Performance Cookies These cookies support analytic services that measure and improve the performance of our site. They help us know which pages are the most and least popular and see how visitors move around the site.
10092	https://www.sciencedirect.com/science/article/pii/S097336981160006X	Varus and valgus deformities in knee osteoarthritis among different ethnic groups (Indian, Portuguese and Canadians) within an urban Canadian rheumatology practice - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Patient Access Search ScienceDirect Article preview Abstract References (13) Cited by (3) Indian Journal of Rheumatology Volume 5, Issue 4, December 2010, Pages 180-184 Original Article Varus and valgus deformities in knee osteoarthritis among different ethnic groups (Indian, Portuguese and Canadians) within an urban Canadian rheumatology practice Author links open overlay panel Raman Joshi a, Nimu Ganguli b, Christopher Carvalho c, Faye de Leon d, Janet Pope e Show more Add to Mendeley Share Cite rights and content Abstract Objective To prospectively evaluate consecutive patients with knee osteoarthritis who presented to a Canadian community rheumatology clinic and determine the prevalence of varus deformities of the knees and the incidence of forefoot overpronation in three ethnically different populations-a Canadian-born population, Indian-born population and Portuguese-born population. Use of different therapies for knee osteoarthritis in the clinic was also evaluated. Methods Data were collected on patient age, sex, body mass index (BMI), visual analog scale (VAS) pain, ethnic background, valgus/varus deformity at the knee and overpronation of the forefoot. Kellgren-Lawrence scores were assigned to plain radiographs. Charts were subsequently reviewed to evaluate rates of intra-articular steroid injection, hyaluronic injection, surgical referral and surgical referral in the first year after being seen in the clinic. Results Eight patients who were Portuguese-born, 26 who were Indian-born and 33 who were Canadian-born were identified. Age was not significantly different. Women had more valgus changes than men (P = 0.04), and VAS pain was not significantly different between men and women. Significantly more varus deformity was noted in the Indian-born group than the Canadian-born group (P=0.002), and more valgus deformity was noted in the Portuguese-born than Canadian-born group (P = 0.009). There was a trend to lower BMI in the Punjabi-born group and lower VAS pain in the Canadian-born group. There was no significant correlation between BMI and VAS pain, nor age and VAS pain (r =−0.192 and −0.050 respectively). There was no significant association with either BMI or age and forefoot overpronation. No ethnicity differences in treatment such as use of intra-articular steroid/hyualuronic acid use, surgical referral or surgery were observed. Conclusions Patient populations differed significantly in terms of varus and valgus deformities at the knee. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Recommended articles References (13) M Tandon et al. A comparative analysis to study dietary habits in person with osteoarthritis knee compared to those without osteoarthritis of knee in Lucknow Indian J Rheumatol (2008) K Reilly et al. The role of foot and ankle assessment of patients with lower limb osteoarthritis Physiotherapy (2009) Statistics Canada [Internet]. Ottawa, Canada L Sharma et al. The role of knee alignment in disease progression and functional decline in knee osteoarthritis JAMA (2001) G Mangat et al. Pattern of osteoarthritis (OA) in India-a hospital based study JIRA (1995) DT Felson et al. High prevalence of lateral knee osteoarthritis in Beijing Chinese compared with Framingham Caucasian subjects Arthritis Rheum (2002) There are more references available in the full text version of this article. Cited by (3) Chronic pain experiences of immigrant Indian women in Canada: A photovoice exploration 2024, Canadian Journal of Pain ### Comparative study of pinless navigation system versus conventional instrumentation in total knee arthroplasty 2021, Cios Clinics in Orthopedic Surgery ### Correlation of serum cartilage oligomeric matrix protein with knee osteoarthritis diagnosis: A meta-analysis 2018, Journal of Orthopaedic Surgery and Research View full text Copyright © 2010 Indian Rheumatology Association. Published by Elsevier, a division of Reed Elsevier India Pvt. Ltd. All rights reserved. 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10093	https://www.shmoop.com/common-core-standards/ccss-hs-n-cn-8.html	Common Core Standards : CCSS.Math.Content.HSN-CN.C.8 We have changed our privacy policy. In addition, we use cookies on our website for various purposes. By continuing on our website, you consent to our use of cookies. You can learn about our practices by reading our privacy policy. See PlansLogin More on Common Core Standards The Standards About See All ### High School: Number and Quantity See All About The Real Number System Quantities The Complex Number System Vector and Matrix Quantities High School: Algebra See All ### High School: Functions See All ### High School: Modeling See All ### High School: Geometry See All ### High School: Statistics and Probability See All ### Grade 8 See All ### Grade 7 See All ### Grade 6 See All ### Grade 5 See All ### Grade 4 See All ### Grade 3 See All ### Grade 2 See All ### Grade 1 See All ### Kindergarten See All High School: Number and Quantity High School: Number and Quantity The Complex Number System HSN-CN.C.8 The Standard Sample Assignments Drills 8. Extend polynomial identities to the complex numbers. For example, rewrite x 2 + 4 as (x + 2i)(x – 2i). Students should know that complex numbers are everywhere. They're in parabolas and quadratics, but they can be found in polynomial identities as well, even in places where you'd never expect them to be. When we say everywhere, we mean it. For instance, students should learn that it's possible to factor x 2 + 4. They'll freak when you tell them. Give them a glass of water and a fluffy pillow and they'll be fine. All they have to do is rewrite x 2 + 4 as x 2 – (-4). Now, as long as they're allowed to use imaginary numbers (and why wouldn't they be?) factoring the difference of perfect squares shouldn't be an issue. We should end up with (x + 2 i)(x – 2 i). Students should also know that you aren't doing this to torture them. (Well, maybe a little.) They should know that we use factoring to help solve quadratic equations. That means we can set these equations to zero and solve rather than use the lengthy quadratic formula, which they're probably sick of already. So it stands to reason that the solutions to the equation x 2+ 4 = 0 are 2 i and -2 i. A nice, pretty little quadratic equation with real coefficients and then all of a sudden, out of nowhere—imaginary roots. Gotta love math; it keeps you guessing. Let's just hope your students feel the same. Drills Factor x 2 + 9. (A) (x + 3)(x – 3) (B) (x – 3 i)(x + 3 i) (C) (x – 3 i)(x – 3 i) (D) Correct Answer: (x – 3 i)(x + 3 i) Answer Explanation: When we rewrite as the difference of squares, our problem becomes x 2 – (-9). Since the square root of -9 is 3 i, choice (B) is correct. It's not (C) because that equals (x – 3 i)2, and not the multiplication of conjugates. That's really all it takes. Factor x 2 + 121. (A) (x + 11 i)(x – 11 i) (B) (x – 11 i)(x – 11 i) (C) (D) Correct Answer: (x + 11 i)(x – 11 i) Answer Explanation: When we rewrite the problem as the difference of perfect squares, it becomes x 2 – (-121). Since , and the only correct pair of multiplied conjugates is in (A). While (C) also has a pair of conjugates, it incorrectly leaves the 11 part under the radical. Solve x 2 + 49 = 0 for x by factoring. (A) ±7 (B) ±7 i (C) (D) ±24.5 Correct Answer: ±7 i Answer Explanation: When we rewrite the equation, it becomes x 2 – (-49) = 0. That factors as (x + 7 i)(x – 7 i). If we set each of those to equal zero, we have x = 7 i and x = -7 i. That means our answer is (B). Solve x 2 + 100 = 0 for x by factoring. (A) ±10 (B) (C) ±10 i (D) ±50 i Correct Answer: ±10 i Answer Explanation: When we rewrite the equation, it becomes x 2 – (-100) = 0. The factors are (x + 10 i)(x – 10 i). Since both of those factors equal zero, our answer becomes x = 10 i and x = -10 i. Solve x 2 + 6 = 0 for x by factoring. (A) ±3 (B) (C) (D) Correct Answer: Answer Explanation: When we rewrite the equation, it becomes x 2 – (-6) = 0. Since 6 isn't a perfect square, the square roots of -6 are . That means our answer is (C). Solve x 2 + 8 = 0 for x by factoring. (A) (B) (C) ±8 i (D) ±8 Correct Answer: Answer Explanation: Rearranging this equation into the difference of two squares, we end up with x 2 – (-8) = 0. This gives us as the two factors. If we solve for x, we end up with . That's (B). Which of the following quadratics has the solutions of ±i? (A)x 2 – 1 = 0 (B)x 2 + 1 = 0 (C)x 2 – 2 = 0 (D)x 2 + 2 = 0 Correct Answer: x 2 + 1 = 0 Answer Explanation: We can disregard (A) and (C) right off the bat since solving those will give us real answers and we want imaginary ones. If we look at (B), we'll get (x – i)(x + i) as our factors. Looks promising, right? Well… it is. Which of the following quadratics has the solutions of ? (A)x 2 + 6 = 0 (B)x 2 – 16 = 0 (C)x 2 + 5 = 0 (D)x 2 + 12 = 0 Correct Answer: x 2 + 12 = 0 Answer Explanation: Since we're looking for imaginary numbers, (B) won't help us here. We want the x values to be , or . If we plug in for x, the only equation that works for is (D). Which of the following quadratics has the solutions of ±5 i? (A)x 2 + 9 = 0 (B)x 2 + 10 = 0 (C)x 2 – 25 = 0 (D)x 2 + 25 = 0 Correct Answer: x 2 + 25 = 0 Answer Explanation: If we say that x = ±5 i, then x 2 = 25 i 2 = -25. That means that one of those equations will be true when we plug in -25 for x 2. In our case, that equation happens to be (D). Which of the following quadratics has the solutions ? (A) 4 x 2 = -13 (B) (C) 2 x 2 = 13 (D) 2 x 2 + 13 = 0 Correct Answer: 4 x 2 = -13 Answer Explanation: These equations are a bit funky, so let's change them to 4 x 2 + 13 = 0, , 2 x 2 – 13 = 0, and 2 x 2 + 13 = 0. That's much better. If we take our value of x and square it, we get , or . One of these equations should be true when we plug in for x 2, and that's (A). If we factored each equation and found its x values, that would have worked too. Aligned Resources PREVIOUS STANDARD NEXT STANDARD More standards from High School: Number and Quantity - The Complex Number System HSN-CN.A.1 HSN-CN.B.4 HSN-CN.C.7 HSN-CN.A.2 HSN-CN.B.5 HSN-CN.C.8 HSN-CN.A.3 HSN-CN.B.6 HSN-CN.C.9 If I'd had to teach to the Common Core, I would have used Shmoop, too. Aristotle The observed gravitational effect between Common Core activities results from their warping of Shmooptime. Albert Einstein Tired of ads? Join today and never see them again. Get Started Site Map Help About Us Jobs Partners Affiliates Colleges Terms of Use Privacy © 2025 Shmoop University. All rights reserved. We speak student® Instagram Facebook Twitter Linkedin Logging out… Logging out... You've been inactive for a while, logging you out in a few seconds... I'm Still Here! W hy's T his F unny? 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10094	https://mathbitsnotebook.com/Algebra1/FunctionGraphs/FNGTransformationRotation.html	Rotation - MathBitsNotebook(A1) Transformations: RotationsMathBitsNotebook.com Topical Outline \| Algebra 1 Outline \| MathBits' Teacher Resources Terms of Use Contact Person:Donna Roberts Rotations are TURNS!! A rotation is a transformation that turns a figure about a fixed point called the center of rotation. • An object and its rotation are the same shape and size, but the figures may be turned in different directions. • Rotations may be clockwise or counterclockwise. When working in the coordinate plane: • assume the center of rotation to be the origin unless told otherwise. • assume a positive angle of rotation turns the figure counterclockwise, and a negative angle turns the figure clockwise (unless told otherwise). The triangle is rotated. The letters used to label the triangle have not been rotated. Rotations can be seen, in a variety of situations: The Earth Windmills Pinwheel The Earth experiences one complete rotation on its axis every 24 hours. The blades on windmills convert the energy of wind into rotational energy. A children's toy that rotates when blown. Amusement Park Swing Ferris Wheel Merry-Go-Round An amusement park rides, such as the swing, allow you to become the of the rotation.Ferris wheels rotate about a center hub. (Yes, the seats tilt to prevent falling.) On the merry-go-round, riders become part of the rotation about the center of the ride. Rotation of 90º:(x,y) becomes (-y,x) Rotation of 180º:(x,y) becomes (-x,-y) Rotation of 270º:(x,y) becomes (y,-x)Remember:Clockwise: Counterclockwise: Rotations in the coordinate plane are counterclockwise. When working with rotations, you should be able to recognize angles of certain sizes. Popular angles include 30º (one third of a right angle), 45º (half of a right angle), 90º (a right angle), 180º, 270º and 360º. You should also understand the directionality of a unit circle (a circle with a radius length of 1 unit). Notice that the degree movement on a unit circle goes in a counterclockwise direction, the same direction as the numbering of the quadrants: I, II, III, IV. Keep this picture in mind when working with rotations on a coordinate grid. Rotations in the coordinate plane: Keep in mind that rotations on a coordinate grid are considered to be counterclockwise, unless otherwise stated. Rotation 90º: Starting with Δ ABC, draw the rotation of 90º. (It is assumed that the center of the rotation is the origin and that the rotation is counterclockwise.) To "see" that this is a rotation of 90º, imagine point B attached to the red arrow. The red arrow is then moved 90º (notice the 90º angle formed by the two red arrows). Look at the new position of point B, labeled B'. This same approach can be used for all three vertices. Rotation of 90º on coordinate axes. (x, y) → (-y, x) Rotation 180º: Starting with Δ ABC, draw the rotation of 180º. (It is assumed that the center of the rotation is the origin and that the rotation is counterclockwise.) As we did in the previous example, imagine point B attached to the red arrow from the center (0,0). The arrow is then moved 180º (which forms a straight line). Notice the new position of B, labeled B'. Rotation of 180º on coordinate axes. (x, y) → (-x, -y) (same as point reflection in origin) Rotation 270º: Starting with quadrilateral ABCD, draw the rotation of 270º. (It is assumed that the center of the rotation is the origin and that the rotation is counterclockwise.) As we did in the previous examples, imagine point A attached to the red arrow from the center (0,0). The arrow is then moved 270º (counterclockwise). Notice the new position of A, labeled A'. Since A was "on" the axis, A' is also on the axis. Rotation of 270º on coordinate axes. (x, y) → (y, -x ) If the rotation angles are giving you trouble, imagine a unit circle with a movable "bug" on a radial arm from the origin. Swing the "bug" around and look at the angle created by the move, and the position of the "bug".For calculator help with rotationsclick here. NOTE:There-posting of materials(in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". Topical Outline \| Algebra 1 Outline \| MathBitsNotebook.com \| MathBits' Teacher ResourcesTerms of UseContact Person:Donna Roberts Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved.
10095	http://www.doylegroup.harvard.edu/edm//amo2013/files/wk2-Foot-Atomic-Physics-5-5.3.pdf	OXFORD MASTER SERIES IN PHYSICS The Odord Maste¡ Series is designed for final year undergraduate and beginning graduate students rn physics and related disciplines. It has been driven by a perceived gap in the literàtuÌe today. While basic undergraduate physics tex-ts often show little or no connection with the huge explosion of resea¡ch over thc last two decades, more adva¡rced and specialized texts tend to be rather daunting for students. In this se¡ies, all topics and their consequences are treated at a simple level, whiÌe pointers to recent developments are provided at va¡ious stages. The emphasis in on clear physical principles like symmetry, quantum mechanics, and electromagnetism v¡hich underlie the whole of physics. At the same time, the subjects are C. J. FOOT related to real measurements and to the experimental techniques and devices currently used by physicists in academe and industr¡ Books in this series are vrritten as course books, and include ample tutoriaÌ Department of Physics material, exa,mples, illustrations, revision points, and problem sets. They can likewise be used as preparation for students starting a doctorate in physics â,rrd related fields, or for recent graduates starting research in one of these fieÌds in industry. . CONDENSED MATTER PHYSICS 1. M. T. Dove: Structure and d11nømics: an atomic uiew of materials 2. J. Singleton: Bønd theory and electron'ic properti,es of solid.s 3. A. M. Fox: Optical properties of soli,d"s 4. S. J. Blundell: Magneti.sm in cond,ensed matter 5. J. F. Annett: Superconducti,ui,ty 6. R. A. L. Jones: ,90fi cond.ensetl matter ATOMIC, OPTICAL, AND LASER PHYSICS 7. C. J. Foot: Atomic physics 8. G. A. Brooker: Mod,ern classical optics 9. S. M. Hooker, C. E. Webb: Laser physi,cs PARTICLE PHYSICS, ASTROPHYSICS, AND COSMOLOGY 10. D. H. Perkins: Par-ti.cle astrophgsi,cs 11. Ta-Pei Cheng: Relatiui,ty, gruút&t¿on, ønd, cosmologg STATISTICAL, COMPUTATIONAL, AND THEORETICAL PHYSICS 12. M. Maggiore: A mod,ern introd,ucti,on to quantum field, theory 13. W. Krauth: St&t¿sti,cal mechanics: ølgorithms and, computati.ons 14. J. P. Sethna: Entropy, ord,er pøranLeters, and, compleri,ty Uniuersity of Orford, OX.FORD UNIVERSITY PRBSS Atomic Physics The LS-coupling scheme In this chapter we shall look at atoms with two valence electrons, e.g. al-kâline eârth metâls such as Mg and Ca. The structures ofthese elements have many simila¡ities with helium, ând we shall also use the central_ field approximation that was introduced for the alkalis in the previous chapter. We sta¡t with the Hamiltonia¡ for 1{ electrons in eqn 4.2 and insert the expression for the cent¡al potential Vsp (r) (eqn 4.3) to give ,. Ë l-$'t -ve.r(r¿),{f+ ,,,",,}] ;=r \| '"' This Hamiltonian can be ¡¡ritten as H : Hcr H,", where the central_ field Hamiltonian llsp is that defined in eqn 4.4 and 5.1 Fine structure in the ZS-coupling scheme a3 5.2 The jj-coupling scheme 84 5,3 Intermediate coupling: the transition between coupling schemes 86 5.4 Selection rul€s in the .tS-coupling scherne 90 5.5 The Zeernan effect 90 5,6 Summary 93 Furtúer reading 94 Ðxercises 94 is the residual electrostat¡c ¡nteraction. This represents that paxt ofthe re_ pulsion not taken into account by the central field. One might think that the field left over is somehow non-central. This is not necessarily true. For configurations such as 1s2s in He, or 3s4s in Mg, both electrons have spherically-symmetric distributions but a central field ca¡not completely account for the repulsion between them-a potential I/6n (r) does not in_ clude the efiect of the correìation of the electrons' positions that leads to the exchange integral.l The residual electrostatic interactioû perturbs the electronic configurations n 1l1n 2l2 that are the eigenstates of the cen_ tral field. These angular momentum eigenstates for the two electrous a¡e products of theù orbital and spin function s ltlm¡,s¡n",) )l2m¡.s2m",) a¡d their energy does not depend on the atom,s orientation soìhat alt the difierent m¿ states are degenerate, e.g. the configuration Bp4p has (2^h .+ 7) (2lz + 1 ) = 9 degenerate combinations ot v¡, ,,v¡. ,^, .2 nactt of these spatial states has four spin functions associãteà itl, it, ¡"t v¡e do not need to consider thirty-six degenerate states since the prob-Iem sepa.rates into spàtial and spin pa.rts, as in helium. Nevertheless, the direct approach would require diagonalising matrices of larger di-mensions than the simple 2 x 2 matrix whose determinant was given in eqn 3.17. Therefore, instead of that b¡ute-force approach, we use the 'look-before-you-leap, method that starts by finding the eigenstates of the_ perturbation I1.". In that representation, fI." is a diagonal matrix with the eigenvalues as its diagonal elements. ¡r,"=Ë{X+ "(",)} (5.1) lChoosing .9(r) to arcount for all the repulsion between the sphe¡ically-symmet c core and the electrons out-side the closed shells, and â,lso within the co¡e, leaves the repulsion between the two valence electrons. i.e. 1{,- -e2f 4o rorr2. Thìs approximation high-lights the sjmjLa¡ity with helium (å,t-though the expectation valüe is e.vãl-uated with difiereût wavefunctionsl. Although it simplifres the equàtio;s nicely, this is not the besi âpproxi-mation for accurate câ.lcülàtjo ns-.9( r ) can be chosen to include most of th€ direct integral (ct Sectjon 3.3.2). For alLalj metal atoms, which we studied ;n the last chapter, the repulsion between electrons gives a sphedcally-syûmetdc potentiâÌ, so that ¡/re = 0. 2For two p-electrons we carìnot igûore n¿¡ aa we did in the treatment of 1sn¿ configuraLions in helium. Conñgura ¡rons wrth one, or mo¡e, s_electrons câlì be treâted in the way already described for helium bùt with the ¡¿dial wave_ firnctions ca.lclrlated mrmerically. The LS -coupling sche¡ne gl The interaction between the electrons, from thei¡ electrostatic repul_ sion, causes their orbital angular momenta to change, i.e. in the vector model 11 and 12 change direction, but their magnituáes remain constant. This internal interâction does not change the total orbital angular mo_ mentum L : lr Al:, so lr and 12 move (or precess) around this vector, as illustrated in Fig- 5.1. When no external torque acts on the atom, L has a fixed orientation in space so its z-component M¿ is also a consia¡t of the motio-n (mh arrdmr2 are not good qua,ntum numbers). This classical picture of conservation of totâl angular momentum coriesponds to the quantum mechanical result that the operators.L2 and Z, bãth commute with I1..:3 lL',H,.]=o and lL.,H".l-0. Since I1"" does not depend on spin it must also be true that Fig. 5.1 The residual electrcstatic ilF leracLion .åusês lt ând 12 ro precêss âround their resultant L = lr ir. 3-, rnê proor rs sLrajghrlorwård lor rhe qù¿ntum operator: L. = lu lIz. since mtr - q always occurs wjlh mt, _ q jn eqn 3.30. 4The H¿miltonian 11 commutes wì r thê exchângê (or swåp) operaro¡ Xir lbal in¡êrchanges thê labpls of rhe Þa,r l;clês i +, j; Lhr¡s s1âLes rhat a¡e siÀrt [aneously eigênfunl^Lions oI bo¡h oper-âLors exist. Thjs is obviously lrue for Lhe Hamilionian of lhe hclium âLom in eqn 3.1 (which looks rhpsameif I ++ 2), buL ir also holds [or eqn S.t. ln gpneral, swâpping particiês wir b rhe sâme ma,ss ând charge does not .húge r be Hamil-Lonian lor the êlectrostâtic interacLions of â svsLem. i . (5.2) ls',H,.1 :o and lS",H".l=0. (58) Actually, 1{." also commutes with the individual spins s1 and s2 but we chose eigenfunctions of S to antisymmetrise the wavefunctions, as in helium-the spin eigenstates for two electrons are ty'$,n and ty'$," for S = 0 and 1, respectively.a The quantum ""-Uerc Z-i,,tZ¿, S a"rr"a ,lZg have well-defined values in this Russell-Saunders or ZS_couplìng scheme. Thus the eigenstates of I1"" are ILMLS Ms). In the tr,S_coupling scheme the energy levels labelled by Z and S are called terms (a,rrd there is degeneracy with respect lo M¡ and M5) We.u- "u-piu. of 1.L and 3.L te¡ms for the 1s7¿l configurations in helium where the ¿S_coopfirrg scheme is a very good approximation. A more complex example is an rzpnlp configuration, e.g. 3p4p in silicon, that has slx terms as follows: h=1, lz=7 11 st=r, s2:t + L:0,7or2, L9:0 or 1; terms: 2s+r¿ = rg, rp, r¡,35,3p,3D. The direct and exchange integrals that determine the energies of these terms are complicated to evaluate (see Woodgate (19g0) for details) and here we shall simply make some empirical observations based on lhe terms diagrams in Figs 5.2 and 5.3. The l2h +7)(2tz+ 1) = 9 degenerare statês of orbital angular momêntum become the 1 + 3 -¡ 5 = I states of M, associated with the S, p and D terms, respectively. As in helium, linear combinations of the four degenerate sp-in slates lead to triplet and one singlet terms but, unlike heium, triplåts do not necessarily lie below singlets. Also, the Jp2 configuration has fewer terms than the 3p4p configuration for equivalent electróns, because ofthe pauli exclusion principle (see Exercise 5.6). I t, u iì In the special case of grourul, configurations of equiualent electrons the spin and orbital angular momentum of the lor¡¡est-energy term follov¡ some empiricaÌ rules, called Hund's rules: the lov/est-energy term has sTwo electrons cânnot both hâve the the largest value of ,S consistent,with the pauli exclusion pîinciple.s If ".;;.;;;r ä;r;iliîffi:,ïl' 82 The Ls-couPling scheme Fig. 5.2 The terms of the 3p4p conlìgu-¡¿.tion ¡n silicoo all lie about 6 eV above the ground stâte. The residual electro-stâtic interâction leads to energy dif ferences of - 0.2 eV between the terms, and the firìe-stmcture splitting is an or-der of mâgnitùde smalle¡, as indicated for the 3P ând 3D terms This structure is well described by the ¿S coupling FiA. 5,3 The ene¡gies of terms of the 3p2 conflguration of silicon. For equiv-alent eìectrons the Pãuli exclusion p n ciple restricts the numbù of terms there are only three compared to the six in Fig. 5-2. The lowest-energy term is 3P, in accordance with Hund's rùÌes, å-nd this is the ground stâ,te of siÌicon atoms, 6lIund's rules are so commonly mis-applied that it is worlh spelling oüt that they only appÌy to the louest term of t}Je ground, confrguration 1o1 cases where there is onÌy one incomplete sub-sheÌl. ?Th" 1..g" total spin has importãnt conseqr¡ences for magretism (Blurdell 2001). 1.8 t.4 t.2 1.0 there are severâl such terms then the one with the largest -L is lowest. The lowest term in Fig. 5.3 is consistent with these rules;6 the rule says nothing about the ordering of the other terms (or about any of the terms in Fig. 5.2). Configurations of equivalent electrons are especially important since they occur in the ground configuration of elements il the periodic table, e.g. for the 3d6 configuration in iron, Hund's rules give the lov/est term as 5D (see Exercise 5.6).7 E (ev) 6.4 6.3 6.1 6.2 6.0 1D :3 3q \'p 5.8 E (eV) 2.0 0.8 0.4 0.2 "P_ EcrlE"+8"-" 5.L Fine structure in the .L,9-coupling scheme Fine structure arises from the spin orbit interaction for each of the un-pàired electrons given by the Hamiltonian H"-o = B1sy .lt I þzsz .12 . For atoms with two valence electrons ¡Is o acts as a perturbation on the states ILMLSM;) In the vector model, this interaction betv/een the spin and orbital angular momentum causes L and S to change di-rection, so that neither L" rror S" remains colstant; but the totâI elec-tronic angulâÀ momentum J = L + S, aûd its z-component "/,, are both constant because no external torque acts on the atom. We shall no,¡' evaluate the effect of the perturbation I1" o on a term using the vec-tor model. In the vector-model description of the -L5-coupling scheme, 11 afld 12 precess around L, as shown in Fig. 5.4; the components per-pendicular to this fixed direction average to zero (over time) so that only the componert of these vectors along L needs to be considered, e.g. l¡--+{(tr . t-, ,,r,.,r". The time average \ . L in the vector model becomes the expectation value (11 .L) in quantum mechanics; also we have to use -L(-L + 1) for the magnitude-squared of the vecto¡. Applying the same projection procedure to the spins leads to tf -a (s\|s) c. (l\|T,\ 1s" s\ 11" 1,\ Hs o = iit s(s + Ds TúL d'ù-ils t'ç't t't¡L =l3LSs.L. (5 4) The derivation of this equation by the vector model that argues by analogy with classical vectors can be fully justified by reference to the theory of angular momentum. It can be shown that, iû the basis .I M7) of the eigenstates of a general angular momentum operator J and its component ,/r, the matrix elements of any vector operator V are pro-portional to those of J, i.e. (J M¡lVlJ M¡) - c(J MrlJlJ Mr).8 Fis-ure 5.5 gives a pictorial representation of why it is only the component of V along J that is s'ell defined. We want to apply this result to the case whe¡e V = 11 or 12 in the basis of eigenstâtes lL M7,1, and analogously for the spins. For (I M¿l11 lL ML) : c(L M¡lLlL M¿) the constant c is determined by taking the dot product of both sides with L to give ^ (L M Lltt .LlL ML) . '- AM¡f LE^4'l' hence (L MLlrllL ML) = #+ Q MLILIL MLt. (5 5) This is an example of the projection theorem and can also be applied to 12 ând to s1 and s2 in the basis of eigenstates lSMs)- It is cÌear that, for diagonal matrix elements, these quâütum mechanical results give the same result of the vector model. 5.7 F'ine structure in the Ls-coupli,ng schenl,e gz Fig. 5.4 In the ¿.9-coupling scheme the orbitàl angulå,r momenta of the two electrons coupÌe to give total â,ngular momentum L = lr lu. ln the vec tor model lr and 12 precess around L; similarly, s1 and s2 precess âround S. L ånd S precess âround the total an-gula-r momentum J (but more slowly than the precession of h and 12 around L becàuse the spin orbit interàction is ¡weaker' than the residual electrosatic iáteraction). SThis is particula¡ case of a more general result called the Wigner Eckart theorem which is the coltrer-stone of the theory of â¡gular momen-tum. This powerful theorem â,lso ap-plies to ofi-diagonal elements such as Q MrlV lJ Mi), and to more com-pÌicated operatoß such as those for quadrupole moments. It is used ex-tensively in advanced atomic physics see the 'Furbher re¿ding' section in this châpter. Fig. 5.5 A pictoriâl reprcsenlation of the p¡oj€ct theorem for an atom, where J defines the à-(is of the sys¿em. 84 The LS-coupling schetne gsi-ilarly, in the one-electron ca,se we found the fine stmcture withoüt deter-mining the eigensbates Isjrnjl explic-itly iû berms of the !,n and spin fünc-tions. Equation 5.4 has the same form as the spin-orbit interaction for the single-electron case but with capital letters rather than s.l. The constant B¿s that gives the spin-orbit interaction for each term is related to that for the individual electrons (see Exercise 5.2). The energy shift is øs o : Brs (S. L) (5 6) To find this energy we need to evaluate the expectation value of the operator L.S = (J.J-L.L-S.S)/2 for each term 2s+1¿. Each term has (25 + 7) (2L + 1) degenerate states. Any linea¡ combination of these states is also an eigenstate with the same electrostatic energy a¡rd we can use this freedom to choose suitable eigenstates and make the calculation ofthe (magnetic) spin orbit interaction straightforwa.rd. We shall use the afes ILSJMJT; these are linear combinations of the basis slates ILM¡SMs) but we do not need to determine their exact form to find the eigenenergies.e Evaluation of eqn 5.6 with the states ILSJMJ) gives ¡ PLS !1.,.--i.tr.t+t)-L(L+r)-.9(,9+1) 2' Thus the energy interval between adjacent J levels is AEps=fi¡-E¡ 1:þ¡sJ. (5.8) This is called the interval rule. For example, a 3P term (, = 1 : li) hâs three,./levels: 2s+rLJ = 'Po,tPr,3P2 (seeFig.5.6); and the separation betv/een -I - 2 and J - 1 is twice that between "I - 1 and J - 0. The existence of a,rt interval rule in the fine structure of a two-electron system genera.lly indicates that the .LS-coupling scheme is a good approximation (see the 'Exercises' in this chapter); however, the converse is not true. The -LS-coupling scheme gives a very accurate description of the energy levels of helium but the fine structure does not exhibit an interval rule (see Example 5.2 later in this chapter). It is irnportant not to confuse -L5-coupling (or Russell Saunders cou-pling) with the interaction between L and S given by B¿e S . L. In this book the word 'interact'ion is used for real physical effects described by a Hamiltonian and coupling refers to the forming of linear-combination wavefunctions that are eigenstates of angular momentum operators, e.g. eigenstates of L and S. The -LS-coupling scheme breaks down as the strength of the interaction þt sS L increases relative to that of 11"".10 5.2 The jj-coupling scheme To calculate the fine structure in the trS-coupling scheme v/e treated the spin-orbit interaction as a perturbation on a term) 2s+1tr. This is valid when .E"u >> E"-", which is generally true in light atoms.11 The spin-orbit interaction increases with atomic number (eqn 4.13) so that it cân be simila¡ to 8". for heavy atoms see Fig. 5.7. However, it is only in cases with particularly small exchange integrals that E"-o exceeds .8 ", so that the spin orbit interaction must be considered before the residual + 'P---# lu l+ 2p + -3D I" I 2ll þ I 3po Fig, 5.6 The frne structure of a 3P term obeys the interval rule. 10In classica.Ì mechanics the ï/ord ,coü pÌing' is commoûly used more loosely, e.g. for coupled pendulums, or coupled oscillators, the rcoùpling between them' is taken to mea¡ the 'itrterâction be-tween them)that leads to their motions being conpled. (This coupling may take the form of a physical Ìinkage srch â6 a ¡od or spring between the two systems.) \rE"-" - þ"s and -Ere is comparable to the excha¡ge integrâ,l. Energy (eV) 102 -r I Crross structure I Residu¿l elechostatic energy ^ l. rne structu¡e z 100 ^ Hg electrostatic interaction. When 11"-o acts directly on a configuration it causes the I and s of each individual electron to be coupled together to Eive i1 : 11 s1 and i2 : lz f sz; in the vector model this corresponds to I and s precessing around j independently of the other electrons. In this jj-coupìing scheme eac.h valence electron acts on its own, as in alkali atoms. For an sp configuration the s-electron can only have ir = ll2 and the p-electron hâs i2 : Il2 or 312; so there are tv/o Ievels, denoted by (jt, j") = (Il2,Il2) anð. (712,312). The residual electrostatic interaction acts as a perturbation on the jj-coupled levels; it causes the angula,r momenta of the electrons to be coupled to give total angular momentum J = jr +jz (as illustrated in Fig. 5.8). Since there is no external torque on the atom, M7 is also a good quântum number. For an sp configuration there are pairs of J levels for each of the two original jj-coupled levels, e.g. (jt, jz)¡ : (U2,712)o, O12,712)1 and (712,312),, (712,312)2. This doublet structure, shown in Fig. 5.10, contrasts ï'ith the singlets a¡d triplets in the -LS-coupling scheme. 5.2 The jj-coupling scheme 85 Fig. 5.7 A ptot of typicaÌ energies âs â firnction of the atomic number Z (on logarithmic scales). A characteristic energy for the gross strüctü¡e is tâ,ken as the energy required to excite an elec tron from the ground stâte to the flßt excited stâte-this is less thâ,n the ior-ization energy but has a simila.r va.ria-tion with Z. The residual electrosLatic . _ interaction is the singlet t plet separa-tion of the lowest excited confi8lllation in some atoms with two valence elec-trcns. The fine structure is the split-ting of the lowest p configuration. For all câses the plotted etrergies ar€ fa.Ìrly close to ihe ma-..imum for that type of structure in neutral atoms higher-Iying conflgu¡atioûs have smaller val-I O 1 Cs IA 4l HHe 86 The LS -coupli,ng scheme Fig. 5.8 The jj coupling scheme. The spin oÌbit interacLion energy is large compared to the -8.. (Cf- Fjg- 5.4 for the ¿S-coupìing scheme.) 12In both of th""u cases s¡e âssume ùr isolated configu¡ation, i.e. that the en-ergy separation of the different conflg-u¡ations in the central field is greater thàr the per¿urbation prcduced by ei-ther ,As o or,Ð.e. In summar¡ the conditions for -L5- and jj-coupling are âs folloì 's:12 trS-coupling scheme: E " ¡> E" o, jj-coupling scheme: E"-" >> .E"". 5.3 Intermediate coupling: the transition tretween coupling schemes In this section we shall look at examples of angular momentum coupling schemes in two-electron systems. Figure 5.9 shows energy-level diagrams of Mg and Hg and the following example looks at the structu¡e of these atoms. Example ó.1 3s3p, Mg 6s6p, Hg 2.1850 3.76 2.1870 3.94 2.191.t 4.40 3.5051 5.40 The table gives the energy levels, in units of 106m 1 measured from the ground state, for the 3s3p configuration in magnesium (Z :12) and 6s6p in mercury (Z = 80). We shall use these data to identify the levels and assign further quantum numbers. For an sp configurâtion we expect lP and 3P terms. In the case of magnesium we see that the spacings between the three lov¡est levels are 2000m 1 and 4100m 1;these are close to the 1 to 2 râtio expected from the interval rule for the levels ¡¡ith ,/ = 0, 1 a¡rd 2 that arise from the triplet. The .L,9-coupling scheme gives an accurâte description because 5.3 Intenrlell,iate coupiing: the transi,t¿on betueen coupling schemes g7 H E (eV) Mg 1s 3s 1P 3P rDsD 1F3F 1S 35 1p3p 1D3D 1F3F -1 -2 -3 -5 _:7 -10 To n=l To ls2 Fig. 5.9 The terms of heliüm, magnesium ànd mercury are plotted on the s¿.Ine energy scâle (with hyd¡ogen on the left fo¡ comparison). The 6ne structure of the lighter atoms is too small to be seen on this scâle and the ¿S-coupling scheme gives a ve¡y âccürate descriptioû. This scheme gives an approximàte description for the towlying terms of mercnry even though it has a much larger frn€ structure, e.g. for the 6s6p configuration the Ð," > Ð"-. but the interval ru-ìe is not obeyed becâuse the spin,orbit i;terâ.ction is not very smaÌl compared to the residùal electrostâtic inte¡artion. The 1s2 confrguration of helium is noi shown; it has à biûding energy of -24.6eV (see Fig. 3.4). The 1s2s and 1s2p configurations of helium lie close to the f¿ = 2 shell in hydrogen, ând similarÌy bhe 1s3l configurations lie close to the n = 3 shell. In mâgnesium, the terms of th€ 3sr¿f configurations bive very similar energies to those in hydrogen, bùt the difierences get larger âs ¿ decreases The energies of the terÃs in mercury have la-rge difiercnces from the hydrogen energy levels. Mùch câû be learnt by carefully studyìng this term diagram, e.g. theie is a lP term which has similar energy in the three confignrations: 1s2p, 3s3p aûd 6s6p in He, Mg a¡d Hg, respóctively thus the efiective quantum nùmber n is similâr despit€ the inc¡ease in ?¿. Complex telms arise when both v¿.lence electrons a¡e exciled in Mg, e.g. the 3p2 configüration, and the 5d96s26p confrguration in Ilg. He 88 The LS-coupti,ng scheme 13This identifrcation of the levels is supported by othe¡ information, e.g. detemination of "/ ùom the Zeeman ef-fect and the theoretically prcdic[ed be-haviou of an sp confrguration shown in Fig.5.10. the fine structure is much smaller than the energy separâtion (E"e -1.3 x 106m..1) betsr'een the 3P term at - 2.2 x106 m-r a,rrd tiru'ip, level at 3.5 x 106m 1. In mercury the spacings of the levels, going dovsn the table, are 0.18, 0.46 and 1.0 (in units of 106 m-1); thesã levels are not so clearly separated into â singlet and triplet. Taking the lowest three levels as 3Pe, 3P1 and 3p2 v¡e see that the interval rule is not well obeyed since 0-46/0.18 = 2.6 (not 2).13 This deviation ftom the .L.9-coupling scheme is hardly surprising since this configuration has a spin-orbit interaction only slightly smaller than the singlet_triplet separation. However, even for this heavy atom, the Z5-coupling scheme gives a closer approximation than the jj-coupling scheme. -2 =0 ..¡.." _ -__ ---> "'.;;" Fig' 5'10 A theoretical pÌot of the energy levels that a.rise ftom an sp configuration âs a function of the strength of the spin-orbit interâction pa¡åmeter P (of the p-electroD defined in eqn 2.55). For B = 0 the tÍ¡o terms, sp and 1p, have an energy sepa¡âtion equal to twice the excb¿ûge integral; this residual electrostatiá ene.gy is assume¿ to ¡e cã.rstanJaJoJff 1"" in the plot As B increâses the fine structure of the triplet becomes observable. Âs B increases fiuther the spin o.tit äria .""iao¡ electrostat¡c interârtjons become cômôùabÌe and the,LS-coupling sctre-e ceases to be a good approximatjon: the inte'-val ¡ule a¡d (¿s-couplinc) selection rules b¡eåt don (as in mercurv, Jnig. s.i.-at ,urge ø the jj-coupring scheme is àpproprìate. The operator J commu¿es wjrh It" o (and-I1.")j th"."f..; ¡i._;;ti ;iíås i"u"t or tt" sa¡ne J, e.g. the two./= l tevets in this case lThe energies ofthe./ = 0 a¡rd 2levóis are straighi ri"". ü"".,r"u irru. .avefunctioûs do not change.) Exercise 5.8 gives an example of this transition ber.ween the two couptin"g scr,""" io. "p(, + t> "";fig";;tl; i;; ;'l'ãìá s"itr,., r'" smâÌl exchange integrals). 5.3 Intermed,i,ate couplìng: the transition betueen ìàupling schemes g9 Example 6.2 The 7s2p conf,guration i,n heliutn J E (m-t) , ,6 gus ,i87 1 16 908 694 0 16 908 793 1 17113 500 The table gives the va.lues of "I and the energy, iir units ofrn 1 measured from ^the ground state, for the levels of the 1ì2p configuration in helium. The 3P term has a fine-structure splitting of u¡o.it IOO- r that is much smaller thân the singlet tdplet separation of 106m-i from the electrostâtic interaction (twice the excha.nge integral). Thus the Z,g_ coupling scheme gives an excellent description of the helium atom and the selection rules in Tabìe 5.1 are well obeyed. But the interval rule is not obeyed-the intervals betv¡een the .I levels are Z m-1 and 99 m-i and the fine structure is inverted. This occurs in helium because spin_spin and spin-other-orbit interactions have an energy comparabÌe witl that of the spin orbit interaction.l4 However, for atoms other than helium, the râpid increase in the strength of the spin orbit interaction v¡ith Z ensures that Il"-" dominates over the othe¡s. Therefore the fine st¡ucture of atoms in the -LS-coupling scheme usually leads to an interval rule. Fìrrther examples of energy levels are given in the exercises at the end of this chapter. Figure 5.10 shows a theoretical plot of the transition from the Z5- to the jj-coupling scheme for an sp configuration. Conser_ vation of the total angula,r momentum means that "I is a good quantum number even in the intermediate coupling regime and can ilwavs be used to label the l"vels. The notalion 2s1tr.r for the LS-coupling scheme is often used even for systems in the intermediate regime and also for one_ electron systems, e.g. 1s 25172 for the ground state of hydrogen. Tãble 5.1 Selectior mles for electric dipoÌe (E1) trânsjtions in the ¿S_coupling scheme. Rules 1-4 appty to âlt €lect c dipole transitions; rules 5 ând 6 a,re obeyeã ollly when ¿ and .9 are good qu¿rÌtum numbers. The right_h¿nd coltmn gives the structure to which the rule applies. 1 2 3 4 5 6 A"r : 0, 1 (J = 0 ç' J¡ :0) Level LM¡ : ¡,¡1 (M¡ = 0 ,n Mr, :0 if LJ :0) State Paxity cha¡ges Configuration ^l: +1 One electron jump Configuration AI:0,+1 (L=\s' L':o) Tà¡m ÂS = 0 Te¡m 14The spi., spi[ interaction ârises from the interaction betreeen two magnetic dipoles (ìtrdependent of any relative mojion). See eqn 6.12 àÌrd its expla-90 The LS-coupling scheme l5The.e is no simple physical ex-planation of why aî MJ = 0 to MJ, - 0 transition does not occur iÎ J = J\|; it is related to the sym-metry of the dipole mâtrix element QJ M¡ = olrlltJ M¡ = 0), where z and ?/ .epresent the other quantum numÞers. The particulâr case of ./ -Jt = 1 a;¡d AMJ = 0 is discussed in Budker ¿¿ ¿¿. (2003). 16This li.r" comes f.om the second level in the tabÌe given in Example 5.1, sjnce 0.254 p.m:1/(3.941 x 106 m 1). 17lnte¡combination lines are not ob-served in magnesium and heÌium. The relã,tive st¡ength of the intercombina tion lines and alÌorved transitions are tâbulated in Kuhn (1969). 18Most quantum mechânics texts de-scribe the ànomalous Zeem¿n efiect for a singÌe valence elect¡on that âpplies to the aÌkalis ând hyd¡ogen. 5.4 Selection rules in the .tS-coupling scheme Table 5.1 gives the selection rules fo¡ electric dipole transitions in the .Ls-coupling scheme (listed approximately in the order of their stdct-ness). The rule for "I ¡eflects the conservation of this quantity and is strictly obeyed; it incorporates the rule fo¡ Aj in eqn 2.59, but with the additional restriction J :0 ,¡, /' - 0 that afiects the levels with ,/ - 0 that occur in atoms with mo¡e thaû one valence eÌectron. The rule for AM7 follows from that for AJ: the emission, or absorption, of a photon cannot change the component along the z-âxis by more than the change in the total atomic angular momentum. (This rule is reÌe-vant when the stàtes are resolved, as in the Zeema,n efiect described in the following section.)15 The requirement for an overall change in parity and the selection rule for orbital angular momentum were discussed in Section 2.2. In a configuration nl1n2l2n3ls ..n1, ortly one electron changes its vâlue of I (and may also change n). The rule for AI al-ìows transitions such as 3p4s 3P1 3p4p 3P1. The selection rule AS:0 arises because the electric dipole operator does not act on spin, as noted in Chapter 3 on helium; as a consequence, singlets and triplets form two unconnected sets of energy levels, as shown in Fig. 3.5. Similarly, the singlet md triplet terms of magnesium sho$/n in Fig. 5.9 could be rearranged. In the mercury atom, hov¡ever, transitions with 4,9 : 1 occur, such as 6s2 1S¡-6s6p 3P1, that gives a so-called intercombination Iine with a v¡avelength of 254nm.16 This arises because this heavy atom is not accurately described by the trS-coupling scheme and the spin-orbit interaction mixes some 1P1 wavefunction into the wâvefunction for the term that has been labelled 3P1 (this being its major corrpo-nent). Although not completely forbidden, the rate of this transition is considerably less than il would be for a fully-allowed transition at the same wavelength; however, the intercombination line from a mercury Iamp is strong because many of the atoms excited to triplet terms will decay back to the ground state via this transition (see Fig. 5.9).17 5.5 The Zeeman effect The Zeeman efiect for atoms s'ith a single valence electron was not presented in earlier chapters to avoid repetition and that case is covered by the genera.l expression derived here for the -LS-coupling scheme.lE The atom's magnetic moment has orbital and spin contributions (see Blundell 2001, Chapter 2): ¡1 :-¡teL-g"¡teS (5 e) The interaction of the atom ¡'ith an external magnetic field is described by E7s - -p.8. The expectation value of this Hamiltonia¿ can be calculated in the basis ILSJMT), provided that EzE 11 E" " << E", i.e- the interaction can be treated âs â perturbation to the fine-structwe levels of the terms in the .L,9-coupling scheme. In the vector model v/e project the magnetic moment onto J (see Fig. 5.11) following the same rules as are used in treating fine structure in the .L5-coupling scheme (and taking B : Bê,). This gives Hz.F. = - ,\1, J),,J . B . (L J) + s' (s J) r lr - t) ffi' 'uut' (5 lo) In the vector model the quantities in angìed brackets are time averages.le In a quantum descdption treatment the quantities (. . .) are expectation va,lues of the form (J M¡1. .lJ M¡).zo In the vector model E2e = g¡¡L,sBM¡, (5.11) where the Landé g-factor is s7 : {(t J) + g, (S..I)} / {-¡ (J + 1)}. As-sumirg that g" = 2 (see Section 2.3.4) gives (5.12) Singlet terms have S - 0 so J - L a¡.'ò. g¡ = 1 (no projection is nec-essary). Thus singlets all have the same Zeeman splitting between M.¡ states and tra¡sitions between singlet terms exhibit the normal Zeeman effect, (shown in Fig. 5. t2). fþc A 4y'.¡ = I I I ransitjons havp lrpqupnries shifted by +psB lh wirh respect to lhe LM¡ = 0 transitions. IIr atoms with two valence electrons the traùsitions between triplet terms exhibit the anomalous Zeemarr effect. The observed pattern de-pends on the values of gJ ând "I for the upper and lower levels, as shown in Fig. 5.13. In both the normal a¡rd a¡omalous efiects the ?¡-transitions (LXt t = g¡ and o-transitions (LMr = +7) have the same polarizations as in the classica,l model in Section 1.8. Other exarnples in Exercises 5.10 to 5.12 show how observation of the Zeeman pattern gives information âbout the angular momentum coupling in the atom. (The Zeeman ef-Tect observed for the2P112 251¡2 and2Ps/2 25172 transitions that arise between the flne-structure components of the alkalis and hydrogen is treâted in Exercise 5.13.) Exercise 5.14 goes through the Paschen Back effect thât occurs in a strong external magnetic field see Fig. 5.14. 5.5 The Zeeman effect g1 Fig. 5.11 The projection of the cont butions to the total magnetic moment ftom the o¡bitâl motion a¡d the spir are projected ãlong J. 19Co.nponents perpendicular to J time average to zero. 20This stàtement isjustified by th€ pro-jection theorem (Section 5-1), derived f¡om the more general Wigner Eckart theorem. The theorem shows thât the expectâtion va,Ìue of the vector opera-tor L is proportional to thât of J in the basis of eigenstates lJ MJ), i.e. (J Mr LIJ Mr) .x (J Mr J J Mr) , and similarly for the expectâtion valne of S. The component a.Ìong the mâg-netic freld is found by taking the dot product Ìvith B: (J MJIL 'B J MJ) x (J M¡lJ .BlJ M¡\ x (J M¡l,l.lJ M¡) = M¡ .
10096	https://www.endotext.org/wp-content/uploads/pdfs/calcium-metabolism-during-pregnancy-and-lactation-2.pdf	Published in WWW.ENDOTEXT.ORG © 2018 CALCIUM AND PHOSPHATE METABOLISM AND RELATED DISORDERS DURING PREGNANCY AND LACTATION Christopher S. Kovacs, MD, FRCPC, FACP, FACE, Faculty of Medicine – Endocrinology, Health Sciences Centre, Memorial University of Newfoundland, 300 Prince Philip Drive, St. John’s, Newfoundland, A1B 3V6, Canada, ckovacs@mun.ca Revised November 26, 2018 ABSTRACT Pregnancy and lactation require women to provide calcium to the fetus and neonate in amounts that may exceed their normal daily intake. Specific adaptations are invoked within each time period to meet the fetal, neonatal, and maternal calcium requirements. During pregnancy, intestinal calcium absorption more than doubles, and this appears to be the main adaptation to meet the fetal demand for mineral. During lactation, intestinal calcium absorption is normal. Instead, the maternal skeleton is resorbed through the processes of osteoclast-mediated bone resorption and osteocytic osteolysis, in order to provide most of the calcium content of breast milk. In women this lactational loss of bone mass and strength is not suppressed by higher dietary intakes of calcium. After weaning, the skeleton appears to be restored to its prior bone density and strength, together with concomitant increases in bone volumes and cross-sectional diameters that may offset any effect of failure to completely restore the trabecular microarchitecture. These maternal adaptations during pregnancy and lactation also influence the presentation, diagnosis, and management of disorders of calcium and bone metabolism such as primary hyperparathyroidism, hypoparathyroidism, and vitamin D deficiency. Pregnancy and lactation can also cause pseudohyperparathyroidism, a form of hypercalcemia that is mediated by parathyroid hormone-related protein, produced in the breasts or placenta during pregnancy, and by the breasts alone during lactation. Although some women may experience fragility fractures as a consequence of pregnancy or lactation, for most women parity and lactation do not affect the long-term risks of low bone density, osteoporosis, or fracture. INTRODUCTION During gestation the average fetus requires about 30 g of calcium, 20 g of phosphorus, and 0.8 g of magnesium to mineralize its skeleton and maintain normal physiological processes. The suckling neonate obtains more than this amount of calcium in breast milk during six months of exclusive lactation. The adaptations through which women meet these calcium demands differ between pregnancy and lactation (Figure 1). Although providing extra calcium to the offspring could conceivably jeopardize the ability of the mother to maintain calcium homeostasis and skeletal mineralization, as this review will make clear, pregnancy and lactation normally do not cause any adverse long-term consequences to the maternal skeleton. The reader is referred to several comprehensive reviews for more details and extensive reference lists for the material covered in this chapter (1-7). Figure 1. Schematic illustration contrasting calcium homeostasis in human pregnancy and lactation, as compared to normal. The thickness of arrows indicates a relative increase or decrease with respect to the normal and non-pregnant state. Although not illustrated, the serum (total) calcium is decreased during pregnancy, while the ionized calcium remains normal during both pregnancy and lactation. Adapted from ref. (8), © 1997, The Endocrine Society. MINERAL PHYSIOLOGY DURING PREGNANCY Calcium provided from the maternal decidua aids in fertilization of the egg and implantation of the blastocyst; from that point onward the rate of transfer from mother to offspring increases substantially. About 80% of the calcium and phosphate present in the fetal skeleton at the end of gestation crossed the placenta during the third trimester and is mostly derived from the maternal diet during pregnancy. Intestinal calcium and phosphate absorption doubles during pregnancy, driven by 1,25-dihydroxyvitamin D (calcitriol) and other factors, and this appears to be the main adaptation through which women meet the mineral demands of pregnancy. Mineral Ions There are several characteristic changes in maternal serum chemistries and calciotropic hormones during pregnancy (Figure 2), which can easily be mistaken as indicating the presence of a disorder of calcium and bone metabolism, especially since it is not common for clinicians to measure calcium, phosphate, and calciotropic hormones during pregnancy (1). The serum albumin and hemoglobin fall during pregnancy due to hemodilution; the albumin remains low until parturition. In turn that fall in albumin causes the total serum calcium to decline to values that can be well below the normal range. The total calcium includes albumin-bound, bicarbonate-and-citrate-complexed, and ionized or free fractions of calcium. The ionized calcium, the physiologically important fraction, remains constant during pregnancy, which confirms that the fall in total calcium is but an artifact that can usually be ignored. However, that artifactual decline in total calcium means that the serum calcium cannot be relied upon to detect hypercalcemia or hypocalcemia. The ionized calcium should be measured or the albumin-corrected total calcium should be calculated to resolve any uncertainty about what the true serum calcium level is in a pregnant woman. Serum phosphate and magnesium concentrations remain normal during pregnancy. Figure 2. Schematic illustration of the longitudinal changes in calcium, phosphate, and calciotropic hormone levels that occur during pregnancy and lactation. Normal adult ranges are indicated by the shaded areas. PTH does not decline in women with low calcium or high phytate intakes, and may even rise above normal. Calcidiol (25OHD) values are not depicted; most longitudinal studies indicate that the levels are unchanged by lactation, but may vary due to seasonal variation in sunlight exposure and changes in vitamin D intake. PTHrP and prolactin surge with each suckling episode, and this is represented by upward spikes. FGF23 values cannot be plotted due to lack of data. Very limited data suggest that calcitriol and PTH may increase during post-weaning, and the lines are dashed to reflect the uncertainty. Adapted with permission from (1). Parathyroid Hormone Parathyroid hormone (PTH) was first measured with assays that reported high circulating levels during pregnancy. The finding of a low total serum calcium and an apparently elevated PTH led to the concept of “physiological secondary hyperparathyroidism in pregnancy.” This erroneous concept persists in some textbooks even today. Those early-generation PTH assays measured many biologically inactive fragments of PTH. When measured with 2-site “intact” assays or the more recent “bio-intact” PTH assays, PTH falls during pregnancy to the low-normal range (i.e. 0-30% of the mean non-pregnant value) during the first trimester, and may increase back to the mid-normal range by term. Most of these recent studies of PTH during pregnancy have examined women from North America and Europe who also consumed calcium-replete diets. In contrast, in women from Asia and Gambia who have very low dietary calcium intakes (and often high intakes of phytate that blocks dietary calcium absorption), the PTH level does not suppress during pregnancy and in some cases it has been found to increase above normal (1). Vitamin D Metabolites 25-hydroxyvitamin D or calcifediol (25OHD) readily crosses the rodent hemochorial placenta (9) and appears to cross hemochorial human placentas just as easily because cord blood 25OHD levels generally range from 75% to near 100% of the maternal value (1,5). A common concern is that the placenta and fetus might deplete maternal 25OHD stores, but this does not appear to be the case. Even in severely vitamin D deficient women there was no significant change in maternal 25OHD levels during pregnancy (1,4,10,11). Total calcitriol levels increase two to five-fold early in pregnancy and stay elevated until parturition, whereas measured free calcitriol levels were reported to be increased only in the third trimester (12). However, when the 20-40% increase in vitamin D binding protein and the decline in serum albumin during pregnancy are considered, that calculated free calcitriol should be increased in all three trimesters (11,13-16).There are several unusual aspects about this situation. PTH is normally the main stimulator of the renal 1alpha-hydroxylase; consequently, elevated calcitriol values are usually driven by high PTH concentrations. An exception to this is the ectopic expression of an autonomously functioning 1alpha-hydroxylase by such conditions as sarcoidosis and other granulomatous diseases. Another exception is pregnancy because the rise in calcitriol occurs when PTH levels are typically falling or quite low. Moreover, this increase in calcitriol occurs despite the ability of high levels of fibroblast growth factor-23 (FGF23) to suppress the synthesis and increase the catabolism of calcitriol, as shown in animal models of X-linked hypophosphatemic rickets (17-19). Evidence from additional animal models suggest that it is not PTH but other factors, such as PTH-related protein (PTHrP), estradiol, prolactin and placental lactogen, which drive the 1alpha-hydroxylase to synthesize calcitriol (1). The placenta expresses 1alpha-hydroxylase and it is often assumed that autonomous placental production of calcitriol explains why the maternal calcitriol level doubles; other sources such as maternal decidua and the fetus itself could conceivably contribute to the maternal value. However, it appears that any contributions of placenta and other extra-renal sources to the maternal calcitriol level are trivial. Animal studies indicate that the maternal renal 1alpha-hydroxylase is markedly upregulated during pregnancy (20,21) and that placental expression of 1alpha-hydroxylase is many-fold less than in the maternal kidneys (17). Clinical studies have revealed that anephric women on dialysis have very low circulating calcitriol levels before and during pregnancy (1,22), confirming that maternal kidneys must be the main source of the normal 2 to 5-fold increase in calcitriol during normal pregnancy. Rodent studies, including pregnancies in mice that lack the 1alpha-hydroxylase, have confirmed that there is a small contribution of fetal or placental calcitriol to the maternal circulation (1,23,24). However, it is not enough to account for the marked increase in maternal calcitriol that normally occurs during pregnancy. Calcitonin Serum calcitonin levels are increased during pregnancy and may derive from maternal thyroid, breast, decidua, and placenta. The importance of these extrathyroidal sites of calcitonin synthesis has been shown by serum calcitonin levels rising from undetectable to normal values in totally thyroidectomized women who become pregnant (25). Whether calcitonin plays an important role in the physiological responses to the calcium demands of pregnancy is unknown. It has been proposed to protect the maternal skeleton against excessive resorption during times of increased calcium demand; however, there are no clinical studies that have addressed this question. Study of pregnant women who lack the gene for calcitonin or the calcitonin receptor would be informative, but no such women have been identified. On the other hand, mice that lack the gene for calcitonin have normal calcium and bone metabolism during pregnancy (26,27). Parathyroid Hormone-related Protein PTHrP concentrations steadily increase in the maternal circulation, reaching the highest levels in the third trimester (1,11). The assays most commonly used in these studies detected PTHrP peptides encompassing amino acids 1-86, but PTHrP is a prohormone. It is cleaved into multiple N-terminal, mid-molecule, and C-terminal peptides, which differ in their biological activities and specificities. None of these peptides have been systematically measured during pregnancy. The commonly available PTHrP1-86 assays do not measure PTHrP1-34, which is likely the most abundant of the active, PTH-like, N-terminal forms of this protein. Moreover, in many clinical studies and case reports it is evident that inappropriate blood samples were used for assaying PTHrP. Special collection and handling are required because PTHrP is rapidly cleaved and degraded in serum. Blood samples should be collected in tubes containing EDTA and aprotinin (a protease inhibitor), kept chilled, and then centrifuged, separated, and frozen within 15 minutes of sample collection. Even with these rigorous standards, PTHrP has been found to begin degrading by 15 minutes after sample collection (28). Many studies did not use this method of sample collection and preparation, but used sera that had been allowed to clot at room temperature for up to 60 minutes. This likely explains why such studies found undetectable serum concentrations of PTHrP, as compared to those that studied the plasma concentration of PTHrP during pregnancy. Individual case reports are also fraught with this problem, since standard blood collection protocols for hospital laboratories do not use the special handling described above. PTHrP is produced by many tissues in the fetus and mother; consequently, it is uncertain which source(s) account for the rise in PTHrP in the maternal circulation. However, the placenta and breasts are likely the major sources of PTHrP. Whether circulating PTHrP has a role in maternal physiology during pregnancy is unclear, but its rise may stimulate the renal 1alpha-hydroxylase and contribute to the increase in calcitriol and, indirectly, the suppression of PTH. However, PTHrP appears less potent than PTH in stimulating the 1alpha-hydroxylase (29,30), which is why its contribution to the rise in calcitriol during pregnancy is uncertain. On the other hand, several case reports have clearly implicated breast- and placental-derived PTHrP as a cause of maternal hypercalcemia with elevated PTHrP and undetectable PTH, a condition called pseudohyperparathyroidism of pregnancy (see below). Since breasts and placenta were sources of excess PTHrP in these cases, those two tissues seem likely to be dominant sources of PTHrP during normal pregnancy. Moreover, since excess PTHrP impacted maternal calcium homeostasis to cause hypercalcemia in these cases, it is possible that the more modest elevations in circulating PTHrP seen during normal pregnancy also affect maternal calcium homeostasis. A carboxyl-terminal form of PTHrP (so-called “osteostatin”) has been shown to inhibit osteoclastic bone resorption in vitro, and thus the notion arises that PTHrP may play a role in protecting the maternal skeleton from excessive resorption during pregnancy (31). Animal studies have shown that PTHrP has other roles during gestation such as regulating placental calcium transport in the fetus (1,32). Maternally produced PTHrP is not likely to regulate placental calcium transport since the protein should not be able to cross the placenta (1,5); instead, it is PTHrP produced within the fetus and placenta that is responsible for regulating placental calcium transport. Fibroblast Growth Factor-23 (FGF23) Intact FGF23 doubles its concentration in the mother’s circulation during rodent pregnancies (17-19), but whether such levels change during human pregnancy has not been reported. Within 24 hours after delivery, mean values in postpartum women were similar to non-pregnant women (33). Other Hormones This section has focused on changes in static concentrations of minerals and the known calciotropic hormones; there are no studies testing hormonal reserves or response to challenges such as hypocalcemia or hypophosphatemia. Pregnancy also induces significant changes in other hormones known to affect calcium and bone metabolism, including sex steroids, prolactin, placental lactogen, oxytocin, leptin, and IGF-1. Each of these – and possibly other hormones not normally associated with mineral and bone metabolism – may have direct or indirect effects on mineral homeostasis during pregnancy. However, this aspect of the physiology of pregnancy has been largely unexplored to date. Prolactin and placental lactogen both increase during pregnancy and activate prolactin receptors. Osteoblasts express prolactin receptors, and prolactin receptor deficient mice show decreased bone formation (34). Suppressing the prolactin level with bromocriptine blunted a pregnancy-related gain in bone mineral content in rats (35). These data are consistent with the notion that prolactin or placental lactogen regulate skeletal metabolism during pregnancy. Furthermore, prolactin can indirectly affect skeletal metabolism by stimulating PTHrP synthesis and release from the breasts (36-38). Circulating oxytocin levels also rise during pregnancy (39), and the oxytocin receptor is expressed by osteoclasts and osteoblasts (40). Male and female mice lacking oxytocin or its receptor have an osteoporotic phenotype with low bone formation (41). Oxytocin has been shown to stimulate osteoblast differentiation and function, stimulate osteoclast formation, but inhibit osteoclast function and skeletal resorption (41,42). Taken together, these data predict that oxytocin may regulate bone metabolism during pregnancy, but this has not been directly studied in vivo. Intestinal Calcium and Phosphate Absorption Intestinal absorption of calcium doubles as early as 12 weeks of human pregnancy, as shown by clinical studies that used stable isotopes of calcium, and by other calcium balance studies (1). This increase in calcium absorption appears to be the major maternal adaptation to meet the fetal need for calcium. It has been generally believed that the doubling or tripling of calcitriol levels explains the increased intestinal calcium absorption and concurrent increases in the intestinal expression of calbindin9k-D (S100G), TRPV6, Ca2+-ATPase (PMCA1), and other genes and proteins involved in calcium transport. However, intestinal calcium absorption doubles in the first trimester, well before the rise in free calcitriol levels during the third trimester. Animal studies have indicated that placental lactogen, prolactin, and other factors may stimulate intestinal calcium absorption (1) and that calcitriol or the vitamin D receptor are not required for intestinal calcium absorption to increase during pregnancy (1,23,43-45). The peak fetal demand for calcium does not occur until the third trimester, and so it is unclear why intestinal calcium absorption should be upregulated in the first trimester. It may allow the maternal skeleton to store calcium in advance of the peak demands for calcium that occur later in pregnancy and lactation; some studies in rodents have shown this to be the case with the bone mineral content rising significantly before term (17,26,45). Women have also been found to be in a positive calcium balance by mid-pregnancy (46), likely due to the effect of increased intestinal calcium absorption on skeletal mineralization. Intestinal phosphate absorption also undergoes a doubling during rodent and other mammalian pregnancies (1), and presumably human pregnancy as well. However, no clinical studies have studied this. Renal Handling of Calcium The doubling of intestinal calcium absorption in the first trimester means that the extra calcium must be passed to the fetus, deposited in the maternal skeleton, or excreted in the urine. Renal calcium excretion is increased as early as the 12th week of gestation and 24-hour urine values (corrected for creatinine excretion) often exceed the normal range. Conversely, fasting urine calcium values are normal or low, confirming that this hypercalciuria is a consequence of the enhanced intestinal calcium absorption (1). This is absorptive hypercalciuria and will not be detected by spot or fasting urine samples that have been corrected for creatinine concentration. Absorptive hypercalciuria contributes to the increased risk of kidney stones during pregnancy. This absorptive hypercalciuria also renders nomograms of fractional calcium excretion invalid for the diagnosis of familial hypocalciuric hypercalcemia during pregnancy (47,48). Pharmacological doses of calcitonin promote renal calcium excretion, but whether the physiologically elevated levels of calcitonin during pregnancy promote renal calcium excretion is unknown. Hypocalciuria during pregnancy has been associated with pre-eclampsia, pregnancy-induced hypertension, and low (equal to non-pregnant values) serum calcitriol (49-52). These changes appear largely secondary to disturbed renal function and reduced creatinine clearance, rather than being causes of the hypertension. However, calcium supplementation reduces the risk of pre-eclampsia in women within the lowest quintile of calcium intake, and so there is a pathophysiological link between calcium metabolism and pregnancy-induced hypertension (1). Skeletal Calcium Metabolism and Bone Density/Bone Marker Changes As mentioned earlier, some studies in rodents indicate that bone mineral content increases during pregnancy, and other studies have shown that histomorphometric parameters of bone turnover are increased at this time. Systematic studies of bone histomorphometry from pregnant women have not been done. However, one study of 15 women who electively terminated a pregnancy at 8-10 weeks found bone biopsy evidence of increased bone resorption, including increased resorption surface and increased numbers of resorption cavities (53). These findings were not present in biopsies obtained from 13 women at term, or in the non-pregnant controls. This study bears repeating but it does suggest that early pregnancy induces skeletal resorption. Bone turnover markers – by-products of bone formation and resorption that can be measured in the serum or urine – have been systematically studied during pregnancy in multiple studies (1). In the non-pregnant adult with osteoporosis these bone markers are fraught with significant intra- and inter-individual variability which limit their utility on an individual basis. There are additional problems with the use of bone markers during pregnancy, including lack of pre-pregnancy baseline values; hemodilution; increased GFR; altered creatinine excretion; placental, uterine and fetal contributions; degradation and clearance by the placenta; and lack of diurnally timed or fasted specimens. Bone resorption has been assessed using urinary (deoxypyridinoline, pyridinoline, and hydroxyproline) and serum (C-telopeptide) markers, and the consistent finding is that bone resorption appears increased from early or mid-pregnancy (1). Conversely, bone formation has been assessed by serum markers (osteocalcin, procollagen I carboxypeptides and bone specific alkaline phosphatase) that were generally not corrected for hemodilution or increased GFR. These bone formation markers are decreased in early or mid-pregnancy from pre-pregnancy or non-pregnant values and rise to normal or above before term (1). The lack of correction for hemodilution and increased GFR means that the apparent decline in bone formation markers may actually occur despite no change or even an increase in bone formation. It should be noted that total alkaline phosphatase rises early in pregnancy due to the placental fraction and is not a useful marker of bone formation during pregnancy. Overall, the scant bone biopsy data and the results of bone turnover markers suggest that bone resorption is increased from as early as the 10th week of pregnancy, whereas bone formation may be suppressed (if the bone formation marker results are correct) or normal (if the bone formation markers are artifactually suppressed due to the aforementioned confounding factors) (1). Notably there is little maternal-fetal calcium transfer occurring in the first trimester, nor is there a marked increase in turnover markers during the third trimester when maternal-fetal calcium transfer is at a peak. These findings may simply underscore that resorption of the maternal skeleton is a minor contributor to calcium homeostasis during pregnancy, whereas the upregulation of intestinal calcium absorption is the main mechanism through which the fetal demand for calcium is met. Another way of assessing whether the maternal skeleton contributes to calcium regulation during pregnancy is to measure bone mineral content or density. A few sequential areal bone density (aBMD) studies have been done using older techniques (single and/or dual-photon absorptiometry, i.e., SPA and DPA), and none with newer techniques (DXA or qCT) due to concerns about fetal radiation exposure. Studies of aBMD are known to be confounded by changes in body composition, weight and skeletal volumes, and all three of these factors change during normal pregnancy. The longitudinal studies used SPA or DPA and found no significant change in cortical or trabecular aBMD during pregnancy (1). Most recent studies examined 16 or fewer subjects with DXA prior to planned pregnancy (range 1-18 months prior, but not always stated) and after delivery (range 1-6 weeks postpartum) [studies reviewed in detail in (54)]. One study found no change in lumbar spine aBMD measurements obtained pre-conception and within 1-2 weeks post-delivery, whereas the other studies reported 4-5% decreases in lumbar aBMD with the postpartum measurement taken between 1-6 weeks post-delivery. A large study from Denmark obtained DXA measurements of hip, spine, and radius at baseline (up to 8 months before pregnancy) and again within 15 days of delivery in 73 women (55). DXA of the radius was also obtained once each trimester. BMD decreased between pre-pregnancy and post-pregnancy by 1.8% at the lumbar spine, 3.2% at the total hip, 2.4% at the whole body, 4% at the ultradistal forearm, and 1% at the total forearm, whereas it increased by 0.5% at the proximal 1/3 forearm (55). All women went on to breastfeed, which means that the final BMD values were confounded by lactation-induced bone loss (see lactation section). These changes in BMD were statistically significant when compared to 57 non-pregnant controls who also had serial measurements done, but the magnitudes of change were small, and would not be considered statistically significant for an individual woman. Ultrasound measurements of the os calcis and fingers have been examined in other longitudinal studies, which reported a progressive decrease in indices that correlate with volumetric BMD (1,54). Whether observed changes in the os calcis accurately indicate a true or clinically meaningful decrease in volumetric BMD, or imply that losses of BMD are occurring in the spine or hip during pregnancy, is not known. The reliability or relevance of data obtained from ultrasound is questionable since this technique failed to detect any change in volumetric BMD at the os calcis during lactation (56), even though substantial bone loss occurs at the spine and hip during lactation (see lactation section). Overall, the existing studies have insufficient power to allow a firm conclusion as to the extent of bone loss that might occur during pregnancy, but it seems likely (especially when data from the Danish study are considered) that modest bone loss occurs, which would be difficult to discern on an individual basis. In the long term, pregnancy does not impair skeletal strength or lead to reduced bone density. Several dozen epidemiological studies of osteoporotic and osteopenic women have failed to find a significant association of parity with bone density or fracture risk (1,57), and many have shown a protective effect of parity (58-75). DISORDERS OF CALCIUM AND BONE METABOLISM DURING PREGNANCY Osteoporosis in Pregnancy The occasional woman will present with a fragility fracture during the third trimester or puerperium, and low bone mineral density may be confirmed by DXA (76). In such cases it is not possible to exclude the possibility that low bone density or skeletal fragility preceded pregnancy. In favor of a genetic predisposition is the report that among 35 women who presented with pregnancy associated osteoporosis, there was a higher than expected prevalence of fragility fractures in their mothers (77). It is conceivable that pregnancy may induce significant skeletal losses in some women and, thereby, predispose to fracture. The normal pregnancy-induced changes in mineral metabolism may cause excessive resorption of the skeleton in selected cases, and other factors such as low dietary calcium intake and vitamin D insufficiency may contribute to skeletal losses (76). A high rate of bone turnover is an independent risk factor for fragility fractures outside of pregnancy, and so the apparently increased bone resorption observed during pregnancy may increase fracture risk. In favor of pregnancy inducing fragility through excess skeletal losses is an observational study of 13 women with pregnancy-associated osteoporosis who were followed for up to eight years. Since the bone mineral density at the spine and hip increased significantly during follow-up in these women, the investigators concluded that a large part of the bone loss must have been related to the pregnancy itself (78). Taken together, fragility fractures in pregnancy or the puerperium may result from the combination of abnormal skeletal microarchitecture prior to pregnancy and increased bone resorption during pregnancy. Osteoporosis in pregnancy usually presents in a first pregnancy and there is no apparent increased risk with higher parity (76,78-80). About 60% of patients present with lower thoracic or lumbar pain that may be quite debilitating due to vertebral collapse (78-80). Most cases show normal serum chemistries and calciotropic hormone levels, but in a few, secondary causes of bone loss could be identified, including anorexia nervosa, hyperparathyroidism, osteogenesis imperfecta, inactivating mutations in LRP5, premature ovarian failure, and corticosteroid or heparin therapy (76,77,79-83). Bone biopsies have confirmed osteoporosis and the absence of osteomalacia, while bone density Z-scores are often lower than expected (78-80). The pain resolves spontaneously over several weeks in most cases while the bone density has been reported to improve in most women following pregnancy. Fractures tend not to recur in subsequent pregnancies. Thus, although myriad medical treatments (bisphosphonates, estrogen, testosterone, calcitonin, teriparatide, denosumab) and surgical interventions (kyphoplasty, vertebroplasty, spinal fusion) have been used in individual cases of pregnancy-associated osteoporosis (76), the tendency for this condition to spontaneously improve may make pharmacological treatment unjustified except for the severest cases. At the least, it may be prudent to wait 12-18 months to determine the extent to which the BMD recovers on its own after a pregnancy-associated vertebral fracture (76). A distinct condition is focal, transient osteoporosis of the hip (76). This is rare, self-limited, and probably not a manifestation of altered calciotropic hormone levels or mineral balance during pregnancy. Instead, it may be a consequence of local factors. A variety of theories have been offered to explain this condition, including femoral venous stasis due to pressure from the pregnant uterus, Sudeck’s atrophy or reflex sympathetic dystrophy (causalgia), ischemia, trauma, viral infections, marrow hypertrophy, immobilization, and fetal pressure on the obturator nerve. These patients present with unilateral or bilateral hip pain, limp and/or hip fracture in the third trimester or puerperium (76,84-86). Radiographs and DXA indicate radiolucency and reduced bone density of the symptomatic femoral head and neck, while MRI demonstrates increased water content of the femoral head and the marrow cavity; a joint effusion may also be present. The differential diagnosis of this condition includes inflammatory joint disorders, avascular necrosis of the hip, bone marrow edema, and reflex sympathetic dystrophy. It is a self-limiting condition with both symptoms and radiological appearance resolving within two to six months post-partum; conservative measures including bed rest are usually all that is required during the symptomatic phase (76). Of course, fractures of an involved femur require urgent arthroplasty or hip replacement. The condition recurs in about 40% of cases (not necessarily during pregnancy), unlike osteoporosis involving the spine, and this has prompted prophylactic hip arthroplasty to be done in a few cases where the opposite hip appears to be affected. Primary Hyperparathyroidism This is probably a rare condition but there are no firm data available on its prevalence. Two case series indicated that parathyroidectomies were done during pregnancy in about 1% of all cases (87,88). The diagnosis will be obscured by the normal pregnancy-induced changes that lower the total serum calcium and suppress PTH; however, finding the ionized or albumin-corrected calcium to be increased, and PTH to be detectable, should indicate primary hyperparathyroidism in most cases. Primary hyperparathyroidism during pregnancy has been reported to cause a variety of symptoms that are not specific to hypercalcemia, and cannot be distinguished from those occurring in normal pregnancy (nausea, vomiting, renal colic, malaise, muscle aches and pains, etc.). Conversely the literature associates primary hyperparathyroidism with an alarming rate of adverse outcomes in the fetus and neonate, including a 10-30% rate for each of spontaneous abortion, stillbirth, and perinatal death, and 30-50% incidence of neonatal tetany (88-92). These high rates were reported in older literature; more recent case series suggest that the rates of stillbirth and neonatal death are each about 2%, while neonatal tetany occurred in 15% (89). The adverse postnatal outcomes are thought to result from suppression of the fetal and neonatal parathyroid glands; this suppression may be prolonged after birth for 3-5 months (89) and in some cases it has been permanent (89,91,93). To prevent these adverse outcomes, surgical correction of primary hyperparathyroidism during the second trimester has been almost universally recommended. Several case series have found elective surgery to be well tolerated, and to dramatically reduce the rate of adverse events when compared to the earlier cases reported in the literature. In a series of 109 mothers with hyperparathyroidism during pregnancy who were treated medically (N=70) or surgically (N=39), there was a 53% incidence of neonatal complications and 16% incidence of neonatal deaths among medically treated mothers, as opposed to a 12.5% neonatal complications and 2.5% neonatal deaths in mothers who underwent parathyroidectomy (88). Choosing the second trimester allows organogenesis to be complete in the fetus and to avoid the poorer surgical outcomes and risk of preterm birth associated with surgery during the third-trimester (89,92,94,95). Many women in the earliest published cases had a more severe form of primary hyperparathyroidism that is not often seen today (symptomatic, with nephrocalcinosis and renal insufficiency). While mild, asymptomatic primary hyperparathyroidism during pregnancy has been followed conservatively with successful outcomes, complications continue to occur, so that, in the absence of definitive data, surgery during the second trimester remains the most common recommendation (96). Milder cases diagnosed during the third trimester may be observed until delivery, although rapid and severe postpartum worsening of the hypercalcemia can occur (95,97-100). This postpartum “parathyroid crisis” occurs because the placental calcium outflow has been lost, while surging PTHrP production in the breasts means an additional factor stimulating bone resorption. There are no definitive medical management guidelines for hyperparathyroidism during pregnancy apart from ensuring adequate hydration and correction of electrolyte abnormalities (96). Pharmacologic agents to treat hypercalcemia have not been adequately studied in pregnancy. Calcitonin does not cross the placenta and has been used safely (96). Oral phosphate has also been used but is limited by diarrhea, hypokalemia, and risk of soft tissue calcifications. Bisphosphonates are relatively contra-indicated because of their potential adverse effects on fetal endochondral bone development, although a review of 78 cases of bisphosphonate use in pregnancy found no obvious problems in most cases (101). Denosumab crosses the placenta and has been shown to cause an osteopetrotic-like phenotype in fetal cynomolgus monkeys and rats (102,103), and so it should be avoided in human pregnancy. High-dose magnesium has been proposed as a therapeutic alternative which should decreases serum PTH and calcium levels by activating the calcium sensing-receptor, but it has not been adequately studied for this purpose (104,105). The calcium receptor agonist cinacalcet, which is used to suppress PTH and calcium in nonpregnant subjects with primary or secondary hyperparathyroidism and parathyroid carcinoma, has also been tried in pregnancy (106-109). However, since the calcium receptor is expressed in the placenta and regulates fetal-placental calcium transfer (110), the possibility of adverse effects of cinacalcet on the fetus and neonate remain a concern. In any case that was followed medically, parathyroidectomy is recommended to be done postpartum, with monitoring in place to detect a postpartum hypercalcemic crisis. Familial Hypocalciuric Hypercalcemia Inactivating mutations in the calcium-sensing receptor cause this autosomal dominant condition which presents with hypercalcemia and hypocalciuria (111). As noted above, fractional excretion of calcium is not reduced during pregnancy in this condition, because it is overridden by the physiological increase in intestinal calcium absorption that in turn causes hypercalciuria (47,48). Pregnancy in women with familial hypocalciuric hypocalcemia may be uneventful for the mother, but the maternal hypercalcemia has caused fetal and neonatal parathyroid suppression with subsequent tetany in both normal and hemizygous children (5,112,113). A hemizygous neonate will later develop benign hypercalcemia, but if the baby has two inactivating mutations of the calcium receptor (most commonly from both parents being hemizygous for FHH), then the neonate may suffer a life-threatening hypercalcemic crisis (5). Hypoparathyroidism Hypoparathyroidism during pregnancy usually presents as a pre-existing condition that the clinician is challenged to manage. The natural history of hypoparathyroidism during pregnancy is confusing due to conflicting case reports in the literature [reviewed in (1,3)]. Early in pregnancy, some hypoparathyroid women have fewer hypocalcemic symptoms and require less supplemental calcium. This is consistent with a limited role for PTH in the pregnant woman, and suggests that an increase in calcitriol and/or increased intestinal calcium absorption occurs in the absence of PTH. However, other case reports clearly indicate that some pregnant hypoparathyroid women required increased calcitriol replacement in order to avoid worsening hypocalcemia. Adding to the confusion is that in some case reports, it appears that the normal, artifactual decrease in total serum calcium during pregnancy was the parameter that led to treatment with increased calcium and calcitriol supplementation; fewer cases reported that dose increments in calcitriol and calcium were made because of maternal symptoms of hypocalcemia or tetany, or objective evidence of true hypocalcemia (ionized or albumin-corrected calcium). It is not possible to know in advance who will improve and who will worsen during pregnancy; the task is to maintain the albumin-corrected serum calcium or ionized calcium in the normal range for the duration of pregnancy. Maternal hypocalcemia due to hypoparathyroidism must be avoided because it has been associated with intrauterine fetal hyperparathyroidism and fetal death. Conversely, overtreatment must be avoided because maternal hypercalcemia is associated with the fetal and neonatal complications described above under Primary Hyperparathyroidism. Calcitriol and 1α-calcidiol are recommended due to their shorter half-lives, lower risk of toxicity, and the clinical experience with these agents. Late in pregnancy, hypercalcemia may occur in hypoparathyroid women unless the calcitriol dosage and supplemental calcium are substantially reduced or discontinued. This effect appears to be mediated by the increasing levels of PTHrP in the maternal circulation in late pregnancy. Conversely, one case report of hypoparathyroidism in pregnancy found that there was a transient interval of increased requirement for calcitriol immediately after delivery and before lactation was fully underway (114). This may be the result of loss of placental sources of PTHrP followed by a surge in production of PTHrP by the lactating breast (see lactation section, below). Pseudohypoparathyroidism Pseudohypoparathyroidism is a genetic disorder causing resistance to PTH and manifest by hypocalcemia, hypophosphatemia, and high PTH levels. In two case reports of pseudohypoparathyroidism during pregnancy, the serum calcium normalized, PTH reduced by half, and calcitriol increased 2- to 3-fold (115). The mechanism by which these changes occur despite pseudohypoparathyroidism remains unclear. If maternal hypocalcemia persists during pregnancy, pseudohypoparathyroidism can lead to the same adverse fetal outcomes that have been associated with maternal hypoparathyroidism, including parathyroid hyperplasia, skeletal demineralization, and fractures (116,117). The maternal calcium concentration must be maintained in the normal range to avoid these fetal outcomes. Pseudohyperparathyroidism As mentioned above, pseudohyperparathyroidism is hypercalcemia that is caused by physiological release of PTHrP driving increased skeletal resorption, akin to how PTHrP also causes hypercalcemia of malignancy. In one such case the breasts were the source of PTHrP because the hypercalcemia and elevated PTHrP did not abate until a bilateral reduction mammoplasty was carried out (118,119). It has occurred in women who simply have large breasts (120,121). In another case the hypercalcemia, elevated PTHrP, and suppressed PTH reversed within a few hours of an urgent C-section, thereby confirming the placenta as the source (122). In all cases of pseudohyperparathyroidism, it should be anticipated that the cord blood calcium will also be increased, and that the baby is at risk for fetal and neonatal hypoparathyroidism with hypocalcemic tetany. Vitamin D Deficiency and Insufficiency There are no comprehensive studies of the effects of vitamin D deficiency or insufficiency on human pregnancy, but the available data from small clinical trials of vitamin D supplementation, observational studies, and case reports suggest that, consistent with animal studies, vitamin D insufficiency and deficiency is not associated with any worsening of maternal calcium homeostasis (this topic is reviewed in detail in (1,4,7). Maternal hypocalcemia is milder with vitamin D deficiency due to the effects of secondary hyperparathyroidism to increase skeletal resorption and renal calcium reabsorption. Consequently, hypocalcemia due to vitamin D deficiency has not been clearly associated with the same adverse fetal outcomes that maternal hypoparathyroidism causes (reviewed in detail in (5,123)). The fetal effects of vitamin D deficiency, inability to form calcitriol, and absence of the vitamin D receptor have been examined across several animal species and all have indicated that the fetus will have a normal serum calcium and fully mineralized skeleton at term (reviewed in detail in (5,123)). Neonatal hypocalcemia and rickets can occur in infants born of mothers with severe vitamin D deficiency, but it is usually in the weeks to months after birth that this presents, after intestinal calcium absorption becomes dependent on calcitriol. There has been much interest in studies that have inconsistently associated third-trimester measurements of 25OHD, or estimated vitamin D intakes during pregnancy or the first year after birth, with possible extraskeletal benefits in the mother (reduced bacterial vaginosis, pre-eclampsia, pre-term delivery) or in the offspring (lower incidence of type 1 diabetes, greater skeletal mineralization, etc.). These associational studies won’t be discussed in detail (some are cited in: (1,5,124)) because they are confounded by factors which contribute to lower 25OHD levels (maternal overweight/obesity, lower socioeconomic status, poor nutrition, lack of exercise, etc.). It is necessary to test these associations in randomized clinical trials that compare higher versus lower intakes of vitamin D during pregnancy. At present the results of the associational studies are insufficient to warrant prescribing higher intakes of vitamin D during pregnancy to prevent these postulated outcomes. Among many clinical trials of vitamin D supplementation that have been carried out (1), only a few have included over a 100 study participants who were vitamin D deficient at entry, while other recent studies that gained press attention did not include many vitamin D deficient subjects at all. Among the trials with over 100 participants (14,125-132), the two largest were from Bangladesh and UK with over 1,000 participants (131,132). Baseline maternal 25OHD levels were lowest (20-29 nmol/L) in trials from Bangladesh, UK, Iran, and UAE, and in the 40-60 nmol/L range in the remainder. Interventions consisted of placebo/no treatment versus low dose (400 IU/day) or high dose (1,000-5,000 IU/day) vitamin D supplementation, initiated before mid-pregnancy, and maintained until delivery. For most trials the primary outcomes were simply maternal and neonatal-cord blood 25OHD and calcium. The most recent and largest study was from Bangladesh, and the primary outcome was pre-specified as infant length-for-age z-scores at 1 year of age (132). Offspring anthropometric parameters and/or bone mineral content were pre-specified only in a few of the remaining studies (128,130,131). In all studies vitamin D supplementation increased maternal serum and cord blood 25OHD, but there was no overall effect on cord blood calcium. The largest achieved difference in a single study was 16 nmol/L (6.4 ng/mL) in the untreated and 168 nmol/L (67 ng/mL) in vitamin D-supplemented mothers at term; however, there was no obstetrical or fetal benefit (125). The incidence of neonatal hypocalcemia was reduced in offspring of vitamin-D treated mothers, reflecting the role of vitamin D/calcitriol to stimulate postnatal intestinal calcium absorption (125). In the large Bangladesh study, there were no significant differences in infant anthropometrics or any other fetal, neonatal or maternal outcomes (132). In one US-based study there was no benefit on mode of delivery, gestational age at delivery, and preterm birth (14), while in another there was no benefit on mode of delivery, C-section rates, adverse events, hypertension, infection, gestational diabetes, still birth, gestational age at delivery, or combinations of these outcomes (127). The UK MAVIDOS trial reported no obstetrical benefit, and no benefit to any of the primary (neonatal bone area, BMC, and BMD within the first 10-14 days after birth) or secondary outcomes (anthropometric and body composition parameters within 48 hours of birth). However, it received much publicity for a demonstrated increase in BMC and BMD in winter-born neonates of vitamin D-supplemented vs. placebo-treated mothers (131). Because the neonatal skeleton accretes 100 mg/day of mineral content after birth, this result may reflect improved intestinal mineral delivery over 14 days after birth, rather than a prenatal effect on skeletal mineralization (1,133,134). Curiously, autumn-born neonates of vitamin D-supplemented vs. placebo-treated mothers showed an adverse trend of similar magnitude on BMC and BMD, which suggests possible harm from vitamin D supplementation, or chance findings due to small numbers within the sub-groups (134). These sub-group analyses of treatment by season interaction were not specified outcomes in the trial registries (ISRCTN 82927713 and EUDRACT 2007-001716-23). In the UK study that achieved the greatest difference in 25OHD levels between untreated and vitamin D-treated mothers and babies, there was a trend for more small for gestational age babies born to mothers who did not receive antenatal vitamin D supplementation (28% vs. 15%, p<0.1), but the study was not powered for this outcome (125). In studies from the UAE, and Iran there was also no benefit on obstetrical outcomes (variably, mode of delivery, C-section rates, adverse events, stillbirths, gestational age at delivery) or neonatal anthropometric measurements and bone mass measurements (126,128,130). The lack of any beneficial effect on maternal, immediate fetal/neonatal and neonatal outcomes (anthropometrics and cord blood calcium), even in studies that included mothers with some of the lowest 25OHD levels (125,128,130,132), suggests that vitamin D supplementation during pregnancy confers no benefit to the neonate. The most recent study was well-powered to demonstrate a beneficial effect on infant length and other fetal/neonatal outcomes, but did not yield any significant results, despite low vitamin D levels in the mothers at study entry (132). Systematic reviews have used these and the results of smaller trials to examine the effect of vitamin D supplementation during pregnancy on maternal, fetal, and neonatal extra-skeletal outcomes (135-140). Vitamin D supplementation had no significant effect on pre-eclampsia in four (136,138-140) and a positive effect in two reviews (135,137), while combined vitamin D and calcium supplementation reduced the incidence of pre-eclampsia in three systematic reviews (135-137). No consistent effect was seen on other outcomes such as preterm birth, low birth weight, small for gestational age, infections, C-section rate, and newborn anthropometrics. Overall, available data are insufficient from the individual clinical trials or these systematic reviews to conclude that vitamin D supplementation during pregnancy confers any proven obstetrical benefits. Genetic Vitamin D Resistance Syndromes Case reports and series have provided insight into the effect of pregnancy on genetic disorders of vitamin physiology. Pregnancies have generally been unremarkable in women with vitamin D-dependent rickets type 1 (VDDR-I) which is due to absence of Cyp27b1, and in women with VDDR-II that is due to absence of functional VDRs (141-143). In one such uneventful VDR-II pregnancy, the pre-pregnancy intake of supplemental calcium (800 mg) and high-dose calcitriol were maintained until her clinicians increased the dose of calcitriol later in pregnancy “because of the knowledge that the circulating 1,25-(OH)2D concentration normally rises during pregnancy,” and not because of any change in albumin-adjusted serum calcium (142). Consequently, it’s unclear that any change was needed. However, it is reasonable to increase the dose of calcitriol to mirror the increase that happens during normal pregnancy. In women with VDDR-I, the dose of calcitriol was unchanged in one-third of pregnancies but increased 1.5 to 2-fold in others (141). 24-Hydroxylase Deficiency Loss of the catabolic effects of 24-hydroxylase causes high calcitriol and mild hypercalcemia in non-pregnant adults, which may be asymptomatic (144). But during pregnancy, the physiological 2 to 5-fold increase in calcitriol is unopposed by catabolism, which causes an exaggerated increase in calcitriol, followed by symptomatic hypercalcemia. Hypercalcemia can be quite marked, with suppressed or undetectable PTH, and calcitriol concentrations that exceed what is expected for pregnancy (145-147). Pregnant patients may also present with nephrolithiasis or acute pancreatitis (147,148). Treatment of the hypercalcemia is difficult because the agents that could be used are not approved for pregnancy. Increased intestinal calcium absorption is the direct cause, and so use of increased hydration and a modestly restricted calcium diet, combined with phosphate supplementation to bind dietary calcium, are relatively safe management approaches. If PTH increases above normal, then dietary calcium restriction should be lessened to prevent maternal bone resorption and fetal secondary hyperparathyroidism. Other pharmacologic therapy should be reserved for the most severe cases and used with caution. This includes oral glucocorticoids, loop diuretics, calcitonin, and bisphosphonates; denosumab should not be used because of teratogenic effects observed in cynomolgus monkeys and mice (102,103). Cinacalcet will not be useful because PTH will already be suppressed due to the combined effects of pregnancy and hypercalcemia. Low or High Calcium Intake Through the doubling of intestinal calcium absorption during pregnancy, women have the ability to adapt to wide ranges of calcium intakes and still meet the fetal demand for calcium. It is conceivable that extremely low maternal calcium intakes could impair maternal calcium homeostasis and fetal mineral accretion, but there are scant clinical data examining this possibility (149). Among women with low dietary calcium intake, there are differing results as to whether or not calcium supplementation during pregnancy improved maternal or neonatal bone density (150). There is short term evidence that bone turnover markers were reduced when 1.2 gm of supplemental calcium was given for 20 days to 31 Mexican woman at 25-30 weeks of gestation; their mean dietary calcium intake was 1 gm (151). In a double-blind study conducted in 256 pregnant women, 2 gm of calcium supplementation improved bone mineral content only in the infants of supplemented mothers who were in the lowest quintile of calcium intake (152). Among cases of fragility fractures presenting during pregnancy, some women had very low calcium intakes (<300 mg per day), and in such cases substantial maternal skeletal resorption must be invoked in order to meet the fetal calcium requirement and maintain the maternal serum calcium concentration (76). Overall the physiological changes in calcium and bone metabolism that usually occur during pregnancy and lactation are likely to be sufficient for fetal bone growth and breast-milk production in women with reasonably sufficient calcium intake (153). However, the use of calcium supplementation for pregnant women with low calcium intake can be defended by the links between low calcium intake and both preeclampsia and hypertension in the offspring (149). Clinical trials and meta-analyses have also demonstrated that supplemental calcium will reduce the risk of preeclampsia in women with low dietary calcium intakes, but not in those with adequate intake (154-157). High calcium intake, similar to primary hyperparathyroidism, can cause increased intestinal calcium absorption, maternal hypercalcemia, increased transplacental flow of calcium, and suppression of the fetal parathyroids. Cases of neonatal hypoparathyroidism have been reported wherein women consumed 3 to 6 grams of elemental calcium daily as antacids or anti-nauseates (1). Hypercalcemia of Malignancy Hypercalcemia of malignancy is usually a terminal condition. When it has been diagnosed during pregnancy, in some cases the baby has been spared from chemotherapy, whereas in other cases the pregnancy was terminated (or ignored) so that chemotherapy could be administered in an attempt to prolong the woman’s life. Half of published case reports haven’t even mentioned the baby’s outcome. A baby born of a mother with humoral hypercalcemia of malignancy may have a high concentration of calcium in cord blood, and is at high risk for fetal and neonatal hypoparathyroidism with hypocalcemic tetany. FGF-23 Disorders X-linked hypophosphatemic rickets (XLH) is caused by inactivating mutations in the PHEX gene, which lead to high circulating levels of FGF23. In turn this causes hypophosphatemia with rickets or osteomalacia. Pregnancies were normal in a mouse model of XLH. In particular, despite very high circulating levels of FGF23, which normally downregulate calcitriol synthesis and increase its catabolism, maternal serum calcitriol increased to the high levels normally seen during pregnancy (19,158). This rise in calcitriol should contribute to increased intestinal calcium and phosphate absorption. Several case reports documented persistent hypophosphatemia during pregnancy in women with XLH, but no adverse outcomes (159,160). Nevertheless, it is generally recommended to supplement with calcitriol and phosphate to keep the serum phosphate near normal during pregnancy. Hyperphosphatemic disorders due to loss of FGF23 action have not been studied during human pregnancy, and animal data are also lacking because these conditions are lethal before sexual maturity. Renal insufficiency or failure causes hyperphosphatemia, and both animal and human data indicate that such renal disorders increase the risks of gestational hypertension, pre-eclampsia, eclampsia, and maternal mortality. However, the extent to which the hyperphosphatemia contributes to these risks is unknown. MINERAL PHYSIOLOGY DURING LACTATION AND POST-WEANING As lactation begins the mother is faced with another demand for calcium in order to make milk. The average daily loss of calcium into breast milk is 210 mg, although daily losses as great as 1000 mg calcium have been reported is some women nursing twins (1). Although women meet the calcium demands of pregnancy by upregulating intestinal calcium absorption and serum concentrations of calcitriol, a different adaptation occurs during lactation. A temporary resorption and demineralization of the maternal skeleton appears to be the main mechanism by which breastfeeding women meet these calcium requirements. This adaptation does not appear to require PTH or calcitriol, but is regulated by the combined effects of increased circulating concentrations of PTHrP and low estradiol levels. Mineral Ions The albumin-corrected serum calcium and ionized calcium are both normal during lactation, but longitudinal studies have shown that both are increased slightly over the non-pregnant values. Serum phosphate levels are also higher and may exceed the normal range. Since reabsorption of phosphate by the kidneys appears to be increased, the increased serum phosphate levels may, therefore, reflect the combined effects of increased flux of phosphate into the blood from diet and from skeletal resorption, in the setting of decreased renal phosphate excretion. Parathyroid Hormone PTH, as measured by 2-site “intact” or newer “bio-intact” assays, may be undetectable or in the lower quarter of the normal range during the first several months of lactation in women from North America and Europe who consume adequate calcium. PTH rises to normal by the time of weaning, and in two case series was found to rise above normal post-weaning. In contrast, and similar to findings during pregnancy, PTH did not suppress in several studies of women from Asia and Gambia who consumed diets that were low in calcium or high in phytate. The low PTH concentrations are an indication that PTH isn’t required for mineral homeostasis during lactation, and this is confirmed by hypoparathyroid and aparathyroid women in whom mineral and skeletal homeostasis normalize while they continue to breastfeed (see Hypoparathyroidism, below). The same is true of mice that lack the gene for parathyroid hormone. They are hypocalcemic and hyperphosphatemic when non-pregnant, but maintain normal serum calcium and phosphate concentrations while lactating and for a time during post-weaning (17). Vitamin D Metabolites A common concern has been that the suckling neonate will deplete maternal 25OHD stores, but this is not the case. 25OHD should not decline because it does not enter breast milk; conversely, although vitamin D can enter milk, it is present at very low concentrations because appreciable amounts exist in the maternal circulation for only a short postprandial interval. In observational studies and in the placebo arms of several clinical trials, there was either no change or at most a nonsignificant decline in maternal 25OHD levels during lactation, even in severely vitamin D deficient women (4). Calcitriol levels were twice normal during pregnancy but both free and bound calcitriol levels fall to normal within days of parturition and remain there in breastfeeding women (a single study found that women breastfeeding twins had higher calcitriol concentrations than women nursing singletons) (161). Animal studies show that severely vitamin D deficient rodents and mice lacking the vitamin D receptor are able to lactate and provide normal milk (4,45), thereby indicating that vitamin D and calcitriol are not required for lactation to proceed normally (at least in rodents). However, a more recent study found that mice lacking calcitriol produced milk with a lower calcium content (23). Calcitonin Calcitonin levels fall to normal during the first six weeks postpartum in women. Mice lacking the gene that encodes calcitonin lose twice the normal amount of bone mineral content during lactation, which indicates that physiological levels of calcitonin may protect the maternal skeleton from excessive resorption during this time period (26). Whether calcitonin plays a similar role in human physiology is unknown. Totally thyroidectomized women are not calcitonin deficient during lactation due to substantial production of calcitonin by the breasts, which in turn leads to systemic calcitonin concentrations that are the same as in women with intact thyroids (25). Consequently, study of totally thyroidectomized women is not the equivalent of studying a calcitonin-null state when they are breastfeeding. PTHrP Plasma PTHrP concentrations are significantly higher in lactating women than in non-pregnant controls. The source of PTHrP appears to be the breast, which secretes PTHrP into breast milk at concentrations that are 1,000 to 10,000 times the level found in the blood of patients with hypercalcemia of malignancy or in normal human controls. The circulating PTHrP concentration also increases after suckling (162,163). Additional evidence that the breasts are the source of PTHrP include that ablation of the PTHrP gene selectively from mammary tissue resulted in reduced circulating levels of PTHrP in lactating mice (164). PTHrP also has an intimate association with breast tissue: in animals it has been shown to regulate mammary development and blood flow, and the calcium and water content of milk in rodents, whereas in humans it is commonly expressed by breast cancers. Furthermore, as described in more detail below, during lactation PTHrP reaches the maternal circulation from the lactating breast to cause resorption of calcium from the maternal skeleton, renal tubular reabsorption of calcium, and (indirectly) suppression of PTH. In support of this hypothesis, deletion of the PTHrP gene from mammary tissue at the onset of lactation resulted in more modest losses of bone mineral content during lactation in mice (164). In humans, PTHrP correlates with the amount of bone mineral density lost, negatively with serum PTH, and positively with the ionized calcium of lactating women (162,165,166). Lastly, clinical observations in hypoparathyroid and aparathyroid women demonstrate the physiological importance of PTHrP to regulate calcium and skeletal homeostasis during lactation (see Hypoparathyroidism, below). Prolactin Prolactin is persistently elevated during early lactation and spikes further upward with suckling. Later during lactation basal prolactin levels are normal but continue to spike with suckling. Prolactin is important for initiating and maintaining milk production (167), but it also alters bone metabolism by stimulating PTHrP production in lactating mammary tissue, inhibiting GnRH and ovarian function, and possibly (as noted earlier) through direct actions in osteoblasts that express the prolactin receptor. Oxytocin Oxytocin induces milk ejection by contracting myoepithelial cells within mammary tissue. If milk is not ejected, the pressure of milk stasis causes apoptosis of mammary cells, and lactation ceases. Oxytocin spikes in the maternal circulation within 10 minutes after the start of suckling (168). As noted earlier, the oxytocin receptor is expressed in osteoblasts and osteoclasts. But whether oxytocin plays a role in bone metabolism during lactation has proven difficult to determine because oxytocin null mice cannot lactate due to the lack of milk ejection (169). Estradiol In lactating women, estradiol levels fall and this stimulates RANKL and inhibits osteoprotegerin production by osteoblasts, thereby stimulating osteoclast proliferation, function, and bone resorption. Studies in mice have shown that increasing the serum estradiol concentration to 7 times the virgin level blunts the magnitude of bone loss during lactation (170), which confirms that estradiol deficiency plays a role in the skeletal resorption that occurs during lactation. FGF23 FGF23 levels during lactation have not been reported. It is possible that FGF23 increases to compensate for the increased serum phosphate and low PTH that occur during lactation, but it’s also possible that FGF23 is low and contributing to the high serum phosphate. Other Hormones Serotonin appears to be involved in regulating PTHrP and its effect to resorb the maternal skeleton (171,172). Lactation induces changes in myriad other hormones, such as luteinizing and follicle stimulating hormone, progesterone, testosterone, inhibins, and activins. Whether these play roles in regulating skeletal metabolism during lactation has not been investigated. Intestinal Absorption of Calcium and Phosphate Although intestinal calcium absorption was upregulated during pregnancy, it quickly decreases post-partum to the non-pregnant rate. This also corresponds to the fall in calcitriol levels to normal. This differs from rodents which maintain increased intestinal calcium absorption during lactation; their large litters sizes mandate the need to provide some of the calcium for milk production through this route. Intestinal phosphate absorption has not been measured during human lactation, whereas in rodents it remains increased. Renal Handling of Calcium and Phosphate Renal excretion of calcium is typically reduced to about 50 mg per 24 hours or lower, and the glomerular filtration rate is also decreased. These findings suggest that the tubular reabsorption of calcium must be increased to conserve calcium, perhaps through the actions of PTHrP. Renal tubular phosphate reabsorption is increased during lactation. Despite this, urine phosphate excretion may be increased, likely due to the large efflux of phosphate from resorbed bone, which exceeds what is needed for milk production. Skeletal Calcium Metabolism and Bone Density/Bone Marker Changes Histomorphometric data from lactating animals have consistently shown increased bone turnover, and losses of 35% or more of bone mineral are achieved during 2-3 weeks of normal lactation in rodents [reviewed in (1)]. There are no histomorphometric data from lactating women; instead, biochemical markers of bone formation and resorption have been assessed in numerous cross-sectional and prospective studies. Confounding factors discussed earlier for pregnancy need to be considered when assessing bone turnover markers in lactating women; in particular, opposing changes from pregnancy include that the glomerular filtration rate is reduced and the intravascular volume is now contracted. Serum and urinary (24-hr collection) markers of bone resorption are elevated 2-3 fold during lactation and are higher than the levels attained in the third trimester. Serum markers of bone formation (not adjusted for hemoconcentration or reduced GFR) are generally high during lactation, and increased over the levels attained during the third trimester. The most marked increase is in the bone resorption markers, suggesting that bone turnover becomes negatively uncoupled, with bone resorption markedly exceeding bone formation, and thereby causing net bone loss. Total alkaline phosphatase falls immediately postpartum due to loss of the placental fraction, but may still remain above normal due to elevation of the bone-specific fraction. Overall, these bone marker results are compatible with significant increased bone resorption occurring during lactation. Serial measurements of aBMD during lactation (by SPA, DPA or DXA) have shown that bone mineral content falls 3 to 10.0% in women after two to six months of lactation at trabecular sites (lumbar spine, hip, femur and distal radius), with smaller losses at cortical sites and whole body (1,57). These aBMD changes are in accord with studies in rats, mice, and primates in which the skeletal resorption has been shown to occur largely at trabecular surfaces and to a lesser degree in cortical bone, and as much as 25-30% of bone mass or aBMD is lost during three weeks of lactation in normal rodents. The loss in women occurs at a peak rate of 1-3% per month, far exceeding the 1-3% per year that can occur in postmenopausal women who are considered to be losing bone rapidly. This bone resorption is an obligate consequence of lactation and cannot be prevented by increasing the calcium intake in women. Several randomized trials and other studies have shown that calcium supplementation does not significantly reduce the amount of bone lost during lactation (173-176). Not surprisingly, the lactational decrease in bone mineral density correlates with the amount of calcium lost in the breast milk (177). The skeletal losses are due in part to the low estradiol levels during lactation which stimulate osteoclast number and activity. However, low estradiol is not the sole cause of the accelerated bone resorption or other changes in calcium homeostasis that occur during lactation. It is worth noting what happens to reproductive-age women who have marked estrogen deficiency induced by GnRH agonist therapy in order to treat endometriosis, fibroids, or severe acne. Six months of GnRH-induced estrogen deficiency caused 1-4% losses in trabecular (but not cortical) aBMD, increased urinary calcium excretion, and suppression of calcitriol and PTH (Figure 3) [reviewed in (1,8)]. In contrast, during lactation women are not as estrogen deficient but lose more aBMD (at both trabecular and cortical sites), have normal (as opposed to low) calcitriol levels, and have reduced (as opposed to increased) urinary calcium excretion (Figure 3). The difference between isolated GnRH-induced estrogen deficiency and lactation appears to be explained by PTHrP. It stimulates osteoclast-mediated bone resorption and stimulates renal calcium reabsorption; by so doing, it complements the effects of low estradiol during lactation. Stimulated in part by suckling and high prolactin levels, PTHrP and estrogen deficiency combine to cause marked skeletal resorption during lactation (Figure 4). Figure 3. Comparison of the effects of acute estrogen deficiency vs. lactation on calcium and bone metabolism. Acute estrogen deficiency (e.g. GnRH analog therapy) increases skeletal resorption and raises the blood calcium; in turn, PTH is suppressed and renal calcium losses are increased. During lactation, the combined effects of PTHrP (secreted by the breast) and estrogen deficiency increase skeletal resorption, reduce renal calcium losses, and raise the blood calcium, but calcium is directed into breast milk. Reprinted from ref. (8), © 1997, The Endocrine Society. Figure 4. Brain-Breast-Bone Circuit. The breast is a central regulator of skeletal demineralization during lactation. Suckling and prolactin both inhibit the hypothalamic gonadotropin-releasing hormone (GnRH) pulse center, which in turn suppresses the gonadotropins (luteinizing hormone [LH] and follicle-stimulating hormone [FSH]), leading to low levels of the ovarian sex steroids (estradiol and progesterone). PTHrP production and release from the breast is controlled by several factors, including suckling, prolactin, and the calcium receptor. PTHrP enters the bloodstream and combines with systemically low estradiol levels to markedly upregulate bone resorption. Increased bone resorption releases calcium and phosphate into the blood stream, which then reaches the breast ducts and is actively pumped into the breast milk. PTHrP also passes into milk at high concentrations, but whether swallowed PTHrP plays a role in regulating calcium physiology of the neonate is unknown. Calcitonin (CT) may inhibit skeletal responsiveness to PTHrP and low estradiol. Not depicted are that direct effects of oxytocin and prolactin on bone cells are also possible. Adapted from ref. (26) © 2006, The Endocrine Society. The mechanism through which the skeleton is resorbed has been shown in rodents to involve two processes, both osteoclast-mediated bone resorption (1) and osteocytic osteolysis, in which osteocytes function like osteoclasts to resorb the bone matrix that surrounds them (178). Both of these processes are dependent upon PTHrP. Conditional deletion of the PTHrP gene from mammary tissue reduced the amount of bone resorbed during lactation, whereas conditional deletion of the PTH/PTHrP receptor from osteocytes appeared to eliminate osteocytic osteolysis (179). Moreover, osteocyte-specific deletion of the PTH/PTHrP receptor resulted in a 50% blunting of the amount of BMD lost during lactation (179), which may indicate that osteocytic osteolysis and osteoclast-mediated bone resorption each contribute about half of the net bone loss achieved during lactation. To date no studies have examined whether osteocytic osteolysis occurs in lactating women. The lactational bone density losses in women are substantially and completely reversed during six to twelve months following weaning (1,57,174). This corresponds to a gain in bone density of 0.5 to 2% per month in a woman who has weaned her infant. The mechanism for this restoration of bone density is unknown, but studies in mice have shown that it is not dependent upon calcitriol, calcitonin, PTH, or PTHrP (17,23,26,45,180,181); nor is it fully explained by restoration of estradiol levels to normal (1). The remarkable ability of the skeleton to recover is exemplified by mice lacking the gene that encodes calcitonin. They lose up to 55% of trabecular mineral content from the spine during lactation but completely restore it within 18 days after weaning (26). Although aBMD appears to be completely restored after weaning in women and all animals that have been studied, more detailed examination of microarchitecture by µCT has shown variable completeness of recovery of microarchitecture by skeletal site. In rodents, the vertebrae recover completely while persistent loss of trabeculae is evident in the long bones (182). Studies in women have similarly shown that the trabecular content of the long bones also appears to be incompletely restored (1,57,174,183,184). However, in both women (74,184,185) and rodents (26,186,187) the cross-sectional diameters and volumes of the long bones may be significantly increased after post-weaning. Such structural changes potentially compensate for any reduction in strength that loss of trabecular microarchitecture might induce, because an increased cross-sectional diameter increases the ability of a hollow shaft to resist bending (cross-sectional moment of inertia) and torsional stress (polar moment of inertia). This is supported by the finding that the breaking strength of rodent bones returns to pre-pregnant values after weaning (1,180), and limited clinical studies that correlated the increased bone volumes achieved after reproductive cycles with increased bone strength (74,185). In women, the vast majority of several dozen epidemiologic studies of pre- and postmenopausal women have found no adverse effect of a history of lactation on peak bone mass, bone density, or hip fracture risk (1,7,54,57). In fact, multiple studies have suggested a protective effect of lactation on the future risk of low BMD or fragility fractures. Consequently, although lactational bone loss can transiently increase risk of fracture (see next section), it is likely unimportant in the long run for most women, in whom the skeleton is restored to its prior mineral content and strength. DISORDERS OF CALCIUM AND BONE METABOLISM DURING LACTATION Osteoporosis of Lactation On occasion a woman will suffer one or more fragility fractures during lactation, and osteoporotic bone density will be found by DXA (76). As with osteoporosis presenting during pregnancy, this may represent a coincidental, unrelated disease; the woman may have had low bone density and abnormal skeletal microarchitecture prior to pregnancy. Alternatively, it is likely that some cases represent an exacerbation of the normal degree of skeletal demineralization that occurs during lactation, and a continuum from the changes in bone density and bone turnover that occurred during pregnancy. It may be somewhat artificial, therefore, to separate “osteoporosis of lactation” from “osteoporosis of pregnancy.” But since lactation normally causes a significant net loss of bone whereas pregnancy does not, it seems more likely for lactation to cause a subset of women to develop low-trauma fractures. For example, excessive PTHrP release from the lactating breast into the maternal circulation could conceivably cause excessive bone resorption, osteoporosis, and fractures. PTHrP levels were high in one case of lactational osteoporosis, and remained elevated for months after weaning (188). The diagnostic and treatment considerations described above for osteoporosis of pregnancy also apply to women who are lactating (76). Primary Hyperparathyroidism When surgical correctional of primary hyperparathyroidism is not possible or advisable during pregnancy, it is normally carried out in the postpartum interval. A hypercalcemic crisis is possible soon after delivery due in part to loss of the placental calcium infusion, which represented a drain on the serum calcium. If a woman with untreated primary hyperparathyroidism chooses to breastfeed, the serum calcium should be monitored closely for significant worsening due to the effects of secretion of PTHrP from the breasts being added to the high concentrations of PTH already in the circulation. The potential impact of this is even more evident in women with hypoparathyroidism, as discussed below. Familial Hypocalciuric Hypercalcemia The calcium-sensing receptor is expressed in mammary epithelial ducts, and it modulates the production of PTHrP and calcium transport into milk during lactation in mice (189,190). Inactivating calcium-sensing receptor mutations increased mammary tissue production of PTHrP but decreased the calcium content of milk (190). These opposing changes meant that there was a further increase in bone resorption during lactation as compared to normal mice, and the serum calcium also became higher because of reduced output of calcium into milk. Conversely, a calcimimetic drug (similar to cinacalcet) caused increased milk calcium content (190). These data predict that women with FHH will have more marked skeletal resorption during lactation, lower milk calcium content, higher serum calcium, and a greater loss of BMD during lactation as compared to normal women. However, the effect of breastfeeding on mineral and skeletal homeostasis in women with FHH has not yet been described. Hypoparathyroidism As noted earlier, in the first day or two after parturition the requirement for supplemental calcium and calcitriol may transiently increase in hypoparathyroid women before secretion of PTHrP surges in the breast tissue (114). The onset of lactation induces an important change in skeletal metabolism because the breasts produce PTHrP at high levels, some of which escapes into the maternal circulation to stimulate bone resorption and raise the serum calcium level. In women who lack parathyroid glands, the release of PTHrP into the circulation during lactation can temporarily restore calcium and bone homeostasis to normal. Levels of calcitriol and calcium supplementation required for treatment of hypoparathyroid women fall early and markedly after the onset of lactation, and hypercalcemia can occur if the calcitriol dosage and calcium intake are not substantially reduced (191-194). This decreased need for calcium and calcitriol occurs at a time when circulating PTHrP levels are high in the maternal circulation (191,194,195). As illustrated in one case, this is consistent with PTHrP reaching the maternal circulation in amounts sufficient to allow stimulation of calcitriol synthesis, and maintenance of normal (or slightly increased) maternal serum calcium (195). Management of hypoparathyroidism during lactation requires monitoring the albumin-corrected calcium or ionized calcium, reducing or stopping the calcitriol and calcium as indicated, and planning to reinstitute both supplements in escalating doses as lactation wanes. However, production of PTHrP doesn’t necessarily promptly cease around the time of weaning. The author is aware of a woman with hypoparathyroidism who required no supplemental calcium or calcitriol at all for about a year after her baby had been weaned. She thought that her hypoparathyroidism had been permanently cured by breastfeeding, until the abrupt recurrence of symptomatic hypocalcemia, and the need for pre-pregnancy doses of calcium and calcitriol, signaled the end of PTHrP production by her breasts. In another woman, lactation appeared to permanently cure her hypoparathyroidism (196), likely because of persistent production of PTHrP by her breasts. Pseudohypoparathyroidism The management of pseudohypoparathyroidism during lactation has been less well documented. Since these patients are likely resistant to the renal actions of PTHrP, and the placental sources of calcitriol are lost at parturition, the calcitriol requirements might well increase and may require further adjustments during lactation. Conversely, these patients do not have skeletal resistance to PTH, and so it is possible that calcium and calcitriol requirements may decrease secondary to enhanced skeletal resorption caused by the combined effects of high PTH levels, PTHrP release from the breast, and lactation-induced estrogen deficiency. Thus, women with pseudohypoparathyroidism might lose more bone density than normal during lactation, but this has not been studied. Pseudohyperparathyroidism Severe, PTHrP-mediated hypercalcemia during lactation was first noted to occur in women with large breasts, but it has also developed in women with average-sized breasts in whom milk let-down took place but the baby’s illness prevented breastfeeding (120). This represents an exaggeration of normal lactational physiology, which benefits hypoparathyroid women, but in some normal women can overwhelm the normal regulatory pathways and cause potentially severe hypercalcemia. Cessation of lactation should reverse the condition, but a reduction mammoplasty or mastectomy has proved necessary for recalcitrant hypercalcemia. Vitamin D Deficiency and Insufficiency, and Genetic Vitamin D Disorders The available data from small clinical trials, observational studies and case reports indicate that lactation proceeds normally regardless of vitamin D status, and breast milk calcium content is unaffected by vitamin D deficiency or supplementation in doses as high as 6,400 IU per day given to the mother, and achieved maternal 25OHD blood levels of 168 nmol/L (topic reviewed in detail in (1,4,5,7,123)). This is likely because maternal calcium homeostasis is dominated by skeletal resorption induced by estrogen deficiency and PTHrP, with vitamin D/calcitriol playing no substantial role in lactational mineral homeostasis. It is the neonate who will suffer the consequences of being born of a vitamin D deficient mother. This is especially true if the infant is exclusively breast fed, since both vitamin D and 25-hyroxyvitamin D are normally present at very low concentrations in breast milk. The high-dose (6,400 IU) vitamin D supplementation strategy raises the maternal vitamin D concentration substantially for hours and, in turn, this increases the penetration of vitamin D into milk. Consequently, breastfed babies whose mothers consumed 6,400 IU per day achieved the same 25OHD level as babies who received a 300 IU dose of vitamin D directly (197). The potential advantage of this approach is that all of the neonate’s nutrition can then come from breast milk, rather than requiring that breastfed babies receive a vitamin D supplement. Further study is needed regarding the safety of this approach for the mothers and their babies. A misconception about vitamin D and milk often arises because marketed forms of cow’s and goat’s milk contain approximately 100 IU of vitamin D per standard serving, but that is a synthetic vitamin D supplement which is added to the milk after the pasteurization stage. It is not put there by the cow or goat. Given that vitamin D deficiency does not affect breast milk content in humans, it is likely that genetic absence of VDR or calcitriol also does not affect milk calcium, but this has not been studied. Whether vitamin D deficiency impairs the ability of the maternal skeleton to recover post-weaning has not been examined in any clinical study. However, studies in mice lacking the vitamin D receptor or Cyp27b1 to synthesize calcitriol, indicate that these mice are able to fully remineralize their skeletons after lactation (23,45). 24-Hydroxylase Deficiency Hereditary absence of Cyp24a1 reduces calcitriol catabolism, which can lead to very high calcitriol concentrations and marked maternal hypercalcemia during pregnancy. But calcitriol production falls to non-pregnant levels during normal lactation, and the same should be true in women with 24-hydroxylase deficiency. Consistent with this, in one affected woman who breastfed, calcitriol was normal and hypercalcemia was milder compared to pregnancy (145). Low and High Calcium Intakes The calcium content of milk appears to be largely derived from skeletal resorption during lactation, a process that cannot be suppressed in women by consuming greater amounts of calcium (however, it can be suppressed in rodents by high calcium intakes). It shouldn’t be surprising, therefore, that low calcium intake does not impair breast milk quality, nor does it accentuate maternal bone loss (153). Even in women with very low calcium intakes, the same amount of mineral was lost during lactation from the skeleton as compared to women who had supplemented calcium intakes, and the breast milk calcium content was unaffected by calcium intake or vitamin D status (198-200). Conversely, since randomized trials and cohort studies have shown that high calcium intakes do not affect the degree of skeletal demineralization that occurs during lactation (173-176), it is unlikely that increasing calcium supplementation well above normal would affect skeletal demineralization either. There is a lingering concern that adolescent mothers with low calcium intakes may not achieve normal peak bone mass as a consequence of lactation-induced bone loss. In fact the adolescent skeleton appears to recover fully from lactation (201), and adolescent women who breastfed have higher BMD than those who did not breastfeed or had not been pregnant as adolescents (202). However, it remains reasonable to give a calcium supplement to adolescents who lactate in order to ensure that the needs of adolescent growth are met and that peak bone mass is achieved (153,201). IMPLICATIONS During pregnancy and lactation, novel regulatory systems specific to these settings complement the usual regulators of mineral homeostasis. Intestinal calcium absorption more than doubles from early in pregnancy in order to meet the fetal demand for calcium. In comparison, skeletal calcium resorption is a dominant mechanism by which calcium is supplied to the breast milk, while renal calcium conservation is also apparent. Calcium supplementation during pregnancy will result in a woman absorbing more calcium, but it is clear from clinical trials and observational studies that calcium supplements have little or no impact on the amount of bone lost during lactation. The skeleton appears to recover promptly from lactation to achieve the pre-pregnancy bone mass through mechanisms that remain unclear. The transient loss of bone mass during lactation can at least temporarily compromise skeletal strength and lead to fragility fractures in some women. Furthermore, full recovery of mineral content and bone strength may not always be achieved after weaning. But the majority of women can be assured that the changes in calcium and bone metabolism during pregnancy and lactation are normal, healthy, temporary, and without adverse consequences in the long-term. REFERENCES 1. Kovacs CS. Maternal Mineral and Bone Metabolism During Pregnancy, Lactation, and Post-Weaning Recovery. Physiol Rev 2016; 96:449-547 2. 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Kwong WT, Fehmi SM. Hypercalcemic Pancreatitis Triggered by Pregnancy With a CYP24A1 Mutation. Pancreas 2016; 45:e31-32 149. Prentice A. Micronutrients and the bone mineral content of the mother, fetus and newborn. J Nutr 2003; 133:1693S-1699S 150. Prentice A. Pregnancy and lactation. In: Glorieux FH, Petifor JM, Jüppner H, eds. Pediatric bone: biology & diseases. New York: Academic Press; 2003:249-269. 151. Janakiraman V, Ettinger A, Mercado-Garcia A, Hu H, Hernandez-Avila M. Calcium supplements and bone resorption in pregnancy: a randomized crossover trial. Am J Prev Med 2003; 24:260-264 152. Koo WW, Walters JC, Esterlitz J, Levine RJ, Bush AJ, Sibai B. Maternal calcium supplementation and fetal bone mineralization. Obstet Gynecol 1999; 94:577-582 153. Prentice A. Calcium in pregnancy and lactation. Annu Rev Nutr 2000; 20:249-272 154. Hofmeyr GJ, Atallah AN, Duley L. Calcium supplementation during pregnancy for preventing hypertensive disorders and related problems. Cochrane Database Syst Rev 2006; 3:CD001059 155. Kumar A, Devi SG, Batra S, Singh C, Shukla DK. Calcium supplementation for the prevention of pre-eclampsia. Int J Gynaecol Obstet 2009; 104:32-36 156. Hiller JE, Crowther CA, Moore VA, Willson K, Robinson JS. Calcium supplementation in pregnancy and its impact on blood pressure in children and women: follow up of a randomised controlled trial. Aust N Z J Obstet Gynaecol 2007; 47:115-121 157. Villar J, Abdel-Aleem H, Merialdi M, Mathai M, Ali MM, Zavaleta N, Purwar M, Hofmeyr J, Nguyen TN, Campodonico L, Landoulsi S, Carroli G, Lindheimer M. World Health Organization randomized trial of calcium supplementation among low calcium intake pregnant women. Am J Obstet Gynecol 2006; 194:639-649 158. Ohata Y, Yamazaki M, Kawai M, Tsugawa N, Tachikawa K, Koinuma T, Miyagawa K, Kimoto A, Nakayama M, Namba N, Yamamoto H, Okano T, Ozono K, Michigami T. Elevated fibroblast growth factor 23 exerts its effects on placenta and regulates vitamin d metabolism in pregnancy of hyp mice. J Bone Miner Res 2014; 29:1627-1638 159. Jonas AJ, Dominguez B. Low breast milk phosphorus concentration in familial hypophosphatemia. J Pediatr Gastroenterol Nutr 1989; 8:541-543 160. Reade TM, Scriver CR. Hypophosphatemic rickets and breast milk. N Engl J Med 1979; 300:1397 161. Greer FR, Lane J, Ho M. Elevated serum parathyroid hormone, calcitonin, and 1,25-dihydroxyvitamin D in lactating women nursing twins. Am J Clin Nutr 1984; 40:562-568 162. Dobnig H, Kainer F, Stepan V, Winter R, Lipp R, Schaffer M, Kahr A, Nocnik S, Patterer G, Leb G. Elevated parathyroid hormone-related peptide levels after human gestation: relationship to changes in bone and mineral metabolism. J Clin Endocrinol Metab 1995; 80:3699-3707 163. Yamamoto M, Duong LT, Fisher JE, Thiede MA, Caulfield MP, Rosenblatt M. Suckling-mediated increases in urinary phosphate and 3',5'-cyclic adenosine monophosphate excretion in lactating rats: possible systemic effects of parathyroid hormone-related protein. Endocrinology 1991; 129:2614-2622 164. VanHouten JN, Dann P, Stewart AF, Watson CJ, Pollak M, Karaplis AC, Wysolmerski JJ. Mammary-specific deletion of parathyroid hormone-related protein preserves bone mass during lactation. J Clin Invest 2003; 112:1429-1436 165. Kovacs CS, Chik CL. Hyperprolactinemia caused by lactation and pituitary adenomas is associated with altered serum calcium, phosphate, parathyroid hormone (PTH), and PTH-related peptide levels. J Clin Endocrinol Metab 1995; 80:3036-3042 166. Sowers MF, Hollis BW, Shapiro B, Randolph J, Janney CA, Zhang D, Schork A, Crutchfield M, Stanczyk F, Russell-Aulet M. Elevated parathyroid hormone-related peptide associated with lactation and bone density loss. JAMA 1996; 276:549-554 167. Kelly PA, Bachelot A, Kedzia C, Hennighausen L, Ormandy CJ, Kopchick JJ, Binart N. The role of prolactin and growth hormone in mammary gland development. Mol Cell Endocrinol 2002; 197:127-131 168. Dawood MY, Khan-Dawood FS, Wahi RS, Fuchs F. Oxytocin release and plasma anterior pituitary and gonadal hormones in women during lactation. J Clin Endocrinol Metab 1981; 52:678-683 169. Nishimori K, Young LJ, Guo Q, Wang Z, Insel TR, Matzuk MM. Oxytocin is required for nursing but is not essential for parturition or reproductive behavior. Proc Natl Acad Sci U S A 1996; 93:11699-11704 170. VanHouten JN, Wysolmerski JJ. Low estrogen and high parathyroid hormone-related peptide levels contribute to accelerated bone resorption and bone loss in lactating mice. Endocrinology 2003; 144:5521-5529 171. Hernandez LL, Gregerson KA, Horseman ND. Mammary gland serotonin regulates parathyroid hormone-related protein and other bone-related signals. Am J Physiol Endocrinol Metab 2012; 302:E1009-1015 172. Horseman ND, Hernandez LL. New concepts of breast cell communication to bone. Trends Endocrinol Metab 2014; 25:34-41 173. Kolthoff N, Eiken P, Kristensen B, Nielsen SP. Bone mineral changes during pregnancy and lactation: a longitudinal cohort study. Clin Sci (Lond) 1998; 94:405-412 174. Polatti F, Capuzzo E, Viazzo F, Colleoni R, Klersy C. Bone mineral changes during and after lactation. Obstet Gynecol 1999; 94:52-56 175. Kalkwarf HJ, Specker BL, Bianchi DC, Ranz J, Ho M. The effect of calcium supplementation on bone density during lactation and after weaning. N Engl J Med 1997; 337:523-528 176. Cross NA, Hillman LS, Allen SH, Krause GF. Changes in bone mineral density and markers of bone remodeling during lactation and postweaning in women consuming high amounts of calcium. J Bone Miner Res 1995; 10:1312-1320 177. Laskey MA, Prentice A, Hanratty LA, Jarjou LM, Dibba B, Beavan SR, Cole TJ. Bone changes after 3 mo of lactation: influence of calcium intake, breast-milk output, and vitamin D-receptor genotype. Am J Clin Nutr 1998; 67:685-692 178. Teti A, Zallone A. Do osteocytes contribute to bone mineral homeostasis? Osteocytic osteolysis revisited. Bone 2009; 44:11-16 179. Qing H, Ardeshirpour L, Pajevic PD, Dusevich V, Jahn K, Kato S, Wysolmerski J, Bonewald LF. Demonstration of osteocytic perilacunar/canalicular remodeling in mice during lactation. J Bone Miner Res 2012; 27:1018-1029 180. Kirby BJ, Ardeshirpour L, Woodrow JP, Wysolmerski JJ, Sims NA, Karaplis AC, Kovacs CS. Skeletal recovery after weaning does not require PTHrP. J Bone Miner Res 2011; 26:1242-1251 181. Collins JN, Kirby BJ, Woodrow JP, Gagel RF, Rosen CJ, Sims NA, Kovacs CS. Lactating Ctcgrp nulls lose twice the normal bone mineral content due to fewer osteoblasts and more osteoclasts, whereas bone mass is fully restored after weaning in association with up-regulation of Wnt signaling and other novel genes. Endocrinology 2013; 154:1400-1413 182. Liu XS, Ardeshirpour L, VanHouten JN, Shane E, Wysolmerski JJ. Site-specific changes in bone microarchitecture, mineralization, and stiffness during lactation and after weaning in mice. J Bone Miner Res 2012; 27:865-875 183. Bjornerem A, Ghasem-Zadeh A, Wang X, Bui M, Walker SP, Zebaze R, Seeman E. Irreversible Deterioration of Cortical and Trabecular Microstructure Associated With Breastfeeding. J Bone Miner Res 2017; 32:681-687 184. Brembeck P, Lorentzon M, Ohlsson C, Winkvist A, Augustin H. Changes in cortical volumetric bone mineral density and thickness, and trabecular thickness in lactating women postpartum. J Clin Endocrinol Metab 2015; 100:535-543 185. Wiklund PK, Xu L, Wang Q, Mikkola T, Lyytikainen A, Volgyi E, Munukka E, Cheng SM, Alen M, Keinanen-Kiukaanniemi S, Cheng S. Lactation is associated with greater maternal bone size and bone strength later in life. Osteoporos Int 2012; 23:1939-1945 186. Bowman BM, Miller SC. Skeletal adaptations during mammalian reproduction. J Musculoskelet Neuronal Interact 2001; 1:347-355 187. Vajda EG, Bowman BM, Miller SC. Cancellous and cortical bone mechanical properties and tissue dynamics during pregnancy, lactation, and postlactation in the rat. Biol Reprod 2001; 65:689-695 188. Reid IR, Wattie DJ, Evans MC, Budayr AA. Post-pregnancy osteoporosis associated with hypercalcaemia. Clin Endocrinol (Oxf) 1992; 37:298-303 189. VanHouten J, Dann P, McGeoch G, Brown EM, Krapcho K, Neville M, Wysolmerski JJ. The calcium-sensing receptor regulates mammary gland parathyroid hormone-related protein production and calcium transport. J Clin Invest 2004; 113:598-608 190. Ardeshirpour L, Dann P, Pollak M, Wysolmerski J, VanHouten J. The calcium-sensing receptor regulates PTHrP production and calcium transport in the lactating mammary gland. Bone 2006; 38:787-793 191. Caplan RH, Beguin EA. Hypercalcemia in a calcitriol-treated hypoparathyroid woman during lactation. Obstet Gynecol 1990; 76:485-489 192. Salle BL, Berthezene F, Glorieux FH, Delvin EE, Berland M, David L, Varenne JP, Putet G. Hypoparathyroidism during pregnancy: treatment with calcitriol. J Clin Endocrinol Metab 1981; 52:810-813 193. Sadeghi-Nejad A, Wolfsdorf JI, Senior B. Hypoparathyroidism and pregnancy. Treatment with calcitriol. JAMA 1980; 243:254-255 194. Shomali ME, Ross DS. Hypercalcemia in a woman with hypoparathyroidism associated with increased parathyroid hormone-related protein during lactation. Endocr Pract 1999; 5:198-200 195. Mather KJ, Chik CL, Corenblum B. Maintenance of serum calcium by parathyroid hormone-related peptide during lactation in a hypoparathyroid patient. J Clin Endocrinol Metab 1999; 84:424-427 196. Rideout KL. When endocrine disorders disrupt pregnancy: perspectives of affected mothers. In: Kovacs CS, Deal CL, eds. Maternal-Fetal and Neonatal Endocrinology: Physiology, Pathophysiology, and Clinical Management. San Diego: Academic Press; 2019:in press. 197. Wagner CL, Hulsey TC, Fanning D, Ebeling M, Hollis BW. High-dose vitamin D3 supplementation in a cohort of breastfeeding mothers and their infants: a 6-month follow-up pilot study. Breastfeed Med 2006; 1:59-70 198. Prentice A, Jarjou LM, Cole TJ, Stirling DM, Dibba B, Fairweather-Tait S. Calcium requirements of lactating Gambian mothers: effects of a calcium supplement on breast-milk calcium concentration, maternal bone mineral content, and urinary calcium excretion. Am J Clin Nutr 1995; 62:58-67 199. Prentice A, Jarjou LM, Stirling DM, Buffenstein R, Fairweather-Tait S. Biochemical markers of calcium and bone metabolism during 18 months of lactation in Gambian women accustomed to a low calcium intake and in those consuming a calcium supplement. J Clin Endocrinol Metab 1998; 83:1059-1066 200. Prentice A, Yan L, Jarjou LM, Dibba B, Laskey MA, Stirling DM, Fairweather-Tait S. Vitamin D status does not influence the breast-milk calcium concentration of lactating mothers accustomed to a low calcium intake. Acta Paediatr 1997; 86:1006-1008 201. Bezerra FF, Mendonca LM, Lobato EC, O'Brien KO, Donangelo CM. Bone mass is recovered from lactation to postweaning in adolescent mothers with low calcium intakes. Am J Clin Nutr 2004; 80:1322-1326 202. Chantry CJ, Auinger P, Byrd RS. Lactation among adolescent mothers and subsequent bone mineral density. Arch Pediatr Adolesc Med 2004; 158:650-656
10097	https://math.stackexchange.com/questions/2405262/or-minimizing-a-sum-of-a-product	nonlinear optimization - or Minimizing a sum of a product - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more or Minimizing a sum of a product Ask Question Asked 8 years, 1 month ago Modified8 years, 1 month ago Viewed 211 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I am trying to minimize a function and I was hoping that someone can point me in the right direction. I have tried using Lagrangian multipliers and experimenting with simple cases, but nothing has worked. I have also tried using Mathematica to no avail. The function is this: ∑i n∏j n(1−b i j x i)∑i n∏j n(1−b i j x i) with the following constraints: ∑i n x i=1∑i n x i=1 0<x i<1 0<x i<1 If it's easier, I can work with this function instead: ∏i n∏j n(1−b i j x i)∏i n∏j n(1−b i j x i) I would like to find a solution for x i x i in terms of the constants b i j b i j, but like I said, I am having some trouble. Since I already know the b i j b i j, if I can just get it in a form that is easy to solve numerically, that would be great. In practice, n n is roughly 10,000 - so it needs to be somewhat efficient. I recently posted a similar question on Math StackExchange. I don't know best practices, so if you would like me to merge this question with that one, then I will try to do so. Thanks! Update: I could also work with this function: ∑i n∏j n(1−b i j x j)∑i n∏j n(1−b i j x j) Here, I substituted x j x j for x i x i. Would this make things easier? nonlinear-optimization numerical-optimization Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 26, 2017 at 4:30 adn bpsadn bps asked Aug 25, 2017 at 2:11 adn bpsadn bps 179 3 3 bronze badges 4 So, just to make sure, you mention "I would like to find a solution for x i x i in terms of...": are you given b i j b i j or is that another parameter of optimization?Guillermo Angeris –Guillermo Angeris 2017-08-25 02:14:34 +00:00 Commented Aug 25, 2017 at 2:14 Also, are you trying to minimize this or maximize it? If maximizing, then this is convex as stated (in your second case) so there are fast, global solutions, if you're trying to minimize it, then I'd throw it into a general local solver, since the problem is pretty non-convex.Guillermo Angeris –Guillermo Angeris 2017-08-25 02:19:56 +00:00 Commented Aug 25, 2017 at 2:19 You goal function can be rewritten as ∑n i=1 f i(x i)∑i=1 n f i(x i) where f i(⋅)f i(⋅) depends only on x i x i. If you apply Lagrange multipliers method to it, you get something like f′i(x i)=λ f i′(x i)=λ, a single λ λ for all i i. You can numerically compute an approximate inverse function for f′i(x)f i′(x) and then solve for λ λ in the equation ∑n i=1(f′i)−1(λ)=1∑i=1 n(f i′)−1(λ)=1 using any root finder. In this way, you should be able to get a rough idea where the optimal x i x i are. If the approximation solution is still not good enough, you can figure out improvement later achille hui –achille hui 2017-08-25 02:31:05 +00:00 Commented Aug 25, 2017 at 2:31 I am given the b i j b i j. I am trying to minimize this function. Achille thanks for the suggestion. I will definitely look into it.adn bps –adn bps 2017-08-25 16:56:51 +00:00 Commented Aug 25, 2017 at 16:56 Add a comment\| 0 Sorted by: Reset to default You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions nonlinear-optimization numerical-optimization See similar questions with these tags. 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10098	https://math.stackexchange.com/questions/3506457/finding-angles-of-right-triangle-without-inverse-trig	trigonometry - Finding angles of right triangle without inverse trig - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding angles of right triangle without inverse trig Ask Question Asked 5 years, 8 months ago Modified5 years, 8 months ago Viewed 528 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I am working through "Basic Mathematics" by Serge Lang, and there is an example in the polar coordinates section that seems incomplete to me. Example. Find polar coordinates for the point whose rectangular coordinates are (1,3–√)(1,3). Their solution is We have x=1 x=1 and y=3–√y=3, so that r=1+3−−−−√=2 r=1+3=2 Also c o s θ=1 2 c o s θ=1 2 and s i n θ=3–√2 s i n θ=3 2 We see that θ=π/3 θ=π/3 At this point in the book, inverse sine and cosine have not been introduced. Was θ=π/3 θ=π/3 just an observation, or is there a method to determine the angles of a right triangle just using the side measurements without the inverse trig functions? trigonometry Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Jan 12, 2020 at 16:41 IyeekeIyeeke 1,038 7 7 silver badges 16 16 bronze badges 2 Why to use complicated advanced methods when simple observations works ?The Demonix _ Hermit –The Demonix _ Hermit 2020-01-12 16:44:43 +00:00 Commented Jan 12, 2020 at 16:44 There are tables available, where you can read the angle if you know the sin sin, cos cos, or tan tan value of that angle.YNK –YNK 2020-01-12 16:49:14 +00:00 Commented Jan 12, 2020 at 16:49 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. No, there isn't. More simply, θ=π/3 θ=π/3 or θ=π/4 θ=π/4 or θ=π/6 θ=π/6 are values for which sin sin and cos cos are mnemonical remembered. Obiuosvly, all the values of sin sin and cos cos are real so there are expression for every value of θ θ. The expression can be very complicated or even not known. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jan 12, 2020 at 17:39 answered Jan 12, 2020 at 16:46 MatteoMatteo 6,661 1 1 gold badge 14 14 silver badges 32 32 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions trigonometry See similar questions with these tags. 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10099	https://khanacademy.fandom.com/wiki/Area_of_triangles	Khan Academy Wiki Sign In Don't have an account? Register Sign In Skip to content Khan Academy Wiki 2,165 pages in: Math exercises, Eureka Math/EngageNY exercises, 6th grade math exercises, and 7 more 6th grade (U.S.): Geometry 6th grade (Eureka Math/EngageNY): Module 5: Area, surface area, and volume problems Algebra basics exercises Algebra basics: Foundations High school geometry exercises High school geometry: Geometry foundations Mathematics I exercises Area of triangles Sign in to edit History Purge Talk (0) The Area of triangles exercise appears under the 6th grade (U.S.) Math Mission, Algebra basics Math Mission, High school geometry Math Mission and Mathematics I Math Mission. This exercise introduces the area of a triangle. Types of Problems[] There are two types of problems in this exercise: Compare the areas of the triangles: This problem provides two triangles draw on lattice grids. The user is asked to determine which, if either, of the triangles has the largest area. Find the area: This problem provides a triangle that has all of its sides and the height labeled. The user is asked to find the area of the triangle and place the answer in the space provided. Strategies[] Knowledge of the general area of a triangle formula is needed to complete this exercise. The formula for the area of a triangle is , or . Several of the Compare the areas of the triangles are equal. All that is needed to see this is to compare the heights and the bases. Some of the triangles can be compared visually instead of numerically to see which is bigger because the pictures are always drawn to scale. Real-life Applications[] Architects use lots of geometry when building bridges, roofs on houses, and other structures. The ancient Egyptians from over 4000 years ago were very good at shapes and geometry. Every time the Nile burst its banks and flooded the planes, they had to use geometry to measure their gardens and fields all over again. Triangles will appear in the Pythagorean Theorem. Categories Community content is available under CC-BY-SA unless otherwise noted.

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