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| Force One of the foundation concepts of physics, a force may be thought of as any influence which tends to change the motion of an object. Our present understanding is that there are four fundamental forces in the universe, the gravity force, the nuclear weak force, the electromagnetic force, and the nuclear strong force in ascending order of strength. In mechanics, forces are seen as the causes of linear motion, whereas the causes of rotational motion are called torques. The action of forces in causing motion is described by Newton's Laws under ordinary conditions, although there are notable exceptions. Forces are inherently vector quantities, requiring vector addition to combine them. The SI unit for force is the Newton, which is defined by Newton = kg m/s2 as may be seen from Newton's second law. | Index |
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| Causes of Motion The influences which cause changes in the motion of objects are forces and torques. The effects of forces on objects are described by Newton's Laws. A force may be defined as any influence which tends to change the motion of an object. The relationship between force, mass, and acceleration is given by Newton's Second Law: Newton's First Law states that an object will continue at rest or in motion in a straight line at constant velocity unless acted upon by an external force. Newton's Third Law states that all forces in nature occur in pairs of forces which are equal in magnitude and opposite in direction. | IndexNewton's laws concepts |
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Category:Functional Equation Problems
This page lists all of the catalogued problems involving functional equations in the AoPSWiki.
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The following 23 pages are in this category, out of 23 total.
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7302 | http://kirkmcd.princeton.edu/examples/2cylinders.pdf | Cylinder Rolling on Another Rolling Cylinder Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (October 2, 2014; updated January 12, 2018) 1 Problem Discuss the motion of a cylinder that rolls without slipping on another cylinder, when the latter rolls without slipping on a horizontal plane. The cylinders have axial moments of inertia Ii = kimir2 i where mi are the masses and ri are the radii of rolling.1 2 Solution This problem was suggested by Bradley Klee. See also sec. 16.2, p. 233, Vol. 1, of , and prob. 2.9, p. 24 of . For the related case of one cylinder rolling inside another, see .
When one cylinder is directly above the other, we define the line of contact of the lower cylinder, 1, with the horizontal plane to be the z-axis, at x = y = 0. Then, the condition of rolling without slipping for the lower cylinder is that when it has rolled (positive) distance x1, the initial line of contact has rotated through angle φ1 = x1/r1, clockwise with respect to the vertical, as shown in the figure below. This rolling constraint can be written as x1 = r1φ1.
(1) Meanwhile, if the upper cylinder, 2, rolls such that the line of centers (in the x-y plane) makes angle θ (positive clockwise) to the vertical, then the initial point of contact of the upper cylinder has rotated through angle φ2, measured counterclockwise from the line of centers, such that for rolling without slipping the arc lengths are equal between the initial 1One of the two dimensionless positive constants ki can be greater than 1 for a “cylinder” in the form of a bobbin that rolls on a narrow cylinder or track.
1 points of contact of the two cylinders and the new point of contact. This second rolling constraint can be written as, r2φ2 = r1(φ1 −θ) , φ2 −θ = r1 r2 φ1 −r1 + r2 r2 θ = r1φ1 −rθ r2 with r ≡r1 + r2. (2) where φ2 −θ is the angle of the initial point of contact of cylinder 2 to the vertical.
Of course, the center of cylinder 1 is at y1 = r1, and so long as the two cylinders are touching, their axes are separated by distance r = r1 +r2. Altogether there are 4 constraints on the 6 degree of freedom (of two-dimensional motion) of the system, such that there are only two independent degrees of freedom, which we take to be the angles φ1 and θ.
Energy E = T + V is conserved, and since neither the kinetic energy T nor the potential energy V (taken to be zero when θ = θ0), V = −m2gr(cos θ0 −cos θ), (3) depend on coordinate φ1 there will be another conserved quantity, the canonical momentum, pφ1 = ∂L ∂˙ φ1 = ∂T ∂˙ φ1 .
(4) where L = T −V is the Lagrangian of the system. However, pφ1 is not a single angular momentum.2 Since there are two conserved quantities and two degrees of freedom, there is no need to evaluate Lagrange’s equations of motion to determine the motion, so long as the cylinders remain in contact and roll without slipping.3 The kinetic energy of cylinder 1, whose axis is at (x1, r1), is, T1 = m1 ˙ x2 1 2 + I1 ˙ φ1 2 = 1 + k1 2 m1r2 1 ˙ φ 2 1, (5) using the rolling constraint (1) and the expression I1 = k1m1r2 1 for the moment of inertia I1 in terms of parameter k1.
The kinetic energy of cylinder 2, whose axis is at (x2, y2), is, using I2 = k2m2r2 2, T2 = m2( ˙ x2 2 + ˙ y2 2) 2 + I2( ˙ φ2 −˙ θ)2 2 = m2( ˙ x2 2 + ˙ y2 2) 2 + k2m2r2 2( ˙ φ2 −˙ θ)2 2 , (6) noting that the separation of kinetic energy into energy of the center-of-mass motion plus energy of rotation about the center of mass requires the angular velocity to be measured 2An example of a system in which there exists a constant of the motion involving angular velocity and moments of inertia, but which is not a single angular momentum, has been given in . See also .
3For the implausible case of n cylinders, one on top of another, there are 3n degrees of freedom, n constraints of touching, and n rolling constraints, leaving n independent degrees of freedom.
Energy is conserved, and if we take the n independent coordinates to be angle φ1 and the n −1 angles θi,i+1 of the lines of centers of adjacent cylinders, then the energy depends on the θi,i+1 but not φ1. Hence, there is one conserved canonical momentum. For n > 2 it is necessary to use some of Lagrangre’s equations of motion to solve for the motion.
2 with respect to a fixed direction in an inertial frame. Then, recalling eqs. (1)-(2), we have, x2 = x1 + r sin θ, ˙ x2 = r1 ˙ φ1 + r cos θ ˙ θ, (7) y2 = r1 + r cos θ, ˙ y2 = −r sin θ ˙ θ, (8) ˙ φ2 −˙ θ = r1 ˙ φ1 −r˙ θ r2 , (9) and the kinetic energy of cylinder 2 can be written as, T2 = m2 2 [r2 1 ˙ φ 2 1 + 2r1r cos θ ˙ φ1˙ θ + r2˙ θ 2] +k2m2 2 [r2 1 ˙ φ 2 1 −2r1r ˙ φ1˙ θ + r2 ˙ θ 2] = 1 + k2 2 m2r2 1 ˙ φ 2 1 + (cos θ −k2)m2r1r ˙ φ1˙ θ + 1 + k2 2 m2r2˙ θ 2.
(10) The total kinetic energy T1 + T2 is, T = (1 + k1)m1 + (1 + k2)m2 2 r2 1 ˙ φ 2 1 + (cos θ −k2)m2r1r ˙ φ1˙ θ + 1 + k2 2 m2r2 ˙ θ 2, (11) and the conserved canonical momentum is, pφ1 = ∂T ∂˙ φ1 = [(1 + k1)m1 + (1 + k2)m2]r2 1 ˙ φ1 + (cos θ −k2)m2r1r˙ θ = constant.
(12) The total horizontal momentum of the system is, using the rolling constraint (1), Px = (m1 + m2) ˙ x1 + m2r cos θ ˙ θ = (m1 + m2)r1 ˙ φ1 + m2r cos θ ˙ θ, (13) while the angular momentum of the cylinder 1 about its axis is, L1 = k1m1r2 1 ˙ φ1, (14) and that of cylinder 2 about its axis is, using the constraint (2), L2 = k2m2r2 2( ˙ φ2 −˙ θ) = k2m2r2(r1 ˙ φ1 −r˙ θ).
(15) Hence, the conserved canonical momentum (12) can be written as, pφ1 = r1Px + L1 + r1 r2 L2.
(16) Equation (12) for the constant pφ1 can be rewritten as, ˙ φ1 = ω0 − (cos θ −k2)m2r [(1 + k1)m1 + (1 + k2)m2]r1 ˙ θ = ω0 −Ar r1 (cos θ −k2) ˙ θ, (17) ¨ φ1 = −Ar r1 (cos θ −k2) ¨ θ −sin θ ˙ θ 2 , (18) where A = m2 (1 + k1)m1 + (1 + k2)m2 .
(19) 3 Equation (17) integrates to give, for θ0(t = 0) = 0, φ1 = ω0t −Ar r1 (sin θ −k2 θ).
(20) A particular solution is that θ is constant, say θ0 with |θ0| < π/2, while φ = ω0t, in which case φ2 = r1(ω0t −θ0)/r2 according to the rolling constraint (2). Here, the two cylinders roll together, with cylinder 2 at fixed angle θ0, but this motion is unstable.4 For k2 < 1 (as for typical cylinders) and motion that starts with ω0 = 0 and x1,0 = φ1,0 = θ0 = 0, after a small perturbation, the motion leads to angles φ1 and θ with opposite signs until sin θ = k2θ after which the signs are the same (if the cylinders remain in contact).
Similarly, the angular velocities ˙ φ and ˙ θ begin with opposite signs, but the signs become the same when cos θ = k2.5 For a bobbin-like cylinder with k2 > 1, angles φ1 and θ (and angular velocities ˙ φ1 and ˙ θ) always have the same signs. The figure on p. 1 corresponds to k2 > 1, in which the system has positive x-momentum, although it started from rest.
From the rolling constraint (2) we now have (for motion starting from rest), φ2 = r1 r2 (φ1 −θ) = r1 r2 ω0 −Ar r2 (sin θ −k2 θ) −r1 r2 θ.
(21) For k2 < 1, angles φ1 and φ2 have the same signs at small times, both opposite to that of θ.
For k2 > 1 the sign of φ2 can be the same as that of θ, but only for a subset of the possible values for the other parameters of the system.
The constant energy E = T + V can now be expressed as a function only of θ and ˙ θ, with the form, E m2r2 = 0 = [1 + k2 −A(cos θ −k2)2] ˙ θ 2 2 −g r(1 −cos θ), (22) for motion that starts from with θ = 0 = φ1 = φ2, and with ω0 = 0.6 2.1 Time Dependence Thus far, we have obtained analytic expressions for angles φ1 and φ2 in terms of angle θ, and from these analytic expressions for x1, x2 and y2 can also be obtained as a function of θ. However, we do not know the time dependence θ(t), from which the time dependence of all other quantities could be inferred.
By differentiating the energy equation (22), we obtain a second-order time-differential equation for θ, ¨ θ = g/r −A(cos θ −k2)˙ θ 2 1 + k2 −A(cos θ −k2)2 sin θ = 1 + k2 −A(cos θ −k2)(3 −2 cos θ) [1 + k2 −A(cos θ −k2)2]2 g r sin θ.
(23) 4If the upper cylinder is a “supercylinder”, making elastic bounces offthe horizontal surface, during which bounces the point of contact of the cylinder comes to rest, the motion of the upper cylinder is a series of pairs of “hops”, with or without net horizontal motion [6, 7].
5For a much simpler example in which a constrained cylinder begins rotation in one sense and later reverses, see .
6The case that m1 = m2, r1 = r2 = a = r/2 and k1 = k2 = 1/2 is considered in ex. 33, p. 492 of . It follows from eq. (22) that ˙ θ 2 = 12(1 −cos θ)/a(17 + 4 cos θ −4 cos2 θ).
4 For the special case that the upper cylinder is a hollow shell, k2 = 1, the equation of motion for small θ simplifies to, ¨ θ ≈g 2rθ, (k2 = 1, θ ≪1).
(24) which is the (Mathieu) equation for an inverted pendulum (of length l = 2r), for which solutions are tabulated in, for example, .
Numerical methods must be used to deduce t(θ) via either eqs. (23) or (24). Strictly, infi-nite time is required to reach any finite value of θ if the system starts from rest, so discussions of such examples usually consider a small, nonzero initial angle or angular velocity. While θ(t) is a monotonic function for the present example, if the axis of the lower cylinder were subject to a periodic horizontal force in the x- (or y-) direction, the system could exhibit stability at θ = 0, as discussed, for example, in sec. 30 of .
2.2 Constraint Forces The various forces on the two rolling cylinders are illustrated in the figure below. Here, we deduce these forces via Newton’s equations of motion, plus the knowledge of the motion obtained above via a variant of Lagrange’s method.7 2.2.1 Forces at the Horizontal Surface The system of two cylinders, whose center of mass is at, xcm = (m1 + m2)x1 + m2r sin θ m1 + m2 , ycm = (m1 + m2)r1 + m2r cos θ m1 + m2 , (25) 7Lagrange’s method was devised to deduce the equations of motion of a system without consideration of constraint forces that do no work. The method can be extended to include such forces by use of appropriate additional coordinates in the Lagrangian, and representing the effects of constraints in terms with Lagrange multipliers. See, for example, sec. 2.4 of and sec. 19 of , as well as the Appendix.
5 is subject to the external force F1 ˆ x + [N1 −(m1 + m2)g] ˆ y, so the equation of motion of the center of mass are, F1 = (m1 + m2)¨ xcm = (m1 + m2)¨ x1 + m2r cos θ ¨ θ −sin θ ˙ θ 2 = (m1 + m2)r1¨ φ1 + m2r cos θ ¨ θ −sin θ ˙ θ 2 , (26) N1 = (m1 + m2)g + (m1 + m2)¨ ycm = (m1 + m2)g −m2r sin θ ¨ θ + cos θ ˙ θ 2 , (27) using the rolling constraint (1). Then, using eqs. (18), (22) and (23) we obtain F1 and N1 as functions of angle θ.
A single cylinder that rolls without slipping on a horizontal plane has constant horizontal speed, and hence the force of friction is zero at the line of contact between the cylinder and plane.
In the present example the horizontal speeds of the two cylinder are not constant, and the force of friction F1, eq. (26), due to the plane is not zero, such that the x-momentum of the system is not constant (as in the figure above).
2.2.2 Friction between the Cylinders The force of friction, F21 = −F12 on cylinder 2 due to cylinder 1, can be determined from the angular acceleration of cylinder 2, using a torque equation and the rolling constraint (2), F21 = I2 r2 (¨ φ2 −¨ θ) = k2m2r2(¨ φ2 −¨ θ) = k2m2(r1¨ φ1 −r¨ θ).
(28) Then, the friction force F1 at the horizontal surface can also be determined from the angular acceleration of cylinder 1, using the torque equation, (F1 + F12)r1 = (F1 −F21)r1 = −I1¨ φ1 = −k1m1r2 1¨ φ1, (29) such that, F1 = −(k1m1 + k2m2)r1¨ φ1 + k2m2r¨ θ.
(30) This is consistent with eq. (26) in view of the relation (18).
These nonzero frictional forces imply that linear momentum Px and angular momenta L1 and L2 are not conserved in this example, although there is a conserved quantity (16).
2.2.3 Normal Force between the Cylinders The normal force N12 = −N21 of cylinder 2 on cylinder 1 can be determined two ways, by consideration of the x- or y-components of the forces on cylinder 1 (or equivalently, on cylinder 2).
The vertical force components on cylinder 1 sum to zero, which implies that, N12 cos θ = N1 −m1g + F12 sin θ = m2 g −r sin θ ¨ θ + cos θ ˙ θ 2 + k2(r1¨ φ1 −r¨ θ) sin θ (31) 6 using eqs. (27) and (28). Likewise, The horizontal force components on cylinder 1 sum to m1¨ x1, which implies that, N12 sin θ = F1 −m1¨ x1 −F12 cos θ = m2 r1¨ φ1 + r cos θ ¨ θ −sin θ ˙ θ 2 −k2(r1¨ φ1 −r¨ θ) cos θ , (32) using eqs. (26) and (28). Then, N12 = N12 cos2 θ + N12 sin2 θ = g cos θ + r1 sin θ ˙ φ1 −r ˙ θ 2.
(33) When N12 goes to zero, the cylinders separate.
2.3 Angle of Separation The above analysis holds only so long as the two cylinders remain in contact, and the normal force N12 between the cylinders is nonzero, i.e., when, r ˙ θ 2 = g cos θ + r1 sin θ ˙ φ1.
(34) For a method that does not use the forces to find the angle θs at which the cylinders separate, we go to the accelerated frame of the lower cylinder, in which there appears to be an effective acceleration due to “gravity” of, geff= −¨ x1 ˆ x −g ˆ y = −r1¨ φ1 ˆ x −g ˆ y.
(35) Cylinder 2 loses contact with cylinder 1 when the component of geffalong the line of centers, ˆ r = −(sin θ, cos θ), of the cylinders equals the instantaneous radial acceleration, r˙ θ 2. That is, separation occurs at angle θs where,8 r˙ θ 2 s = ˆ r · geff= g cos θs + r1 sin θs ¨ φ1 = g cos θs −rA sin θs (cos θs −k2) ¨ θs −sin θs ˙ θ 2 s , (37) using eq. (18). This confirms eq. (34).
Even for the special case of identical cylinders, m1 = m2, r1 = r2 = r/2 and k1 = k2, the expression (37) remains intricate.
8When the lower cylinder is fixed, geff= g, and eq. (37) reduces to r ˙ θ 2 s = g cos θs. The energy expression (22) simplifies to (1 + k2)r ˙ θ 2/2 = g(1 −cos θ), for motion that starts with θ = 0 = φ1 = φ2. Separation occurs when r ˙ θ 2 s = g cos θs, such that, cos θs = 2 3 + k2 .
(36) In the limit that the upper cylinder is a line/point, k2 →0 and cos θs →2/3, as in the well known “freshman physics” problem of a bug sliding offa log. For a solid cylinder, k2 = 1/2 and cos θs = 4/7, for a solid sphere k2 = 2/5 and cos θs = 10/17, etc.
7 3 Variants Thus far we have assumed that both cylinders roll without slipping. Variants include the three cases in which it is assumed instead that there is no friction at one or both lines of contact, and the cases where either one or two of the coordinates x1, φ1 and φ2 are held fixed with either no friction anywhere or rolling without slipping where rolling is possible. Here, we consider only the first of these examples.
In all cases the potential energy v is given by eq. (3) and the kinetic energy T by a variant of eq. (11). We only consider systems that start from rest with cylinder 2 directly above cylinder 1.
3.1 No Friction Anywhere In the case of no friction anywhere, the cylinders do not rotate.9 Energy is conserved, and the conserved total horizontal momentum is always zero. The system has two degrees of freedom, which we take to be x1, the coordinate of the center of the lower cylinder, and the angle θ of the line of centers between the two cylinders.
The total kinetic energy can be obtained from eq. (11) by setting k1 and k2 to zero, and replacing factors of r1 ˙ φ1 by ˙ x1 (undoing the rolling constraint (1), so to speak), T = m1 + m2 2 ˙ x2 1 + m2r cos θ ˙ x1˙ θ + m2 2 r2˙ θ 2.
(38) The conserved canonical momentum is, for motion starting from rest with φ1 = φ2 = θ = 0, px1 = ∂T ∂˙ x1 = (m1 + m2) ˙ x1 + m2r cos θ ˙ θ = Px = 0, (39) which is just the total horizontal momentum.10 Using eq. (39) to eliminate ˙ x1 from the kinetic energy, we obtain the total energy as, E m2r2 = 0 = m1 + m2 sin2 θ m1 + m2 ˙ θ 2 2 −g r(1 −cos θ).
(40) To find the angle θs at which the cylinders separate, we again go to the accelerated frame of the lower cylinder, in which there appears to be an effective acceleration due to “gravity”, geff= −¨ x1 ˆ x −g ˆ y = m2r m1 + m2 cos θ ¨ θ −sin θ ˙ θ 2 ˆ x −g ˆ y, (41) where ˙ θ and ¨ θ can be deduced in terms of θ from eq. (40).
Cylinder 2 loses contact with cylinder 1 when the component of geffalong the line of centers, ˆ r = −(sin θ, cos θ), of the cylinders equals the instantaneous radial acceleration, r˙ θ 2. That is, separation occurs at angle θs where, r˙ θ 2 s = ˆ r · geff= g cos θs −m2r sin θs m1 + m2 cos θs ¨ θs −sin θs ˙ θ 2 s .
(42) 9The case of two spheres with no friction was discussed on p. 260 of .
10This example includes other conserved/zero generalized momenta such as the z-component of the linear momentum, and the angular momentum about the x- and y-axes.
8 After considerable effort, one can verify that eqs. (40) and (42) combine to give, m2 cos θ2 s = (m1 + m2)(3 cos θs −2), (43) as noted in ex. 6, p. 121 of .
When the lower cylinder is fixed, we can set k2 = 0 and the result (36) again becomes cos θs = 2/3 as for a point mass sliding on a cylinder/sphere.
3.2 No Friction at the Horizontal Plane In the case of no friction at the horizontal plane, but rolling without slipping of cylinder 2 on cylinder 1, both cylinders rotate, being torqued by the friction along the line of contact of the cylinders. Energy is conserved, and the conserved total horizontal momentum is always zero. The system has three degrees of freedom, which we take to be x1, φ1 and θ.
The total kinetic energy can be obtained from eq. (11) by replacing factors of r1 ˙ φ1 not associated with k1 or k2 by ˙ x1 (again undoing the rolling constraint), T = m1 + m2 2 ˙ x2 1 + k1m1 + k2m2 2 r2 1 ˙ φ 2 1 + m2r cos θ ˙ x1˙ θ −k2m2r1r ˙ φ1˙ θ + 1 + k2 2 m2r2 ˙ θ 2. (44) There are now two the conserved canonical momenta, px1 = ∂T ∂˙ x1 = (m1 + m2) ˙ x1 + m2r cos θ ˙ θ = Px = 0, (45) which is the total horizontal momentum (for motion starting from rest with φ1 = φ2 = θ = 0), and, pφ1 = ∂T ∂˙ φ1 = (k1m1 + k2m2)r2 1 ˙ φ1 −k2m2r1r ˙ θ = L1 + r1 r2 L2 = 0.
(46) Using eqs. (45)-(46) to eliminate ˙ x1 and ˙ φ1 from the kinetic energy, we obtain the total energy as,11 E m2r2 = 0 = 1 + k2 −m2 cos2 θ m1 + m2 − k2 2m2 k1m1 + k2m2 ˙ θ 2 2 −g r (1 −cos θ).
(47) To find the angle θs at which the cylinders separate, we again go to the accelerated frame of the lower cylinder, in which there appears to be an effective acceleration due to “gravity”, geff= −¨ x1 ˆ x −g ˆ y = m2r m1 + m2 cos θ ¨ θ −sin θ ˙ θ 2 ˆ x −g ˆ y, (48) where ˙ θ and ¨ θ can be deduced in terms of θ from eq. (47).
11This case is considered in ex. 32, p. 492 of for k1 = k2 = 1/2, where eq. (47) takes the form (3m1 + 2m2 sin2 θ)˙ θ 2 = 4(m1 + m2)g(1 −cos θ)/r.
9 Cylinder 2 loses contact with cylinder 1 when the component of geffalong the line of centers, ˆ r = −(sin θ, cos θ), of the cylinders equals the instantaneous radial acceleration, r˙ θ 2. That is, separation occurs at angle θs where, r˙ θ 2 s = ˆ r · geff= g cos θs − m2r m1 + m2 cos θs ¨ θs −sin θs ˙ θ 2 s sin θs.
(49) This has the same form as eq. (42), but since the energy expressions (40) and (47) are different, the value of θs will be different.12 When the lower cylinder is fixed, we again have cos θs = 2/(3 + k2) as in eq. (36).
3.3 No Friction between the Cylinders In the case of no friction between the cylinders, but cylinder 1 rolls without slipping on the horizontal plane, only cylinder 1 rotates, being torqued by the friction at the horizontal surface. Energy is conserved, but the total horizontal momentum is not. The system has two degrees of freedom, which we take to be φ1 and angle θ, using the rolling constraint (1) to eliminate x1 from the energy, and the rolling constraint (2) to eliminate φ2 in favor of φ1 and θ.
The total kinetic energy can be obtained from eq. (11) by setting k2 to zero, T = (1 + k1)m1 + m2 2 r2 1 ˙ φ 2 1 + m2r1r cos θ ˙ φ1˙ θ + m2 2 r2 ˙ θ 2, (51) and the conserved canonical momentum is, for motion starting from rest with φ1 = φ2 = θ = 0, pφ1 = ∂T ∂˙ φ1 = [(1 + k1)m1 + m2]r2 1 ˙ φ1 + m2r1r cos θ ˙ θ = 0.
(52) Using eq. (52) to eliminate ˙ φ1 from the kinetic energy, we obtain the total energy as, E m2r2 = 0 = 1 − m2 cos2 θ (1 + k1)m1 + m2 ˙ θ 2 2 −g r(1 −cos θ).
(53) To find the angle θs at which the cylinders separate, we again go to the accelerated frame of the lower cylinder, in which there appears to be an effective acceleration due to “gravity”, geff= −¨ x1 ˆ x −g ˆ y = −r1¨ φ1 ˆ x −g ˆ y = m2r (1 + k1)m1 + m2 cos θ ¨ θ −sin θ ˙ θ 2 ˆ x −g ˆ y, (54) 12An amusing special case has been noted in ex. 5, p. 121 of . Suppose the two cylinders are identical, m1 = m2, r1 = r2 = a = r/2, and k1 = k2 = k . Then, eq. (46) becomes ˙ φ1 = ˙ θ, such that φ1 = θ, and then by the rolling constraint (2), φ2 = φ1 −θ = 0. That is, the two cylinder roll together as if they were a single rigid body – until they separate.
Routh also claims that the cylinders separate at angle θ related by (k + 1 + sin2 θ)a ˙ θ 2 = 2g(1 −cos θ), where a = r1 = r2 = r/2, and we note that Routh’s k2 equals our ka2. However, this is just the energy relation (47), which holds for any angle at which the cylinders touch. If I evaluated eq. (49) correctly, 2[(2 + k) cos θs −1] = cos θs(1 −cos θs)2(1 + cos θs).
(50) 10 where ˙ θ and ¨ θ can be deduced in terms of θ from eq. (53).
Cylinder 2 loses contact with cylinder 1 when the component of geffalong the line of centers, ˆ r = −(sin θ, cos θ), of the cylinders equals the instantaneous radial acceleration, r ˙ θ 2. That is, separation occurs at angle θs where, r ˙ θ 2 s = ˆ r · geff= g cos θs − m2r sin θs (1 + k1)m1 + m2 cos θs ¨ θs −sin θs ˙ θ 2 s .
(55) When the lower cylinder is fixed, we again have cos θs = 2/3 as in sec. 3.1.
A Appendix: Constraint Forces via Lagrange Multipliers In general, two rigid bodies, such as the two cylinders of the present example, are to be described by six coordinates per body (say, the spatial coordinates of the center of mass of a body, the two angular directions of some fixed body axis, and the angle of orientation of the body about this axis), for a total of twelve coordinates. In the present example, only two of these twelve coordinates are independent, as there are ten constraints: the axes of the cylinders lie along the z-axis (4 constraints), the centers of mass of the cylinders are at z = 0 (2 constraints), the lower cylinder lies on the plane y = 0 (1 constraint), the two cylinders touch one another (1 constraint), the lower cylinder rolls without slipping on the plane y = 0 (1 constraint), and the upper cylinder rolls without slipping on the lower cylinder (1 constraint).
Furthermore, there is no dissipation of energy in this problem.
Given these constraints/conditions, Lagrange’s method consists of computing the kinetic energy T and the potential energy V in terms of the independent coordinates (taken above to be φ1 and θ). The total energy E(φ1, θ) = T + V is conserved, so the time derivative dE/dt = 0 provides one relation between φ1 and θ.From the Lagrangian L = T −V we can, in principle, deduce the equations of motion via Lagrange’s equations, d dt ∂L ∂˙ qi = ∂L ∂qi , (56) If ∂L/∂qi = 0 (as for qi = φ1 in the present example), then ∂L/∂˙ qi is constant (as in eq. (12) for the present example), and may be called a conserved quantity.13 Thus, it may be (as in the present example) that there are as many conserved quantities as independent coordinates, and Lagrange’s equations (56) are not needed to determine the motion.
In Lagrange’s method, for examples like the present with no dissipation of energy and “simple” constraints on the coordinates, no mention is made of forces. If desired, expressions for various forces can be deduced from Newton’s F = ma with the acceleration a being obtained from Lagrange’s equations (56). A subclass of the forces are those associated with the various constraints on the coordinates of the systems; these are the so-called constraint forces, which do no work (if no energy is dissipated).
13Astonishingly, a paper was published claiming that this in not “well known to instructors and students of physics”.
11 We can also deduce the constraint forces via a method in which more than the minimum number of coordinates are used, as apparently first proposed by Routh [15, 16] for holonomic constraints,14 as a special case of a method for problems with nonholonomic constraints given by Ferrers . See also .
In this method, the minimum number n of independent coordinates is augmented with m additional coordinates, so that the total set of coordinates is qi, i = 1, . . . , n + m, and for which the m constraint equations fj(qi) = 0, j = 1, . . . , m, are known, but not explicity enforced initially. Then, we consider the n + m modified Lagrange equations, d dt ∂L ∂˙ qi −∂L ∂qi = m j=1 λj ∂fj ∂qi , (57) where the λj are so-called Lagrange multipliers (which have the physical significance of being the j constraint force if the dimensions of the constraint equation fj = 0 are chosen appropriately).
In the present example with 12 coordinates, of which only 2 are independent, there are 10 constraint equations. Any number of these can be ignored in an implementation of eq. (57), so there are 210 = 1024 different possible variations of the analysis of the present problem.
Here, we consider the problem to be two dimensional, in which case the first six constraints are automatically satisfied. The remaining four constrains are: 1. That the lower cylinder rolls without slipping on the plane y = 0, eq, (1), f1 = x1 −r1φ1 = 0, (58) 2. That the upper cylinder rolls without slipping on the lower cylinder, eq. (2), f2 = r2φ2 −r1(φ1 −θ) = 0, (59) 3. That the two cylinders touch, f3 = r −r1 −r2 = 0, (60) where r is the distance in between the axes of the two cylinders, 4. That the lower cylinder touches the plane y = 0, f4 = y1 −r1 = 0.
(61) That is, we consider as many as six coordinates, x1, y1, φ1, φ2, θ and r, rather than the minimal set φ1, θ used in the main body of this note.
We now consider the 15 analyses based on temporarily relaxing various subsets of the constraints f1, f2, f3 and f4.
14The term “holonomic” was introduced by Hertz on p. 91 of .
12 A.1 Relax the Rolling Constraint on the Lower Cylinder If we imagine that the constraint (58) on the lower cylinder is relaxed, then we need three coordinates, x1, φ1 and θ to describe the system.
Constraints (59)-(61) are still enforced, so the kinetic energy of the lower cylinder is given by the first form of eq. (5), while the kinetic energy of the upper cylinder becomes, T2 = m2 2 ˙ x2 1 + m2 ˙ x1r cos θ ˙ θ + (1 + k2)m2 2 r2 ˙ θ 2 + k2m2r2 1 ˙ φ 2 1 2 −k2m2r1r ˙ φ1˙ θ, (62) and the potential energy is still given by eq. (3).
The Lagrangian L = T1 + T2 −V does not depend on x1 or φ1, so it is useful to identify the canonical momenta, px1 = ∂L ∂˙ x1 = (m1 + m2) ˙ x1 + m2r cos θ ˙ θ = Px, (63) which is the total horizontal momentum, eq. (13), of the system, and, pφ1 = ∂L ∂˙ φ1 = (k1m1 + k2m2)r2 1 ˙ φ1 −k2m2r1r˙ θ = k1m1r2 1 ˙ φ1 + r1[k2m2(r1 ˙ φ1 −k2m2r˙ θ)] = L1 + r1 r2 L2, (64) where L1 and L2 are the angular momenta, eqs. (14)-(15), of the two cylinders about their axes.
The derivatives of the constraint equation (58) are, ∂f1 ∂x1 = 1, ∂f1 ∂φ1 = −r1, ∂f1 ∂θ = 0.
(65) The extended Lagrange method for this case involves a single multiplier λ1 associated with the rolling constraint (58), such that the three Lagrange equations are now, dpx1 dt = λ1 ∂f1 ∂x1 = λ1, (66) dpφ1 dt = λ1 ∂f1 ∂φ1 = −r1λ1, (67) d dt ∂L ∂˙ θ −∂L ∂θ = λ1 ∂f1 ∂θ = 0.
(68) Combining eqs. (66) and (67), we have that, d dt px1 + pφ1 r1 = 0, px1 + pφ1 r1 = Px + L1 r1 + L2 r2 = 0, (69) supposing that the system starts with x1 = φ1 = θ = 0, which is eq. (16) divided by r1.
The force λ1 associated with the constraint f1 that the lower cylinder rolls without slipping on the plane y = 0 is related by, −λ1 = 1 r1 dpφ1 dt = (k1m1 + k2m2)r1¨ φ1 −k2m2r¨ θ, (70) which is the force F1 found in eq. (30).
13 A.2 Relax the Rolling Constraint on the Upper Cylinder If we imagine that the constraint (59) on the upper cylinder is relaxed, then we need three coordinates, φ1, φ2 and θ to describe the system.
Constraints (58) and (60)-(61) are still enforced, so the kinetic energy of the lower cylinder is given by the second form of eq. (5), while the kinetic energy of the upper cylinder is given by, T2 = m2 2 r2 1 ˙ φ 2 1 + 2r1r cos θ ˙ φ1˙ θ + r2˙ θ 2 + k2m2r2 2 2 ˙ φ 2 2 −2 ˙ φ2˙ θ + ˙ θ 2 , (71) and the potential energy is still given by eq. (3).
The Lagrangian L = T1 + T2 −V does not depend on φ1 or φ2, so it is useful to identify the canonical momenta, pφ1 = ∂L ∂˙ φ1 = [(1 + k1)m1 + m2]r2 1 ˙ φ1 + m2r1r cos θ ˙ θ, (72) and, pφ2 = ∂L ∂˙ φ2 = k2m2r2 2( ˙ φ2 −˙ θ).
(73) The derivatives of the constraint equation (59) are, ∂f2 ∂φ1 = −r1, ∂f2 ∂φ2 = r2, ∂f2 ∂θ = r1.
(74) The extended Lagrange method for this case involves a single multiplier λ2 associated with the rolling constraint (59), such that the three Lagrange equations are now, dpφ1 dt = λ2 ∂f2 ∂φ1 = −r1λ2, (75) dpφ2 dt = λ2 ∂f2 ∂φ2 = r2λ2, (76) d dt ∂L ∂˙ θ −∂L ∂θ = λ2 ∂f2 ∂θ = r1λ2.
(77) Combining eqs. (75) and (76), we have that, d dt pφ1 r1 + pφ2 r2 = 0, pφ1 r1 + pφ2 r2 = 0 (78) supposing that the system starts with x1 = φ1 = θ = 0.
After we enforce the rolling constraint (59), this becomes Px + L1/r1 + L2/r2 = 0, as previously noted.
The force λ2 associated with the constraint f2 that the upper cylinder rolls without slipping on the lower cylinder is related by, F2 = λ2 = 1 r2 dpφ2 dt = k2m2r2(¨ φ2 −¨ θ) = k2m2(r1¨ φ1 −r¨ θ), (79) which was previously found as F21 in eq. (28), 14 A.3 Relax the Constraint that the Cylinders Touch If we imagine that the constraint (60) between the cylinders is relaxed, then we need four coordinates, φ1, φ2, θ and r to describe the system.
Constraints (58) and (61) are still enforced, so the kinetic energy of the lower cylinder is given by the second form of eq. (5), while the kinetic energy of the upper cylinder is given by, T2 = m2 2 r2 1 ˙ φ 2 1 + 2r1(r cos θ ˙ θ + ˙ r sin θ) ˙ φ1 + r2 ˙ θ 2 + ˙ r2 + k2m2r2 2 2 ˙ φ 2 2 −2 ˙ φ2˙ θ + ˙ θ 2 , (80) while the potential energy should now be written as V = m2g(r cos θ −r1 −r2) (to be zero when cylinder 2 sits directly on top of cylinder 1).
The Lagrangian L = T1 + T2 −V does not depend on φ1 or φ2, so it is useful to identify the canonical momenta, pφ1 = ∂L ∂˙ φ1 = [(1 + k1)m1 + m2]r2 1 ˙ φ1 + m2r1(r cos θ ˙ θ + ˙ r sin θ) = r1Px + L1 + m2r1 ˙ r sin θ,(81) and pφ2 = ∂L ∂˙ φ2 = k2m2r2 2( ˙ φ2 −˙ θ) = L2.
(82) The derivatives of the constraint equation (60) are, ∂f3 ∂φ1 = 0, ∂f3 ∂φ2 = 0, ∂f3 ∂θ = 0, ∂f3 ∂r = 1.
(83) The extended Lagrange method for this case involves a single multiplier λ3 associated with the touching constraint (60), such that the four Lagrange equations are, dpφ1 dt = λ3 ∂f3 ∂φ1 = 0, (84) dpφ2 dt = λ3 ∂f3 ∂φ2 = 0, (85) d dt ∂L ∂˙ θ −∂L ∂θ = λ3 ∂f3 ∂θ = 0, (86) d dt ∂L ∂˙ r −∂L ∂r = λ3 ∂f3 ∂r = λ3.
(87) The force λ3 associated with the constraint f3 that the upper cylinder touches the lower cylinder is related by, λ3 = d dt ∂L ∂˙ r −∂L ∂r = m2r1 sin θ ¨ φ1 + cos θ ˙ φ1˙ θ −m2 r1 cos θ ˙ θ ˙ φ1 + r ˙ θ 2 + m2g cos θ = m2 r1 sin θ ¨ φ1 −r ˙ θ 2 + g cos θ , (88) 15 on setting ¨ r = 0, as this expression makes physical sense only after constraint (60) is enforced.
A case of particular interest is when this force goes to zero, at the angle θs of separation, which is now related by, r ˙ θ 2 s = g cos θ + r1 sin θs ¨ φ1 (89) = g cos θ −Ar sin θs (cos θs −k2)¨ θs −sin θs ˙ θ 2 s , using eq. (17). This relation was previously found in eq. (37).
A.4 Relax the Constraint that the Cylinder 1 Touches the Plane y = 0 If we imagine that the constraint (61) is relaxed, then we need three coordinates, y1, φ1 and θ to describe the system.
Constraints (58)-(60) are still enforced, so the kinetic and potential energies of the system are given by eqs. (11) and (3) with the additional terms, ΔT = m1 + m2 2 ˙ y2 1 −m2r sin θ ˙ y1˙ θ, ΔV = (m1 + m2)g(y1 −r1).
(90) The Lagrangian L = T1 + T2 −V does not depend on φ1, so it is useful to identify the canonical momentum, pφ1 = ∂L ∂˙ φ1 = [(1 + k1)m1 + (1 + k2)m2]r2 1 ˙ φ1 + (cos θ −k2)m2r1r˙ θ = r1Px + L1 + r1 r2 L2. (91) The derivatives of the constraint equation (61) are, ∂f4 ∂φ1 = 0, ∂f4 ∂θ = 0, ∂f4 ∂y1 = 1.
(92) The extended Lagrange method for this case involves a single multiplier λ4 associated with the touching constraint (61), such that the four Lagrange equations are, dpφ1 dt = λ4 ∂f3 ∂φ1 = 0, (93) d dt ∂L ∂˙ θ −∂L ∂θ = λ4 ∂f4 ∂θ = 0, (94) d dt ∂L ∂˙ y1 −∂L ∂y1 = λ4 ∂f4 ∂r = λ4.
(95) The force λ4 associated with the constraint f4 that the lower cylinder touches the plane y = 0 is related by, F4 = λ4 = d dt ∂L ∂˙ y1 −∂L ∂y1 = (m1 + m2)¨ y1 −m2r sin θ ¨ θ + cos θ ˙ θ 2 + (m1 + m2)g.
(96) This makes physical sense only after the constraint (61) is enforced, such that ¨ y1 = 0, and the constraint force is just the normal force upward on cylinder 1, F4 = N1 = (m1 + m2)g −m2r sin θ ¨ θ + cos θ ˙ θ 2 = (m1 + m2)g + m2¨ y2, (97) which was previously found as N1 in eq. (27).
16 A.5 Relax All Constraints If we imagine that all constraints (58)-(61) are relaxed, then we consider the six coordinates x1, y1, φ1, φ2 θ and r.
The kinetic energy is now, T = m1 + m2 2 ( ˙ x2 1 + ˙ y2 1) + m1k2r2 1 2 ˙ φ 2 1 + k2m2r2 2 2 ˙ φ 2 2 −2 ˙ φ2˙ θ + ˙ θ 2 + m2 2 ˙ r2 + r2˙ θ 2 +m2 ˙ r( ˙ x1 sin θ + ˙ y1 cos θ) + m2r˙ θ( ˙ x1 cos θ −˙ y1 sin θ), (98) and the potential energy is, V = −m2gr(1 −cos θ) + (m1 + m2)g(y1 −r1).
(99) The Lagrangian L = T −V does not depend on coordinates x1, φ1, or φ2, so we identify the canonical momenta, px1 = ∂L ∂˙ x1 = (m1 + m2) ˙ x1 + m2( ˙ r sin θ + r cos θ ˙ θ) = Px, (100) pφ1 = ∂L ∂˙ φ1 = m1k1r2 1 ˙ φ1 = L1, (101) pφ2 = ∂L ∂˙ φ2 = m2k2r2 2( ˙ φ2 −˙ θ) = L2, (102) The extended Lagrange method for this case involves four multipliers λ1-λ4 associated with the four constraints (58)-(61), such that the six Lagrange equations are, dpx1 dt = λ1 ∂f1 ∂x1 + λ2 ∂f2 ∂x1 + λ3 ∂f3 ∂x1 + λ4 ∂f4 ∂x1 = λ1, (103) dpφ1 dt = λ1 ∂f1 ∂φ1 + λ2 ∂f2 ∂φ1 + λ3 ∂f3 ∂φ1 + λ4 ∂f4 ∂φ1 = −r1λ1 −r1λ2, (104) dpφ2 dt = λ1 ∂f1 ∂φ2 + λ2 ∂f2 ∂φ2 + λ3 ∂f3 ∂φ2 + λ4 ∂f4 ∂φ2 = r2λ2, (105) d dt ∂L ∂˙ θ −∂L ∂θ = λ1 ∂f1 ∂θ + λ2 ∂f2 ∂θ + λ3 ∂f3 ∂θ + λ4 ∂f4 ∂θ = r1λ2, (106) d dt ∂L ∂˙ r −∂L ∂r = λ1 ∂f1 ∂r + λ2 ∂f2 ∂r + λ3 ∂f3 ∂r + λ4 ∂f4 ∂r = λ3, (107) d dt ∂L ∂˙ y1 −∂L ∂y1 = λ1 ∂f1 ∂y1 + λ2 ∂f2 ∂y1 + λ3 ∂f3 ∂y1 + λ4 ∂f4 ∂y1 = λ4, (108) using the derivatives (65), (74), (83) and (92). We can combine eqs. (103)-(105) to find, d dt px1 + pφ1 r1 + pφ2 r2 = d dt Px + L1 r1 L2 r2 = 0, Px + L1 r1 + L2 r2 = 0, (109) for a system that starts with φ1 = φ2 = θ = 0. This form is suggestive, but its content is only understandable if one writes it out in detail, as in eq. (12), which integrates to (20).
Then, we have a description of the motion in terms of a single variable, θ.
17 We now enforce the constraints, and evaluate the multipliers.
The force λ1 associated with the constraint f1 that the lower cylinder rolls without slipping on the plane y = 0 is related by eq. (103), λ1 = dpx1 dt = (m1 + m2)¨ x1 + m2 r cos θ ¨ θ −r sin θ ˙ θ 2 , (110) after setting ˙ r = 0, which is the force F1 found in eq. (26).
The force λ2 associated with the constraint f2 that the upper cylinder rolls without slipping on the lower cylinder is related by eq. (105), F2 = λ2 = 1 r2 dpφ2 dt = k2m2r2(¨ φ2 −¨ θ) = k2m2(r1¨ φ1 −r¨ θ), (111) which was previously found as F21 in eq. (28).15 The force λ3 associated with the constraint f3 that the upper cylinder touches the lower cylinder is related by eq. (107), λ3 = d dt ∂L ∂˙ r −∂L ∂r = m2(¨ x1 sin θ + ˙ x1 cos θ ˙ θ) −m2 ˙ x1 cos θ ˙ θ + r ˙ θ 2 + m2g cos θ = m2 ¨ x1 sin θ −r ˙ θ 2 + g cos θ , (112) on setting ¨ r = 0 and ˙ y1 = 0, as this expression makes physical sense only after constraints (60)-(61) are enforced.
The force λ4 associated with the constraint f4 that the lower cylinder touches the plane y = 0 is related by eq. (108), F4 = λ4 = d dt ∂L ∂˙ y1 −∂L ∂y1 = (m1 + m2)¨ y1 −m2r sin θ ¨ θ + cos θ ˙ θ 2 + (m1 + m2)g.
(113) This makes physical sense only after the constraint (61) is enforced, such that ¨ y1 = 0, and the constraint force is just the normal force upward on cylinder 1, F4 = N1 = (m1 + m2)g −m2r sin θ ¨ θ + cos θ ˙ θ 2 = (m1 + m2)g + m2¨ y2, (114) which was previously found as N1 in eq. (27).
We return to the description of the motion, and note that since the Lagrangian does not depend on time, energy is conserved. After enforcing the constraints (58)-(61), and using the integral (20) of the conserved quantity (109), we arrive at the expression (22) for the (conserved) energy as a function of angle θ only. The time derivative of this expression16 (as well as Lagrange’s equations) provides a second-order differential equation for θ, which can in principle be integrated to describe the motion in detail, as discussed in sec. 2.1.
Thus, the method of relaxing constraints and adding Lagrange multipliers eventually recovers the description of the motion that was obtained more directly via the basic method of Lagrange, which utilizes only the minimum number of independent coordinates (2 in this example).
15The forces F1 and F12 could also be determined via eqs. (104) and (106).
16This approach is called the principle of vis viva in sec. 141 of .
18 References A.S. Ramsey, Dynamics, 2 Vols. (Cambridge U. Press, 1929, 1933, 1962), D.F. Lawden, Analytical Mechanics (Allen & Unwin, 1972), K.T. McDonald, Cylinder Rolling inside Another Rolling Cylinder, (Oct. 21, 2014), W.K. Robinson and B.P. Watson, A misuse of angular momentum conservation, Am.
J. Phys. 53, 82 (1985), A. Tort, F.C. Santos and O.M. Ritter, An extra constant of motion for the N-disc problem, Eur. J. Phys. 10, 217 (1989), R.L. Garwin, Kinematics of an Ultraelastic Rough Ball, Am. J. Phys. 37, 88 (1969), F.S. Crawford, Superball and time-reversal invariance, Am. J. Phys. 50, 856 (1982), K.T. McDonald, Motion of a Cylinder Tied to a Slope by a String, (Jan. 12, 2018), L.A. Pars, Introduction to Dynamics (Cambridge U. Press, 1953).
H. Goldstein, C.H. Poole and J. Safko, Classical Mechanics, 3rd ed. (Addison-Wesley, 2002), A.L. Fetter and J.D. Walecka, Theoretical Mechanics of Particles and Continua (McGraw-Hill, 1980), M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions (National Bureau of Standards, 1964), L.D. Landau and E.M. Lifshitz, Mechanics, 3rd ed. (Pergamon, 1976), B.F. Plybon, Conservation Laws for Undergraduates, Am. J. Phys. 39, 1372 (1971), E.J. Routh, The Elementary Part of a Treatise on the Dynamics of a System of Rigid Bodies, 6th ed. (Macmillan, 1897), Arts. 400 and 429, 19 E.J. Routh, The Advanced Part of a Treatise on the Dynamics of a System of Rigid Bodies, 6th ed. (Macmillan, 1905), Art. 47, S.A. Loney, An Elementary Treatise on the Dynamics of a Particle and of Rigid Bodies, 2nd ed. (Cambridge U. Press, 1913), H. Hertz, Die Prinzipien der Mechanik (Barth, Leipzig, 1894), N.M. Ferrers, Extension of Lagrange’s Equations, Quart. J. Pure Appl. Math. 12, 1 (1872), K.T. McDonald, Calculus of Variations, Princeton U. Ph205 Lecture Notes (1980), More about Lagrange’s Equations, 20 |
7303 | https://www.chegg.com/homework-help/questions-and-answers/13-determine-speed-sound-20-c-air-b-helium-c-natural-gas-methane-isentropic-sound-waves-b--q91655369 | Solved 13. Determine the speed of sound at 20 °C in (a) air, | Chegg.com
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13. Determine the speed of sound at 20 °C in (a) air, (b) helium, and (c) natural gas (methane) for a) isentropic sound waves, b) isothermal sound waves. Express your answer in m/s. Use Table 1.8 or similar for constants.
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Question: 13. Determine the speed of sound at 20 °C in (a) air, (b) helium, and (c) natural gas (methane) for a) isentropic sound waves, b) isothermal sound waves. Express your answer in m/s. Use Table 1.8 or similar for constants.
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Determine the speed of sound at 20 °C in (a) air, (b) helium, and (c) natural gas (methane) for a) isentropic sound waves, b) isothermal sound waves. Express your answer in m/s. Use Table 1.8 or similar for constants.
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7304 | https://www.amboss.com/us/knowledge/male-reproductive-organs/ | Male reproductive organs - Knowledge @ AMBOSS
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Male reproductive organs
Last updated: March 28, 2025
Summary
The male reproductive system consists of the penis, testes, epididymis, ejaculatory ducts, prostate, and accessory glands. These organs function together to produce sperm and deliver semen from the testes via ejaculation. For more information on the embryological development of the male reproductive system, see “Development of the reproductive system”.
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Penis
Function
Urination and release of semen
Gross anatomy
Primarily composed of erectile tissue, blood vessels, nerves, and connective tissue
Structure
Root: composed of the bulb of the penis and two crura
Two crura anchor the corpora cavernosa to the pubic arch.
The bulb is a continuation of the corpus spongiosum.
Body (shaft): contains the erectile tissue
Corpus spongiosum
Location: lies ventrally between the corpora cavernosa
Function: keeps the urethra open during erection
Corpora cavernosa (paired)
Location: dorsal and proximal to the corpus spongiosum
Function: enable erection
Glans penis: the distal end of the corpus spongiosum
Attached to the frenulum and covered by foreskin
Runs from the deep surface of the body to the prepuce
Medial fold located ventrally
Corona: margin of the glans penis
Contains the external urethral meatus
Fossa navicularis: opening of the external urethra
Urethral glands (glands of Littre): present along the length of the penile urethra
Muscles
Bulbospongiosus muscle (paired)
Cover the bulb of the penis
Empty the urethra of residual semen/urine and contribute to erection
Ischiocavernosus muscle (paired)
Cover the crura of the penis
Compress the crura to move blood into the body of the penis and maintain erections
Arteries
| Arterial supply to the penis |
| Artery | Course | Supplies |
| Internal pudendal artery | Branch of the internal iliac artery Exits the pelvis through the greater sciatic foramen Enters the perineum through the lesser sciatic foramen | External genitalia |
| Bulbourethral artery | Branch of the internal pudendal artery Travels between superior and inferior fascia of the urogenital diaphragm Exits the inferior fascia Travels to the bulb of the urethra and the posteriorcorpus spongiosum | Bulb of the penis Bulbourethral gland |
| Deep artery of the penis | Branch of the internal pudendal artery Travels between superior and inferior fascia of the urogenital diaphragm Exits the inferior fascia Enters the crus penis Travels to corpus cavernosumpenis | Erectile tissue |
| Dorsal artery of the penis | Branch of the internal pudendal artery Travels between the crus penis and pubic symphysis Exits inferior fascia of the urogenital diaphragm Travels in the suspensory ligament of the penis Runs on the dorsum of the penis to the glans | Glans penis Prepuce |
| Urethral artery | Travels through the corpus spongiosum | Glans of the penis |
| External pudendal artery | Branch of the femoral artery Enters through the saphenous ring | Skin above the pubis, penis, and scrotum |
Veins
| Venous drainage of the penis |
| Vein | Anatomy | Drains into |
| Deep dorsal vein of the penis | Single vein on the midline, in the dorsum of the penis Travels deep to Buck fascia of the penis Superficial to the tunica albuginea | Prostatic and pelvic venous plexuses |
| Superficial dorsal vein of the penis | Travels along the superficial and deep fascia | External pudendal veins → greater saphenous vein |
Lymphatics
| Lymphatic drainage of the penis |
| Structure | Lymph nodes | Course |
| Perineum, penis, and scrotum | Superficial inguinal lymph nodes | External iliac nodes →paraaortic nodes |
| Glans penis | Deep inguinal lymph nodes | Internal iliac nodes →common iliac nodes →paraaortic nodes |
| Corpora cavernosa | Internal iliac nodes | Common iliac nodes →paraaortic nodes |
Innervation
| Innervation of the penis |
| Innervation | Structures | Function |
| Parasympathetic | Pelvic splanchnic nerves (S2–S4) | Erection |
| Sympathetic | Hypogastric nerve (T11–L4) | Emission |
| Sensory | Dorsal nerve of the penis (a branch of the pudendal nerve) | Innervates the skin, prepuce, and glans |
Microscopic anatomy
Tunica albuginea: thick fibrous layer of connective tissue surrounding the corpora cavernosa and the corpus spongiosum
Corpora cavernosa: composed of connective tissue and smooth muscle
Corpus spongiosum: composed of connective tissue (primarily elastic fibers) and smooth muscle
Embryology
Genital tubercle →glans penis, corpora cavernosa, corpus spongiosum
Urogenital folds →ventral shaft of the penis
See “Development of the reproductive system”.
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Testes, scrotum, and spermatic cord
The testes are paired organs composed of seminiferous tubules, the site of spermatogenesis. They are also responsible for the secretion of male sex hormones.
The scrotum encases the testes and is connected to the abdominal wall via the spermatic cord.
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Gross anatomy
Structure
Testes
Ellipsoid-shaped
Composed of ∼ 600 seminiferous tubules, which connect to the rete testis
Surrounded by a thick capsule (tunica albuginea)
Connect to the abdominal wall via the spermatic cord
Testicular lobules
Interlobular septa divide the testis into testicular lobules.
Each testicular lobule contains 1–4 seminiferous tubules.
Spermatic cord
Extension of the abdominal wall into the scrotum
Contents of the spermatic cord
Three arteries: testicular artery, ductus deferensartery, cremasteric artery
Three nerves: genital branch of genitofemoral, cremasteric nerve, sympathetic nerve fibers
Three other structures: ductus deferens, pampiniform plexus, lymphatic vessels
Scrotum
Distalsac-like end of the spermatic cord that is continuous with the anterior abdominal wall
Layers of the scrotum (from internal to external)
Tunica albuginea: highly vascular, fibrous layer of connective tissue
Tunica vaginalis: visceral layer; derived from peritoneum
Internal spermatic fascia: derived from the transversalis fascia
Cremasteric fascia and cremaster muscle: originates from the internal oblique muscle
External spermatic fascia: derived from the aponeurosis of the external oblique muscle
Dartos fascia and dartos muscle: contraction causes skinwrinkles → regulates temperature
Scrotal skin: wrinkles →↑ skin thickness →↓ heat loss
“TIE turns into ICE”:
T ransversalis fascia → I nternal spermatic fascia
I nternal oblique →C remasteric muscle and fascia
E xternal oblique →E xternal spermatic fascia
Vasculature, lymphatics, and innervation of the testes and scrotum
| Vasculature, lymphatics, and innervation of the testes and scrotum |
| Testes | Scrotum |
| Arteries | Testicular arteries (branches of the abdominal aorta) Pass through the inguinal canal | Anterior scrotal arteries (a branch of the external pudendal artery) Posterior scrotal arteries (a branch of the internal pudendal artery) |
| Veins | Pampiniform venous plexus drains to testicular vein →renal vein (left) and inferior vena cava (right) High risk of left-sidedvaricocele because of the perpendicular orientation of the left gonadal vein to the left renal vein (see “Varicocele”) | Scrotal veins drain to external pudendal vein |
| Lymphatics | Lumbar lymph nodes drain to paraaortic lymph nodes See “Lymphatic drainage” for more information. | Drain to superficial inguinal nodes |
| Autonomic Innervation | Abdominal aortic plexus T10 →testicular plexus | Sympathetic fibers from the genital branch of the genitofemoral nerve →dartos muscle contraction |
| Motor innervation | None | Genital branch of the genitofemoral nerve → cremasteric muscle contraction See “Cremasteric reflex”. |
| Sensory innervation | Genitofemoral nerve → genital branch (L2) →tunica vaginalis of the testes | Anterior scrotal nerves (genital branch of the genitofemoral nerve and ilioinguinal nerve) Posterior scrotal nerves (perineal branch of the internal pudendal nerve and posterior femoral cutaneous nerve) |
The L eft G onadal vein is the L on G est.
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Microscopic anatomy
Microscopic anatomy
Spermatogonia
Description: undifferentiated germ cells lining the seminiferous tubules
Type A (dark): do not undergo mitosis
Type A (pale): form type Bspermatogonia
Type B: undergo mitosis → mature into primary spermatocytes
Function: differentiate into primary spermatocytes
Sertoli cells
Description
Columnar epithelial cells lining seminiferous tubules (non-germ cells)
Homologous to granulosa cells in females
Connected to one another viatight junctions
Form the blood-testis barrier: separation of gametes in seminiferous tubules from the immune system
Sensitive to temperature: ↑ temperature (e.g., due to cryptorchidism or varicocele) →↓ inhibin B secretion and ↓ production of spermatozoa
Function
Produce inhibin B
Stimulated by FSH
Increased inhibin B levels lead to decreased FSH secretion (negative feedback loop).
Contain aromatase, which is responsible for the conversion of testosterone and androstenedione to estrogen
Secreteandrogen-binding protein
Produce Mullerian inhibitory factor
Regulate spermatogenesis
Provide protection and nourishment for developing spermatocytes
The S-rule for S ertoli cells: line the S eminiferous tubes, S ustain S perm S ynthesis, and S uppress F S H.
Leydig cells
Description
Interstitialendocrine cells with cytoplasm rich in cholesterol (used for testosterone production)
Homologous to theca interna cells in females
Not affected by temperature
Function
Produce testosterone
Stimulated byLH
Increased serum testosterone levels lead to decreased LH secretion (negative feedback loop).
Testosterone also diffuses to Sertoli cells, where it stimulates spermatogenesis.
Secrete androgens during the embryonic period
During puberty, testosterone production increases, which allows for the start/maintenance of full spermatogenesis.
Conversion of testosterone to estrogen via aromatase
“Leydis L ove Testosterone”: Leydig cells are L H-stimulated and secreteTestosterone.
Sperm
Head: formed by the acrosome and the nucleus
Tail: formed by the midpiece (mitochondria and microtubules), principal piece (flagellum), and the end piece
Blood-testis barrier
Formed by tight junctions between Sertoli cells
Separates gametes in seminiferous tubules from the immune system
Divides seminiferous tubules into:
Basal compartment
Contains spermatogonia and immature primary spermatocytes
In contact with blood and lymph
Adluminal compartment
Contains mature spermatocytes and spermatids
Not in contact with blood and lymph
“Sertoli fence for sperm defense”: Sertoli cells form the blood-testes barrier.
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Spermatogenesis and spermiogenesis
Spermatogenesis and spermiogenesis
Spermatogenesis
The process by which spermatogonia (see “Microscopic anatomy” above) become spermatids
Begins in puberty
Occurs in the seminiferous tubules
As spermatogonia mature, they move from the basal layer of the seminiferous tubule to its lumen
Spermatogonia (2n, diploid) cross the blood-testis barrier → primary spermatocytes (2n, diploid) → secondary spermatocytes (1n, haploid) → spermatids (1n, haploid)
Duration: ∼ 2 months
Spermiogenesis
The process by which spermatids become spermatozoa
Requires testosterone
Formation of the acrosome
Derived from the Golgi apparatus
Contains acid phosphatase, neuraminidase, and hyaluronidase
Loss of cytoplasmic content
Condensation of genetic material
Spermatid elongation
Development of the flagellum (dysfunctional or absent in primary ciliary dyskinesia/Kartagener syndrome)
GONIUM is GOING to be sperm: Spermato gonia becomes a sper matid.
ZOON ZOOMS to the egg: Mature spermato zoon travel to the egg.
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Embryology
Testes: SRYgene on Y chromosome encodes for testis determining factor → primitive sex cords development →testes formation
Testes develop in the retroperitoneum, then descend into the scrotum:
Transabdominal migration of both testes begins during week 7 and is usually complete by 32 weeks'gestation.
Testes descend through the inguinal canal, along with their associated vasculature, nerves, and ductus deferens.
The descending testes are supported by a layer of the peritoneum (tunica vaginalis), which is normally secured to the superior and inferior posterior pole of the testis.
With the assistance of the gubernaculum at the lower pole, the testes are pulled into the scrotum.
The spermatic cord, which holds the vas deferens and testicular vasculature, meets the testis within the scrotum.
Play a role in further sexual development
Mullerian inhibitory factor (Sertoli cells) suppresses the development of the paramesonephric ducts.
Androgens (Leydig cells) stimulate the development of the mesonephric ducts.
Scrotum
Derived from the labioscrotal swelling
Development driven by dihydrotestosterone (DHT)
See “Development of the reproductive system” for more information.
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Epididymis, ductus deferens, and accessory glands
Epididymis
Gross anatomy: long, coiled duct along the posterior aspect of the testis →distal end is continuous with ductus deferens
Function
Storage and maturation of spermatozoa
Propulsion of spermatozoa into the ductus deferens
Embryology: derived from the mesonephric duct (differentiation requires testosterone)
Ductus deferens
Gross anatomy
Long, muscular duct that enters the pelvis at the deep inguinal ring (lateral to inferior epigastric artery)
Crosses the umbilical artery and obturator nerve → passes superior to the ureter → expands at the distal end to form the ampulla → joins the ejaculatory duct
Function: transports spermatozoa from the epididymis to the ejaculatory duct and provides fructose to the spermatozoa
Embryology: derived from the mesonephric duct (differentiation requires testosterone)
Ejaculatory ducts
Gross anatomy: formed by the ductus deferens and seminal vesicleducts → open into the prostatic urethra
Function: propel spermatozoa with seminal fluid into the urethra
Embryology: derived from the mesonephric duct (differentiation requires testosterone)
Accessory glands
Prostate (see below)
Seminal vesicles
Gross anatomy
Paired glands located on the inferoposterior surface of the bladder
Connect with the ductus deferens to form the ejaculatory ducts → empty into the prostatic urethra
Microscopic anatomy
Lobulated glandular tissue
Lined with pseudostratifiedcolumnar epithelium
Contain secretory granules
Function: secrete an alkaline fluid that makes up the greater part of seminal fluid
Contains sorbitol dehydrogenase which converts sorbitol into fructose and thus provides energy for spermatozoa motility
Contains citrate, fructose, prostaglandins, and other proteins
Embryology: derived from the mesonephric duct (differentiation requires testosterone)
Bulbourethral gland (Cowper gland)
Gross anatomy: located in the deep perineal pouch within the urogenital diaphragm
Function: secretes mucus into the bulbar urethra for lubrication during ejaculation
Embryology: : derived from the pelvic portion of the urogenital sinus (differentiation requires DHT)
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Prostate gland
An accessory gland of reproduction located at the base of the bladder and composed primarily of glandular, fibrous, and smooth muscle tissue.
Function
Secretion of:
Components of semen
Proteolytic enzymes (regulated by DHT) to maintain the fluidity of semen
Acid phosphatase, citric acid, prostaglandins, fibrinolysins, lipids, amylase
Prostate-specific antigen (PSA)
Gross anatomy
Location
Posterior to the pubic symphysis, inferior to the bladder, anterior to the rectum, superior to the perineal membrane
The base wraps around the neck of the urinary bladder.
The apex surrounds the proximalurethra (prostatic urethra).
Structure
Capsules
External: continuation of the pelvicfascia (false capsule)
Internal: true capsule
Five anatomic lobes: Lobe location is described relative to the urethra (e.g., anterior =anterior to the urethra).
Anterior (isthmus)
Anterior to the urethra
Does not contain glandular tissue
Posterior
Below the ejaculatory ducts and behind the urethra
Contains glandular tissue
Most common anatomic location of malignant transformation →prostate cancer
Middle (median): between the ejaculatory ducts and the urethra
Latera l (paired; right and left lobe)
On both sides of the urethra
Most common anatomic site of benign prostatic hyperplasia (BPH) → causes obstruction of urinary flow
Vasculature, lymphatics, and innervation of the prostate gland
| Vasculature, lymphatics, and innervation of the prostate gland |
| Arteries | Prostaticarteries (branches of the internal iliac artery) |
| Veins | Prostatic venous plexus → internal iliac vein → internal vertebral plexus (Batson plexus) Located between the true and false capsule Covered by the anteriorprostaticfascia and the endopelvic fascia |
| Lymphatics | Periprostatic subcapsular network →internal iliac nodes |
| Sympathetic innervation | Superior hypogastric plexus: contraction of smooth muscle during ejaculation |
| Parasympathetic innervation | Pelvic splanchnic nerves (sacral levels S2–S4) →pelvic plexus →cavernous nerves (located within the lateralprostaticfascia; also contain sympathetic fibers) Prostatic secretion Pelvic plexus and cavernous nerves are at risk of damage during radical prostatectomy. |
Microscopic anatomy
General
Composed of glandular, fibrous, and smooth muscle tissue
Glandular epithelium contains foamy cytoplasm and abundant secretory granules, RER, and lysosomes
The fibrous capsule forms septae that penetrate the gland and divide it into lobes.
Functional zones
Peripheral zone
Largest area of glandular tissue with some smooth muscle
Palpable on digital rectal examination (DRE)
Most common site of prostate cancer
Central zone (periurethral zone)
Fibromuscular tissue
Surrounds the ejaculatory ducts
Accounts for 20–25% of the prostate's tissue
Site of origin for 1–5% of prostate cancer
Location: from the base of the prostate to the seminal colliculus (verumontanum)
Transitional zone
Smallest area of glandular tissue: tubuloalveolar glands with columnar epithelium
Surrounds the urethra
Most common site of benign prostatic hypertrophy
Accounts for 10% of the prostate's tissue
Site of origin for 20% of prostate cancer
Location: proximal to the seminal colliculus (verumontanum)
Contains two lateral lobes and a median lobe
Urethral crest → on the posteriormidline
Surrounds the distal end of the preprostatic urethra
Prostatic ducts: drain into the prostatic sinus, a groove on the sides of the urethral crest
Ejaculatory ducts: open into the prostatic urethra, on the seminal colliculus
Glands
There are ∼ 30–50 branched tubuloalveolar glands
Main prostatic glands are located peripherally
Periurethral submucosal glands
Embryology
Prostate derives from the urogenital sinus (differentiation depends upon the presence of DHT)
See “Development of the reproductive system.”
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Male sexual response
There are three parts of the male sexual response: erection, emission, and expulsion.
Erection
Description: The penis becomes rigid as its corpora cavernosa fill with blood due to sensory and/or mental stimulation of primarily parasympathetic nerves.
Autonomic control
Parasympathetic: pelvic splanchnic nerves (S2–S4) →cavernous nerves → nitric oxide (NO)release →↑cGMP → cavernous muscle relaxation →vasodilation →erection
Sympathetic: superior hypogastric plexus (T11–L2) → fibers travel along pelvic plexus →norepinephrinerelease →↑ calciuminflux → cavernous muscle contraction →vasoconstriction→ antagonizes erection (detumescence)
Types of erection
Reflex erection (mostly sensory stimulation)
Genital stimulation →pudendal nerveafferents → sacral erection center S2–S4 (parasympathetic nervous system) → erection
Injuries to the cervical or thoracic spinal cord may result in an erection.
Psychogenic erection (mostly mental stimulation)
Cortical stimulation via visual, auditory, and sensory stimuli or fantasy → impact on the thoracolumbar erection center T11–L2 (sympathetic nervous system) → inferior mesenteric plexus and superior hypogastric plexus → some sympatheticneurons act synergistically with sacral parasympathetic signals (S2–S4) to increase blood flow to corpora cavernosa →erection
Sympathetic innervation to the smooth muscle of vas deferens, seminal vesicle, and internal sphincter of the bladder
Nocturnal penile tumescence
Spontaneous erection while sleeping or upon waking up
The cause is not entirely understood.
Nocturnal tumescence monitoring serves as a means of differentiating psychological from physiological erectile dysfunction.
PDE-5 inhibitors (e.g., sildenafil) inhibit hydrolysis of cGMP →↑ NO →vasodilation → prolonged erection
S2, S3, and S4 keep the penis off the floor.
Ejaculation
Emission
During the first phase of ejaculation, sperm and secretions from the reproductive glands are transported through the reproductive ducts (epididymis, vas deferens, and ejaculatory duct) into the internal urethra.
Ejaculate consists of prostatic secretion, seminal vesicle fluid, and spermatozoa
Secretion of fluid from accessory glands promotes sperm survival.
Stimulated primarily by the sympathetic nervous system (hypogastric nerve; T11–L2) → closure of the bladder neck and contraction of reproductive glands (seminal vesicle, bulbourethral, and prostate glands)
Expulsion
During the second phase of ejaculation, the semen is transported through the urethra and expelled from the body.
Stimulated by:
Sympathetic nervous system: hypogastric nerve (T11–L2)
Somatic nervous system: pudendal nerve → rhythmic contraction of the bulbocavernosus muscle → expulsion of semen from the urethra
Remember the ejaculatory pathway of sperm with SEVE n UP: S eminiferous tubules →E pididymis →V as (ductus) deferens →E jaculatory duct →U rethra →P enis
Innervation of male sexual response: P oint, S queeze, and S hoot.
P arasympathetic → Point (erection)
S ympathetic → Squeeze (emission)
S omatic (pudendal nerve) → Shoot (expulsion)
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Clinical significance
Penis
Priapism
Erectile dysfunction
Phimosis
Paraphimosis
Balanitis
Carcinoma of the penis
Penile fracture
Sexually transmitted infections
Peyronie disease
Testes and scrotum
Testicular tumors
Testicular torsion
Hydrocele
Varicocele
Cryptorchidism
Prostate
Benign prostatic hyperplasia
Prostate cancer
Prostatitis
Other
Epididymitis
Anterior urethral injury
Posterior urethral injury
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Antonym) An ADVOCATE for women's rights
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Terms in this set (15)
Tawdry
Synonym) A closet filled with GARISH outfits
Turncoat
Synonym) Denounced as a TRAITOR
Excise
Synonym) DELETE an irrelevant cause
Haggard
Synonym) Looking EXHAUSTED after a long trek
Ravage
Synonym) DEMOLISHED by a series of storms
Waver
Synonym) Seldom FALTERS under pressure
Stance
Synonym) Had an unpopular POINT OF VIEW
Menial
Synonym) Left the cleaning up to the SUBORDINATES
Doctrine
Synonym) The POLICY of equal justice for all
Parry
Synonym) Sought to DEFLECT the force of the assault
Excise
Antonym) INSERT the word crucial
Menial
Antonym) undertaking a GRAND endeavor
Turncoat
Antonym) An ADVOCATE for women's rights
Parry
Antonym) Someone who can ATTRACT a large audience
Haggard
Antonym) A RADIANT expression on her face
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7306 | https://math.stackexchange.com/questions/553960/extended-stars-and-bars-problemwhere-the-upper-limit-of-the-variable-is-bounded | combinatorics - Extended stars-and-bars problem(where the upper limit of the variable is bounded) - Mathematics Stack Exchange
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Extended stars-and-bars problem(where the upper limit of the variable is bounded)
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The problem of counting the solutions (a 1,a 2,…,a n)(a 1,a 2,…,a n) with integer a i≥0 a i≥0 for i∈{1,2,…,n}i∈{1,2,…,n} such that
a 1+a 2+a 3+…+a n=N a 1+a 2+a 3+…+a n=N
can be solved with a stars-and-bars argument.
What is the solution if one adds the constraint that a i≤r i a i≤r i for certain integers r 1,…,r n r 1,…,r n?
e.g. for n=3 n=3, N=6 N=6 and (r 1,r 2,r 3)=(3,3,2)(r 1,r 2,r 3)=(3,3,2), the tuple (a 1,a 2,a 3)=(2,3,1)(a 1,a 2,a 3)=(2,3,1) is a solution, but (2,1,3)(2,1,3) is not a solution because a 3=3>2=r 3 a 3=3>2=r 3.
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edited Feb 28, 2022 at 19:03
TShiong
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asked Nov 6, 2013 at 6:10
Niaz Mohammad KhanNiaz Mohammad Khan
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If you limit each a i≤1 a i≤1, then you don't need stars and bars in the first place, as it just becomes counting combinations of N N (nonzero a i a i) chosen from a set of r r. However if your imposed limit is larger than 1 1, then the problem is not quite as easy. I suggest that you choose a more representative example in the question.Marc van Leeuwen –Marc van Leeuwen 2013-11-06 06:17:35 +00:00 Commented Nov 6, 2013 at 6:17
There are general solution via both iteration & recursion, check answers to this question: stackoverflow.com/questions/76058880Eric –Eric 2023-04-20 17:30:54 +00:00 Commented Apr 20, 2023 at 17:30
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There is as far as I know no closed formula for this general problem, but there is a formula that allows the number of solutions to be computed in a number of operations independent of N N. Consider first the case that all limits are equal r 1=r 2=⋯=r n=r r 1=r 2=⋯=r n=r. Then the number is the coefficient of X N X N in the polynomial (1+X+⋯+X r)n(1+X+⋯+X r)n. By writing this as a rational function of X X
(1+X+⋯+X r)n=(1−X r+1 1−X)n=(1−X r+1)n(1−X)n(1+X+⋯+X r)n=(1−X r+1 1−X)n=(1−X r+1)n(1−X)n
the coeffiecient of X k X k in the numerator is zero unless k k is a multiple q(r+1)q(r+1) of r+1 r+1, in which case it is (−1)q(n q)(−1)q(n q), and the coefficient of X l X l in the inverse of the denominator is (−1)l(−n l)=(l+n−1 l)(−1)l(−n l)=(l+n−1 l), which is zero unless l≥0 l≥0 and otherwise equal to (l+n−1 n−1)(l+n−1 n−1). It remains to sum over all k+l=N k+l=N, which gives
∑q=0 min(n,N/(r+1))(−1)q(n q)(N−q(r+1)+n−1 n−1),∑q=0 min(n,N/(r+1))(−1)q(n q)(N−q(r+1)+n−1 n−1),
where the summation is truncated to ensure that N−q(r+1)≥0 N−q(r+1)≥0 (the condition l≥0 l≥0). Although the summation looks complicated, it has at most n+1 n+1 easily computed terms, for any N N. Just to illustrate, the coefficient for n=5 n=5, r=100 r=100 and N=243 N=243 is easily computed to be 62018665 62018665. An interesting point to remark is that if the summation had not been truncated, then the result would clearly have been a polynomial function of N N of degree<n<n (because binomial coefficients (x k)(x k) are polynomial functions of x x of degree k k). But on one hand that polynomial function gives the exact values of this problem for N≥n(r+1)N≥n(r+1) where no truncation takes place, while on the other hand, given the original problem, those values are all 0 0; so the polynomial function will be identically zero! So an alternative formula for the result is to compute the negative of the truncated terms, which formula becomes after some massaging
∑q=⌈N+n r+1⌉n(−1)n−q(n q)(q(r+1)−1−N n−1),∑q=⌈N+n r+1⌉n(−1)n−q(n q)(q(r+1)−1−N n−1),
which is easier to use for large N N. For instance in the above example this formula gives a single term (78 4)=1426425(78 4)=1426425 for N=426 N=426; it is the same value as obtained for N=74=500−426 N=74=500−426 (from the first formula) which can be understood by the fact that the "remainders" r i−a i r i−a i add up to n r−N n r−N.
In the general case of distinct limits r i r i, the approach is the same, but the formula becomes a bit messy. Instead of a numerator (1−X r+1)n(1−X r+1)n one gets a product P=(1−X r 1+1)…(1−X r n+1)P=(1−X r 1+1)…(1−X r n+1) which in general has more nonzero terms (the number of terms can be up to min(Σ r i+n+1,2 n)min(Σ r i+n+1,2 n)), but which can be computed once and for all. With P=∑i c i X e i P=∑i c i X e i, the formula for the result becomes
∑i c i(N−e i+n−1 n−1),∑i c i(N−e i+n−1 n−1),
which still is a sum of a number of terms independent of N N. But of course computing the polynomial P(1−X)n P(1−X)n beforehand, and then for any N N just looking up the coefficient of X N X N, is another essentially constant-time (in N N) solution.
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edited Mar 25, 2019 at 9:12
answered Nov 6, 2013 at 11:32
Marc van LeeuwenMarc van Leeuwen
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I posted a question a couple of days ago which appears to be essentially the same question as this one - at least I think it is. If they are the same problem, is the upshot of your post that the most effective way to calculate the numbers in the general case is to check the coefficients of the polynomial? I'm having a little trouble understanding what number the last formula in your post calculates.Geoffrey –Geoffrey 2013-11-07 05:57:32 +00:00 Commented Nov 7, 2013 at 5:57
1 The last formula just computes the coefficient of X N X N in (∑i c i X e i)(1−X)−n(∑i c i X e i)(1−X)−n. It is interesting as a computational method only if the left factor P P (which was computed as a product of sparse polynomials) is still sparse. If it is dense, better compute the product polynomial right away, to get the answer for all N N at once.Marc van Leeuwen –Marc van Leeuwen 2013-11-07 06:13:30 +00:00 Commented Nov 7, 2013 at 6:13
Nice answer. An alternative method for the case where all r i=r r i=r is to use inclusion exclusion. Also, a perhaps clearer way to arrive at the "remainder" formula is to just make the change of variables a i=r i−b i a i=r i−b i. Then you simply need to count the number of solutions to b 1+…+b n=(∑r i)−N b 1+…+b n=(∑r i)−N where each b i≤r i b i≤r i.Andrew Szymczak –Andrew Szymczak 2015-02-22 01:31:15 +00:00 Commented Feb 22, 2015 at 1:31
There are general solution via both iteration & recursion, check answers to this question: stackoverflow.com/questions/76058880Eric –Eric 2023-04-20 17:30:40 +00:00 Commented Apr 20, 2023 at 17:30
@Eric: in this question the problem is one of enumerative combinatorics, in other words counting the number of solutions, not generating them by an algorithm. The idea is to be able to count the solutions even when their number is beyond what can be practically enumerated. When I say there is no closed formula (not something easy to define, but in any case summations and products without a priori bound on the number of terms/factors are excluded), I mean just that. My solution is not a closed formula, even though it is reasonably efficient. The linked question is just (related but) different.Marc van Leeuwen –Marc van Leeuwen 2023-04-22 08:26:49 +00:00 Commented Apr 22, 2023 at 8:26
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For future reference, to people who are unfamiliar with generating functions, here is a solution using the principle of inclusion exclusion.
Ignoring the constraint a i≤r i a i≤r i, the number of solutions is (N+n−1 n−1)(N+n−1 n−1), by stars and bars. To incorporate these constraints, we subtract the "bad" solutions where some a i>r i a i>r i. To count solutions where a 1>r 1 a 1>r 1, we instead count solutions to the equation
(a 1−r 1−1)+a 2+a 3+⋯+a n=N−r 1−1(a 1−r 1−1)+a 2+a 3+⋯+a n=N−r 1−1
Now, all summands on the left hand side are nonnegative integers, so the number of solutions is (N−r 1−1+n−1 n−1)(N−r 1−1+n−1 n−1). Therefore we subtract (N−r i−1+n−1 n−1)(N−r i−1+n−1 n−1) for each i=1,2,…,n i=1,2,…,n.
However, solutions with two variable which were too large have now been subtracted twice, so these must be added back in. Solutions where a i>r i a i>r i and a j>r j a j>r j can be counted by subtracting r i+1 r i+1 from a i a i and r j+1 r j+1 from a j a j, leaving a list of integers summing to N−(r i+1)−(r j+1)N−(r i+1)−(r j+1), the number of which is (N−(r i+1)−(r j+1)+n−1 n−1)(N−(r i+1)−(r j+1)+n−1 n−1).
We must then correct for the solutions with three variables which are too large, then four, and so on. This can be handled systematically using the principle of inclusion exclusion. The result is
∑S⊆{1,2,…,n}(−1)|S|(N+n−1−∑i∈S(r i+1)n−1)∑S⊆{1,2,…,n}(−1)|S|(N+n−1−∑i∈S(r i+1)n−1)
Here, we define (m k)=0(m k)=0 whenever m<0 m<0.
For the special case r 1=r 2=⋯=r n=r r 1=r 2=⋯=r n=r where the upper limit is the same for each variable, the result is
∑k=0⌊N/(r+1)⌋(−1)k(n k)(N−k(r+1)+n−1 n−1).∑k=0⌊N/(r+1)⌋(−1)k(n k)(N−k(r+1)+n−1 n−1).
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edited Apr 14, 2019 at 23:35
answered Apr 10, 2019 at 19:59
Mike EarnestMike Earnest
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I transplanted this answer from a posted question that is actually a duplicate. So, the duplicate question can now be deleted. This answer, which essentially covers the same analysis as answers already posted to this question, is more long-winded, and is intended for readers totally new to the topics of Stars and Bars and Inclusion Exclusion.
Added an Addendum-2, to make the answer more generic. Addendum-2 covers the situation where the lower bound on each variable is 1 1, rather than 0 0.
Identify all solutions to x 1+⋯+x k=n x 1+⋯+x k=n subject to the following constraints:
x 1,⋯,x k x 1,⋯,x k are all non-negative integers.
c 1,⋯,c k c 1,⋯,c k are all fixed positive integers.
x i≤c i:i∈{1,2,⋯,k}.x i≤c i:i∈{1,2,⋯,k}.
To the best of my knowledge, there are two general approaches:
Generating functions : which I know nothing about.
Stars and Bars which is also discussed here combined with Inclusion-Exclusion. See also this answer for an explanation of and justification for the Inclusion-Exclusion formula.
First, I will provide the basic Inclusion Exclusion framework. Then, within this framework, I will apply Stars and Bars theory.
Inclusion Exclusion framework:
Let A A denote the set of all solutions to the equation x 1+⋯+x k=n:x 1,⋯,x k x 1+⋯+x k=n:x 1,⋯,x k are all non-negative integers.
For i∈{1,2,⋯,k}i∈{1,2,⋯,k}, let A i A i denote the subset of A, where x i x i is restricted to being >c i>c i. That is, A i A i represents the specific subset where the upper bound on x i x i is violated.
For any finite set S S, let |S||S| denote the number of elements in the set S S.
Then it is desired to enumerate |A|−|A 1∪⋯∪A k|.|A|−|A 1∪⋯∪A k|.
Let T 0 T 0 denote |A||A|.
For j∈{1,2,⋯,k}j∈{1,2,⋯,k} let T j T j denote
∑1≤i 1<i 2<⋯<i j≤k|A i 1∩A i 2∩⋯∩A i j|.∑1≤i 1<i 2<⋯<i j≤k|A i 1∩A i 2∩⋯∩A i j|.
That is, T j T j represents the sum of (k j)(k j) terms.
Then, in accordance with Inclusion - Exclusion theory, the desired enumeration is
∑i=0 k(−1)i T i.∑i=0 k(−1)i T i.
This concludes the description of the Inclusion-Exclusion framework.
All that remains is to provide a systematic algorithm for enumerating
T 0,T 1,⋯,T k T 0,T 1,⋯,T k.
Stars and Bars theory:
(1)
T 0=(n+[k−1]k−1).T 0=(n+[k−1]k−1).
(2)
To enumerate A i A i, you have to enumerate the number of solutions to x 1+⋯+x k=n:x i>c i x 1+⋯+x k=n:x i>c i.
The standard approach is to set y i=x i−(c i+1),y i=x i−(c i+1), with all of the rest of the variables y 1,⋯,y k=x 1,⋯,x k,y 1,⋯,y k=x 1,⋯,x k, respectively.
Then, there is a one to one correspondence between the solutions that you are trying to enumerate and the solutions to
y 1+⋯+y i+⋯+y k=n−(c i+1):y 1+⋯+y i+⋯+y k=n−(c i+1): each variable is restricted to the non-negative integers.
Here, n−(c i+1)<0 n−(c i+1)<0 implies that there are 0 0 solutions.
Otherwise, per (1) above, the number of solutions is (n−[c i+1]+[k−1]k−1).(n−[c i+1]+[k−1]k−1).
As defined, T 1=∑k i=1|A i|T 1=∑i=1 k|A i|, so now, the procedure for enumerating T 1 T 1 is clear.
(3)
Consider how to enumerate T 2 T 2 which denotes
∑1≤i 1<i 2≤k|A i 1∩A i 2|.∑1≤i 1<i 2≤k|A i 1∩A i 2|.
That is, T 2 T 2 denotes the summation of (k 2)(k 2) terms.
I will illustrate how to specifically enumerate |A 1∩A 2||A 1∩A 2|, with the understanding that this same approach should also be used on the other
[(k 2)−1][(k 2)−1] terms.
The approach will be very similar to that used in (2) above.
Let y 1=x 1−(c 1+1).y 1=x 1−(c 1+1).
Let y 2=x 2−(c 2+1).y 2=x 2−(c 2+1).
Let y i=x i:3≤i≤k y i=x i:3≤i≤k.
Then, there is a one to one correspondence between |A 1∩A 2||A 1∩A 2| and the number of non-negative integer solutions to
y 1+y 2+y 3+⋯+y k=n−(c 1+1)−(c 2+1).y 1+y 2+y 3+⋯+y k=n−(c 1+1)−(c 2+1).
Again, if n−(c 1+1)−(c 2+1)<0 n−(c 1+1)−(c 2+1)<0, then the number of solutions =0=0.
Otherwise, again per (1) above, the number of solutions is given by
(n−[c 1+1]−[c 2+1]+[k−1]k−1).(n−[c 1+1]−[c 2+1]+[k−1]k−1).
(4)
Now, consider how to enumerate T j:3≤j≤k T j:3≤j≤k which denotes
∑1≤i 1<i 2<⋯<i j≤k|A i 1∩A i 2∩⋯∩A i j|.∑1≤i 1<i 2<⋯<i j≤k|A i 1∩A i 2∩⋯∩A i j|.
That is, T j T j denotes the summation of (k j)(k j) terms.
I will illustrate how to specifically enumerate |A 1∩A 2∩⋯∩A j||A 1∩A 2∩⋯∩A j|, with the understanding that this same approach should also be used on the other
[(k j)−1][(k j)−1] terms, when j<k j<k.
The approach will be very similar to that used in (3) above.
Let y 1=x 1−(c 1+1).y 1=x 1−(c 1+1).
Let y 2=x 2−(c 2+1).y 2=x 2−(c 2+1).
⋯⋯
Let y j=x j−(c j+1).y j=x j−(c j+1).
Let y i=x i:j<i≤k.y i=x i:j<i≤k.
Then, there is a one to one correspondence between |A 1∩A 2∩⋯∩A j||A 1∩A 2∩⋯∩A j| and the number of non-negative integer solutions to
y 1+y 2+y 3+⋯+y k=n−(c 1+1)−(c 2+1)−⋯−(c j+1).y 1+y 2+y 3+⋯+y k=n−(c 1+1)−(c 2+1)−⋯−(c j+1).
Again, if n−(c 1+1)−(c 2+1)−⋯−(c j+1)<0 n−(c 1+1)−(c 2+1)−⋯−(c j+1)<0, then the number of solutions =0=0.
Otherwise, again per (1) above, the number of solutions is given by
(n−[c 1+1]−[c 2+1]−⋯−(c j+1)+[k−1]k−1).(n−[c 1+1]−[c 2+1]−⋯−(c j+1)+[k−1]k−1).
This completes the Stars and Bars theory.
Addendum
Shortcuts
Suppose, for example, that c 1=c 2=⋯=c k c 1=c 2=⋯=c k.
Then, to enumerate T j T j, all that you have to do is enumerate
|A 1∩A 2⋯∩A j||A 1∩A 2⋯∩A j|.
When j<k j<k, the other [(k j)−1][(k j)−1] terms will be identical.
Therefore, T j T j will enumerate as
(k j)×|A 1∩A 2⋯∩A j|(k j)×|A 1∩A 2⋯∩A j|.
Further, if only some of the variables c 1,c 2,⋯,c k c 1,c 2,⋯,c k are equal to each other, you can employ similar (but necessarily sophisticated) intuition to presume that some of the (k j)(k j) terms will have the same enumeration.
Also, in practice, you typically don't have to manually enumerate each of T 1,T 2,⋯,T k.T 1,T 2,⋯,T k.
The following concept is based on the assumption (which may be false) that
(c 1+1)+(c 2+1)+⋯+(c k+1)>n.(c 1+1)+(c 2+1)+⋯+(c k+1)>n.
Re-order the scalars c 1,⋯c k c 1,⋯c k in ascending order, as
m 1≤m 2≤⋯≤m k.m 1≤m 2≤⋯≤m k.
Find the smallest value of j j such that (m 1+1)+⋯+(m j+1)>n(m 1+1)+⋯+(m j+1)>n.
Then T j,T j+1,⋯,T k T j,T j+1,⋯,T k must all equal 0 0.
Addendum-2
This section is being added to make the answer more generic. This section is not on point re the originally posted problem, because the originally posted problem (presumably) intended that each variable have a lower bound of 0 0.
Suppose that you are asked to identify the number of solutions to the following problem:
x 1+x 2+⋯+x k=n:n∈Z+.x 1+x 2+⋯+x k=n:n∈Z+.
x 1,x 2,⋯,x k∈Z+.x 1,x 2,⋯,x k∈Z+.
For i∈{1,2,⋯,k},x i≤c i:c i∈Z+.i∈{1,2,⋯,k},x i≤c i:c i∈Z+.
Basic Stars and Bars theory assumes that the lower bound of (for example) x 1,x 2,⋯,x k x 1,x 2,⋯,x k is 0 0 rather than 1.1.
The easiest adjustment is to take the preliminary step of converting the problem into the desired form, through the change of variables
y i=x i−1:i∈{1,2,⋯,k}.y i=x i−1:i∈{1,2,⋯,k}.
Then, the revised problem, which will have the same number of solutions, will be:
y 1+x 2+⋯+y k=(n−k).y 1+x 2+⋯+y k=(n−k).
y 1,y 2,⋯,y k∈Z≥0.y 1,y 2,⋯,y k∈Z≥0.
For i∈{1,2,⋯,k},y i≤(c i−1):c i∈Z+.i∈{1,2,⋯,k},y i≤(c i−1):c i∈Z+.
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answered Nov 16, 2023 at 22:21
user2661923user2661923
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7307 | https://www.quora.com/How-do-you-convert-mph-to-feet-per-second-and-vice-versa | Something went wrong. Wait a moment and try again.
Feet per Second
Unit Converter
Conversion (math)
Unit of Measurement
Kilometre per Hour
5
How do you convert mph to feet per second and vice versa?
S. Patrick Maiorca
Lives in Oklahoma (2017–present)
·
Author has 30.5K answers and 70.7M answer views
·
2y
How do I convert feet/second to miles/hour?
To convert feet per second to miles per hour you start by multiplying by 3600 which is the number of seconds in an hour
1.467 feet per second
3600 x 1.4667 = 5280.12 feet per hour
Now divide by 5280 I am dropping the .12 as we're dealing with an approximate number
5280 feet per hour /5280 = 1 mph
Now to go the other way you multiply
1× 5280 to get feet per hour
So 5280 feet per hour Now we divide by 3600 to get
1.46666666667 feet per second
And we round to the desired level of accuracy
You can also type in feet per second to miles per hour in a search
How do I convert feet/second to miles/hour?
To convert feet per second to miles per hour you start by multiplying by 3600 which is the number of seconds in an hour
So
1.467 feet per second
Becomes
3600 x 1.4667 = 5280.12 feet per hour
Now divide by 5280 I am dropping the .12 as we're dealing with an approximate number
5280 feet per hour /5280 = 1 mph
Now to go the other way you multiply
1× 5280 to get feet per hour
So 5280 feet per hour Now we divide by 3600 to get
1.46666666667 feet per second
And we round to the desired level of accuracy
You can also type in feet per second to miles per hour in a search engine and get it automatically
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Related questions
60 mph is how many feet per second?
How do I convert feet per second to miles per hour?
What is 30 mph miles per hour in FPS feet per second?
How many feet per second do you travel at 25 mph?
How fast is 10 feet per second?
Steve Peterson
Registered Professional Engineer (1984–present)
·
Author has 176 answers and 71.3K answer views
·
2y
Originally Answered: How do you convert miles per hour into feet per second or kilometers per hour into meters per second?
·
You do it by units analysis: look at the units you have and the units you want. Handle the numerator units separately from the denominator units. If you multiply by 1 you do not change the value. for example, you know that 60 seconds = 1 minute. If you divide both sides of that equation by 1 minute, you get (60sec/min) = 1. Likewise, 60 min/hr = 1.
for distance, 5280ft/mile = 1 and 1000 m/km = 1. Also, 1mile/5280 ft = 1 and 1 km/1000 m =1
Example-
X miles/hr = ??? Ft/sec
Miles are on top and I want feet on top. If I multiply the X miles by a version of 1 that has feet in the numerator and miles in
You do it by units analysis: look at the units you have and the units you want. Handle the numerator units separately from the denominator units. If you multiply by 1 you do not change the value. for example, you know that 60 seconds = 1 minute. If you divide both sides of that equation by 1 minute, you get (60sec/min) = 1. Likewise, 60 min/hr = 1.
for distance, 5280ft/mile = 1 and 1000 m/km = 1. Also, 1mile/5280 ft = 1 and 1 km/1000 m =1
Example-
X miles/hr = ??? Ft/sec
Miles are on top and I want feet on top. If I multiply the X miles by a version of 1 that has feet in the numerator and miles in the denominator, miles top and bottom cancel, and I am left with feet - just what I want. Similar for getting the hrs to become seconds. Here is the entire calculation:
(X miles/hr)(5280 ft/ mile)(1 hr/60 min) (1 min/ 60 sec) —-> You cancel all the like units and end up with feet/sec. Numerically (X5280) / (6060) ft/sec.
Y km/hr (1000 m/km) (1 hr/60 min) (1 min/ 60 sec) = 1000Y/3600 m/sec
this was a very wordy answer to a fairly simple question - I hope you can see that you can do your own conversion without consulting the internet: just make an expression for 1 that gets you where you need to go.
Shalashaska Adamska
interested and strive to learn in this field
·
5y
Originally Answered: How do I convert miles per hour to meters per second?
·
Alright, easy as pie. We got 1 mile / 1 hour to begin
First we should transform hours to seconds, so there are 3600 seconds in an hour (60 seconds x (60 minutes / 60 seconds)) with the seconds cancelling. As of now our equation is 1 mile / 3600 seconds.
Now we aim for the mile. We will turn the mile into kilometers, then into meters to keep things simple. 0.621 kilometers are in 1 mile, so we now have 0.621 kilometers / 3600 seconds (1 mile x (0.621 kilometers / 1 mile) with the miles cancelling.
Then we transform the kilometers into meters. We know there are 1000 meters in one kilometers, so our
Alright, easy as pie. We got 1 mile / 1 hour to begin
First we should transform hours to seconds, so there are 3600 seconds in an hour (60 seconds x (60 minutes / 60 seconds)) with the seconds cancelling. As of now our equation is 1 mile / 3600 seconds.
Now we aim for the mile. We will turn the mile into kilometers, then into meters to keep things simple. 0.621 kilometers are in 1 mile, so we now have 0.621 kilometers / 3600 seconds (1 mile x (0.621 kilometers / 1 mile) with the miles cancelling.
Then we transform the kilometers into meters. We know there are 1000 meters in one kilometers, so our equation becomes (0.621 kilometers x (1000 meters / 1 kilometers) with the kilometers cancelling, so now we got 621 meters / 3600 seconds.
Now we have to turn the equation from meter per 3600 seconds to meter per second, so once again we write another equation (621 meters / 3600 seconds = x meters / 1 seconds) then we cross multiply and get (621 meters x seconds / 3600 meters x seconds) which equals 0.1725, which amounts to 0.1725 meters / 1 second. If you intend to apply the laws of significant figures then the answer becomes 0.173 meters / 1 seconds.
Cheers!
Ken Price
Former Non-Licenced Operator at Bruce Power (1978–2010)
·
Author has 1.4K answers and 1.2M answer views
·
2y
Originally Answered: How does one convert miles/hour into feet/second or vice versa?
·
How does one convert miles/hour into feet/second or vice versa?
One mile equals 5,280 feet.
One hour equals 60 minutes or 3,600 seconds.
That should be enough to allow you to find your way.
Good Luck!!!
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Related questions
How many miles per hour is 150 feet per second?
How can you convert miles per hour into feet per second without using a conversion chart?
How do I convert MPH to feet per second?
How many feet per second will a car travel at 45 mph?
How do you calculate speed in feet per second?
Loring Chien
Using measurement tools for 50 years
·
Author has 67.9K answers and 250M answer views
·
7y
Originally Answered: How do I convert feet per second to miles per hour?
·
My way depends on a equivalence I know by heart.
That is 60 mph is the same as 88 ft/sec
So if given mph I multiply by 88/60 to get fps.
or if given fps I multiply by 60/88 to get mph.
To make sure you did the right direction, the quick check is that fps should be a little larger than the equivalent mph.
Anil Bapat
Lives in Mumbai, Maharashtra, India
·
Author has 2.8K answers and 3.8M answer views
·
6y
Originally Answered: How do I convert miles/h to feet/sec?
·
How do I convert miles/h to feet/sec?
There’s a formula to convert miles / hour into feet / second. The formula is to multiply miles per hour quantity by the factor 22/15.
So, if some vehicle is moving at 60 Miles / Hour it’s basically going at the rate of 88 Feet / Second.
How do you get the formla? There’s no magic.
1 Mile = 5280 Feet and 1 Hour is 3600 Seconds. We have Miles in the numerator and Hour(s) in the denominator. So simply divide 5280 by 3600. This gives you fraction - 5280 / 3600, which is nothing but 22 / 15 after cancelling 240 from both the terms.
Thus, the conversion formula is to
How do I convert miles/h to feet/sec?
There’s a formula to convert miles / hour into feet / second. The formula is to multiply miles per hour quantity by the factor 22/15.
So, if some vehicle is moving at 60 Miles / Hour it’s basically going at the rate of 88 Feet / Second.
How do you get the formla? There’s no magic.
1 Mile = 5280 Feet and 1 Hour is 3600 Seconds. We have Miles in the numerator and Hour(s) in the denominator. So simply divide 5280 by 3600. This gives you fraction - 5280 / 3600, which is nothing but 22 / 15 after cancelling 240 from both the terms.
Thus, the conversion formula is to multiply Miles / Hour by 22/15 to obtain Feet / Second.
For different types of conversion, various multiplication factors could be determined, by following first principle as described above.
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Don Flint
Former Retired General Contractor
·
Author has 181 answers and 410.7K answer views
·
6y
Originally Answered: How do I convert miles/h to feet/sec?
·
Miles per hour (MPH) and feet per second (FPS) are both representative of the amount of distance you would travel in a given amount of time. When you are driving with your speedometer indicating 60 MPH you would cover 60 miles in one hour. Converting miles per hour(MPH) to feet to second (FPS) is done by first converting the miles to feet and the hour to seconds. This is done by multiplying your MPH by 5280, the number of feet in one mile, in order to have the total number of feet. Next you multiply 60 ,the number of seconds in one minute by 60, the number of minutes in one hour, in order to h
Miles per hour (MPH) and feet per second (FPS) are both representative of the amount of distance you would travel in a given amount of time. When you are driving with your speedometer indicating 60 MPH you would cover 60 miles in one hour. Converting miles per hour(MPH) to feet to second (FPS) is done by first converting the miles to feet and the hour to seconds. This is done by multiplying your MPH by 5280, the number of feet in one mile, in order to have the total number of feet. Next you multiply 60 ,the number of seconds in one minute by 60, the number of minutes in one hour, in order to have the number of seconds in one hour. then you divide the feet by the time in order to get feet per second.
For example 4 MPH converted to FPS would be: 4 x 5280 (21120) divided by (60 x 60) or 3600 which is 5.86667 feet per second or normal walking speed.
Driving down the road at 70 MPH would be: 70 x 5280 (369,600) divided by 3600 which is 102.6667 FPS. So when you try to text while driving and you look at your phone for 3 SECONDS you have traveled over the length of a FOOTBALL field.
Anh Duc Tran
Versatile Engineer
·
Author has 148 answers and 63K answer views
·
2y
Originally Answered: How do you convert km/hr to m/s and vice versa?
·
It's very simple. 1(km/h) = 1000(m)/3600(s) = 1/3.6(m/s), or 3.6km/h = 1m/s.
So to convert from km/h to m/s, we just divide by 3.6, for quick calculation: divide by 4 then add 11% of initial result (not accurate but acceptable tolerance 0.1%); or mutiple by 3.6, quick calculation: multiple by 4 then subtract 10% of initial result to convert from m/s to km/h.
For example:
10km/h -> 10:3.6 = 2.777, 10:4 =2.5, 2.5 + 0.25 + 0.025 = 2.775m/s
12m/s -> 123.6 = 43.2, 124 = 48, 48 - 4.8 = 43.2km/h
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Omar Jetté
Have been driving and repairing for 55+ years
·
Author has 642 answers and 1.7M answer views
·
4y
Originally Answered: How do I convert MPH to feet per second?
·
The simplest and easiest to remember method is (mph, i.e., statute miles/hour) x 88/60 = feet per second. Example: 60 mph x 88/60 = 88 feet/sec. Otherwise, 88/60 gives an awkward repeating decimal (mph x 1.46666…) which would be reasonably accurate for most purposes.
With a pocket calculator one could also do (mph x 22/15 = ft/sec), however, if working “longhand” then long division by 15 is not as handy as dividing (mph x 88)/6 and shifting the decimal point one space left.
Bheema Mudda
Former Rtd Director at Government of India (2002–2007)
·
Author has 4.8K answers and 2.8M answer views
·
4y
Originally Answered: How do I convert feet per second to miles per hour?
·
1 MILE/HOUR = 5280 FT/3600 SEC =1320/900 =330 /225 =66/45 =22/15
EXAMPLE CONVERT 77 fps TO MPH
22/15 FPS=1 MPH FOR 77 FPS = 771 15/22 =715/2 =105/2 =52.5 MPH
RULE TO CONVERT FPS TO MPH : MULTIPLY FPS BY 15/22
RULE TO CONVERT MPH TO FPS:MULTIPLY MPH BY 22/15
CONVERT 80 MPH TO FPS = 80 22/15 =117.33 FPS
CONVERT 60 MPH TO FPS =60 22/15 =422= 88 FPS
Mark Bodenstein
BS in Mathematics, City College of New York (Graduated 1969)
·
Author has 1.5K answers and 3.4M answer views
·
3y
Multiply feet per second by 3600 (60 x 60 = seconds per hour) to get feet per hour. Divide the result by 5280 (feet per mile) to get miles per hour.
If we’re doing the math by hand rather than with a calculator or computer, we can simplify this a bit. Since 3600 and 5280 have 240 as their greatest integer common factor, we can get the same result by multiplying feet per second by 15 (=3600/240) and dividing by 22 (=5280/240).
Bill Spencer
Enjoys mensuration & metrology!
·
Author has 1.4K answers and 7.1M answer views
·
7y
Just convert each unit.
1 mile = 5280 feet (exactly)
0.3048 m = 1 foot (exactly)
1 hour = 3600 seconds (exactly)
X mph becomes
[math]X \frac{mile}{hour} \frac{5280 feet}{1 mile} \frac{0.3048 m}{1 foot} \frac{1 hour}{3600 seconds} = X\frac{52800.3048}{3600} \frac{m}{sec} [/math]
Gregory Kusiak
Lived in Earth (1972–0)
·
2y
How many feet are there in a statute mile? 5000 or so, (I’m too lazy to Google it), and then we divide that by 3600, the number of seconds in an hour (60 minutes times 60 seconds). That works out to somewhere in the range of 1.4 feet per second, allowing for all the rounding and guessing- Close enough to show you how to math the first part of your question.
Feet per second back to mph..well if something can travel at X feet per second, multiply that by 3600 (see above) to see how many feet it can travel in an hour and then divide that by 5000-odd feet per mile.
Related questions
60 mph is how many feet per second?
How do I convert feet per second to miles per hour?
What is 30 mph miles per hour in FPS feet per second?
How many feet per second do you travel at 25 mph?
How fast is 10 feet per second?
How many miles per hour is 150 feet per second?
How can you convert miles per hour into feet per second without using a conversion chart?
How do I convert MPH to feet per second?
How many feet per second will a car travel at 45 mph?
How do you calculate speed in feet per second?
55 feet per second is how many mph?
How do you convert feet per second into meters per second?
How does one convert miles/hour into feet/second or vice versa?
How do you convert mph to meters per second?
How do I convert ft per second to mph?
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2022 AMC 8 Problems/Problem 17
Contents
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1 Problem
2 Solution 1
3 Solution 2 (Solution 1 worded differently)
4 Video Solution 1
5 Video Solution 2
6 Video Solution 3
7 Video Solution 4
8 Video Solution 5
9 Video Solution 6
10 Video Solution 7
11 Video Solution 8
12 Video Solution 9
13 Video Solution 10
14 See Also
Problem
If is an even positive integer, the notation represents the product of all the even integers from to . For example, . What is the units digit of the following sum?
Solution 1
Notice that once the units digit of will be because there will be a factor of Thus, we only need to calculate the units digit of We only care about units digits, so we have which has the same units digit as The answer is
~wamofan
Solution 2 (Solution 1 worded differently)
We can see that after in the sequence, the units digit is always (every value after is a multiple of ). Therefore, our answer is the sum of the units digits of and respectively. This sum is equal to , or
~Irfans123
Video Solution 1
~hsnacademy
Video Solution 2
~Math-X
Video Solution 3
~Education, the Study of Everything
Video Solution 4
Video Solution 5
~Interstigation
Video Solution 6
~STEMbreezy
Video Solution 7
~savannahsolver
Video Solution 8
~Jamesmath
Video Solution 9
Video Solution 10
~Ismail.Maths
See Also
2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 16Followed by
Problem 18
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7309 | https://girlsangle.wordpress.com/2014/07/15/marion-walters-theorem-via-mass-points/ | Girls' Angle
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← Girls’ Angle Bulletin, Volume 7, Number 5
Girls’ Angle Bulletin, Volume 7, Number 6 →
Marion Walter’s Theorem Via Mass Points
Posted on July 15, 2014 by girlsangle
I recently had the good fortune of learning Marion Walter’s theorem from Marion Walter herself:
Marion Walter’s Theorem: In a triangle, draw line segments from each vertex to the trisection points on the opposite side. The six drawn line segments will form the edges of a central hexagon. The ratio of the area of the hexagon to that of the whole triangle is 1/10.
An efficient way to prove Marion Walter’s theorem is to use mass points.
In this post, I’ll give details because the proof is a model example of the mass points technique. If you’re having difficulty learning the technique, I hope this post will help it all come together for you. As always, try to use mass points to prove the theorem yourself, and, only after you have tried, read on. If you’ve never heard of mass points before, please google it or check out Volume 7, Number 3 of the Girls’ Angle Bulletin (in particular, please read pages 12-14 and pages 19-22 of that issue) before reading further, because here, we’re going to show how the mass point technique is used to prove Marion Walter’s theorem and only sketch how the method itself works.
Spoiler Alert! Proof Below!
It all begins with Archimedes’ Law of the Lever, which we will apply over and over.
Two point masses connected by a massless rod will balance at their center of mass. Archimedes tells us that the distances of each mass from the center of mass satisfy the equation .
The center of mass will lie along the line segment connecting the two masses and sit nearer the heavier one. If the masses are equal, then the center of mass will be exactly halfway between. If one mass is twice the other, then the center of mass will be 2/3 of the way from the lighter to the heavier. (In general, the center of mass is the weighted average of the positions of each object weighted by the mass of that object.)
For a proof of this “piecemeal” way to compute the center of mass and for more information about the center of mass and other applications, see, for instance, Section 19.1 of the Feynman Lectures on Physics, Volume 1, or checkout Volume 7, Number 3 of the Girls’ Angle Bulletin, particularly page 20.
In addition to the Law of the Lever, a beautiful property of the center of mass that we will use repeatedly is that the center of mass can be computed piecemeal. That is, we can split the objects into different sets, compute the center of mass of the objects in each set separately and pretend that each set is replaced by a single point mass equal to the total mass of the objects in that set and located at that set’s center of mass. Then we can compute the center of mass of these pretend point masses to learn the location of the center of mass of the original collection of objects.
For point masses, this “piecemeal” fact enables us to find the center of mass of any number of point masses with repeated application of Archimedes’ Law of the Lever. All we have to do is compute 2 masses at a time, as illustrated in the following figures.
Here are 3 point masses.
To find their center of mass, we can proceed 2 masses at a time. If we start with the first 2 masses, their center of mass will be somewhere along the line segment connecting them.
We pretend there’s a point mass at the center of mass of the first 2 masses with mass equal to their sum. The exact location of the center of mass is determined using Archimedes’ Law of the Lever.
To find the center of mass of the original 3 point masses (indicated by the red dot), we find the center of mass of the imagined green point mass that we computed in the previous step and the 3rd mass. The exact location of the red dot can be determined again using Archimedes’ Law of the Lever.
Proof of Marion Walter’s Theorem
Here’s a figure illustrating Marion Walter’s theorem with the 3 medians of the triangle added in.
We’ll focus on the green triangle and compute how much of the tip of that triangle is inside the orange hexagon:
Let be the measure of .
The area of triangle XOY is .
The area of triangle AOB is .
Therefore, the ratio of the area of triangle XOY to the area of triangle AOB is:
.
To compute these two ratios of lengths, we will use the technique of mass points to compute each length as a fraction of the length of the cevian that it lies on.
Back to the larger figure:
Let’s start with the indicated lengths (which correspond to OY and OB in the labeling of the prior diagram). We’ll use mass points to figure out their lengths as a fraction of the median they’re both on. Let’s start with the longer length.
If we knew what masses to place at the endpoints of the median so that their center of mass sat at the intersection of the medians, then we could use Archimedes’ Law of the Lever to compute the ratio of the lengths that the median is split into. But the problem is, we don’t know that ratio. That’s essentially the ratio we’re trying to find.
What we do know is the ratios of key lengths along the sides of the triangle, because the whole problem is set up by trisecting the sides. This suggests using 3 masses, one at each vertex, as shown below:
Can we figure out how to assign masses to the 3 vertices so that the center of mass of all 3 masses will be where the medians intersect?
Notice that if the two leftmost masses are equal, then, by Archimedes’ Law of the Lever, their center of mass will be at one end of the red median, since the median bisects the side. The importance of assigning the 2 leftmost masses so that their center of mass is at one end of the red median is that it guarantees that the center of mass of all 3 masses will lie on the red median (because of the “piecemeal” property for computing the center of mass).
When the unknown mass (indicated by “?”) is zero, the center of mass of all 3 masses will be midway between the two leftmost masses. As the unknown mass increases, the center of mass of all 3 masses will move along the red median toward the unknown mass. What must the unknown mass be so that the center of mass of all 3 masses will be right where the medians intersect?
We know we can compute the center of mass of the 3 masses piecemeal, and we used this fact to see that by making the 2 leftmost masses equal, the center of mass of all 3 masses will be somewhere on the red median. And now comes a key idea: if we cleverly arrange it so that the center of mass of the 2 lower masses is exactly halfway between them, then the center of mass of all 3 masses will also lie along a second median, and, therefore, it will be at the intersection of both medians!
In other words, to make the center of mass of all 3 masses be at the intersection of the medians, we now only have to concentrate on making the center of mass of the lower 2 masses be at their midpoint. Using Archimedes’ Law of the Lever, we know this happens when the unknown mass is also m.
Thus, the center of mass of 3 equal masses placed at the vertices of a triangle is located at the intersection of the triangle’s medians.
Now, we can compute the ratio of the lengths that the red median is split into by the other medians. We go back to computing the center of mass piecemeal by first computing the center of mass of the 2 leftmost masses and imagining a single point mass of mass 2m placed there. We’ve purposefully assigned the masses so that the center of mass of this 2m-mass and the m-mass in the lower right corner will be at the intersection of the medians, hence the ratio of the lengths that the red median is split into, by Archimedes’ Law of the Lever, must be 1 : 2. We deduce that the length we are looking for (indicated by the double-headed arrow) is 1/3 the length of the red median.
Now let’s turn our attention to the other length:
This time, we try to assign masses to the vertices so that their center of mass will be where the red and orange cevians intersect in the figure below.
Since the red cevian is a median, we know to place equal masses at the 2 leftmost vertices, just as before. This time, however, we want to assign a mass to the lower right vertex in such a way that the center of mass of the 2 lower masses will be at the foot of the orange cevian, which is 1/3 of the way from left to right. Using Archimedes’ Law of the Lever, we compute that the lower right mass must be m/2.
We now compute the center of mass piecemeal, starting with the 2 leftmost masses. Their center of mass is at the foot of the red median (by design), so we imagine a point mass of mass 2m placed there. The red median is split into two pieces by the orange cevian. Archimedes’ Law of the Lever tells us that the ratio of the lengths of these two pieces is equal to the ratio m/2 to 2m, which is 1/4. Hence, the center of mass of these 3 point masses is 1/5 of the way along the red median, measured from its foot.
Therefore, the distance we’re interested in (indicated by the double-headed arrow in the figure above) is 1/3 – 1/5 = 2/15 of the length of the red median.
Next, we turn our attention to the other 2 important lengths indicated below:
What fraction of the purple median are they?
The longer length stretches from the vertex to the intersection of the medians, and we already know how to balance the triangle there. That’s achieved with equal point masses at each vertex and we conclude that the longer length is 2/3 of the length of the purple median.
For the shorter length, we try to assign point masses to the vertices so that their center of mass is where the purple median and orange cevian intersect. The figure above shows what the masses need to be to achieve that. Computing this center of mass location piecemeal, starting with the 2 rightmost masses and then applying Archimedes’ Law of the Lever, we find that the orange cevian actually bisects the purple median! Hence, the shorter length is 2/3 – 1/2 = 1/6 of the length of the purple median.
The numbers are the fraction of the lengtheach segment is of the cevian it sits upon.
Putting it all together, we find that:
.
(Note that the numbers are fractions of lengths, not absolute lengths. That is OX is not 1/6 nor is OA equal to 2/3, but since we are taking ratios, OX/OA is equal to (1/6)/(2/3).)
To find the fraction of the triangle’s area that the hexagon occupies, we note that the tips of the 6 green triangular sections that dip into the hexagon are all 1/10 of their respective green triangles. That’s because the computations for each green triangle will look just like the computation we just went through by symmetry.
Therefore, the area of the hexagon, which consists of the 6 tips of the green triangles, is exactly 1/10 the area of the whole, and the theorem is proven.
To show how efficient this technique is, I’ve scanned in my actual scratch work. When you’ve gotten the hang of the mass point technique, you can compute the necessary masses “as you go.”
One set of mass assignments is uncircled and the other set is circled. After the fact, I tried to indicate which parts of the computation correspond to which line segments in the figure.
For more practice with mass points, there is a Summer Fun problem set devoted to the technique by Johnny Tang in the latest issue of the Girls’ Angle Bulletin (Volume 7, Number 5). Also, check out these lecture notes by Tom Rike for a talk he gave at the Berkeley Math Circle. At MathWorld, there’s a slightly different approach to proving Marion Walter’s theorem that also essentially uses mass points. There are many other proofs that use different techniques, such as the one outlined at the end of this handout by James King.
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7310 | https://www.youtube.com/playlist?list=PLQfiOKXnQpw-fDd830qvJolGvFIfz_XNn | Stewart Calculus Chapter 6 || Applications of Integration - YouTube
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7311 | https://openstax.org/books/biology-2e/pages/7-4-oxidative-phosphorylation | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Biology 2e
7.4 Oxidative Phosphorylation
Biology 2e7.4 Oxidative Phosphorylation
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Learning Objectives
By the end of this section, you will be able to do the following:
Describe how electrons move through the electron transport chain and explain what happens to their energy levels during this process
Explain how a proton (H+) gradient is established and maintained by the electron transport chain
You have just read about two pathways in glucose catabolism—glycolysis and the citric acid cycle—that generate ATP. Most of the ATP generated during the aerobic catabolism of glucose, however, is not generated directly from these pathways. Instead, it is derived from a process that begins by moving electrons through a series of electron carriers that undergo redox reactions. This process causes hydrogen ions to accumulate within the intermembranous space. Therefore, a concentration gradient forms in which hydrogen ions diffuse out of the intermembranous space into the mitochondrial matrix by passing through ATP synthase. The current of hydrogen ions powers the catalytic action of ATP synthase, which phosphorylates ADP, producing ATP.
Electron Transport Chain
The electron transport chain (Figure 7.12) is the last component of aerobic respiration and is the only part of glucose metabolism that uses atmospheric oxygen. Oxygen continuously diffuses into plant tissues (typically through stomata), as well as into fungi and bacteria; however, in animals, oxygen enters the body through a variety of respiratory systems. Electron transport is a series of redox reactions that resembles a relay race or bucket brigade in that electrons are passed rapidly from one component to the next, to the endpoint of the chain where the electrons reduce molecular oxygen and, along with associated protons, produces water. There are four complexes composed of proteins, labeled I through IV in Figure 7.12, and the aggregation of these four complexes, together with associated mobile, accessory electron carriers, is called the electron transport chain. The electron transport chain is present with multiple copies in the inner mitochondrial membrane of eukaryotes and within the plasma membrane of prokaryotes.
Figure 7.12 The electron transport chain is a series of electron transporters embedded in the inner mitochondrial membrane that shuttles electrons from NADH and FADH2 to molecular oxygen. In the process, protons are pumped from the mitochondrial matrix to the intermembrane space, and oxygen is reduced to form water.
Complex I
First, two electrons are carried to the first complex via NADH. This complex, labeled I, is composed of flavin mononucleotide (FMN) and an iron-sulfur (Fe-S)-containing protein. FMN, which is derived from vitamin B2 (also called riboflavin), is one of several prosthetic groups or cofactors in the electron transport chain. A prosthetic group is a nonprotein molecule required for the activity of a protein. Prosthetic groups are organic or inorganic, nonpeptide molecules bound to a protein that facilitate its function. Prosthetic groups include coenzymes, which are the prosthetic groups of enzymes. The enzyme in complex I is NADH dehydrogenase and is composed of 44 separate polypeptide chains. Complex I can pump four hydrogen ions across the membrane from the matrix into the intermembrane space, and it is in this way that the hydrogen ion gradient is established and maintained between the two compartments separated by the inner mitochondrial membrane.
Q and Complex II
Complex II directly receives FADH2—which does not pass through complex I. The compound connecting the first and second complexes to the third is ubiquinone B. The Q molecule is lipid soluble and freely moves through the hydrophobic core of the membrane. Once it is reduced (QH2), ubiquinone delivers its electrons to the next complex in the electron transport chain. Q receives the electrons derived from NADH from complex I, and the electrons derived from FADH2 from complex II. This enzyme and FADH2 form a small complex that delivers electrons directly to the electron transport chain, bypassing the first complex. Since these electrons bypass and thus do not energize the proton pump in the first complex, fewer ATP molecules are made from the FADH2 electrons. The number of ATP molecules ultimately obtained is directly proportional to the number of protons pumped across the inner mitochondrial membrane.
Complex III
The third complex is composed of cytochrome b—another Fe-S protein, a Rieske center (2Fe-2S center), and cytochrome c proteins. This complex is also called cytochrome oxidoreductase. Cytochrome proteins have a prosthetic group of heme. The heme molecule is similar to the heme in hemoglobin, but it carries electrons, not oxygen. As a result, the iron ion at its core is reduced and oxidized as it passes the electrons, fluctuating between different oxidation states: Fe++ (reduced) and Fe+++ (oxidized). The heme molecules in the cytochromes have slightly different characteristics due to the effects of the different proteins binding to them, giving slightly different characteristics to each complex. Complex III pumps protons through the membrane and passes its electrons to cytochrome c for transport to the fourth complex of proteins and enzymes. (Cytochrome c receives electrons from Q; however, whereas Q carries pairs of electrons, cytochrome c can accept only one at a time.)
Complex IV
The fourth complex is composed of cytochrome proteins c, a, and a3. This complex contains two heme groups (one in each of the two cytochromes, a, and a3) and three copper ions (a pair of CuA and one CuB in cytochrome a3). The cytochromes hold an oxygen molecule very tightly between the iron and copper ions until the oxygen is completely reduced by the gain of two electrons. The reduced oxygen then picks up two hydrogen ions from the surrounding medium to make water (H2O). The removal of the hydrogen ions from the system contributes to the ion gradient that forms the foundation for the process of chemiosmosis.
Chemiosmosis
In chemiosmosis, the free energy from the series of redox reactions just described is used to pump hydrogen ions (protons) across the mitochondrial membrane. The uneven distribution of H+ ions across the membrane establishes both concentration and electrical gradients (thus, an electrochemical gradient), owing to the hydrogen ions’ positive charge and their aggregation on one side of the membrane.
If the membrane were continuously open to simple diffusion by the hydrogen ions, the ions would tend to diffuse back across into the matrix, driven by the concentrations producing their electrochemical gradient. Recall that many ions cannot diffuse through the nonpolar regions of phospholipid membranes without the aid of ion channels. Similarly, hydrogen ions in the matrix space can only pass through the inner mitochondrial membrane by an integral membrane protein called ATP synthase (Figure 7.13). This complex protein acts as a tiny generator, turned by the force of the hydrogen ions diffusing through it, down their electrochemical gradient. The turning of parts of this molecular machine facilitates the addition of a phosphate to ADP, forming ATP, using the potential energy of the hydrogen ion gradient.
Visual Connection
Figure 7.13 ATP synthase is a complex, molecular machine that uses a proton (H+) gradient to form ATP from ADP and inorganic phosphate (Pi). (Credit: modification of work by Klaus Hoffmeier)
Dinitrophenol (DNP) is an “uncoupler” that makes the inner mitochondrial membrane “leaky” to protons. It was used until 1938 as a weight-loss drug. What effect would you expect DNP to have on the change in pH across the inner mitochondrial membrane? Why do you think this might be an effective weight-loss drug?
Chemiosmosis (Figure 7.14) is used to generate 90 percent of the ATP made during aerobic glucose catabolism; it is also the method used in the light reactions of photosynthesis to harness the energy of sunlight in the process of photophosphorylation. Recall that the production of ATP using the process of chemiosmosis in mitochondria is called oxidative phosphorylation. The overall result of these reactions is the production of ATP from the energy of the electrons removed from hydrogen atoms. These atoms were originally part of a glucose molecule. At the end of the pathway, the electrons are used to reduce an oxygen molecule to oxygen ions. The extra electrons on the oxygen attract hydrogen ions (protons) from the surrounding medium, and water is formed. Thus, oxygen is the final electron acceptor in the electron transport chain.
Visual Connection
Figure 7.14 In oxidative phosphorylation, the pH gradient formed by the electron transport chain is used by ATP synthase to form ATP. Credit: Rao, A., Ryan, K., Fletcher, S. and Tag, A. Department of Biology, Texas A&M University.
Cyanide inhibits cytochrome c oxidase, a component of the electron transport chain. If cyanide poisoning occurs, would you expect the pH of the intermembrane space to increase or decrease? What effect would cyanide have on ATP synthesis?
ATP Yield
The number of ATP molecules generated from the catabolism of glucose varies. For example, the number of hydrogen ions that the electron transport chain complexes can pump through the membrane varies between species. Another source of variance stems from the shuttle of electrons across the membranes of the mitochondria. (The NADH generated from glycolysis cannot easily enter mitochondria.) Thus, electrons are picked up on the inside of mitochondria by either NAD+ or FAD+. As you have learned earlier, these FAD+ molecules can transport fewer ions; consequently, fewer ATP molecules are generated when FAD+ acts as a carrier. NAD+ is used as the electron transporter in the liver and FAD+ acts in the brain.
Another factor that affects the yield of ATP molecules generated from glucose is the fact that intermediate compounds in these pathways are also used for other purposes. Glucose catabolism connects with the pathways that build or break down all other biochemical compounds in cells, and the result is somewhat messier than the ideal situations described thus far. For example, sugars other than glucose are fed into the glycolytic pathway for energy extraction. In addition, the five-carbon sugars that form nucleic acids are made from intermediates in glycolysis. Certain nonessential amino acids can be made from intermediates of both glycolysis and the citric acid cycle. Lipids, such as cholesterol and triglycerides, are also made from intermediates in these pathways, and both amino acids and triglycerides are broken down for energy through these pathways. Overall, in living systems, these pathways of glucose catabolism extract about 34 percent of the energy contained in glucose, with the remainder being released as heat.
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7312 | https://www.learnalberta.ca/content/memg/division03/prism/index.html | Prism
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Prism
Initial Definition
A prism is a three-dimensional object with two congruent, parallelpolygon bases. The remaining faces are parallelograms coresponding to each side of a base. The faces which are not bases are called lateral faces.
A right prism is a prism with one of the bases aligned directly above the other so that all of the lateral faces are rectangles.
Usually right prisms with triangular bases, rectangular bases or bases which are regular polygons are studied. In the example below, the base is a regular pentagon. Note that each lateral face of a right prism is a rectangle.
Volume and Surface Area of a Prism
In general, the volume of a prism is the area of the base times the height of the prism.
In general, the surface area of a prism is the sum of the areas of all of the faces of the prism.
Prism with a Regular Polygon Base
a: The length of an apothem of the base of the prism
n: The number of sides of the base of the prism
s: The length of a side of the base of the prism
h: The height of the prism
V: The volume of the interior of the prism
S: The surface area of the prism
Explanation of volume formula (regular polygon base)
Explanation of surface area formula (regular polygon base) |
7313 | https://www.uv.es/normas/2012/ANEJOS/Libro%20Gimenez_2012.pdf | ANEJO 3 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI DAVID GIMÉNEZ FOLQUÉS LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI Anejo n.º 3 de Normas. Revista de Estudios Lingüísticos Hispánicos 2012 Los extranjerismos en el español académico del siglo XXI Anejo n.º 3 de Normas. Revista de Estudios Lingüísticos Hispánicos [Recibido: 1-06-2012. Evaluado: 15-07-2012. Aprobado: 30-09-2012] ISBN: 978-84-608-4985-8 ISSN: 2174-7245 Dpto. de Filología Española Universitat de València Primera edición: septiembre de 2012 Libro electrónico de acceso gratuito Texto en línea: Monografías científicas Diseño de portada: María Estellés Arguedas ANEJO N.º 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS Recibido: 1-06-2012. Evaluado: 15-07-2012. Aprobado: 30-09-2012. ÍNDICE Prefacio 9 CAPÍTULO 1. EL DICCIONARIO DE LA REAL ACADEMIA ESPAÑOLA 2001 11 0. Introducción 11 1. Listado de extranjerismos 12 CAPÍTULO 2. EL DICCIONARIO PANHISPÁNICO DE DUDAS 27 0. Introducción 27 1. Listado de extranjerismos 29 2. Diferencias entre el Diccionario panhispánico de dudas y el Diccionario de la Real Academia Española 2001 47 CAPÍTULO 3. LA NUEVA GRAMÁTICA DE LA LENGUA ESPAÑOLA 51 0. Introducción 51 1. Estudio del número 52 CAPÍTULO 4. LA ORTOGRAFÍA DE LA LENGUA ESPAÑOLA 57 0. Introducción 57 1. Las grafías en los extranjerismos 59 1.1. Las grafías b y v 59 1.2. Las grafías g y j en representación del fonema /y/ 60 1.3. La grafía k 62 1.4. La secuenciación gráfica ll 63 1.5. Las grafías m y n 64 1.6. La grafía q 65 1.7. La secuenciación «s inicial + consonante» 65 1.8. El dígrafo sh 66 8 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS 1.9. La grafía w 68 1.10. La grafía z ante e, i 69 1.11. Las terminaciones con -age/-aje 70 1.12. Las terminaciones con -ing 70 1.13. Las voces terminadas en -y precedidas de consonante 72 1.14. Las consonante dobles 73 2. Conclusiones 74 BIBLIOGRAFÍA 77 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS PREFACIO Desde el inicio de este siglo, los extranjerismos han representado un fenómeno muy importante en la evolución de la lengua española. Por ello, ante las oleadas de estas voces extranjeras, las principales obras académicas las han recogido y ordenado con el fin de mantener la integridad de nuestra lengua. Obras como el Diccionario de la Real Academia Española, el Diccionario panhispánico de dudas, la Nueva gramática de la lengua española y la Ortografía de la lengua española revisan todo este caudal léxico con el objetivo de mantener el ideal de unidad lingüística entre las diferentes zonas hispanohablantes. En el horizonte se vislumbra el tono panhispánico que tanto se ha cuidado desde la Real Academia Española y la Asociación de Academias de la Lengua Española. Ante esta situación, es importante que la lengua no se fragmente con la incorporación excesiva de estos extranjerismos, ya que la situación de unidad es la que asegura un futuro panhispánico uniforme en nuestra lengua. Las grafías y los sonidos ajenos a la estructura de nuestra lengua representan el principal campo de batalla en la revisión de las mismas. Ciertamente, la acogida de nuevos extranjerismos supone para la lengua una riqueza lingüística obvia, ya que muchas de estas nuevas palabras cubren realidades que no se podían cubrir con palabras propias del español. Aun así, parece que la Real Academia, junto con la Asociación de Academias de la Lengua Española, pese a no rechazarlos en su totalidad, prefiere derivar el fenómeno al uso de una forma española. De esta manera, las obras académicas han intentado que las grafías de estas voces extranjeras se adapten a formas propias de la lengua española. El primer cambio fuerte, en cuanto a adaptación se refiere, lo hizo el Diccionario panhispánico de dudas en el 2005. En este Diccionario, se adaptan muchos de los extranjerismos que aparecían en el Diccionario de la Real Academia Española1. Para ello, sigue una serie de criterios, que como hemos comentado anteriormente, van a derivar en la utilización de una forma española. Ejemplos de ello serían los lemas del Diccionario panhispánico de dudas, escritos en letra redonda en el diccionario, del siguiente listado: 1 Edición de 2001. 10 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS DLE: body. DPD: bodi. DLE: boom. DPD: bum. DLE: bourbon. DPD: burbon. DLE: brandy. DPD: brandi. DLE: marketing. DPD: márquetin. DLE: slip. DPD: eslip. DLE: spray. DPD: espray. DLE: sprint. DPD: esprín. DLE: striptease. DPD: estriptis, estriptís. Todos estos cambios han ido llegando a nuestros medios de comunicación y a nuestros hablantes; aunque como pudimos comprobar en Giménez Folqués (2001), no todas estas adaptaciones están siendo seguidas por estos sectores. Muchos de los hablantes están muy familiarizados con muchos de los extranjerismos originales; por ese motivo, les resulta extraño usar algunas de las formas adaptadas. La asimilación de estas adaptaciones por parte de los hablantes llevará su tiempo, aunque no todas ellas triunfarán. Más recientemente, contamos con la aparición de obras como la Nueva gramática de la lengua española y la Ortografía de la lengua española. En la primera, los extranjerismos son tratados, principalmente, desde el punto de vista del número. En la segunda, son revisados los extranjerismos desde el punto de vista ortográfico. Por ello, muchos de los extranjerismos que aparecían el DPD son revisados y cambiados siguiendo criterios ortográficos. Ejemplos de ello sería el cambio de wiski por güisqui, de vodka por vodca o de júnior por yúnior (todos ellos en letra redonda). Así pues, en este libro vamos a observar cómo los extranjerismos han evolucionado a lo largo del siglo XXI, y ello lo haremos fijándonos en las principales obras académicas. Además, aportaremos los listados de extranjerismos que aparecen en ellas para analizarlos y compararlos. Para llevar a cabo todo este trabajo, utilizaremos anteriores trabajos como Giménez Folqués (2010) o Giménez Folqués (2011b). David Giménez Folqués Universitat de València ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS CAPÍTULO 1. EL DICCIONARIO DE LA REAL ACADEMIA ESPAÑOLA 2001 0. INTRODUCCIÓN En 2001, aparece una nueva edición del Diccionario de la Real Academia Española. En esta nueva edición, se realiza una profunda revisión con respecto a la última edición nueve años antes. Esta revisión se basó en dos ideas principales según DRAE (2001: I, § Introducción): De una parte, la necesidad de mantener actualizado el cuerpo de la obra en cuanto a los términos en ella registrados, trabajo indispensable para que el repertorio académico pudiera mantener su función unificadora del español. Por otro lado, la acomodación, en lo posible, de todo ese contenido a la estructura fijada por la Nueva planta del Diccionario, un conjunto de normas que los académicos, reunidos en Pleno, aprobaron en junio de 1997. En total, más de 55 000 artículos de la edición anterior fueron revisados y, por lo tanto, enmendados. En cuanto a los extranjerismos, el Diccionario incluye, aproximadamente, 223 extranjerismos en letra cursiva, como por ejemplo los siguientes: affaire (voz francesa). m. Negocio, asunto o caso ilícito o escandaloso, body (voz inglesa). m. Prenda interior femenina, elástica y ajustada, de una sola pieza, que cubre el tronco, camping (voz inglesa). m. Campamento (lugar al aire libre). 2. Actividad que consiste en ir de acampada a este tipo de lugares, full time (expresión inglesa). loc. adj. Con dedicación exclusiva, hobby (voz inglesa). m. Pasatiempo, entretenimiento que se practica habitualmente en los ratos de ocio, paparazzi (voz italiana, del nombre propio Paparazzo, un fotógrafo en el filme La dolce vita). m. pl. Fotógrafos de prensa que se dedican a hacer fotografías a los famosos sin su permiso, 12 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS show (voz inglesa). m. Espectáculo de variedades. 2. Acción o cosa realizada por motivo de expedición. montar un show. fr. Organizar o producir un escándalo. Todos estos extranjerismos aparecen en su forma original, escritos en cursiva. Aunque algunos de ellos, si su escritura o pronunciación se ajustan a la estructura española, como es el caso de club, réflex o airbag, aparecen escritos en letra redonda. Por otro lado, si estos extranjerismos cuentan con derivados en una forma española, como por ejemplo las voces pizzería y flaubertiano, aunque presenten dificultades gráficas o de pronunciación, se representan en letra redonda. A continuación vamos a incluir un listado de los extranjerismos originales que aparecen en el Diccionario de la Real Academia Española. Aparecen en orden alfabético y se incluye el origen, el género y la definición. Si aparecen varias acepciones también serán incluidas. 1. LISTADO DE EXTRANJERISMOS adagio (voz italiana). adv. m. Mús. Con movimiento lento. 2. Mús. Composición o parte de ella que se ha de ejecutar con este movimiento, affaire (voz francesa). m. Negocio, asunto o caso ilícito o escandaloso, alzheimer (de A. Alzheimer, 1864-1915, neurólogo alemán). m. Enfermedad de Alzheimer, ampère. m. Fís. Amperio, angstrom (de A. J. Ǻngstrom, 1814-1874, físico sueco). m. Fís. Medida de longitud que equivale a la diezmilmillonésima (10-º) parte del metro, apartheid (voz afrikáans). m. Segregación racial, especialmente la establecida en la República de Sudáfrica por la minoría blanca. U. t. en sent. Fig, baby-sitter (voz inglesa). com. Canguro (persona que atiende a los niños pequeños), balimbing. m. Filip. Camias (árbol de Filipinas), ballet (voz francesa). m. Danza clásica de conjunto, representada sobre un escenario. 2. Música de esta danza. -3. Compañía que interpreta este tipo de danza, becquerel (de A. H. Becquerel, 1852-1908, físico francés). m. Fís. Unidad LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 13 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS de radiactividad del Sistema Internacional, que equivale a una desintegración nuclear por segundo. (Símb. Bq), beige. adj. Beis, best seller (voz inglesa). m. Libro o disco de gran éxito y mucha venta, big bang (voz inglesa). m. Gran explosión en que una teoría cosmogónica sitúa el origen del universo, blazer (voz inglesa). amb. Chaqueta deportiva, originariamente utilizada en los uniformes de colegios y equipos, blues (voz inglesa). m. Forma musical del folclore de la población de origen africano de los Estados Unidos de América, Bock (voz alemana). m. Jarro de cerveza de un cuarto de litro de capacidad. 2. Contenido de este jarro, body (voz inglesa). m. Prenda interior femenina, elástica y ajustada, de una sola pieza, que cubre el tronco, boiserie (voz francesa). f. Revestimiento de madera aplicado a paredes. 2. Mueble de madera empotrado en una pared, boîte (voz francesa). f. Sala de fiestas o discoteca, boom (voz inglesa). m. Éxito o auge repentino de algo, especialmente de un libro. El boom de la novela hispanoamericana, boshito, ta. adj. Méx. Boxito, bourbon (voz inglesa, y esta del condado de Bourbon, en Kentucky). m. Variedad de wiski que se obtiene de una mezcla de maíz, malta y centeno, originaria del sur de los Estados Unidos de América, boutique (voz francesa). f. Tienda de ropa de moda. 2. Tienda de productos selectos, brandy (voz inglesa, y esta del neerlandés brandewijn, vino quemado). m. Nombre que, por razones legales, se da hoy comercialmente a los tipos de coñac elaborados fuera de Francia y a otros aguardientes, brassavola (del latín científico brassavola, y este de A. Brassavola, 1500-1555, médico y naturalista italiano). f. Hond. Planta epifita, variedad de orquídea, cuya flor tiene los pétalos de color verde amarillento, el labio de color crema con un tinte verdoso y los márgenes lacerados. Es la flor nacional de Honduras, bulldozer (voz inglesa). m. Máquina automóvil de gran potencia, provista 14 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS de una pieza delantera móvil, de acero, que le permite abrirse camino removiendo obstáculos, bungalow (voz inglesa). m. Casa pequeña de una sola planta que se suele construir en parajes destinados al descanso, bushido (voz japonesa). m. Código de honor por el que debían regirse los samuráis, byte (voz inglesa). m. Inform. Octeto (unidad de información), caddie (voz inglesa). com. Dep. Persona que lleva los palos a un jugador de golf, cafisho (etimología discutida). m. Ur. Proxeneta, camping (voz inglesa). m. Campamento (lugar al aire libre). 2. Actividad que consiste en ir de acampada a este tipo de lugares, carpaccio (voz italiana). m. Plato compuesto de lonchas de carne o pescado, cortadas muy finas y condimentadas con diversas especias, que se consume crudo, casting (voz inglesa). m. Selección de actores o de modelos publicitarios para una determinada actuación, catering (voz inglesa). m. Servicio de suministro de comidas y bebidas a aviones, trenes, colegios, etc, chartreuse (voz francesa). m. Licor verde o amarillo de hierbas aromáticas fabricado por los monjes cartujos, christmas (voz inglesa). m. Tarjeta ilustrada de felicitación navideña. cicca. f. Bot. Arbusto de la familia de las Euforbiáceas, cuyas semillas son purgantes, clown (voz inglesa). m. Payaso de circo, y especialmente el que, con aires de afectación y seriedad, forma pareja con el augusto, collage (voz francesa). m. Técnica pictórica consistente en pegar sobre lienzo o tabla materiales diversos. 2. Obra pictórica efectuada con este procedimiento, coulis (voz francesa). m. Salsa hecha a base de vegetales, como el tomate o el pepino. 2. Puré o crema de frutas crudas que se emplea para acompañar un postre, coulomb (de Ch. de Coulomb, 1736-1806, físico francés). m. Fís. Culombio, crack (voz inglesa). m. Droga derivada de la cocaína. 2. Deportista de extraordinaria calidad. 3. Caballo que destaca en las carreras, crescendo (voz italiana, y esta del latín crescendo, gerundio de crescĕre, crecer). m. Aumento gradual de la LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 15 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS intensidad del sonido. 2. Aumento progresivo.in crescendo. loc. adv. Con aumento gradual. El enfado del público fue in crescendo, cricket (voz inglesa). m. Juego de pelota que se practica con paletas de madera, cross (del inglés cross-country, de cross, cruz, cruzar, y country, campo). m. Dep. Carrera de larga distancia a campo traviesa, cueshte (del nahua cuechtic, muy machacado, molido, amasado), adj. El Salv. Bien molido, muy fino, curie (voz francesa, y esta de M. Curie, 1867-1934, y P. Curie, 1859-1906, científicos franceses). m. Fís. Curio (unidad de radiactividad), curry (voz inglesa, y esta del tamil kai). m. Condimento originario de la India compuesto por una mezcla de polvo de diversas especias, cyclo-cross (voz inglesa). m. Deporte consistente en correr en bicicleta a campo traviesa o por circuitos accidentados, czarda (voz húngara). f. Danza húngara de movimiento muy vivo, generalmente de compás binario, a la que suele anteceder una introducción lenta y patética, dancing (voz inglesa). m. Sala pública de baile. U. m. en América, delicatessen (voz inglesa, y esta del alemán Delikatessen). f. pl. Alimentos selectos. 2. amb. Tienda donde se venden delicatessen, déshabillé (voz francesa). m. Salto de cama (bata ligera de mujer para el momento de levantarse de la cama), disc-jockey (voz inglesa). com. Pinchadiscos, dossier (voz francesa). m. Informe o expediente, dumping (voz inglesa). m. Econ. Práctica comercial de vender a precios inferiores al costo, para adueñarse del mercado, con grave perjuicio de este, ertzaina (voz euskera). com. Miembro de la Policía territorial del País Vasco español, ertzaintza (voz euskera). f. Policía territorial dependiente del Gobierno autónomo del País Vasco español, ferry (voz inglesa). m. Transbordador (embarcación que enlaza dos puntos), flash (voz inglesa). m. Aparato que, mediante un destello, da la luz precisa para hacer una fotografía instantánea. 2. Destello producido por dicho aparato. 3. Noticia breve que, con carácter urgente, transmite un medio de comunicación, 16 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS flash-back (voz inglesa). m. En una película, interrupción de la acción en curso para insertar la mostración de hechos ocurridos en un tiempo anterior que afectan a dicha acción. -2. Esta interrupción en un relato literario, foie-gras o foie gras (voz francesa). m. Paté de hígado, generalmente de ave o cerdo, fondue (voz francesa). f. Comida de origen suizo, a base de queso que se funde dentro de una cazuela especial, en el momento de comerla. Por extensión, la que se hace con otros ingredientes como carne, chocolate, etc. 2. Conjunto de utensilios para preparar esta comida, footing (voz francesa, y esta con cambio de sentido del inglés footing, posición). m. Paseo higiénico que se hace corriendo con velocidad moderada al aire libre, force (voz francesa). V. tour de forcé, forfait (voz francesa). m. Contrato hecho à forfait. loc. adv. Mediante el procedimiento de comprar o vender un conjunto de cosas o servicios conviniendo anticipadamente un precio global. U. t. c. loc. adj., free lance (voz inglesa). adj. Dicho de una persona: Que realiza por su cuenta trabajos periodísticos escritos o gráficos y los ofrece en venta a los medios de comunicación. 2. Se aplica a quien trabaja independientemente por este sistema en otras actividades, full time (expresión inglesa). loc. adj. Con dedicación exclusiva, gang (voz inglesa). m. Banda organizada de malhechores, gauss (de C. F. Gauss, 1777-1855, físico y astrónomo alemán). m. Fís. Unidad de inducción magnética en el Sistema Cegesimal, equivalente a una diezmilésima (10-) de tesla. (Símbolo Gs), geisha (voz japonesa). f. En el Japón, muchacha instruida para la danza, la música y la ceremonia del té, que se contrata para animar ciertas reuniones masculinas, gentleman (voz inglesa). m. Caballero inglés de cierto rango social u hombre que se le asemeja en porte, comportamiento y actitud, gilbert (del inglés gilbert, y este de W. Gilbert, 1544-1603, físico inglés). m. Fís. Unidad de fuerza magnetomotriz en el Sistema Cegesimal de unidades, equivalente a 10 : 4π ampervueltas. (Símbolo Gi), gillete o gillette (de K. C. Gillette, 1855-1932, industrial estadounidense LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 17 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS que la inventó; marca registrada). f. Hoja de afeitar desechable, gin (voz inglesa). m. Ginebra (bebida alcohólica), gin-fizz (voz inglesa). m. Combinación de ginebra, zumo de limón, azúcar y soda, ginger-ale (voz inglesa). m. Bebida refrescante elaborada con jengibre, gin-tonic (voz inglesa). m. Combinación de tónica con ginebra, glamour. m. Encanto sensual que fascina, gong (del inglés gong, y este del malayo gong). m. Instrumento de percusión formado por un disco que, suspendido, vibra al ser golpeado por una maza. 2. Campana grande de barco, gospel (voz inglesa). m. Música religiosa propia de las comunidades afronorteamericanas, gouache (voz francesa). m. Aguada (color diluido en agua). 2. Pintura realizada con esta técnica, gourmet (voz francesa). com. Gastrónomo, green (voz inglesa). m. Dep. En el campo de golf, zona de césped bajo y muy cuidado situada alrededor de cada hoyo, guashpira. f. El Salv. mentira (expresión contraria a lo que se sabe), hall (voz inglesa). m. Vestíbulo, recibidor, hándicap (voz inglesa). m. En hípica y en algunos otros deportes, competición en la que se imponen desventajas a los mejores participantes para igualar las posibilidades de todos. 2. Circunstancia desfavorable, desventaja, hardware (voz inglesa). m. Inform. Conjunto de los componentes que integran la parte material de una computadora, hassio (de Hassia, nombre latino del Estado de Hesse, en Alemania). m. Quím. Elemento químico transuránico de núm. atóm. 108. Se obtiene artificialmente por bombardeo de plomo con iones de hierro, y su vida media es tan corta que se mide en milisegundos. (Símbolo Hs), Henry (de J. Henry, 1797-1878, físico estadounidense). m. Fís. Henrio, hertz. m. Fís. Hercio, hippie o hippy (voz inglesa). adj. Se dice del movimiento contracultural juvenil surgido en los Estados Unidos de América en la década de 18 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS 1960 y caracterizado por su pacifismo y su actitud inconformista hacia las estructuras sociales vigentes. 2. com. Partidario o simpatizante de este movimiento, o que adopta alguna de las actividades que le son propias. U. t. c. adj, hobby (voz inglesa). m. Pasatiempo, entretenimiento que se practica habitualmente en los ratos de ocio, hockey (voz inglesa). m. Dep. Juego entre dos equipos, consistente en introducir en la portería contraria una pelota o un disco impulsado por un bastón curvo en su parte inferior, y que se practica en un campo de hierba o con patines en una pista de hielo u otra superficie dura, holding (voz inglesa) m. Sociedad financiera que posee o controla la mayoría de las acciones de un grupo de empresas, hooligan (voz inglesa). m. Hincha británico de comportamiento violento y agresivo. U. t. c. adj, input (voz inglesa). m. Econ. Elemento de la producción, como un terreno, un trabajo o una materia prima. U. t. en sent. fig. 2. Inform. Conjunto de datos que se introducen en un sistema informático. 3. Dato, información, jacuzzi (voz inglesa, marca registrada). m. Bañera para hidromasaje, jazz (voz inglesa). m. Género de música derivada de ritmos y melodías afronorteamericanos. 2. Orquesta especializada en la ejecución de este género de música, jet (voz inglesa). m. Reactor (avión), jet2 (voz inglesa, abreviación de jet set). f. Clase social internacional, rica y ostentosa, jet set (voz inglesa). f. Clase social internacional, rica y ostentosa, jogging (voz inglesa). m. Paseo higiénico que se hace corriendo con velocidad moderada al aire libre, joule (de J. P. Joule, 1818-1889, físico inglés). m. Fís. Julio² (unidad de trabajo del Sistema Internacional), junior (voz inglesa). adj. Que es más joven que otra persona, generalmente su padre, y tiene el mismo nombre, kétchup (voz inglesa, y esta del chino k’ēchap, zumo de tomate). m. Salsa de tomate condimentada con vinagre, azúcar y especias, kilohertz. m. Electr. kilohercio, Kirsch (voz alemana). m. Aguardiente de cerezas, Kitsch (voz alemana). adj. Dicho de un objeto artístico: Pretencioso, pasado LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 19 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS de moda y considerado de mal gusto. U. t. c. s. m., klystron (del inglés klystron, marca registrada). m. Fís. Tubo electrónico empleado para generar o amplificar microondas en comunicaciones y radares, lady (voz inglesa). f. Título de honor que se da en Inglaterra a las señoras de la nobleza, leasing (voz inglesa). m. Econ. Arrendamiento con opción de compra del objeto arrendado, Leitmotiv (voz alemana, derivada de leiten, guiar, dirigir, y Motiv, motivo). m. Tema musical dominante y recurrente en una composición. 2. Motivo central o asunto que se repite, especialmente de una obra literaria o cinematográfica, lifting (voz inglesa). m. Operación de cirugía estética consistente en el estiramiento de la piel, generalmente de cara y cuello, para suprimir las arrugas. U. t. en sent. fig. Los teatros de la ciudad necesitan un lifting, light (voz inglesa). adj. Dicho de una bebida o de un alimento elaborado: Con menos calorías de las habituales. 2. Dicho de un cigarrillo: Que se presenta como portador de menos elementos nocivos. 3. irón. Que ha perdido gran parte de sus caracteres esenciales. Un comunista light, living (voz inglesa). m. cuarto de estar, lobby (voz inglesa). m. Grupo de personas influyentes, organizado para presionar en favor de determinados intereses. 2. Vestíbulo de un hotel y de otros establecimientos como cines, teatros, restaurantes, etc., especialmente si es grande, look (voz inglesa). m. Imagen o aspecto de las personas o de las cosas, especialmente si responde a un propósito de distinción, lunch (voz inglesa). m. Comida ligera que se sirve a los invitados en una celebración, lycra (marca registrada). f. Tejido sintético elástico, utilizado generalmente en la confección de prendas de vestir, maître (voz francesa). com. Jefe de comedor en un restaurante, majorette (voz francesa). f. Muchacha vestida con uniforme militar de fantasía que, en ocasiones festivas, desfila junto con otras agitando rítmicamente un bastón y al son de una banda de música, maquilishuat (del nahua macuilli, cinco, e ishuat, hoja, pétalo). m. El Salv. y Hond. Árbol de la familia de las 20 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS Bignonáceas, de 30 m de altura, hojas opuestas, compuestas de hojuelas elípticas o lanceoladas, flores rosadas y fruto en vaina. Crece silvestre en los bosques desde México a Ecuador. La madera se utiliza para la construcción de barcos y muebles. El cocimiento de la corteza y de la hoja se utiliza en medicina tradicional, marketing (voz inglesa). m. mercadotecnia, mashca (del quichua machka). f. Ecuad. máchica (harina de cebada tostada), mass media (voz inglesa). m. pl. Conjunto de los medios de comunicación, maxwell (de J. C. Maxwell, 1831-1879, físico y matemático escocés). m. Fís. Unidad de flujo de inducción magnética en el Sistema Cegesimal, equivalente a 10- webers. (Símbolo Mx), megahertz. m. Electr. megahercio, mezzosoprano (voz italiana) f. Voz media entre la de soprano y la de contralto. 2. com. Persona que tiene voz de mezzosoprano, miss (voz inglesa, acortamiento de mistress, señorita). f. Ganadora de un concurso de belleza. Miss Universo. Se casó con una miss italiana, motocross (voz francesa, acrónimo de motocyclette, motocicleta, y el inglés cross country, cross). m. Dep. Carrera de motocicletas a través del campo o por circuitos accidentados, mousse (voz francesa). amb. Plato preparado con claras de huevo que dan consistencia esponjosa a los ingredientes dulces o salados que lo componen, mozzarella (voz italiana). f. Queso de procedencia italiana, hecho originalmente con leche de búfala, que se come muy fresco, music hall (voz inglesa). m. Sala que ofrece espectáculos de variedades. 2. Espectáculo de variedades. 3. Género al que pertenece este tipo de espectáculos, newton (de I. Newton, 1642-1727, científico inglés). m. Fís. Unidad de fuerza del Sistema Internacional, equivalente a la fuerza que, aplicada a un cuerpo cuya masa es de un kilogramo, le comunica una aceleración de un metro por segundo cada segundo. (Símbolo N), oersted (de H. C. Oersted, 1777-1851, físico danés). m. Fís. Unidad de intensidad de campo magnético en el Sistema Cegesimal, equivalente a 79,58 amperios por metro. (Símbolo Oe), LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 21 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS offset (voz inglesa). m. Impr. Procedimiento de impresión en el que la imagen entintada es traspasada a un rodillo de caucho que, a su vez, la imprime en el papel. 2. Impr. Máquina que imprime por este procedimiento, ossobuco (voz italiana). m. Estofado de carne de vacuno, cortada del jarrete, con el hueso y su caña incluidos. 2. Arg. y Chile. Corte del hueso del jarrete vacuno, con su tuétano y la carne que lo rodea, output (voz inglesa). m. Econ. Producto resultante de un proceso de producción. U. t. en sent. fig. 2. Inform. Información que sale procesada por un sistema informático o por una computadora. U. t. en sent. Fig, overbooking (voz inglesa). m. Venta de plazas, especialmente de hotel y de avión, en número superior al disponible, paddle (voz inglesa). m. Juego de pelota entre cuatro paredes, en el que aquella se golpea con una pala de mango corto, pajla (voz quechua). adj. Bol. Dicho de una persona: Calva, sin pelo. U. t. c. s., panty (voz inglesa). m. Prenda femenina, a modo de leotardo de tejido fino y muy elástico. U. m. en pl. con el mismo significado que en sing, paparazzi (voz italiana, del nombre propio Paparazzo, un fotógrafo en el filme La dolce vita). m. pl. Fotógrafos de prensa que se dedican a hacer fotografías a los famosos sin su permiso, partenaire (voz francesa). com. Persona que interviene como compañero o pareja de otra en una actividad, especialmente en un espectáculo, ping-pong (voz inglesa, marca registrada). m. Juego semejante al tenis, que se practica sobre una mesa de medidas reglamentarias, con pelota ligera y con palas pequeñas de madera a modo de raquetas, pishishe. m. El Salv. pijije (cierta ave americana). 2. El Salv. Botella de forma especial que se usa para recoger la orina del hombre que guarda cama, pizza (voz italiana). f. Especie de torta chata, hecha con harina de trigo amasada, encima de la cual se pone queso, tomate frito y otros ingredientes como anchoas, aceitunas, etc. Se cuece en el horno, pizzicato (voz italiana). adj. Mús. Dicho de un sonido: Que se obtiene en los instrumentos de arco pellizcando las 22 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS cuerdas con los dedos. 2. m. Mús. Trozo de música que se ejecuta de esta forma, punk (voz inglesa). adj. Perteneciente o relativo al movimiento punk. 2. Seguidor o partidario de ese movimiento. U. m. c. s. 3. m. Movimiento musical aparecido en Inglaterra a fines de la década de 1970, que surge con carácter de protesta juvenil y cuyos seguidores adoptan atuendos y comportamientos no convencionales, quark (voz inglesa). m. Fís. Tipo de partículas elementales, componentes de otras partículas subatómicas, como el protón y el neutrón, y que no existen de manera aislada, quasar (voz inglesa, acrónimo de quasi stellar [radio source]). m. Astr. Cuerpo celeste de pequeño diámetro y gran luminosidad, que emite grandes cantidades de radiación en todas las frecuencias. Es el tipo de astro más alejado en el universo, quiche (voz francesa, y esta del alemán Kuchen). f. Pastel hecho con una base de pasta sobre la que se pone una mezcla de huevos, leche y otros ingredientes y se cuece al horno, rabassa morta (expresión catalana; literalmente, ‘muerta la cepa’). f. En el derecho foral catalán, contrato parecido al de censo, en que el dueño del terreno lo cede, mediante renta, para plantación principalmente de viñas al cultivador que disfruta el predio durante la vida de las primeras plantas, ragtime o rag-time (voz inglesa). m. Ritmo musical sincopado de origen afroamericano, rally (voz inglesa). m. Competición deportiva de resistencia, de automóviles o motocicletas, celebrada fuera de pista y generalmente por etapas, ranking (voz inglesa). m. Clasificación de mayor a menor, útil para establecer criterios de valoración, reggae (voz inglesa). m. Música de origen jamaicano, caracterizada por un ritmo sencillo y repetitivo, ring (voz inglesa). m. Cuadrilátero (espacio limitado por cuerdas), rock (voz inglesa). m. Género musical de ritmo muy marcado, derivado de una mezcla de diversos estilos del folclore estadounidense, y popularizado desde la década de 1950. U. t. c. adj. Música rock. La era rock. 2. Cada uno de los diversos estilos musicales derivados del rock and roll. U. t. c. adj. Los Beatles son el grupo rock más famoso de la historia. 3. Baile de pareja que se ejecuta con esta música, LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 23 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS rock and roll (voz inglesa). m. Género musical de ritmo muy marcado, derivado de la mezcla de diversos estilos del folclore estadounidense, y popularizado desde la década de 1950. U. t. c. adj. Música rock and roll. La era rock and roll. 2. Baile que se ejecuta con esta música, roentgen (de W. C. von Roentgen, 1845-1923, físico alemán descubridor de los rayos X). m. Fís. y Med. Unidad electrostática cegesimal de poder ionizante con relación al aire. Se emplea en las aplicaciones terapéuticas de los rayos X, rouge (voz francesa). m. Pintalabios, roulotte (voz francesa). f. caravana (vehículo acondicionado para cocinar y dormir en él), rugby (voz inglesa, de Rugby, escuela pública de Warwickshire, en Inglaterra, donde se inventó). m. Deporte que se practica, con las manos y los pies, entre dos equipos de quince jugadores cada uno, con un balón ovalado que se debe depositar tras la línea que marca el final del campo o introducir entre un travesaño y dos postes que se elevan sobre los extremos de este, scooter (voz inglesa). m. Motocicleta ligera o ciclomotor, con ruedas pequeñas, que tiene una plataforma para apoyar los pies y una plancha protectora en su parte delantera, self-service (voz inglesa). m. Autoservicio, sex-appeal (voz inglesa). m. Atractivo físico y sexual, sexy (voz inglesa). adj. Que tiene atractivo físico y sexual. Es muy sexy. 2. m. Atractivo físico y sexual. Tiene sexy, shaurire. m. Hond. Sotorrey (cierta ave), sheriff (voz inglesa). m. En los Estados Unidos de América y ciertas regiones o condados británicos, representante de la justicia, que se encarga de hacer cumplir la ley, sherpa (voz inglesa, y esta del tibetano sharpa, habitante de la zona oriental del país). adj. Perteneciente o relativo a un pueblo de Nepal, cuyos habitantes suelen participar como guías y portadores en las expediciones en el Himalaya. Apl. a pers., u. t. c. s. 2. m. Guía o porteador sherpa, short (voz inglesa). m. Pantalón muy corto, usado principalmente para practicar deportes, show (voz inglesa). m. Espectáculo de variedades. 2. Acción o cosa realizada por motivo de expedición. -24 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS montar un show. fr. Organizar o producir un escándalo, shuar. adj. Se dice del individuo de un pueblo amerindio que habita en las selvas del sur de la región oriental ecuatoriana. U. t. c. s. 2. Perteneciente o relativo a los shuar. -3. m. Lengua hablada por los shuar, shunte (voz lenca). m. Hond. Variedad de aguacate (árbol lauráceo). 2. Hond. Fruto de este árbol, siemens (de C. W. Siemens, 1823-1883, ingeniero alemán). m. Fís. Unidad de conductancia del Sistema Internacional, equivalente a la conductancia de un conductor que tiene una resistencia eléctrica de un ohmio. (Símbolo S), sievert (de R. Sievert, 1896-1966, físico sueco). m. Fís. Unidad de dosis equivalente de radiación del Sistema Internacional, igual a un julio por kilogramo. (Símbolo Sv), sioux (voz francesa, este de nadouesssioux, y este del algonquino nātowēssiwak, con cambio de la terminación de plural algonquina -ak por la terminación de pl. fr. –x). adj. Se dice del individuo de un pueblo amerindio oriundo de los valles del norte del Misisipí. U. t. c. s. 2. Perteneciente o relativo a este pueblo, slip (voz inglesa). m. Calzoncillo ajustado que cubre el cuerpo desde debajo de la cintura hasta las ingles, software (voz inglesa). m. Conjunto de programas, instrucciones y reglas informáticas para ejecutar ciertas tareas en una computadora, soufflé (voz francesa). adj. Dicho de un alimento: Preparado de manera que quede inflado. Patatas soufflé. m. Alimento preparado con claras de huevo a punto de nieve y cocido en el horno para que adquiera una consistencia esponjosa. Soufflé de coliflor, souvenir (voz francesa). m. Objeto que sirve como recuerdo de la visita a algún lugar determinado, speech (voz inglesa). m. despect. irón. Pequeño discurso, sponsor (voz inglesa, y esta del latín sponsor, fiador). com. Patrocinador, sport (voz inglesa). adj. Dicho de una prenda: Que es informal, con respecto a la de vestir. Una chaqueta sport. 2. m. Am. p. us. Deporte. de sport. loc. adj. Deportivo (cómodo e informal). Un traje de sport. U. t. c. loc. adv. Vestir de sport, spot (voz inglesa). m. Película de muy corta duración, generalmente de carácter publicitario, LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 25 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS spot² (voz inglesa, y esta, acortamiento de spotlight). m. Foco de luz potente y directa que se utiliza en fotografía, cine, teatro, etc., para iluminar una zona pequeña, spray (voz inglesa). m. Envase con un dispositivo especial para pulverizar los líquidos que contiene. 2. Sustancia líquida contenida en este envase, sprint (voz inglesa). m. Dep. Aceleración que realiza un corredor en un tramo determinado de la carrera, especialmente en la llegada a meta para disputar la victoria a otros corredores. 2. Esfuerzo final que se realiza en cualquier actividad, stand (voz inglesa). m. Instalación dentro de un mercado o feria, para la exposición y venta de productos, standing (voz inglesa). m. Posición económica y social. Una urbanización de alto standing, stock (voz inglesa). m. Cantidad de mercancías que se tienen en depósito, striptease (voz inglesa). m. Espectáculo en el que una persona se va desnudando poco a poco, y de una manera insinuante. 2. Local en que se realiza este tipo de espectáculos, suite (voz francesa). f. En los hoteles, conjunto de sala, alcoba y cuarto de baño. 2. Mús. Composición instrumental integrada por movimientos muy variados, basados en una misma tonalidad. Suite en re mayor, swahili (del árabe sawāḥil, pl. de sāḥil, costa). m. Lengua del grupo bantú hablada en África oriental, toffee (voz inglesa). m. Caramelo masticable de café con leche, topless o top-less (voz inglesa). m. Modo de vestir femenino que deja los pechos al aire. 2. Bar o local de espectáculos en el que las empleadas trabajan con los pechos al aire, top-model (voz inglesa). com. Modelo de alta costura, especialmente el muy cotizado, tory (voz inglesa). adj. Perteneciente o relativo al partido conservador de Gran Bretaña. 2. com. Miembro de este partido, tour (voz francesa). m. Excursión, gira o viaje por distracción. 2. Gira, serie de actuaciones sucesivas de un cantante, grupo musical, etc., por diferentes localidades. 3. Mil. p. us. Período o campaña de servicio obligatorio de un marinero. tour de force. m. Esfuerzo grande, físico o anímico, tournée (voz francesa). f. Gira (serie de actuaciones de una compañía teatral 26 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS o de un artista). 2. Excursión, gira o viaje por distracción. 3. Viaje profesional de un político, un viajante de comercio, etc., de itinerario y visitas predeterminados. La tournée del presidente. Una tournée de inspección, troupe (voz francesa). f. Grupo de artistas, especialmente de teatro, de cine o de circo que trabajan juntos, desplazándose de un lugar a otro. 2. Grupo de personas que van juntas o que obran de forma similar, vedette (voz francesa). f. Artista principal en un espectáculo de variedades. 2. Persona que destaca o quiere hacerse notar en algún ámbito. Ese saltador es una vedette del atletismo. U. t. c. adj, vendetta (voz italiana). f. Venganza derivada de rencillas entre familias, clanes o grupos rivales. U. t. en sent. Fig, vichy (de Vichy, ciudad de Francia). m. Tejido fuerte de algodón, de rayas o cuadros, voyeur (voz francesa). com. Persona que disfruta contemplando actitudes íntimas o eróticas de otras personas, watt (de J. Watt, 1736-1819, ingeniero escocés). m. Electr. Vatio, weber (de W. E. Weber, 1804-1891, físico alemán). m. Fís. Unidad de flujo de inducción magnética del Sistema Internacional, equivalente al flujo magnético que, al atravesar un circuito de una sola espira, produce en ella una fuerza electromotriz de un voltio si se anula dicho flujo en un segundo por decrecimiento uniforme. (Símbolo Wb), western (voz inglesa). m. Género de películas del Lejano Oeste. 2. Película del Lejano Oeste, whisky. m. Güisqui, windsurf o wind surf (voz inglesa). m. Deporte que consiste en deslizarse por el agua sobre una tabla especial provista de una vela, windsurfing o wind surfing (voz inglesa). m. Deporte que consiste en deslizarse por el agua sobre una tabla especial provista de una vela, yang (voz china). m. En la filosofía china, especialmente en el taoísmo, fuerza activa o masculina que, en síntesis con el yin, pasiva o femenina, constituye el principio del orden universal.
ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS CAPÍTULO 2. EL DICCIONARIO PANHISPÁNICO DE DUDAS 0. INTRODUCCIÓN La aparición del Diccionario panhispánico de dudas supuso un gran impacto en el terreno de los extranjerismos en el español. Este fenómeno provocó una profunda revisión de los mismos. Se realiza una reforma de estas voces extranjeras con respecto a la aparición del último Diccionario académico en el 2001. En el prólogo del DPD se indica que se ha respetado la inclusión de extranjerismos en el Diccionario académico pero se han añadido algunas voces (DPD: 19, § Introducción): Con el fin de recomendar soluciones que se ajusten a las pautas señaladas, este diccionario comenta un grupo numeroso, aunque necesariamente limitado, de voces extranjeras habitualmente empleadas por los hispanohablantes. Concretamente, los extranjerismos crudos incluidos en la última edición del Diccionario académico (2001), así como los extranjerismos adaptados que allí se registran cuando aún es frecuente encontrarlos escritos en textos españoles con las grafías originarias. Además, se han añadido algunos extranjerismos no recogidos por el Diccionario académico, pero que son hoy de uso frecuente en el español de América o de España. Muchos de estos extranjerismos son adaptados a una grafía y pronunciación española. De entre los factores que existen en la adaptación de estas voces extranjeras, sobresale, como señala el DPD, el de mantener la «esencia» de la lengua. Algunos ejemplos de esta adaptación serían jacuzzi, piercing o blues, que han sido adaptados por yacusi, pirsin o blus (escritos en letra redonda). Aunque los extranjerismos originales de estas adaptaciones son conocidos internacionalmente, se ha tomado la decisión de adaptarlos igualmente. Otro factor con el que contamos para la ampliación de voces extranjeras es la participación de América en el Diccionario. El hecho de que la Academia haya ampliado su campo de estudio a América ha provocado que los extranjerismos que aparecen allí sean contemplados también por el Diccionario. El objetivo de esta ampliación ha sido el de mantener la unidad lingüística entre el español de España y el de América. Este hecho se ve reflejado en RAE (2005: XVI, § Introducción): 28 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS Por la misma razón, se reconocen, cuando existen, las divergencias entre la norma española y la norma americana, o entre la norma de un determinado país o conjunto de países y la que rige en el resto del ámbito hispánico, considerando en pie de igualdad y plenamente legítimos los diferentes usos regionales, a condición de que estén generalizados entre los hablantes cultos de su área y no supongan una ruptura del sistema de la lengua que ponga en riesgo su unidad. Solo se desaconsejan los particularismos dialectales que pueden impedir la comprensión mutua, por ser fuente de posibles malentendidos; nos referimos a los pocos casos en que una estructura lingüística adquiere en un área concreta un valor o significado diferente, e incluso opuesto, al que tiene en el español general. Finalmente, el DPD sigue una serie de pautas para la adaptación de estos extranjerismos. Para ello, clasifica los extranjerismos en dos tipos. En primer lugar el Diccionario habla de extranjerismos superfluos o innecesarios; se refiere a aquellos que tienen equivalente en español, e incluye los ejemplos de consulting (en español consultora o consultoría) o back-up (en español, copia de seguridad). En este caso, la Academia aconseja que se utilice el equivalente en español. Por otro lado nos habla de extranjerismos necesarios o muy extendidos; evidentemente se refiere a aquellos extranjerismos que no tienen equivalente en español o es muy complicado encontrarlo. En este último caso se aplican dos criterios, el primero el de mantener la grafía y la pronunciación originaria (como sería en el caso de software o ballet) y el segundo el de adaptar la pronunciación o la grafía originaria (como el caso de airbag o master). En este segundo caso, se recomienda utilizar la adaptación. En ambos casos, la Academia nos deriva a una forma española. A continuación, vamos a incluir un listado de los extranjerismos que aparecen en el DPD. Van a aparecer por orden alfabético. Se incluyen el origen del extranjerismo y el plural, siempre que el DPD lo incluya. Respetamos la grafía de los lemas (cursiva o redonda) tal como aparecen en el diccionario. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 29 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS 1. LISTADO DE EXTRANJERISMOS a capela (< a cappella), (en el ejemplo: canciones a capela, plural invariable) abstract, acimut o azimut (plural: acimuts, azimuts), acmé adagio agur aeróbic o aerobic (< aerobics), afer (< affaire), afiche (< affiche), airbag (< air bag o air-bag), (plural : airbags), alegro (< allegro), alegreto (< allegretto), alemanda (< allemande), aligátor (< alligator), alioli (< all i oli), amateur ambigú (< ambigu) (plural: ambigús), ampáyer (< umpire) (plural: ampáyeres), anorac o anorak (plural: anoraks). Se desaconseja por minoritaria anorac. apartotel (< acrónimo de apart[ment] + [h]otel), apartheid archivolta o arquivolta armañac (< armagnac), (plural: armañacs), armonio (< harmonium, voz francesa de grafía latinizante). Es preferible a armónium (plural: armonios) asistente (assistant). Calco del inglés (- asistente, 2), aspaventar, áspic (plural: áspics), atrezo (< attrezzo), attachment (-adjunto), autofoco (< autofocus), autoestop (< auto-stop o autostop), baby-sitter, bacará (< baccarat), bacará o bacarrá (< baccara), background, back-up, bádminton (< badminton), bafle (< baffle), baguete (< baguette) (plural: baguetes), baipás (< by-pass) (plural: baipases), 30 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS balé (< ballet), (plural: balés), balonvolea, voleibol o vóleivol, vóley (< volleyball), balotaje (< ballottage), bambú (plural: bambús, bambúes), banyo (variante recomendada), banjo (< banjo), barman (< barman) (se utiliza el plural <barmen) (plural: bármanes), bartender (barman), básquetbol o basquetbol, básquet (< basketball). baloncesto (calco), bate (< bat), batiscafo, bazuca (< bazooka), bebé o bebe (epiceno en España y en el Río de la plata, en el resto de América: el/la bebé), beicon (variante preferible), bacón (< bacon), beis (< beige) (plural invariable), béisbol o béisbol (< baseball), bencina o benzina, besamel (variante preferible), bechamel, besamela, bechamela (< béchamel); best seller, bibelot (< bibelot) (plural: bibelots), bidé (< bidet), big bang, bikini (grafía mayoritaria); biquini (< Bikini), billón (< billion), bíper (< beeper), biscote (< biscotte), bísquet (< biscuit) (plural: bísquets), bistec, bisté (< beefsteak) (plurales respectivos: bistecs, bistés). Se desaconseja por minoritaria la variante biftec. También bife (< beef) (plural: bifes), bistró (< bistro) (plural: bistrós), bit, bitio (< bit, acrónimo de bi[nary digi]t), bíter (< bitter) (plural: bíteres), bivac o bivouac (vivac), bléiser (< blazer), (plural: bléiseres), blíster (< blister (pack)) (plural: blísteres), bloc (< block) (plural: blocs), blof, bluf (< bluff) (plurales: blofs, blufs), blog, blue jeans, blues, bluesman, blúmer (< bloomer) (plural: blúmeres), LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 31 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS blus (< blues) (plural: bluses), blusero (< bluesman), bluyín, yin (< blue jeans) (plural: bluyines, yines), bodi (< body) (plural: bodis), bóer (< boer) (plural: bóeres), boicot (< boycott) (plural: boicots), boicoteo, boiler, bol (< bowl) (plural: boles), bonsái, bonsay (plural: bonsáis); Sin indicación de origen, box (< box) (plural: boxes), boy scout, brandi (< brandy) (plural: brandis), breque (< brake), bricolaje (< bricolage), brocheta (< brochette), brócoli, brécol, bróculi (minoritaria) (< broccoli, plural de broccolo) (plural del primero: brócolis), bróker (< broker) (plural: brókeres), budín (pudin o pudín), bufé (< buffet) (plural: bufés), bufete (< buffet) (plural: bufetes), buganvilia, buganvilla (< Bougainville), buldócer (< bulldozer) (plural: buldóceres), bulevar (< boulevard) (plural: bulevares), bullabesa (< bouillabaisse), bum (< boom) (plural: bums), búmeran o bumerán (< boomerang) (plurales respectivos: los búmeran, bumeranes), búngalo o bungaló (< bungalow) (plurales respectivos: búngalos, bungalós), búnker (< Bunker) (plural: búnkeres), buqué (< bouquet) (plural: buqués), burbon (< bourbon) (plural: búrbones), buró (< bureau) (plural: burós), business, butade (< boutade) (plural: butades), byte (se utiliza también el plural bytes), cabaré (< cabaret) (plural: cabarés), cabeza rapada (calco), cabriolé (< cabriolet) (plural: cabriolés), caché (< cache [memory]), caché (< cachet) (plural: cachés), cácher (< catcher) (plural: cácheres), cadi (< caddie o caddy) (plural: cadis), cameraman (camarógrafo), campin (< camping) (plural: cámpines), campus (latinismo tomado del inglés) (plural invariable), 32 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS cancán (< cancan), canelón (< cannellone) (plural: canelones), cantábile, cantable (empleo raro) (< cantabile), capó (< capot) (plural: capós), capuchino (< cappuccino), carcaj (plural: carcajes) o carcax, aunque esta carece de uso en la actualidad, carné (< carnet) (plural: carnés), carpacho (< carpaccio), carrusel (< carrousel), cartel o cártel (< Kartell), casba (< qasabah), casete (< cassette) (plural: casetes), cash, cash flow, cast, castin (< casting) (plural: cástines), cayac o kayak, cáterin (< catering) (plural invariable), CD (< CD, sigla de compact disc), CD-ROM (< CD-ROM, sigla de compact disc read-only memory), cedé (< CD), cederrón (< CD-ROM), chacolí (plural: chacolís), chalé (plural: chalés), champán, champaña (< champagne) (plural del primero: champanes), champú (< shampoo) (plural: champús), chance (< chance), chándal (plural: chándales), chantillí (< chantilly) (plural: chantillís), chapó (< chapeau), chaqué (< jaquette) (plural: chaqués), charme, chárter (< charter) (plural: chárteres), chasis o chasís (< châssis) (plurales respectivos: chasis y chasises), chat (< chat) (plural: chats), chef (< chef), chequear, checar (< to check), chic (< chic) (plural: chics), chifonier (variante más usada), sifonier (< chiffonnier) (plurales respectivos: chifonieres, sifonieres), chip (< chip) (plural: chips), chofer o chófer (< chauffeur) (plural: choferes y chóferes), chóped (< chopped) (plural: chópedes), chor (short) (plural: chores), chovinismo (< chauvinisme), chucrut (< choucroute < Sauerkraut) (plural: chucruts), LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 33 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS chutarse (< to shoot), chut, chute (plural:chuts, chutes), cicerone (género : el/la cicerone), ciclocrós (< cyclo-cross) (En América bicicrós), clac, claque (< claque) (plural del primero: clacs), claqué (< claquette), cliché (variante más usada), clisé (< cliché), clínex (< kleenex) (plural invariable), clip (< clip) (plural: clips), clíper (< clipper) (plural: clíperes), cloche (< clutch), clóset (< closet) (plural: clósets), club (< club) (plural: clubs, clubes), coach, coctel o cóctel (< cocktail) (plurales respectivos: cocteles y cócteles), colaje (< collage), cómic (< comic) (plural: cómics), commodity, compact (disc), compacto, disco compacto, (disco) compacto (< compact (disc)). También: compacto, CD, cedé, CD-ROM, cederrón; REPE, complot (< complot) (plural: complós), conducir, conductor (calco del inglés), confeti (< confetti) (plural: confetis), confort (de origen francés), consulting, container, convoy (plural: convoyes), coñac (< cognac) (plural: coñacs), copyright, coque, cok (< coke), coque, cok (variante menos frecuente) (< coke), coqueluche (tomada del francés), córner (< corner) (plural: córneres), corsé (< corset) (plural: corsés), cowboy, coy (plural: coyes), crac (< crack) (plural: cracs), crepé, crep (< crêpe ‘tejido’) (plurales: crepés, creps), crepe, crep, crepa (< crêpe ‘tortita’) (plurales de los dos primeros: crepes, creps), crescendo, in crescendo, críquet (< criket), croché (< crochet), crol (< crawl), 34 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS crómlech (< origen bretón o galés introducido a través del francés) (plural invariable), cros (< cross, forma abreviada de cross-country). S. v. campo, 2; cross, cross-country, cruasán (< croissant) (plural: cruasanes), crupier (< croupier) (plural: crupieres), cuché (< (papier) couché), culi o culí (< coolie) (plurales respectivos: culis, culíes o culís), culote. 1 (< culotte), culote. 2 (< culot), cupé (< coupé) (plural : cupés), cuplé (< couplet) (plural: cuplés), curri (< curry,voz inglesa de origen tamil) (plural: curris), cuscús (grafía mayoritaria), cuzcuz, alcuzcuz (< couscous) (plural del primero: cuscuses), cúter (< cutter) (plural: cúteres), dancing, dandi (< dandy) (plural: dandis), debacle (< débâcle), debut (< début) (plural: debuts), déficit (latinismo tomado del francés) (plural: déficits), delicatesen (< delicatessen) (plural invariable), demo (< demo, acortamiento de demonstration), derbi (< derby) (plural: derbis), déshabillé, destroyer, dirham, dírhem (origen árabe) (plurales respectivos: dírhams, dírhems), (disco) compacto (< compact (disc)). También: compacto, CD, cedé, CD-ROM, cederrón, display, disquete (< diskette), disyóquey (< disc jockey) (plural: disyoqueis), dólar (< dollar) (plural: dólares), domo (< dôme), doping (dopaje), dosier (< dossier) (plural: dosieres), driblar, driblear (< to dribble), dribbling, dumpin (< dumping), dúplex (< duplex) (plural invariable), DVD (< DVD, sigla de digital versatil disc) (plural invariable). También: devedé, deuvedé (< DVD) (plurales respectivos: devedés, deuvedés), echarpe (< écharpe), LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 35 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS electroshock, elepé (forma preferible), LP (< LP, sigla de long play) (plural: elepés). Larga duración (calco), Élite o elite, e-mail, emoticono (variante preferible), emoticón (< emot[ion] + icon) (plurales respectivos: emoticonos, emoticones), empoderar(se) (calco de to empower), enervar (toma el sentido del francés‘excitar o irritar’, sentido que el francés añadió a esta voz en el siglo xix, la forma ya existía en castellano con el sentido de ‘debilitar o relajar’), entrecot (uso mayoritario), entrecote (< entrecôte) (plurales respectivos: entrecots, entrecotes), escáner (< scanner) (plural: escáneres), escay (< scay), escúter (< scooter) (plural: escúteres), eslalon (< slalom) (plural: eslálones), eslip (< slip) (plural: eslips), eslogan (< slogan) (plural: eslóganes), esloti (< zloty), esmog (< smog, acrónimo de sm[oke] + [f ]og), esmoquin (< smoking) (plural: esmóquines), esnifar (< to sniff), esnob (< snob) (plural: esnobs), espagueti (< spaghetti) (plural: espaguetis), esparrin (< sparring) (plural: espárrines), esplín (< spleen), espray (< spray) (plural: espráis), esprín (< sprint) (plural: esprines), esprínter (< sprinter) (plural: esprínteres), esquí (< ski) (plural: esquís, esquíes), estándar (< standard) (plural: estándares), estárter (< starter) (plural: estárteres), estatus (< status) (plural invariable), estor (< store) (plural: estores), estrés (< stress) (plural: estreses), estríper (< stripper) (plural: estríperes). S. v. estriptis o estriptís, estriptis o estriptís (< striptease) (plurales respectivos: invariable, estriptises), exprés, expreso (< express), facsímil o facsímile (del latín: facere + simile) (plural: facsímiles), fair play, 36 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS fan (< fan, acortamiento de fanatic) (plural: fanes), fast food, fax (< fax) (plural: faxes), feedback, feeling, ferri (< ferry) (plural: ferris), file, filin (< feeling), filme (variante preferible), film (< film) (plural: films, flimes), flas (<flash) (plural: flases), flashback, flirt, folclore, folclor, folklor, folklore (< folklore), fólder (< folder), (plural: fólderes), folk (< folk) (plural: folks), footing, forfait (< forfait) (plural: forfaits), forinto (< forint), fovismo (< fauvisme), frac, fraque (de uso minoritario) (< frac) (plurales respectivos: fracs, fraques), free lance, frízer (< freezer) (plural: frízeres), fuagrás (< foie-gras, foie gras) (plural: fuagrases), fueloil, fuel, fuelóleo (< fuel oil), fular (< foulard) (plural: fulares), full-time, fútbol o futbol (< football), gag (< gag) (plural: gags), gang, gánster (< gangster) (plural: gánsteres), gap (< gap) (plural: gaps), garaje (< garage), gasoil, gasóleo (< gas oil), gay (< gay) (plural: gais), géiser (< geyser) (plural: géiseres), geisha, gentleman, gillette, girl scout, glamur, glamor (< glamour), glas (< glace), glasé (< glacé), glasear, goal average, góspel (< gospel) (plural: góspeles), gouache, grafito (< graffito), grafiti ( < graffiti) (plural del segundo: grafitis), grapa (< grappa), gratín (al gratín, al gratén) (< au gratin), grill, LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 37 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS grogui (< groggy) (plural: groguis), gruyer (< gruyère) (plural: gruyeres), gueisa (< geisha), gueto (< ghetto), guipur (< guipure), güisqui (< whisky o whiskey) (plural: güisquis), gulag (plural: gulags), gurmé (< gourmet) (plural: gurmés), gurú (< del hindi, a través del francés o inglés) (plural: gurús, gurúes), hábitat (< habitat, latinismo introducido a través del inglés) (plural: hábitats), hachís (origen árabe), hacker, half time, hall, hámster (< Hamster) (plural: hámsteres), hándicap (< handicap) (plural: hándicaps), hardware, harén (forma mayoritaria y preferible), harem (< harem. MIRAR DLE) (plural del segundo: harems), head-hunter, highball, hippie, hippy, hit, hobby, hockey, holding, home run, hooligan, iceberg (origen reerlandés, incorporada a través del inglés) (plural: icebergs), idéntikit o identikit (< identikit (picture)) (plurales respectivos: idéntikits, identikits), iglú (< igloo) (plural: iglús, iglúes), impasse, indexar (de base latina, se ha incorporado a través del inglés o del francés), influenciar (< influencer), input, interfaz (< interface) (plural: interfaces), Internet interviú (< interview) (plural: interviús), jaibol (< highball) (plural: jaiboles), jazz, jean, jeep, jersey, yérsey, yersi (< jersey) (plurales respectivos: jerséis, yerseis, yersis), 38 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS jetlag, jipi (< hippy o hippie), jit (< hit) (plural: jits). En béisbol, jockey, jogging, jonrón (< home run) (plural: jonrones), jóquey (< hockey), júnior (< junior) (plural: júniores), junk food, junkie, junky, káiser (plural: káiseres), kamikaze, camicace (variante minoritaria) (del japonés) (plural del primero: kamikazes), kebab (plural: kebabs), kermés (variante mayoritaria), quermés, kermese, quermese (< kermese) (plurales respectivos: kermeses, quermeses, kermeses, quermeses), kétchup (< ketchup), cátchup (< catchup), cátsup (< catsup), kibutz (< kibbutz o kibboutz) (plural invariable), kindergarten, kínder (< kindergarten) (plurales respectivos: kínderes, kindergártenes), kiosco, quiosco, kipá (< kippab) (plural: kipás), kit (< kit) (plural: kits), kitsch, kiwi, kivi (desaconsejada) (plural: kiwis), kleenex, klystron, knock-out, know-how, kril (< krill), kung-fu (< kung-fu), lady, lasaña (< lasagna), láser (< laser, acrónimo de l[ight a[mplification by] s[timulated] e([mission of] r([adiation]) (plural: láseres), leitmotiv, licra (< lycra), lifting, light, linier (< quizás del catalán) (plural: linieres), link, lipstick, lobby, lobbyst, lock-out, lonche (< lunch), long play, LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 39 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS look, lord (< lord) (plural: lores), lutier (< luthier) (plural: lutieres), macadán (forma recomendable), macadam (< Mac Adam), magacín, magazine (< magazine), mail, e-mail, mailing, maillot (< maillot) (plural: maillots), mall, mamut (< mammouth) (plural: mamuts), management (evitable, se aconseja la forma «administración»), mánager (< manager) (plural invariable), márquetin (< marketing), marron glace, mass media, máster (< master) (plural: másteres), match, matiné (< matinée), meeting, menaje (< ménage), metre (< maître), mezzosoprano, microchip, microfilm (e), mildiu o mildiú, mildéu (menor uso) (< mildew), milord (< milord), minué (< menuet), (plural: minués), minueto (< minuetto), misil o mísil (< missile), miss, mitin (< meeting) (plural: mítines), mobbing, moca (grafía tradicional), moka (de la ciudad yemení del mismo nombre), módem (< modem, acrónimo de mo[dulator] + dem([odulator]) (plural: módems), mofle (< muffler), moiré, motocrós (< motocross), motor-home, mouse, mousse, mozarela (< mozzarella), muaré, moaré (< moiré), musaca, musaka (menos recomendable) (< moussaka o mousaka, transcripciones de la voz de origen árabe), music hall, 40 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS muyahidín (árabe < plural de muyahid) (plural: muyahidines), naíf o naif (< naïf) (plurales respectivos: naífs, naifs), nailon (variante mayoritaria), nilón (< nylon), neroli (más usada) o nerolí (< néroli), nocaut (< knock-out) (plural: nocauts), noquear (< to knock out), novel, nurse (< nurse), nursery, off the record, ófset (< offset) (plural: ófsets), ombudsman, on line, open, optimizar (< to optimize), osobuco (< ossobuco), output, outillage, overall, overbooking, overol (< overall) (plural: overoles), pachulí o pachuli (< patchouli), pack, pádel (< paddle (tennis)), paipái (forma asentada), paipay (plural de la primera: paipáis), pancake, panqueque (< pancake). S.v. crepe, panti (< panty) (plural: pantis), paparazi (< paparazzi, plural de paparazzo) (plural: paparazis), paper, páprika o páprika (< páprika), paquebote (< paquebot), paralímpico –ca (< parlympic, acrónimo de para(plegic)+(o)lympic), parkinsonismo, párkinson (< Parkinson), parqué (< parquet) (plural: parqués), parquin (< parking) (plural: párquines), partenaire, part-time, paspartú (< passe-partout) (plural: paspartús), passe-partout, password, pasteurizar (< pasteuriser >Pasteur), paté (< pâté) (plural: patés), pay per view, pedigrí (< pedigree) (plural: pedigríes, pedigrís), LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 41 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS penalti (< penalty) (plural: penaltis), (en América: penal, plural: penales), performance, petiso (forma mayoritaria y preferible), petizo (< petiz), photo finish, pichar (< to pitch), pícher (< pitcher) (plural: pícheres), pick-up, picop (< pick-up), (plural: picops), pijama (variante más frecuente), piyama (< pyjamas), pimpón (< ping-pong), pin (< pin) (plural: pines), piolet, piolé (mayoritaria) (< piolet) (plural: piolets, piolés), pipermín (< peppermint) (plural: pipermines), pirsin (< (body) piercing) (plural: pírsines), pívot, pivote (< pivot) (plural: pívots), píxel o píxel (< pixel) (plural: píxeles, pixeles), pizzicato, placar (< placard) (plural: placares), planning (se considera innecesaria al existir la forma «plan»), playback, play-off, pogromo (rusa pogrom) (plural: pogromos), poliéster (< polyester) (plural: poliésteres), ponche (< punch), poni (< pony), póney (menos recomendable) (< poney), (plurales respectivos ponis, poneis), pool, pop (< pop, acortamiento coloquial de popular) (plural: pops), popurrí (< pot pourri) (plural: popurrís), póquer, póker (< poker), porche (< porch), póster (< poster) (plural: pósteres), premier, premier (< première), prime time, presión (< pressing), pudin o pudín, budín (< pudding) (plurales respectivos: púdines, pudines, budines), puf (< pouf) (plural: pufs), pulman (< pullman) (plural: púlmanes), pulóver (< pullover) (plural: pulóveres), púlsar o pulsar (< pulsar, acrónimo de puls[ating st]ar) (plural: púlsares o pulsares), 42 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS punch, punk (< punk rock), punki (< punky) (plurales respectivos: punks, punkis), puzle (< puzzle), quark (tomada de inglés) (plural: quarks), quásar, cuásar (menos frecuente) (< quasar, acrónimo de quas[i stell] ar [radio source]) (plurales respectivos: quásares, cuásares), quepis, quepí, (< képi) (plurales respectivos: invariable, quepís), queroseno, querosene, querosén, kerosene, kerosén, keroseno (< kérosène o kerosene); querosín, kerosín (< kerosine o kérosine); quiche (< quiche) (plural: quiches), quiosco, kiosco, rafting, raglan o raglán, ranglan, ranglán (< Raglan), ragú (< ragoût) (plural: ragús), raid (< raid) (plural: raides), raid (< ride), raíl (< rail) (plural: raíles), rali (< rally o rallye) (plural: ralis), ranquin (< ranking) (plural: ránquines), rapel o rápel (< rappel) (plurales respectivos: rapeles, rápeles), rating, ratón (calco semántico de mouse), ravioli, raviol (< ravioli, plural de raviolo) (plurales respectivos: raviolis, ravioles), razia (< razzia), récord, récor (más recomendable) (< record) (plurales respectivos: récords, récores), recordman, recordwoman, réferi o referí (< referee) (plurales respectivos: réferis, referís), remake, rentrée, restaurante (<restaurant), restaurant, revival, ricota (< ricotta), ricotta, riel (< riell), rímel (< rimmel) (plural: rímeles), ring, ripostar (< riposter), ritornelo, retornelo (de menor uso) (< ritornello), roast beaf, rock, LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 43 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS rock and roll, rol (< role), rosbif (< roast beef o roast-beef) (plural: rosbifs), rotisería (< rôtisserie); rosticería (posiblemente de rosticceria ), rouge, roulotte, round, royalty, rugbi (<rugby), salacot (origen tagalo) (plural: salacots), salami, salame (< del plural de salame) (plural: salamis, salames), sándwich (< sandwich) (plural: sándwiches), satén, satín (< satin), sauerkraut, savoir faire, scanner, scooter, score, script (innecesario, se aconseja la forma «guión»), secuoya, secoya (de uso minoritario) (< sequoia), self-service, sénior (< del latín revitalizado por el inglés) (plural: seniores), serpa (< sherpa), set (< set) (plural: sets), sexapil (< sex-appeal), sexi (< sexy) (plural: sexis), shampoo, share, sheij, sheik, sheikh, sheriff, sherpa, shock, shoot, short (se usa con más frecuencia la forma shorts), short stop, shot, show (innecesario, se aconseja la forma «espectáculo»), showman, show-woman, sifonier (chiffonier), sij (plural: sijes, sijs) se recomienda el uso del plural sijes, single, siux (< sioux) (plural invariable), skai, 44 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS skin, skinhead, slalom, slip, slogan, smog, smoking, snack bar, snob, sofa (plural: sofas), software, sottovoce, soufflé, sóviet o soviet (origen ruso) (plural: sóviets, soviets), soya, soja (< shoyu), spaghetti, sparring, speaker, speech (innecesario, se aconseja la forma «discurso»), spiritual, spleen, sponsor, sport, spot, spray, sprint, staff, stand, standard, standing, starter, status, stock, stop, stopper, store, strike-out, stripper, striptease, stud, suajili (< swahili) (plural: suajilis), suéter (< sweater) (plural: suéteres), suflé (< soufflé) (plural: suflés), sumiller (< sommelier) (plural: sumilleres), surf (< to surf o acortamiento de surfing), suvenir (< souvenir) (plural: suvenires), swahili, swaskika, sweater, symposium, tabloide (< tabloid), LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 45 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS taekuondo (forma preferible), taekwondo (< tae kown do, coreano), taichí (< tàijí, chino), tándem (< tandem) (plural: tándems), tape, masking tape (innecesario, se aconseja la forma «cinta» o «casete»), tarantela (< tarantella), team (innecesaria, se aconseja la forma «equipo»), ténder (< tender) (plural: ténderes), test (< test) (plural invariable), testar, testear (< to test) (calco). S. v. test, 2, thriller, tique, tiquete (< ticket), tofi, tofe (< toffe), tóner (< toner) (plural: tóneres), top (< top) (plural: tops), top model, top secret, toples (< topless) (plural invariable), topless, tory, tour, tour de force, tournée (innecesario, se aconseja la forma «gira»), tráiler (< trailer) (plural: tráileres), training (innecesario, se aconseja la forma «adiestramiento»), transfer, trávelin (< travelling) (plural invariable), travesti o travestí (del francés) (purales respectivos: travestis o travestís), trekking, trillón (< trillion), troika (de uso mayoritario), troica (origen ruso), trol (< troll) (plural: troles), trole (< trolley), trolebús, (< trolleybus < trolley bus), (plural: trolebuses), troupe (innecesario, se aconseja la forma «compañía»), trust (< trust) (plural invariable), valquiria, valkiria (< walkyria), vals, valse (plural único: valses), vampiro -ra (de origen húngaro, a través del francés), varietés (< variétés), váter, wáter (< water(-closet)) (plural del primero: váteres), vedet, vedete (< vedette) (plurales respectivos: vedets, vedetes), vendetta, 46 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS vermú, vermut (< vermout o vermouth) (plurales respectivos: vermús, vermuts), versus, vichí (< vichy), video o vídeo, videoclip (< videoclip) (plural: videoclips), vip (< vip = v[ery ] i[mportant ] p[erson ]) (plural: vips), vis a vis (calco del francés), vivac, vivaque (menos frecuente) (< bivac) (plurales respectivos: vivacs, vivaques), vodevil (< vaudeville) (plural: vodeviles), vodka, vodca (grafía minoritaria) (del ruso ), voleivol o vóleivol, volibol o vólivol, vóley (< volleyball). Balonvolea (calco), volován (< vol-au-vent) (plural: volovanes), voyerista (< voyeur), waterpolo (< water polo), web (< web) (plural: webs), wéstern (< western) (plural: wésterns), windsurf, windsurfing (existe la forma «tablavela»), wolframio (variante mayoritaria y preferible), volframio, wólfram (origen germánico), yacusi (< jacuzzi) (plural: yacusis), yanqui (< yankee), yate (< yacht), yidis (< yiddish < jüdisch), yincana (< gymkhana), yogur (origen turco) (plural: yogures), yonqui (< junkie o junky), yoqueta (< jockette), yóquey, yoqui (de uso minoritario) (<jockey) (plurales respectivos: yoqueis, yoquis), yudo, judo (origen japonés), zapeo (< zapping), zarda (< csárdás), zarévich (< tsarevich) (plural invariable), zigurat (< ziggurat) (plural: zigurats), zíper (< zipper) (plural: zíperes), zombi (zombie, africano occidental, a través del inglés) (plural: zombis), zum (< zoom) (plural: zums).
LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 47 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS 2. DIFERENCIAS ENTRE EL DICCIONARIO PANHISPÁNICO DE DUDAS Y EL DICCIONARIO DE LA REAL ACADEMIA ESPAÑOLA 2001 El Diccionario de la Real Academia Española (2001 y posteriormente 2003) incluye muchos extranjerismos, pero todos ellos originales y en cursiva1. Sin embargo, el Diccionario panhispánico de dudas castellaniza la mayoría de ellos. Además de castellanizar parte de los extranjerismos, el DPD incluye términos que no aparecían en la edición del DLE2 2001 y 2003. En Aleza (2006: 291) podemos observar una reproducción de este hecho3: De origen galorrománico: DLE: affaire. DPD: afer. DLE: ballet. DPD: balé. DLE: collage. DPD: colaje. DLE: dossier. DPD: dosier. DLE: foie-gras o foie gras. DPD: fuagrás. DLE: forfait. DPD: forfait. DLE: gourmet. DPD: gurmé. DLE: limusina. DPD: limusina, limosina (< limousine). DLE: maître. DPD: metre. DLE: motocross. DPD: motocrós. DLE: narguile. DPD: narguile o narguilé. DLE: sioux. DPD: siux. DLE: soufflé. DPD: suflé. DLE: souvenir. DPD: suvenir. DLE: vedette. DPD: vedet, vedete. DLE: voyeur. DPD: voyerista. 1 Los denominados «extranjerismos crudos» 2 Forma abreviada del Diccionario de la Real Academia Española 3 Los extranjerismos van a ser ordenados alfabéticamente. Además, incluimos en negrita los extranjerismos que añade el DPD con respecto al DLE 48 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS De origen italiano: DLE: brécol, brócol, brócul, bróculi. DPD: brócoli, brécol, bróculi o broccoli. (< broccoli). DLE: carpaccio. DPD: carpacho. DLE: mezzosoprano. DPD: mesosoprano. DLE: mozzarella. DPD: mozarela. DLE: ossobuco. DPD: osobuco. DLE: paparazzi. DPD: paparazi. DLE: raviole, ravioli. DPD: ravioli o raviol4. (< ravioli). De origen inglés: DLE: básquet, basquetbol. DPD: básquet, básquetbol, basquetbol. (< basketball). DLE: béisbol. DPD: béisbol, beisbol (< baseball). DLE: blazer. DPD: bléiser. DLE: blues. DPD: blus. DLE: body. DPD: bodi. DLE: boom. DPD: bum. DLE: bourbon. DPD: burbon. DLE: brandy. DPD: brandi. DLE: bulldozer. DPD: buldócer. DLE: bumerán. DPD: búmeran, bumerán (< boomerang). DLE: bungalow. DPD: búngalo, bungaló. DLE: caddie. DPD: cadi. DLE: camping. DPD: campin. DLE: casting. DPD: castin. DLE: catering. DPD: cáterin. DLE: CD. DPD: CD o cedé. DLE: chutar. DPD: chutar, chutear. DLE: crack. DPD: crac. DLE: cricket. DPD: críquet. DLE: cross. DPD: cros. 4 Ha desaparecido raviole en el DPD LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 49 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS DLE: culi. DPD: culi, culí o coolie. (< coolie). DLE: curry. DPD: curri. DLE: delicatessen. DPD: delicatesen. DLE: disc-jockey. DPD: disyóquey. DLE: display. DPD: display5. DLE: dumping. DPD: dumpin. DLE: DVD. DPD: devedé o DVD. DLE: elepé, LP. DPD: elepé6 (< long play). DLE: ferry. DPD: ferri. DLE: flash. DPD: flas. DLE: flash-back. DPD: flashback. DLE: gong, gongo. DPD: gong o gongo. DLE: gospel. DPD: góspel. DLE: handicap. DPD: hándicap. DLE: hippie o hippy. DPD: jipi. DLE: no consta. DPD: jit. DLE: hockey. DPD: jóquey. DLE: jacuzzi. DPD: yacusi. DLE: junior. DPD: júnior. DLE: ketchup. DPD: kétchup, cátchup, cátsup. DLE: klystron. DPD: klistrón. DLE: lunch. DPD: lonche. DLE: marketing. DPD: márquetin. DLE: mass media. DPD: medio. DLE: miss. DPD: mis. DLE: offset. DPD: ófset. DLE: pin-pong. DPD: pimpón. DLE: piolet. DPD: piolé o piolet (< piolet). DLE: píxel. DPD: píxel o pixel (< pixel). DLE: póquer. DPD: póquer o póker (< poker). DLE: punk. DPD: punk, punki. DLE: puzle. DPD: puzle. DLE: quasar. DPD: quásar o cuásar. 5 Se rectifica en el DPD al considerarlo extranjerismo crudo. 6 No se recomienda «LP». 50 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS DLE: ranking. DPD: ranquin. DLE: récord. DPD: récord o récor (< record). DLE: rock. DPD: rock. DLE: rock and roll. DPD: rocanrol. DLE: rugby. DPD: rugbi. DLE: scooter. DPD: escúter. DLE: sex-appeal. DPD: sexapil. DLE: sexy. DPD: sexi. DLE: sherpa. DPD: serpa. DLE: slip. DPD: eslip. DLE: spray. DPD: espray. DLE: sprint. DPD: esprín. DLE: striptease. DPD: estriptis, estriptís. DLE: toffe. DPD: tofe, tofi. DLE: topless o top-less. DPD: toples. DLE: váter. DPD: váter, wáter. DLE: voleibol. DPD: voleibol, vóleibol, volibol, vólibol, vóley (< volleyball). DLE: western. DPD: wéstern. De origen alemán: DLE: volframio, wólfram, wolframio. DPD: wolframio, volframio o wólfram (< Wolfram). De origen japonés: DLE: geisha. DPD: gueisa. De origen ruso: DLE: sóviet. DPD: sóviet o soviet (< soviet). ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS CAPÍTULO 3. LA NUEVA GRAMÁTICA DE LA LENGUA ESPAÑOLA 0. INTRODUCCIÓN El 10 de diciembre de 2009, se presentó la Nueva gramática de la lengua española. La obra se completó el 20 de diciembre de 2011, con un tercer volumen, Fonética y fonología. Según la RAE (2009: I, § Introducción) la Nueva gramática es el resultado de un gran esfuerzo de colaboración panhispánico: Es la primera gramática académica desde 1931 y ofrece el resultado de once años de trabajo de las veintidós Academias de la Lengua Española, que aquí fijan la norma lingüística para todos los hispanohablantes. Esta acción conjunta de las Academias significa la articulación de un consenso que fija la norma común para todos los hispanohablantes, armonizando la unidad del idioma con la fecunda diversidad en que se realiza. La obra se presenta en tres versiones: Nueva gramática de la lengua española. Recoge el texto completo y detallado. Puede usarse como obra de consulta general y como texto de estudio en el nivel universitario. Manual. Un volumen de cerca de 1000 páginas, conciso y didáctico, dirigido especialmente a los profesores y estudiantes de español en los niveles no universitarios y a todos los hispanohablantes de nivel culto medio. Gramática básica. Un volumen de 305 páginas, fácilmente adaptable al ámbito escolar, que presenta, muy simplificados, los conceptos fundamentales. En cuanto a los objetivos de la obra, la RAE (2009: I, § Introducción) pretende: Describir las construcciones gramaticales propias del español general, así como reflejar adecuadamente las variantes fónicas, morfológicas y sintácticas. Ofrecer recomendaciones de carácter normativo. Ser obra de referencia para el conocimiento y la enseñanza del español. 52 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS El objetivo de las Academias es lograr que la Nueva gramática llegue a todos los hispanohablantes: a los usuarios y a los especialistas e investigadores; a los que tienen el español como primera o segunda lengua, y a los profesores de español en los diversos niveles académicos. En el año 2010 aparece Nueva gramática de la lengua española. En cuanto a los extranjerismos se refiere, la obra los trata principalmente desde el punto de vista del número. En la Nueva gramática de la lengua española se hace referencia al proceso de adaptación de los extranjerismos y se señala que si el extranjerismo está adaptado, evidentemente será más fácil extender su plural. En el caso contrario, en los extranjerismos crudos puede que se mantenga el plural en la grafía originaria, como son los ejemplos de man y woman que se forman como men y women. La NGLE, cuando habla de extranjerismos se centra en su plural; indica que la mayoría de plurales de los extranjerismos son adaptados y siguen la normativa española. Pero señala también las excepciones que se producen, además de las citadas en los extranjerismos crudos poco extendidos. A continuación, vamos a observar los tipos de plurales que recoge la Nueva gramática en el uso de los diferentes extranjerismos con los que nos encontramos. 1. EL ESTUDIO DEL NÚMERO En primer lugar, las palabras acabadas en –y precedida de consonante forman su plural cambiando la y por –is. Son excepción, por no gozar de la suficiente aceptación y así mantener el plural de su idioma de origen, las siguientes voces: body, brandy, caddy, curry, ferry, rally, rugby, sexy. Por otro lado, los préstamos acabados en las consonantes –n, -l, -r, -d, -j, -z hacen el plural con –es. Siguiendo esta regla en nuestro listado hemos señalado que las formas LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 53 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS «fan» y «pin» forman el plural con –es. Sin embargo, la NGLE indica que aunque se recomienda este plural, hoy en día son más frecuentes las formas pins y fans (en redonda). Lo mismo sucede con las voces siguientes: búnker (plural: búnkers o búnkeres), córner (plural: córners o córneres), chándal (plural: chándals o chándales), chárter (plural: chárters o chárteres), claxon (plural: claxons o cláxones), escáner (plural: escáners o escáneres), eslalon (plural: eslalons o eslálones), eslogan (plural: eslogans o eslóganes), gánster (plural: gánsters o gánsteres), hámster (plural: hámsters o hamsters), máster (plural: másters o másteres), póster (plural: pósters o pósteres), tráiler (plural: tráilers o tráileres). Aunque tenemos las dos variantes, se sigue prefiriendo la terminación en –es para los casos anteriores. La Nueva gramática de la lengua española señala que es importante utilizarlas en la variante normativa para regularizarlos. En el caso de los sustantivos procedentes de otras lenguas terminados en grupo consonántico se pluralizan con –s. Es el caso de: camembert (plural: camemberts), cíborg (plural: cíborgs), folk (plural: folks), ginseng (plural: ginsengs), gong (plural: gongs), iceberg (plural: icebergs), punk (plural: punks), récord (plural: récords). Otro caso diferente es el de test. Su plural se registra en tests, pero dada la dificultad que supone pronunciar el grupo /sts/ en español, la NGLE recomienda dejarla sin variación: los tests. Por esta misma situación permanecen invariables en plural otros sustantivos: 54 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS el compost (plural: los compost), el karst (plural: los karst), el kibutz (plural: los kibutz), el trust (plural: los trust). El sustantivo club forma los plurales clubs y clubes, ambos igualmente válidos. Esta duplicidad se extiende a aeroclubs o aeroclubes, videoclubs o videoclubes y teleclubs o teleclubes (todos ellos en redonda). Plantea alguna dificultad la formación del plural de voces adaptadas que son ya plurales en su lengua de origen. Así, los sustantivos confeti, espagueti o ravioli proceden de nombres plurales italianos. No se perciben como tales en español, por lo cual, se les añade –s: los confetis, los espaguetis o los raviolis. Finalmente, la Nueva gramática de la lengua española señala un grupo de extranjerismos crudos que forman su plural al igual que lo hacen en su idioma de origen: ballet (plural: ballets), best-seller (plural: best-sellers), camping (plural: campings), catering (plural: caterings), crack (plural: cracks), dancing (plural: dancings), flash-back (plural: flash-backs), gentleman (plural: gentlemen), gourmet (plural: gourmets), hall (plural: halls), hobby (plural: hobbies), holding (plural: holdings), hooligan (plural: hooligans), input (plural: inputs), jet (plural: jets), lady (plural: ladies), leitmotiv (plural: leitmotivs), mailing (plural: mailings), parking (plural: parkings), LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 55 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS party (plural: parties), pub (plural: pubs), ranking (plural: rankings), resort (plural: resorts), roulotte (plural: roulettes), sex-shop (plural: sex-shops), short (plural: shorts), show (plural: shows), slip (plural: slips), sport (plural: sports), stand (plural: stands), stock (plural: stocks), tour (plural: tours), woman (plural: women). Para concluir, la Nueva gramática de la lengua española indica que si el estudio de un extranjerismo supone una tarea un tanto compleja, el de su plural puede resultarlo doblemente. En el corpus que acabamos de estudiar apreciamos que, exceptuando algunos casos especiales, los plurales con los que nos encontramos siguen la norma aconsejada. ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS CAPÍTULO 4. LA ORTOGRAFÍA DE LOS EXTRANJERISMOS EN LA LENGUA ESPAÑOLA 0. INTRODUCCIÓN El 17 de diciembre de 2010, bajo el amparo de la Real Academia Española y de la Asociación de Academias de la Lengua Española, se presenta la nueva edición de la Ortografía de la lengua española, que viene a sustituir a la anteriormente vigente, de 1999. Aparece bajo la perspectiva de la política lingüística panhispánica, como viene siendo habitual en las publicaciones académicas actuales. Su objetivo es el de describir el sistema ortográfico de la lengua española y realizar una exposición pormenorizada de las normas que rigen su correcta escritura en la actualidad. Dentro de esta perspectiva ortográfica, tiene cabida el uso de los extranjerismos que aparecen tanto en España como en Hispanoamérica. Se habla de ambas zonas hispanohablantes por el ideal de unidad lingüística que se ha perseguido con la aparición de las últimas obras académicas, tales como las mencionadas anteriormente; la Nueva gramática de la lengua española, el Diccionario panhispánico de dudas o la misma Ortografía de la lengua española. Así pues, la RAE y la Asociación de Academias de la Lengua Española se han marcado una serie de objetivos que ven su reflejo en esta Ortografía revisada. Estos objetivos se resumen en la ya comentada necesidad de mantener la unidad de la lengua; el principio de economía lingüística y la resistencia a cambios drásticos que supongan una ruptura entre la forma gráfica de las palabras y su pronunciación; esta última postura la podemos observar mejor en las páginas de la Real Academia Española (2010c: 22): […] el peso de la tradición ortográfica heredada […] establece un fuerte vínculo entre las palabras y su forma gráfica fijada. Así, cualquier cambio drástico en la grafía de una palabra se siente más como una deformación que desfigura su identidad visual que como una simplificación beneficiosa, lo que explica la fuerza que el criterio del uso constante ha tenido y tiene en la fijación de la ortografía de las lenguas. Una ruptura radical con la tradición gráfica anterior dificultaría, además, la lectura de textos de otras épocas, a los que habría que sumar los costes económicos que supondría la adaptación a las nuevas normas de todas aquellas obras escritas conforme al sistema ortográfico precedente, y el sinfín de 58 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS cambios que habría que realizar en todos aquellos ámbitos relacionados de algún modo con el lenguaje natural (diccionarios, bases de datos, aplicaciones informáticas, etc.). Como podemos observar, la Ortografía se basa en el peso de la historia para fortalecer su ideal de identidad. Además, añaden el criterio de economía lingüística, ya que el hecho de mantener la tradición lingüística permitirá no tener que realizar cambios en la mayoría de obras ya escritas. Un punto importante en esta cuestión es el uso de extranjerismos; ya que muchos de ellos contienen grafías ajenas a la de la lengua española. Por ello, la RAE considera que ante la masiva llegada de préstamos a nuestra lengua durante las últimas décadas era necesaria una revisión bastante amplia1, donde hubiera un control lingüístico de los mismos. De esta manera, se intenta que estas nuevas voces lleguen a adaptarse al sistema de la lengua española tanto en forma como en pronunciación2. De no producirse este control, la Real Academia Española (2010c: 598) señala que se produciría un proceso de desestabilización: Aunque, como se ha visto, el fenómeno del préstamo lingüístico es algo natural que no cabe censurar en modo alguno, es necesario tener en cuenta que la proliferación indiscriminada de extranjerismos crudos o semiadaptados en textos españoles puede resultar un factor desestabilizador de nuestro sistema ortográfico, especialmente cuando se ponen en circulación grafías que se apartan del sistema de correspondencias entre grafemas y fonemas propio de nuestra lengua. De ahí que la Real Academia Española, junto con el resto de las que con ella integran la Asociación de Academias de la Lengua Española, siendo las instituciones encargadas de preservar la coherencia y la unidad del español, procuren orientar los procesos de adopción de extranjerismos para que su incorporación responda, en lo posible, a nuevas necesidades expresivas y se produzca dentro de los moldes propios de nuestra lengua. Así pues, la Ortografía trata de adaptar las formas extranjeras al sistema ortográfico español. Por ello, en el presente trabajo incluimos las referencias que hace la obra hacia estas formas extranjeras, donde aparecen muchas novedades; algunas de estas ya las encontramos en otras obras académicas como el Diccionario panhispánico de 1 Revisión ya realizada por el Diccionario panhispánico de dudas de una manera pormenorizada y ampliada en la Ortografía de la lengua española. 2 Ejemplos de este proceso de adaptación serían los siguientes: balé por ballet, sérif por sheriff y gueisa por geisha. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 59 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS dudas. Las más novedosas también las recoge Gómez Torrego (2011), novedades que citamos en cada uno de los apartados que adjuntamos a continuación. Finalmente, distribuiremos los apartados en grafías para una mayor clarificación de la información, e incluiremos la situación ortográfica actual de cada una de ellas, en cuanto a usos extranjeros se refiere. Es decir, los criterios que la RAE considera más apropiados para la adaptación de estas grafías y las excepciones que puedan surgir. Todo ello, con la mayor cantidad de ejemplos posible3. 1. LA ORTOGRAFÍA DE LOS EXTRANJERISMOS 1. 1. Las grafías b y v En general, el uso de las grafías b y v depende casi siempre de criterios etimológicos, es decir, de la configuración gráfica de su étimo. De este modo, la mayoría de palabras extranjeras mantienen la b o v etimológicas: boceto (del italiano bozzetto), convoy (del francés convoi), overol4 (del inglés overall), taburete (del francés tabouret), tobogán (del inglés toboggan), valija (del italiano valigia). Aunque también podemos encontrar factores antietimológicos en la adaptación de algunos extranjerismos a la lengua española: arquitrabe5 (del italiano architrave), arribista (del francés arriviste), esbelto (del italiano svelto) o rendibú6 (del francés rendez vous). La Real Academia Española (2010c: 93) señala que la aparición de estas excepciones tienen una explicación difícil: Las razones que explican la existencia de estas grafías antietimológicas impuestas por el uso son distintas en cada caso y responden a factores muy diversos, difíciles de sistematizar, como pueden ser, entre otros, la analogía con palabras semánticamente relacionadas (caso, por ejemplo, de arribista ‘persona que progresa por medios rápidos y sin escrúpulos’, cuya b antietimológica se explica por analogía con arribar y arriba). 3 Extraídos de la Real Academia Española (2010c) y de Gómez Torrego (2011). 4 Con el significado de ‘mono como prenda de vestir’. 5 Con el significado de ‘parte inferior del entablamento, la cual descansa inmediatamente sobre el capitel de la columna’. 6 Con el significado de ‘acatamiento, agasajo que se hace a alguien, por lo general con la intención de adularlo’. 60 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS Con la presencia de estas excepciones, muchas veces resulta difícil determinar el uso de b y v. Sin embargo, la Ortografía señala algunas pautas orientativas. Entre ellas, hay algunas que afectan a los extranjerismos como: a) Se emplea b en posición final de palabra: club, esnob etc. Son excepción los extranjerismos de procedencia eslava molotov y lev7, así como las transcripciones al alfabeto latino de ciertos nombres propios eslavos, ya sean topónimos, como Kiev, o antropónimos, como los apellidos que contienen las terminaciones patronímicas -ev, -ov: Prokófiev, Romanov. b) Se emplea b después de las sílabas iniciales ra-, ri-, ro-, ru.: rabel8, ribete9, robinsón10, etc. Son excepciones: raviol, ravioli o raviole y algunos topónimos y sus gentilicios, como Ravena o Rávena (ciudad de Italia) y ravenés. c) Se emplea b en las palabras que empiezan por alb-: albornoz, álbum etc. d) Se emplea b en las palabras que empiezan por bar-: barbacoa, barman etc. e) Se emplea b en las palabras que empiezan por bu-, bur-, y bus-: budín11, bufé o bufete, buganvilia, buldócer, bulevar, bumerán, burbon etc. f) Se emplea v en las palabras que contienen la forma vídeo o video: video o vídeo, videoclip, videoclub etc. 1. 2. Las grafías g y j en representación del fonema /y/ En algunos préstamos encontramos las letras j y g (ante e, i) en representación del fonema /y/. Sería el caso de manager o el de junior. Este fenómeno es ajeno al sistema gráfico español. Por lo tanto, cuando se quiera adaptar una voz extranjera a la lengua española se debe hacer sustituyendo estas grafías por la letra y, según aconseja la 7 Con el significado de ‘moneda de Bulgaria’; variante de la forma preferida leva. 8 Con el significado de ‘instrumento musical pastoril, pequeño, de hechura como la del laúd y compuesto de tres cuerdas solas, que se tocan con arco y tienen un sonido muy agudo’. 9 Con el significado de ‘cinta o cosa análoga con que se guarnece y refuerza la orilla del vestido, calzado, etc’. 10 Con el significado de ‘hombre que en la soledad y sin ayuda ajena llega a bastarse a sí mismo’. 11 Variante de pudin o pudín. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 61 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS Ortografía de la lengua española: yúnior como adaptación de junior y mánayer como adaptación de manager. Otros ejemplos serían los siguientes: bluyín por blue jean, yacusi por jacuzzi, yincana por gymkhana, yonqui por junkie y yudo por judo. Si por el contrario, se emplea la grafía y pronunciación originarias, estas formas deben ser consideradas como extranjerismos crudos o no adaptados y escribirse, por lo tanto, en cursiva. La Ortografía incluye ejemplos de ello: gigolo [yigoló] (‘hombre joven mantenido por una mujer mayor a cambio de favores sexuales’); gin [yín] (‘ginebra’); ginseng [yínsen] o [yinsén] (‘sustancia tónica y estimulante, extraída de una planta del mismo nombre originaria de Corea’); manager [mánayer] (‘gerente o directivo de una empresa o sociedad, o representante de un artista o un deportista’); pidgin [pídyin] (‘lengua formada por la mezcla de elementos de otras varias, usada entre hablantes de diferente origen lingüístico’); jazz [yás] (‘cierto tipo de avión’) Una tercera opción sería la de mantener las grafías originales pero pronunciándolas según nuestro sistema gráfico-fonológico: júnior [júnior] y mánager [mánajer] como explica mejor la Real Academia Española (2010c: 107): Cuando en un mismo término existen grafías con -j- y con -y-, cada una debe pronunciarse según el fonema que representa en español el grafema que contiene, como en soja [sója] y soya [sóya]. Lo que no resultaría posible sería considerar préstamos adaptados como formas en las que existe discordancia entre grafía y pronunciación de acuerdo con nuestro sistema ortográfico; así, la forma judo solo puede considerarse como extranjerismo adaptado si se pronuncia [júdo]; si se pronuncia [yúdo] ha de escribirse en cursiva, como corresponde a los extranjerismos crudos, aunque el hablante tiene también la opción de emplear, en este caso, la grafía adaptada yudo. La Real Academia Española (2010c: 107) señala casos donde se dan los dos fenómenos: Hay casos en que un mismo extranjerismo se ha adaptado al español siguiendo ambas pautas: tanto manteniendo la pronunciación originaria y modificando la grafía como manteniendo la grafía originaria y modificando la pronunciación. Así, procedente del inglés 62 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS jersey existen en español las adaptaciones jersey [jerséi], usada sobre todo en España, y yérsey [yérsei] o yersi [yersi], usadas en América; del mismo modo, del inglés pyjamas proceden las adaptaciones pijama [pijáma], en España, y piyama [piyáma], en América. 1. 3. La grafía k Tradicionalmente, la grafía k en los préstamos se solía sustituir por grafías patrimoniales como c o qu en favor de la total adaptación de estos al sistema gráfico español. Ejemplos de esta adaptación serían: canguro, del francés kangourou; cinc o zinc, del alemán Zink; o esmoquin, del inglés smoking. Sin embargo, en los préstamos de nueva o más reciente incorporación, es más normal que la letra k se mantenga por estar ya integrada en el abecedario del español. Serían los casos de: kamikaze, kayak, kilo, kiwi, etc. Es posible también encontrar la forma etimológica y la adaptada. Así pues, encontraríamos: bikini / biquini, moka / moca, póker / póquer, etc. De esta manera, aunque en muchos casos, especialmente en los préstamos más actuales, se tiende a mantener la grafía etimológica, no se pueden censurar las grafías que se adapten a las pautas gráficas tradicionales en español12. Por ello, la existencia de diversas posibilidades gráficas para representar el fonema /k/ ha dado lugar a numerosos casos de variantes gráficas en la escritura de los préstamos. Algunos casos destacados por la Ortografía donde se observa algunas variantes serían13: bikini (‘traje de baño femenino de dos piezas’), mejor que biquini; caqui (‘color que varía entre el amarillo ocre y el verde grisáceo’ y ‘árbol oriental y su fruto comestible’), mejor que kaki; euskera (‘lengua vasca’), mejor que eusquera; folclor(e) (‘conjunto de costumbres, tradiciones y manifestaciones artísticas de un pueblo’), folclórico -ca (‘del folclore’), folclorista (‘estudioso del folclore’), mejor que folklor(e), folklórico -ca, folklorista; harakiri (‘suicidio ritual japonés que consiste en abrirse el vientre’), mejor que haraquiri; Irak (‘país árabe’), mejor que Iraq; 12 Para más información consultar la Real Academia Española (2010c: 116). 13 Muchos de estos préstamos difieren de las adaptaciones propuestas por la Real Academia Española (2005). LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 63 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS kamikaze (‘piloto suicida japonés’ y ‘persona temeraria o que comete un atentado que puede implicar su propia muerte’), mejor que camicace; kimono (‘túnica japonesa’), mejor que quimono; koiné (‘lengua estándar común, resultado de la unificación de distintas variantes dialectales’), mejor que coiné; moka (‘cierta variedad de café’), mejor que moca; musaka (‘plato típico de la cocina griega’), mejor que musaca; neoyorquino -na (‘de Nueva York’), mejor que neoyorkino -na; pakistaní (‘de Pakistán’), mejor que paquistaní; pekinés (‘de Pekín’), mejor que pequinés -sa; póker (‘cierto juego de naipes’), mejor que póquer; polca (‘danza folclórica de Bohemia’), mejor que polka; queroseno o en América, querosén, querosene, querosín (‘combustible derivado del petróleo’), mejor que keroseno, kerosén, kerosene, kerosín; quiosco (‘templete para celebrar conciertos al aire libre’ y ‘puesto de venta en la calle’), quiosquero -ra (‘persona que atiende un quiosco’), mejor que kiosco, kiosquero -ra; telequinesia o telequinesis (‘desplazamiento de objetos por la fuerza de la mente’), mejor que telekinesia o telekinesis; vodka (‘cierto aguardiente’), mejor que vodca Gómez Torrego (2011: 27) indica que puede haber variantes entre una forma y sus derivados en lo que se refiere a la grafía k. Sería la situación de rock. Considera a esta forma como extranjerismo no adaptado por su grafía ck, alejada del sistema español. Por ello, debe escribirse con resalte tipográfico, es decir, en cursiva. Sin embargo, su variante será roquero, -a y no rockero, -a. Por el contrario, el mismos autor señala que los derivados foráneos de nombres propios de persona deben mantener la consonante de estos nombres: steimbeckiano de Steimbeck; kafkiano de Kafka; trotskista de Trotski; kantiano de Kant, etc. Se mantiene, de esta forma, la k original. 1. 4. La secuencia gráfica ll Los extranjerismos con la grafía ll que representan un sonido idéntico a nuestro fonema /l/, si se adaptan al español tomando como referencia la pronunciación, deben transformar la ll etimológica en l: a capela por a cappella, balotaje por ballotage, 64 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS béisbol o beisbol por baseball, dril por drill, pulóver por pullover, salmonela por Salmonella, etc. Además, el dígrafo ll no aparece nunca en español en posición final de sílaba. Así pues, los extranjerismos que se adapten al español deben sustituir la ll final de palabra o de sílaba por una l: buldócer por bulldozer, kril por krill, overol por overall etc. 1. 5. Las grafías m y n En español, el sistema gráfico señala que ante p o b se debe escribir m y no n. De este modo, esta misma norma debe aplicarse a los extranjerismos cuando son adaptados al español. Esta norma se debe seguir aunque en su forma original aparezca una n. Así se señala en la Real Academia Española (2010c: 90), donde se nos da ejemplos de ello: […] debe aplicarse a los extranjerismos, incluidos los topónimos foráneos y sus gentilicios, si se adaptan al español: lumpemproletariado (del alemán Lumpenproletaria), Brandemburgo (en alemán Brandenburg), brandemburgués, Camberra (del inglés Canberra), camberrano. Sin embargo, desde la Ortografía se nos indica también que la secuencia nb se mantiene en los apellidos extranjeros y sus derivados, ya que en este caso se suele mantener la grafía original: Gutenbert, Hartzenbusch, Schönberg, schönbergiano, Steinbeck, steinbeckiano. Se puede escribir n ante el fonema /b/ cuando se representa con la letra v. Este sería el caso de convoy (del francés convoi con el significado principal de ‘acompañamiento o séquito’ y ‘tren como medio de transporte’) o el de tranvía, procedente de la palabra inglesa tramway. Por otro lado, la nasal que se articula ante el fonema /f/ es /n/ y no /m/. Con lo cual, todos los extranjerismos que originalmente contengan -mph- o -mf- deben adaptarse al español con n y no con m: anfetamina por amphetamine, anfibol14 por amphibole, panfleto por pamphlet, etc. De esta manera, formas como comfort no serían correctas en español, debería aparecer como confort. 14 Con el significado de ‘mineral compuesto de sílice, magnesia, cal y óxido ferroso, de color por lo común verde o negro, y brillo anacarado’. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 65 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS 1. 6. La grafía q Hablamos ahora de la letra q (sin formar dígrafo con la vocal u), para la representación del fonema /k/. En un intento de regularizar la escritura española, en 1815, se determinó que anglicismos como quark y quásar, que poseían la letra q, se escribieran con «cu + vocal». Así, se aconseja escribir estos préstamos de la siguiente forma: cuark, cuásar. Así pues, estas formas ya están plenamente difundidas en la lengua española15. La Real Academia Española (2010c: 616) incluye los siguientes ejemplos: Según algunos expertos, la poderosa luz del cuásar sería el alarido de muerte de estrellas devoradas por un agujero negro. Los investigadores han presentado pruebas decisivas de este nuevo estado de la materia en el que los cuarks, que son probablemente los componentes más pequeños de la materia […], se desplazan en forma libre. El hecho de añadir una grafía más, la q, a las tres que ya existen para representar el fonema /k/ (el dígrafo qu y las letras c y k16), provocaba dispersión en la utilización de los mismos. Por ello, como dictaba la norma establecida en 1815, citada más arriba, para mantener la coherencia y simplicidad del sistema ortográfico español, la Real Academia Española (2010c: 115) recomienda: Todos aquellos préstamos de otras lenguas (sean latinismos o extranjerismos) cuya grafía etimológica incluya una q con valor fónico independiente se adapten por completo al español sustituyendo dicha q por las grafías hoy asentadas en nuestra lengua para representar el fonema /k/ […]. En caso de mantener la q etimológica, estas voces deben considerarse extranjerismos o latinismos no adaptados y escribirse, por ello, en cursiva y sin tildes: quadrivium, quark, quasar, quiorum, exequatur. 1. 7. La secuenciación «s inicial + consonante» En español no resulta natural empezar una palabra con s + consonante. Por ello, cuando una voz extranjera con esta secuencia se adapta a nuestra lengua, lo hacen 15 En especial cuásar. Incluso hoy más frecuente que la forma etimológica. 16 Para representar el fonema /k/ existe también los dígrafos ck y cq. Estos no pertenecen a ningún sistema gráfico del español. Así pues, si los extranjerismos que los incluyen se adaptan al español lo 66 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS añadiendo una e inicial que sirve como apoyo en la pronunciación: escáner por scanner, eslalon por slalom, esmoquin por smoking, espagueti por spaguetti, estándar por standard, estor por store, estrés por stress etc17. También contamos con topónimos en español adaptados de otras lenguas que incluyen la e- inicial de apoyo, serían los casos de: Estocolmo (en sueco Stockholm) o Estrasburgo (en francés Strasbourg). Aun así, la Real Academia Española (2010c: 126) señala que la s líquida inicial se mantiene en la forma de topónimos extranjeros. El problema viene en sus derivados, donde encontramos algunas excepciones18: La s líquida inicial se mantiene, no obstante, en la grafía de algunos topónimos foráneos, como Skopie (capital del país cuya denominación oficial provisional es Antigua República Yugoslava de Macedonia) o Sri Lanka (nombre actual de la antigua Ceilán), aunque en los derivados, cuando existen, sí se añade la e- de apoyo: esrilanqués (gentilicio de Sri Lanka). En los derivados de antropónimos, en cambio, aunque en algunos casos se han fijado en el uso grafías con e- inicial, como estalinismo, estalinista (de Stalin, dirigente de la antigua URSS) o estajanovismo, estajanovista (de Stajanov, minero soviético), lo normal es que se mantenga sin cambios la grafía del nombre propio: spengleriano (de Spengler, filósofo alemán), stendhaliano (de Stendhal, escritor francés). Finalmente, esta e- de apoyo que encontramos en la secuencia con s, también la usamos en español en otras secuencias consonánticas iniciales extranjeras como: embayá por mbayá19. 1. 8. El dígrafo sh El dígrafo sh que representa el fonema /sh/ no pertenece al sistema gráfico español actual. Así pues, las voces que los poseen deben considerarse como extranjerismos crudos y deben escribirse en cursiva. Ejemplos de ello son: flash, show, sushi, etc. deben hacer sustituyendo estos dígrafos por las grafías propias de nuestro sistema. Por ejemplo: bloc por block; coctel o cóctel por cocktail; críquet por cricket etc. 17 Según señala la Ortografía, se trata de un fenómeno bastante común en español, desde sus orígenes, como se puede observar en las numerosas voces patrimoniales que proceden de palabras latinas con s líquida: escala por scala, escena por scena, espejo por specŭlum etc. 18 También encontramos la s líquida en las locuciones latinas que se usan en español con su grafía originaria, como statu quo o stricto sensu. 19 Del guaraní. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 67 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS Cuando estas voces se han adaptado al español, lo hacen sustituyendo el dígrafo por alguna forma de nuestro sistema gráfico, normalmente ch o s: chute por shoot, hachís por hashish, champú por shampoo etc. Un ejemplo significativo es el anglicismo short(s) que según la Real Academia Española (2010c: 127) en América «comienza a escribirse en la forma adaptada chor (pl. chores) e incluso en la forma diminutiva chorcito(s); o con el anglicismo flash, que en España puede verse escrito en la forma adaptada flas» Excepcionalmente, aparece la sh etimológica en algunos topónimos que mantienen la grafía propia de su lengua de original, como por ejemplo Washington (capital de los Estados Unidos) o Shanghái (ciudad de China). Este uso se extiende al de los gentilicios, como sería el caso de washingtoniano. El caso de estos ejemplos no significa que la Real Academia Española (2010c: 128) recomiende adaptar en lo posible este dígrafo a una forma española: La existencia de estas grafías no obsta para que se recomiende escribir con simple s muchos topónimos que contienen sh en su grafía originaria, o en la de otras lenguas como el inglés o el francés, pero que en español se pronuncian normalmente con /s/, como es el caso, entre otros, de Bangladés o Ingusetia, grafías más recomendables que Bangladesh e Ingushetia. Por otra parte, en los nombres de persona y en su derivado sí se mantienen la sh en formas extranjeras: shakespeariano (de Shakespeare, dramaturgo inglés), sherlockiano (de Sherlock Holmes, famoso detective de ficción). Finalmente, Gómez Torrego (2011: 32) se hace eco de este fenómeno y, además, señala que el fenómeno de la adaptación del grupo sh en ch o s podría crear un precedente en las voces siguientes: chou (de show), aunque es preferible la palabra espectáculo; chouman (de showman), si no triunfaran las propuestas del Diccionario panhispánico de dudas: hombre espectáculo, presentador o animador. 68 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS 1. 9. La grafía w Según la Ortografía, la letra w, que no existía en latín, entró en español por la vía del préstamo20. Se emplea en voces procedentes de otras lenguas, en las que representa dos fonemas distintos, el vocálico /u/ y el consonántico /b/. Como la w se consideró una letra extranjera durante mucho tiempo en nuestro sistema ortográfico, cuando se pretendía adaptar una voz extranjera con esta grafía a nuestro sistema gráfico se realizaba con otras grafías, tales como gu, u o v: la obra señala como ejemplos los casos de güelfo21, del alemán Welf; vagón, del inglés wagon; o suéter, del inglés sweater. Por otro lado, en la actualidad, al ser considerada como letra perteneciente al abecedario español, se entiende que se conserve en los préstamos la grafía w. Por ello, encontramos los casos de: waterpolo, sándwich, web, etc. Algunas veces, en el fonema vocálico /u/, se dan por válidas ambas grafías; sería el caso de taekwondo o taekuondo, lawrencio o laurencio22, etc. Así pues, hasta que la letra w fue considerada como parte del sistema gráfico español, se adaptaban los préstamos sustituyendo la w por letras españolas como v, u o gu. En la actualidad, desde la Ortografía se aboga por el uso de la w etimológica, ya que todos los hablantes hispanos la consideramos hoy en día como propia. Aun así, en el transcurso de estos dos períodos hay préstamos que identificamos como propios por haberse adaptado ya plenamente a nuestra lengua, estos serían los casos de váter, vagón, suéter o vatio. Como ejemplo de este fenómeno, se explicaría que durante algún tiempo el diccionario académico mostrase únicamente la forma darvinismo como adaptación del inglés Darwinism. Aun así, hoy en día se admite solo la forma darwinismo con la grafía etimológica. Gómez Torrego (2011: 30) recoge también este fenómeno y habla, en particular, de dos casos: el de kiwi y el de wiski. En el primer caso, el autor señala que se desecha la forma kivi en favor de kiwi, aunque añade que se prefiere la pronunciación con u semiconsonántica [kíui]. 20 La letra w se incorporó oficialmente al abecedario español en 1969. En origen fue un dígrafo por duplicación de la v latina. En español, durante la Edad Media, se empleó inicialmente para determinados nombres propios de origen germánico. 21La Real Academia Española (2001) indica que su uso viene de la Edad Media y lo define como: ‘partidario de los papas contra los gibelinos, defensores de los emperadores de Alemania'. 22 Según la Real Academia Española (2001) ‘elemento químico transuránico de núm. atóm. 103. Se obtiene artificialmente por bombardeo de californio con iones de boro, pertenece a la serie de los actínidos, y su vida media es de ocho segundos’. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 69 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS En el segundo caso, Gómez Torrego señala que se propone por primera vez la forma wiski como adaptación de whisky o whiskey, en lugar de la anterior adaptación güisqui23. La justificación la encuentra por su aparición en documentos literarios. 1. 10. La grafía z ante e, i En español, algunas palabras extranjeras introducen términos que empiezan con z ante e y i24. Estas formas se mantienen porque estaban en su grafía originaria o en su transcripción a nuestro alfabeto. Ejemplos de este fenómeno sería: kamikaze, nazi, zéjel, zen, zepelín, zigurat25, zigzag, zíper, etc. Entre estas formas, también encontramos topónimos y antropónimos originarios de otras lenguas y sus derivados. La Real Academia Española (2010c: 124) aporta ejemplos sobre este fenómeno: Azerbaiyán (y sus gentilicios azerbaiyano y azerí), Nueva Zelanda (y su gentilicio neozelandés), Suazilandia (y su gentilicio suazí), Zimbabue (y su gentilicio zimbabuense), Elzevir (apellido de una célebre familia de impresores holandeses, y sus derivados elzevir o elzevirio y elzeviriano), Ezequiel, Zenón, Zeus. Pese a ello, estas formas son ajenas a la lengua española. En las voces patrimoniales españolas suele aparecer en este contexto la letra c. Por lo tanto, según la Real Academia Española (2010c: 618) «no resulta extraño encontrar variantes gráficas en muchos extranjerismos. Junto a la grafía que mantiene la z etimológica, presentan grafías con c, plenamente adaptadas a la ortografía del español». Así pues, estas variantes han aparecido por la convivencia en el uso de grafías etimológicas con z, junto a grafías adaptadas al sistema gráfico del español que utiliza c. La Ortografía incluye algunos ejemplos de esta alternancia: acimut/azimut (‘ángulo que con el meridiano forma el círculo vertical que pasa por un punto de la esfera celeste o del globo terráqueo’); cigoto/zigoto (‘célula resultante de la unión del gameto masculino con el femenino’); 23 Se refiere a la adaptación que propuso la RAE (2001) y el DPD. 24 Especialmente cultismos griegos, arabismos y préstamos de otras lenguas. 70 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS cinc/zinc (‘cierto metal’); cíngaro -ra/zíngaro –ra (‘gitano’); circón/zircón (‘silicato de circonio’); eccema/eczema (‘afección cutánea que produce descamación en la piel’); magacín/magazín (‘revista ilustrada sobre temas diversos’ y ‘programa de radio o televisión de contenido muy variado’). 1. 11. Las terminaciones con -age/-aje En la lengua española, hemos contado con voces francesas terminadas en -age (pronunciación [áʒ]), que se han adaptado al español con la grafía -aje (pronunciación [áje]). Actualmente, cuando se adaptan al español este tipo de galicismos también lo hacemos con la terminación -aje: bricolaje por bricolage, garaje por garage, etc. La Real Academia Española (2010c: 102) incluye más ejemplos de esta adaptación: «aterrizaje por atterrissage, brebaje por breuvage, camuflaje por camouflage, chantaje por chantage, espionaje por espionnage, masaje por massage, potaje por potage, tatuaje por tatouage». La Ortografía señala que si se utilizan estas voces francesas sin adaptar, con su grafía y pronunciación originarias, deben escribirse en cursiva, para destacar así su situación de extranjerismo crudo. Así pues, en muchas zonas de América, se prefiere la forma cruda a la adaptación española. Por ejemplo, en la voz garage (pronunciado [garáӡ]), en lugar de la adaptación garaje (pronunciado [garáje]). 1. 12. Las terminaciones con -ing Tradicionalmente, las voces inglesas con la terminación -ing se han adaptado a la lengua española cambiando esta terminación por -in. Este hecho se ha dado porque la grafía no se representaba en la pronunciación de los hispanohablantes. Ejemplos de esta adaptación serían: mitin por meeting, pudin o pudín por pudding, esmoquin por smoking, etc. 25 En arquitectura: ‘Torre escalonada y piramidal, característica de la arquitectura religiosa asiria y caldea’. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 71 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS También encontramos esta adaptación en los anglicismos actuales acabados en -ing. Se puede dar por su falta de equivalentes en la propia lengua, o por su arraigo y generalización en el uso actual: campin por camping, castin por casting26, cáterin por catering, márquetin por marketing27, pirsin por piercing etc. Finalmente, Gómez Torrego (2011: 31) señala que este es el camino que podrían seguir otros anglicismos si se decidiera su adaptación al español. E incluye los siguientes ejemplos: bulin (de bullying), si no triunfara un posible sustituto como el de acoso escolar (aparece también la forma bulling en los medios); estrechin (de stretching28); fisin (de phishing29); futin (falso anglicismo formado en español con foot e -ing), si no triunfaran las formas aerobismo o el infinitivo trotar propuestas en el Diccionario panhispánico de dudas; lisin (de leasing30); mobin (de mobbing), si no triunfara la propuesta del Diccionario panhispánico de dudas acoso laboral; puentin (forma híbrida formada con el sustantivo español puente y la terminación inglesa -ing), en el caso de que no triunfe el sustituto español puentismo propuesto en el DPD; rentin (de renting); surfin (de surfing), como variante de surf; yoguin (de jogging). Gómez Torrego, en algunas de estas formas, cuando indica una hipótesis de adaptación, lo hace diciendo que esa hipótesis se dará si no triunfa el equivalente español propuesto por la Real Academia Española. Esto lo señala porque la RAE aconseja siempre el uso del equivalente español, por encima de la propia adaptación, para mantener la integridad de la lengua. Por ello, por ejemplo, si la forma española puentismo acaba triunfando, no será necesario utilizar ni la forma cruda puenting, ni su 26 Aunque se prefiere el equivalente español audición. 27 Aunque se prefiere el equivalente mercadotecnia. 28 Con el significado de ‘estiramiento’. 29 Con el significado de ‘engaño informático’. 30 Con el significado de ‘contrato de arrendamiento de medios de producción’. 72 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS hipotética adaptación puentin. Lo mismo sucedería con las forma mobbing (o su adaptación mobin), innecesarias si triunfa la voz española que aconseja el DPD acoso laboral, o el extranjerismo bullying (con su adaptación bullin) en favor de la forma española acoso escolar. Llamativo es el caso de footing (plenamente adaptada al español) donde el DPD aconseja las formas aerobismo o trotar. El tiempo dirá qué formas acaban triunfando31. 1. 13. Las voces terminadas en -y precedidas de consonante Las palabras terminadas en y precedida de consonante son propias de los préstamos, ya que en español, solo se admite la terminación en y cuando va precedida de vocal. Las palabras terminadas en y precedidas de consonante suelen pertenecer a la lengua inglesa: curry, dandy, derby, ferry, panty, penalty, pony etc. Ya que este fenómeno es ajeno al español, cuando se quieren adaptar estos préstamos, lo deben hacer sustituyendo la y final por i: curri por curry, dandi por dandy, derbi por derby, ferri por ferry, panti por panty, penalti por penalty, poni por pony, sexi por sexy etc. Por otro lado, tampoco es propio del español actual las palabras con y con valor vocálico en posición interior de palabra. De esta forma, también es normal que cuando se adapten estas formas al español, lo hagan con i en lugar de la y: géiser por geyser, pijama o piyama por pijamas. Resulta una excepción el término byte, ya que según la Real Academia Española (2010c: 81): «El término byte (pronunciado [báit]) es un extranjerismo crudo o no adaptado, motivo por el cual conserva su grafía y su pronunciación originarias». Por el mismo motivo, cuando un término acabado en y forma su plural (añadiendo una s), lo debe hacer, también, cambiando la y por i, ya que conserva en el plural su valor vocálico: así el plural de jersey debe ser jerséis; el de espray, espráis; el de gay, gais, etc. En cambio, si para añadir el plural se utiliza -es, se conserva la y del singular, ya que en el plural representa el fonema consonántico /y/: bueyes (de buey), convoyes (de convoy), virreyes (de virrey), etc. La Ortografía sí permite que se mantenga la y final si esta va precedida de una o dos vocales con las que forma un diptongo o un triptongo: bocoy, tepuy, yóquey etc. 31 Normalmente, según Giménez Folqués (2010), suelen influir factores como la moda lingüística, la comodidad en la pronunciación y la influencia de los medios de comunicación. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 73 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS En ciertos términos extranjeros se admiten ambas grafías, con -y y con -i: bonsái/bonsay, paipái/paipay, samurái/samuray y tipói32/tipoy. 1. 14. Las consonantes dobles Las consonantes dobles o geminadas en español son rechazadas si dicha duplicación no tiene reflejo en la pronunciación. Así, los préstamos de otras lenguas en cuya grafía originaria contengan dos consonantes iguales seguidas, si son adaptadas al español lo hacen simplificando las dos consonantes a una. La Ortografía incluye varios ejemplos de esta simplificación: driblar de to dribble, brócoli de broccoli, racor de raccord, adenda de addenda, pudin o pudín de pudding, chofer o chófer de chauffeur, esnifar de to sniff, grogui de groggy, zigurat de ziggurat, a capela de a cappella, chambelán de chambellan, consomé de consommé, canelón de cannellone, escáner de scanner, chóped de chopped, dosier de dossier, estrés de stress, confeti de confeti, cúter de cutter, esbozar de sbozzare, o puzle de puzzle. Por ello, en las nuevas adaptaciones debe aplicarse la misma pauta: cadi de caddie, chédar de cheddar, pádel de paddle, suflé de soufflé, rali de rally, osobuco de ossobuco, mozarela de mozzarella. Así pues, los extranjerismos que no aparecen adaptados, como es el caso de pizza o jazz, se considerarán extranjerismos crudos. También utilizamos este tipo de adaptación en los topónimos foráneos: Adís Abeba (capital de Etiopía, de Addis Ababa), Búfalo (ciudad de EE.UU., de Buffalo), Lausana (ciudad de Suiza, de Lausanne), Mesina (ciudad de Italia, de Messina), Misisipi (río y estado de EE.UU., de Mississippi) o Róterdam (ciudad de los Países Bajos, de Rotterdam). Según la Real Academia Española (2010c: 179) existen algunas excepciones donde se mantienen las consonantes dobles: Excepcionalmente, se mantienen las secuencias de dos grafemas consonánticos iguales en los nombres de algunas letras del alfabeto griego (y en sus derivados), que son transcripción literal del original griego: kappa, digamma y gamma (y sus derivados gammagrafía y gammaglobulina). También se conserva la doble consonantes etimológica en antropónimos o en topónimos foráneos no adaptados, y en sus derivados. Aunque no necesariamente la 32 En América: ‘túnica larga de manga muy corta y escote cuadrado’. 74 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS grafía tenga reflejo en la pronunciación: heideggeriano (de Heidegger, filósofo alemán), popperiano (de Popper, filósofo austriaco), picassiano (de Picasso, pintor español), Ottawa (capital de Canadá) y su gentilicio ottawense, etc. Por otro lado, nombres propios de persona pueden presentar alternancia con la doble consonante y su simplificación: Gemma/Gema, Emma/Ema o Emmanuel/Emanuel33. 2. CONCLUSIONES Como hemos podido comprobar, la primera conclusión que podemos extraer del trabajo de la Ortografía en el uso de extranjerismos es la necesidad de acercar estas voces al sistema ortográfico español. Por ello, se intenta dejar de lado las grafías que resultan ajenas, como por ejemplo las consonantes dobles (broccoli), la terminación -ing (renting), el dígrafo sh (shampoo), o la secuenciación «s + consonante» (spaguetti); por otras grafías más cercanas al sistema español: brócoli, rentin, champú, espagueti. Con el paso de las últimas obras académicas, hemos podido ir observando cómo se ha ido radicalizando la postura en torno a los extranjerismos crudos. No prohíben su uso, pero sí se inclinan hacia los equivalentes en español, si los hubiere, o en su defecto a las adaptaciones a formas españolas. De esta manera, tenemos como resultado dos clases de extranjerismos, los que se adaptan al español y los que se resisten a adaptarse manteniendo su forma original. Esta resistencia suele provenir de varios factores, como el prestigio de la palabra, su extensión en el uso o por la influencia de los medios de comunicación. Los que sí son adaptados lo hacen por medio de varios procedimientos según señala la Real Academia Española (2010c: 602): La mayor parte de las veces, la adaptación de los extranjerismos se realiza modificando la grafía originaria para adecuarla, según nuestras reglas ortográficas, a la pronunciación de esas voces en español, que suele aproximarse a la que tienen en la lengua de origen. En la grafía adaptada se prescinde normalmente de los grafemas del original que no tienen reflejo en la dicción española y se aplican las reglas de acentuación gráfica propias de nuestro idioma […]. 33 También, según la Real Academia Española (2010c: 179) conserva la doble consonante el prefijo del sistema internacional de unidades atto-, proveniente de la palabra noruega y danesa atten LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 75 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS A veces la adaptación solo requiere la aplicación de la tilde, como en ambigú (del fr. ambigu), bádminton (del ingl. badminton), blíster (del ing. blister) o máster (del ingl. master). Y, en algunas ocasiones, el extranjerismo no plantea ningún problema de inadecuación entre grafía y pronunciación de acuerdo con la ortografía del español, y se incorpora a nuestra lengua con la misma grafía que tiene en el idioma de origen, como ocurre con box, kit o set, voces tomadas del inglés que se pronuncian en español tal como se escriben: [bóks], [kit] y [sét]. En otros casos, se mantiene la grafía originaria sin cambios o con leves modificaciones, y es la pronunciación de los hispanohablantes la que se acomoda a dicha grafía, aunque al hacerlo se aparte de la pronunciación original de la voz extranjera. Lo que sí queda claro es que si se utiliza la grafía extranjera original, esta debe ir marcada en cursiva. Mientras que, cuando el extranjerismo se ha adaptado con la grafía propia española, esta ya puede aparecer en letra redonda. Por otro lado, en cuanto a estas adaptaciones se refiere, Gómez Torrego (2011), además de recogerlas junto con las propuestas de la Ortografía de la lengua española, también propone previsiones de futuras adaptaciones según los criterios observados en la Real Academia Española. Sería el caso de la voz show, donde según la Real Academia las palabras que contienen el dígrafo sh se deben adaptar con grafías propias españolas, así pues, el autor propone la voz chou. Lo mismo sucede con las palabras terminadas en -ing y con la palabra renting en favor de rentin (voz propuesta por Gómez Torrego). Aunque se trate de seguir unas pautas en cuanto a la adaptación de ortografías extranjeras al sistema ortográfico español, muchas veces, es inevitable encontrarse con excepciones que complican esta sistematización. Sería el caso de los topónimos Washington y Shanghái que mantienen el dígrafo sh; o cuando se emplea b en posición final de palabra pero nos encontramos con las voces eslavas: molotov o lev que acaban en v. Algunas de estas excepciones vienen por la permisividad que da la Real Academia Española (2010c: 37) a algunas formas etimológicas: En la configuración del sistema ortográfico del español ha operado también, aunque con menor incidencia que en lenguas como el inglés o el francés, el criterio etimológico, según el cual, en la escritura de las palabras, debe respetarse en alguna medida la forma (‘dieciocho’), que se aplica a nombres de unidades de medida para formar los nombres de los submúltiplos un trillón de veces inferiores, como en attosegundo. 76 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS gráfica de su étimo, es decir, del término del cual derivan. Este criterio funciona, en muchos casos, en sentido opuesto al fonológico, y explica por qué la forma escrita de determinadas palabras contradice el principio básico de adecuación entre grafía y pronunciación […]. La aplicación del criterio etimológico explica también muchas de las excepciones a las reglas generales (por ejemplo, la presencia de z ante las vocales e, i en la escritura de ciertas palabras por figurar dicha letra en su étimo, como en zeugma o nazi), y es asimismo la causa de la mayor parte de las desviaciones del ideal de correspondencia biunívoca entre grafemas y fonemas que hay en nuestro sistema ortográfico, desde la presencia de un grafema, como la h, sin valor fonológico, hasta la existencia de varias posibilidades gráficas para representar un mismo fonema (b, v y w para /b/; j y g para /j/, etc.) De hecho, el criterio etimológico sigue operando hoy a la hora de fijar la grafía de aquellas palabras que contienen fonemas que admiten varias representaciones gráficas, pues lo habitual es respetar, en esos casos, los grafemas etimológicos; así, voces como káiser, anorak o búnker se escriben con k (y no con c ni con qu) por ser esa la letra que aparece en su étimo. Así pues, están claras las posturas de la Real Academia Española y la Asociación de Academias en todas las obras académicas que acabamos de estudiar en torno al uso de los extranjerismos en un futuro. El control y la revisión de los mismos serán la constante que encontraremos. Dentro de este control, la españolización de estas voces será el objetivo primordial, ya que de esta forma, se aseguran la integridad y el equilibrio de la lengua en el afán de mantener el ideal de unidad lingüística entre las diferentes zonas hispanohablantes. ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS BIBLIOGRAFÍA ALEZA IZQUIERDO, Milagros (2001): «Sobre la presencia de voces de origen amerindio en la última edición del Diccionario de la Lengua Española», en HERNÁNDEZ, César, Actas del VI Congreso Internacional, El español de América, Instituto Interuniversitario de Estudios de Iberoamérica y Portugal, Universidad de Valladolid (Tordesillas, octubre 2005). ALEZA IZQUIERDO, Milagros y José María ENGUITA UTRILLA (2002): El español de América: aproximación sincrónica, Valencia, Tirant lo Blanch. ALEZA IZQUIERDO, Milagros (2005): «Vacilaciones gráficas en el uso de los extranjerismos en la prensa de la Comunidad Valenciana», Quaderns de Filologia. Estudis Lingüístics 10, 12-29. ALEZA IZQUIERDO, Milagros, coord.ª (2006): Lengua española para los medios de comunicación: usos y normas actuales, Valencia, Tirant lo Blanch. ALEZA IZQUIERDO, Milagros (2008): «Sobre la presencia de voces de origen extranjero en el DPD» en Álvarez Tejedor, Antonio et alii (eds.), Lengua viva. Estudios ofrecidos a César Hernández Alonso, Valladolid, Universidad de Valladolid y Diputación de Valladolid, 255-270. ALEZA IZQUIERDO, Milagros coord.ª (2010a): Normas y usos correctos en el español actual, Valencia, Tirant lo Blanch. ALEZA IZQUIERDO, Milagros, coord.ª (2010b): La lengua española en América: normas y usos actuales en línea, Valencia, Universitat de València. ALVAR EZQUERRA, Manuel (1999): Manual de redacción y estilo, Madrid, Istmo. BOSQUE, I. & DEMONTE, V., drs. (1999): Gramática descriptia de la lengua española. Real Academia Española, Colección Nebrija y Bello, Madrid, Espasa Calpe. Cano Aguilar, R. (1999): El español a través de los tiempos, Madrid, Arco/Libros. CABRÉ, M. Teresa (2002): «La neologia efímera», Lèxic i neologia, Barcelona, Observatori de Neologia, Institut Universitari de Lingüística Aplicada, Universitat Pompeu Fabra, 13-28. CANO AGUILAR, R. (1999): El español a través de los tiempos, Madrid, Arco/Libros. 78 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS ENGUITA UTRILLA, José María (2002): «Léxico», en Aleza, Milagros y Enguita, José María, El Español de América: aproximación sincrónica, Valencia, Tirant lo Blanch, 203-234. ESTORNELL PONS, María (2009): Neologismos en la prensa. Criterios para reconocer y caracterizar las unidades neológicas, Valencia, Universitat de València. GIMÉNEZ FOLQUÉS, David (2010): «Introducción a los extranjerismos del Diccionario panhispánico de dudas y criterios de adaptación» en Aleza Izquierdo, Milagros coord.ª , Normas y usos correctos en el español actual, Valencia, Tirant lo Blanch, págs. 217-228. GIMÉNEZ FOLQUÉS, David (2011a): Normativa académica, adaptación y uso de los extranjerismos en el español actual, Valencia, Universitat de València (Tesis doctoral). GIMÉNEZ FOLQUÉS, David (2011b): «Innovaciones académicas actuales en la ortografía de los extranjerismos en la lengua española » en Aleza Izquierdo, Milagros coord.ª , Normas. Revista de estudios lingüísticos hispánicos. Número 1. Valencia, Universitat de Valéncia. GIMENO MENÉNDEZ, Francisco y GIMENO MENÉNDEZ, M. Victoria (2003): El desplazamiento lingüístico del español por el inglés, Madrid, Cátedra. GÓMEZ CAPUZ, Juan (1992): «La problemática de los extranjerismos en los libros de estilo: purismo y defensa de la lengua», en Laguna, A. y A. López García., eds. Dos-cents anys de promesa valenciana. I Congrés Internacional de Periodisme. Actes, Valencia, Generalitat Valenciana, 899-909. GÓMEZ CAPUZ, Juan (1996): «Observaciones sobre la función de los extranjerismos en español coloquial: valores estilísticos, semánticos y pragmáticos», en Briz, Antonio; J. Ramón Gómez y M. J. Martínez, Pragmática y gramática del español hablado. Actas del II Simposio sobre análisis del discurso oral, Departamento de Filología Española/Libros Pórtico, Valencia/Zaragoza, 305-310. GÓMEZ CAPUZ, Juan (1997a): Anglicismos en español actual: su estudio en el registro coloquial, Facultat de Filologia de la Universitat de València. Servei de Publicacions de la Universitat de València. GÓMEZ CAPUZ, Juan (1997b): «Aspectos etimológicos y diacrónicos de los préstamos en la lexicografía española», leída en las Jornadas La lengua española: pasado y presente, celebradas en la Universidad de Valencia. GÓMEZ CAPUZ, Juan (1998): El préstamo lingüístico (conceptos, problemas y métodos), Anejo XXIX de la Revista «Cuadernos de Filología», Valencia, Universitat. GÓMEZ CAPUZ, Juan (2004): Préstamos del español: lengua y sociedad, en la colección «Cuadernos de Lengua Española», Madrid, Arco/Libros. LOS EXTRANJERISMOS EN EL ESPAÑOL ACADÉMICO DEL SIGLO XXI 79 ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS GÓMEZ CAPUZ, Juan (2005): La inmigración léxica, Madrid, Arco/Libros, S.L. GÓMEZ TORREGO, Leonardo (1995): El léxico en el español actual: Uso y norma, Madrid, Arco. GÓMEZ TORREGO, Leonardo (2000): Ortografía del uso del español actual, Madrid, Ediciones SM, 2000. GÓMEZ TORREGO, Leonardo (2007): Hablar y escribir correctamente, Gramática normativa del español actual (II), Madrid, Arco / Libros, S.L. GÓMEZ TORREGO, Leonardo (2011): Ortografía y gramática. Las normas académicas: últimos cambios, Madrid, Ediciones SM. GUERRERO SALAZAR, Susana y NÚÑEZ CABEZAS, Emilio Alejandro (2002): Medios de comunicación y español actual, Málaga, Aljibe. HERNANDO CUADRADO, L. A. (2006): «La creación neológica y las nuevas tecnologías» In: Vilches, F., coord., Creación léxica y nuevas tecnologías. Nuevos discursos, Madrid: Universidad Rey Juan Carlos, 311-362. LORENZO, EMILIO (1996): Anglicismos hispánicos, Madrid, Gredos. MARTÍNEZ DE SOUSA, José (2000): Manual de estilo de la lengua española, Gijón, Trea. MARTÍNEZ DE SOUSA, José (2001): Diccionario de edición, tipografía y artes gráficas, Gijón, Trea. MARTÍNEZ DE SOUSA, José (2004): Ortografía y ortotipografía del español actual, Gijón, Ediciones Trea. MARTÍNEZ DE SOUSA, José (2005): Manual de edición y autoedición, Madrid, Pirámide, 2.ª ed. MARTÍNEZ DE SOUSA, José (2007): Manual de estilo de la lengua española, Gijón, Ediciones TREA, 3.ª ed. MEDINA LÓPEZ, Javier (1996): El anglicismo en el español actual, Madrid, Arco. MEDINA LÓPEZ, Javier (1997): Lenguas en contacto, Madrid, Arco/Libros, S.L. MOLINER, María (2007): Diccionario de uso del español, Madrid: Gredos. MONTOLÍO, Estrella, coord.ª (2000): Manual práctico de escritura académica, Barcelona, Ariel Practicum. OBSERVATORI DE NEOLOGIA (2002a): La neologia en el tombant de segle, Institut Universitari de Lingüística Aplicada, Barcelona, Papers de iula. OBSERVATORI DE NEOLOGIA (2002b): Lèxic i neologia, Barcelona, Institut Universitari de Lingüística Aplicada, Universitat Pompeu Fabra. REAL ACADEMIA ESPAÑOLA (1999): Ortografía de la lengua española, Madrid, Espasa. 80 DAVID GIMÉNEZ FOLQUÉS ANEJO 2 DE NORMAS. REVISTA DE ESTUDIOS LINGÜÍSTICOS HISPÁNICOS REAL ACADEMIA ESPAÑOLA (2001): Diccionario de la Lengua Española, Madrid, Espasa. REAL ACADEMIA ESPAÑOLA Y ASOCIACIÓN DE ACADEMIAS DE LA LENGUA ESPAÑOLA, (2005): Diccionario panhispánico de dudas, Madrid, Santillana. REAL ACADEMIA ESPAÑOLA Y ASOCIACIÓN DE ACADEMIAS DE LA LENGUA ESPAÑOLA, (2009): Nueva gramática de la lengua española, Madrid, Santillana. Volúmenes 1 (Morfología y Sintaxis) y 2 (Sintaxis). REAL ACADEMIA ESPAÑOLA Y ASOCIACIÓN DE ACADEMIAS DE LA LENGUA ESPAÑOLA, (2010a): Nueva gramática de la lengua española, Madrid, Santillana. Volúmenes de Fonética y fonología. REAL ACADEMIA ESPAÑOLA Y ASOCIACIÓN DE ACADEMIAS DE LA LENGUA ESPAÑOLA, (2010b): Diccionario de americanismos, Madrid, Santillana. REAL ACADEMIA ESPAÑOLA Y ASOCIACIÓN DE ACADEMIAS DE LA LENGUA ESPAÑOLA (2010c): Ortografía de la lengua española, Madrid, Espasa. REAL ACADEMIA ESPAÑOLA Y ASOCIACIÓN DE ACADEMIAS DE LA LENGUA ESPAÑOLA (2011): Nueva gramática básica de la lengua española, Madrid, Espasa Libros. SANMARTÍN SÁEZ, Julia (2000): «Creación léxica (I): Neologismos semánticos: las metáforas de cada día» en Briz, A & Grupo Valesco, eds., ¿Cómo se comenta un texto coloquial?, Barcelona: Ariel, Practicum, 125-142. SANMARTÍN SÁEZ, Julia (2004): Diccionario de argot, Madrid, Espasa. SANMARTÍN SÁEZ, Julia (2008): «El neologismo en un corpus de prensa valenciana. ¿Un hecho diferencial?» (en prensa). SECO, Manuel et alii (1999): Diccionario del español actual, Madrid, Santillana. SECO, Manuel (2001): Diccionario de dudas y dificultades de la lengua española, Madrid, Espasa Calpe. TORRES TORRES, Antonio (2000): «El español en los Estados Unidos de América» en El español de América, Barcelona, Edicions de la Universitat de Barcelona, 87-116. VV.AA. (1999): El neologismo necesario, Madrid, Agencia efe. |
7314 | https://www.mathway.com/popular-problems/Calculus/596772 | Enter a problem...
Calculus Examples
Popular Problems
Graph y=(x^2)/(x-1)
Step 1
Find where the expression is undefined.
Step 2
Consider the rational function where is the degree of the numerator and is the degree of the denominator.
If , then the x-axis, , is the horizontal asymptote.
If , then the horizontal asymptote is the line .
If , then there is no horizontal asymptote (there is an oblique asymptote).
Step 3
Find and .
Step 4
Since , there is no horizontal asymptote.
No Horizontal Asymptotes
Step 5
Find the oblique asymptote using polynomial division.
Tap for more steps...
Step 5.1
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of .
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Step 5.2
Divide the highest order term in the dividend by the highest order term in divisor .
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Step 5.3
Multiply the new quotient term by the divisor.
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Step 5.4
The expression needs to be subtracted from the dividend, so change all the signs in
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Step 5.5
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
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Step 5.6
Pull the next terms from the original dividend down into the current dividend.
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Step 5.7
Divide the highest order term in the dividend by the highest order term in divisor .
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Step 5.8
Multiply the new quotient term by the divisor.
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The expression needs to be subtracted from the dividend, so change all the signs in
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After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
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Step 5.11
The final answer is the quotient plus the remainder over the divisor.
Step 5.12
The oblique asymptote is the polynomial portion of the long division result.
Step 6
This is the set of all asymptotes.
Vertical Asymptotes:
No Horizontal Asymptotes
Oblique Asymptotes:
Step 7
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7315 | https://dlmf.nist.gov/18.3 | Published Time: Wed, 16 Jul 2025 17:43:16 GMT
DLMF: §18.3 Definitions ‣ Classical Orthogonal Polynomials ‣ Chapter 18 Orthogonal Polynomials
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18 Orthogonal PolynomialsClassical Orthogonal Polynomials18.2 General Orthogonal Polynomials18.4 Graphics
§18.3 Definitions
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Keywords:Bochner’s theorem, Chebyshev polynomials, Hermite polynomials, Jacobi polynomials, Laguerre polynomials, Legendre polynomials, Rodrigues formulas, characterizations, classical orthogonal polynomials, degree lowering and raising, differentiation formulas, of coefficients, tables, tables of coefficients, ultraspherical polynomialsReferenced by:§1.17(iii), §1.17(iii), §1.17(iii), §10.23(ii), §10.54, §10.59, §12.7(i), §13.18(v), §13.6(v), §15.9(i), §15.9(ii), §15.9(iii), §18.2(ii), §18.34(iii), §18.35(iii), §18.38(ii), §28.8(ii), §28.9, §29.15(ii), §29.15(ii), §3.5(v), §3.5(v), §3.5(v), §3.5(v), §3.5(iv), §3.5(v), §31.11(iv), §31.16(ii), §32.10(iv), §34.3(vii), §7.10, §7.18(iv), §7.6(ii), §8.7, Erratum (V1.0.10) for References, Erratum (V1.2.0) §18.3Permalink: (effective with 1.2.0): Some descriptive text was added under Table 18.3.1. See also:Annotations for Ch.18
The classical OP’s comprise the Jacobi, Laguerre and Hermite polynomials. There are many ways of characterizing the classical OP’s within the general OP’s {p n(x)}, see Al-Salam (1990). The three most important characterizations are:
1.
As eigenfunctions of second order differential operators (Bochner’s theorem, Bochner (1929)). See the differential equations A(x)p n′′(x)+B(x)p n′(x)+λ np n(x)=0, in Table 18.8.1.
2.
With the property that {p n+1′(x)}n=0∞ is again a system of OP’s. See §18.9(iii).
3.
As given by a Rodrigues formula (18.5.5).
Table 18.3.1 provides the traditional definitions of Jacobi, Laguerre, and Hermite polynomials via orthogonality and standardization (§§18.2(i) and 18.2(iii)). This table also includes the following special cases of Jacobi polynomials: ultraspherical, Chebyshev, and Legendre.
Table 18.3.1: Orthogonality properties for classical OP’s: intervals, weight functions, standardizations, leading coefficients, and parameter constraints. In the second row 𝒜 n denotes 2 α+β+1Γ(n+α+1)Γ(n+β+1)/((2n+α+β+1)Γ(n+α+β+1)n!), with 𝒜 0=2 α+β+1Γ(α+1)Γ(β+1)/Γ(α+β+2). For further implications of the parameter constraints see the Note in §18.5(iii).
| Name | p n(x) | (a,b) | w(x) | h n | k n | k~n/k n | Constraints |
--- --- --- --- |
| Jacobi | P n(α,β)(x) | (−1,1) | (1−x)α(1+x)β | 𝒜 n | (n+α+β+1)n 2 nn! | n(α−β)2n+α+β | α,β>−1 |
| Ultraspherical(Gegenbauer) | C n(λ)(x) | (−1,1) | (1−x 2)λ−1 2 | 2 1−2λπΓ(n+2λ)(n+λ)(Γ(λ))2n! | 2 n(λ)n n! | 0 | λ>−1 2,λ≠0 |
| Chebyshev of first kind | T n(x) | (−1,1) | (1−x 2)−1 2 | {1 2π,n>0 π,n=0 | {2 n−1,n>0 1,n=0 | 0 | |
| Chebyshev of second kind | U n(x) | (−1,1) | (1−x 2)1 2 | 1 2π | 2 n | 0 | |
| Chebyshev of third kind | V n(x) | (−1,1) | (1−x)−1 2(1+x)1 2 | π | 2 n | −1 2 | |
| Chebyshev of fourth kind | W n(x) | (−1,1) | (1−x)1 2(1+x)−1 2 | π | 2 n | 1 2 | |
| Shifted Chebyshev of first kind | T n∗(x) | (0,1) | (x−x 2)−1 2 | {1 2π,n>0 π,n=0 | {2 2n−1,n>0 1,n=0 | −1 2n | |
| Shifted Chebyshev of second kind | U n∗(x) | (0,1) | (x−x 2)1 2 | 1 8π | 2 2n | −1 2n | |
| Legendre | P n(x) | (−1,1) | 1 | 2/(2n+1) | 2 n(1 2)n/n! | 0 | |
| Shifted Legendre | P n∗(x) | (0,1) | 1 | 1/(2n+1) | 2 2n(1 2)n/n! | −1 2n | |
| Laguerre | L n(α)(x) | (0,∞) | e−xx α | Γ(n+α+1)/n! | (−1)n/n! | −n(n+α) | α>−1 |
| Hermite | H n(x) | (−∞,∞) | e−x 2 | π 1 22 nn! | 2 n | 0 | |
| Hermite | 𝐻𝑒 n(x) | (−∞,∞) | e−1 2x 2 | (2π)1 2n! | 1 | 0 | |
ⓘ
Symbols:T n(x): Chebyshev polynomial of the first kind, W n(x): Chebyshev polynomial of the fourth kind, U n(x): Chebyshev polynomial of the second kind, V n(x): Chebyshev polynomial of the third kind, Γ(z): gamma function, H n(x): Hermite polynomial, 𝐻𝑒 n(x): Hermite polynomial, P n(α,β)(x): Jacobi polynomial, L n(α)(x): Laguerre (or generalized Laguerre) polynomial, P n(x): Legendre polynomial, (a)n: Pochhammer’s symbol (or shifted factorial), π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, !: factorial (as in n!), (a,b): open interval, T n∗(x): shifted Chebyshev polynomial of the first kind, U n∗(x): shifted Chebyshev polynomial of the second kind, P n∗(x): shifted Legendre polynomial, C n(λ)(x): ultraspherical (or Gegenbauer) polynomial, p n(x): polynomial of degree n, w(x): weight function, n: nonnegative integer, x: real variable and h nKeywords:Chebyshev polynomials, Hermite polynomials, Jacobi polynomials, Laguerre polynomials, Legendre polynomials, classical orthogonal polynomials, definition, definitions, leading coefficients, of the first, second, third, and fourth kinds, orthogonality properties, orthogonality property, parameter constraint, parameter constraints, shifted, standardization, standardizations, ultraspherical polynomials, weight function, weight functions, with respect to integrationA&S Ref:22.2.1, 22.2.3, 22.2.4, 22.2.5, 22.2.8, 22.2.9, 22.2.10, 22.2.11, 22.2.12, 22.2.13, 22.2.14, 22.2.15, and 22.3.1, 22.3.2, 22.3.4, 22.3.6, 22.3.7, 22.3.8, 22.3.9, 22.3.10, 22.3.11 (see column for k n)Notes:In Table 18.3.1, for Row 2 see Szegő (1975, §2.4, Item 1, (4.3.3), and (4.21.6)): the entry in the last column follows from (18.5.7); for Row 3 see Szegő (1975, (4.7.1), (4.7.14), and (4.7.9)): the entry in the last column follows from the symmetry in the fourth row of Table 18.6.1; for Rows 4–7 see Andrews et al. (1999, §5.1 and Remark 2.5.3) and Mason and Handscomb (2003, §4.2.2); for Row 10 specialize Row 2 to α=β=0; for Row 12 see Szegő (1975, §2.4, Item 2, (5.1.1), and (5.1.8)): the entry in the last column follows from (18.5.12); for Row 13 see Szegő (1975, §2.4, Item 3, (5.5.1), and (5.5.6)): the entry in the last column follows from the symmetry in the tenth row of Table 18.6.1.Referenced by:§1.18(v), §18.15(vi), (18.17.34_5), §18.2(xi), §18.2(iii), §18.3, Table 18.3.1, §18.3, §18.3, §18.3, §18.30(ii), §18.36(v), §18.39(i), §18.5(i), §18.5(ii), §18.5(iv), §18.7(i), §3.5(v), §3.5(vi), Erratum (V1.0.28) for Table 18.3.1, Erratum (V1.0.28) for Table 18.3.1, Erratum (V1.0.7) for Table 18.3.1, Erratum (V1.0.7) for Table 18.3.1Permalink: (effective with 1.0.28): The DLMF now adopts the definitions for the Chebyshev polynomials of the third and fourth kinds V n(x), W n(x) used in Mason and Handscomb (2003). Therefore V n(x), W n(x), having been interchanged, Rows 5 and 6 have been modified accordingly. For further details see Errata. Errata (effective with 1.0.7): The special case of 𝒜 n when n=0 has been introduced explicitly in the caption for this table. This was needed because the general formula for 𝒜 n is undefined at n=0 when α+β+1=0.
Reported 2013-06-13 by Howard Cohl
See also:Annotations for §18.3 and Ch.18
For expressions of ultraspherical, Chebyshev, and Legendre polynomials in terms of Jacobi polynomials, see §18.7(i). For representations of the polynomials in Table 18.3.1 by Rodrigues formulas, see §18.5(ii). For finite power series of the Jacobi, ultraspherical, Laguerre, and Hermite polynomials, see §18.5(iii) (in powers of x−1 for Jacobi polynomials, in powers of x for the other cases). Explicit power series for Chebyshev, Legendre, Laguerre, and Hermite polynomials for n=0,1,…,6 are given in §18.5(iv). For explicit power series coefficients up to n=12 for these polynomials and for coefficients up to n=6 for Jacobi and ultraspherical polynomials see Abramowitz and Stegun (1964, pp.793–801).
Chebyshev
ⓘ
Keywords:Chebyshev polynomials, orthogonality properties, with respect to summation, zerosReferenced by:§18.3Correction (effective with 1.0.28): The DLMF now adopts the definitions for the Chebyshev polynomials of the third and fourth kinds V n(x), W n(x), used in Mason and Handscomb (2003), therefore we have interchanged the expressions for V n(x), W n(x). Since the current definitions are now consistent with Mason and Handscomb (2003), the sentence and reference added in Version Version 1.0.10 (August 7, 2015) has been removed. For further details see Errata. Addition (effective with 1.0.10): A sentence and reference to Mason and Handscomb (2003) were added at the end of this paragraph, namely Paragraph Chebyshev. See also:Annotations for §18.3 and Ch.18
In this chapter, formulas for the Chebyshev polynomials of the second, third, and fourth kinds will not be given as extensively as those of the first kind. However, most of these formulas can be obtained by specialization of formulas for Jacobi polynomials, via (18.7.4)–(18.7.6).
In addition to the orthogonal property given by Table 18.3.1, the Chebyshev polynomials T n(x), n=0,1,…,N, are orthogonal on the discrete point set comprising the zeros x N+1,n,n=1,2,…,N+1, of T N+1(x):
18.3.1∑n=1 N+1 T j(x N+1,n)T k(x N+1,n)=0,
0≤j≤N, 0≤k≤N, j≠k,
ⓘ
Symbols:T n(x): Chebyshev polynomial of the first kind, N: positive integer, n: nonnegative integer and x: real variableReferenced by:§18.3, §18.3Permalink: pMML, pngSee also:Annotations for §18.3, §18.3 and Ch.18
where
18.3.2 x N+1,n=cos((n−1 2)π/(N+1)).
ⓘ
Symbols:π: the ratio of the circumference of a circle to its diameter, cosz: cosine function, N: positive integer, n: nonnegative integer and x: real variablePermalink: pMML, pngSee also:Annotations for §18.3, §18.3 and Ch.18
When j=k≠0 the sum in (18.3.1) is 1 2(N+1). When j=k=0 the sum in (18.3.1) is N+1.
For proofs of these results and for similar properties of the Chebyshev polynomials of the second, third, and fourth kinds see Mason and Handscomb (2003, §4.6).
For another version of the discrete orthogonality property of the polynomials T n(x) see (3.11.9).
Formula (18.3.1) can be understood as a Gauss-Chebyshev quadrature, see (3.5.22), (3.5.23). It is also related to a discrete Fourier-cosine transform, see Britanak et al. (2007).
Legendre
ⓘ
See also:Annotations for §18.3 and Ch.18
Legendre polynomials are special cases of Legendre functions, Ferrers functions, and associated Legendre functions (§14.7(i)). In consequence, additional properties are included in Chapter 14.
Jacobi on Other Intervals
ⓘ
See also:Annotations for §18.3 and Ch.18
For −1−β>α>−1 a finite system of Jacobi polynomials P n(α,β)(x) is orthogonal on (1,∞) with weight function w(x)=(x−1)α(x+1)β. For ν∈ℝ and N>−1 2 a finite system of Jacobi polynomials P n(−N−1+iν,−N−1−iν)(ix) (called pseudo Jacobi polynomials or Routh–Romanovski polynomials) is orthogonal on (−∞,∞) with w(x)=(1+x 2)−N−1e 2νarctanx. For further details see Koekoek et al. (2010, (9.8.3) and §9.9).
Bessel polynomials
ⓘ
See also:Annotations for §18.3 and Ch.18
Bessel polynomials are often included among the classical OP’s. However, in general they are not orthogonal with respect to a positive measure, but a finite system has such an orthogonality. See §18.34.
18.2 General Orthogonal Polynomials18.4 Graphics
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Volume 37, Issue 5 pp. 1074-1087
INVITED REVIEW
Open Access
Update on the management of vitamins and minerals in cystic fibrosis
Correction(s) for this article
###### Corrigendum for Update on the management of vitamins and minerals in cystic fibrosis. Nutr Clin Pract. 2022;37(5):1074-1087
Senthilkumar Sankararaman MD,
Sara J. Hendrix MS, RD,
Terri Schindler MS, RD,
Volume 39Issue 2Nutrition in Clinical Practice
pages: 506-506
First Published online: October 21, 2022
Senthilkumar Sankararaman MD,
Corresponding Author
Senthilkumar Sankararaman MD
senthilkumar.sankararaman@uhhospitals.org
orcid.org/0000-0003-3094-9703
Department of Pediatrics, Division of Pediatric Gastroenterology, UH Rainbow Babies & Children's Hospital, Case Western Reserve University School of Medicine, Cleveland, Ohio, USA
Correspondence Senthilkumar Sankararaman, MD, Department of Pediatrics, Division of Pediatric Gastroenterology, UH Rainbow Babies & Children's Hospital, Case Western Reserve University School of Medicine, Cleveland, OH 44106, USA.
Email: senthilkumar.sankararaman@uhhospitals.org
Search for more papers by this author
Sara J. Hendrix MS, RD,
Sara J. Hendrix MS, RD
orcid.org/0000-0002-7089-7074
Department of Nutrition Services, Medical University of South Carolina, Charleston, South Carolina, USA
Search for more papers by this author
Terri Schindler MS, RD,
Terri Schindler MS, RD
orcid.org/0000-0003-3001-4588
Department of Pediatrics, Division of Pediatric Pulmonology, UH Rainbow Babies & Children's Hospital, Cleveland, Ohio, USA
Search for more papers by this author
Senthilkumar Sankararaman MD,
Corresponding Author
Senthilkumar Sankararaman MD
senthilkumar.sankararaman@uhhospitals.org
orcid.org/0000-0003-3094-9703
Department of Pediatrics, Division of Pediatric Gastroenterology, UH Rainbow Babies & Children's Hospital, Case Western Reserve University School of Medicine, Cleveland, Ohio, USA
Correspondence Senthilkumar Sankararaman, MD, Department of Pediatrics, Division of Pediatric Gastroenterology, UH Rainbow Babies & Children's Hospital, Case Western Reserve University School of Medicine, Cleveland, OH 44106, USA.
Email: senthilkumar.sankararaman@uhhospitals.org
Search for more papers by this author
Sara J. Hendrix MS, RD,
Sara J. Hendrix MS, RD
orcid.org/0000-0002-7089-7074
Department of Nutrition Services, Medical University of South Carolina, Charleston, South Carolina, USA
Search for more papers by this author
Terri Schindler MS, RD,
Terri Schindler MS, RD
orcid.org/0000-0003-3001-4588
Department of Pediatrics, Division of Pediatric Pulmonology, UH Rainbow Babies & Children's Hospital, Cleveland, Ohio, USA
Search for more papers by this author
First published: 23 August 2022
Citations: 26
Abstract
Advancements in respiratory and nutrition management have significantly improved the survival of patients with cystic fibrosis (CF). With the availability of several nutrition interventions such as oral/enteral nutrition supplements, enteric-coated pancreatic enzymes, and water-miscible CF-specific vitamin supplements, frank vitamin deficiencies—with the exception of vitamin D—are rarely encountered in current clinical practice. Whereas they were previously considered as micronutrients, our current understanding of fat-soluble vitamins and minerals as antioxidants, immunomodulators, and disease biomarkers has been evolving. The impact of highly effective modulators on the micronutrient status of patients with CF remains elusive. This narrative review focuses on the updates on the management of fat-soluble vitamins and other micronutrients in CF in the current era and identifies the gaps in our knowledge.
INTRODUCTION
When cystic fibrosis (CF) was first recognized as a distinct entity in the 1930s by Dr Dorothy Andersen, most infants died from consequences of severe malnutrition.1 Advancements primarily in respiratory and nutrition therapies have led to significant improvements in the median life span of individuals with CF over the past five decades. The Cystic Fibrosis Foundation (CFF) Registry data are released annually, and the latest data for the year 2020, published in 2021, showed a predicted median reported survival of approximately 50 years.2 The role of registered dietitians is pivotal in the multidisciplinary CF team, and common nutrition interventions include individualized nutrition counseling utilizing weight, body mass index, and/or body composition to frame recommendations; exercise/activity to promote optimal body composition; pancreatic enzyme replacement therapy (PERT) for individuals with exocrine pancreatic insufficiency (EPI); and appropriate micronutrient supplementation.3-5 The CFF 2020 Registry data revealed approximately 40% of patients with CF required oral nutrition supplements and 10% received enteral nutrition.2
The nutrition state of individuals with CF has significantly improved and is mainly attributed to early CF diagnosis through newborn screening, use of effective enteric-coated pancreatic enzymes, optimization of fat-soluble vitamin supplementation, implementation of systematic quality-improvement interventions, and availability of newer CF-targeted therapies such as the highly effective modulators (HEMs).6, 7 HEMs are postulated to positively impact nutrition status in several ways, such as reduction in energy requirement, decrease in intestinal inflammation, improvement of intestinal dysbiosis, and improved fat absorption.8 Lately, the landscape of nutrition management of CF, especially in adult patients, is drifting toward newer nutrition challenges such as overweight/obesity and associated comorbidities, such as cardiovascular disease, hypertension, and metabolic syndrome.6, 9-11 In the past, there were several reports of deficiencies in fat-soluble vitamins in CF, including Dr Andersen's first description of CF.1 In the current era, frank micronutrient deficiencies of vitamin A and vitamin E are rarely encountered in patients with CF, presumably owing to early diagnosis and initiation of pancreatic enzymes, CF-specific vitamin supplementation, and possibly improved absorption with HEMs. On the contrary, instances of elevated vitamin levels are now seen.12-17 Also, once solely considered as micronutrients, our latest understanding of the role of fat-soluble vitamins and trace minerals has evolved. Currently, they are increasingly recognized for their various functions as antioxidants, immunomodulators, and disease biomarkers.18-26 In other words, rather than a focus on their status as deficient vs sufficient in CF, there is recognition of optimal vs suboptimal status impacting disease outcome in CF.18, 27, 28
Ivacaftor (VX-770 [Kalydeco; Vertex Pharmaceuticals Inc]) was the first HEM approved in 2012 for gating mutations, which account for approximately 15% of patients with CF. Two trials (ARRIVAL and KIWI studies) noted improving pancreatic function in young children with gating mutations.29, 30 In 2019, the US Food and Drug Administration (FDA) approved the first triple combination therapy (elexacaftor/tezacaftor/ivacaftor [TRIKAFTA; Vertex Pharmaceuticals Inc]) for patients with at least one delta F508 mutation, which is present in up to 90% of individuals with CF. Currently, elexacaftor/tezacaftor/ivacaftor is approved for patients ≥6 years of age. It is yet unclear whether this triple combination will significantly impact pancreatic function; studies are underway. Currently, PERT dosage is not adjusted in patients after the initiation of elexacaftor/tezacaftor/ivacaftor, owing to concerns of complications such as micronutrient deficiencies and worsening of gastrointestinal (GI) symptoms (including distal intestinal obstruction syndrome) and lack of data to guide any changes in PERT. However, there is a concern that individuals with CF may lower PERT dosage on their own after starting HEMs for reasons such as GI symptoms and concerns of excessive weight gain.31 It is important to keep the line of communication open and approach these topics in a supportive manner.
A recent study by Sommerburg and colleagues has shown that lumacaftor/ivacaftor in CF could significantly alter vitamin levels.13 In a small cohort of patients who have CF and are homozygous for delta F508, the median plasma vitamin A level improved from 1.2 to 1.6 µmol/L in a 2-year period. None of the patients had levels in the toxicity range. This change also correlated with reduction in lung inflammation, as demonstrated by significant decrease in serum immunoglobulin G. Here, the authors postulated that improved intestinal absorption of vitamin A along with reduction in lung exacerbations could be the reason for this increase. At the same time, the plasma vitamin E/cholesterol ratio decreased from 6.2 to 5.5 µmol/mmol, and a competitive decrease in intestinal absorption, redistribution of vitamin E from plasma to adipose tissue, and increased degradation were proposed as causes of this decrease. The effect of HEMs such as elexacaftor/tezacaftor/ivacaftor on fat-soluble vitamin and other micronutrient status is unclear, and the literature remains sparse. Preliminary evidence has shown that elexacaftor/tezacaftor/ivacaftor could improve vitamin D levels.32 The exact reason for these improving levels is unclear but could be related to improved intestinal absorption and/or optimization of vitamin D utilization with concomitant decrease in inflammation in CF patients.32 This review focuses on the basic principles in current practice of vitamins and minerals in CF in the era of HEMs and identifies the gaps in our knowledge.
Fat-soluble vitamins
Reports of deficiencies in fat-soluble vitamins A, D, E, and K in CF are commonly attributable to fat malabsorption, particularly in patients with poorly controlled EPI. The original description of CF by Dr Andersen in 1938 included references to vitamin A deficiency.1 In a multicenter study, deficiencies in fat-soluble vitamins were documented in up to one-third of children with CF and EPI but rarely noted in patients with pancreatic sufficiency.33 CF-specific vitamins are available, most of which contain fat-soluble vitamins in a water-miscible form, as well as water-soluble vitamins and zinc. Regular monitoring of fat-soluble vitamins and appropriate supplementation are necessary to prevent these deficiency states. In people with CF and EPI, fat-soluble vitamins can be deficient owing to uncontrolled fat malabsorption, suboptimal adherence to therapy, and comorbid issues such as short bowel syndrome and/or CF-related liver disease (CFLD).34, 35 Recommendations for doses of fat-soluble vitamins in CF are mostly from expert consensus rather than evidence-based practice.3 The CFF recommends regular screening for deficiencies in fat-soluble vitamins at the time of diagnosis and then annually and after any dose change. The important functions of fat-soluble vitamins and minerals and their interpretation are noted in Table 1. Also, Table 2 outlines the micronutrient status in specific populations affected by CF.
Table 1.
Summary of functions and evaluation of vitamins and minerals in cystic fibrosis
| Fat-soluble vitamins | Functions | Evaluation and interpretation of serum levels4, 36, 37 | Daily vitamin and mineral recommendations3, 4, 38, 39 |
--- --- |
| Vitamin A | Vision, immune function, epithelial integrity; beta-carotene is an antioxidant | Negative acute-phase reactant (falsely decreased during illness and inflammatory states).40, 41 Interpret with retinol-binding protein in setting of liver disease. Check zinc level if vitamin A (serum retinol) is persistently low. Toxicity possible; can cause hepatotoxicity and bone toxicity. | Infants: 1500 IU; toddlers: 5000 IU; 4–8 years: 5000–10,000 IU; >8 years: 10,000 IU (Dosage recommendations based on retinol form) 10,000 IU retinol = 3000 mcg RAE; 15 mg beta-carotene = 7500 mcg RAE = 25,000 IU retinol |
| Vitamin D | Bone health/calcium absorption, immune function | Negative acute-phase reactant.28 Serum 25-hydroxyvitamin D is used to measure vitamin D status. Levels may be influenced by season (higher in late summer or early fall because of increased sun exposure). Toxicity increases risk for hypercalciuria and hypercalcemia. | Infants: 400–500 IU; 1–10 years: 800–1000 IU; >10 years: 800–2000 IU (Cholecalciferol or vitamin D3 is the preferred form for supplementation in CF. Minimum daily doses are depicted here. Dosage can be increased based on serum levels and after ensuring compliance.) 1 IU = 0.025 mcg 400 IU/ml = 10 mcg/ml |
| Vitamin E | Antioxidant, cellular membrane stability, important for cognitive function | Level may be decreased in pulmonary exacerbations.41 Serum alpha-tocopherol levels reflects supplementation. Lipid level abnormalities may influence level. | Infants: 40–50 IU; toddlers: 80–150 IU; 4–8 years: 100–200 IU; >8 years—200–400 IU. 1 IU of the synthetic form is equivalent to 0.45 mg of alpha-tocopherol. |
| Vitamin K | Blood clotting, bone formation, cell growth regulation | Serum vitamin K levels do not reflect stores. PT/INR are late and nonspecific indicators of vitamin K deficiency. PIVKA-II or uc-OC level is a sensitive indicator of early vitamin K deficiency. | 0.3–0.5 mg for all age groups |
| Salt | Hyponatremic dehydration; salt loss is one of the reasons for poor weight gain in infants | Patients with CF have 2–4 times higher sodium chloride in the sweat, resulting in enhanced loss. Urine sodium: creatinine ratio can be utilized to evaluate enhanced sodium loss. | Historically, infants <6 months of age are provided one-eighth teaspoon of table salt (approximately 11 mEq sodium), and infants beyond 6 months of age are given one-fourth teaspoon of table salt. Patients with CF who exercise or play outside in hot weather also may need one-eighth teaspoon of salt added to 12 ounces (360 ml) of beverage. |
| Zinc | Immune function, growth, tissue healing, component of almost 300 enzymes, and structural role as zinc fingers in certain proteins. | Plasma zinc levels may not reflect deficiency. Consider empiric supplementation if deficiency is suspected or persistent poor weight gain despite adequate calorie and PERT supplementation. Patients with ileostomy are at increased risk of zinc deficiency. Patients at risk of zinc deficiency can be empirically supplemented for 6 months. | No consensus on routine zinc supplementation. Dosing: infants up to 2 years receive 1 mg/kg/day, children receive 15 mg/day, and adults receive 25 mg/day. |
| Calcium | Bone health, muscle and nerve functions, clotting; functions as coenzyme in many metabolic processes | In patients with a low serum albumin level, calcium levels should be corrected for low serum albumin level status. Calcium levels should be screened annually, and recommended calcium intake is similar to that in patients without CF. | Dietary reference intake of calcium for general population is recommended for CF population. |
| Magnesium | Muscle and nerve function, bone health, and a cofactor for many enzymatic reactions. | Low magnesium levels are increasingly recognized in patients with CF due to many factors. | No formal guidelines available regarding evaluation or treatment of magnesium in CF. |
Abbreviations: CF, cystic fibrosis; PERT, pancreatic enzyme replacement therapy; PIVKA, protein induced in vitamin K antagonism/absence; PT/INR, prothrombin time/international normalized ratio; RAE, retinol activity equivalent; uc-OC, undercarboxylated osteocalcin.
Table 2.
Micronutrients status in specific populations affected by CF
| Special populations | Micronutrient considerations |
--- |
| Lung transplant recipients | Elevated levels of vitamin A and vitamin E have been documented.42 Renal dysfunction due to the use of immunosuppressive medications can further complicate the vitamin levels and should be monitored closely. CF-specific vitamins should be discontinued, and over-the-counter multivitamins can be given. Vitamin D and vitamin K status should be optimized. Higher vitamin A can adversely affect bone health and can can liver fibrosis. |
| Pancreatic-sufficient CF | Vitamin D deficiency is noted to be similar to that in populations with pancreatic insufficiency.18 There is sparse literature on routine supplementation of other fat-soluble vitamins. |
| Pregnancy | Fat-soluble vitamin levels ideally should be tested before conception and monitored every trimester to ensure levels are optimized. Vitamin A supplementation should continue at <10,000 IU/day.43, 44 Also, clinicians should ensure patients are taking adequate amounts of folic acid, iron, calcium, and phosphorus. |
| Liver disease | Concomitant liver disease further predisposes patients to deficiencies in fat-soluble vitamins and needs close monitoring. Higher vitamin A levels can worsen liver disease. |
| Short bowel syndrome | Pancreatic enzymes, specifically trypsin, also play a significant role in cleaving R-binders produced in the salivary glands. This enhances vitamin B12–intrinsic factor coupling and later aids in B12 absorption in the ileum.45 In patients with extensive ileal resection related to meconium ileus, B12 deficiency has been reported.46 |
Abbreviation: CF, cystic fibrosis.
Vitamin A
Vitamin A is a crucial component in many physiological functions, including embryonic development, vision, bone health, cellular proliferation and differentiation, immunity, and antioxidant function. Importantly in CF, vitamin A protects respiratory epithelial cells against oxidation.47 Better vitamin A status is linked with better pulmonary status in people with CF.48 Vitamin A can be consumed in two forms, preformed vitamin A (retinol, retinyl esters) and provitamin A carotenoids such as beta-carotene. Preformed vitamin A is found in dairy, liver, eggs, and fortified foods, such as breakfast cereal. Provitamin A is found in plant-based foods, such as fruits and vegetables. Provitamin A is only converted to retinol if needed by the body; and this is an important difference between the two forms of vitamin A, given that there is concern for vitamin A deficiency as well as vitamin A toxicity in people with CF.27
Vitamin A supplementation in CF is standard clinical practice, and the recommended dosage in the US varies from 450 to 3000 mcg retinol activity equivalents per day (equivalent to 1500–10,000 IU)49; however, these guidelines are not based on clinical studies.50 Even though vitamin A deficiency has been reported to be independent of pancreatic function,51 most studies reported vitamin A deficiency in patients with EPI either before initiation of PERT and supplementation of fat-soluble vitamins or due to poor adherence to therapy.33, 52 Vitamin A deficiency is becoming rare in people with CF, whereas elevated serum retinol levels have become more prevalent.15 Elevated serum vitamin A is associated with liver and bone disease and increased intracranial pressure.53, 54
A serum retinol value below a cutoff of 20 mcg/dl represents biochemical vitamin A deficiency.47 Although obtaining fasting annual laboratory values may be burdensome, nonfasting levels may reflect recent intake of vitamin A. Thus, serum retinol levels are ideally assessed in the fasting state.36 Vitamin A is a negative acute-phase reactant and therefore can result in falsely low levels when assessed during acute illness.55 Vitamin A is transported bound to retinol-binding protein (RBP), which is produced in the liver and often low in patients with liver disease and/or malnutrition. When there is not enough RBP to transport vitamin A, it accumulates in the liver. Therefore, preformed vitamin A can cause toxicity if RBP is low. In patients with liver disease or malnutrition, assessing the molar ratio of retinol to RBP can help guide the need for vitamin A supplementation. A ratio of <0.8 indicates true vitamin A deficiency that requires vitamin A supplementation.56 Measurement of serum retinyl esters can be more useful for estimating vitamin A toxicity if >10% of total vitamin A is in the form of retinyl esters.36, 57
Zinc is necessary for synthesis of RBP, and therefore, zinc deficiency may lead to inadequate RBP available to circulate retinol.58 Vitamin A absorption and conversion of beta-carotene take place in the small intestine; therefore, a history of small bowel resection may lead to low serum levels of vitamin A. CF-specific multivitamins are commonly the first choice for vitamin A supplementation with the content predominantly as beta-carotene,59 which decreases the risk of toxicity. Individualized vitamin A supplementation based on estimation of serum levels is important to prevent vitamin A deficiency and toxicity.60 Higher retinol levels with possible toxicity may be encountered in patients treated predominantly with retinol-based supplements, patients who have chronic kidney disease, and patients posttransplant.15, 42, 61-63 There are case reports of hypervitaminosis A in patients treated with CFTR modulator therapy, and more studies are needed to elucidate the effect of HEMs on vitamin A status in patients with CF.12-14
Vitamin D
Vitamin D serves an important role in bone health by regulating circulating calcium and phosphorus levels to promote normal bone mineralization.64 There is also evidence for its role in immunity,64 microbiome,65 inflammation, and pulmonary health.28, 66 Vitamin D deficiency and insufficiency are common in people with CF.18, 67 Factors that contribute to deficiency in CF include decreased intestinal absorption, inadequate intake of vitamin D–containing foods or supplements, low sunlight exposure, treatment with glucocorticoids, impaired hydroxylation of vitamin D in the liver or kidneys, and reduced fat stores.67
Vitamin D can be obtained orally through diet or supplementation with ergocalciferol or cholecalciferol and by skin production through sunlight exposure. Vitamin D is not naturally present in many foods, with the exception of the flesh of fatty fish and fish liver and in smaller amounts in beef liver, dairy products, and egg yolk in the form of cholecalciferol and its metabolite 25-hydroxyvitamin D3 [25(OH)D]. Some mushrooms are a source of ergocalciferol.68 Supplemental cholecalciferol has been shown to be superior at improving serum 25(OH)D levels compared with ergocalciferol,69 and current CFF guidelines recommend treatment with cholecalciferol and offer specific dosing guidelines.38 Ergocalciferol may be considered for individuals who avoid the use of animal products, although higher doses may be required.38 Vitamin D absorption may be improved when taken with pancreatic enzymes before meals. The CFF recommends that individuals with difficult-to-treat vitamin D deficiency be referred to a specialist with expertise in vitamin D therapy.38
Additionally, levels will be influenced by time of year/exposure to sunlight, with peak levels likely to be during the summer months.70 The CFF recommends yearly screening for vitamin D status using serum 25(OH)D, preferably at the end of winter when sun exposure is lowest.38 Vitamin D status is considered sufficient when 25(OH)D concentrations are ≥30 ng/ml, insufficient when concentrations are ≥20 and <30 ng/ml, and deficient when concentrations are <20 ng/ml. Optimal vitamin D status is defined as a 25(OH)D concentration between 30 and 50 ng/ml, and the CFF recommends a minimum 25(OH)D level of 30 ng/ml (75 nmol/L) for individuals with CF. Levels should not exceed 100 ng/ml given the increased risk for associated hypercalcemia. Vitamin D status should be reevaluated roughly 3 months after changes to vitamin D dosing.38 Serum 25(OH)D is a negative acute-phase reactant and is an unreliable marker of vitamin D status during acute illness.71
Despite routine supplementation with vitamin D, serum vitamin D levels remain insufficient in many individuals with CF.72, 73 A recent study found that reported sunlight exposure was significantly associated with higher serum vitamin D levels at admission for pulmonary exacerbation, whereas total oral vitamin D intake was not significantly associated with vitamin D levels, suggesting that sun exposure is a major source of vitamin D production in individuals with CF and malabsorption.74 Given these findings, sunlight may provide an alternative or synergist benefit to oral vitamin D treatment when feasible. There are limitations to sun-induced synthesis; some patients may not be able to tolerate sun exposure because of increased photosensitivity to certain medications. Other factors, including season, time of day, latitude, skin pigmentation, and sunscreen use, influence the amount of vitamin D absorbed by the skin.16, 75 In the general population, exposing 20% of the body surface to sunlight for half the time it would take to cause mild sunburn is equivalent to ingesting roughly 1400–2000 IU (35–50 mcg) of vitamin D3 and is effective for all skin types.75
Numerous studies have demonstrated the impact of vitamin D deficiency and insufficiency status in CF outcomes. Simoneau and colleagues noted increased association of Pseudomonas aeruginosa colonization in vitamin deficient/insufficient patients.18 Increased serum vitamin D levels were found to be associated with better lung function tests in both children and adults with CF.28, 76, 77 However, a recent meta-analysis evaluated eight randomized controlled trials and showed that the intervention group (receiving vitamin D supplementation) had no difference in bone disease, pulmonary status, and immunological outcomes when compared with the placebo group.78
Vitamin E
Vitamin E is a fat-soluble antioxidant that is found in foods such as nuts, seeds, and oils, but it is also an additive to some foods and available as a supplement. There are eight chemical forms of vitamin E found naturally in foods; however, alpha-tocopherol is the only form recognized to meet human requirements, has the highest biological activity, and is the form most commonly found in supplements.79, 80 Alpha-tocopherol serves as an antioxidant preventing the deleterious effects of free oxygen radicals on unsaturated fatty acids.81 It is unclear whether there are subclinical benefits of vitamin E supplementation for individuals with CF, including those with pancreatic sufficiency; there may be increased need from higher oxidative stress from chronic inflammation and bacterial infections.81 A recent Polish study involving young children and adults with CF identified vitamin E deficiency in 8% of participants and high levels in 11.4%.17
Severe hemolytic anemia can result from vitamin E deficiency and can be seen in infants with CF as young as 6 weeks of age who are not receiving supplements.82 Vitamin E deficiency can also manifest as neurological problems, including sensorimotor neuropathy, and cognitive impairment.83, 84 Deficiencies are uncommon in individuals treated with enzymes and CF vitamins, and supplementation may not be needed for individuals with pancreatic sufficiency.84, 85 Supplementation is associated with improved vitamin E levels; however, there is a lack of research indicating improved clinical outcomes associated with supplementation.84
Annual measurement of alpha-tocopherol levels in individuals with CF is recommended, with more frequent follow-up if abnormalities are noted. Aggressive supplementation may lead to elevated serum alpha-tocopherol vitamin levels and/or suppression of gamma-tocopherol levels.86 The clinical significance of this suppression in CF is unknown, and there are no recommendations to assess other tocopherol levels besides alpha-tocopherol. Elevated levels of vitamin E are particularly common in individuals after transplant and may predispose them to risk of hemorrhage.42 Altered lipid levels may be associated with elevated or low vitamin E levels. Vitamin E circulates in the blood bound to lipoproteins. A more accurate assessment of vitamin E deficiency can be obtained using the vitamin E to total lipid ratio, which has a sensitivity of 95% and a specificity of 99%.87 A lipid panel is recommended in situations when abnormal vitamin E levels are noted. High-dose vitamin E supplementation has been noted to antagonize the effect of vitamin K in populations without CF.88 When taken with pancreatic enzyme replacement,89 a water-soluble form of vitamin E does not appear to have any advantage over fat-soluble vitamin E in terms of absorption.81
Vitamin K
Vitamin K is the least studied of all fat-soluble vitamins in terms of dosing (type and amount) and optimal methods for monitoring vitamin K status. Vitamin K is also unique because it is produced by intestinal bacteria (menaquinones or vitamin K2) in addition to food (phylloquinone or vitamin K1) and supplemental sources.81, 90 Dark green and leafy vegetables such as broccoli, kale, and spinach are good sources of vitamin K. Vitamin K is a cofactor for the enzyme gamma-glutamyl carboxylase, which is involved in the posttranslational modification of peptidyl gamma-carboxyglutamic acid (Gla) from specific glutamyl residues. Gla residues are found in coagulation factors (II, VII, IX, and X), and these proteins are synthesized in the liver.90, 91 Whereas vitamin K is most commonly known for its importance in synthesizing proteins necessary for blood clotting, the other proteins involved in bone metabolism and cell growth regulation such as osteocalcin, matrix Gla protein, transmembrane Gla proteins, and proline-rich Gla proteins also contain Gla.90, 91 In vitamin K deficiency, undercarboxylated proteins are formed, which are functionally defective because they cannot bind either calcium or phospholipids. These abnormal coagulation factors are called protein induced by vitamin K absence or antagonism (PIVKA) and des-gamma-carboxy-prothrombin (otherwise known as PIVKA-II). PIVKA-II and undercarboxylated osteocalcin (uc-OC) are sensitive markers of vitamin K deficiency and are available mostly for research settings and not for clinical utility.3, 91-93 Serum vitamin K levels reflect intake over the prior 24 h and are not useful in estimating deficiency status. Prothrombin time (PT) is not useful to identify vitamin K deficiency in the early stages, and PT levels become abnormal when prothrombin levels fall below 50% of normal values.94 Also, the liver is capable of utilizing vitamin K better than bone in early stages of vitamin K deficiency, and thus, deficiency status may increase risk of poor bone health and cancer.90, 95, 96 Rashid and colleagues demonstrated mild vitamin K deficiency in 75% of their patients with CF, both children and adults. They demonstrated increased levels of PIVKA-II.97 Similarly, Wilson and colleagues showed that vitamin K at a mean daily dose of 0.18 mg demonstrated a significant decrease in PIVKA-II.98 In a study by Hubert and colleagues, decreased bone mineral density and high PIVKA-II levels were noted both before and after lung transplant.99
Patients with concomitant CFLD have increased susceptibility to vitamin K deficiency.98 Similarly, patients with short bowel syndrome and patients with frequent antibiotic use are at increased risk of vitamin K deficiency.100 The CFF consensus guidelines recommend a vitamin K dose of 0.3–0.5 mg daily for the pediatric age group and 2.5–10 mg weekly for adults.3, 101 The guidelines from the European Society for Clinical Nutrition and Metabolism; European Society for Paediatric Gastroenterology, Hepatology and Nutrition; and the European Cystic Fibrosis Society (ESPEN-ESPGHAN-ECFS) recommended a daily dose of 0.3–1 mg in infants and 1–10 mg in patients beyond infancy.4 Adverse effects from excessive vitamin K supplementation are not reported.102 Also, vitamin K intake should be carefully monitored in patients on medications such as warfarin.103
Iron
Iron deficiency anemia is more commonly seen in individuals with CF who have more advanced lung disease and may be related to iron losses into the airway, which may facilitate Pseudomonas infection.104 Incidence of hypoferremia varies widely based on publication, likely owing to the difficulty in interpreting iron studies because iron studies are influenced by inflammation.105 Despite the association between iron and Pseudomonas, oral iron supplementation does not seem to lead to an increase in pulmonary exacerbations or negatively affect the lung microbiome.105 Also, patients can be encouraged to consume foods that are good sources of iron, such as liver, meat, and lamb. Further studies are warranted to examine the incidence of iron deficiency in CF, as well as an optimal way to monitor and treat deficiency.
Iron deficiency may be a concern among some people with CF and is associated with anemia, deficiencies in fat-soluble vitamins, and worse lung function.106, 107 Loss of iron in the GI tract and sputum may contribute to the prevalence of iron deficiency in CF.104 There are currently no guidelines for screening, diagnosing, and treating iron deficiency and anemia in CF. Assessing iron status is challenging because of the lack of standardized values used to define iron deficiency. Furthermore, infection and inflammation are known to affect iron studies, leading to elevated of levels of ferritin and transferrin and low levels of iron and percent transferrin saturation.108 Management of iron deficiency in CF is challenging because it is hard to distinguish between anemia of chronic disease and anemia due to iron deficiency. Another concern is that supplementing iron may cause harm. When surveyed, about a quarter of CF clinicians reported concern that supplemental iron could enhance the growth of lung bacteria despite lack of evidence between iron supplementation and pulmonary status.105, 109
Because iron studies are affected by inflammation, it is best to use multiple laboratory values to assess iron status, including serum iron, transferrin saturation, and soluble transferrin receptor. Checking C-reactive protein may help improve the interpretation of results.110 Studies of iron status in patients with CF found that a ferritin cutoff of 12 ng/ml, which is the World Health Organization (WHO) classification for iron deficiency, showed poor sensitivity and underestimated iron deficiency.106, 111 It may be more reasonable to use a ferritin cutoff of 30 ng/ml, which is used in guidelines for other chronic inflammatory diseases. The American Gastroenterology Association recommends using a ferritin cutoff of 45 ng/ml in patients with anemia (hemoglobin <13 g/dl in men and <12 g/dl in women, as defined by WHO).111 Iron supplementation has been shown to be helpful in adults with CF who suffer from restless leg syndrome regardless of iron status.112
When deciding how to treat iron deficiency, the clinician should assess for anemia of chronic disease and treat the underlying disease first. In most patients, an initial trial of oral iron should be given because it is readily available, inexpensive, and safe.110 Available evidence does not support any available formulation as being more effective or better tolerated than the others.113 However, GI intolerance to oral iron supplements is common, and patients with malabsorption syndromes may have limited response.110 Although there are no dosing guidelines specific to CF, traditionally, a daily elemental iron dose of 150–200 mg has been recommended for adults, and a dose of 3–6 mg/kg/day elemental iron divided two or three times per day for children. Recent studies suggest that lower dosing or every-other-day dosing may improve tolerability and absorption.114, 115 Oral iron supplements acutely increase serum hepcidin that persists for about 24 h. Hepcidin is a regulator of iron balance and negatively correlates with iron bioavailability, therefore decreasing iron absorption from supplements given later that day or the next day. One study found that hepcidin was significantly increased at 24 h with oral iron doses of 60 mg or more and returned to baseline after 48 h.115 Vitamin C can enhance iron absorption, and phytates, calcium, and tannins can reduce absorption. In patients who have intolerance to oral iron, intravenous iron infusions have been reported in CF, but the efficacy and adverse effects of this intervention remain unclear.116
Salt (sodium)
An old adage depicts the fate of an infant born with CF before identification of the diagnosis: “Woe to the child who tastes salty from a kiss on the brow, for he is cursed and soon will die.” Sodium is important in fluid balance and maintenance of blood volume. It is found ubiquitously in food because it is used as a preservative and flavor enhancer. Abnormal transport of sodium and chloride across sweat gland epithelial cells has long been recognized as a consequence of CF. Di Sant'Agnese and colleagues documented a twofold to fourfold higher sweat sodium and chloride content in individuals with CF compared with that of controls.117 Sodium chloride deficits can be particularly problematic in infancy because of increased requirements due to rapid growth as well as the low sodium content of first foods and breast milk. Infants also have increased salt losses through their skin, especially in warm climates.3
Chronic clinical consequences of salt include poor growth and failure to thrive, particularly in young children.118 Sodium deficit can be detected via urine sodium excretion and may be particularly useful in infants with persistent growth issues.119 Past the age of 2 years, the CFF has no specific recommendations for salt intake for individuals with CF, other than using the salt shaker at meals and providing additional sources of salt when exercising and/or exposed to excessive heat. However, other guidelines outside of the USA have more specific recommendations for salt intake.120 Measurement of chloride levels during sweat tests has traditionally been the method of confirming the diagnosis of CF.117, 121 Cases of hyponatremic dehydration have been described in individuals with CF during infancy and during periods of excessive sweating, mainly during heat waves.122, 123 Changes in sweat chloride levels have been the basis of ongoing research, used as a secondary outcome examining the efficacy of modulator therapy. Although ivacaftor causes an overall significant decrease in sweat chloride levels, the changes are variable and, so far, do not seem to be associated with improvements in lung function.124 It is yet unknown whether salt recommendations should be modified in individuals with CF after starting HEM therapy. Additionally, some patients experienced only mild decreases in chloride levels, and it is unknown whether there is variation in losses from day to day or during periods of increased sweating.119, 120 There is some evidence that ivacaftor raises blood pressure in adults with CF,125 and hypertension may be problematic in the posttransplant population as well. Individuals with portal hypertension may be advised to limit salt intake owing to ascites; however, this recommendation is met with controversy.126 Additionally, not all individuals are “salt sensitive,” and it can be challenging to identify those who have a significant blood pressure response to excess sodium intake.127 Individualized recommendations may be needed in these situations, focusing on extra salt intake only in situations in which excess salt loss is expected to occur (ie, exercise, heat).
Zinc
Zinc is essential for many metabolic pathways and several enzymes. It has significant antioxidant and anti-inflammatory activity.128 Zinc has been shown to influence growth and development, protein and DNA synthesis, wound healing, oxidation, and cellular immune responses.128, 129 Common dietary sources of zinc include meat, fish, and fortified foods. Most CF-specific vitamins contain zinc. Zinc deficiency is a risk in infants who are exclusively breastfed beyond 6 months of age, and meat/fortified cereal can be considered as a first food in this scenario. Young children who consume excessive soy-based beverages may have lower bioavailability of zinc due to the presence of phytates.130, 131
Zinc deficiency can occur in the setting of diarrhea, including steatorrhea caused by untreated or undertreated pancreatic insufficiency.132 Acrodermatitis enteropathica–like presentation has been reported in infants newly diagnosed with CF who have not yet started enzyme replacement; however, reports of this condition are rare since the advent of newborn screening.133 Zinc deficiency has been associated with growth failure and reduced growth velocity.132 In a cross-sectional study of 30 infants with CF diagnosed by newborn screen, plasma zinc was significantly lower in one-third of the infants and improved after initiating PERT.134, 135 In adults with CF, low zinc levels were associated with adverse clinical outcomes.136 Measurement of zinc (Red blood cell or plasma zinc level) does not necessarily reflect zinc status.137 In 62 children with CF, plasma zinc did not correlate with growth and pulmonary status.138
Both CFF and European guidelines recommend an empiric trial of zinc for 6 months in infants < 2 years of age with unexplained growth failure despite adequate caloric intake and PERT.3, 39 However, studies also demonstrated no improvement of growth with zinc supplementation.139 In a double-blinded placebo-controlled study of 26 children with CF, intake of 30 mg of zinc daily for a year decreased the requirement of oral antibiotics for pulmonary infections.140 The effect of zinc supplementation was higher in children who had low plasma zinc levels at baseline compared with those children who had higher levels at baseline.140 Also, prolonged high doses of zinc supplementation can lead to copper deficiency and should be carefully monitored.141 Serum zinc levels are not sensitive in early stages of deficiency, and lower levels are noted in cases of prolonged and severe zinc deficiency.142
Calcium and bone health
Along with vitamin D and vitamin K, calcium intake should be optimized for bone health.143 Patients with CF have poor bone health secondary to many factors such as CFTR dysfunction, corticosteroids use, poor nutrition, decreased physical activity (particularly, reduced weight-bearing exercise), pubertal delay, and systemic inflammation.144 Patients with CF may have deficient calcium status due to multiple reasons, such as vitamin D deficiency or reduced calcium intake or absorption.4, 36 Recommended calcium intake depends on age and sex.4, 36 Some patients with respiratory problems are hesitant to include milk in the diet because of perceived concerns regarding increased mucus production, and clinicians should thoroughly counsel against this perception.145 In patients with ongoing fat malabsorption, the unabsorbed intestinal fat can chelate unbound calcium, which decreases the calcium available to sequestrate intestinal oxalate, thereby promoting intestinal oxalate absorption.146 This enhanced oxalate absorption can possibly predispose patients with CF to renal stones. Also, clinicians should be aware that calcium is not included in any of the CF-specific vitamin products and calcium intake should be screened annually.36
Magnesium
Magnesium is essential for muscle and nerve function and bone health and is a cofactor for many enzymatic reactions. Low magnesium levels in patients with CF are increasingly recognized, and the causes could be multifactorial—eg, increased renal loss secondary to proximal convoluted tubular damage from frequent aminoglycosides (used for Pseudomonas infections), decreased absorption, malnutrition, CF-related diabetes, and medications such as calcineurin inhibitors in transplant recipients.147, 148 Recombinant human deoxyribonuclease efficacy can be decreased in the presence of concomitant magnesium deficiency.149 Higher doses of oral magnesium supplements can result in diarrhea, and intravenous magnesium may be needed to rapidly replenish the magnesium levels if the levels are severely low. Hypomagnesemia could predispose critically ill patients to recalcitrant hypokalemia and hypocalcemia.150 There is no evidence to recommend routine monitoring and evaluation of magnesium status in CF.
Water-soluble vitamins
There are no recommendations for specific doses of water-soluble vitamin supplementation in CF.3 Most CF vitamin preparations contain multivitamins, including water-soluble vitamins, listed as daily values. As nutrition intake and nutrition status improve in CF owing to HEMs, it would be beneficial to reexamine the CF vitamin preparations and necessity to include water-soluble vitamins.
CONCLUSIONS AND FUTURE DIRECTIONS
Except for vitamin D deficiencies, overt deficiencies in fat-soluble vitamins and trace minerals are not seen commonly in routine clinical practice. Higher levels of vitamins, particularly vitamin A, are increasingly recognized in selective populations such as those after lung transplant and after initiation of HEMs. Currently, the FDA has approved the elexacaftor/tezacaftor/ivacaftor combination for children ≥6 years of age. If this modulator is approved for younger children, one may expect some restoration of pancreatic function specifically in patients with heterozygous delta F508 mutation and the presence of a concomitant milder mutation.151 In this scenario, the macronutrient recommendations, dosing of PERT, and dosing of fat-soluble vitamins may also need to be revised. Vitamin K is the least-studied fat-soluble vitamin, and an accurate test for detecting the deficient status such PIVKA-II or uc-OC is needed for clinical use. The role of many vitamins and minerals as antioxidants and immunomodulators is emerging, and further studies are needed to explore these roles.
AUTHOR CONTRIBUTIONS
Senthilkumar Sankararaman, Sara J. Hendrix, and Terri Schindler contributed to the conception and design of this review. All authors drafted the manuscript, critically revised the manuscript, agree to be fully accountable for ensuring the integrity and accuracy of the work, and read and approved the final manuscript.
CONFLICT OF INTEREST
Senthilkumar Sankararaman is a consultant for Nestlé. Sara J. Hendrix has no conflicts to declare. Terri Schindler has the following to declare: Chiesi (advisor and speaker's bureau), AbbVie (speaker's bureau), and Nestlé (advisor).
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Volume37, Issue5
Acid-Base Balance, Fluids, & Electrolytes
October 2022
Pages 1074-1087
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7317 | http://dr.iiserpune.ac.in:8080/xmlui/bitstream/handle/123456789/188/zuhair_ms_thesis.pdf?sequence=1&isAllowed=y | Published Time: Fri, 30 Sep 2022 05:33:21 GMT
Equidistribution and related Ergodic methods in Number Theory
A thesis submitted to Indian Institute of Science Education and Research Pune in partial fulfillment of the requirements for the BS-MS Dual Degree Programme Thesis Supervisor: Ritabrata Munshi by Mohammed Zuhair. M. M April, 2012
Indian Institute of Science Education and Research Pune Sai Trinity Building, Pashan, Pune India 411021 This is to certify that this thesis entitled ”Equidistribution and related Ergodic methods in Number Theory” submitted towards the partial fulfillment of the BS-MS dual degree programme at the Indian Institute of Science Education and Research Pune, represents work carried out by Mohammed Zuhair. M. M under the supervision of Ritabrata Munshi. Mohammed Zuhair. M. M Thesis committee: Ritabrata Munshi Baskar Balasubramanyam Anupam Kumar Singh A. Raghuram Coordinator of Mathematics Acknowledgments
I express my sincere thanks to Ritabrata Munshi for kindly accepting me to guide me through my thesis and making it an exciting learning experience. I thank the Tata Institute of Fundamental Research, Mumbai, for the intellectually stimulating atmosphere, and for their warm hospitality. Many thanks are also due to Prof. S. G. Dani and Dr. Baskar Balasubramanyam for the invaluable help I received from them during the course of this thesis. It is my pleasure to thank Dr. Anupam Kumar Singh for his advice and encouragement. Last, but not the least, I thank the Kishore Vaigyanik Protsahan Yojana for financial support. vvi Abstract
Equidistribution and related Ergodic methods in Number Theory
by Mohammed Zuhair. M. M There has been a recent surge of interest in distributional problems related to number theory. Equidistribution has been widely recognized as a ubiquitous phenomenon in the subject. Solutions to equidistribution problems often involve techniques from several distinct areas of mathematics, and as such is the meeting ground of number theory, analysis and ergodic theory. In this thesis, we study several equidistribution problems and the techniques used for their resolution. We also study some ergodic methods relevant to the subject. In chapter 1, we introduce the notion of equidistribution and proceed to study equidistribution modulo 1 through the Weyl criterion. The Weyl criterion is an important and effective tool in proving equidistribution. We then give a generalized version of the Weyl criterion and use this to look at the distribution of Farey fractions in [0 , 1]. It is shown that the rate of their equidistribution is intimately related to the distribution of zeros of the Riemann zeta function. The chapter ends with the study of “randomness” in the map x 7 → x (mod p).
In chapter 2 we explore the Linnik’s problem - a classical problem regarding the distribution of integral solutions to the equation x2 + y2 + z2 = n as n → ∞ . We will see that the problem is inextricably linked to bounds for the size of fourier coefficients of modular forms of half integral weight. We also study the Linnik’s problem for squares, an easier version of the problem, using the Shimura correspondence. In chapter 3, we begin by establishing the equidistribution of ( n2α) modulo 1 using ergodic theory. We then study the dynamics of unimodular lattices under the action of the diagonal torus and prove the isolation of periodic orbits. We then connect these results with the Littlewood conjecture and Minkowski’s theorem on ideal classes. vii viii Contents
Abstract vii 1 Equidistribution in Number theory 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Equidistribution mod 1 and the Weyl criterion . . . . . . . . . . . . . . . . . . . . . 21.3 Equidistribution of rationals and the Riemann hypothesis . . . . . . . . . . . . . . . 71.4 The map x 7 → x mod p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2 Linnik’s Problem 15
2.1 Rational points on the sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2 Modular forms of half integral weight . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.3 Sali´ e sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.4 Iwaniec’s bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3 Ergodic methods in number theory 39
3.1 Measure rigidity and equidistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.2 Dynamics of Lattices and the Littlewood conjecture . . . . . . . . . . . . . . . . . . 48 3.3 Application: Strengthening Minkowski’s theorem . . . . . . . . . . . . . . . . . . . . 59 ix x CONTENTS Chapter 1
Equidistribution in Number theory
1.1 Introduction
The notion of equidistribution is of fundamental importance to number theory. Many results in number theory are best described as equidistribution of certain sequences in appropriate spaces. Before illustrating this, we first make the much needed definition.
Definition 1.1. Let ( X, B, μ ) be a Borel probability space, where X is a topological space, B the
σ-algebra of Borel sets and μ a normalized probability measure on ( X, B). A sequence of points ( xn)in X is said to be equidistributed with respect to μ if for every open set U , we have lim
N→∞
{n ≤ N : xn ∈ U }
N −→ μ(U ). (1.1) The notion of equidistribution is a ubiquitous one in number theory. Several examples will be provided, in the course of this dissertation, to illustrate this fact. We first begin with two important examples: Dirichlet’s theorem on arithmetic progressions: Let q be any positive integer. Our space of inter-est is the set of units modulo q, i.e. ( Z/q Z)×, with uniform probability measure. The famous prime number number theorem of Dirichlet on arithmetic progressions is equivalent to the equidistribution of the sequence of prime numbers mod q, on ( Z/q Z)×.Let L/K be a Galois extension of number fields with Galois group G. To every unramified prime
p of OK we can associate a conjugacy class in G, namely the Frobenius element of p. Our space of interest X is the set of conjugacy classes in G with measure μ of a class proportional to its size. Now, if we arrange the (unramified) primes in K with increasing order of their norm, the equidistribution of the corresponding Frobenius elements in X with respect to μ amounts to the famous Chebotarev density theorem. 12 CHAPTER 1. EQUIDISTRIBUTION IN NUMBER THEORY
1.2 Equidistribution mod 1 and the Weyl criterion
We say a sequence of real numbers ( xn) is equidistributed modulo 1 if their fractional parts {xn}
is equidistributed in [0 , 1) with respect to the lebesgue measure. Let α be a real number. We are interested in the distribution of {nα }, the fractional part of nα , in [0 , 1). One finds quickly that there is a dichotomy depending on whether α is in Q or not. If α ∈ Q, then {nα } repeats periodically. On the other hand if α is irrational, the points {nα } are all distinct. More over, by Kronecker’s theorem, the points {nα } from a dense subset of [0 , 1). What is more interesting, is that the points
{nα } equidistribute in [0 , 1) with respect to the Lebesgue measure, a fact first proved by Weyl. Originally, Weyl was interested in the the distribution modulo 1 of the sequence xn = f (n), where
f is a polynomial with real coefficients. In Weyl introduced a criterion, now bearing his name, which has since become a fundamental tool in establishing equidistribution results. Before stating Weyl criterion we observe that [0 , 1) can be identified with the unit circle T via the natural identification t 7 → e2πit , which also identifies the the respective Lebesgue measures. This has its advantages because T is a compact topological group. From now on we shall use e(x) to denote exp(2 πix ). With the above identification, it is clear that a sequence of real numbers ( xn) is equidistributed mod 1 if and only if e(xn) is equidistributed on T.Our first lemma, essentially a restatement of the equivalence between measure and integration, is a step towards the Weyl criterion.
Lemma 1.2. Let ( T, l ) be the unit circle with the normalized Lebesgue measure on it. A sequence (xn) is equidistributed in T with respect to l if and only if, for every continuous f : T → R, we have lim
N→∞
1
N
∑
n≤N
f (xn) −→
∫
T
f dl. (1.2)
Proof. (⇐) Suppose ( xn) is a sequence that is equidistributed in T and let f be a continuous real valued function on T. Let I1 ∪ · · · ∪ Im be a finite partition of T into intervals. Let U and L be the upper and lower Riemann sum with respect to this partition. Now for any interval Ij in the partition we have that lim
N→∞
{n ≤ N : xn ∈ Ij }/N → l(Ij ).
It follows that
L ≤ lim
N→∞
1
N
∑
n≤N
f (xn) ≤ U.
Since this is true for any partition we conclude that lim N →∞ 1
N
∑
n≤N
f (xn) → ∫
T
f dl.
(⇒) Let I be an open set in T. It is enough to consider the case when I is an interval. It is easy to construct a sequence of continuous functions fm and gm (using Urysohn’s lemma, say) such that
fm ≤ χI ≤ gm and lim m→∞
∫ 10 gm dx = lim m→∞
∫ 10 fm dx = l(I). It follows that 1
N
∑
n≤N
fm(xn) ≤ 1
N
∑
n≤N
χ(xn) ≤ 1
N
∑
n≤N
g(xn).1.2. EQUIDISTRIBUTION MOD 1 AND THE WEYL CRITERION 3Taking the limit N → ∞ , we obtain for any m,
∫
I
fmdl ≤ lim
N→∞
1
N
∑
n≤N
χI (xn) ≤
∫
I
gmdl.
Now, by our choice fm and gm, as m → ∞ , we have that lim
m→∞
∫
I
fmdl = l(I) = lim
m→∞
∫
I
gmdl.
It follows that lim N →∞
∑
n≤N
χI (xn)/N = l(I). But observe that, 1
N
∑
n≤N
χI (xn) = #{n ≤ N : xn ∈ I}
N . (1.3) This completes the proof. We remark that for a general space ( X, B, μ ) conditions (1.1) and (1.2) are not equivalent. However under mild assumptions on the space X (normality, locally compactness etc), they are equivalent. Since all spaces of our interest (smooth manifolds, homogeneous spaces for Lie groups etc) satisfy these conditions, we shall take the analog of (1.2) as our definition of equidistribution. More explicitly
Definition 1.3. Let X be a locally compact topological space and μ a Borel probability measure on it. A sequence of points ( xn) in X is said to be equidistributed with respect to μ if lim
N→∞
1
N
∑
n≤N
f (xn) −→
∫
X
f dμ
for every f ∈ Cc(X) (recall Cc(X) is the space of compactly supported continuous functions on X). The above definition is more useful than the former as it is easier to work with the space of continuous functions than with open sets. In order to state the Weyl criterion in a succinct manner we introduce the little-o notation from analytic number theory. By the notation f (n) = o(g(n)) we mean that lim n→∞ f (n)/g (n) = 0.
Theorem 1.4. (Weyl criterion) A sequence of real numbers ( un) is equidistributed modulo 1 if and only if for every h ∈ Z, h 6 = 0 , we have
∑
n≤x
e(hu n) = o(x).
Proof. Suppose ( un) is equidistributed modulo 1. Observe that, for all h 6 = 0, lim
N→∞
1
N
∑
n≤N
cos(2 πhu n) =
∫ 10
cos(2 πhx )dx = 0 ,4 CHAPTER 1. EQUIDISTRIBUTION IN NUMBER THEORY
where the first equality is due to the previous lemma. A similar statement holds for sin( x). We conclude that ∑
n≤x
e(hu n) = o(x).
(⇒) Conversely, suppose ∑
n≤x
e(hu n) = o(x) for every h 6 = 0. Let f : [0 , 1] → R be a continuous function. We know from basic Fourier theory that finite linear combination of trigonometric poly-nomials are dense in C([0 , 1]) in the uniform metric. That is, for any > 0, there exists a function
p(t) = ∑Mk=−M cke(kt ) such that |f (t) − p(t)| < , for all t ∈ [0 , 1]. From the given condition, it follows that lim
N→∞
1
N
∑
n≤N
p(un) = c0 =
∫ 10
p(t)dt.
Therefore, we have that
∣∣∣ 1
N
∑
n≤N
f (un) −
∫ 10
f (t)dt
∣∣∣ ≤
∣∣∣ 1
N
∑
n≤N
f (un) − 1
N
∑
n≤N
p(un)
∣∣∣ +
∣∣∣∫ 10
(p(t) − f (t)) dt
∣∣∣+
∣∣∣ 1
N
∑
n≤N
p(un) −
∫ 10
p(t)dt
∣∣∣
≤ 2 +
∣∣∣ 1
N
∑
n≤N
p(un) −
∫ 10
p(t)dt
∣∣∣
This implies
∫ 10
f (t)dt − 2 ≤ lim inf
N→∞
1
N
∑
n≤N
f (un) ≤ lim sup
N→∞
1
N
∑
n≤N
f (un) ≤
∫ 10
f (t)dt + 2 .
We deduce that lim
N→∞
1
N
∑
n≤N
f (un) =
∫ 10
f (t)dt.
Corollary 1.5. If α ∈ R is irrational, then ( nα ) is equidistributed modulo 1.
Proof. Let h be a non-zero integer. Since α is irrational, it follows that e(hα ) 6 = 1.
∣∣ ∑
n≤x
e(hnα )∣∣ = |e(hα )|
∣∣∣ e(xhα ) − 1
e(hα ) − 1
∣∣∣ ≤ 2
|e(hα ) − 1| .
By Weyl criterion we obtain the required equidistribution result. Weyl criterion makes establishing equidistribution results considerably easy. (The above equidis-tribution result would have been a lot harder to prove directly). We illustrate this with yet another example (based on ).
Theorem 1.6. For any increasing sequence of integers a1, a 2, . . . , the sequence {anx : n ≥ 1} is uniformly distributed mod 1 for almost all x ∈ R.1.2. EQUIDISTRIBUTION MOD 1 AND THE WEYL CRITERION 5
Proof. Let h ∈ Z \ { 0}. We have that
∫ 10
∣∣ 1
N
∑
n≤N
e(ha nx)∣∣2dx = 1
N 2
∑
m,n ≤N
∫ 10
e(hx (am − an)) dx = 1
N .
Therefore, by replacing N by m2 and summing over m, we get
∫ 10
∑
m≥1
∣∣ 1
m2
∑
n≤m2
e(ha nx)∣∣2dx = ∑
m≥1
1
m2 = π2/6.
If we call the above integrand as f , we have ∫ 10 f (x)dx < ∞. Since f is non-negative, we must have
f (x) < ∞ for almost all x in [0 , 1]. But note that f (x) = f (x + 1), so f (x) < ∞ for almost all x.That is, we have ∑
m≥1
∣∣ 1
m2
∑
n≤m2
e(ha nx)∣∣2 < ∞
for almost all x, and hence lim
m→∞
∣∣ 1
m2
∑
n≤m2
e(ha nx)∣∣ = 0 .
Now if m2 ≤ N < (m + 1) 2 then ∑
n≤N
e(ha nx) = ∑
n≤m2
e(ha nx) + O(m), hence 1
N
∑
n≤N
e(ha nx) = m2
N
1
m2
∑
n≤m2
e(ha nx) + O( mN )and the theorem follows.
Corollary 1.7. Almost all x ∈ R are normal.
Proof. Recall a real number α is normal to base b ∈ N if the sequence ( bnα) is equidistributed distributed mod 1. (That is, each sequence of digits appears, in the expansion of α to the base b,about as often as in a random sequence). And α is normal if it is normal in every base b ≥ 2. To see the proof of the corollary, take an = bn in the above theorem, by noting that the exceptional set has measure zero, as it is the countable union of measure zero sets.
1.2.1 Weyl criterion in greater generality
If one looks carefully at the proof of Weyl criterion for equidistribution mod 1, it will become clear that the crux of the proof has to do with the fact that the linear span of the set of functions
{e(nx ) : n ∈ Z} is dense in the space of continuous functions on T. This idea can be used in many other situations and an analogous Weyl criterion can be devised to prove equidistribution. For concreteness, let us focus on the case of Compact metric spaces. Let X be a compact metric space and let P(X) be the set Borel probability measures on X. Let C(X) be the space of continuous functions on X endowed with the uniform norm (i.e. ‖f ‖ = sup x∈X |f (x)|).
Definition 1.8. A sequence of measures μn ∈ P (X) is said to be equidistributed with respect to 6 CHAPTER 1. EQUIDISTRIBUTION IN NUMBER THEORY
μ ∈ P (X) if they converge to μ in the weak ∗ topology, i.e. for every f ∈ C(X) we have
μn(f ) =
∫
X
f dμ n → μ(f ) =
∫
X
f dμ as n → ∞ . (1.4) When this is the case, we write μn → μ. (More about measures on compact metric spaces can be read in section 3.1). The Weyl criterion generalizes as follows:
Weyl criterion. Let X be a compact metric space and let φn ∈ C(X) be a sequence of functions with the property that their linear combination is dense in C(X). Then μn → μ if and only if
μn(φm) → μ(φm) for all m ∈ N.
Proof. (⇒) Let f ∈ C(X). Given > 0 there exists g = ∑Nk=1 ckφk such that ‖f − g‖ < . (We assume ck’s are non-zero). Write f − g = h. Then
|ν(h)| < for any ν ∈ P (X).
Now,
|μn(f ) − μ(f )| = |μn(g + h) − μ(g + h)| = |μn(g) + μn(h) − μ(g) + μ(h)|≤ | μn(g) − μ(g)| + 2 = ∣∣ N∑
k=1
ckμn(φk) −
N
∑
k=1
ckμ(φk)∣∣ + 2
Let c = max {| c1|, . . . , |cN |} . Now, for each k ∈ { 1, 2, . . . , N } there exists an M (k) ∈ N such that
|μn(φk) − μ(φk)| < /cN for all n > M (k). Let M = max {M (1) , . . . , M (N )}. Then, for n > M we have N∑
k=1
|ck| | μn(φk) − μ(φk)| <
and therefore |μn(f ) − μ(f )| < 3. Thus we have shown μn(f ) → μ(f ) for every f ∈ C(X). Let us look at some examples of spaces for which nice Weyl criterion exists. By the Weyl criterion, as in the above form, there could be plenty of choices for the system of test functions φn; any set that generates a dense subset of C(X) will do, for example any orthonormal basis for L2(X, μ ). But very often there is a natural choice. The Stone-Weierstrass theorem provides us with a tool to find such a system of functions. We recall this theorem, which, we shall have the opportunity to invoke later.
Stone-Weierstrass theorem. Let X be a compact metric space and let A ⊂ C(X) be a linear subspace with the following properties:
• A is closed under multiplication.
• A contains the constant functions.
• A separates points in X.Then A is dense in C(X). Now, let us look again at the example of T. What is so special about the functions e(nx )? T
is a compact abelian Lie group and e(nx ) are precisely the characters of irreducible representations 1.3. EQUIDISTRIBUTION OF RATIONALS AND THE RIEMANN HYPOTHESIS 7(over C) of T. They form an orthonormal basis for L2(T, l ), the space of square integrable functions with respect to l.More generally, let G be a compact Lie group. The famous Peter-Weyl theorem asserts that matrix coefficients of irreducible representations of G, form an orthonormal basis for L2(G). Similarly for X = G#, the space of conjugacy classes of G, the characters of irreducible representations of G
form an orthonormal basis for L2(X, dg ) where dg is the measure derived from the Haar measure on
G. Also, the integral of any non-trivial character over X is zero. If ( un) is a sequence of points on
X, the Weyl criterion then reads: ∑
n≤x
T r (ρ(un)) = o(x)for all irreducible representations ρ of G.Although we won’t have the occasion to work with general Lie groups, it is important to note that the above choices of test functions provides a potent tool for establishing equidistribution results.
1.3 Equidistribution of rationals and the Riemann hypothe-sis
Any rational number α can be uniquely represented as a fraction a/b in it lowest terms, that is, a pair of integers a and b with b > 0 and the greatest common divisor ( a, b ) = 1. We define the height of α by
H(α) = max {| a|, b }.
For each Q > 0, we can look at the rational numbers in [0 , 1) of height at most Q. Let us denote them by x1, . . . , x N , where the xi’s are arranged in the increasing order of their magnitude. As we shall soon see, the distribution these numbers is of fundamental importance. But before that we need a lemma.
Lemma 1.9. (Ramanujan sum)
q
∑
a= 1 (a,q ) = 1
e
( ah q
)
= ∑
c|hc|q
cμ
( qc
)
,
where μ(n) is the M¨ obius function.
Proof. We have that ∑
d|n
μ(d) = 0 for n > 1. Also the common factors of ( a, q ) are the common factors of a and q. Therefore the required sum is
q
∑
a=1
∑
d|ad|q
μ(d)e
( ah q
)
= ∑
d|q
μ(d)
q/d
∑
b=1
e
( bh q/d
)
,
where we have put a = bd . The sum over b is zero unless h is a multiple of q/d . Writing c = q/d ,8 CHAPTER 1. EQUIDISTRIBUTION IN NUMBER THEORY
we get the sum to be ∑
d|qq/d |h
μ(d) qd = ∑
c|h
∑
dcd =q
cμ (d) = ∑
c|hc|q
cμ
( qc
)
.
Lemma 1.10. (Weyl sum for the rational numbers) Let Q be a positive integer and let x1, . . . , x N
be the rational numbers of height at most Q in [0 , 1). Then
N = Q2
2ζ(2) + O(Qloq Q),
and for each integer h N∑
1
e(hx n) = ∑
d|h
dM (Q/d ), (1.5) where M (x) = ∑
n≤x
μ(n) is the sum function of the M¨ obius function.
Proof. Put h = 0 in the previous lemma. We get,
N =
Q
∑
q=1
∑
a=1 (a,q )=1
1 =
Q
∑
q=1
∑
cd =q
cμ (d)=
Q
∑
q=1
∑
d|q
qd μ(d) = ∑
d≤Q
μ(d)
d
Q
∑
q=1
q≡0(mod d)
q
= ∑
d≤Q
μ(d)
d
[ 12
Q2
d + O(Q)
]
= Q2
2
∞
∑
d=1
μ(d)
d2 + O
[
Q2
∞
∑
d=Q+1
|μ(d)|
d2 + Q ∑
d≤Q
|μ(d)|
d
]
.
Now, it is easy to see that,
∞
∑
d=Q+1
|μ(d)|
d2 =
∫ ∞
Q
dx x2 + O( 1
Q2 ) = 1
Q + O( 1
Q2 ),
and ∑
d≤Q
|μ(d)|
d ≤ ∑
d≤Q
1
d = log Q + O(1) .
This completes the proof of the first assertion. 1.3. EQUIDISTRIBUTION OF RATIONALS AND THE RIEMANN HYPOTHESIS 9For the second assertion, the sum can be rearranged to give
Q
∑
q=1
q
∑
a=1 (a,q )=1
e
( ah q
)
=
Q
∑
q=1
∑
d|hd|q
dμ
( qd
)
= ∑
d|h
d ∑
r≤Q/d
μ(r) = ∑
d|h
dM (Q/d ).
We had previously defined the notion of equidistribution of a sequence of points in a space X.Similarly one can define the notion of equidistribution for a sequence of finite subsets of X. Given a finite subset E of X we define μE ∈ P (X) as
μE = 1
|E|
∑
x∈E
δx
where δx is the Dirac measure at x.
Definition 1.11. A sequence of finite subsets En ⊂ X, is said to be equidistributed with respect to μ if μEn → μ as n → ∞ .It is time to introduce yet another notation from analytic number theory. Let f be a complex valued function on N and g a non-negative real valued function on N. By the notation f g, we mean that there is a c > 0 such that |f (n)| ≤ cg (n), for all n large enough.
Theorem 1.12. Let EQ be the set of rationals of height at most Q in [0 , 1). Then, as Q → ∞ , E Q
is equidistributed in [0 , 1) with respect to the Lebesgue measure.
Proof. Let x1, . . . , x N correspond to the complete set rationals in [0 , 1) of height at most Q. Using the trivial bound |M (x)| ≤ x in (1.5) we get, for non-zero h,
N
∑
1
e(hx n) ∑
d|h
Q d(h)N 1/2,
where d(h) is the number of divisors of h. That is (generalized) Weyl criterion holds true for each non-zero h. The theorem follows. We now prove a lemma on Dirichlet series which we shall have multiple occasions to use.
Lemma 1.13. Let f (s) = ∑
n≥1
ann−s be a general Dirichlet series. Let A(x) = ∑
x≤n
a(n). Suppose A(x) = O(xα+) for every > 0. Then the Dirichlet series f (s) converges for any s with Re s > α .
Proof. This follows directly from Abel’s partial summation formula
∑
n≤x
a(n)
ns = A(x)
xs + s
∫ x
1
A(u)
u1+ s du 10 CHAPTER 1. EQUIDISTRIBUTION IN NUMBER THEORY
by noting that if Re s > α , as x → ∞ , the first term in the right side goes to zero by assumption, and the integral above is (absolutely) convergent. We now state an important theorem in the converse direction, which we shall use later.
Wiener-Ikehara theorem. Let f be a non-negative, non-decreasing real valued function on [1 , ∞)and suppose that the Mellin transform
g(s) := s
∫ ∞
1
f (x)x−(s+1) dx
exists for Re s > 1. Also, suppose that for some constant α, the function
g(s) − αs − 1has continuous extension to the closed half-plane Re s ≥ 1. Then lim
x→∞
f (x)
x = α.
Remark 1.14. Of special interest is the case when f (x) = A(x), the sum function of a sequence of non-negative integers an. Suppose that the corresponding Dirichlet series φ(s) = ∑
n≥1
ann−s
converges on the half-plane Re s > 1 and has analytic continuation except for a simple pole at s = 1 with residue α. First, note that φ(s) is precisely the Mellin transform of A(x). To see this, in the Abel’s formula quoted above, take s = σ > 1. Then, all the terms involved are positive. Since the series converge for any σ > 1, we must have that A(x)/x σ goes to zero as x → ∞ , for any σ > 1. Now, we can apply the Wiener-Ikehara theorem to deduce that lim x→∞ A(x)/x = α. If φ(s) has analytic continuation with a simple pole at s = b with residue α, we can shift s to s − b and can apply the Wiener-Ikehara theorem to get
A(x) ∼ αx b
b .
We now make some remarks pertinent to the title of this subsection. From the general theory of Dirichlet series mentioned above we know that
M (x) xθ+ (1.6) holds for all > 0 if and only if the Dirichlet series for 1 /ζ (s) converges in the half plane Re( s) > θ ,that is when ζ(s) is non-zero in the region Re( s) > θ . Now, if (1.6) holds true for some θ ≥ 1/2(θ < 1/2 is ruled out by the existence of zeros of ζ(s) on the critical line 1 /2 + it ), we see using (1.5) that N∑
1
e(hx n) ∑
d|h
d(Q/d )θ+ f (h)N θ/ 2+ / 2,
where f (h) = ∑
d|h
d1−θ− (a function polynomially bounded in h). Therefore, the extent to which the rationals are equidistributed depends on the truth or falsity of the Riemann hypothesis. Quoting 1.4. THE MAP X 7 → X MOD P 11 from , “...perhaps this is what the Riemann hypothesis really means”.
1.4 The map x 7 → x mod p
1
Take a large prime. The function inverse modulo p in the interval [1 , 2, . . . , p − 1] has certain “randomness” which is of great importance in analytic number theory (see ), and is also ex-ploited in cryptography. We shall explain what this “randomness” is using the notion of effective equidistribution. Let T2 = R2/Z2. Through out this section, x will mean the inverse of x mod p. Define
S(p) =
{( xp , xp
)
: x ∈ (Z/p Z)×}
⊆ T2
and let
νp = 1
p − 1
∑
w∈S(p)
δw
be the uniform measure on S(p). Let λ be the normalized Lebesgue measure on T2. We shall prove that
νp → λ as p → ∞ . (1.7) The above limit is taken over primes p and the convergence is in the weak ∗ topology. The qualitative (1.7) is an immediate consequence of the following theorem
Theorem 1.15. Let f be a smooth function on T2, define S(f ) by
S(f )2 = ‖f ‖22 + ‖∂2f /∂x 2‖22 + ‖∂2f /∂y 2‖22.
Then there exists a κ > 0 such that for all f ∈ C∞(T2) one has
|νp(f ) −
∫
f dλ | p−κS(f ). (1.8) We shall prove theorem 1.15 with the exponent κ = 1 /4. Clearly, (1.8) is an effective version of (1.7), as any qualitative statement that can be deduced from (1.7) can be made quantitative using (1.8). For example from (1.7), we can conclude that the set S(p) intersects , the box [0 .5, 0.51] × [0 .7, 0.71] for p sufficiently large, whereas using (1.8) we can compute how large p should be for this to happen. We say a sequence of measures μn is effectively equidistributed with respect to an ambient measure
λ when an estimate of the type (1.8) holds true.
Kloosterman Sums . Since f is a smooth function on T2, write f in its Fourier series.
f (x, y ) = ∑
n∈Z2
ˆf (n)en(x, y ), (1.9)
1
Based on a lecture by Farrell Brumely, at the Summer School on Analytic Questions in Arith-metic (AQUA), 2010. 12 CHAPTER 1. EQUIDISTRIBUTION IN NUMBER THEORY
where if n = ( n1, n 2) we have put
en(x, y ) = e(n1x + n2y).
A direct computation using the above Fourier expansion will show that the Sobalev norm of f
as defined above can be written as
S(f )2 = ∑
n
(1 + (2 π)2‖n‖4)| ˆf (n)|2. (1.10) As f is continuously differentiable, the series (1.9) is absolutely convergent. We may therefore interchange the order of summation to obtain
νp(f ) = ∑
n
ˆf (n)ˆ νp(n), (1.11) where ˆ νp(n) = νp(en), is the Fourier coefficient of νp at the harmonic en. More explicitly we have, ˆνp(n) =
∫
en(w)dν p(w) = ( p − 1) −1 ∑
x∈(Z/p Z)×
e
( n1x + n2xp
)
.
The last sum is called the classical Kloosterman sum .Note that ˆf (0) = ∫
T2
f dλ and also that if n ∈ pZ2 then ˆ νp(n) = 1. Therefore the series (1.11) may be rewritten as
νp(f ) −
∫
T2
f dλ = ∑
n6∈pZ2
ˆf (n)ˆ νp(n) + E(p), (1.12) where
E(p) = ∑
n∈pZ2{ 0}
ˆf (n) = ∑
n∈pZ2{ 0}
1
‖n‖2 ‖n‖2 ˆf (n)
p−2( ∑
n∈Z2
‖n‖4| ˆf (n)|2)1/2
≤ p−2S(f ).
We have used the Cauchy-Schwartz inequality and (1.10) to arrive at the above result. Now, as we shall see shortly , and as first demonstrated by Kloosterman in , if n 6 ∈ pZ2 then ˆνp(n) p−1/4 (1.13) uniformly in n. Inserting this in (1.12), we get
|νp(f ) −
∫
T2
f dλ | p−1/4 ∑
n∈Z2
| ˆf (n)| + p−2S(f ).1.4. THE MAP X 7 → X MOD P 13 But ∑
n∈Z2
| ˆf (n)| ≤ S (f ) using the Cauchy-Swartz trick as above. Hence we have
|νp(f ) −
∫
T2
f dλ | p−1/4S(f ).
Therefore theorem 1.15 follows from Kloosterman’s estimate (1.13).
Proof of Kloosterman bound (1.13)
Let n = ( n1, n 2), be such that n 6 ∈ pZ2. Put Kp(n) = ( p − 1)ˆ νp(n). If a is relatively prime to
p, then as x varies through residues modulo p, ax also varies through residues modulo p. Therefore we have,
Kp(n1, n 2) =
p−1
∑
x=1
e( n1ax + n2ax p ) = Kp(an 1, an 2).
Now, if p|n2 then p - n1, in which case Kp(n) = 0. Therefore we may assume p - n1, n 2, so that the sum
M4(p) =
p−1
∑
r=0
p−1
∑
s=0
|Kp(r, s )|4
contains p − 1 copies of |Kp(n1, n 2)|4. Hence, (p − 1) |Kp(n1, n 2)|4 ≤ M4(p). (1.14) Now, we may expand |Kp(r, s )|4 as
p−1
∑
m1=1
p−1
∑
m2=1
p−1
∑
m3=1
p−1
∑
m4=1
e( rA + sB p ),
where A = m1 + m2 − m3 − m4, B = m1 + m2 − m3 − m4.
On rearranging the order of summation, it follows that
M4(p) = ∑
mi
∑
r,s
e( rA + sB p ) = ∑
mi
{∑
r
e( rA p )}{ ∑
s
e( sB p )}.
The innermost sums are easy to evaluate. In fact we have
p−1
∑
r=0
e( rA p ) =
0, p - A, p, p | A,
and similarly for the sum over s. We therefore conclude that
M4(p) = p2.#{(m1, m 2, m 3, m 4) : p | A, B }.
Let m1 + m2 = m3 + m4 (mod p) and m1 + m2 = m3 + m4 (mod p). Multiplying the two equations we get m1m2 + m1m2 = m3m4 + m3m4 (mod p). Put α = m1m2 and β = m3m4.We have α + α = β + β = c(mod p) (say), i.e. α and β both satisfy the quadratic equation
x2 − cx + 1 = 0(mod p). We conclude that α = β or α = β. Consider the case α = β. Then the 14 CHAPTER 1. EQUIDISTRIBUTION IN NUMBER THEORY
equations m1 + m2 = m3 + m4 mod p and m1m2 = m3m4(mod p) holds true. Multiplying both, we get, m1(m1m2 + 1) = m3(m3m4 + 1)(mod p). This implies that either m1 = m3(mod p) or (m1m2 + 1) = ( m3m4 + 1) = 0(mod p). The latter is equivalent to m1 + m2 = m3 + m4 = 0 mod p.
The case α = β can be treated similarly. We have just shown that if p | A, B then either m3, m 4 is a permutation of m1, m 2 or m1 + m2 =
m3 + m4 = 0(mod p). Thus there are at most 3( p − 1) 2 available sets of values for ( m1, m 2, m 3, m 4). It follows that
M4(p) ≤ 3p2(p − 1) 2 < 3p3(p − 1) .
We now deduce from (1.14) that
|Kp(n)| < 31/4p3/4 (n 6 ∈ pZ2).
The Kloosterman bound (1.13) follows. Chapter 2
Linnik’s Problem
Let α = ( x1, x 2, x 3) ∈ Z3 with |α|2 = x21 + x22 + x23. For n ∈ Z+ the set
Vn = {x = α/ |α| : α ∈ Z3; |α|2 = n}
lies on the unit sphere S2. As was first observed by Legendre, Vn is non-empty if and only if
n 6 = 4 a(8 b+7), for a, b integers, a non-negative. In Linnik asked if the set Vn gets equidistributed on S2 with respect to the Lebesgue measure dσ as n → ∞ , subject to the condition that n ≡
1, 2, 3, 5, 6(mod 8). He was able to prove this using an “ergodic method”, under an additional hypothesis on n that ( np
) = 1, for a small fixed prime p. Much later (in 1987) Iwaniec made a breakthrough in the estimation of Fourier coefficients of modular forms of half-integral weight in , which allowed this condition to be removed. We shall see soon that modular forms enters the picture via the Weyl criterion. What is a good choice for the system of test functions to apply Weyl criterion? One should be looking for an orthonormal basis for L2(S2). We shall see that there is indeed a ‘natural’ choice for such a basis. We begin by noting the following; if we let e(θ) = ( x + iy )/|x + iy , then for m > 0
e(mθ ) =
( x + iy
|x + iy |
)m
and
e(−mθ ) =
( x − iy
|x − iy |
)m
.
Now ( x+iy )m and ( x−iy )m are homogeneous harmonic polynomials on R2. This example generalizes nicely to R3 (in fact to Rn). Let H ⊂ C(S2) be the subspace of finite sum of homogeneous harmonic polynomials in R3 restricted to S2. Clearly, H is multiplicatively closed and contains the constant functions. Further, it can be shown that they separate points in S2. Therefore, by the Stone-Weierstrass theorem H is dense in C(S2). Also, since they are eigen functions of the spherical laplacian, their integral over S2 with respect to the Lebesgue measure σ is zero. Let P be a homogeneous harmonic polynomial of degree l ≥ 1. From the preceding discussion, in order to show a sequence of measures μn equidistribute to σ, it is sufficient to show that μn(P ) → 015 16 CHAPTER 2. LINNIK’S PROBLEM
as n → ∞ for every homogeneous harmonic polynomial P . Linnik’s problem asks if μVn → σ as
n → ∞ through admissible values of n. By Weyl criterion it is sufficient to show 1#Vn
∑
x∈Vn
P (x) → 0 as n → ∞
through admissible values. Equivalently, we require
∑
α∈Z3
|α|2=n
P
( α
|α|
)
= o(r3(n)) (2.1) where r3(n) = # {α ∈ Z3 : |α|2 = n} = |Vn|.The above bound is established by noting that the left side of (2.1) is essentially the fourier coefficient of a modular form. Define
θP (z) = ∑
α∈Z3
P (α)e(|α|z ) =
∞
∑
n=1
r(n, P )e(nz ).
Shimura in proves the following theorem (in a vastly more general form).
Theorem 2.1. The function θP (z) is a holomorphic cusp form of weight 3 /2 + l for Γ 0(4). Also
θP (z) = 0, for l odd.
Proof. See . Note that
r(n, P ) = ∑
|α|2=n
P (α).
Since P is homogeneous of degree l, we have P (α/ |α|) = |α|−lP (α), so that
r(n, P ) = nl/ 2 ∑
|α|2=n
P
( α
|α|
)
. (2.2) To prove (2.1), we will also need bounds on r3(n)(= |Vn|). It was Gauss (in ) who first discovered some remarkable algebraic structure in the set Vd, when d is square free. In modern language, he had proved that the ideal class group of the quadratic order Z[√−d] acts transitively on the quotient SO 3(Z)\Vd. (See for the details, and for an exposition of Linnik’s original approach in modern language). From this, Gauss was able to prove
r3(n) = 24 h(d)
w(d)
(
1 −
( d
2
))
, (2.3) where d = disc( Q(√−n)), h(d) the class number of Q(√−n) and w(d) the number of roots of unity in this field. ( d
2
), of course, is the quadratic symbol. Now, it is a famous (alas non-effective!) theorem of Siegel that h(d) |d|1/2−. It follows that
r3(n) n1/2−. (2.4) 2.1. RATIONAL POINTS ON THE SPHERE 17 Suppose we had a theorem of the following form: If f (z) = ∑ ane(nz ) is a cusp form of half integral weight k for Γ 0(N ) (where 4 | N ), then there exists a δ > 0 such that for each n we have
|an| nk/ 2−1/4−δ . (2.5) Then with k = 3 /2 + l, combining (2.2), (2.4) and (2.5), we would have 1
r3(n)
∑
|α|2=n
P (α/ |α|) = O(n−δ+) (2.6) and Linnik’s conjecture would follow. In fact, Iwaniec had proven precisely an estimate of type (2.6) with δ = 1 /28. We shall settle for a weaker estimate with δ = 1 /222 (as given in ). We shall see later that the bound (2.5), as stated, is false for general n. (One has to make an assumption like
n is square; but we shall ignore it for the time being and return to it later.) The bound (2.5), may at first, seem out of the blue. But we shall, in due course, explain the naturality in (asking for) it. At this stage, rather than embarking on proving Iwaniec’s bound, we shall consider an ε-modification of the Linnik’s problem which will (hopefully) shed some light on the original problem. Our presentation is based on .
2.1 Rational points on the sphere
Let R be the set of rational points on the unit sphere, i.e. R = Q3 ∩ S2. Define the height h(x)of a rational point x ∈ R as the least common denominator of its coordinates in reduced form. We shall show that the rational points of height ≤ T become equidistributed on S2 with respect to σ as
T → ∞ .For a function ρ on S2, define
A(T, ρ ) = ∑
x∈R
h(x)≤T
ρ(x).
Thus A(T, 1) is the number of rational points on on S2 of height ≤ T .
Theorem 2.2. As T → ∞ , A(T, 1) ∼ 32κ T 2, where κ = 1 /12 − 1/32 + 1 /52 − 1/72 + · · · ' 0.9159 is the Catalan’s constant. For any continuous function ρ : S2 → C we have
A(T, ρ )/A (T, 1) →
∫
S2
ρ dσ as T → ∞ .
Proof. As before we may restrict our attention to homogeneous harmonic polynomials. Let P be a such a polynomial of degree l ≥ 0 (if l = 0 then P is just the constant function 1). Define
a(n, P ) = ∑
x∈R
h(x)= n
P (x)18 CHAPTER 2. LINNIK’S PROBLEM
and consider the Dirichlet series
φ(s, P ) = ∑
n≥1
a(n, P )n−s.
We shall soon see that this Dirichlet series has analytic continuation and a functional equation, owing to the fact that a(n, P )’s are related to r(n, P )’s, which are Fourier coefficients of modular forms. We now derive this relation. Put b(n, P ) = r(n, P )n−l/ 2, i.e. the sum in (2.2). Consider the set of integral vectors
V = {(x1, x 2, x 3, y ) ∈ Z4 : x21 + x22 + x23 = y2, y > 0 and gcd (x1, x 2, x 3, y ) = 1 }
and observe that the map (x1, x 2, x 3, y ) → (x1/y, x 2/y, x 3/y )gives a bijection from V onto R, where y is the height of the image of ( x1, x 2, x 3, y ). It follows that
b(n2, P ) = ∑
d|n
a(d, P )which on M¨ obius inversion gives
a(n, P ) = ∑
d|n
b(d2, P )μ(n/d ).
This is equivalent to the following identity of Dirichlet series
φ(s, P ) = ζ(s)−1 ∑
n≥1
b(n2, P )n−s. (2.7) For the first part of the theorem, put P = 1, and note that b(n2, 1) = r3(n2). By a classical result of Hurwitz we have the following identity
∑
n≥1
b(n2, 1) n−s = 6(1 − 21−s) ζ(s)ζ(s − 1)
L(s, χ −4)where χ−4(p) = ( −4
p
) is the Kronecker symbol. This gives
φ(s, 1) = 6(1 − 21−s) ζ(s − 1)
L(s, χ −4) , (2.8) which is holomorphic for Re( s) > 1, except for a simple pole at s = 2 with residue 3 /κ , where κ =
L(2 , χ −4)−1. By applying the Wiener-Ikehara theorem (see remark 1.14) we derive the asymptotic relation
A(T, 1) ∼ 32κ T 2 as T → ∞ .2.1. RATIONAL POINTS ON THE SPHERE 19 To finish the proof of the theorem, by Weyl criterion, we need to show that
T −2A(T, P ) = T −2 ∑
n≤T
a(n, P ) → 0 (2.9) as T → ∞ for any P with degree l > 0. Furthermore, we may assume that l is even (as for odd l,
P (−x) = −P (x) and A(T, P ) = 0). The analog of Hurwitz’s result (2.8) for a general P is given by the Shimura’s correspondence. In , Shimura introduced a family of correspondence between modular forms of half-integral weight and modular forms of even integral weight. We state his theorem and specialize it to our case.
Shimura Map. Let t be a positive square-free integer, and suppose f (z) = ∑∞
n=1
a(n)e(nz ) ∈
Sk+1 /2(Γ 0(4 N ), ψ ), where k is a positive integer. If the numbers A(n) are defined by
∑
n≥1
A(n)n−s = L(s − k + 1 , ψχ k
4
χt) ∑
n≥1
a(tn 2)n−s, (2.10) where χt = ( t ) is the usual Kronecker symbol modulo t, then F (z) = ∑∞
n=1
A(n)e(nz ) ∈ M2k(2 N, ψ 2). Moreover, if k > 1 then F(z) is a cusp form. We shall take t = 1, f (z) = θP (z) = ∑
n≥1
r(n, P )e(nz ). In this case N = 1 , ψ = 1 and k = l + 1. Since l is even, k is odd and χk
4
= χ4.
A(n)’s are defined by
∑
n≥1
A(n)n−s = L(s − l, χ 4) ∑
n≥1
r(n2, P )n−s.
Substituting b(n2, P )nl for r(n2, P ) and using (2.7), we get
∑
n≥1
A(n)n−s = L(s − l, χ 4)φ(s − l, P )ζ(s − l). (2.11) Shimura’s theorem says, F (z) := ∑
n≥1
A(n)e(nz ) is a cusp form of weight 2 l + 2 for Γ 0(2). If we define the (normalized) L-function associated F by
L(s, F ) = ∑
n≥1
A(n)n−l− 12 n−s, (2.12) it is clear from (2.12) that
φ(s, P ) = L(s − 1/2, F )
ζ(s)L(s, χ 4) . (2.13) Now, by the famous Deligne bound (originally a conjecture by Ramanujan),
A(n) nl+ 12 +
for any > 0. It follows from lemma 1.13 that the Dirichlet series (2.12) converges absolutely for Re( s) > 1, hence that for φ(s, P ) in (2.13) converges absolutely for Re( s) > 3/2. A consideration 20 CHAPTER 2. LINNIK’S PROBLEM
using Abel’s summation formula, similar to the one in remark 1.14, will imply that
∑
n≤T
a(n, P ) T 32 +
for any > 0. This completes the proof, as we have proved (2.9).
2.2 Modular forms of half integral weight
Jacobi’s theta series
θ(z) = ∑
n∈Z
e(n2z),
has the following remarkable transformation property for any γ = ( a b c d
) ∈ Γ0(4):
θ(γz ) =
( cd
)
ε−1
d
(cz + d)1/2θ(z),
where ( cd
) is the extended Legendre symbol (see ) and εd is 1 or i depending on whether d is 1 or 3 mod 4. (As usual √z denotes the branch which is positive on R+). Define
j(γ, z ) :=
( cd
)
ε−1
d
(cz + d)1/2.
Definition 2.3. Let 4 |N and k be a 1/2 integer. A modular form f (z) of weight k for Γ 0(N ) is a holomorphic function on H satisfying (i) f (γz ) = ( j(γ, z )) 2kf (z) for γ ∈ Γ0(N )(ii) f (z) is holomorphic at each cusp. (See [16, Chapter 4] for the precise meaning of (ii).) We are interested in finding ‘good’ bounds for Fourier coefficients of cusp forms of half integral weight. Firstly, we derive what is called as the ‘trivial’ bound.
Proposition 2.4. Let f ∈ Sk(Γ) and with Fourier coefficients an, then
an = O(nk/ 2).
Proof. Since Im( γz ) = Im( z)/ |(cz + d)|2, F (z) = |f (z)|yk/ 2 is Γ invariant. Also, since f vanishes at cusps, F (z) is bounded on H, i.e. |F (z)| < M (this condition on F is equivalent to f being a cusp form, see ). Now, the coefficient an may be expressed as
an = e2πny
∫ 10
e(−nx )f (x + iy )dx.
Hence |an| ≤ M e 2πny y−k/ 2. Put y = 1 /n to get the required result. It is an important result in the theory of modular forms that, for any congruence subgroup Γ,
Mk(Γ)-modular forms of weight k, form a finite dimensional vector space over C. Furthermore, 2.2. MODULAR FORMS OF HALF INTEGRAL WEIGHT 21
Sk(Γ)-the space of cusp forms of weight k, is a finite dimensional Hilbert space. For f, g ∈ Sk(Γ), the inner product, called the Peterson inner product, is defined as
〈f, g 〉 =
∫
Γ\H
f (z)g(z)yk dxdy y2 .
For any congruence subgroup Γ there is a canonical way to generate cusp forms of weight k. For given cusp p of Γ, the idea is to average out the factor of automorphy over coset representatives of Γp (the stabilizer of p in Γ). This is called the Poincare series at p. We shall be only considering the case p = ∞. More precisely, the m-th Poincare series of weight k at ∞ is defined by
Pm(z, k ) = ∑
γ∈Γ∞\Γ
(j(γ, z )) −2ke(mγz ).
If we assume k > 2, then it can be shown that the above series converges absolutely. The importance of Poincare series lies in the fact that Pm(z, k ) ∈ Sk(Γ) (which we assume) and that
Proposition 2.5. The Pm(z, k ), m ≥ 1 span the space of cusp forms Sk(Γ).
Proof. Let f ∈ Sk(Γ) then
〈Pm, f 〉 =
∫
Γ\H
Pm(z)f (z)yk dxdy y2
=
∫
Γ\H
∑
γ∈Γ∞\Γ
j(γ, z )) −2ke(mγz )f (z)yk dxdy y2
=
∫ ∞
0
∫ 10
e(mz )f (z)yk dxdy y2
= am
(4 πm )k−1 Γ( k − 1) ,
(2.14) where f (z) = ∑∞
n=1
ane(nz ) and Γ is the gamma function. It follows that if 〈Pm, F 〉 = 0 for all m,then am = 0 for all m, i.e. f ≡ 0. Hence Pm’s span Sk(Γ). From now on, we shall exclusively focus on Poincare series, as bounds on fourier coefficients of
Pm’s will imply similar bounds for general cusp forms. A computation will show that if we write
Pm(z, k ) = ∑
n≥1
ˆPm(n)e(nz )then ˆPm(n) = 2
( nm
)(k−1) /2
δm,n + 2 πi −k ∑
c≡0( N)
c> 0
Jk−1
( 4π√mn c
) K(m, n, c )
c
(2.15) where
Jk−1(z) = ∑
l≥0
(−1) l
l!Γ( l + k)
( z
2
)k−1+2 l
(2.16) 22 CHAPTER 2. LINNIK’S PROBLEM
is the J-Bessel function and
K(m, n, c ) = ∑
dmod c
(d,c )=1
( cd
)2k
ε−2kd e
( md + nd c
)
(2.17) is a Kloosterman sum closely related to the one we have met in section 1.2. Let f1, . . . , f R be an orthonormal basis for Sk(N ) (with respect to the Petersson inner product). If we write fj = ∑
n≥1
ˆfj (n)e(nz ), then (2.14) gives
Pm(z, k ) =
R
∑
j=1
〈Pm, f j 〉fj
= Γ( k − 1) (4 πm )k−1
R
∑
j=1
ˆf j (m)fj (z).
Hence ˆPm(n) =
R
∑
j=1
〈Pm, f j 〉 ˆfj (n) = Γ( k − 1) (4 πm )k−1
R
∑
j=1
ˆf j (m) ˆfj (n).
Substituting for ˆPm(n) from (2.15) and setting m = n we get ˆPn(n) = Γ( k − 1) (4 πm )k−1
R
∑
j=1
| ˆfj (n)|2 = 1 + 2 πi −k ∑
c≡0( N)
Jk−1
( 4πn c
) K(n, n, c )
c . (2.18) The above equation is called the Petersson Formula , and it will form the backbone for our estimates of ˆfj (n)’s. In order to do so, we must evaluate K(m, n, c ) in a form in which we can see cancelation in the sum in (2.18).
2.3 Sali´ e sums
In the case of our interest (i.e. for Linnik’s problem) k = 3 /2 + l, where l is even, so ε−2kd = ε−3
d
= εd
(recall εd takes 1 or i as its value). Substituting this in (2.17) gives
K(m, n, c ) = ∑
dmod c
(d,c )=1
εd
( cd
)
e
( md + nd c
)
.
An interesting thing about this sum is that it can be evaluated in a simpler form. After using Chinese remainder theorem and quadratic reciprocity we are led to the following
Lemma 2.6. If c = qr with ( q, r ) = 1 and 4 |r then
Kk(m, n, c ) = Kk−q+1 (mq, nq, r )S(mr, nr, q ),2.3. SALI ´E SUMS 23 where for q odd S(m, n, q ) is the Sali´ e sum defined as
S(m, n, q ) = ∑
xmod q
( xq
)
e
( mx + nx q
)
.
Proof. Expand the RHS. Set d = xrr + yqq with x, y ranging over the residue classes modulo q
and r respectively. So d = x (mod q), d = y (mod r). Also d = xrr + yqq (mod c) (as can be directly verified by multiplying d and d and using CRT). By the quadratic reciprocity we have
( cd
)
= ( −1) y−12
q−12
( ry
) ( xq
)
.
We also have εd = εy and ( −1) (y−1) /2( q−1) /2 = εq−1
y
. Combining all this,we get the LHS. We now prove some properties satisfied by S(m, n, q ).
Lemma 2.7. Let ( m, q ) = 1 = ( n, q ).
S(m, n, q ) =
( mq
)
S(1 , mn, q ) (i)
S(1 , n 2, q ) = εq
√q ∑
x2≡1( q)
e
( 2xn q
)
(ii)
Proof. (i) Since ( m, q ) = 1, do a change of variable by setting y = mx , (hence x = my, x = my ). We get
S(m, n, q ) = ∑
y(q)
( my q
)
e
( y + mny q
)
=
( mq
)
S(1 , mn, q )as ( mq
) = ( mq
) .(ii) The proof is via Gauss sum. Let
G(a, b ; q) = ∑
x(q)
e
( ax 2 + bx q
)
.
It is a classical evaluation that
G(a, 0; q) = εq
√q
( aq
)
.
Set A = ∑
x2≡n2(q)
e(2 x/q ). Since ( n, q ) = 1, it follows that we need to show
S(n2, 1; q) = εq
√q A. 24 CHAPTER 2. LINNIK’S PROBLEM
Using the trivial fact that ∑
a(q)
e(ay/q ) = q or 0 depending on y ≡ 0( q) or not, we get that
A = 1
q
∑
x(q)
e
( 2xq
) ∑
a(q)
e
( a(x2 − n2)
q
)
= 1
q
∑
a(q)
G(a, 2; q)e
( −an 2
q
)
Claim : G(a, b ; q) = 0 if ( a, q ) - b.
Proof : Let d = ( a, q ). Write a = a′d, q = q′d and x = x1 + q′x2. Then the sum over x(mod q)splits as a double sum over x1(mod q′) and x2(mod d). Clearly, we have x2 ≡ x21(mod q′). With this observation, we write
e
( ax 2 + bx q
)
= e
( a′x2
q′
)
e
( b(x1 + q′x2)
q
)
= e
( a′x21
q′
)
e
( bx 1
q
)
e
( bx 2
d
)
.
Now the sum ∑
x2(q)
e( bx 2
d
) is zero, unless d | b.
In the present case, since q is odd and b = 2, the above sum then is
A = 1
q
∑
(a,q )=1
G(a, 2; q)e
( −an 2
q
)
.
But if ( a, q ) = 1 then ax 2 + 2 x = a(( x + a)2 − a2) = a(x + a)2 − a. Therefore,
G(a, 2; q) = e
( −aq
)
G(a, 0; q)= e
( −aq
)( aq
)
εq
√q,
so
A = εq
√q
∑
a(q)
( aq
)
e
( −an 2 − aq
)
.
From the definition of A, note that A = A (as A is obtained by changing x to −x and a to −a), which when combined with the fact that εq = 1 /ε q gives that
εq
√qA = ∑
a(q)
( aq
)
e
( an 2 + aq
)
= S(n2, 1, q )A word on notation. We shall use x to denote inverse of x with respect to some modulus which shall be implicit, or sometimes clear from the context. If x appears in an expression with a denominator y, then generally x is the inverse with respect to y.2.3. SALI ´E SUMS 25
Corollary 2.8. Let q be odd and ( a, q ) = 1, we have
S(n, n, q ) = εq
√q
( nq
) ∑
ab =q
(a,b )=1
e
(
2n
( ab − ba
))
. (2.19)
Proof. Since S(n, n, q ) = ( nq
)S(n2, 1, q ), it is enough, by lemma 2.7(ii) to show that
∑
x2=1( q)
e
( 2xn q
)
= ∑
ab =q
(a,b )=1
e
(
2n
( ab − ba
))
.
Consider y = aa − bb . y mod b = 1 and y mod a = −1 . Hence y2 = 1 mod ( ab )(by CRT). Conversely it can be shown that every solution to y2 = 1( q) can be written in the form aa − bb for some ab = q, (a, b ) = 1.
Corollary 2.9. If q is odd and ( n, q ) = 1, the Sali´ e sums S(n, n, q ) satisfy the bound
|S(n, n, q )| ≤ τ (q)q1/2 (2.20) where τ is the divisor function.
Proof. Clear from (2.19). For any k, the Kloosterman sums (2.17) also satisfies the bound analogous to (2.20). This is a consequence of the deep work of A. Weil on Riemann hypothesis for curves over finite fields.
Weil Bound on Kloosterman sums. The Kloosterman sums K(m, n, c ) satisfy the bound
K(m, n, c ) ≤ (m, n, c )1/2τ (c)c1/2. (2.21) In particular if ( n, c ) = 1 we have
K(n, n, c ) c1/2+
So we have ∑
c> 0
c−σ |K(n, n, c )| n
for any > 0, if σ > 3/2. Also the J-Bessel function satisfies the following bound for x > 0
Jk−1(x) min {xk−1, 1
√x
} ≤ xν , if − 1/2 ≤ ν ≤ k − 1.
Choosing ν = 1 /2 + δ with δ arbitrarily small, we get by putting these bounds in (2.18) that ˆPn(n) n1/2+ δ ∑
c> 0
|K(n, n, c )|
c3/2+ δ n1/2+ δ+.
Since and δ are arbitrary, we conclude that ˆPn(n) n1/2+ for any > 0. (We are abusing the notation by using the same everywhere). It follows readily from (2.18) that ˆfj (n) nk/ 2−1/4+ 26 CHAPTER 2. LINNIK’S PROBLEM
for any > 0, for j = 1 , . . . , R (recall f1, . . . , f R is an orthonormal basis for Sk(N ) ). Since the Fourier coefficient of any f ∈ Sk(N ) is a linear combination of Fourier coefficient of fj ’s we have
Proposition 2.10. Let k be half an odd integer and let f ∈ Sk(N ), (4 | N ), then the n-th Fourier coefficient of f satisfies
an = O(nk/ 2−1/4+ ). (2.22) We now make some important remarks. The bound (2.22) just falls short of (2.5), the bound required to solve the Linnik’s problem. However, we note that (2.22) is essentially the best possible bound for a general n. To see this, consider the theta series θ(z, ψ ) = ∑
m∈Z
ψ(m)e(m2z), where
ψ is a character (mod 4) with ψ(−1) = −1. It is a cusp form of weight 3 /2 for Γ 0(8). Clearly,
|am2 | = m and k/ 2 − 1/4 = 1 /2. Hence a bound of the type (2.5) is simply false in general. So how do we proceed to settle Linnik’s problem now? The best way out is to assume that n’s are square free. This is not a major restriction as we have already solved the Linnik’s problem for squares (this is essentially the content of section 2.1). The case for a general n can be handled similarly (as we did for squares) using the Shimura correspondence. Suppose an’s are Fourier coefficients of a modular form of half integral weight k. Then from Shimura correspondence (2.10) we get (upon inversion) that
a(tn 2) = a(t) ∑
d|n
χ(d)μ(d)dk−3/2A
( nd
)
,
where A(n)’s are Fourier coefficients of a modular form of even integral weight 2 k − 1 (and χ some character). By the Deligne bound, alluded to previously, we have
A(n) nk−1+ .
Hence
|a(tn 2)| | a(t)| ∑
d|n
dk−3/2
∣∣∣ nd
∣∣∣k−1+
= |a(t)|nk−1+ ∑
d|n
d−1/2 | a(t)|nk−1+ .
So if at’s satisfied a bound of the form
a(t) tk/ 2−1/4−δ+,
for all square free t (recall t’s in the Shimura map are square free) for some 0 < δ ≤ 1/4, we would have
|a(tn 2)| (tn 2)k/ 2−1/4−δ+.
With this insight, we shall restrict ourselves to the case of am’s where m is square free. Notice that in proving proposition 2.10 we did not use any cancellation that might occur in the sum in (2.18). As we shall see, improvement in (2.22) for square free n is obtained by exploiting this cancellation. 2.4. IWANIEC’S BOUND 27
2.4 Iwaniec’s bound
Theorem 2.11. Let k ≥ 5/2, 4 | N and f ∈ Sk(N ). Then for n square free the n-th Fourier coefficient of f satisfies the bound
a(n) n k
2−14−1222
τ (n)log (n), (2.23) where the implied constant depends on f
We shall break the proof into several small steps. We first outline the strategy. We know that if f ∈ Sk(N ), then f ∈ Sk(qN ) for any q. Also the Petersson norm of f with respect to Γ 0(qN )(denoted by ‖f ‖2
q
) is related to ‖f ‖2 by the relation
‖f ‖2
q
= [Γ 0(N ) : Γ 0(qN )] ‖f ‖ (2.24) where the index is determined by [Γ 0(N ) : Γ 0(qN )] = qN ∏
p|qN
(1 + p−1)
N ∏
p|N
(1 + p−1) . (2.25) If an is the n-th Fourier coefficient of f , then the n-th Fourier coefficient of normalized form fq =
f / ‖f ‖q (normalized with respect to Γ 0(qN )) is given by ˆf (n) = a(n)/‖f ‖q (2.26) Take an orthonormal basis for Sk(qN ) which contains fq . Using the positivity of Petersson formula (2.18) and substituting for ˆf (n) from (2.26) we get Γ( k − 1) |a(n)|2
(4 πn )k−1‖f ‖2[Γ 0(N ) : Γ 0(qN )] ≤ 1 + 2 πi −k ∑
c≡0( qN )
c−1K(n, n, c )Jk−1
( 4πn c
)
(2.27) The main idea in the proof of Iwaniec is to sum the above inequalities by varying q through primes in some interval of the form ( P, 2P ), exploiting cancellations that occur in the right side due to change in the arguments of the Kloosterman sums. We now prove some lemmas which shall be used in the course of the proof. It is time to introduce yet another notation from analytic number theory. For two real valued functions f and g, by the notation f (x) ∼ g(x) we mean lim x→∞ f (x)/g (x) = 1
Lemma 2.12. Let τ be the divisor function.
∑
n≤x
τ (n) log n ∼ x(log x)2
∑
n≥x
τ (n) log nn2 ∼ 3(log x)2
x28 CHAPTER 2. LINNIK’S PROBLEM
Proof. Recall that the sum of divisor function T (x) = ∑
n≤x
τ (n) satisfies the bound T (x) = x log x+
O(x). Also, recall Abel’s partial summation formula
∑
y<n ≤x
anf (n) = A(x)f (x) − A(y)f (y) −
∫ xy
A(t)f ′(t)dt
where A is the sum function of an’s and f a continuously differentiable function on [0 , 1]. Applying this formula with f (t) = log t we get
∑
n≤x
τ (n) log n = T (x) log( x) −
∫ x
1
T (u)
u du
= ( x log x + O(x)) log x +
∫ x
1
(log u + O(1)) du
= x(log x)2 + O(log x)For the second statement, we use the above result and put f (t) = t−2 in Abel’s formula. We get
∑
n≥x
τ (n) log nn2 = (log x)2
x + 2
∫ ∞
x
(log u)2
u2 du + O
( log xx2 +
∫ ∞
x
log uu2 du
)
.
After evaluating the integrals we get
∑
n≥x
τ (n) log nn2 = 3(log x)2
x + O
( log xx .
)
Lemma 2.13. Let F : Z → C be a periodic function with period m. Let ˆF be its fourier transform. Then ∑
u≤X
F (u) X| ˆF (0) |
m + ‖ ˆF ‖∞ log m
Proof.
∑
u≤X
F (u) = ∑
u≤X
1
m
∑
r(mod m)
ˆF (r)e
( ru m
)
= Xm ˆF (0) + ∑
r(mod m)
r6=0
1
m ˆF (r) ∑
u≤X
e
( ru m
)
Xm | ˆF (0) | + 1
m
∑
r(mod m)
r6=0
| ˆF (r)||1 − e(r/m )| Xm | ˆF (0) | + ‖ ˆF ‖∞
∑
1≤r≤m/ 2
1
r
X| ˆF (0) |
m + ‖ ˆF ‖∞ log m2.4. IWANIEC’S BOUND 29
Corollary 2.14.
∑
u≤X
e
( νu m
)
Xτ (m)( m, ν )
m + m1/2(m, ν )1/2τ (m) log m. (2.28)
Proof. Take F (u) = e ( νu m
). It is a periodic function with period m/ (m, ν ). Its Fourier transform is ˆF (r) = ∑
u(mod m)
e
( νu − ru m
)
,
the classical Kloosterman sum. The corollary follows by using the Weil bound (2.21). The above technique is called completing the sum and is often employed in analytic number theory. Note that since the bound is linear in X, the estimate is also valid for intervals.
Lemma 2.15. (Polya-Vinogradov inequality. ) Let χ be a primitive character modulo D, with
D square free and let r < D . Then, for any a, M, N with M < N
∑
l≡a(mod r)
M≤n≤N
χ(l) D1/2 log D (2.29)
Proof. Define the Gauss sum associated to χ by
g(χ) =
D
∑
m=1
χ(m)e
( mD
)
.
If ( n, D ) = 1, then
χ(n)g(χ) =
D
∑
m=1
χ(m)χ(n)e
( mD
)
=
D
∑
h=1
χ(h)e
( nh D
)
, (2.30) where we have put m ≡ nh (mod D). It is easy to show that if ( n, D ) > 1, then the right side of (2.30) vanishes (by a simple change of variable, as in lemma 2.17.(ii) , and using the primitivity of
χ). Hence (2.30) holds true for any n. From this, we get
|χ(n)|2|g(χ)|2 =
D
∑
h1=1
D
∑
h2=1
χ(h1)χ(h2)e
( n(h1 − h2)
D
)
.
Summing over n over a complete set of residues (mod D), the left side gives φ(D)|g(χ)|2. While on 30 CHAPTER 2. LINNIK’S PROBLEM
the right side, the sum over the exponentials is zero unless h1 ≡ h2. So we get
φ(D)|g(χ)|2 = D ∑
h(D)
|χ(h)|2 = Dφ (D).
We deduce that
|g(χ)| = D1/2. (2.31) Let ( r, D ) = d. Write D = dq . Since r < D , we have d < D , so q > 1. Also, since D is square free, we have ( d, q ) = 1. Hence χ = χ1χ2, where χ1 and χ2 are primitive characters modulo d and q
respectively. Note that χ2 is non-trivial as q > 1. Also, d|r, so the congruence l ≡ a (mod r) implies
l ≡ a (mod d). Therefore, the factor χ1(a) comes out of the sum in (2.29) and what is left is a sum over χ2(l). Further, ( q, r ) = 1, so we could have as well started by assuming ( r, D ) = 1, which we do now. Put l = a + kr . The condition M ≤ l ≤ N gives ⌈ M −ar
⌉ ≤ k ≤ ⌊ N −ar
⌋ . We denote these new bounds by M ′ and N ′ respectively. Then, expressing χ(n) using (2.30), we get
∑
l≡a(mod r)
M≤n≤N
χ(l) = ∑
M′≤k≤N′
χ(a + kr )= 1
g(χ)
D
∑
h=1
χ(h)e
( ah D
) ∑
M′≤k≤N′
e
( hkr D
)
(2.32) Now the inner sum is a GP which satisfies the trivial bound
| ∑
M′≤k≤N′
e
( hkr D
)
| ≤ 2
∣∣1 − e ( hr D
)∣ ∣ .
Note that, as r is relatively prime to D, hr will be a multiple of D only if h = D, in which case there is no contribution to the sum (2. 32) as χ(h) = 0. Hence taking absolute values in (2.32) and using (2.31), we get
D1/2
∣∣∣∣∑
l≡a(mod r)
M≤n≤N
χ(l)
∣∣∣∣ ≤
D−1
∑
h=1
2
∣∣1 − e ( hr D
)∣ ∣ =
D−1
∑
h=1
2
∣∣1 − e ( hD
)∣ ∣ =
D−1
∑
h=1
1
| sin( πh/D )| . (2.33) The first equality above is due to the fact that hr and h runs through the same set of residues (mod D), as ( r, D ) = 1. We now estimate the last sum. For any convex function f (x) we have that
f (x) ≤ 1
δ
∫ x+δ/ 2
x−δ/ 2
f (t)dt.
Taking f (x) = (sin πx )−1, δ = 1 /D we see that
D−1
∑
h=1
1sin( πh/D ) ≤ D
D−1
∑
h=1
∫ hD + 12DhD − 12D
dt
sin πt = D
∫ 1− 12D
12D
dt
sin πt = 2 D
∫ 1212D
dt
sin πt .2.4. IWANIEC’S BOUND 31 Now sin πt > 2t for 0 < t < 1/2, so that 2D
∫ 1212D
dt
sin πt < 2D
∫ 1212D
dt
2t = D log D.
Substitute this back in (2.33) to complete the proof. We now get on with the proof of Iwaniec’s bound. There will be four independent parameters
C, D, P and R whose values will be chosen at the end. We shall be taking q to a prime, in which case the index in (2.25) is just p or p + 1. We average the inequality (2.27) over prime p - n in the interval P < p < 2P , each such p being weighted by log p. We shall choose P later subject to
N (log n)2 < P ≤ n1/6. The resulting left side is estimated with help of the following
Proposition 2.16. ∑
P <p< 2Pp-n
log pp + 1 ∼ log 2
Proof. We may omit the restriction p - n without affecting the result. Define a function θ on N by,
θ(m) = log m if m is a prime and 0 otherwise. We are interested in the asymptotics of
∑
P <k< 2P
θ(k)
k + 1 By the prime number theorem we have that
∑
P <k< 2P
θ(k) ∼ P.
A simple partial summation using the above gives the required result. Therefore, the resulting left side is asymptotically μn 1−k|a(n)|2, where μ is a positive constant depending only on f . On the right hand side of our basic inequality we obtain a sum of Kloosterman sums to moduli c ≡ 0(mod N ) weighted by
ω(c) = ∑
P <p< 2Pp-n,p |c
log p.
We clearly have ω(c) ≤ log c if c > 0. The contribution from the constant term 1 on the right side is ω(0) ∼ P (by the prime number theorem). Therefore we have
n1−k|a(n)|2 P + |S| (2.34) where S is the weighted sum of Kloosterman sums,
S = ∑
c≡0(mod N)
ω(c)c−1K(n, n, c )Jk−1
( 4πn c
)32 CHAPTER 2. LINNIK’S PROBLEM
The Bessel function satisfies the bound
Jk−1
( 4πn c
)
min
{( cn
)1/2
,
( nc
)3/2}
.
This along with Weil bound (2.21) implies that terms with c ≤ C contribute to S at most
S1 n−1/2 ∑
c≤C
(c, n )1/2τ (c) log c Cn −1/2(τ (n) log n)2, (2.35) where the last inequality is a consequence of
Proposition 2.17. If C = nα, then
S∗
1
:= ∑
c≤C
(c, n )1/2τ (c) log c C(τ (n) log n)2. (2.36)
Proof. The required sum is nothing but
∑
c≤C
∑
d|n
(c,n )= d
d1/2τ (c) log c = ∑
d|n
d1/2 ∑
c≤C
(c,n )= d
τ (c) log c.
But ∑
d|n
d1/2 ∑
c≤C
(c,n )= d
τ (c) log c ≤ ∑
d|n
d1/2 ∑
c≤Cd|c
τ (c) log c
Writing c = kd , and using τ (kd ) ≤ τ (k)τ (d), the last sum is
∑
d|n
d1/2τ (d) log d ∑
k≤C/d
τ (k) log k
Estimating the inner sum using lemma 2.12 we get that
S∗
1
≤ C(log C)2 ∑
d|n
τ (d)(log d)3
d1/2 C(τ (n) log n)2.
For the last bound we have used the fact that C is a power of n and that τ (d)(log d)3 d1/4(say), and hence the sum over d is τ (n) (τ (n)) 2.Similarly, using lemma 2.12, terms with c ≥ D contribute to S at most
S3 n3/2 ∑
c≥D
(c, n )3/2τ (c)c−2 log c D−1n3/2(τ (n) log n)2. (2.37) We shall later choose C = nα < n < D = nβ so that the above bound will be valid. In fact we shall 2.4. IWANIEC’S BOUND 33 choose α and β quite close to 1. We are then left with estimating the central term
S2 = ∑
C<c<D c≡0(mod N)
ω(c)c−1K(n, n, c )Jk−1
( 4πn c
)
.
We set c = qr where q is the largest factor of c coprime with nN , so r has all its prime factors in
nN and is divisible by N . Therefore ω(c) = ω(q) and S2 splits into
S2 = ∑
r|(nN )∞
r≡0(mod N)
r−1Tr (2.38) where
Tr = ∑
C<qr<D
(q,nN )=1
ω(q)q−1K(n, n, qr )Jk−1
( 4πn qr
)
.
We first estimate Tr as follows:
Tr (n, r )1/2r1/2τ (r) ∑
q
τ (q)q−1/2 min
{( qr n
)1/2
,
( nqr
)3/2}
log q,
where we have used the fact that ( q, r ) = 1 and hence ( n, qr ) = ( n, r ), τ (qr ) = τ (q)τ (r).
Proposition 2.18.
T ∗
r
:= ∑
q
τ (q)q−1/2 min
{( qr n
)1/2
,
( nqr
)3/2}
log q r−1/2n1/2(log n)2. (2.39)
Proof.
T ∗
r
= r1/2n−1/2 ∑
q<n/r
τ (q) log q + n3/2r−3/2 ∑
q≥n/r
q−2τ (q) log q.
Applying lemma 2.12 we get the required result (recall that qr < D hence r < D ; therefore we can replace log r terms with log n terms). We conclude that
Tr τ (r)( n, r )1/2n1/2(log n)2.
We shall use this bound only for r sufficiently large, say r > R . Hence the contribution to S2 of terms with r > R is bounded by
∑
r>R
r−1Tr R−1/2n1/2(log n)2 ∑
r
τ (r)( n, r )1/2r1/2.
Recall that all primes factors of r are in nN . Hence the complete sum over r above is given by a finite product over primes in nN which is estimated by O(τ (n)2). So, we get
∑
r>R
r−1Tr R−1/2n1/2(τ (n) log n)2. (2.40) 34 CHAPTER 2. LINNIK’S PROBLEM
It remains estimate Tr for r ≤ R. By lemma 2.6 the Kloosterman sum in Tr factors into
K(n, n, c ) = K(nq, nq, r )S(nr, nr, q )where qq = 1(mod r), rr = 1(mod q) and S(nr, nr, q ) is the Sali´ e sum. Since the Kloosterman sum
K(nq, nq, r ) depends only on q(mod r), we can split the sum over q in Tr into residue classes mod r
to obtain
Tr ≤ ∑∗
s(mod r)
|K(ns, ns, r )Trs | (2.41) where
Trs = ∑
C<qr<D q≡s(mod r),(q,nN )=1
ω(q)q−1S(nr, nr, q )Jk−1
( 4πn qr
)
.
The Sali´ e sum was evaluated in Corollary 2.8; we have
S(nr, nr, q ) = εq q1/2
( nr q
) ∑
ab =q
(a,b )=1
e
(
2nr
( ab − ba
))
.
If ( a, b ) = 1, then we clearly have aa + bb = 1(mod ab ) (by CRT). From this we get the following ‘reciprocity formula’
ab + ba ≡ 1
ab (mod 1) From now on we shall assume ( a, b ) = 1 without explicit mention. Using the above reciprocity formula we write
∑
ab =q
e
(
2nr
( ab − ba
))
= 2 Re ∑
ab =qa<b
e
(
2n bar + 2 n br a − 2nabr
)
.
Recall r ≡ s(mod r) and N | r and 4 | N , so ( rq
) = ( rs
). Using this, and inserting the above expression into Trs we get
Trs = εs
( rs
) ( 2rn
)1/2
Re ∑
C<abr<D a<b, ab ≡s(mod r)
ω(ab )
( nab
)
e
(
2n bar + 2 n br a
)
j
( 2nabr
)
where
j(x) = x1/2e(−x)Jk−1(2 πx ).j(x) satisfies the estimates j(x) x1/2, j(x) 1 and j′(x) 1. Using these estimates we remove
j(2 n/abr ) using partial summation in b to obtain
Trs C−1(nr )1/2 ∑
a<A
(a,nr )=1
∣∣∣ ∑
a<b<B ab ≡s(mod r)
ω(ab )
( nb
)
e
(
2n(1 + rr ) bar
)∣∣∣2.4. IWANIEC’S BOUND 35 where A = ( D/r )1/2 and some B which depends on a such that C < arB < D . Since ω(ab ) =
ω(a) + ω(b), the innermost sum splits into Va + ω(a)V ′(a), where
Va = ∑
a<b<B ab ≡s(mod r)
ω(b)
( nb
)
e
(
2n(1 + rr ) bar
)
,
and V ′(a) is given by the same sum without the weight ω(b). Thus
Trs C−1(nr )1/2 ∑
a<A
(a,nr )=1
(|Va| + ω(a)|V ′
a
|).
We now turn to estimating Va, the most non-trivial part of the proof. First, observe that if we put
b = pl , with P < p < 2P , (if there is no such prime then ω(b) = 0 by definition) the sum over b
splits into sum over p and l as follows
Va = ∑
P <p< 2P
(log p)
( np
) ∑
a/p<l<B/p apl ≡s(mod r)
( nl
)
e
(
2n(1 + rr ) pl ar
)
. (2.42) Suppose a ≤ P . We now make the following observation: The character ( nl
) is non-trivial on any arithmetic progression to modulus ar because n is square free and larger than ar . Let us look at the sum over l more closely. l is varying over some interval subject to the condition l ≡ aps (mod r). If we further impose the condition that l ≡ λ(mod a), by Chinese remainder theorem (as ( a, r ) = 1) we can split the sum over l into residue classes (mod ar ) as follows
∑
λ(mod a)
∑
a/p<l<B/p l≡aps (mod r)
l≡λ(mod a)
( nl
)
e
(
2n(1 + rr ) pl ar
)
= ∑
λ(mod a)
e
(
2n(1 + rr ) pl ar
) ∑
a/p<l<B/p l≡λ′(mod ar )
( nl
)
.
Put L = B/P , then l < B/p < L . We now apply the Polya-Vinogradov estimate for character sums (lemma 2.15) ∑
l<L l≡λ(mod ar )
( nl
)
n1/2 log n
on the inner sum and the trivial estimate on the outer sum to get that in (2.42) ∑
l
an 1/2 log n.Substituting this back in (2.42) and performing the sum over P we get.
Va aP n 1/2 log n. (2.43) If P < a < A we interchange the sums in Va to get
Va = ∑
l<L
( nl
) ∑
P1<p<P 2
apl ≡s(mod r)
log p
( np
) (
2n(1 + rr ) pl ar
)
,36 CHAPTER 2. LINNIK’S PROBLEM
where P1 = max( P, a/l ) and P2 = min(2 P, B/l ). We now use Cauchy-Schwarz inequality in the outer sum on l (in the form | ∑ni=1 xi|2 ≤ n ∑ni=1 |xi|2 ) to get
V 2
a
≤ L ∑
l<L
∣∣ ∑
P1<p<P 2
apl ≡s(mod r)
log p
( np
) (
2n(1 + rr ) pl ar
)∣∣2.
Squaring out and changing the order of summation back gives
V 2
a
L(log P )2 ∑
P <p 1≤p2<2Pp1≡p2(mod r)
∣∣ ∑
a/p 2<l<B/p 1
ap 1l≡s(mod r)
e
(
2n(1 + rr )( p1 − p2) p1p2lar
)∣∣. (2.44) We have used the trivial bound log p < log P to take it out of the summation. As for the term in the exponential, originally we have ( p2 − p1). But ( p2 − p1) ≡ (p1 − p2)p1p2 (mod a) (recall p|b
and hence p - a, thus we can talk of p1 and p2). Since p1 ≡ p2 (mod r) we also have ( p2 − p1) ≡
(p1 − p2)p1p2 (mod r), and hence ( p2 − p1) ≡ (p1 − p2)p1p2 (mod ar ) which justifies the replacement. The contribution of the terms with p1 = p2 to the sum is O(L2(log P )2 P/ log P ) = O(L2P log P )Suppose p1 6 = p2. We now take a closer look at the inner sum in (2.44). Firstly, the congruence condition in the sum that l ≡ aps (mod r) imposes no restriction on l(mod a) as a and r are relatively prime. Also, r | (p1 − p2) and rr = 1(mod a) (recall it is 1 (mod q) and q = ab ), so the argument in the exponential is essentially 4 n p1−p2
rp1p2la
. Therefore the sum over l is precisely an incomplete sum of the form (2.28). Since ( r, a ) = 1 = ( n, a ), we have ( 4n(p1−p2)
r
, a ) = ( p1 − p2, a ). Applying corollary 2.14 we deduce that the sum over l in (2.44) satisfies
∑
l
(p1 − p2, a )a−1L + ( p1 − p2, a )1/2a1/2τ (a) log a.
Summing over p1 6 = p2, we get
V 2
a
(
L2P + a−1L2P 2 + a1/2P 2)
(τ (a) log n)2.
Since a > P the middle term can be ignored. After substituting L = B/P and taking square we obtain
Va
(
BP −1/2 + a1/4B1/2P 1/2)
τ (a) log n. (2.45) The same estimates hold for V ′
a
by similar arguments. We now use these estimates to bound Trs .By (2.43), terms with a ≤ P contribute to Trs at most
C−1(nr )1/2P n 1/2 log n ∑
a≤P
a C−1(nr )1/2P 3n1/2(log n)2. (2.46) And, by using (2.45) terms with P < a < A contribute to Trs at most
C−1(nr )1/2 ∑
P <a<A
(
BP −1/2 + a1/4B1/2P 1/2)
τ (a) log n. 2.4. IWANIEC’S BOUND 37 Using the fact that B < D/ra and that A = ( D/r )1/2, the above sum can be estimated using Abel’s formula as we did in lemma 2.12. Combining the result with (2.46) we get
Trs C−1(nr )1/2 (
n1/2P 3 + DP −1/2 + D7/8P 1/2)
(log n)2.
We may drop the term n1/2P 3 in comparison to other terms, a step which can be justified after we have chosen P and D. Inserting the rest into (2.41) and summing over s(mod r) using the trivial estimate |K(ns, ns, r | ≤ r we deduce that
Tr r5/2C−1 (
DP −1/2 + D7/8P 1/2)
n1/2(log n)2.
Next summing over r ≤ R, we get
∑
r≤R
r−1Tr R2C−1 (
DP −1/2 + D7/8P 1/2)
n1/2(log n)2 ∑
r
r−1/2.
The sum over r can be estimated to be O(τ (n)) 2. We deduce
∑
r≤R
r−1Tr R2C−1 (
DP −1/2 + D7/8P 1/2)
n1/2(τ (n) log n)2. (2.47) Combining the bounds (2.47),(2.40),(2.37) and (2.35) we get that
S
[
Cn − 12 + D−1n 32 + R− 12 n 12 + R2C−1 (
DP − 12 + D 78 P 12
)
n 12
]
(τ (n) log n)2.
We choose C = n 110 111 , D = n 112 111 , P = n 14 111 , R = n 2111 to obtain the much sought out bound
S n 12 − 1111 (τ (n) log n)2. (2.48) Finally, substituting (2.48) into (2.34), we deduce theorem 2.
Remark 2.19. As mentioned before, Iwaniec in proves the bound
a(n) n k
2−14−128 +
for all > 0, for square free n. It is believed that the analogue of Ramanujan’s conjecture holds for the half-integral modular forms as well, i.e.
a(n) n k
2−12+
,
for square free n. But this still remains a conjecture. 38 CHAPTER 2. LINNIK’S PROBLEM Chapter 3
Ergodic methods in number theory
Ergodic methods have been successfully employed to solve a variety of problems in number theory, especially problems involving diophantine approximations and equidistribution. Just as the study of diophantine equations had opened up new vistas of mathematics over the course of centuries, the study of diophantine inequalities, a subject still in its infancy, holds a similar promise. As a case in point, we give the example of the Oppenheim conjecture. Let Q be an indefinite quadratic form on Rn which is not a multiple of a rational form, for example x2 + y2 − √2z2.Oppenheim conjectured in 1929 that if n ≥ 3, the set of vales taken by Q on Zn is dense in R. In other words
Q(x) < is solvable, with x ∈ Zn for any > 0.
The conjecture was solved for large values of n using techniques from analytic number theory. However, the case n = 3 remained open for a long time (it was known that if the conjecture is true for a particular n then it is true for any larger n, hence the case n = 3 was of prime interest). The conjecture was finally solved in 1987 by G. Margulis using deep techniques from the theory of unipotent flows. More specifically, he considered the action of H = SO (Q) on SL n(R)/SL n(Z) and showed that relatively compact orbits of H are necessarily compact. The Oppenheim conjecture then followed as a consequence of a remarkable observation made by M. S. Raghunathan. In this chapter we shall try to explore some aspects ergodic methods which have number theoretic relevance, especially the dynamics of lattices under the action of the diagonal torus.
3.1 Measure rigidity and equidistribution
We saw in chapter 1 that the sequence ( nα ) is equidistributed modulo 1 when α is irrational. We had also remarked that Weyl had proved a similar result for ( p(n)) where p is a polynomial with at least one irrational coefficient. The purpose of this section is to prove the equidistribution of ( n2α)mod 1, using ergodic theory. Such a proof was first given by Furstenberg, but we shall follow and in our exposition. 39 40 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
3.1.1 Measures on Compact Metric spaces
In this section we will recall some results from measure theory and functional analysis and use these to establish certain results needed in the proof of equidistribution on ( n2α). Through out this section and the next, X will denote a compact metric space and B the Borel σ-algebra on X. Let
μ be a measure on ( X, B). For a measurable function f : X → R, we shall use ∫ f dμ and μ(f )interchangeably to denote the integral of f over whole of X.A map T : X → X is said to be measurable if T −1A ∈ B for all A ∈ B , and is measure preserving
if it is measurable and if μ(T −1A) = μ(A) for all A ∈ B . If T is measure preserving we say μ is
T -invariant (to put the emphasis on μ when T is fixed). Let L1
μ
denote the space of (equivalence classes) of measurable functions f : X → R with ∫ |f |dμ < ∞.
Proposition 3.1. A measure μ on X is T invariant if and only if
∫
f dμ =
∫
f ◦ T dμ (3.1) for all f ∈ L1
μ
.
Proof. If (3.1) holds, then for any A ∈ B , we have
μ(A) =
∫
χAdμ =
∫
χA ◦ T dμ =
∫
χT −1Adμ = μ(T −1A).
Conversely suppose μ is T -invariant, then (3.1) holds for functions of the form χA with A ∈ B ,and hence for any simple function (a finite linear combination of characteristic functions). Let f
be a non-negative real valued function in L1
μ
. Choose a sequence of functions simple functions ( fn)increasing to f . Then ( fn ◦ T ) increases to f ◦ T . By the dominated convergence theorem we have
∫
f ◦ T dμ = lim
n→∞
∫
fn ◦ T dμ = lim
n→∞
∫
fndμ =
∫
f dμ.
From now on, we shall restrict our attention to Borel probability measures on X. As in section 1.2, let P(X) be the space of Borel probability measures on X and let PT (X) denote the space of
T -invariant measures in P(X). As usual, C(X) denotes the space of continuous functions on X.Recall the following important result:
Riesz representation theorem. There is a one-to-one correspondence between Borel probability measures on X and positive linear functionals Λ : C(X) → R, with Λ(1) = 1. More specifically, given such a Λ, there exists a unique μ ∈ P (X) such that Λ( f ) =
∫
f dμ.
(A functional Λ : C(X) → R is positive if f ≥ 0 ⇒ Λ( f ) ≥ 0.) 3.1. MEASURE RIGIDITY AND EQUIDISTRIBUTION 41 An immediate corollary is that if μ1, μ 2 ∈ P (X), then
μ1 = μ2 ⇐⇒
∫
f dμ 1 =
∫
f dμ 2, for all f ∈ C(X). (3.2) Recall that the weak ∗ topology on P(X) is the smallest topology on P(X) making each of the maps μ 7 → ∫ f dμ continuous for every f ∈ C(X)
Corollary 3.2. P(X) endowed with the weak ∗ topology is compact.
Proof. The Riesz representation theorem enables us to identify P(X) as a subspace of C∗(X) en-dowed the weak ∗ topology. Under this identification P(X) is mapped into the unit sphere in C∗(X). (as ‖Λ‖ = |Λ(1) | = | ∫ dμ | = 1). The Banach-Alaoglu theorem says that the closed unit ball in
C∗(X) is weak ∗ compact. Being a closed subset of this, P(X) is compact. Given any continuous map T : X → X, it induces a map T∗ : P(X) → P (X) defined by
T∗(μ)( A) = μ(T −1A)for any A ∈ B .
Proposition 3.3. Let f ≥ 0 be a measurable map and μ ∈ P (X). Then
∫
f dT ∗μ =
∫
f ◦ T dμ. (3.3)
Proof. Clearly, (3.3) holds true for functions of the form χA for any A ∈ B , and hence for simple functions. We now proceed as in proposition 3.1 by taking a sequence of simple functions ( fn)increasing to f . We then have
T∗(μ)( f ) = lim
n→∞
T∗(μ)( fn) = lim
n→∞
μ(fn ◦ T ) = μ(f ◦ T ).
In particular, (3.3) holds for any continuous f . As a consequence we get that T∗ : P(X) → P (X)is a continuous map. To see this, suppose μn → μ. Then, for f ∈ C(X), T∗(μn)( f ) = μn(f ◦ T ) →
μ(f ◦ T ) = T∗(μ)( f ), so T∗(μn) → T∗(μ). Using these observations, we strengthen proposition 3.1.
Lemma 3.4. Let μ be measure in P(X). Then μ ∈ P T (X) if and only if ∫ f dμ = ∫ f ◦ T dμ for all
f ∈ C(X).
Proof. (⇒) Indeed, by the previous proposition, T∗(μ)( f ) = μ(f ◦ T ) = μ(f ), for all f ∈ C(X). We conclude from (3.2) that T∗(μ) = μ, i.e. μ ∈ P T (X). Of course, the other implication follows directly from proposition 3.1. The next theorem demonstrates that for continuous maps on X, we can always find invariant measures. 42 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
Theorem 3.5. Let T : X → X be a continuous map of a compact metric space and let ( νn) be any sequence in P(X). Then any weak ∗ limit point of the sequence ( μn) defined by μn = 1
n
∑n−1
k=0
T k
∗
(νn)is a member of PT (X).
Proof. First note that the sequence of measures ( μn) will have a limit point in P(X) as P(X) is weak ∗ compact by corollary 3.2. Let μ be a limit point and let μnj → μ. For a continuous function
f , let ‖f ‖∞ denote the supremum of f on X, which is finite as X is compact. By the definition of
T∗μn as in (3.2), we get
∣∣∫
f ◦ T dμ nj −
∫
f dμ nj
∣∣ = 1
nj
∣∣∫ nj −1
∑
k=0
(f ◦ T k+1 − f ◦ T k)dν nj
∣∣
= 1
nj
∣∣∫
(f ◦ T nj +1 − f )dν nj
∣∣
≤ 2
nj
‖f ‖∞ −→ 0as j → ∞ . It follows that ∫ f ◦ T dμ = ∫ f dμ for all f ∈ C(X), so μ ∈ P T (X) by lemma 3.4.
3.1.2 Ergodicity and Unique ergodicity
Definition 3.6. A measure preserving transformation T : X → X on a probability space ( X, B, μ )is said to be ergodic if for any B ∈ B
T −1(B) = B =⇒ μ(B) = 0 or μ(B) = 1 .
In words, a transformation is ergodic if there is no non-trivial way to divide the space into smaller
T -invariant ones.
Proposition 3.7. Let T : X → X be a measure preserving transformation on ( X, B, μ ). T is ergodic if and only if, for any measurable function f : X → R
f ◦ T = f a.e = ⇒ f is a constant a.e.
Proof. (⇒) Let B ∈ B be T -invariant, i.e. χB ◦ T = χT −1(B) = χB . We take f = χB and conclude that χB is constant almost everywhere. It follows that μ(B) = 0 or 1. (⇐) Conversely suppose T is ergodic. Let f : X → R be such that f ◦ T = f almost everywhere. We can redefine f on a measure zero set such that the new function, which we still call f , is invariant under T . Define
Bt = {t ∈ R : f (x) > t }.
Since f is T -invariant, we have that T −1(Bt) = Bt. Thus, for any t, μ(Bt) = 0 or 1. Now, if f
were not constant almost everywhere, there would be some t0 such that 0 < μ (Bt0 ) < 1, which contradicts our previous statement.
Example 3.8. The circle rotation Rα : T → T given by z 7 → ze (α) is ergodic with respect to the Lebesgue measure l if and only if α is irrational. 3.1. MEASURE RIGIDITY AND EQUIDISTRIBUTION 43
Proof. It is easy to construct non-trivial subsets invariant under Rα when α is rational. So, suppose
α is irrational. Then it follows from Dirichlet’s approximation theorem that Zα, considered in T,is dense in T. Now, let B ⊆ T be invariant under Rα. By Lusin’s theorem, for any > 0 we can choose a function f ∈ C(T) such that ‖f − χB ‖1 < . By invariance of B we have
‖f ◦ Rnα − f ‖ < 2
for all n. Since f is continuous it follows that
‖f ◦ Rt − f ‖ < 2 (3.4) for all t ∈ R. Thus, since l is rotation invariant, we have
‖f −
∫
f (t)dt ‖1 =
∫
|
∫
(f (x) − f (x + t)) dt |dx
≤
∫ ∫
‖f (x) − f (x + t)|dxdt ≤ 2
by Fubini’s theorem and (3.4). We deduce that
‖χB − μ(B)‖1 ≤ ‖ χB − f ‖1 + ‖f −
∫
f (t)dt ‖1 + ‖
∫
f (t)dt − μ(B)‖1 ≤ 4.
Since this holds true for any > 0, we conclude that χB is equal to μ(B) all most everywhere. But this can only be true if μ(B) = 0 or μ(B) = 1. Therefore Rα is ergodic. We now give a second proof using Fourier series. Let ∑
n∈Z
ane(nx ) be the Fourier expansion of χB . Since χB is invariant under Rα, this expansion is invariant under the change of variable
x → x + α. From the uniqueness of Fourier series we get an = ane(nα ). Since α is irrational this can only be true if an = 0 for n 6 = 0, i.e. χB is a constant almost everywhere. We conclude as in the first proof that Rα is ergodic. We now state a fundamental result in ergodic theory.
Birkhoff ’s ergodic theorem. Let ( X, B, μ ) be a probability space and let T : X → X be an ergodic measure preserving transformation. Then for any f ∈ L1(X, μ )lim
n→∞
1
n
n−1
∑
k=0
f (T kx) =
∫
f dμ for almost every x ∈ X.
We shall not prove this result as it is not entirely in line with the spirit of our results; a proof could be found in any standard book in ergodic theory or dynamical systems. Instead we shall deduce some notable corollaries from it. Before that, we remark that ∫ f dμ is the space average of
f , and (lim n→∞
∑n−1
k=0
f (T kx)) /n can be thought of the time average of f along the T -trajectory of x. The Birkhoff ergodic theorem then says that if T is ergodic, then the time average of f along the trajectory of almost any point in X is a constant equal to the space average of f . In fact, this interpretation is what originally prompted Boltzmann to introduce, albeit in an implicit way, the 44 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
notion of ergodicity through his ‘Ergodic hypothesis’ on thermodynamical systems.
Corollary 3.9. Let ( X, B, μ ) be a Borel probability space. If T : X → X is an ergodic measure pre-serving transformation, then for almost every x ∈ X the sequence of points ( T nx) is equidistributed with respect to μ.
Proof. Let U be an open set. Note that #{0 ≤ k < n : T kx ∈ U }
n = 1
n
n−1
∑
k=0
χU (T kx).
We apply the Birkhoff ergodic theorem to f = χU , which is in L1(X, μ ) to get 1
n
n−1
∑
k=0
χU (T kx) =
∫
X
χU dμ = μ(U ).
Therefore (1.1) holds true for any open set U and the corollary follows.
Definition 3.10. Let T : X → X be a continuous μ-invariant transformation. We say a point
x ∈ X is generic (with respect to T and μ) if lim
n→∞
1
n
n−1
∑
k=0
f (T kx) =
∫
f dμ,
for any f ∈ C(X).
Remark 3.11. Birkhoff’s ergodic theorem says that generic points with respect to an ergodic transformation ( T, μ ) have full measure. Also, notice that by (3.2), if x ∈ X is generic with respect to some μ ∈ P T (X), then it cannot be generic with respect to any other measure in PT (X).
Definition 3.12. Let X be a compact metric space. A transformation T : X → X is said to be
uniquely ergodic if there is exactly one T -invariant probability measure on X.
Remark 3.13. We now make an observation. Let X and T be as in the above definition and let μ be the unique T -invariant measure. Since X is compact, for any f ∈ C(X), the limit lim n→∞ n−1 ∑n−1
k=0
f (T kx) exists for any x ∈ X. Therefore we can define a measure νx on X as follows
νx(f ) =
∫
f dν x = lim
n→∞
1
n
n−1
∑
k=0
f (T kx).
By construction νx(f ◦ T ) = νx(f ), i.e. νx is T -invariant. So νx = μ for all x. In other words if T
is uniquely ergodic, then statement of the Birkhoff ergodic theorem is true for all x ∈ X. It follows, as in corollary (3.3), that ( T kx) is equidistributed for all x ∈ X.
Theorem 3.14. For a continuous map T : X → X on a compact metric space the following are equivalent. 3.1. MEASURE RIGIDITY AND EQUIDISTRIBUTION 45 (1) T is uniquely ergodic. (2) For every f ∈ C(X),
Sn(f ) = 1
n
n−1
∑
k=0
f (T kx) → Cf , (3.5) where Cf is a constant independent of x.(3) The convergence (3.5) holds for every f in a dense subset of C(X).
Proof. (1) ⇒ (2). Let μ be the unique invariant measure for T . We apply theorem 3.5 to the constant sequence ( δx). Since P(X) is compact and since μ is the only possible limit point, we deduce that 1
n
n−1
∑
k=0
δT k x −→ μ
in the weak ∗ topology, so for any f ∈ C(X) we have 1
n
n−1
∑
k=0
f (T kx) −→
∫
f dμ.
(2) ⇒ (1). Let μ ∈ P T (X). (3.5), by the dominated convergence theorem, implies that
∫
f dμ =
∫
lim
n→∞
1
n
n−1
∑
k=0
f (T kx)dμ = Cf
for all f ∈ C(X). It follows that Cf is the integral of f with respect to any measure in PT (X). Hence by (3.2), PT (X) can contain only a single measure. (3) ⇒ (1). Suppose μ, ν ∈ P T (x). Then, as in the the previous step we get
∫
f dμ = Cf =
∫
f dν
for any f in a dense subset of C(X). In other words the functionals μ 7 → ∫ f dμ and ν 7 → ∫ f dν
agrees on a dense subset of C(X). From continuity, it follows that they agree on the whole of C(X), hence μ = ν.(2) ⇒ (3) is trivial.
Example 3.15. Let α ∈ R be irrational. Then the circle rotation Rα : T → T is uniquely ergodic. The unique invariant measure is the Lebesgue measure l.
Proof. We shall prove this using property (3) of theorem (3.14). We already know that l is invariant
Rα. Let f (t) = e(ht ) for some h ∈ Z. We have 1
n
n−1
∑
k=0
f (Rkα(t)) = 1
n
n−1
∑
k=0
e(h(t + kα )) =
1 if h = 0 1
n e(ht ) e(nhα ) − 1
e(hα ) − 1 if h 6 = 0 46 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
The above equation clearly shows that 1
n
n−1
∑
k=0
f (Rkα(t)) →
∫
f dl =
1 if h = 0 0 if h 6 = 0 .
By linearity, the above convergence holds true for any trigonometric polynomial, which are dense in
C(X) is be the Stone-Weierstrass theorem. As a corollary, we obtain yet another proof of the equidistribution of ( nα ) mod 1, for α irrational.
3.1.3 Equidistribution of (n2α) mod 1
We now turn towards proving the equidistribution of ( n2α). Let α ∈ R be irrational. We first define a map T on T2 = R2/Z2 using α which we shall show to be uniquely ergodic with λ as the unique invariant measure . Define T : T2 → T2 by
T (x, y ) = ( x + α, y + 2 x + α).
A simple induction argument will give us that
T n(x, y ) = ( x + nα, y + 2 nα + n2α).
Now, if ( T2, T ) were uniquely ergodic it would follow from remark 3.13 that all T orbits would equidistribute in T2 with respect to λ. In particular the orbit of (0, 0) would be equidistributed. Since T n(0 , 0) = ( nα, n 2α), this would imply that ( n2α) is equidistributed modulo 1. As a first step, we will prove that T is ergodic with respect to the Lebesgue measure λ on T2.Clearly, T preserves the Lebesgue measure as it is a unimodular transformation ( 1 0 2 1 ) followed by a translation.
Proposition 3.16. Lebesgue measure, λ, on T2 is ergodic under T
Proof. Let f ∈ L2(λ) be T -invariant. We expand f to a Fourier series
f (x, y ) = ∑
m,n
ˆfm,n e(mx + ny ).
By T -invariance, we have ˆfm,n = ˆfm+2 n,n e(( m + n)α). (3.6) In particular | ˆfm,n | = | ˆfm+2 n,n | (= | ˆfm+2 kn,n | for any k). By the Riemann-Lebesgue lemma, ˆfm,n → ∞ as ( m, n ) → ∞ . Hence ˆfm,n = 0 if n 6 = 0 (by taking k → ∞ ). On the other hand if
n = 0, (3.6) gives ˆfm, 0 = e(mα ) ˆfm, 0, from which we conclude ˆfm, 0 = 0 as α is irrational. It follows that f is constant almost everywhere and that T is ergodic. We now turn towards the proof of unique ergodicity. 3.1. MEASURE RIGIDITY AND EQUIDISTRIBUTION 47
Theorem 3.17. Let g : T → T be a continuous function, and Tg : T2 → T2 be the map
Tg (x, y ) = ( x + α, y + g(x)) with α irrational. If the Lebesgue measure λ is Tg -ergodic then it is the only Tg -invariant probability measure, i.e. ( T2, T g ) is uniquely ergodic.
Proof. Let l denote the Lebesgue measure on T. Assume that Tg is ergodic with respect to λ = l × l.Let
E = {(x, y ) | (x, y ) is generic w.r.t λ}.
As Tg is ergodic with respect to λ, we have that λ(E) = 1 (see remark 3.11). We claim that E is invariant under the map ( x, y ) 7 → f (x, y + a). To see this, notice that ( x, y ) ∈ E means 1
k
n−1
∑
k=0
f (T kg (x, y )) −→
∫
f dλ
for all f ∈ C(T2). Define fa to be the map ( x, y ) 7 → (x, y + a). It follows that 1
n
n−1
∑
k=0
f (T kg (x, y + a)) = 1
n
n−1
∑
n=0
fa(T kg (x, y ))
−→
∫
fadλ =
∫
f dλ,
since l is invariant under rotations. So ( x, y + a) ∈ E also. This means that E = E1 × T for some set
E1 ⊆ T with μ(E1) = 1. Now, suppose ν is a Tg -invariant ergodic measure on T2. Write π : T2 → T
for the projection ( x, y ) → x. Then π∗ν is Rα invariant, so by unique ergodicity of Rα, π∗ν = l.In particular, ν(E) = ν(E1 × T) = l(E1) = 1. By ergodicity of ν, ν-almost every point in T2 is generic with respect to ν. Thus there must be a point ( x, y ) ∈ E generic with respect to ν. But we have already seen (remark 3.11) that a point in X cannot be generic with respect to more than one invariant probability measure. We conclude that ν = λ.
Corollary 3.18. If α is irrational then the sequence ( n2α) is equidistributed modulo one.
Proof. Take g(x) = 2 x+α, we have already proved that Tg = T is ergodic for the Lebesgue measure λ.From the last theorem we conclude that Tg uniquely ergodic, therefore every T orbit equidistributes with respect to λ. The orbit of (0 , 0) is ( nα, n 2α). Projecting to the second coordinate gives the required result.
Remark 3.19. The word measure rigidity has no precise meaning. It used to refer to the dearth of invariant measures under specific situations as in theorem 3.17. Sometimes it provides a quick route to equidistribution. 48 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
3.2 Dynamics of Lattices and the Littlewood conjecture
For number theoretic applications, one of the most important and interesting space to study is the space of lattices. By a lattice in Rn we mean a discrete subgroup of Rn whose R-span is the whole of Rn. If L is a lattice in Rn, fixing a basis for L gives us an element A of GL n(R), and changing the basis takes A to AM where M ∈ GL n(Z). Therefore, the space of lattices in Rn is naturally identified with the quotient GL n(R)/GL n(Z). A lattice L is said be unimodular if vol (Rn/L ) = 1, which is the same as saying |A| = 1 where A is any matrix in GL N (R) representing L. So the space of oriented unimodular lattices is naturally identified with SL n(R)/SL n(Z). We shall denote this space by Xn. Two lattices L1 and L2 are said to be homothetic (or similar) if one is obtained by scaling the other, i.e. L2 = λL 1 where λ is a non-zero real number. And the space of lattices up to homothety, denoted by Yn, is identified with P GL n(R)/P GL n(Z). The space Yn is essentially the same as Xn (as can be seen by scaling a lattice to have covolume one), except that in Yn we do not keep track of the orientation. Almost anything that can be said in the context of Xn can also be said about Yn and vice versa. So, if necessary, we shall switch back and forth between these spaces. From now on, we shall refer SL n(R) by G and SL n(Z) by Γ. The topology on Xn can be described as follows: a sequence of lattices Li in Xn converges to L if and only if each Li has a basis (b(i)1 , . . . , b (i)
n
) converging, as i → ∞ , to ( b1, . . . , b n) - a basis for L. For n ≥ 2, the space Xn is not compact, as can be seen by considering the sequence of lattices
Li = span Z(i−1, i −1, . . . , i n−1).
What makes the space Xn amenable to ergodic methods is the fact that there is a natural
probability measure on Xn which is invariant under the action of G. (See ). We denote it by μ.Further the precise way in which Xn fails to be compact is described by the Mahler’s compactness criterion. Before we get to this, let us introduce some notations. For x ∈ Rn let |x| denote the usual Euclidean norm of x. Let L ∈ Rn be a unimodular lattice. We define |L| to be
|L| = min {| x| : x ∈ L, x 6 = 0 }.
Theorem 3.20. (Mahler) Let r > 0. Define Ω r = {L ∈ Xn : |L| ≥ r}. Then Ω r is compact.
Proof. Let L ∈ Ωr . Choose v1, v 2, . . . , v n in the following way. v1 is the shortest non-zero vector in
L (if there are multiple such vectors choose any). Having chosen ( v1, . . . , v i), vi+1 is the shortest vector in L linearly independent from the previous ones. Thus 0 < r ≤ | v1| ≤ . . . ≤ | vn|. (3.7) It is easy to show that the Z span of the set {v1, . . . , v n} has bounded index in L (a bound independent of L). If we show that |vk| ≤ fk(r), for k = 1 , . . . , n , then the theorem would follow. For then, vi’s would vary in a compact set and span a lattice of bounded covolume (because of (3.7)) which is contained in L with bounded index. We prove this by induction on k. By applying Minkowski’s convex body theorem with the closed 3.2. DYNAMICS OF LATTICES AND THE LITTLEWOOD CONJECTURE 49 ball of radius r, we see that we take f1(r) = cr , where c is a constant(depending only on n). For the inductive case, let V be the subspace spanned by ( v1, . . . , v k). This subspace determines a k-torus
Y = V / (V ∩ L) ⊂ X = Rn/L.
Clearly, Y contains an embedded k-ball of radius r/ 2, so the volume of Y is bounded below by in terms of r. Also, by our hypothesis, Y has diameter at most fk(r). Suppose |vk+1 | diam( Y ). Then X would contain an embedded product of Y with a large (n − k) ball. This will lead to a contradiction as X has volume 1.
Corollary 3.21. (Mahler’s compactness criterion) Let E ⊂ Xn. Then E is compact if and only if inf {| L| : L ∈ E} > 0.
Proof. From the previous theorem we have that ⋃
r
Ωr , r > 0 is an increasing union of compact sets in Xn which cover the whole of Xn (as r decreases to zero). So E is compact if and only it is contained in some Ω r for some r > 0. This is precisely a restatement of what we want to prove. There is a natural left action of G on Xn given by [ X] 7 → [M X ]. Now, let H be a closed subgroup of G. Clearly μ is invariant for the action of H on Xn. As with the Oppenheim conjecture mentioned in the introduction, very often, we are interested in knowing the properties of H-orbits (closedness, cocompactness, compactness etc). Let A be the diagonal torus in G. That is
A = {diag ( t1, . . . , t n), t i > 0, t 1t2 · · · tn = 1 }.
In this section we shall be chiefly concerned with the action of A on Xn. We now define the notion of periodic orbits for A.
Definition 3.22. An A-orbit Ax Γ is said to be periodic if it is compact. More generally, for a closed subgroup H of G, an H-orbit is defined to be periodic if it carries a finite H-invariant measure. When H is abelian this definition coincides with the above one. We shall give a complete characterization of periodic A-orbits in this section.
3.2.1 Lattices arising from number fields
Let K be a number field of degree n over Q.
Definition 3.23. A lattice in K is the set of all integral linear combination of n Q-linearly inde-pendent elements of K. In other words it is a Z-submodule of K of rank n.Let M be a lattice and let μ1, . . . , μ n be a Z-basis for M . We define the discriminant of M , de-noted by disc( M ), to be the discriminant of μ1, . . . , μ n. (Recall discriminant of an n-tuple α1, . . . , α n
in K is the determinant of the matrix (Tr( αiαj )).) Different bases for M are obtained by a unimod-ular transformation (i.e. an element of GL N (Z)) and they have the same discriminant, so there is no ambiguity involved in the definition. Note that the discriminant of M is non-zero as μ1, . . . , μ n
are linearly independent over Q.50 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
Two lattices M1 and M2 are said to be similar if there exists a non-zero α ∈ K such that
M1 = αM 2. α ∈ K is called a coefficient of M if αM ⊆ M , i.e. for any ξ ∈ M, αξ ∈ M . It is easily verified that the set of all coefficients of M form a ring. It is denoted by OM and called the
coefficient ring ring of M . We now prove that OM is itself a lattice. We only need to prove that it has full rank. Let μ1, . . . , μ n be a basis for M . It is also a Q-basis for K. Let α ∈ K be non-zero. Write
αμ i = ∑
j
aij μj
where aij ’s are in Q. Let c be the least common denominator of all the aij . Then for each i,
cαμ i ∈ M , and hence cα ∈ O M . So if we start with a basis α1, . . . , α n for K, then we can find an integer c such that each of cα i is in OM . It follows that OM has full rank and hence a lattice. For any γ 6 = 0 the condition αM ⊆ M is equivalent to the condition αγM ⊆ γM . This implies that similar lattices have same coefficient ring, i.e. OM = OγM .Notice that any element α of OM is actually an algebraic integer (as the condition αM ⊆ M
is equivalent to saying that α satisfies a monic polynomial with coefficients in Z). Hence OM is actually an order in the ring of integers, OK , of K. If M ⊂ O M , then M is actually an ideal in OM .On the other hand given any lattice M it is easy to find (using the least common denominator trick) a non-zero integer b such that bM ⊂ O M . We summarize these results.
Theorem 3.24. The coefficient ring of any lattice in a number field K is an order in this field. Coefficient rings of similar lattices coincide. Any lattice is similar to an ideal contained in its coefficient ring. Let O be an order in OK . Then O is the coefficient ring for some lattice in K (for example one could take the lattice to be O itself). We are only interested in lattices up to similarity. We now state a basic and fundamental result in algebraic number theory, proof of which could be found in .
Theorem 3.25. Let O be any order in a number field K. Then there only finite many equivalence class of similar modules with O as their coefficient ring. From now on, we shall assume that K is a totally real field of degree n. (Recall: a totally real field is a field all whose embeddings into C is actually into R). Let σ1, . . . , σ n be the distinct embeddings K into R. Recall the geometric embedding of θ : K ↪ → Rn given by
θ(x) = ( σ1x, . . . , σ nx). (3.8)
Proposition 3.26. Let M be a lattice in K. Then θ(M ) is a lattice in Rn.
Proof. Let α1, . . . , α n be a Z-basis for M . It is enough to show that θ(α1), . . . , θ(αn) are linearly independent over R. Consider the matrix, B, corresponding these n vectors, namely bij = σiαj .(det B)2 is nothing but the discriminant of M and hence non-zero. The proposition follows. The logarithmic embedding of K into Rn given by
Log (x) = (log |σ1x|, . . . , log |σnx|). (3.9) 3.2. DYNAMICS OF LATTICES AND THE LITTLEWOOD CONJECTURE 51 If η is a unit in OK , then Log (η) lies in the n − 1 dimensional subspace
a = {(x1, . . . , x n) : x1 + . . . + xn = 0 }.
The converse is true as well, as Log (x) being in a is equivalent to |N K
Q
(x)| = 1. Let O be an order in K and let O× be the group of units in O. We look at the restriction of the map Log on O×. The Dirichlet unit theorem states that image of O× under Log is a lattice in
a. In particular, the group of units, as an abelian group, has rank n − 1. We define the regulator of
O to be the volume of the quotient a/(Log O×). (As can easily be seen, this definition of regulator is equivalent to the classical one). A unit η in O is said to be totally positive if σi(η) > 0 for all i.Clearly, totally positive units in O, denoted by O×
+
, form a group and the Dirichlet unit theorem also holds for O×
+
. That is Log (O×
+
) is a lattice in a. Our interest in totally positive units is partly due to the fact that if η is such a unit then diag( θ(η)) ∈ A, the diagonal group. Let M be a lattice in K with coefficient ring OM . If γ ∈ O ×
M
, then γM ⊆ M and γ−1M ⊆ M
hence γM = M . So the units in O are precisely the elements α ∈ K which stabilize M , i.e.
αM = M . Given a lattice M in K, we shall use the notation LM to denote the unimodular lattice in Rn defined by
LM = αθ (M ), where α = |det θ(M )|−1/n . (3.10)
Lemma 3.27. The A-orbit of LM is compact in Xn.
Proof. Let O be the coefficient ring of M . We have from the previous discussion that M is stabilized by each element in O×
+
. Therefore, LM is stabilized by elements of A of the form diag( θ(γ)) where
γ ∈ O ×
+
. Let ALM denote the stabilizer of LM under the action of A. There is a natural map log : A → a (the inverse of exp : a → A) which takes ALM to Log (O×
+
). But, by Dirichlet’s unit theorem, a/Log (O×
+
) is compact (isomorphic to Tn−1). We conclude that A/A LM is compact. As
AL M ' A/A LM (the orbit stabilizer theorem), we deduce that AL M is compact. We say that a lattice L in Rn arises from a number field if AL = AL M , where M is a lattice in a totally real number field K/ Q of degree n. Observe that if M1 = γM 2 with γ ∈ K{ 0}, then LM1
and LM2 are in the same A-orbit. More specifically, we have
LM1 = β diag( σ1γ, . . . , σ nγ) LM2 , β = |N K
Q
(γ)|−1/n . (3.11) Let α1, . . . , α n be a basis for M . Define the norm form , F , on M by
F (x1, . . . , x n) = N K
Q
(x1α1 + . . . + xnαn). (3.12) It is a easy to see that F is a homogeneous polynomial in x1, . . . , x n of degree n with integral coefficients. Choosing a different basis for M will only change F (x1, . . . , x n) to F (x′
1
, . . . , x ′
n
) , where (x1, . . . , x n) is related to ( x′
1
, . . . , x ′
n
) by a unimodular transformation. So up to this equivalence, F
is uniquely determined by M . We use the notation ( M, F ) to denote the norm form associated to M .With a little more work it can be shown that F is irreducible over Q. Conversely any homogenous 52 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
polynomial in x1, . . . , x n of degree n irreducible over Q which splits over R is of the form (3.12) (see ). More generally, given any lattice L ⊂ Rn, we can associate a form to it as follows. Choose a basis v1, . . . , v n for L. Let Li be the linear form defined as
Li(x1, . . . , x n) = (
n
∑
j=i
xj vj )i,
where the notation ( y)i means the i-th coordinate of y. We define the product form associated to
L, denoted as ( L, F ), by
F (x1, . . . , x n) = ∏
i
Li(x1, . . . , x n). (3.13) It is easily verified that if L is a lattice arising from a number field, then the product form agrees with the norm form. The next theorem is rather interesting in that it shows every A-periodic orbit comes from a number field.
Theorem 3.28 () . For any unimodular lattice L ⊂ Rn the following conditions are equivalent. 1. AL is periodic. 2. L arises from a number field. 3. The pair ( L, F ) is equivalent to ( Zn, αf ) where α ∈ R and f is an integral form that is irreducible over Q.
Proof. (2 ⇒ 1). Suppose A.L is compact and let AL denote the stabilizer of L in A. Then L ⊗ Q
is a module over the commutative algebra R = Q[AL] ⊂ Mn(R). The matrices in A have only real eigen values and so are semi-simple. By the structure theorem for semi-simple rings we get that R is the direct sum of m totally real fields and therefore the rank of O×
R
is n − m. Since A.L is compact
AL ∼= Zn−1. (Otherwise AL will have infinite index in A and A.L ∼= A/A L cannot be compact). Now, since AL ⊂ O ×
R
we conclude that m = 1 and R itself is a totally real field. Thus L ⊗ Q is a one dimensional vector space over R, so the lattice L itself is obtained from a full module M ⊂ R
by the process described above. We have already seen from lemma 3.27 that (1 ⇒ 2). (2) and (3) are equivalent by our preceding discussion. This theorem allows us to associate two important invariants to periodic A-orbits.
Definition 3.29. Let x ∈ Xn be such that Ax is periodic. Then by the theorem above, Ax = AL M
where M is a lattice in a totally real number field of degree n. We define the discriminant of Ax to be disc( Ax ) = disc( OM ), and we define the volume of Ax to be vol( Ax ) = reg( OM ).
Remark 3.30. We see that periodic A-orbits in Xn come naturally in packets. Each packet corre-sponds to an order O in a totlally real field K of degree n and consists of orbits of the form AL M
where M is a lattice in K, up to similarity, with O as its coefficient ring. 3.2. DYNAMICS OF LATTICES AND THE LITTLEWOOD CONJECTURE 53
3.2.2 Discreteness of periodic orbits.
The main purpose of this section is to prove that if n ≥ 3, periodic orbits of A of discriminant less than D cannot be too close in Xn. We shall also explain the duality between lattices and linear forms and use this to put the Littlewood conjecture (or rather a generalization of it due to Cassels and Swinnerton-Dyer) in the perspective of the A action on Xn. Our treatment is based on , and . Let N : Rn → R denote norm function defined by N (x) = ∏n
1
xi. For a unimodular lattice
L ⊂ Rn we define norm of the lattice by
N (L) = inf {| N (w)| : w ∈ L, w 6 = 0 }.
Clearly, the function N is constant on A-orbits.
Proposition 3.31. The function N : Xn → R is semicontinuous, i.e. if Ln → L in Xn, then lim sup N (Ln) ≤ N (L).
Proof. Define the star body of radius > 0 by S = {x ∈ Rn : |N (x)| < }. In terms of star bodies,
N (L) = inf { : S ∩L 6 = {0}} . Since Ln → L, given any > 0, all but finitely many Ln’s will intersect
SN (L)+ . In other words N (Ln) ≤ N (L) + , for all n large enough. The proposition follows.
Proposition 3.32. Let L ⊂ Rn be a unimodular lattice. Then AL is compact if and only if
N (L) > 0.
Proof. By AM-GM inequality we have that |x| ≥ √nN (x)1/n for any x ∈ Rn; also, if N (x) 6 = 0, the equality holds for some y ∈ Ax . So, if N (L) > r > 0 we would also have that |L| > √nr 1/n = r′.But N (L) = N (aL ) for any a ∈ A, so |aL | > r ′ for any a. By Mahler’s criterion we conclude that
AL is compact. Conversely suppose AL is compact. Then there exists r > 0 such that |aL | > r for all a ∈ A. If
x 6 = 0 in L we can find an a such that
N (x) = N (ax ) = |ax |n/n n/ 2 ≥ rn/n n/ 2 > 0.
So N (L) > 0.
Remark 3.33. AL being compact is equivalent to saying that AL is bounded. So the proposition may be restated as AL is bounded if and only if N (L) > 0. We now turn towards computing the norm of lattices arising from number fields. Observe that if θ(α) = x then N (x) = N K
Q
(α). Using (3.10), we find that if M is a lattice in K then N (LM ) =
αnN (θ(M )). But αn = |det θ(M )| = |disc( M )|−1/2. And N (θ(M )) = inf {N K
Q
(β) : β ∈ M, β 6 = 0 }.Since similar lattices are related as in (3.11), N (LM ) depends only on the similarity class of M .Assume then that M = I, an ideal in its coefficient ring OI . We have that disc( I) = disc( OI )N (I)2,where N (I) is the familiar multiplicative function [ OI : I] on ideals. Set N ∗(I) = min {| N K
Q
(β)| :54 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
β ∈ I, β 6 = 0 }. We then see that
N (LI ) = N ∗(I)
N (I)√|disc( OI )| . (3.14) Suppose β ∈ I be an element such that |N K
Q
(β)| = N ∗(I). But |N K
Q
(β)| = N (β) = N (I)[ I : ( α)]. We deduce that
N (LI ) ≥ 1
√|disc( OI )| . (3.15) We some times use N (L) to denote the set of values N (l) taken by l ∈ L. The meaning will be clear from the context. We now turn towards the main result in this section: isolation of periodic orbits. First, we construct the necessary set up for the proof. For each pair 1 ≤ i 6 = j ≤ n, we define the root
αij : a → R to be the linear functional
t 7 → ti − tj .
The set of roots will be denoted by Φ. Recall the logarithmic embedding (3.9) Log : O×
K
→ a. For any order O in K, let us denote the image of O× under Log by Ω O .
Lemma 3.34. () Let n ≥ 3. Then, for any root αij ∈ Φ, the set {αij (a) : a ∈ ΩO } is dense in reals.
Proof. Since Ω O has finite index in Ω K , it is enough to justify why α(Ω K ) is dense in R. As Ω K is a lattice is a, this is equivalent Ω K ∩ ker( αij ) not being a lattice in ker( αij ). If this is indeed the case then the field L = {θ ∈ K : σi(θ) = σj (θ)} is a subfield of K with a group of units containing a copy of Zn−2. Since L has degree at most n/ 2, by Dirichlet’s unit theorem, the degree of the group of units in L is at most n/ 2 − 1. This means that n/ 2 − 1 ≥ n − 2, or equivalently n ≤ 2, a contradiction.
Corollary 3.35. Let O be an order in a totally real field K of degree n ≥ 3. Then for any i 6 = j,the set { σi(η)
σj (η) : η ∈ O ×
+
}
is a dense subset of R+.
Proof. Log (O×
+
) has finite index in Ω K , so αij (O×
+
) is dense in R. Applying exp : R → R+ and using the definition of Log map (3.9) gives the desired result. An element a in A is said to be regular if α(a) 6 = 0 for any root α ∈ Φ; equivalently all entries of
a are distinct. Let b ∈ A be regular. We define the stable horospherical subgroup corresponding to b
to be
U −(b) = {g ∈ G : bngb −n → e as n → ∞} ,
and the unstable horospherical subgroup to be
U +(b) = {g ∈ G : b−ngb n → e as n → ∞} .3.2. DYNAMICS OF LATTICES AND THE LITTLEWOOD CONJECTURE 55 The following lemma explains their usefulness.
Lemma 3.36. Let b ∈ A be regular. Any element g ∈ A which is close enough to e has a unique decomposition g = au +u−, where a ∈ A, u + ∈ U +(b), u − ∈ U −(b).
Proof. (Sketch). Let b = diag( b1, . . . , b n). The ij -th entry in bgb −1 is gij bi/b j . As there is no change to diagonal entries on conjugation, we assume from now on that i 6 = j. If gij is non-zero, repeated conjugation will take the corresponding entry to 0 or ∞ depending on whether bi/b j < 1or bi/b j > 1. This is equivalent to the condition αij (b) < 0 or αij (b) > 0 respectively. ( αij (b) = 0 is ruled out as b is regular). In particular, u belongs to U −(b) if and only of all its diagonal entries are 1 and uij = 0 for any ij such that αij (b) > 0. We then see that U −(b) is of rank (n
2
), half the number of roots. Similar statement holds true for U +(b) (in fact, U +(b) is the transpose of U −(b)). Consider the map from A × U +(b) × U −(b) −→ G, given by (a, u +, u −) 7 → au +u−.
It can be shown, by using the aforementioned facts, that the Jacobian of the above map is non-singular at e. By the inverse function theorem, there exists neighborhoods W 0(b), W +(b), W −(b) of the identity elements in the groups A, U +(b) and U −(b) respectively, such that the above map is a diffeomorphism onto its image. From now on, for simplicity, we shall assume n = 3. We shall later explain how the results generalize to any n ≥ 3.
Theorem 3.37. (Isolation of periodic orbits ). Let L0 ∈ X3 be periodic (i.e. AL 0 is compact). Suppose L ∈ Xn be another lattice such that AL intersects AL 0. Then either AL 0 = AL or N (L)is dense in R.
Proof. Suppose AL 0 6 = AL , then L0 ∈ AL \AL . We will show that N (L) is dense in R. By semicontinuity of N , it is sufficient to show that N (L′) is dense for some L′ ∈ AL . Note that both
AL 0 and AL are A-invariant. Let V ⊆ G = SL 3(R) denote the set of g such that gAL 0 ∩ AL 6 = ∅. As AL 0 is compact, V is closed. For a, b in A we have
agbAL 0 ∩ AL = a(gAL 0 ∩ AL ),
so V is invariant under the action of A from both left and right. Since L0 is periodic, from theorem 3.24 , it is of the form aL I for some ideal I in a totally real order O. Further, since AL is invariant under A, we may as well assume a = 1. Let A0 = stab A(LI ). Elements in A0 are precisely of the form diag( σ1η, σ 2η, σ 3η) where η ∈ O ×
+
, the group of totally positive units in O. Fix a regular element b ∈ A0.Now, choose a sequence of lattices Ln → L0 from the orbit AL . Write Ln = gnL0 where
gn ∈ G and gn → e. From the previous lemma we have a decomposition gn = anu+
n
u−
n
, where
an ∈ A, u +
n
∈ U +(b), u −
n
∈ U −(b), and an, u +
n
, u −
n
→ e. Replacing gn by a−1
n
gn, if necessary, we may further assume that an = 1. Since gnL0 ∈ AL , we must have that gn 6 = e for any n (as we have 56 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
assumed AL 0 6 = AL ). This condition is equivalent to saying that the pairs ( u+
n
, u −
n
) are non-trivial. We can then write
gn = ( I + v+
n
)( I + v−
n
),
where v+
n
and v−
n
are matrices with diagonal entries zero. Further, we also have that diagonal entries of v+
n
v−
n
is zero (owing to the complementarity in structure of the groups U +(b) and U −(b)). We deduce that
gn = I + vn,
where vn is a non-zero matrix with diagonal entries zero. Since gn → e, we must have vn → 0. Let
‖vn‖ denote the maximum of the absolute values of the entries of vn. As n varies from 1 to ∞,there must be some entry ij which achieves ‖vn‖ infinitely often. We pass to this subsequence and abusing notation, we denote it again by vn. For concreteness, assume that the maximum is achieved at ( vn)12 , i.e., ‖vn‖ = |(vn)12 |. Further assume, passing to a subsequence if necessary, that all ( vn)12
are of the same sign, say positive. Conjugating by a = diag( a1, a 2, a 3) sends ( vn)ij to ( vn)ij ai/a j . In particular, conjugating by diag( a, 1/a, 1) multiplies ( vn)ij by the ij -th entry of the matrix
1 a2 aa−2 1 a−1
a−1 a 1
.
Given t > 0, take an such that
a2
n
(vn)12 = t.
Since ‖vn‖ → 0 we have that an → ∞ . Further, as ( vn)12 is the largest entry, we conclude that the matrix obtained by conjugating gn with diag( an, 1/a n, 1) converges to
ut =
1 t 00 1 00 0 1
.
As V is closed we deduce that V contains the semigroup U = {ut : t > 0}. Note that U is normalized by A, i.e. aU a −1 = U for all a ∈ A.Since ut ∈ V , we have that uta(L0) ∈ AL for some a ∈ A. But U is normalized by A, and thus
ut0 (L0) ∈ AL for some t0 > 0. In order to complete the proof, it suffices to show that N (ut0 L0) is dense in R. Let x ∈ L0.
N (ut0 x) = ( x1 + tx 2)x2x3 = N (x)(1 + tx 2/x 1).
Let c = diag( c1, c 2, c 3) ∈ A0, where A0 is the stabilizer of L0 in A. Then cx ∈ L0 and N (cx ) = N (x). We infer that
N (ut0 cx ) = N (x)(1 + t c2
c1
x2
x1
).
From corollary 3.35 we have that {c2/c 1 : c ∈ A0} is a dense subset of R+. Choosing yet another x3.2. DYNAMICS OF LATTICES AND THE LITTLEWOOD CONJECTURE 57 with a different sign for x2/x 1, we conclude that N (ut0 L0) is dense in R.By semicontinuity of N , it follows that N (L) is dense in R.
Remark 3.38. The above theorem is not true for n = 2. That is, we can construct bounded A
orbits in X2 which spiral in or oscillate between periodic orbits. The reason for the failure of the above theorem is that corollary 3.35 is not valid for n = 2. In fact, for any order O is a real quadratic field, the set {σ1η/σ 2η : η ∈ O ×} is a discrete subset of R.We can strengthen theorem (3.37), using techniques similar to the ones used in its proof, as follows:
Theorem 3.39. (Discreteness of periodic orbits ) Let n ≥ 3. Suppose AL 0 ∈ Xn is periodic. Let ( Lk) ∈ Xn\AL 0 be a sequence of lattices such that Lk → L0. Then given any > 0, we can find, for all k ≥ M (), lk ∈ Lk such that 0 < |N (lk)| < . In particular, as k → ∞ , N (Lk) → 0. We now describe how the above theorem can be interpreted as the “discreteness of periodic orbits”. Let AL 0 be periodic, and let LIk be a sequence of lattices arising from totally real cubic number fields which converge to L0. In the light of (3.15) and the theorem above, we conclude that
|disc( OIk )| → ∞ . In other words, qualitatively speaking, periodic orbits whose discriminants are bounded from above cannot be too close in Xn.We remark that in only n = 3 is treated, but the general case ( n ≥ 3) is similar. The following is a conjecture first stated in , for n = 3, which was later generalized by Margulis.
Conjecture 3.40. (Margulis) Let L ∈ Xn, n ≥ 3. If N (L) > 0, then L arises from a number field. Equivalently, bounded A-orbits in Xn are periodic. We now state the Littlewood conjecture, an outstanding open problem concerning simultaneous Diophantine approximation of real numbers.
Conjecture 3.41. (Littlewood) For any α, β ∈ R we have lim inf
n→∞
n ‖nα ‖ ‖ nβ ‖ = 0 ,
where ‖x‖ = min n∈Z |x − n|
Theorem 3.42. Margulis’s conjecture implies the Littlewood conjecture.
Proof. Suppose ( α, β ) ∈ R2 be a counterexample to the Littlewood conjecture, i.e.
|n(nα − a)( nβ − b)| ≥ δ > 0, (3.16) for all integers n 6 = 0, a, b . Consider the unimodular lattice L0 ∈ R3 generated by
{e1, e 2, e 3} = {(1 , 0, 0) , (0 , 1, 0) , (α, β, 1) }.
Let M0 = Ze1 ∪ Ze2. From (3.16) we see that the norm is bounded away from zero on L0 − M0 and vanishes exactly on M0. In particular N (L0) is not dense. 58 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
We say a lattice L ∈ R3 is admissible for a region V ∈ R3 if L ∩ V = {0}. From (3.16), we see that L0 is admissible for the region
|xyz | < δ, max( |x|, |y|) < 1.
Let an = diag( n, n, n −2). Then Ln = anL0 is admissible for the region
|xyz | < δ, max( |x|, |y|) < n.
Note that all these lattices Ln are admissible for Bδ′ , the open ball of radius δ′, for some δ′ > 0 . (One could take δ′ = 3 δ2/3, assuming δ < 1). By Mahler’s compactness criterion, the sequence ( Ln)lies in a compact subset of X3. Let L∞ be a limit point of this sequence. Clearly, L∞ is admissible for
|xyz | < δ.
In other words, N (L∞) ≥ δ. By Margulis’s conjecture, L∞ ∈ T for some periodic orbit T . But
L∞ ∈ AL 0 and AL 0 is unbounded (as N (L0) = 0). This contradicts theorem 3.37 as N (L0) is not dense We now state yet another conjecture by Margulis. In the next section we shall explain some recent progress made towards it and its implications for strengthening the Minkowski bound on class numbers.
Conjecture 3.43. For any compact set Ω ⊂ Xn, n ≥ 3, there are only finitely many periodic
A-orbits contained in Ω.
Theorem 3.44. Margulis’s conjecture implies conjecture 3.43.
Proof. Suppose there were infinitely many lattices Ln ∈ Ω such that AL n is periodic. From corollary 3.21 we have that Ω ⊂ Ωδ for some δ > 0. Then N (Ln) > r > 0, for some r = r(δ) for all n. Let L0
be a limit point of Ln’s. Clearly, N (L0) > r . If Margulis’s conjecture is true, then AL 0 is periodic. But now there is subsequence of lattices, whose norms are bounded away from zero, converging to
L0. This is in flat contradiction with theorem 3.39, and we conclude that if conjecture 3.43 is false then Margulis’s conjecture is false.
Remark 3.45. Margulis’s conjecture says that if n ≥ 3, A-orbits in Xn are either compact or unbounded. Using theorem 3.28 one can reformulate the conjecture as follows: Let F (x1, . . . , x n) be a product of linear forms in n-variables over R with n ≥ 3. If F is not proportional to a homogeneous polynomial with integer coefficients, then inf
06=x∈Zn
|F (x)| = 0 .
In fact, for n = 3, it is in this form that Cassels and Swinnerton-Dyer states this conjecture in . 3.3. APPLICATION: STRENGTHENING MINKOWSKI’S THEOREM 59
3.3 Application: Strengthening Minkowski’s theorem
In this section we shall describe some recent progress made towards conjecture 3.43 in . Following , we then explain an interesting application of this work in strengthening Minkowski’s theorem regarding ideal classes, which we now recall.
Theorem 3.46. (Minkowski) Let K be a number field of degree n with maximal order OK . Then any ideal class in OK possesses a representative J ⊂ O K of norm N (J) = O(√disc K), where the
O-constant depends only on n.Note that finiteness of the ideal-class group follows from the above theorem. We shall be working with the group P GL n(R). As mentioned before, the proofs could easily be translated back to SL n(R). Throughout this section, we shall denote P GL n(R) by G, P GL n(Z) by Γ and P GL n(R)/P GL n(Z) by Yn. For x ∈ Rn, let ‖x‖∞ denote the sup norm of x. We continue to denote the group of diagonal matrices by A. For δ > 0, let Ω ′
δ
denote the set of homothety classes of lattices Λ ⊂ Rn containing no vectors v with ‖v‖n
∞
< δ covol(Λ), i.e., Ω′
δ
= {gΓ ∈ Yn : ‖gx ‖n
∞
≥ δdet( g) for every x ∈ Z}.
This set is compact by Mahler’s compactness criterion. Also, Ω ′
a
⊂ Ω′
b
if a > b .Let K be a totally real number field, with OK as its ring of integers, and let [ J] be an ideal class in OK . We denote the regulator of OK by RK . Define
m([ J], K ) = min
J′∈[J],J ′⊂O K
N (J′)
m(K) = max
[J]
m([ J], K ),
where in the later definition the maximum is taken over all ideal classes in OK . As before, let
θ : K ↪→ Rn be the geometric embedding of K into Rn. We shall now see that m([ J], K ) is intimately related to how far the A-orbit of the homothety class of the lattice θ(J−1) penetrates the cusp of Yn.
Lemma 3.47. Let J be a fractional ideal of K- a totally real number field of degree n. Let Y be the periodic A-orbit corresponding to the ideal class [ J−1], i.e., Y = A.θ (J−1). Then the following are equivalent: (1) m([ J], K ) < δ disc( K)1/2 ;(2) Y is not contained in Ω ′
δ
.
Proof. It is clear from the definition of Ω ′
δ
that if Λ ⊂ Rn is a lattice, then AΛ ⊂ Ω′
δ
if and only if, for all non-zero x ∈ Λ, we have ∏
i
|xi| = |N (x)| ≥ δcovol(Λ) .
We apply this statement to the lattice Λ = θ(J−1). Recall that on this lattice the norm N (θx )agrees with field norm N K
Q
(x) and thus, that N is a multiplicative function on the fractional ideals of K. Hence, the covolume Λ is N (J)−1(disc K)1/2, where N (J) is the norm of the ideal J. With this, we see that (2) is equivalent to the following There exists x ∈ J−1 with |N K
Q
(x)| < δ (disc K)1/2. (3.17) 60 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY
Consider the map x 7 → xJ form J−1 to ideal classes I ⊂ O K equivalent to J. This map is surjective. We conclude that (3.17) is equivalent to condition (1). Next, we describe a consequence of conjecture 3.43.
Conjecture 3.48. Let n ≥ 3 be fixed. Then any ideal class in a totaly real number field of degree
n has a representative of norm o(√disc K).
Theorem 3.49. Conjecture 3.43 implies conjecture 3.48.
Proof. Suppose conjecture 3.48 was false. Then for some δ > 0 there would an infinite sequence of totally real fields Ki and ideals Ji ⊂ O Ki with m([ Ji], K i) ≥ δ√disc Ki. By lemma 3.47, this gives us an infinite sequence of periodic A-orbits all contained inside the compact set Ω ′
δ
, in contradiction to conjecture 3.43. Finally we describe the theorem of Einsiedler, Lindenstrauss, Michel and Venkatesh mentioned in the beginning of this section.
Theorem 3.50. () For any fixed compact set Ω ⊂ Yn, n ≥ 3, and for any > 0, the total volume of all periodic A-orbits contained in Ω of discriminant ≤ D is at most O(D). We shall not go into the proof of this theorem. We merely mention that the proof involves deriving a relationship between total volume of a collection of A-periodic orbits Yi and the entropy of any weak limit of A-invariant probability measures, μi = μYi , supported on Yi. The idea goes back to Linnik and the authors call it Linnik’s principle. Let hδ (K) be the number of ideal classes in OK with m([ J], K ) > δ disc( K)1/2. Substituting conjecture 3.43 with theorem 3.50, we get the following unconditional result towards conjecture 3.48.
Theorem 3.51. Let n ≥ 3, and let K denote a totally real number field of degree d. For all ε, δ > 0we have ∑
disc K<X
RK hδ (K) ε,δ Xε. (3.18) In particular, conjecture 3.48 is true for almost all totally real fields. That is, the number of fields
K with discriminant ≤ X for which m(K) ≥ δ disc( K)1/2 is Oε(Xε), for any ε, δ > 0.
Proof. Suppose that theorem 3.51 were false. Then we have constants C, ε, δ > 0 and a sequence of integers Di → ∞ such that
∑
disc( K)<D i
RK hδ (K) > CD εi , for all i, (3.19) the summation being over totally real fields of fixed degree n. For every totally real field for which
m(K) ≥ δ√disc K, let [ JK,j ] j = 1 , . . . , h δ (K), be the ideal classes of K with m([ JK,j ], K ) ≥
δ√disc K.Let YK,j be the periodic A-orbit corresponding to θ(J−1
K,j
). The volume of YK,j is proportional to RK and the discriminant is proportional to disc( K). By lemma 3.47 we have that YK,j ⊂ Ω′
δ
.3.3. APPLICATION: STRENGTHENING MINKOWSKI’S THEOREM 61 The assumption (3.19) implies that the collection of periodic A-orbits
Ci = {YK,j : disc( K) < D i, 1 ≤ j ≤ hδ (K)} ⊂ Ω′
δ
have discriminant ≤ Di and total volume Dεi . Clearly, this contradicts theorem 3.50. 62 CHAPTER 3. ERGODIC METHODS IN NUMBER THEORY Bibliography
M. B. Bekka and M. Mayer. Ergodic theory and topological dynamics of group actions on ho-mogeneous spaces . London Mathematical Society Lecture Notes Series-269. Cambridge Univ. Press, 2000. Z. I. Borevich and I. R. Shaferevich. Number theory . Academic Press, 1966. J. W. S. Cassels and H. P. F. Swinnnerton Dyer. On the product of three homogeneous linear forms and indefinite ternary quadratic forms . Phil. Tran. Roy. Soc. London, Ser.A. 248: 73-96, 1955. W. Duke. Rational points on the sphere . Ramanujan J. No.1-3, 235-239, 2003. W. Duke. An Introduction to Linnik Problems . Equidistribution in Number theory, An Intro-duction, NATO Science Series, Springer, 2007. M. Einsiedler, E. Lindenstruass, P. Michelle, A. Venkatesh. Distribution of periodic torus orbits on homogeneous spaces . Duke Math. J. Vol. 148, No. 1, 2009, 119-174. J. Ellenberg, P. Michelle, A. Venkatesh. Linnik’s ergodic method and the distribution of integer points on the sphere . Preprint. M. Einsiedler and T. Ward. Ergodic theory with a view towards number theory . Graduate Texts in Mathematics-259. Springer, 2011. C. F. Gauss. Disquisitiones Arithmeticae . Translated by A. Clarke. Springer-Verlag, New York, 1986. A. Granville and Z. Rudnick. Uniform Distribution . Equidistribution in Number theory, An Introduction, NATO Science Series, Springer, 2007. D.R. Heath-Brown. Arithmetic applications of Kloosteman sums . Nieuw Arch. Wiskd. (2000), no. 4, 380-384. M. N. Huxley. Area, Lattice Points and Exponential Sums , LMS Monographs. New Series, 13. Oxford University Press, 1996. H. Iwaniec. Fourier coefficients of modular forms of half-integral weight , Invent. Math. 1987, 385-401. H. Iwaniec. Topics in automorphis forms Graduate Studies in Mathematics, AMS, 1997. H. Kloosterman. On the representation of numbers in the form ax 2 + by 2 + cz 2 + dt 2. Acta. Math. 46 (1926), 407- 464. N. Kolitz. Introduction to Elliptic curves and modular forms . Graduate texts in mathematics-97. Springer-Verlag, 1984. 63 64 BIBLIOGRAPHY
E. Lindenstrauss. Some examples how to use measure classification in number theory . Equidis-tribution in Number theory, An Introduction, NATO Science Series, Springer, 2007. Yu. V. Linnik. Ergodic properties of algebraic fields , Translated from Russian by M. S. Keane. Springer-Verlag, 1968. C. T. McMullen. Minkowski’s conjecture, well-rounded lattices and topological dimension . J. Amer. Math. Soc. Volume 18, no 3, 711-734. C. T. McMullen. Hyperbolic manifolds, discrete groups and ergodic theory . Harvard Course notes, 2007. T. Miyake. Modular Forms , Springer Monographs in Mathematics. Springer-Verlag, 1989. U. Shapira. A solution to a problem of Cassels and diophantine properties of cubic numbers ,Ann. of Math.(2) 173 (2011), 543-557. G. Shimura. On modular forms of half-integral weight , Ann. of Math. 97 (1973), 440-481. P. Sarnak. Some applications of Modular Forms , Cambridge Tracts in Mathematics-99, Cam-bridge Univ. Press, 1990. G. Watson. A treatise on Bessel functions , Cambridge Univ. Press. H. Weyl. ¨Uber die Gleichverteilung von Zahlen mod. Eins . Math. Ann. 77 (1916), 313-352. |
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1.26: Solving Fractional Equations
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: Jun 8, 2024
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1.25: Adding and Subtracting Rational Expressions
1.27: Rectangular Coordinate System
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: 48329
( \newcommand{\kernel}{\mathrm{null}\,})
A fractional equation is an equation involving fractions which has the unknown in the denominator of one or more of its terms.
Example 24.1
The following are examples of fractional equations:
a) 3x=920
b) x−2x+2=35
c) 3x−3=4x−5
d) 34−18x=0
e) x6−23x=23
The Cross-Product property can be used to solve fractional equations.
Cross-Product Property
If AB=CD then A⋅D=B⋅C.
Using this property we can transform fractional equations into non-fractional ones. We must take care when applying this property and use it only when there is a single fraction on each side of the equation. So, fractional equations can be divided into two categories.
I. Single Fractions on Each Side of the Equation
Equations a), b) and c) in Example 24.1 fall into this category. We solve these equations here.
a) Solve 3x=920
Cross-ProductLinear EquationDivide by 9 both sides3⋅20=9⋅x60=9x609=x
The solution is x=609=203.
b) x−2x+2=35
Cross-ProductRemove parenthesesLinear Equation: isolate the variableDivide by 2 both sides5⋅(x−2)=3⋅(x+2)5x−10=3x+65x−3x=10+62x=162x2=162
the solution is x=8.
c) 3x−3=4x−5
Cross-ProductRemove parenthesesLinear Equation: isolate the variableDivide by 2 both sides3⋅(x−5)=4⋅(x−3)3x−15=4x−123x−4x=15−12−x=3−x−1=3−1
The solution is x=−3
Note: If you have a fractional equation and one of the terms is not a fraction, you can always account for that by putting 1 in the denominator. For example:
Solve
3x=15
We re-write the equation so that all terms are fractions.
3x=151
Cross-ProductLinear Equation: isolate the variableDivide by 15 both sides3⋅1=15⋅x3=15x315=15x15
The solution is x=315=3⋅13⋅5=15.
II. Multiple Fractions on Either Side of the Equation
Equations d) and e) in Example 24.1 fall into this category. We solve these equations here.
We use the technique for combining rational expressions we learned in Chapter 23 to reduce our problem to a problem with a single fraction on each side of the equation.
d) Solve 34−18x=0
First we realize that there are two fractions on the LHS of the equation and thus we cannot use the Cross-Product property immediately. To combine the LHS into a single fraction we do the following:
Find the LCM of the denominatorsRewrite each fraction using the LCMCombine into one fractionRe-write the equation so that all terms are fractionsCross-ProductRemove parenthesesLinear Equation: isolate the variableDivide by 6 both sides8x3⋅2x8x−18x=06x−18x=06x−18x=01(6x−1)⋅1=8x⋅06x−1=06x=16x6=16
The solution is x=16.
e) Solve x6+23x=23
Find the LCM of the denominators of LHSRewrite each fraction on LHS using their LCMx2+46x=23Combine into one fractionCross-ProductRemove parenthesesQuadratic Equation: Standard formQuadratic Equation: FactorDivide by 3 both sidesQuadratic Equation: Zero-Product Property6xx⋅x6x+2⋅26x=23(x2+4)⋅3=6x⋅23x2+12=12x3x2−12x+12=03x2−12x+12=03⋅x2−3⋅4x+3⋅4=03(x2−4x+4)=03(x−2)(x−2)=03(x−2)(x−2)3=03(x−2)(x−2)=0(x−2)=0 or (x−2)=0
Since both factors are the same, then x−2=0 gives x=2. The solution is x=2
Note: There is another method to solve equations that have multiple fractions on either side. It uses the LCM of all denominators in the equation. We demonstrate it here to solve the following equation: 32−92x=35
FindtheLCMofalldenominatorsintheequationMultiply every fraction (both LHS and RHS) by the LCMSimplify every fractionSee how all denominatiors are now 1, thus can be disregardedSolve like you would any other equationLinear equation: islolate the variable10x10x⋅32−10x⋅92x=10x⋅3510x⋅32−10x⋅92x=10x⋅355x⋅31−5⋅91=2x⋅315x⋅3−5⋅9=2x⋅315x−45=6x15x−6x=459x=45x=459x=5
The solution is x=5
Exit Problem
Solve: 2x+13=12
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Nodal Analysis in Electric Circuits (Example With Step by Step Analysis)
Contents
💡
Key learnings:
Nodal Analysis Definition: Nodal analysis is a method for analyzing electric circuits by calculating voltages at different points, or nodes, in the circuit.
Kirchhoff’s Current Law (KCL): KCL is used in nodal analysis to ensure the total current entering a node equals the total current leaving, which is fundamental for setting up the equations.
Reference and Non-Reference Nodes: Reference nodes are used as a baseline or zero voltage point, while non-reference nodes are the points where voltage is measured and calculated.
Handling of Components: In nodal analysis, components like resistors and current sources are used to formulate equations based on Ohm’s Law.
Supernode Analysis: A supernode occurs when a voltage source connects two non-reference nodes, requiring an extended analysis approach that combines KCL and KVL.
Definition of Nodal Analysis
Nodal analysis, also known as the Node-Voltage Method, is a technique for analyzing circuits by focusing on the voltages at various nodes.
Some Features of Nodal Analysis are as
Nodal Analysis is based on the application of the Kirchhoff’s Current Law (KCL).
Having ‘n’ nodes there will be ‘n-1’ simultaneous equations to solve.
Solving ‘n-1’ equations all the nodes voltages can be obtained.
The number of non reference nodes is equal to the number of Nodal equations that can be obtained.
Types of Nodes in Nodal Analysis
A Non-Reference Node – is any node with a specific voltage, such as Node 1 and Node 2 in this example.
Reference Node – It is a node which acts a reference point to all the other node. It is also called the Datum Node.
Types of Reference Nodes
Chassis Ground – This type of reference node acts a common node for more than one circuits.
Earth Ground – When earth potential is used as a reference in any circuit then this type of reference node is called Earth Ground.
Solving of Circuit Using Nodal Analysis
Basic Steps Used in Nodal Analysis
Choose one node as the reference node and assign voltages V1, V2… Vn-1 to the others, measuring all relative to the reference node.
Apply KCL to each of the non reference nodes.
Use Ohm’s law to express the branch currents in terms of node voltages.
Node Always assumes that current flows from a higher potential to a lower potential in resistor. Hence, current is expressed as follows
IV. After the application of Ohm’s Law get the ‘n-1’ node equations in terms of node voltages and resistances.
V. Solve ‘n-1’ node equations for the values of node voltages and get the required node Voltages as result.
Nodal Analysis with Current Sources
Nodal analysis with current sources is very easy and it is discussed with a example below.
Example: Calculate Node Voltages in following circuit
In the following circuit we have 3 nodes from which one is reference node and other two are non reference nodes – Node 1 and Node 2.
Step I. Assign the nodes voltages as v1 and 2 and also mark the directions of branch currents with respect to the reference nodes
Step II. Apply KCL to Nodes 1 and 2
KCL at Node 1
KCL at Node 2
Step III. Apply Ohm’s Law to KCL equations
• Ohm’s law to KCL equation at Node 1
Simplifying the above equation we get,
• Now, Ohm’s Law to KCL equation at Node 2
Simplifying the above equation we get
Step IV. Now solve the equations 3 and 4 to get the values of v1 and v2 as,
Using elimination method
And substituting value v2 = 20 Volts in equation (3) we get-
Hence node voltages are as v1 = 13.33 Volts and v2 = 20 Volts.
Nodal Analysis with Voltage Sources
Case I. If a voltage source is connected between the reference node and a non reference node, we simply set the voltage at the non-reference node equal to the voltage of the voltage source and its analysis can be done as we done with current sources. v1 = 10 Volts.
Case II. If the voltage source is between the two non reference nodes then it forms a supernode whose analysis is done as following
Supernode Analysis
Definition of Super Node
A Super node is formed when a voltage source, either independent or dependent, connects two non-reference nodes. It encompasses the voltage source and both connected nodes.
In the above Figure 5V source is connected between two non reference nodes Node – 2 and Node – 3. So here Node – 2 and Node – 3 form the Super node.
Properties of Supernode
Always the difference between the voltage of two non reference nodes is known at Supernode.
A supernode has no voltage of its own
A supernode requires application of both KCL and KVL to solve it.
Any element can be connected in parallel with the voltage source forming the supernode.
A Supernode satisfies the KCL as like a simple node.
How Solve Any Circuit Containing Supernode
Let’s take a example to understand how to solve circuit containing Supernode
Here 2V voltage source is connected between Node-1 and Node-2 and it forms a Supernode with a 10Ω resistor in parallel.
Note – Any element connected in parallel with the voltage source forming Super node doesn’t make any difference because v2– v1 = 2V always whatever may be the value of resistor. Thus 10 Ω can be removed and circuit is redrawn and applying KCL to the supernode as shown in figure gives,
Expressing and in terms of the node voltages.
From Equation 5 and 6 we can write as
Hence, v1 = – 7.333V and v2 = – 5.333V which is required answer.
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7320 | https://www.ck12.org/flexi/math-grade-8/standard-form-of-linear-equations/identify-the-slope-and-the-y-intercept-of-the-given-linear-function-5x-2y4-and-use-the-answer-to-draw-and-characterize-the-graph-of-the-function/ | Flexi answers - Identify the slope and the y-intercept of the given linear function, 5x-2y=4, and use the answer to draw and characterize the graph of the function. | CK-12 Foundation
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Math Grade 8
Standard Form of Linear Equations
Question
Identify the slope and the y-intercept of the given linear function, 5x-2y=4, and use the answer to draw and characterize the graph of the function.
Flexi Says:
Slope-Intercept Form
Rewriting the Equation in Slope-Intercept Form
First, let's rewrite the equation in slope-intercept form, which is y=m x+b, where m is the slope and b is the y-intercept.
Starting with the given equation 5 x−2 y=4, we can isolate y:
−2 y=−5 x+4
y=5 2 x−2
So, the slope m is 5 2 and the y-intercept b is −2.
The graph of this function would be a straight line that crosses the y-axis at −2 and rises 5 units for every 2 units it moves to the right. This is because the slope 5 2 indicates a "rise over run" of 5 over 2.
m=5 2,b=−2
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Graph the linear functionRotate the axes to eliminate the xy-term in the equation. Then write the equation in standard form. Sketch the graph of the resulting equation, showing both sets of axes. xy – 2y – 4x = 0What is the standard form of a simple linear equation?
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Square and Square Root
Math and Algebra
Number Theory
Square Equations
Square Numbers
Square Roots (functions)
5
What are the properties of square and square roots?
·
The properties of squares and square roots are fundamental concepts in mathematics. Here's an overview of each:
Properties of Squares
Definition: The square of a number x is the result of multiplying the number by itself:
x2=x×x
Non-negativity: The square of any real number is always non-negative:
x2≥0for all x∈R
Even and Odd Properties:
The square of an even number is even:
(2n)2=4n2(even)
The square of an odd number is odd:
(2n+1)2=4n(n+1)+1(odd)
Distributive Property:
(a + b)^2 = a^
The properties of squares and square roots are fundamental concepts in mathematics. Here's an overview of each:
Properties of Squares
Definition: The square of a number x is the result of multiplying the number by itself:
x2=x×x
Non-negativity: The square of any real number is always non-negative:
x2≥0for all x∈R
Even and Odd Properties:
The square of an even number is even:
(2n)2=4n2(even)
The square of an odd number is odd:
(2n+1)2=4n(n+1)+1(odd)
Distributive Property:
(a+b)2=a2+2ab+b2
Difference of Squares:
a2−b2=(a−b)(a+b)
Properties of Square Roots
Definition: The square root of a number y is a value x such that:
x2=y
This is denoted as x=√y.
Non-negativity: The principal square root (the non-negative root) of a non-negative number is always non-negative:
√y≥0for y≥0
Square Root of a Square:
√x2=|x|(absolute value)
Product Property:
√a⋅b=√a⋅√b
Quotient Property:
√ab=√a√b(for b≠0)
Sum of Roots:
The sum of two square roots cannot be simplified in a general form, but:
√a+√b cannot be simplified further without specific values.
Additional Notes
The square root function is defined only for non-negative numbers in the real number system.
In the context of complex numbers, square roots can be defined for negative numbers as well.
These properties are essential for solving equations, simplifying expressions, and understanding the relationships between numbers in algebra and geometry.
Praveenkumar Kalikeri
Engineer by Choice, Maths Educator by passion
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8y
Some of properties of squares are as follows:
Square number always ends with either 0, 1, 4, 9, 5, 6
The number of zeros at the end of a perfect square is always even.
Squares of even numbers are always even numbers and square of odd numbers are always odd.
Square of a natural number other than one is either a multiple of 3 or exceeds a multiple of 3 by 1.
The unit’s digit of the square of a natural number is the unit’s digit of the square of the digit at unit’s place of the given natural number.
A perfect square number is never negative.
If n is a perfect square, then 2n can never be a perfect squar
Some of properties of squares are as follows:
Square number always ends with either 0, 1, 4, 9, 5, 6
The number of zeros at the end of a perfect square is always even.
Squares of even numbers are always even numbers and square of odd numbers are always odd.
Square of a natural number other than one is either a multiple of 3 or exceeds a multiple of 3 by 1.
The unit’s digit of the square of a natural number is the unit’s digit of the square of the digit at unit’s place of the given natural number.
A perfect square number is never negative.
If n is a perfect square, then 2n can never be a perfect square.
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What are the properties of a square root?
From side to side they typically are the same length and width. However, they will extend downward for a much greater distance.
From side to side they typically are the same length and width. However, they will extend downward for a much greater distance.
Related questions
What are the properties of a square root?
What are the properties of a square?
What are some properties of square roots that we can use to find them quickly?
What are some properties of cube roots and square roots?
What are square roots? What are some examples?
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What is the square root of a number, and what are its properties?
The square root of a number, essentially, is the number, when multiplied with itself will result in the number it is the square root of.
For example, the square root of 16 is 4. This is because 4 4 is 16. Like this, the square root of 25 is 5 because 5 5 is 25.
There are many properties to square roots. For example, any positive number has two square roots.
For example, 16 has 4 and -4 as its square roots. This is because 4 4 = 16 AND -4 -4 = 16. This is because minus into minus is plus. The two square roots are just the positive and negative of each other eg. 4 & -4 (for 16) and 5 & -5 (
The square root of a number, essentially, is the number, when multiplied with itself will result in the number it is the square root of.
For example, the square root of 16 is 4. This is because 4 4 is 16. Like this, the square root of 25 is 5 because 5 5 is 25.
There are many properties to square roots. For example, any positive number has two square roots.
For example, 16 has 4 and -4 as its square roots. This is because 4 4 = 16 AND -4 -4 = 16. This is because minus into minus is plus. The two square roots are just the positive and negative of each other eg. 4 & -4 (for 16) and 5 & -5 (for 25).
Unfortunately, negative numbers do not have any square roots.
This is because they are negative - they have a minus. Unfortunately, plus into plus is plus and minus into minus is plus. This is why no two of the same number can result in a negative number unless someone finds the square root of -1.
This is why the square roots of negative numbers are called “imaginary numbers” - they do not exist.
These are some of the properties of negative numbers. Hope this helps!
Jorj Kowszun
Former Senior Lecturer in Mathematics (Now Retired) at University of Brighton (2010–2020)
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What are the properties of a square root?
If this is a mathematical question then there is only one property that matters. The square root of a number when squared is the original number!
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Related
What is the square root of -4?
What is the square root of -4?
The square root of −4 does not exist as an Integer or as a Real number, but treating it as the Complex number (−4,0) it has two square roots, namely (0,2) and (0,−2). Or (0,±2).
Complex numbers are pairs of Real numbers, (x,y), where addition and multiplication are defined by:
(a,b)+(c,d)=(a+c,b+d)
(a,b)(c,d)=(ac−bd,bc+ad)
You will note that (0,1)(0,1)=(0−1,0+0)=(−1,0) so that
(0,1)2=(−1,0)
For notational convenience we write this as i2=−1.
You will also note that
(x,0)+(0,1)(y,0)=(x,0)+(0−0,y+0)=(x,y)
so you will also see (x,y) written as x+iy.
So the answer
What is the square root of -4?
The square root of −4 does not exist as an Integer or as a Real number, but treating it as the Complex number (−4,0) it has two square roots, namely (0,2) and (0,−2). Or (0,±2).
Complex numbers are pairs of Real numbers, (x,y), where addition and multiplication are defined by:
(a,b)+(c,d)=(a+c,b+d)
(a,b)(c,d)=(ac−bd,bc+ad)
You will note that (0,1)(0,1)=(0−1,0+0)=(−1,0) so that
(0,1)2=(−1,0)
For notational convenience we write this as i2=−1.
You will also note that
(x,0)+(0,1)(y,0)=(x,0)+(0−0,y+0)=(x,y)
so you will also see (x,y) written as x+iy.
So the answer to the question can also be written as ±2i.
You will also see i referred to as the “imaginary unit” which, unfortunately, makes people think it is less real than some other pairs of numbers, but it is just as good a number as 12 for example.
Related questions
What is the definition of squares and square roots in mathematics?
What is a square root’s importance in math?
Who discovered the square and square roots?
What are some properties of square roots that can be used to determine whether a given number has a real square root?
What are some properties of square roots that can be used to find them by taking the difference between their sums and products?
Prakhar Agarwal
Studied at Ryan International School
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Author has 53 answers and 157.6K answer views
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8y
Related
Why do we take two values of square root?
No we do not!
Square roots are defined only for positive numbers (and zero) and are always positive (or zero) only.
See the graph of √x below. It only outputs positive numbers.
Any calculator would also give you positive outputs for any input of yours. (Example √9=3 and not 3,−3)
The confusion arises due to equations like these
x2=9
⟹x=3,−3
This is correct!
Here you were looking for values of x which upon squaring will give you 9. Hence 3 and −3 both do the job.
Square roots are different. While solving them you musn't look for values whose square will give you the thing inside
No we do not!
Square roots are defined only for positive numbers (and zero) and are always positive (or zero) only.
See the graph of √x below. It only outputs positive numbers.
Any calculator would also give you positive outputs for any input of yours. (Example √9=3 and not 3,−3)
The confusion arises due to equations like these
x2=9
⟹x=3,−3
This is correct!
Here you were looking for values of x which upon squaring will give you 9. Hence 3 and −3 both do the job.
Square roots are different. While solving them you musn't look for values whose square will give you the thing inside your square root.
If this were true √x2 should have been equal to x.
That is not the case!
√x2=|x|
x can be both positive and negative but |x| is always positive.
See the difference in their graphs (|x| is green and x is blue)
x=|x| ∀ x≥0 (green and blue overlap)
But
x=−|x| ∀ x<0
Hope this helped!
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What is the formula for square root?
A handy one is the “Babylonian method.”
Step 1: take a guess. Call it S0
Step 2: Refine your guess with this formula.
Sn+1=12(Sn+NSn)
Step 3: Repeat step 2 until you have achieved a sufficient level of precision.
Since the square root of most numbers is irrational and Sn is rational, this process must be repeated infinitely to get true convergence.
Here is an alternative using Geometry.
Make a circle with diameter N+1. Make a vertical one unit from the edge as above. The length of the vertical =√N
A handy one is the “Babylonian method.”
Step 1: take a guess. Call it S0
Step 2: Refine your guess with this formula.
Sn+1=12(Sn+NSn)
Step 3: Repeat step 2 until you have achieved a sufficient level of precision.
Since the square root of most numbers is irrational and Sn is rational, this process must be repeated infinitely to get true convergence.
Here is an alternative using Geometry.
Make a circle with diameter N+1. Make a vertical one unit from the edge as above. The length of the vertical =√N
Alan Bustany
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Author has 9.8K answers and 58.3M answer views
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Aug 14
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What is the definition of squares and square roots in mathematics?
Given a set,
S
, and an infix multiplication operator,
S
×
S
→
S
, then:
The square of
x
∈
S
is
x
×
x
denoted
x
2
; and
The square root of
x
∈
S
is
y
∈
S
where
x
=
y
×
y
, if it exists, denoted
√
x
.
S
might be an ordinairy set of numbers like the set of Integers,
Z
, or some other algebraic structure like, say, the Field of Integers modulo seven,
F
7
, where:
3
2
=
2
; so
√
2
=
3
.
Or it could be some more exotic non-numeric algebraic structure like, say, a Rubik's cube, with "multiplication" defined appropriately.
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Related
What is square and square root?
Square and square root are opposites of each other, just like multiply and divide, or plus and minus.
The “square” of a number Y (written as Y to the power of 2), is calculated by multiplying Y by itself => Y x Y. So, 4 squared is equal to 4 x 4 = 16.
If you raise the number Y to a different power other than 2, say 4, then it is calculated by multiplying Y four times => Y x Y x Y x Y.
The square root is the opposite. The square root of 16 is found by asking the question, “what, times itself, is equal to 16?” As we see above, 4 x 4 = 16, so, the square root of 16 is 4.
A fourth root is where you wo
Square and square root are opposites of each other, just like multiply and divide, or plus and minus.
The “square” of a number Y (written as Y to the power of 2), is calculated by multiplying Y by itself => Y x Y. So, 4 squared is equal to 4 x 4 = 16.
If you raise the number Y to a different power other than 2, say 4, then it is calculated by multiplying Y four times => Y x Y x Y x Y.
The square root is the opposite. The square root of 16 is found by asking the question, “what, times itself, is equal to 16?” As we see above, 4 x 4 = 16, so, the square root of 16 is 4.
A fourth root is where you would ask yourself, “what, times itself four times, is equal to the number. Since 2 x 2 x 2 x 2 = 16, the fourth root of 16 is 2.
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What is the difference between square and square root?
Square is the opposite of square root.
Square is the a number multiplied by itself. The square root of a number is the number that is multiplied by itself to get that number.
David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979)
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What are square roots?
If you’re only considering positive numbers, the square root of a positive number is the positive number whose square is the given number.
For example, the square root of 1.21 is 1.1. That’s because if you square 1.1 (that is, multiply 1.1 by itself) you get 1.21.
The standard notation for the square root of a positive number x is √x. The √ symbol has been used for that since the 1600s. Apparently it’s a stylized letter “r”, the first letter of “radix”, which is a Latin word which means the same as “root” in English.
If you’re also considering negative numbers, then there are two squar
If you’re only considering positive numbers, the square root of a positive number is the positive number whose square is the given number.
For example, the square root of 1.21 is 1.1. That’s because if you square 1.1 (that is, multiply 1.1 by itself) you get 1.21.
The standard notation for the square root of a positive number x is √x. The √ symbol has been used for that since the 1600s. Apparently it’s a stylized letter “r”, the first letter of “radix”, which is a Latin word which means the same as “root” in English.
If you’re also considering negative numbers, then there are two square roots of a positive number. When you square –1.1 you also get 1.21, therefore –1.1 is also a square root of 1.21. The negative square root of a positive number x is denoted −√x. If you want to name both the positive and negative roots of x, use the notation ±√x.
The number 0 has only one square root, and that is 0 itself.
If you’re considering complex numbers, there will be two square roots. For example, √i=±12(√2+i√2).
Bernard Leak
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Horst H. von Brand
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What is the square root of 1 + square root of 1 + square root of 1 infinitely?
I suppose you mean
√1+√1+√1+√1+…
… and by that you mean
where [math]a_0 = \sqrt{1}[/math] and [math]a_{i + 1} = \sqrt{1 + a_i}[/math]
(and the [math]a_i[/math] are all real numbers).
It's easy to see that [math]a_{i + 1} > a_i[/math]. The question is, does the sequence [math]a_i[/math] increase without bound, or is there an upper bound? In the latter case, there must be a limit.
Well, if [math]a_i < 1.7[/math], [math]a_{i + 1} = \sqrt{1 + a_i} < \sqrt{2.7} < 1.7[/math]
[math]a_0 < 1.7[/math], so all the [math]a_i[/math] are also less than [math]1.7[/math], by induction.
[math]1.7[/math] is therefore an upper bound, and the sequence [math]a_i[/math] has a least upper bound, which is the lim
I suppose you mean
[math]\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}}[/math]
… and by that you mean
[math]\lim\limits_{n \to \infty} a_i[/math]
where [math]a_0 = \sqrt{1}[/math] and [math]a_{i + 1} = \sqrt{1 + a_i}[/math]
(and the [math]a_i[/math] are all real numbers).
It's easy to see that [math]a_{i + 1} > a_i[/math]. The question is, does the sequence [math]a_i[/math] increase without bound, or is there an upper bound? In the latter case, there must be a limit.
Well, if [math]a_i < 1.7[/math], [math]a_{i + 1} = \sqrt{1 + a_i} < \sqrt{2.7} < 1.7[/math]
[math]a_0 < 1.7[/math], so all the [math]a_i[/math] are also less than [math]1.7[/math], by induction.
[math]1.7[/math] is therefore an upper bound, and the sequence [math]a_i[/math] has a least upper bound, which is the limit.
Having got that far, call the limit [math]a[/math]. We must have [math]\sqrt{1 + a} = a[/math] (why?)
or [math]a + 1 = a^2[/math]. Glory be, it's [math]\phi[/math] again.
[math]a = \frac{1 + \sqrt{5}}{2}[/math]
Related questions
What are the properties of a square root?
What are the properties of a square?
What are some properties of square roots that we can use to find them quickly?
What are some properties of cube roots and square roots?
What are square roots? What are some examples?
What is the definition of squares and square roots in mathematics?
What is a square root’s importance in math?
Who discovered the square and square roots?
What are some properties of square roots that can be used to determine whether a given number has a real square root?
What are some properties of square roots that can be used to find them by taking the difference between their sums and products?
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Who invented square roots and cube roots?
What are two consecutive square roots?
What is the value of a square root?
Can exact square roots not be found?
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7322 | https://www.merriam-webster.com/word-of-the-day/meander-2022-02-15 | Est. 1828
Word of the Day
: February 15, 2022
meander
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verb mee-AN-der
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What It Means
Meander means "to wander aimlessly or casually" or "to follow a winding or intricate course."
// The couple spent the afternoon meandering around the seaside town.
// The stream meanders through the canyon.
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meander in Context
"Seeking a respite from the frantic pace of the digital world, we turned to 'The Unwinding: And Other Dreamings.' … [We] meandered through the pages, absorbed in peaceful fragments of prose and Morris's elaborate watercolors." — Julian Wray and Sam Mathisson, The Michigan Daily, 10 Dec. 2021
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7323 | https://www.jsums.edu/nmeghanathan/files/2016/08/CSC323-Fall2016-Module-3-Greedy-Strategy.pdf | Module 3 Greedy Strategy Dr. Natarajan Meghanathan Professor of Computer Science Jackson State University Jackson, MS 39217 E-mail: natarajan.meghanathan@jsums.edu Introduction to Greedy Technique • Main Idea: In each step, choose the best alternative available in the hope that a sequence of locally optimal choices will yield a (globally) optimal solution to the entire problem.
• Example 1: Decimal to binary representation (objective: minimal number of 1s in the binary representation): Technique – Choose the largest exponent of 2 that is less than or equal to the unaccounted portion of the decimal integer. 1 2 4 8 16 32 64 To represent 75: 1 0 0 1 0 1 1 • Example 2: Coin Denomination in US – Quarter (25 cents), Dime (10 cents), Nickel (5 cents) and Penny (1 cent). • Objective: Find the minimum number of coins for a change • Strategy: Choose the coin with the largest denomination that is less than or equal to the unaccounted portion of the change.
• For example, to find a change for 48, we would choose 1 quarter, 2 dimes and 3 pennies. The optimal solution is thus 6 coins and there cannot be anything less than 6 coins for US coin denominations.
Greedy Technique: Be careful!!!
• Greedy technique (though may appear to be computationally simple) cannot always guarantee to yield the optimal solution. It may end up only as an approximate solution to an optimization problem.
• For example, consider a more generic coin denomination scenario where the coins are valued 25, 10 and 1. To make a change for 30, we would end up using 6 coins (1 coin of value 25 and 5 coins of value 1 each) following the greedy technique. On the other hand, if we had used a dynamic programming algorithm for this generic version, we would have end up with 3 coins, each of value 10.
Fractional Knapsack Problem (Greedy Algorithm): Example 1 • Knapsack weight is 6lb.
• Item 1 2 3 4 5 • Value, $ 25 20 15 40 50 • Weight, lb 3 2 1 4 5 • Value/Weight 8.3 10 15 10 10 • Greedy Strategy: Pick the items in the decreasing order of the Value/Weight.
• Break the tie among the items the same Value/Weight by picking the item with the lowest Item index • An optimal solution would be: • Item 3 (1 lb), Item 2 (2 lb), and 3 lbs of Item 4. • The maximum total Value of the items would be: $65 • Item 3 ($15), Item 2 ($20) and Item 4( (3/4)40 = $30) • Dynamic Programming: If the items cannot be divided, and we have to pick only either the full item or just leave it, then the problem is referred to as an Integer (a.k.a. 0-1) Knapsack problem, and we will look at it in the module on Dynamic Programming.
Fractional Knapsack Problem (Greedy Algorithm): Example 2 Knapsack weight = 5 lb.
Item 1 2 3 4 Value, $ 12 10 20 15 Weight, lb 2 1 3 2 Solution: Compute the Value/Weight for each item Item 1 2 3 4 Value/Weight 6 10 6.67 7.5 Re-ordering the items according to the decreasing order of Value/Weight (break the tie by picking the item with the lowest Index) Item 2 4 3 1 Value/Weight 10 7.5 6.67 6 Value, $ 10 15 20 12 Weight, lb 1 2 3 2 Weight collected 1 2 2 Items collected: Item 2 (1 lb, $10); Item 4 (2 lb, $15); Item 3 (2 lb, (2/3)20 = $13.3); Total Value = $38.3 Variable Length Prefix Encoding • Encoding Problem: We want to encode a text that comprises of symbols from some n-symbol alphabet by assigning each symbol a sequence of bits called the codeword.
• If we assign bit sequences of the same length to each symbol, it is referred to as fixed-length encoding, we would need log2n bits per symbol of the alphabet and this is also the average # bits per symbol.
– The 8-bit ASCII code assigns each of the 256 symbols a unique 8-bit binary code (whose integer values range from 0 to 255).
– However, note that not all of these 256 symbols appear with the same frequency.
• Motivation for Variable Code Assignment: If we can come up with a code assignment such that symbols are assigned a bit sequence that is inversely related to the frequency of their occurrence (i.e., symbols that occur more frequently are given a shorter bit sequence and symbols that occur less frequently are given a longer bit sequence), then we could reduce the average number of bits per symbol.
• Motivation for Prefix-free Code: However, care should be taken such that if a given sequence of bits encoding a text is scanned (say from left to right), we should be able to clearly decode each symbol. In other words, we should be able to tell how many bits of an encoded text represent the ith symbol in the text?
Huffman Codes: Prefix-free Coding • Prefix-free Code: In a prefix-free code, no codeword is a prefix of a code of another symbol. With a prefix-free code based encoding, we can simply scan a bit string until we get the first group of bits that is a codeword for some symbol, replace these bits by this symbol, and repeat this operation until the bit string’s end is reached.
• Huffman Coding: – Associate the alphabet’s symbols with leaves of a binary tree in which all the left edges are labeled by 0 and all the right edges are labeled by 1.
– The codeword of a symbol can be obtained by recording the labels on the simple path (a path without any cycle) from the root to the symbol’s leaf.
• Proof of correctness: The binary codes are assigned based on a simple path traversed from the root to a leaf node representing the symbol. Since there cannot be a simple path from the root to a leaf node that leads to another leaf node (then we have to trace back some intermediate node – meaning a cycle). Hence, Huffman codes are prefix codes.
Huffman Algorithm • Assumptions: The frequencies of symbol occurrence are independent and are known in advance.
• Optimality: Given the above assumption, Huffman’s encoding yields a minimum-length encoding (i.e., the average number of bits per symbol is the minimum). This property of Huffman’s encoding has lead to its use one of the most important file-compression methods.
– Symbols that occur at a high-frequency have a smaller number of bits in the binary code, compared to symbols that occur at a low-frequency. • Step 1: Initialize n one-node trees (one node for each symbol) and label them with the symbols of the given alphabet. Record the frequency of each symbol in its tree’s root to indicate the tree’s weight.
• Step 2: Repeat the following operation until a single tree is obtained: – Find two trees with the smallest weight (ties can be broken arbitrarily). – Make them the left and right sub trees of a new tree and record the sum of their weights in the root of the new tree as its weight.
Huffman Algorithm and Coding: Example • Consider the five-symbol alphabet {A, B, C, D, -} with the following occurrence frequencies in a text made up of these symbols. – Construct a Huffman tree for this alphabet. – Determine the average number of bits per symbol.
– Determine the compression ratio achieved compared to fixed-length encoding.
Initial Iteration - 1 Break any tie by preferring to include the node with a smaller height to the right If the height is not distinguishable, use Node ID, if possible; otherwise, break the ties arbitrarily.
Huffman Algorithm and Coding: Example Iteration - 2 Iteration - 3 Iteration – 4 (Final) Avg. # bits per symbol = 20.35 + 30.1 + 20.2 + 20.2 + 30.15 = 2.25 bits per symbol.
A fixed-length encoding of 5 symbols would require = 3 symbols. Hence, the Avg. Compression ratio is 1 – (2.25/3) = 25%.
5 log2 A 0.4 B 0.2 C 0.25 D 0.1 -0.05 -0.05 A 0.4 D 0.1 B 0.2 C 0.25 Initial Iteration 1 -0.05 D 0.1 0.15 A 0.4 B 0.2 C 0.25 Iteration 2 -0.05 D 0.1 0.15 A 0.4 B 0.2 C 0.25 0.35 Huffman Coding: Example 2 Iteration 3 -0.05 D 0.1 0.15 A 0.4 B 0.2 C 0.25 0.35 0.6 Iteration 4 -0.05 D 0.1 0.15 A 0.4 B 0.2 C 0.25 0.35 0.6 1.0 -0.05 D 0.1 0.15 A 0.4 B 0.2 C 0.25 0.35 0.6 1.0 Huffman Tree 0 1 0 1 0 1 0 1 A 0.4 0 B 0.2 111 C 0.25 10 D 0.1 1101 -0.05 1100 Huffman Codes Average # bits per symbol (generic) = (0.4)(1) + (0.2)(3) + (0.25)(2) + (0.1)(4) + (0.05)(4) = 0.4 + 0.6 + 0.5 + 0.4 + 0.2 = 2.1 bits/symbol Average (Generic) Compression Ratio 1 – (2.1/3) = 0.3 = 30% where 3 is the # bits/symbol under fixed encoding scheme.
A 0.4 0 B 0.2 111 C 0.25 10 D 0.1 1101 -0.05 1100 Huffman Codes Specific Character/Symbol Sequence: A A B C A C D – A B 0 0 111 10 0 10 1101 1100 0 111 Total # bits in the above sequence = 22 bits Average # bits / symbol in the above sequence = 22 / 10 = 2.2 bits/symbol where 10 is the number of symbols in the above sequence If we had used fixed-length encoding, we would have used: 3 bits/symbol 10 symbols = 30 bits Compression ratio = 1 – (22/30) = 26.7% |
7324 | https://mathresearch.utsa.edu/wiki/index.php?title=Intervals | Intervals
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The addition x + a on the number line. All numbers greater than x and less than x + a fall within that open interval.
In mathematics, a (real) interval is a set of real numbers that contains all real numbers lying between any two numbers of the set. For example, the set of numbers x satisfying 0 ≤ x ≤ 1 is an interval which contains 0, 1, and all numbers in between. Other examples of intervals are the set of numbers such that 0 < x < 1, the set of all real numbers , the set of nonnegative real numbers, the set of positive real numbers, the empty set, and any singleton (set of one element).
Real intervals play an important role in the theory of integration, because they are the simplest sets whose "size" (or "measure" or "length") is easy to define. The concept of measure can then be extended to more complicated sets of real numbers, leading to the Borel measure and eventually to the Lebesgue measure.
Intervals are central to interval arithmetic, a general numerical computing technique that automatically provides guaranteed enclosures for arbitrary formulas, even in the presence of uncertainties, mathematical approximations, and arithmetic roundoff.
Intervals are likewise defined on an arbitrary totally ordered set, such as integers or rational numbers. The notation of integer intervals is considered in the special section below.
Contents
1 Terminology
1.1 Note on conflicting terminology
2 Notations for intervals
2.1 Including or excluding endpoints
2.2 Infinite endpoints
2.3 Integer intervals
3 Classification of intervals
4 Properties of intervals
5 Resources
6 References
Terminology
An open interval does not include its endpoints, and is indicated with parentheses. For example, (0,1) means greater than 0 and less than 1. This means (0,1) = {x | 0 < x < 1}.
A closed interval is an interval which includes all its limit points, and is denoted with square brackets. For example, [0,1] means greater than or equal to 0 and less than or equal to 1.
A half-open interval includes only one of its endpoints, and is denoted by mixing the notations for open and closed intervals. For example, (0,1] means greater than 0 and less than or equal to 1, while [0,1) means greater than or equal to 0 and less than 1.
A degenerate interval is any set consisting of a single real number (i.e., an interval of the form [a,a]). Some authors include the empty set in this definition. A real interval that is neither empty nor degenerate is said to be proper, and has infinitely many elements.
An interval is said to be left-bounded or right-bounded, if there is some real number that is, respectively, smaller than or larger than all its elements. An interval is said to be bounded, if it is both left- and right-bounded; and is said to be unbounded otherwise. Intervals that are bounded at only one end are said to be half-bounded. The empty set is bounded, and the set of all reals is the only interval that is unbounded at both ends. Bounded intervals are also commonly known as finite intervals.
Bounded intervals are bounded sets, in the sense that their diameter (which is equal to the absolute difference between the endpoints) is finite. The diameter may be called the length, width, measure, range, or size of the interval. The size of unbounded intervals is usually defined as +∞, and the size of the empty interval may be defined as 0 (or left undefined).
The centre (midpoint) of bounded interval with endpoints a and b is (a + b)/2, and its radius is the half-length |a − b|/2. These concepts are undefined for empty or unbounded intervals.
An interval is said to be left-open if and only if it contains no minimum (an element that is smaller than all other elements); right-open if it contains no maximum; and open if it has both properties. The interval [0,1) = {x | 0 ≤ x < 1}, for example, is left-closed and right-open. The empty set and the set of all reals are open intervals, while the set of non-negative reals, is a right-open but not left-open interval. The open intervals are open sets of the real line in its standard topology, and form a base of the open sets.
An interval is said to be left-closed if it has a minimum element, right-closed if it has a maximum, and simply closed if it has both. These definitions are usually extended to include the empty set and the (left- or right-) unbounded intervals, so that the closed intervals coincide with closed sets in that topology.
The interior of an interval I is the largest open interval that is contained in I; it is also the set of points in I which are not endpoints of I. The closure of I is the smallest closed interval that contains I; which is also the set I augmented with its finite endpoints.
For any set X of real numbers, the interval enclosure or interval span of X is the unique interval that contains X, and does not properly contain any other interval that also contains X.
An interval I is subinterval of interval J if I is a subset of J. An interval I is a proper subinterval of J if I is a proper subset of J.
Note on conflicting terminology
The terms segment and interval have been employed in the literature in two essentially opposite ways, resulting in ambiguity when these terms are used. The Encyclopedia of Mathematics defines interval (without a qualifier) to exclude both endpoints (i.e., open interval) and segment to include both endpoints (i.e., closed interval), while Rudin's Principles of Mathematical Analysis calls sets of the form [a, b] intervals and sets of the form (a, b) segments throughout. These terms tend to appear in older works; modern texts increasingly favor the term interval (qualified by open, closed, or half-open), regardless of whether endpoints are included.
Notations for intervals
The interval of numbers between a and b, including a and b, is often denoted [a, b]. The two numbers are called the endpoints of the interval. In countries where numbers are written with a decimal comma, a semicolon may be used as a separator to avoid ambiguity.
Including or excluding endpoints
To indicate that one of the endpoints is to be excluded from the set, the corresponding square bracket can be either replaced with a parenthesis, or reversed. Both notations are described in International standard ISO 31-11. Thus, in set builder notation,
Each interval (a, a), [a, a), and (a, a] represents the empty set, whereas [a, a] denotes the singleton set {a}. When a > b, all four notations are usually taken to represent the empty set.
Both notations may overlap with other uses of parentheses and brackets in mathematics. For instance, the notation (a, b) is often used to denote an ordered pair in set theory, the coordinates of a point or vector in analytic geometry and linear algebra, or (sometimes) a complex number in algebra. That is why Bourbaki introduced the notation ]a, b[ to denote the open interval. The notation [a, b] too is occasionally used for ordered pairs, especially in computer science.
Some authors use ]a, b[ to denote the complement of the interval (a, b); namely, the set of all real numbers that are either less than or equal to a, or greater than or equal to b.
Infinite endpoints
In some contexts, an interval may be defined as a subset of the extended real numbers, the set of all real numbers augmented with −∞ and +∞.
In this interpretation, the notations [−∞, b] , (−∞, b] , [a, +∞] , and [a, +∞) are all meaningful and distinct. In particular, (−∞, +∞) denotes the set of all ordinary real numbers, while [−∞, +∞] denotes the extended reals.
Even in the context of the ordinary reals, one may use an infinite endpoint to indicate that there is no bound in that direction. For example, (0, +∞) is the set of positive real numbers, also written as . The context affects some of the above definitions and terminology. For instance, the interval (−∞, +∞) = is closed in the realm of ordinary reals, but not in the realm of the extended reals.
Integer intervals
When a and b are integers, the notation ⟦a, b⟧, or [a .. b] or {a .. b} or just a .. b, is sometimes used to indicate the interval of all integers between a and b included. The notation [a .. b] is used in some programming languages; in Pascal, for example, it is used to formally define a subrange type, most frequently used to specify lower and upper bounds of valid indices of an array.
An integer interval that has a finite lower or upper endpoint always includes that endpoint. Therefore, the exclusion of endpoints can be explicitly denoted by writing a .. b − 1 , a + 1 .. b , or a + 1 .. b − 1. Alternate-bracket notations like [a .. b) or [a .. b[ are rarely used for integer intervals.
Classification of intervals
The intervals of real numbers can be classified into the eleven different types listed below, where a and b are real numbers, and :
: Empty:
: Degenerate:
: Proper and bounded:
: Open:
: Closed:
: Left-closed, right-open:
: Left-open, right-closed:
: Left-bounded and right-unbounded:
: Left-open:
: Left-closed:
: Left-unbounded and right-bounded:
: Right-open:
: Right-closed:
: Unbounded at both ends (simultaneously open and closed): :
Properties of intervals
The intervals are precisely the connected subsets of . It follows that the image of an interval by any continuous function is also an interval. This is one formulation of the intermediate value theorem.
The intervals are also the convex subsets of . The interval enclosure of a subset is also the convex hull of .
The intersection of any collection of intervals is always an interval. The union of two intervals is an interval if and only if they have a non-empty intersection or an open end-point of one interval is a closed end-point of the other (e.g., ).
If is viewed as a metric space, its open balls are the open bounded sets (c + r, c − r), and its closed balls are the closed bounded sets [c + r, c − r].
Any element x of an interval I defines a partition of I into three disjoint intervals I1, I2, I3: respectively, the elements of I that are less than x, the singleton , and the elements that are greater than x. The parts I1 and I3 are both non-empty (and have non-empty interiors), if and only if x is in the interior of I. This is an interval version of the trichotomy principle.
Resources
Intervals and Interval Notation, Khan Academy
References
"List of Arithmetic and Common Math Symbols". Math Vault. 2020-03-17. Retrieved 2020-08-23.
"Intervals". www.mathsisfun.com. Retrieved 2020-08-23.
Weisstein, Eric W. "Interval". mathworld.wolfram.com. Retrieved 2020-08-23.
"Interval and segment - Encyclopedia of Mathematics". www.encyclopediaofmath.org. Archived from the original on 2014-12-26. Retrieved 2016-11-12.
Rudin, Walter (1976). Principles of Mathematical Analysis. New York: McGraw-Hill. pp. 31. ISBN 0-07-054235-X.
"Why is American and French notation different for open intervals (x, y) vs. ]x, y[?". hsm.stackexchange.com. Retrieved 28 April 2018.
"Compendium of Mathematical Symbols". Math Vault. 2020-03-01. Retrieved 2020-08-23.
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7325 | https://fiveable.me/ap-calc/unit-6/fundamental-theorem-calculus-accumulation-functions/study-guide/TyDEzN9M5wieA4Kw | printables
♾️AP Calculus AB/BC Unit 6 Review
6.4 The Fundamental Theorem of Calculus and Accumulation Functions
♾️AP Calculus AB/BC Unit 6 Review
6.4 The Fundamental Theorem of Calculus and Accumulation Functions
Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025
♾️AP Calculus AB/BC
Unit & Topic Study Guides
Unit 6 Overview: Integration and Accumulation of Change
6.1 Integration and Accumulation of Change
6.2 Approximating Areas with Riemann Sums
6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
6.4 The Fundamental Theorem of Calculus and Accumulation Functions
6.5 Interpreting the Behavior of Accumulation Functions Involving Area
6.6 Applying Properties of Definite Integrals
6.7 The Fundamental Theorem of Calculus and Definite Integrals
6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation
6.9 Integrating Using Substitution
6.10 Integrating Functions Using Long Division and Completing the Square
6.11 Integrating Using Integration by Parts
6.12 Integrating Using Linear Partial Fractions
6.13 Evaluating Improper Integrals
6.14 Selecting Techniques for Antidifferentiation (AB)
6.4 The Fundamental Theorem of Calculus and Accumulation Functions
Now that we have introduced integrals, you may be wondering how they connect with derivatives. Today, we’ll introduce the Fundamental Theorem of Calculus which connects these two topics.
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∫ Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation and integration. It states that the derivative of an integral is just the function inside the integral. 😊
We can define a new function, F, that represents the antiderivative of f.
F(x)=∫axf(t)dt
The Fundamental Theorem of Calculus states that if f is continuous on the interval (a,b) then for every x in the interval:
dxd[∫axf(t)dt]=F′(x)=f(x)
This means that we can use the Fundamental Theorem of Calculus to find derivatives. Let’s walk through an example to see what this means.
✏️ Fundamental Theorem of Calculus Example
Find g′(16) if g(x) is the function below.
g(x)=∫5x4tdt
How should we proceed? If we ignore what g(x) is specifically defined as for a minute and think about how we would proceed for a function that is not defined with integrals, we would do so by first finding the derivative of g(x) and then plugging 16 in. Well for functions defined by definite integrals, also called accumulation functions, we do the same!
How do we differentiate ∫5x4tdt?
This is where the Fundamental Theorem of Calculus comes in. Since it states that the derivative of an integral is just the function inside the integral…
g′(x)=4x
Now all we have to do is substitute 16.
g′(16)=416=2
Not too bad right? 🪄
✏️ Fundamental Theorem of Calculus Example 2
In the above example, the upper bound function was just x, making it super easy! All we had to do was substitute x into the integrated and multiply by the derivative of x.
Questions become slightly more complicated when the functional bound is something other than x, such as the following function. Let’s find F′(x).
F(x)=∫3x2(t+4)dt
The upper bound in this question is x2, so we have to multiply our answer by the derivative of the functional bound.
F′(x)=dxd∫3x2(t+4)dt
Let’s use the chain rule.
F′(x)=(x2+4)dxdx2 F′(x)=(x2+4)(2x)
So, when the upper bound of the definite integral is a function of x, you need to multiply the integral's integrand by the derivative of the upper bound. It’s not too difficult, but it does mean you have to keep an eye on the upper bound! 👀
📝 Practice Problems
Time to try some practice on our own!
❓ Problems
Question 1
Let g(x)=∫0x8+cos(t)dt. Find g′(0).
Question 2
Let g(x)=∫1x(5t2+2t)dt. Find g′(3).
Question 3
Let g(x)=∫0xsin(t)+15dt. Find g′(2π).
Question 4
Let F(x)=∫3x1sec2(t)dt. Find F′(x).
✅ Answers and Solutions
Question 1
To find g′(0), we simply have to find g′(x) and evaluate it at x=0 since the upper bound is already x!
By the Fundamental Theorem of Calculus, g′(x)=8+cos(x).
Therefore, g′(0)=8+cos(0)=8+1=9=3.
Question 2
To find g′(3), we simply have to find g′(x) and evaluate it at x=3.
By the Fundamental Theorem of Calculus, g′(x)=5x2+2x.
Therefore, g′(3)=5⋅32+2⋅3=45+6=51.
Question 3
To find g′(2π), we simply have to find g′(x) and evaluate it at x=2π.
By the Fundamental Theorem of Calculus, g′(x)=sin(x)+15.
Therefore, g′(2π)=15+sin(2π)=15+1=16=4.
Question 4
To find F′(x), we first have to notice that the upper bound is not x.
The upper bound needs to have the variable, so we have to use integral rules to switch the bounds. If you’re going through our guides in order, you haven’t come across integral rules yet. We’re covering that in two key topics, 6.6! Here’s the integral rule you’d be using in this question:
∫abf(x)dx=−∫baf(x)dx
Here’s how we apply it…
F′(x)=dxd(−∫13xsec2(t)dt)
And here’s the rest of the problem, using what we already know!
F′(x)=−sec2(3x)dxd3x =−3sec2(3x)
💫 Closing
Great work! You've learned about the Fundamental Theorem of Calculus, which, remember, bridges the gap between differentiation and integration.
You've also seen how to apply this theorem to find derivatives of functions defined by definite integrals, including cases where the upper bound is a function of x. When the upper bound is not a simple x, you multiply the integrand by the derivative of the upper bound.
Keep practicing and working through the practice problems to solidify your understanding. Calculus can be challenging, but with dedication and practice, you'll master it! Good luck. 📚
Frequently Asked Questions
How do I find the derivative of an integral like d/dx of the integral from 0 to x of f(t) dt?
If g(x) = ∫_0^x f(t) dt and f is continuous on an interval containing 0, then by the Fundamental Theorem of Calculus (FTC, FUN-5.A.2) g′(x) = f(x). In words: the derivative of an accumulation (area) function with a variable upper limit is just the integrand evaluated at that upper limit. Notes you should know: - The variable of integration (t) is a dummy variable—don’t confuse it with x. - Continuity of f on the interval containing 0 is required for FTC to apply. - If the upper limit is a function, say h(x), use the chain rule: d/dx ∫_0^{h(x)} f(t) dt = f(h(x))·h′(x). - If the lower limit is variable, e.g. ∫_{a(x)}^{b(x)} f(t) dt, differentiate using: f(b(x))b′(x) − f(a(x))a′(x). For a quick review of these ideas (and AP-aligned examples), see the Topic 6.4 study guide ( and practice problems (
What's the formula for the Fundamental Theorem of Calculus part 1?
Fundamental Theorem of Calculus, Part 1 (FTC1)—formula and notes: If f is continuous on an interval containing a, then the accumulation function g(x) = ∫_a^x f(t) dt satisfies g′(x) = f(x). So the derivative of the integral with a constant lower limit and variable upper limit x equals the integrand evaluated at x (the variable of integration t is a “dummy” variable). If the upper limit is a function u(x), use the chain rule: d/dx ∫_a^{u(x)} f(t) dt = f(u(x)) · u′(x). Why this matters for AP: this is FUN-5.A.2 in the CED—you’ll use FTC1 to link definite integrals and derivatives, build accumulation functions, and solve net-change problems on the exam. For a quick refresher, see the Topic 6.4 study guide ( and practice problems (
When do I use FTC1 vs FTC2 and what's the difference between them?
Think of the Fundamental Theorem as two complementary tools: - FTC1 (derivative-of-an-integral): Use this when you need the derivative of an accumulation function. If f is continuous near a, then d/dx [∫_a^x f(t) dt] = f(x). If the top (or bottom) limit is a function g(x), use the chain rule: d/dx [∫_a^{g(x)} f(t) dt] = f(g(x))·g′(x). This is the one tied to "accumulation functions" (FUN-5.A.2) on the CED. - FTC2 (evaluation/antiderivative form): Use this when you need the value of a definite integral. If F is any antiderivative of f (F′ = f), ∫_a^b f(x) dx = F(b) − F(a). This is how you compute net change or area from an antiderivative (also called the Evaluation Theorem). When to pick which: - Differentiate an integral-expression → use FTC1 (with chain rule if limits depend on x). - Compute a definite integral (find area or net change) → use FTC2 by finding an antiderivative. Both require the integrand to be continuous on the interval. For AP review, see the Topic 6.4 study guide ( and practice problems at (
I'm confused about accumulation functions - how do they work exactly?
An accumulation function is a new function defined by a definite integral: g(x) = ∫_a^x f(t) dt. It literally "accumulates" the net signed area under f from the fixed start a up to x (FUN-5.A.1). If f is continuous, the Fundamental Theorem of Calculus says g′(x) = f(x) (FUN-5.A.2)—so the slope of the accumulation function at x equals the integrand value at x. That makes g increasing where f>0 and decreasing where f<0 (signed area). Changing the lower limit a just shifts the vertical starting value g(a)=0; the shape is driven by f. If the upper limit is a function, g(x)=∫_a^{h(x)} f(t) dt, use the chain rule: g′(x)=f(h(x))·h′(x) (Leibniz notation). Use graphical, numerical, or analytic info about f to sketch or compute g (FUN-5.A.3). For more examples and AP-style practice, see the Topic 6.4 study guide ( and Unit 6 resources ( For extra practice, try problems at (
How do I solve problems where the upper limit of integration is a function of x?
Use the FTC with the chain rule. If g(x) = ∫_a^{u(x)} f(t) dt and f is continuous, then by FUN-5.A.2 g'(x) = f(u(x)) · u'(x). So you evaluate the integrand at the upper limit (replace t with u(x)) and multiply by the derivative of that limit. Example: if g(x)=∫_0^{x^2+1} e^{-t^2} dt, then g'(x)=e^{-(x^2+1)^2}·2x. If both limits depend on x, use the same idea: h(x)=∫_{p(x)}^{q(x)} f(t) dt ⇒ h'(x)=f(q(x))·q'(x) − f(p(x))·p'(x). Make sure f is continuous on an interval containing the limits (CED requirement). On the AP exam you’ll often see accumulation functions and must use Leibniz notation/chain rule to find derivatives quickly (Topic 6.4). For extra practice and worked examples, check the Topic 6.4 study guide ( and more practice problems (
What does it mean when they say the definite integral defines a new function?
Saying “the definite integral defines a new function” means you use the integral with a variable upper limit to make a function of x. For a continuous f on an interval and a fixed a, define g(x) = ∫_a^x f(t) dt. For each input x, g(x) equals the signed area under f from a to x—so the integral isn’t just a number, it’s a rule that produces a number for every x. By the Fundamental Theorem of Calculus, g′(x) = f(x), so g is an antiderivative (and its rate of change at x equals the original integrand). Graphically, if f is positive on an interval, g increases there; if f is negative, g decreases. This is exactly what FUN-5.A in the CED covers. For a clear walkthrough and examples, see the Topic 6.4 study guide ( Want more practice? Try the AP problem bank (
Can someone explain step by step how to find g'(x) when g(x) equals an integral?
Quick recipe—step-by-step—for finding g′(x) when g(x) is defined by a definite integral. 1. Identify the integrand and the variable of integration. If g(x) = ∫[a]^{x} f(t) dt and f is continuous, FTC part 2 (FUN-5.A.2) gives g′(x) = f(x). The variable inside the integral (t) is a dummy variable—replace it if needed. 2. If the upper limit is a function u(x): g(x) = ∫[a]^{u(x)] f(t) dt, use the chain rule: g′(x) = f(u(x)) · u′(x). 3. If the lower limit is a function v(x): g(x) = ∫[v(x)]^{b} f(t) dt, rewrite as −∫[b]^{v(x)] f(t) dt or use g′(x) = −f(v(x))·v′(x). 4. If both limits depend on x: g(x)=∫[v(x)]^{u(x)] f(t) dt, then g′(x)=f(u(x))u′(x) − f(v(x))v′(x). 5. If the integrand itself depends on x (e.g., ∫[a]^{x} f(t,x) dt), differentiate under the integral sign: g′(x)=f(x,x)+∫[a]^{x} ∂f/∂x (t,x) dt (use only if you see x inside the integrand). Example: g(x)=∫_{2}^{x^2} sin(t) dt → g′(x)=sin(x^2)·2x. These are exactly the FTC/accumulation rules tested on the AP (Topic 6.4 FUN-5.A). For more examples and practice, check the Topic 6.4 study guide ( and extra problems (
Why does the derivative of the integral from a to x of f(t) dt just equal f(x)?
Think of G(x) = ∫_a^x f(t) dt as an "area so far" function. If you increase x by a tiny h, the extra area is the integral from x to x+h: G(x + h) − G(x) = ∫_x^{x+h} f(t) dt. If f is continuous at x, for very small h the values f(t) on [x, x+h] are all very close to f(x), so the new area ≈ f(x)·h. Dividing by h and taking the limit gives G′(x) = lim_{h→0} [G(x+h)−G(x)]/h = lim_{h→0} (1/h)∫_x^{x+h} f(t) dt = f(x). That's the Fundamental Theorem of Calculus (FUN-5.A.2 in the CED). In Leibniz notation this is d/dx ∫_a^x f(t) dt = f(x); if the upper limit is a function u(x), use the chain rule: d/dx ∫_a^{u(x)} f(t) dt = f(u(x))·u′(x). For AP review, see the Topic 6.4 study guide ( and try practice problems (
How do I know if a function is an accumulation function or just a regular function?
Short answer: an accumulation function is one defined by a definite integral with a variable limit—usually written g(x) = ∫_a^x f(t) dt (or with x in an upper/lower limit). If you see integral notation with a dummy variable (like t) and a limit that depends on x, it’s an accumulation function. If you just see an algebraic/closed form (like g(x)=x^2+3) with no integral, it’s a regular function (though it might be an antiderivative of something). Quick checks you can use - Notation: look for ∫ and a limit that is x (or some function of x). That’s the giveaway. - Dummy variable: integrals use a dummy variable (t, s); x should not appear inside the integrand except as the limit. - FTC test: if g(x)=∫_a^x f(t) dt then g′(x)=f(x) (FUN-5.A.2). If the derivative equals the integrand, that confirms accumulation behavior. - Chain-rule case: if g(x)=∫_a^{h(x)} f(t) dt, then g′(x)=f(h(x))·h′(x)—still an accumulation function with a variable limit. For more examples and AP-aligned practice, see the Topic 6.4 study guide ( and try problems at (
What's the difference between integrating f(x) and finding an accumulation function of f?
Integrating f(x) usually means evaluating a definite integral over a fixed interval (like ∫_a^b f(x) dx) to get a number (signed area, net change). An accumulation function is a new function defined by a definite integral with a variable upper limit: g(x) = ∫_a^x f(t) dt. Key differences: - Output: ∫_a^b f(x) dx → a number. g(x)=∫_a^x f(t) dt → a function of x (an “area function” or accumulation function). - Variable of integration: use a dummy variable (t) for g so x is free; that matters for differentiation. - FTC connection: If f is continuous, g′(x) = f(x) (Fundamental Theorem of Calculus, FUN-5.A.2). So accumulation functions link integrals and derivatives. - Interpretation: ∫_a^b f = total/net change from a to b. g(x) = net accumulation from a up to x; its slope equals the instantaneous rate f(x). On the AP exam expect tasks asking you to set up accumulation functions, use FTC to differentiate them (including chain rule if limits are functions of x), and interpret behavior from graphs (see the Topic 6.4 study guide on Fiveable for examples) ( For more practice, use Fiveable’s problem set (
I don't understand why we use t as the variable inside the integral instead of x.
Think of the variable inside the integral as a “dummy” variable that gets summed away—it’s just a placeholder for the values you add up. If you write g(x) = ∫a^x f(t) dt, t is the variable of integration; the result g depends on x only through the upper limit. Using t avoids confusing the integrand with the outside variable x. If you used x inside the integral like ∫a^x f(x) dx, it looks like the same x plays two roles (both the dummy and the limit), which is ambiguous and can break rules when you do substitutions or apply the chain rule. This is exactly what FUN-5.A.2 in the CED states: d/dx ∫a^x f(t) dt = f(x). The t makes clear that f is being integrated over a dummy variable and then evaluated at the limit. For more examples and practice on accumulation functions and the FTC, check the Topic 6.4 study guide ( and Unit 6 overview ( For lots of practice problems, see (
How do I graph an accumulation function when I'm only given the original function f(x)?
Think of g(x) = ∫_a^x f(t) dt as an “area so far” function. Two quick rules from the FTC you’ll use when sketching g from the graph of f: - g′(x) = f(x). So the slope of g at x equals the height of f at x. Where f is positive, g is increasing; where f is negative, g is decreasing. - g′′(x) = f′(x). So g is concave up where f is increasing and concave down where f is decreasing. How to sketch step-by-step: 1. Pick the starting point g(a)=0. 2. For selected x-values, compute g(x) as the net signed area under f from a to x (add areas above the axis, subtract below). Mark those points. Use simple geometric areas (triangles/rectangles) or estimate from the graph. 3. Between those plotted points, draw g so its slope at each x matches f(x) (steeper where f is large, flat where f crosses 0). Make concavity follow whether f is rising or falling. 4. Note corners: if f has cusps or discontinuities, g’s slope will reflect that (FTC requires continuity of f for simple form). This is exactly what AP wants in Topic 6.4: represent accumulation functions and use g′(x)=f(x) to interpret behavior. For more examples and guided practice, see the Topic 6.4 study guide ( and lots of practice problems (
When the problem gives me information about f(x), how do I find properties of g(x) = integral of f(t) dt?
When you’re given info about f(x) and asked about g(x) = ∫_a^x f(t) dt, use the Fundamental Theorem of Calculus: g'(x) = f(x) (if f is continuous). From that you can get lots of properties quickly: - Increasing/decreasing: g is increasing where f(x) > 0 and decreasing where f(x) < 0. - Critical points/extrema: where f(x) = 0 (and changes sign) are candidates for local max/min of g. - Concavity: g''(x) = f'(x). So g is concave up where f' > 0, concave down where f' < 0. - Values: g(x) = g(a) + ∫_a^x f(t) dt; with the usual choice g(a)=0 you can compute net (signed) area. - If the upper limit is a function, use chain rule: d/dx ∫_a^{u(x)} f(t) dt = f(u(x))·u'(x). - Use the Mean Value Theorem for Integrals to relate average values: (1/(b−a))∫_a^b f = f(c) for some c. On the AP exam they expect you to connect graphical/numerical/analytic info about f to these properties of g (CED FUN-5.A.2–3). For a focused review, see the Topic 6.4 study guide ( and try practice problems (
What happens to the derivative when the lower limit isn't a constant in the integral?
If the lower limit is a function of x, you use the Fundamental Theorem of Calculus together with the chain rule. For a continuous f: - If g(x) = ∫_{a}^{x} f(t) dt then g′(x) = f(x) (FTC, FUN-5.A.2). - If the lower limit is a function u(x), and the upper limit is a constant or another function, use d/dx ∫_{u(x)}^{v(x)} f(t) dt = f(v(x))·v′(x) − f(u(x))·u′(x). So if g(x) = ∫_{x}^{2} f(t) dt, g′(x) = −f(x) (because v(x)=2 gives v′=0, u(x)=x gives u′=1). If both limits vary, you get the difference of the two contributions. Note: f must be continuous on the interval for these rules to apply (CED continuity requirement). For more practice and examples, see the Topic 6.4 study guide ( and the Unit 6 overview (
How do I use the Fundamental Theorem of Calculus on my calculator for AP exam problems?
You won’t use your calculator to “prove” the Fundamental Theorem of Calculus—you use it to compute definite integrals and to get numerical derivatives of accumulation functions when the problem allows calculator work. How to use it on the AP exam - Use the integrate/numeric ∫ feature (MATH → fnInt or ∫ on TI/Nspire) to evaluate ∫_a^b f(x) dx quickly—useful on Part B MC and the calculator-required FR (see exam format above). - If you’re asked for g(x) = ∫_a^x f(t) dt and need g(x) numerically, define g(x) = fnInt(f(t), t, a, x) and evaluate. For g′(x) you can either apply FTC analytically (g′(x)=f(x)) or get a numeric derivative with nDeriv(g(x), x, x0) if allowed. - For integrals with variable limits like ∫_{a}^{h(x)} f(t) dt, use chain rule: d/dx = f(h(x))·h′(x). Use your calculator to evaluate f(h(x)) and h′(x) numerically when needed. - Watch for continuity: FTC requires the integrand continuous on the interval; if it isn’t, interpret signed area carefully. - Always show the analytic FTC reasoning when the free-response asks for justification even if you used the calculator for numbers. For quick review and examples, check the Topic 6.4 study guide ( unit overview ( and practice problems ( |
7326 | https://journals.lww.com/jhypertension/fulltext/2023/11000/sympathetic_modulation_as_a_goal_of.2.aspx | Journal of Hypertension
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November 2023 - Volume 41 - Issue 11
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Outline
INTRODUCTION
RATIONALE FOR SYMPATHOMODULATION IN HYPERTENSION TREATMENT
METHODOLOGICAL DRAWBACKS OF THE PUBLISHED STUDIES
SYMPATHOMODULATORY EFFECTS OF NON-PHARMACOLOGICAL INTERVENTIONS
SYMPATHOMODULATORY EFFECTS OF PHARMACOLOGICAL INTERVENTIONS
Monotherapy
Old and new drugs
Combination drug treatment
SYMPATHOMODULATORY EFFECTS OF DEVICES-BASED INTERVENTIONS
OPEN QUESTIONS AND CONCLUSION
ACKNOWLEDGEMENTS
Conflicts of interest
REFERENCES
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REVIEWS
Sympathetic modulation as a goal of antihypertensive treatment: from drugs to devices
Grassi, Guido
Author Information
Clinica Medica, Department of Medicine and Surgery, University of Milano-Bicocca, Milan, Italy
Correspondence to Prof. Guido Grassi, Clinica Medica, University Milano-Bicocca, Via Pergolesi 33, 20052 Monza, Italy. Tel: +39 039 2333357; e-mail: guido.grassi@unimib.it
Abbreviation: ESH, European Society of Hypertension
Received 19 June, 2023
Revised 26 July, 2023
Accepted 28 July, 2023
This is an open access article distributed under the Creative Commons Attribution License 4.0 (CCBY), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Journal of Hypertension 41(11):p 1688-1695, November 2023. | DOI: 10.1097/HJH.0000000000003538
Open
Abstract
The present study aims to examine the effects of nonpharmacological, pharmacological and devices-based treatment on hypertension-related sympathetic overactivity. This will be done by analyzing the results of different published studies, in which sympathetic activity has been assessed via indirect or direct techniques. After examining the rationale for sympathomodulatory interventions in antihypertensive treatment, the study will discuss the methodological intrinsic limitations of the studies aimed at assessing different therapeutic interventions. The core of the study will be then focused on the effects of nonpharmacological (dietary restriction of sodium intake, physical exercise training, weight reduction), pharmacological (monotherapy, combination drug treatment, new drugs such as sodium glucose co-transport protein-2 inhibitors and angiotensin receptor neprilysin inhibitors), as well as devices-based interventions (renal sympathetic nerves ablation and carotid baroreceptor activation therapy) on the hypertension-related sympathetic overdrive. Finally, the areas worthy of future research as well as the debated issues in the field will be highlighted.
INTRODUCTION
Original studies, frequently included in meta-analyses, based on evaluation of direct as well as on indirect markers of sympathetic cardiovascular drive, have conclusively shown that neuroadrenergic influences to the heart and the peripheral circulation are markedly activated in hypertensive patients, the resulting sympathetic overdrive representing a hallmark of the hypertensive disease [1–13]. These studies and meta-analyses have also documented that the neuroadrenergic overactivity is already manifest in the prehypertensive state and becomes progressively more pronounced, as the severity of the blood pressure elevation becomes more marked [2,4–6,9–13]. This process will follow the chain of events, which participate at the cardiovascular continuum, favoring the progression of the disease, the occurrence of cardiovascular complications and the development of the hypertension-related target organ damage (Fig. 1). Both pharmacological and nonpharmacological blood pressure lowering interventions have been shown to exert, along with their antihypertensive properties, additional favorable effects known as “ancillary properties.” They may occur throughout a variety of mechanisms, including the drug-related sympathomodulatory properties aimed at reducing the degree of sympathoexcitation almost invariably characterizing the essential hypertensive state.
FIGURE 1:
Chain of events which participate at the cardiovascular continuum, favoring the progression of the disease, the occurrence of cardiovascular complications and the development of the hypertension-related end organ damage. HT, hypertension; SNS, sympathetic nervous system; TOD, target organ damage. ↑: increase; ↑↑: marked increase; ↑↑↑: very marked increase.
The present study, which is based on the European Society of Hypertension (ESH) Presidential Lecture given at the 32 nd ESH Meeting in Milan, will be focused on the effects of different nonpharmacological, pharmacological and devices-based interventions on the hypertension-related sympathetic overactivity. Following an introductory paragraph focused on the rationale for sympathomodulatory interventions in antihypertensive treatment, the most critical methodological aspects of the published studies aimed at assessing the antisympathetic therapeutic approaches adopted in the disease will be reviewed. The core of the study will be focused on the effects of nonpharmacological, pharmacological (monotherapy, combination drug treatment, new drugs), as well as devices-based interventions (renal sympathetic nerves ablation and carotid baroreceptor activation therapy) on the hypertension-related sympathetic overdrive. The final part of the manuscript will offer to the readers an outlook on the open questions related to this area of research that will need to be addressed in the forthcoming years. These include, among others, the new potential methodologies to assess sympathetic function during antihypertensive treatment and the link between adrenergic overdrive, adherence to antihypertensive treatment and controlled or uncontrolled blood pressure status.
RATIONALE FOR SYMPATHOMODULATION IN HYPERTENSION TREATMENT
Recent years have seen a relevant increase in our know-how on the role of the sympathetic nervous system in hypertension pathophysiology and, more in general, in the determination of cardiovascular risk. The consistent evidence collected can be summarized as follows. First, different methodological approaches to investigate human sympathetic neural function have shown that a state of sympathetic overactivity characterizes the hypertensive condition, being detectable in the very initial hypertensive stages and becoming more and more pronounced, as the severity of the hypertensive state results more and more evident [1–13]. The relationship between blood pressure values and sympathetic overactivity is strengthened by the significant independent direct relationship detectable between these two variables in different studies and in a recent meta-analysis including data collected in more than 1000 hypertensive patients [5,6,9,12,13]. Second, the neuroadrenergic activation not only concurs at the blood pressure elevation, but it is also involved in favoring, along with the contribution of hemodynamic variables and other humoral factors, the development and progression of target organ damage, particularly at the level of the heart, the kidneys, the macro-circulation and the micro-circulation as well [12,14–17]. Of special interest is the participation of sympathetic factors to the occurrence of the endothelial dysfunction, which represents an early vascular/systemic alteration typical of the hypertensive state . Third, metabolic alterations frequently detected in hypertensive patients, such as glucose intolerance, insulin resistance, hypertriglyceridemia, hypercholesterolemia and hyperuricemia, may be facilitated in their development by the adrenergic overdrive, particularly when prediabetes, overweight, metabolic syndrome, obesity and diabetes are detected as comorbidities of the hypertensive state [10,12]. Finally, an increased neuroadrenergic drive to the heart and the peripheral circulation has been shown to be associated with an augmented risk of future nonfatal and fatal cardiovascular events. The adverse prognostic impact of an elevated sympathetic cardiovascular drive, with close relationships with fatal events, has been documented in a variety of disease, such as chronic heart failure, acute thrombolembolic stroke, myocardial infarction, hepatic cirrhosis and end-stage renal failure, but not in hypertension [19–24]. In the hypertensive state, however, evidence exists that an indirect sympathetic marker such as an elevated heart rate value assessed at rest may predict the future development of fatal and nonfatal cardiovascular events .
Altogether, the above-mentioned adverse consequences of the hypertension-related sympathetic activation represent a strong rationale for adopting in the treatment of this disease therapeutic approaches capable to modulate the adrenergic overdrive.
METHODOLOGICAL DRAWBACKS OF THE PUBLISHED STUDIES
A significant number of published studies investigating the sympathetic effects of therapeutic interventions in hypertension suffer from intrinsic methodological limitations, which are summarized in Table 1. Four leading drawbacks will be worthy to be discussed here. First, in some studies, the effects of a given drug on sympathetic tone have been assessed after acute administration of the compound. In this case, a paradoxical increase in sympathetic tone has been almost invariably reported, which vanished when the evaluation with the same drug was done on a chronic basis, namely after days or weeks of regular daily treatment. This was the case even when the drug tested was a central sympatholytic agent . The explanation for this paradoxical response is that the acute blood pressure reduction induced by the administration of a single dose of the antihypertensive agent tested triggers, almost invariably and independently on the pharmacological properties of the drug, a reflex sympathetic activation. This response appears to be largely unavoidable in its occurrence, because of the inability of the arterial baroreflex, which physiologically modulates sympathetic cardiovascular drive, to undergo an effective downward resetting before chronic treatment is established . It should be emphasized that this limitation does apply not only to the studies investigating antihypertensive drugs, but also to the investigations aimed at exploring the effects of nonpharmacological or devices-based interventions. This is the case, for example, for the evaluation of the acute effects of physical exercise training on the adrenergic nervous system, which appear to be remarkably different from those reported during a physical training program prolonged for weeks or months [29,30]. This is also the case for renal denervation, whose sympathetic effects are different according to the time of performing the investigation after the procedure (days vs. weeks or vs. months) . The second limitation refers to the fact that not rarely the studies have been carried out in normotensive individuals, which are characterized by a sympathetic function, which is already normal and thus impossible to be antagonized for obtaining an already present normalization. A further limitation is represented by the evidence that in the vast majority of the published studies the data on the blood pressure lowering effects of the therapeutic interventions were based on office rather than on 24-h blood pressure measurement, with all the well known limitations of this approach.
TABLE 1 - Methodological limitations of published studies
• Intrinsic limitations of various indirect techniques to assess sympathetic function.
• Difficulties in assessing directly human sympathetic cardiovascular drive.
• Study population frequently including normotensive individuals.
• Lack of placebo-control group.
• Different pharmacokinetic profile of various drugs tested.
• Evaluation after acute drug administration.
• Blood pressure lowering effects frequently assessed on clinic rather than on 24-h ambulatory blood pressure measurements.
• Study population small.
Finally, a critical issue refers to the method used for assessing the sympathomoderating properties of a given therapeutic intervention. All the techniques available, which include the assay of the circulating venous plasma levels of the adrenergic neurotransmitter norepinephrine, the norepinephrine spillover technique, the power spectral analysis of the heart rate signal in specific bands as well as the microneurographic recording of efferent postganglionic sympathetic nerve traffic, display advantages and limitations already highlighted in previous studies [10,12,32]. Combination of two techniques to assess adrenergic function may allow to overcome some of these limitations, although the complexity of the investigation raises difficulties in repeating it various times before and during the treatment phase.
SYMPATHOMODULATORY EFFECTS OF NON-PHARMACOLOGICAL INTERVENTIONS
The three main nonpharmacological approaches adopted in current clinical practice in the treatment of hypertension for which extensive information have been collected for their effects on sympathetic function include dietary restriction of salt intake, regular physical exercise and body weight reduction . An additional nonpharmacological intervention, which is employed in specific clinical conditions characterized by the presence of the sleep apnea syndrome, is continuous positive airway pressure . All these procedures display well defined sympathoinhibitory effects, the exception being represented by dietary restriction of salt intake, even when of moderate degree (80 mmol NaCl/day). Indeed, different methodological approaches employed to assess sympathetic function have almost univocally shown an increase in sympathetic cardiovascular drive during long-term dietary restriction of sodium intake. This is the case for the circulating venous plasma levels of the adrenergic neurotransmitter norepinephrine, renal norepinephrine spillover and muscle sympathetic nerve traffic, the increase in these sympathetic markers during sodium restriction averaging to 25–30% of the baseline prediet values [35–38]. Several mechanisms are likely to participate in the sodium restriction-related sympathoexcitation. These include an impairment of the inhibitory effects exerted by the arterial baroreceptors on the adrenergic function and the development of an insulin resistance state, the resulting hyperinsulinemia triggering a marked increase in sympathetic cardiovascular drive [39,40].
As far as physical exercise training is concerned, there is evidence (although not always homogeneous) that the procedure may not only exert blood pressure lowering effects but also sympathomodulatory properties. These have been documented via the heart rate variability approach and the regional norepinephrine spillover technique . They have been more recently confirmed by microneurographic recording of muscle sympathetic nerve traffic, with an average reduction in the resting values approaching 40% [30,43]. The sympathoinhibitory effects, when evaluated via the microneurographic technique, are quite similar for magnitude to the ones detected in heart failure, according to the data presented in a recent meta-analysis .
Of a similar consistent nature are the results of the studies investigating the effects of body weight reduction obtained via dietary interventions alone or associated with bariatric surgery. Both the procedures induce a reduction in body weight, whose magnitude is extremely variable between studies and it was less, as expected, for the dietary interventions than for surgery . The sympathoinhibitory properties of these interventions have been documented by different methoodological approaches to evaluate human adrenergic function, including the norepinephrine radiolabeled technique and the microneurographic recording of muscle sympathetic nerve traffic [45–48]. An analysis of the results of the microneurographic studies based on dietary interventions shows, together with a body weight reduction, a decrease in sympathetic neural outflow to muscle circulation amounting on average to 25%. This was associated with an office blood pressure lowering effect, which achieved statistical significance, however, for the SBP component only.
Finally, also the sympathoinhibitory effects of continuous positive airway pressure have been documented by different methodologies. The results of eight microneurographic studies with longitudinal design have been very recently included in a meta-analysis, showing a reduction amounting on average to about 25% of the baseline preintervention values . The sympathoinhibition was associated with an office blood pressure reduction of quite consistent magnitude, that is, about 10.0 mmHg for systolic and 7.0 mmHg for diastolic .
SYMPATHOMODULATORY EFFECTS OF PHARMACOLOGICAL INTERVENTIONS
Monotherapy
All the five classes of the antihypertensive drugs recommended by current guidelines for antihypertensive treatment have been extensively evaluated as monotherapies for their effects on sympathetic activity. The results can be outlined as follows. Beta-adrenergic blockers exert their blood pressure lowering effects by inducing cardiac and peripheral sympathetic modulation, which usually translates in clinical practice in a reduction of resting heart rate together with a peripheral vasodilation (particularly evident with beta-adrenergic blocking agents without sympathomimetic activity) [33,50]. The majority of the studies performed with these compounds evaluated the sympathetic effects only indirectly, such as measuring heart rate or circulating plasma levels of the adrenergic neurotransmitter norepinephrine. The few studies done with direct recording of efferent postganglionic sympathetic nerve traffic not always have shown, as should be expected from the pharmacological and clinical profile of the drug class, a marked sympathoinhibition, at variance from what has been reported in chronic heart failure . It is likely that these discrepant results depend on the different levels of resting adrenergic overdrive characterizing these two conditions, which is more consistent in heart failure than in hypertension . In contrast, a more effective degree of sympathoinhibition has been reported with drugs acting on the renin-angiotensin system, such as ACE-inhibitors and angiotensin II receptors blockers . The data collected so far strongly support the notion that ACE-inhibitors and angiotensin II receptor antagonists reduce central sympathetic neural discharge in essential hypertensive individuals and may also cause inhibitory effects at the level of peripheral nerve terminals, modulating the spillover rate of the adrenergic neurotransmitter and improving its tissue clearance . The drugs may also improve both vagal and sympathetic baroreflex control of the cardiovascular system, partially restoring this homeostatic function in treated hypertensive individuals . Similar sympathoinhibitory effects have been documented for the renin inhibitor aliskiren .
Considering their sympathetic effects, calcium antagonists represent a heterogeneous group of compounds. Short-acting dihydropyridines exert clearcut sympathoexcitatory effects (likely dependent on a baroreflex impairment), which are reflected by the marked increase in heart rate, low-frequency component of heart rate variability, venous plasma norepinephrine and muscle sympathetic nerve traffic values detected following administration of these compounds . These effects are hampered in the case of long-acting dihydropyridines, which however may display remarkable differences between different drugs belonging to the same pharmacological class .
Diuretic agents, particularly chlortalidone at a daily dosage greater than 25 mg, have been shown to elicit an increase in sympathetic drive, the magnitude of the occurring sympathoexcitation being however much less evident with chlorothiazide and indapamide. Along with volume-dependent sympathetic effects, diuretics may favor the occurrence of hyperinsulinemia, a metabolic alteration which, as already mentioned, may favor an increase in adrenergic cardiovascular drive . In contrast, potent sympathoinhibitory effects have been described with the antialdosterone drug spironolactone, presumably dependent on baroreflex mechanisms .
Old and new drugs
Clonidine, moxonidine and rilmenidine represent old and relatively new central sympatholytic drugs capable to inhibit various markers of adrenergic function, such as plasma norepinephrine, norepinephrine spillover and sympathetic nerve traffic . As it will be mentioned in the following paragraph, central agents are now preferably used in the treatment of hypertension only as third or fourth agent in combination with other drugs . In this case, the data available suggest that they can be of help in normalizing sympathetic neural function of the treated hypertensive patients . Another class of drugs, statins, have been shown to exert significant sympathoinhibition , as also documented by a recent meta-analysis, which included five studies for a total of more than 80 patients . The average reduction in sympathetic nerve traffic amounted to about 15% of baseline values . The statins-related sympathoinhibition may depend on the improvement of baroreflex modulation of adrenergic drive described with these compounds and may be responsible for the slight blood pressure reduction reported in recent studies and meta-analyses .
Two new classes of compounds which have been recently introduced in the therapeutic approach to cardiovascular and metabolic disease have been shown to exert modulatory effects on sympathetic neural function. These include sodium glucose co-transport protein-2 inhibitors and angiotensin receptor neprilysin inhibitors, which have been successfully employed in major clinical trials in the treatment of diabetic patients and patients with congestive heart failure, respectively. As far as the first group of compounds, conclusive evidence exists that they trigger favorable effects on glucose metabolism and blood pressure, which are associated with no change or even a reduction in different sympathetic markers [60,61]. The neprilysin inhibitor sacubitril, in association with an angiotensin ii receptor antagonist (valsartan), has been shown in chronic heart failure patients with a reduced left ventricular ejection fraction to be associated with an incidence of cardiovascular complications and events significantly lower than the one detected in the patients displaying a superimposable severity of the disease and under standard drug treatment without sacubitril . These effects are accompanied by consistent sympathoinhibitory effects, as documented by the significant reduction (-18.0% on average) in muscle sympathetic nerve traffic values observed during prolonged administration of the drug .
Combination drug treatment
The vast majority of the studies aimed at providing information on the sympathomodulatory effects of antihypertensive drug combinations have been based on indirect methods to assess the sympathetic responses to a given therapeutic intervention, namely the assay of the venous plasma levels of norepinephrine and the power spectral analysis of the heart rate signal. As discussed in previous studies [12,32], both the approaches have major limitations. In the case of plasma norepinephrine assay, the fact that the vasodilating properties of antihypertensive agents may reduce plasma norepinephrine levels increasing the tissue clearance of the sympathetic neurotransmitter [64,65]. This may produce misleading information on the impact of the drugs on sympathetic drive, wrongly pointing toward a sympathomodulation. In the case of the power spectral approach, it should be emphasized that the technique provides information only on sympathetic/parasympathetic balance at the level of the heart, which frequently does not reflect the effects of a given therapeutic intervention on systemic sympathetic neural function [12,32].
An analysis of the results of the studies aimed at assessing the sympathetic responses to therapeutic interventions via the direct microneurographic recording of sympathetic nerve traffic has provided the following information. Results of the nine studies enrolling on the whole more than 170 hypertensive patients under combination treatment have shown a consistent significant blood pressure reduction (17.0 mmHg for systolic and 11.0 mmHg for diastolic) coupled with a reduction in sympathetic nerve traffic (10–15% of baseline values) close to achieve statistical significance (Fig. 2). The effects on the sympathetic function were more marked when the drugs combination was based on an ACE-inhibitor and an angiotensin II receptor blocker or a central sympatholytic agent.
FIGURE 2:
Effects of two antihypertensive drugs combinations on SBP (S), DBP (D) and muscle sympathetic nerve traffic (MSNA) in 176 patients enrolled in different microneurographic studies. Data are shown as means ± SD. Asterisks (∗∗P < 0.01) refer to the level of statistical significance between values recorded before (baseline) and during combination drugs treatment (combo).
As a general rule, with the exception of the study mentioned above, which made use of the combination treatment between an angiotensin II receptor blocker and a central sympatholytic agent , no published study reported during treatment a reduction in sympathetic nerve traffic at values similar to the ones detected in pure normotensive control individuals. The lack of a full sympathetic normalization during antihypertensive drug treatment has been hypothesized as to be one of the mechanisms responsible for the so-called ‘residual risk’ , that is, for the fact that blood pressure lowering drugs, despite reducing blood pressure to normal values, do not allow to bring back to ‘normal’ the cardiovascular risk of the treated hypertensive patient .
SYMPATHOMODULATORY EFFECTS OF DEVICES-BASED INTERVENTIONS
There are two main assumptions supporting the analysis of the effects of devices-based interventions on sympathetic cardiovascular drive. The first one is represented by the evidence that drug-resistant hypertension, that is, the hypertensive phenotype identified by earlier studies as preferred target of the devices-based interventions, such as bilateral renal nerve ablation and carotid baroreceptor activation therapy, is characterized by a remarkable sympathetic overdrive [11,13]. The second assumption, on the contrary, is that the sympathetic activation is responsible for the development of the resistant hypertensive state, its reduction by devices being associated with a clearcut decrease in blood pressure values . The two assumptions have received throughout the years experimental and clinical supports, although not always homogeneous in their nature, however. Two main successful approaches have been used, that is, the radiofrequency or, more recently, endovascular ultrasound bilateral ablation of renal nerves and carotid baroreceptor activation therapy . In a recent meta-analysis done by our group including 11 microneurographic studies published so far for a total of about 400 patients, bilateral renal nerves ablation elicited a blood pressure lowering effect associated with a reduction in muscle sympathetic nerve traffic at the sixth month follow-up (Fig. 3, upper) . Interestingly, the blood pressure and the sympathetic effects do not appear to follow the same time course; however, the occurrence of the blood pressure reduction preceding by weeks the sympathetic one . In addition, the blood pressure reduction and the sympathoinhibition did not show any significant quantitative relationships among each other . The hypothesis has been therefore advanced that the blood pressure lowering effects of the procedure may be driven by extrasympathetic mechanisms and not necessarily only by the sympathetic ones .
FIGURE 3:
Effects of bilateral renal nerves ablation (RDN, upper) and carotid baroreceptor activation therapy (BAT, lower) on SBP (S), DBP (D) and muscle sympathetic nerve traffic (MSNA) in patients enrolled in different microneurographic studies. Data are shown as means ± SD. Asterisks (∗∗P < 0.01, ∗ P < 0.05) refer to the level of statistical significance between values recorded before (BL) and during RDN or BAT treatment.
Figure 3, lower, illustrates the effects of carotid baroreceptor activation therapy obtained in four studies on blood pressure and sympathetic neural drive. Similar to what is described in the case of renal nerves ablation, also in this case, the procedure caused a significant blood pressure reduction combined with a sympathoinhibition. However, at variance from renal denervation, the magnitude of the blood pressure effect was quantitatively and often qualitatively related to the degree of the sympathetic inhibition . As a final consideration of the analysis of the data obtained with renal nerves ablation and carotid baroreceptor activation therapy, it should be worthy to mention that, similarly to what has been already described for combination drug treatment, both the procedures, although effective in reducing elevated sympathetic nerve traffic values, fail to obtain their full normalization.
OPEN QUESTIONS AND CONCLUSION
Results of recent studies have allowed to prompt the hypothesis that the clinical condition characterized by uncontrolled blood pressure values may be linked with, and possibly dependent on, an elevated degree of sympathetic activation [70–72]. This alteration would prevent antihypertensive drug treatment to achieve blood pressure control, thus determining the clinical phenotype defined as ‘refractory hypertension.’ It should be mentioned that as these studies were based on indirect (and frequently inadequate) methods for assessing sympathetic cardiovascular drive, future investigations assessing directly adrenergic drive via the microneurographic technique should be planned. A further question which is still unanswered is whether and to what extent adherence to treatment is related to sympathetic activation. In other words, it is unknown whether hypertensive patients who display a poor compliance to treatment are characterized by a sympathetic overactivity greater for magnitude than the one detected in patients, which display a full treatment adherence.
Finally, in a next future, the investigation of the sympathomodulatory effects of nonpharmacological, pharmacological or device-based antihypertensive treatment should be simplified by the use of techniques capable to assess neuroadrenergic function in a less complex fashion. In this context, promising approaches are represented by the assessment of skin blood flow responses to antihypertensive treatment during mental stress via a tissue flowmeter . Another approach, already tested in some studies investigating the effects of calcium channel blockers on adrenergic cardiovascular drive, is pupillometry, assessing pupillary diameters changes induced by light stimulus . These new approaches to assess human sympathetic function, once validated and adequately tested for their main features, will hopefully allow to collect information on the sympathomodulatory effects of different therapeutic interventions adopted in hypertension treatment in large-scale clinical studies. Indeed, it should be emphasized that the information collected so far on this issue are obtained in small studies. This represents an important limitation, making it difficult to extrapolate the data collected in small groups of hypertensive patients to the large hypertensive population.
ACKNOWLEDGEMENTS
Conflicts of interest
There are no conflicts of interest.
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Keywords:
bariatric surgery; baroreceptor activation therapy; combination drug treatment; continuous positive airway pressure; dietary salt restriction; dietary weight reduction; monotherapy; physical exercise training; refractory hypertension; renal nerves ablation; residual risk; sympathetic nervous system
Copyright © 2023 The Author(s). Published by Wolters Kluwer Health, Inc.
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Sympathetic modulation as a goal of antihypertensive treatment: from drugs to devices
Journal of Hypertension41(11):1688-1695, November 2023.
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7327 | http://en.people.cn/n3/2024/0718/c98649-20195168.html | NW China's Shaanxi builds 'green miracle' in Yellow River Basin - People's Daily Online
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Home>>People's Daily Online Exclusives
NW China's Shaanxi builds 'green miracle' in Yellow River Basin
(People's Daily Online) 16:34, July 18, 2024
Situated in the middle reaches of the Yellow River, northwest China's Shaanxi Province has promoted ecological conservation and high-quality development along the river, building a "green miracle" in the Yellow River basin in the province through relentless efforts.
Photo shows a green shelterbelt near watch tower Zhenbeitai in Yulin city, northwest China's Shaanxi Province. (Xinhua/Shao Rui)
Greening the Loess Plateau
A vast woodland combining trees, shrubs, and grass stretches out as far as the eye can see at the Mu Us Desert afforestation base in Shenmu, a county-level city administered by Yulin city, located on the Loess Plateau in Shaanxi.
It's hard to imagine that the woodland was once a barren desert area plagued by sandstorms years ago. Elders of the area shake their heads remembering the past, stating that sandstorms would sometimes last for days, leaving a layer of sand in their yards that had to be scooped out with dustpans. Today, the main color palette has transformed from “yellow” to “green”.
"Now, we systematically monitor wind speed, moisture, microorganisms, etc. in the woodland," said Zhang Yinglong, president of the Shenmu ecological protection and construction association and a veteran in desertification control. He added that combating desertification relies on science and technology.
Over the past 21 years, Zhang and his team have developed a series of afforestation techniques and models. Their process involves three steps. First, planting psammophytes, which are especially adept at growing in sand, to reduce wind erosion; second, planting shrubs to stop migrating sand; and finally planting trees to gradually form a forest. The team, working on an area of 428,000 mu (28,533 hectares), has seen vegetation coverage increase from 3 percent to 65 percent.
Building barriers to protect soil and water
Yulin city has brought 8.6 million mu of shifting sand under control and expanded its forest area from 600,000 mu in the early days of New China to 23.6 million mu today, creating a "green miracle."
A similar story has unfolded in Yan'an city, also located on the Loess Plateau in Shaanxi, thanks to over two decades of afforestation efforts. Data shows that the city's vegetation coverage has increased from 46 percent in 2000 to 81.3 percent in 2023.
Over the past few years, Shaanxi has been able to move the border between green and desert northward 400 kilometers. Since 2020, the province has planted nearly 18.36 million mu of trees and tackled over 3.89 million mu of desertified land in the Yellow River basin.
An excavator works on enhancing a warp-land dam at the Xindiangou water and soil conservation demonstration park in Suide county, Yulin city, northwest China's Shaanxi Province. (Xinhua/Li He)
The rugged terrain of the Loess Plateau at the Xindiangou water and soil conservation demonstration park in Suide county, Yulin is now covered in lush greenery.
Xindiangou, which features loess hills and ravines, was once plagued by water and soil erosion, with huge amounts of mud and sand flowing into the Yellow River.
To combat soil erosion, a "three lines of defense" management model was formulated.
The first line of defense involves constructing terraced fields on the ridges and upper regions of loess hills where the slopes are relatively gentle. Fruit trees and a rotation of grass and crops are planted on the terraced fields to modify the terrain and prevent water and soil erosion.
The second line focuses on the cultivation of shrubs along with timber and grass on steep ravine slopes to stabilize hillsides and prevent erosion.
The third line emphasizes the construction of warp-land dams, or silt dams, at the bottom of ravines for further flood control, sediment trapping and farming.
Using an analogy, the “three lines of defense” equates to putting hats on the ridges by building terraces, dressing the slopes in coats of tree and grass, and boots on the valley bottoms with slit-damns, explained Gao Jianjian, head of the Ecological Engineering Department at the Water and Soil Conservation Bureau of the Yellow River in Suide. Gao went on to say that through decades of effort, 80 percent of the area facing water and soil erosion has been treated, with a forest and grass coverage of more than 75 percent.
Photo shows the lush vegetation of Yan'an city, northwest China's Shaanxi Province. (Xinhua/Zhang Bowen)
Xindiangou is a good example of the comprehensive treatment of water and soil erosion taking place in the rest of northern Shaanxi. Data reveals that in 2023, areas in Shaanxi facing water and soil erosion decreased by 1,261 square kilometers compared to the previous year, and its water and soil conservation rate increased to 70.15 percent.
The water quality of the Shaanxi section of the Yellow River’s mainstream has reached the Level II standard for two consecutive years, according to an official with the Shaanxi Provincial Development and Reform Commission. An annual average of 4,000 square kilometers of land in the Yellow River basin in the province has been saved from water and soil erosion. The Loess Plateau has become the region with the largest increase in green coverage nationwide.
This "green miracle" has yielded both ecological and economic benefits. Yan'an city, for example, has 3.32 million mu of apple orchards with an annual apple output of 4.32 million tonnes, accounting for one-third of Shaanxi's apple production and one-ninth of China's total output. Out of the city's population of over 2 million, more than 1 million people are engaged in the apple industry. The sector benefits 800,000 farmers, with apple sales contributing to 61 percent of their operational income.
Dali county in Weinan city, located in the Yellow River basin in Shaanxi, has seen similar success with fruit cultivation. In 2023, the county produced 500,000 tonnes of winter jujubes, and over 300,000 people worked in the winter jujube industry, with farmers earning an average annual income of more than 100,000 yuan.
It's not just apples and jujubes. Across the Yellow River Basin in Shaanxi, vegetables, kiwi fruit, and various grains from northern Shaanxi are thriving. From "ecological beauty" to "ecological wealth," an increasing number of people are turning green into gold.
"As clear waters and green mountains are invaluable assets," an official in the Shaanxi Development and Reform Commission affirms, "we will continue our efforts to comprehensively promote the ecological protection and high-quality development of the Yellow River Basin."
(Web editor: Hongyu, Liang Jun)
Photos
Farmers in SE China's Fujian rejoice at bumper harvest of late-maturing lychees
Yacht tourism thrives in Sanya, S China's Hainan
Scenery of grassland in north China's Inner Mongolia
In pics: Slow train adds new vitality to rural revitalization in NE China's Jilin
Related Stories
The greening of China in 2023
County once tormented by desertification builds "green miracle" in NE China's Liaoning
China makes notable achievements in afforestation in 2023
China's Jiangsu to complete afforestation of 40,000 hectares by 2025
Extending "green Great Wall" in north China
Turning deserts into forests: How China is helping the world go green
County in SW China's Yunnan afforests stony mountains, develops green industries
A villager's four decades of commitment to afforestation
China's Inner Mongolia to pool 100 mln yuan for major sandland afforestation
Inner Mongolia ranks first in China in afforestation
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7328 | https://www.ixl.com/math/algebra-1/identify-arithmetic-and-geometric-sequences | IXL - Identify arithmetic and geometric sequences (Algebra 1 practice)
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A restaurant used 4 onions on Thursday, 24 onions on Friday, 144 onions on Saturday, and 864 onions on Sunday. What kind of sequence is this?
arithmetic
geometric
both
neither
key idea
In an arithmetic sequence, there is a constant difference between consecutive terms. This means that you can always get from one term to the next by adding or subtracting the same number.
In a geometric sequence, there is a constant multiplier between consecutive terms. This means that you can always get from one term to the next by multiplying or dividing by the same number.
The only way a sequence can be both arithmetic and geometric is if it repeats the same number over and over again.
solution
First check if the sequence is arithmetic. There is not a constant difference between consecutive terms. So, the sequence is not arithmetic.
4,24,144,864,...
+ 20+ 120+ 720
Next check if the sequence is geometric. There is a constant multiplier of 6 between consecutive terms. So, the sequence is geometric.
4,24,144,864,...
×6×6×6
So, the sequence is only geometric.
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7330 | https://openstax.org/books/organic-chemistry/pages/17-7-oxidation-of-alcohols | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Organic Chemistry
17.7 Oxidation of Alcohols
Organic Chemistry17.7 Oxidation of Alcohols
Search for key terms or text.
17.7 • Oxidation of Alcohols
Perhaps the most valuable reaction of alcohols is their oxidation to give carbonyl compounds—the opposite of the reduction of carbonyl compounds to give alcohols. Primary alcohols are oxidized either to aldehydes or carboxylic acids, and secondary alcohols are oxidized to ketones, but tertiary alcohols don’t normally react with most oxidizing agents.
Primary and secondary alcohols can be oxidized by any of a number of reagents, including CrO3 in aqueous acetic acid and KMnO4 in aqueous NaOH, but chromium-based reagents are rarely used today because of their toxicity and fire danger. Today, primary and secondary alcohols are oxidized to aldehydes and ketones, respectively, using the iodine-containing Dess–Martin periodinane in dichloromethane solution.
Primary alcohols are oxidized to carboxylic acids by heating with KMnO4 in a basic aqueous solution. An aldehyde is involved as an intermediate in the KMnO4 reaction but can’t usually be isolated because it is further oxidized too rapidly.
All these oxidations occur by a mechanism that is closely related to the E2 reaction (Section 11.8). In the Dess–Martin oxidation, for instance, the first step involves a substitution reaction between the alcohol and the I(V) reagent to form a new periodinane intermediate, followed by expulsion of reduced I(III) as the leaving group. Similarly, when a Cr(VI) reagent, such as CrO3, is the oxidant, reaction with the alcohol gives a chromate intermediate followed by expulsion of a reduced Cr(VI) species.
Biological alcohol oxidations are the opposite of biological carbonyl reductions and are facilitated by the coenzymes NAD+ and NADP+. A base removes the –OH proton, and the alkoxide ion transfers a hydride ion to the coenzyme. An example is the oxidation of sn-glycerol 3-phosphate to dihydroxyacetone phosphate, a step in the biological metabolism of fats (Figure 17.9). Note that addition occurs exclusively on the Re face of the NAD+ ring, adding a hydrogen with pro-R stereochemistry.
Figure 17.9 The biological oxidation of an alcohol (sn-glycerol 3-phosphate) to give a ketone (dihydroxyacetone phosphate). This mechanism is the exact opposite of the ketone reduction shown previously in Figure 17.5.
Problem 17-14
What alcohols would give the following products on oxidation?
(a)
(b)
(c)
Problem 17-15
What products would you expect from oxidation of the following compounds with chromium trioxide in hydronium? In the Dess–Martin periodinane?
(a)
1-Hexanol
(b)
2-Hexanol
(c)
Hexanal
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© Jul 9, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. |
7331 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7671473/ | Lentigo Maligna: Clinical Presentation and Appropriate Management - PMC
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Clin Cosmet Investig Dermatol
. 2020 Nov 13;13:837–855. doi: 10.2147/CCID.S224738
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Lentigo Maligna: Clinical Presentation and Appropriate Management
Helena Iznardo
Helena Iznardo
1 Dermatology Service, Hospital de la Santa Creu i Sant Pau, Universitat Autònoma de Barcelona, Barcelona, Spain
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1,, Cristina Garcia-Melendo
Cristina Garcia-Melendo
1 Dermatology Service, Hospital de la Santa Creu i Sant Pau, Universitat Autònoma de Barcelona, Barcelona, Spain
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1,, Oriol Yélamos
Oriol Yélamos
1 Dermatology Service, Hospital de la Santa Creu i Sant Pau, Universitat Autònoma de Barcelona, Barcelona, Spain
2 Dermatology Service, Centro Médico Teknon – Quirónsalud, Barcelona, Spain
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1 Dermatology Service, Hospital de la Santa Creu i Sant Pau, Universitat Autònoma de Barcelona, Barcelona, Spain
2 Dermatology Service, Centro Médico Teknon – Quirónsalud, Barcelona, Spain
✉
Correspondence: Oriol Yélamos Dermatology Service, Hospital de la Santa Creu i Sant Pau, Universitat Autònoma de Barcelona, C/Mas Casanovas 90, Block A, 5th Floor, Module 3, Barcelona, 08041, Spain, Phone: Tel +34 935537007, Fax +34 935537008 Email oyelamos@gmail.com
These authors contributed equally to this work
Received 2020 Jun 25; Accepted 2020 Oct 15; Collection date 2020.
© 2020 Iznardo et al.
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PMCID: PMC7671473 PMID: 33223843
Abstract
Lentigo maligna (LM) is a type of melanoma in situ that has distinctive characteristics regarding epidemiology, risk factors and clinical features. In addition, LM has a potential to progress to an invasive tumor with potentially aggressive behavior: lentigo maligna melanoma (LMM). Overall, LM has a very good prognosis, whereas LMM has the same prognosis as other invasive melanomas with similar Breslow thickness. LM/LMM represents a challenging entity not only regarding the diagnosis but also regarding the management. Diagnostic criteria are not well established, and there is an overlap of clinical, dermoscopic and pathological features with other benign pigmented skin lesions such as lentigines, pigmented actinic keratoses or macular seborrheic keratoses. LM/LMM’s common appearance within photodamaged skin makes lesion border identification difficult. Wide excisions are often required, but since LM/LMM typically appears on cosmetically sensitive areas such as the face, sometimes large excisions are not possible nor desirable. In this sense, specialized approaches have been developed such as margin-controlled surgery or image-guided treatment using reflectance confocal microscopy. Other treatments for LM such as cryosurgery, imiquimod, radiotherapy or photodynamic therapy have been proposed, although recurrence/persistence is common. The current manuscript reviews extensively the published data regarding the diagnosis, treatment and management of both complex entities LM and LMM.
Keywords: lentigo maligna, melanoma, lentigo maligna melanoma, dermoscopy, dermatoscopy, reflectance confocal microscopy, staged excision, Mohs surgery, imiquimod
Introduction and Epidemiology
Lentigo maligna (LM) is a type of melanoma in situ (MIS), recognized both by the American Committee on Cancer and the World Health Organization. Despite being classified as a type of MIS, LM has distinctive characteristics regarding epidemiology, risk factors and clinical features. In addition, LM has a potential to progress to an invasive tumor with potentially aggressive behavior: lentigo maligna melanoma (LMM).1,2 The lifetime risk of progression from LM to LMM ranges from 5% to 50% and increases with time.3 Overall, LM has a very good prognosis, whereas LMM has the same prognosis as other invasive melanomas with similar Breslow thickness.4
Regarding incidence, LM and LMM account for approximately 4–15% of all melanomas, representing up to 10–26% of head and neck melanomas, as they most commonly present on these areas.4 They typically appear on chronically sun-damaged skin in elderly people, as opposed to the most common subtype of melanoma, the superficial spreading type, which typically occurs on areas acutely exposed to UV radiation.5 Mean presentation age of LM/LMM ranges from 66 to 72 years, approximately a decade older than for other melanoma subtypes.1 Men are more likely to be diagnosed with LM/LMM, although in some series a predominance in females has been observed.6,7
Incidence has increased over the past few decades: Swetter et al8 found that LM was the most prevalent (79–83%) form of MIS in North Carolina. In a recent study in Catalonia (northeast Spain) with 4999 MM cases (282 cases of LM and 136 of LMM), a significant increase in incidence was observed during an 8-year period, from 6.9% in 2000 to 13.1% in 2007. Several factors may explain this increase such as increased chronic sun-exposure. Because LM appears on chronically sun-exposed skin, there is a higher incidence in southern latitudes compared to northern latitudes (1.3 cases/100.000 person-years in Australia vs. 0.8 cases/100.000 person-years in the United States). Other factors such as Fitzpatrick skin types and skin cancer awareness must be taken into account as they can influence the reported incidence of LM. This increase has been attributed to aging of the population and changes in sun exposure patterns,6–12 as well as lack of recognition of LM as a distinct histologic subtype in the past.9 It must be noted that the increase in the number of LM cases was significantly higher than the increase of the invasive ones, and mean Breslow thickness of LMM has remained stable.7
Risk factors associated to LM/LMM include increased age, chronically sun-exposed areas of sun-damaged skin, increased number of lentigines, increased number of actinic keratosis and history of previous keratinocyte carcinomas.13 Genetic conditions such as xeroderma pigmentosum, oculocutaneous albinism, Werner syndrome and porphyria cutanea tarda are also associated with LM.13 Larger size of LM has been proposed as a risk factor for transformation to LMM.
The pathogenesis of LM/LMM is complex. Whiteman et al14 proposed in 2003 that cutaneous melanomas arise through 2 distinct pathways, with chronic solar exposure being responsible for LMM. Elder et al15 have recently expanded on this concept suggesting a new classification of melanoma with different genomic characteristics. Contrary to the observed in superficial spreading melanoma, BRAF V600E mutations are infrequent in skin with marked solar elastosis.16,17 Driver mutations in LMM include NF1, BRAF V600K, NRAS and KIT. In addition, gain- or loss- of function mutations such as CCND1, MITF and TP53 have also been implicated in melanomas on sun-damaged skin.18
LM represents a challenging entity not only regarding the diagnosis but also regarding the management. Diagnostic criteria are not well established, and there is an overlap of clinical, dermoscopic and pathological features with other benign pigmented skin lesions such as lentigines, pigmented actinic keratoses and macular seborrheic keratoses.12,19 LM common appearance within photodamaged skin makes lesion border identification difficult and they often extend beyond what is visible with the naked eye. Wide excisions are often required, with appropriate surgical margins appearing to be around 9 mm on the trunk and extremities, and greater than 1 cm on the head and neck.20,21 However, sometimes these large excisions are not possible nor desirable in cosmetically-sensitive areas such as the face. In this sense, margin-controlled surgery such as Mohs micrographic surgery or staged-excision offers the highest cure rates with the minimal scarring necessary to achieve curation. However, the location of LM/LMM on the head and neck region and their often large size frequently represent obstacles to these therapeutic options. In addition, margin-controlled surgical techniques are not always available since they require trained surgeons and dedicated facilities, thus limiting its availability. Other treatments such as cryosurgery, imiquimod, radiotherapy and photodynamic therapy have been proposed in order to decrease morbidity and recurrence rates. Furthermore, a wait-and-see strategy has also been suggested in select patients such as patients with numerous comorbidities and patients with limited life expectancy.
The current manuscript reviews the published data on the diagnosis, treatment and management of LM/LMM.
Clinical Presentation
LM/LMM arises on chronically sun damaged skin predominantly of the head and neck area, which hosts 78.3% of the cases22 and with the cheeks accounting for 53.7% of the cases.23
However, extrafacial LM/LMM can also be encountered. According to a study analyzing 71 cases of LM,24 extrafacial lesions account for 17.5% of all LM/LMM cases, with patients being significantly younger when compared to head and neck LM/LMM patients. Extrafacial LM/LMM is usually seen on the trunk in men or extremities in women.25,26
LM/LMM usually presents as a slowly growing isolated large pigmented macule or patch with irregular borders. At times it can have a discontinuous appearance on clinical examination with ill-defined borders. Additionally, LM/LMM can also present as an amelanotic/hypomelanotic macule or patch, especially in fair-skinned individuals (Figure 1). Repigmentation of previous white or gray hair may rise suspicion towards the possibility of LM in the scalp, and it has been described as an early sign of LM.27
Figure 1.
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Lentigo maligna(LM)/LM melanoma (LMM) clinical presentations. (A) LM presenting as a large irregularly pigmented patch in the forehead with central regression and ill-defined borders. (B) LM arising on the right cheek clinically simulating a seborrheic keratosis. (C) Large LM occurring on the nose and surrounded by chronically sun damaged skin. (D) Light brown patch with ill-defined borders in a fair-skinned individual corresponding to a LMM that can clinically be easily mistaken for solar lentigo. (E) Extrafacial LMM affecting the upper back presenting as an isolated large homogeneous brown patch with irregular borders. (F) Nodular melanoma arising on a previous LM (note the brown pigment surrounding the nodule corresponding to the in situ component).
Because of this slow growth and sometimes hypomelanotic presentation, it is not uncommon that LM/LMM diagnosis is delayed, with 48.8% of lesions being >10mm in diameter at the time of biopsy.23 With time, LM may evolve into papules, nodules or thick plaques which correspond to invasive foci thus becoming LMM and conferring risk of metastatic disease (Figure 1).
The clinical differential diagnosis includes solar lentigo, seborrheic keratosis, lichen planus-like keratosis (LPLK) and pigmented actinic keratosis which are summarized in Table 1.
Table 1.
Lentigo Maligna (LM)/LM Melanoma (LMM) Differential Diagnosis: Clinical Presentation and Dermoscopy
| | Clinical Presentation | Dermoscopy |
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| LM/LMM47,151 | Appears on chronically sun damaged skin (predominantly head and neck) Pigmented macule or patch with ill-defined borders | Polygons/rhomboids/zig-zag pattern (angulated lines) Annular-granular pattern Asymmetric pigmented follicular openings Circle within a circle |
| Solar lentigo152 | Multiple lesions Brown papule/plaque Darker areas can be elevated or verrucous | Moath-eaten borders Pseudonetwork Comedo-like openings Diffuse opaque-brown pigmentation Lightbrown fingerprint-like structures Milia-like cysts |
| Seborrheic keratosis153–155 | Appears on the torso and face Multiple lesions Sharply demarcated brown-to-black papules/plaques Can have verrucous/rough texture | Milia-like cysts Comedo-like openings Hairpin vessels Sharp demarcation and moth-eaten borders Fat fingers |
| LPLK156,157 | Solitary lesion Slightly reddish papule/plaque with smooth or verrucous surface Appears on sun-exposed trunk and upper extremities | Localized or diffuse annular granular pattern |
| Pigmented AK49,151 | Predominantly on the face Scaly and rough surface | Annular pattern Grayish short strikes and granules Pseudoreticular structures Grayish brown dots and globules White and evident follicles Scales Brown-to-gray rhomboidal lines (without invading follicular openings) Background erythema |
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Abbreviations: LM/LMM, Lentigo maligna/lentigo maligna melanoma; LPLK, lichen planus like keratosis; AK, actinic keratosis.
Non-Invasive Diagnostic Tools
Non-invasive imaging methods including Wood’s lamp, dermoscopy and reflectance confocal microscopy can improve the diagnostic accuracy for LM/LMM, improve margin delineation, aid in biopsy site selection and also become an important tool for monitoring treatment.
Wood’s Lamp
The Wood’s light emits ultraviolet (UV) light in a wavelength of 320–400nm with a peak irradiance at 365nm. Epidermal melanin absorbs most of the light emitted from the Wood’s light with little reflectance, making superficially pigmented lesions appear darker than the surrounding normal epidermis.
Rather than for the diagnosis, Wood’s lamp is especially useful for estimating LM/LMM size – particularly areas of subclinical disease invisible to the naked-eye- and therefore becomes an essential part of physical examination.28 Wood’s light examination has been demonstrated to be useful for margin assessment of melanoma in situ (MIS) before biopsy,29 and is particularly useful in determining margins of LM/LMM before surgical excision.30,31
However, Walsh et al30 prospectively studied the accuracy of preoperative Wood’s light examination for margin assessment of MIS after excisional biopsy. They concluded that it is not reliable to assess subclinical disease with Wood's light in these patients since many false positive and false negatives can occur. This is not surprising since surrounding photodamaged skin can harbor numerous activated melanocytes. Hence, this may limit Wood’s lamp utility to delineate the LM/LMM margins as these melanocytes either isolated or within benign lesions (seborrheic keratoses, pigmented actinic keratoses) may be highlighted as well. Also, it must be born in mind that dermal melanin is not accentuated by the Wood’s light and may lead to false negatives regarding a deeper atypical melanocytic component.32
Dermoscopy
Dermoscopy is a non-invasive technique that allows for the visualization of skin structures not visible to the naked eye and therefore improves diagnostic accuracy for both pigmented and non-pigmented skin lesions. Dermoscopy consists of a handheld magnifier lens (normally around 10x) which is coupled to a light source that can be polarized or non-polarized. Polarized and non-polarized dermoscopy provide complementary information for the diagnosis of LM/LMM and it has been found to be superior to Wood’s lamp examination to delineate the borders of LM/LMM.28
When assessing facial LM/LMM it is important to take into account that facial skin has numerous terminal hair follicles, sweat gland ostia, and attenuated rete ridges. These special features in facial skin create a pseudo-network appearance: a structureless pigment area interrupted by nonpigmented adnexal openings.33–35
According to the third consensus conference of the International Society of Dermoscopy36 both metaphoric and descriptive terminology can be used to describe lentigo maligna dermoscopy. LM/LMM dermoscopy will be briefly discussed herein by using metaphoric and descriptive terminology, as well as their histopathological correlates.
Annular-Granular Pattern
Dots and structureless brown to blue-gray areas scattered throughout the lesion or arranged adnexal openings (Figure 2A). Brown dots correspond on histopathology to aggregates of melanocytes and small nests at the dermoepidermal junction between the follicles. Blue-gray granularity corresponds to melanophages in the upper dermis.
Figure 2.
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Lentigo maligna(LM)/LM melanoma (LMM) dermoscopic findings. (A) Facial LM with an annular-granular pattern consisting of brown dots scattered around adnexal openings. (B) LM showing a large blotch with the blue-black sign. (C) LM on the scalp with follicle obliteration and central regression (gray peppering or granularity). (D) Facial LM with an annular-granular pattern around adnexal openings and follicle obliteration. (E) Brown lines coalescing to form rhomboids (angulated lines) around adnexal ostial openings at the periphery of this LM. Also, an asymmetric distribution of pigment surrounding the follicular openings can be seen. (F) Extrafacial LM with presence of network and atypical dots. Note the absence of adnexal openings in extrafacial skin.
Polygons/Rhomboids/Zig-Zag Pattern (angulated lines)
They correspond to gray-brown lines that are connected at an angle or coalescing to form polygons (Figure 2E). The term “rhomboids” was introduced to describe gray-brown angulated lines around adnexal ostial openings in LM. This term is reserved for facial skin, and they are more frequently seen in LM located on the upper part of the face.23,36,37 However, the term “zig-zag pattern” can be regarded as a variant of rhomboids on facial skin.38
Whereas the terms “rhomboids” and “zig-zag pattern” are reserved for facial skin, angulated lines39 and polygons40 are the appropriate terms to characterize pigmented lines that form angles in melanomas on nonfacial skin,39,41 although many use “angulated lines” as a single term that encompasses polygons and rhomboids, regardless of their location.
These structures correspond histologically to confluent junctional atypical melanocytes together with underlying melanophages in the papillary dermis.33,42,43
Asymmetric Pigmented Follicular Openings
The presence of asymmetric distribution of pigment surrounding the follicular openings is characteristic, especially in LM located on the lower part of the face23 (Figure 2). A fine circle, semicircle, signet ring-like circle, irregular circle and double circle can be seen.
On histopathology, it corresponds to atypical melanocytes as single units or small nests in the epidermis that are surrounding and/or extending down hair follicles.35
Circle Within a Circle
When concentric pigmented circles surrounding follicular openings are present it is called the circle-within-a-circle structure, which is identified in 5% and 25.4% of cases.23,41,44
Regarding extrafacial LM/LMM, the most common dermoscopic structures are granularity (67.7%), angulated lines (44.1%), and atypical dots (36.6%).45 However, because of preserved rete ridges and significantly fewer adnexal openings in extrafacial skin, focal islands of network without an annular granular pattern can be observed46 (Figure 2F).
Schiffner et al33 described a model of the dermoscopic progression for LM/LMM (Figure 3). Initially, asymmetric pigmentation and dots around the follicle are seen, which later evolve to rhomboidal structures and finally form homogeneous pigmented areas and cause obliteration of the follicular openings. Pralong et al47 found that at least one of the structures previously described are present in 87% of LMM. The combination of asymmetrically pigmented follicular openings, gray dots, gray globules, or rhomboidal structures located anywhere within a lesion has a sensitivity of 89% and a specificity of 96% for LM/LMM.33 Another diagnostic clue for LM/LMM diagnosis is the darkening at dermoscopic examination.47 In fact, in 25% of LM/LMM the pigment appears darker and with different shades of brown and gray when viewed with dermoscopy compared to naked-eye examination.47 Moreover, it should be taken into account that increased density of the vascular network within the lesion when compared to adjacent skin is found in 58% of cases,47 and should rise suspicion for invasive melanoma.48
Figure 3.
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Schiffner progression model fo LM/LMM. Initially asymmetric pigmentation and dots around the follicle are seen (A), which later evolve to rhomboidal structures (B) and finally form homogeneous pigmented areas and cause obliteration of the follicular openings (C). Adapted with permission from Dermoscopedia (
Differential Diagnosis
It must be born in mind that gray color, gray circles, and annular granular structures can be seen in both pigmented actinic keratoses (PAK) and LM/LMM. Similarly, the hyperpigmented rim of follicular openings of LM/LMM can be mistaken for pseudofollicular openings of seborrheic keratoses.49 The differential diagnosis among these entities is further assessed in Table 1.
Some algorithms have been proposed to differentiate pigmented lesions of the face.
Micantonio et al50 proposed an algorithm to distinguish between facial LM and PAK with an excellent accuracy and a high positive predictive value for early facial LM. Eight dermoscopic criteria achieved the highest discriminative power to distinguish LM from PAK: light brown color, brown-to-gray incomplete circles, brown-to-black structureless zone obscuring the hair follicles, structureless brown eccentric zone, structureless blue zone, brown-to-black structureless zone, in-focus brown discontinued lines and scales. Tschandl et al51 focused on the evaluation of non-melanoma dermatoscopic features in order to differentiate LM/LMM from other pigmented lesions in the face. These features include scales, white follicles, erythema/reticular vessels, reticular and/or curved lines/fingerprints, structureless brown color, sharp demarcation and classic criteria of seborrheic keratosis. If no prevalent non-melanoma patterns are seen then the suspicion for melanoma increases.
Reflectance Confocal Microscopy
Reflectance confocal microscopy (RCM) is a non-invasive imaging tool that uses near-infrared laser light obtaining horizontal quasi-histological images.52 RCM improves diagnostic accuracy of multiple skin tumors.52,53 It is very useful for diagnosing and monitoring LM/LMM since it has cellular resolution and allows the visualization of very small amounts of melanin which are invisible to the naked eye or dermoscopy.52 Therefore, RCM is an ideal tool to differentiate LM/LMM from benign macules and solar damage.54,55
RCM increases the physician’s diagnostic confidence and increases diagnostic sensitivity, therefore improving the management of difficult lesions.53 In fact, RCM is more sensitive (overall sensitivity of RCM 0.93) and specific (overall specificity 0.89) than dermoscopy (overall sensitivity 0.73, and overall specificity was 0.84) for the diagnosis of LM.56 Moreover, the integration of dermoscopy and RCM increases the diagnostic accuracy of both these techniques used alone for facial tumors.57
RCM is remarkably useful when assessing lesions located on the head and neck area,58 and is particularly suitable for the identification of hypomelanotic/amelanotic and recurrent LM/LMM.59,60
Menge et al found the RCM and histopathology results were consistent in 89% of the cases when evaluating suspected LM, but skin damage may limit the specificity of the diagnosis.61
The proliferation of atypical melanocytes at the DEJ may be visualized on RCM as atypical round or dendritic cells, typically large (twice the size of the adjacent keratinocytes)59 (Figure 4). As LM becomes more extensive, pagetoid spread of large pleomorphic cells are seen through all layers of the epidermis, causing epidermal disarray. Poorly defined dermal papillae and atypical cells may be seen at the dermal-epidermal junction and can form bridges resembling mitochondrial structures.62 Other characteristic findings include junctional swelling with infiltration of the hair follicle, resembling caput medusae (Figure 4), which was found to be indicative of LM/LMM compared to nonmelanocytic skin neoplasms, with an overall sensitivity of 96% and specificity of 83%.59,63,64
Figure 4.
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Reflectance confocal microscopy images of a lentigo maligna. The epidermis (depth ~50 µm) shows numerous pleomorphic pagetoid cells composed of large round cells (arrowheads, A) as well as dendritic cells (star, A), which can invade the hair follicles (star, B). At the dermal-epidermal junction (depth ~150 µm), one can also identify disarrangement of the basal layer as well as large plump cells which correlate with melanophages (arrowhead, B), and junctional thickenings than can radiate from the hair follicles adopting a medusa head-like () structure (arrowhead, C), or a mitochondria-like structure (asterisk, D). (white scale bars: 250 µm).
Guitera et al59 developed a LM score to distinguish LM from benign lesions of the face. The score consists of 2 major and 4 minor criteria, represented in Table 2. A LM score of greater than 2 resulted in a sensitivity of 85% and specificity of 76% for the diagnosis of LM, and the algorithm was equally effective in the diagnosis of amelanotic lesions. Some of the features included in the algorithm are the presence of round and large pagetoid cells and the presence of more than three atypical cells at the dermoepidermal junction in five images. However, Champin et al65 consider that the presence of a single large bright dendritic cell-predominantly found around follicle openings – should be considered part of the tumor when assessing LM margins. Moreover, Yélamos et al66 believe atypical melanocytes continuing from the center of a LM/LMM, regardless of their size, should be considered as positive for LM as they may reflect the trailing edge of a LM/LMM (Table 2).
Table 2.
In vivo Confocal Features for LM Diagnosis
Guitera et al, 2010 (LM Score)59
Major features (2 points each)
Nonedged dermal papillae
Round and large pagetoid cells (>20 microm)
Minor features (1 point each)
More than three atypical cells at the dermoepidermal junction in five images
Follicular localization of pagetoid cells
Nucleated cells within the dermal papillae
Minor negative feature (1 point)
Broadened honeycomb pattern
Champin et al, 201465
Single large round cell or large dendritic cell
Yélamos et al, 201766
Atypical dendritic cell (any size) continuing from the LM trailing edge.
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RCM is also useful for margin definition in ill-defined lesions45 and to map the extent of LM/LMM before treatment.67,68 Champin et al65 described a new procedure that combined the“spaghetti” technique (described in the treatment section) with in vivo RCM to evaluate LM margins, with better results than with the “spaghetti” technique alone.
Handheld RCM (HRCM) together with the use of videomosaics (static mosaics obtained from dynamic videos) has allowed an accurate evaluation of large lesions in curved areas of the body, including the face. HRCM has been reported to be useful for detecting subclinical margins, and presence of invasion,69 thus becoming a valuable tool for deciding the optimal treatment. Yélamos et al66 combined adhesive rings and radial videomosaics obtained with HRCM to calculate clinical extension and presurgical margins of LM/LMM, and to identify LM/LMM extending beyond the dermoscopic and Wood’s lamp margins showing excellent accuracy when compared to margin-controlled surgery using staged excision.
A prospective study evaluating the correlation of LM/LMM subclinical extension defined by RCM compared to the gold standard histopathology showed an overall agreement of 85.9% between RCM imaging and histopathology of staged excision margins.70 Diagnostic accuracy for detection of residual melanoma in the tumor debulking (after biopsy) had a sensitivity of 96.7% and a specificity of 66.7% when compared to the histopathology.
Future Directions
Machine learning (ML) is a form of artificial intelligence which uses computer algorithms to help with clinical decisions. An interesting subfield of ML is deep learning, in which large datasets are scaled, allowing to improve themselves with more data.71 Deep learning convolutional neural networks (CNNs) have further improved the accuracy of ML in melanoma screening, even exceeding some dermatologists.71,72 These algorithms could improve LM/LMM diagnosis in the future,73 although some limitations have to be addressed. Winkler et al74 investigated the diagnostic performance of a CNN across different melanoma subtypes, including LMM. The authors used a dermatoscopic image set containing 30 LMM and 100 benign lesions (flat, macular solar lentigines, seborrheic keratoses and nevi), matched for localization and morphology. A “malignancy score” ranging from 0 to 1 (higher scores indicating a higher probability of melanoma) was obtained by the CNN. Average score was 0.98 for LMM and 0.37 for solar lentigines, seborrheic keratoses and nevi. Sensitivity and specificity (95% confidence interval) for diagnosis of LMM (a priori cut-off of >0.5) was 100% (88.7–100%) and 65.0% (55.3–73.6%), respectively. These good results are promising, although the authors admit that their dermatoscopic images were of greater quality than those obtained in a clinical routine setting. In addition, most images were derived from fair-skinned patients. Images of LMM in individuals of other ethnic background are scarce, therefore these can suppose additional limitations for pattern recognition by CNNs. In addition, some characteristics of pigmented skin lesions make them unavailable for ML analysis.75–77 The most relevant is the difficulty in assessing the border of the lesion (lack of pigmentation, lack of surrounding normal skin, presence of hair and lesions appearing in volar skin). Another important limitation is large lesions that do not fit into the field of view of the dermatoscopic camera. In addition, a recent study78 evaluated the limitations in image selection for ML analysis. Authors found that 66.7% of the LM included in the study showed exclusion criteria, with only 33.3% of them being eligible for ML analysis.78 Therefore, although ML and CNN will probably play an important role in the future management of LM/LMM, there are still limitations that need to be addressed by the use of larger image datasets that better represent different skin types, include benign lesions as well as images obtained with consumer cameras in a rather uncontrolled manner.
Histopathology
Biopsy Technique
When a lesion has been identified as suggestive of LM/LMM, a skin biopsy typically allows for a definitive diagnosis. Different biopsies can be performed regarding the purpose (incisional/excisional) or the technique (shave/punch/ellipse). The choice between one or another depends on the size of the lesion, location, cosmesis and patient’s and physician’s preferences.
Agarwal-Antal et al80 reported that 16% of LM contain invasive melanoma. Diagnostic excisional biopsy with narrow margins has been considered the gold standard for diagnosing melanoma because incisional biopsy can underestimate the depth of the lesion due to sampling error.81 Also, a broad shave biopsy extending into the deep papillary dermis or superficial reticular dermis can be acceptable for LM/LMM since it allows the evaluation of a large piece of tissue.82,83
However, RCM enables the physician to assess the invasive components of LMM and target biopsies to the thickest areas, thus potentially reducing sampling error.68 In a retrospective study by Mataca et al,84 RCM selected sites showed more histopathologic criteria when compared to dermoscopy selected sites.
HRCM can guide mapping biopsies by detecting features of LM with high sensitivity, thus reducing sample bias inherent to blind mapping biopsies.67 Also, mapping of LM/LMM with HRCM-videomosaicing can reduce the number of biopsies needed in doubtful areas.69
Histopathology Analysis
The diagnosis of LM/LMM is usually made by histopathologic examination. The dermis may show solar elastosis1 and may display a patchy or band-lymphocytic infiltrate. Some authors suggest that the presence of melanophages is a helpful to differentiate LM from melanocytic hyperplasia of chronically sun-damaged skin.86 The epidermis of LM/LMM is often atrophic and it can have hyperpigmentation of basal keratinocytes.4,87 It is important to highlight that no histological differences have been found between facial and extrafacial LM/LMM.88
Initially, LM is characterized by an increased density of melanocytes along the dermal-epidermal junction, although they may also be found in the spinous layer1 (Figure 5). These melanocytes can have variable nuclear atypia1 and can show dendritic appearance.89 Some melanocytes can be multinucleated and have prominent dendritic processes, known as “starburst giant cell”,90 which is not specific for LM and can also be seen in benign melanocytic nevi.91 The adnexa can be involved by neoplastic melanocytes,92 and is first seen in the infundibular part of the follicle. At the next stage, pagetoid spread, melanocytes with hyperpigmented and angulated nucleus in the basal layer, and involvement of the deeper portion of the follicles can be seen.93
Figure 5.
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(A) Histological pictures of lentigo maligna. Lentigo maligna in the radial growth phase. Solar elastosis (arrow), a patchy lymphocytic infiltrate and melanophages are seen in the dermis (+ sign). An atrophic epidermis (asterisk) and an increased density of melanocytes along the dermal-epidermal junction can be seen, some of them with variable nuclear atypia (arrowheads). Hematoxylin-eosin stain10x. (B) Extrafacial lentigo maligna on chronically sun-damaged skin. Atrophic epidermis (asterisk), prominent solar elastosis (arrow) and increased density of junctional pleomorphic melanocytes forming asymmetric nests with pagetoid growth of single melanocytes (arrowheads). Dermal inflammatory infiltrate with melanophages (+ sign). Hematoxylin-eosin stain10x.
Regarding LMM, some histologic variables have been identified. These include melanocytes forming rows, subepidermal clefts, a lesser degree of solar elastosis and nests.94 Nests of pleomorphic melanocytes – resembling dysplastic junctional nevus – predominate in 43% of LM/LMM.95
The vertical growth phase often shows spindle cell morphology. Up to two-thirds of desmoplastic melanomas are associated with LM,96 and neurotropism is not uncommon in deeply invasive LMM86 Immunohistochemical markers such as MART-1 may help identify a dermal invasive component.
Treatment
Even though LM has good prognosis, treatment is necessary in order to reduce the potential for LMM transformation and its associated mortality. The classic treatment which remains the mainstay is surgical excision with safety margins. However, due to the cosmetically-sensitive location on sun-exposed areas such as the face, surgery is not always possible or desirable. Herein we will review the different treatment modalities with their grades of recommendation and levels of evidence (Table 3).81,97 A big limitation is that there have been no randomized controlled trials to date comparing the efficacy of all LM treatments. Treatment choice is complex and depends on factors such as anatomic location and previous treatments, as well as patient comorbidities, age and treatment adherence.98 In fact, Fosko et al have proposed a personalized treatment plan in which patients with no comorbidities are suited for surgical excision, whereas patients who do not tolerate surgery are offered off-label imiquimod or radiotherapy.99 Patient care would benefit from a multidisciplinary team composed by dermatologists, surgeons and both medical and radiation oncologists.100
Table 3.
Evidence-Based Recommendations for Lentigo Maligna (LM)
| | Strength of Recommendation | Level of Evidence |
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| Complete surgical removal of LM with 5- to 10-mm clinical margins is the preferred management, when possible | C | III |
| Peri-operative reflectance confocal microscopy margin assessment should be considered when available | C | III |
| Mohs micrographic surgery is recommended for LM when available | B | II/III |
| Topical imiquimod 5% cream as second-line treatment for LM when surgery is not possible at the outset (primary setting) or when optimal surgery has been performed (adjuvant setting) | B | II/III |
| Radiotherapy as a second-line therapy for nonsurgical candidates | C | II/III |
| Superficial brachytherapy is not recommended | C | III |
| Cryotherapy is not recommended | C | III |
| Laser therapy is not recommended | C | III |
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Notes: Adapted from Swetter et al81 and Robinson et al.97 Strength of Recommendations. A. Recommendation based on consistent and good quality patient-oriented evidence. B. Recommendation based on inconsistent or limited-quality patient-oriented evidence. C. Recommendation based on consensus, opinion, case studies, or disease-oriented evidence. Level of Evidence. I: Good-quality patient-oriented evidence (ie, evidence measuring outcomes that matter to patients: morbidity, mortality, symptom improvement, cost reduction, and quality of life). II. Limited-quality patient-oriented evidence. III. Other evidence, including consensus guidelines, opinion, case studies, or disease-oriented evidence (ie, evidence measuring intermediate, physiologic, or surrogate end points that may or may not reflect improvements in patient outcomes).
Surgical Excision
Current guidelines suggest that surgical excision is the treatment of choice for both LM and LMM.81 The primary goal of surgical excision of LM/LMM is to achieve histologically-negative margins and prevent local recurrence because of persistent disease, as stated by the latest guidelines for the treatment of melanoma published by the American Academy of Dermatology and the National Comprehensive Cancer Network.81,101 Therefore, a surgical excision technique that carefully examines the entire tumor to exclude areas of invasion and complete surgical margins for accurate tumor clearance must be chosen.102 A standard elliptical wide excision examines a very small percentage of the total margin, with the risk of false-negative margins increasing as the number of bread loaf sections decreases. Margin-controlled surgical techniques such as staged excision or Mohs micrographic surgery offer low recurrence rates of 0.5% to 5%, and they also provide tissue-sparing in these cosmetically-sensitive sites.102
Conventional Surgery
The conventional melanoma margins of 5 mm for in situ disease and 10 mm for invasive disease have been demonstrated to be inadequate in LM and LMM,21,81,102–104 especially in the head and neck region. Kunishige et al21 observed clearance rates of 79.4%, 94.4%, 97.4% and 98.7% for 6, 9, 12 and 15 mm margins, respectively, in overall LM. Microscopic assessment of LM is difficult because adjacent actinically damaged skin frequently shows actinic melanocytic hyperplasia, which can simulate atypical junctional melanocytic proliferations.105
Mohs Micrographic Surgery (MMS)
It consists of excision of the lesion with immediate microscopic frozen section examination of 100% of the peripheral and deep excision margin. The main limitations are its high costs, that it’s a time-consuming technique, and that histologic interpretation of melanocytic atypia is difficult since freezing leads to artefactual changes in the melanocytes.106 Sensitivity and specificity rates for diagnosing LM/LMM on frozen sections vary from 100% and 90% to 73% and 68%, respectively.107,108 However, immunohistochemical stains such as MART-1 or MITF can be used in frozen sections, thus improving the diagnostic accuracy. Local recurrence rates for melanoma treated with MMS with MART-1 are between 0.2% and 0.49%.109,110 If tumor is detected at the margin, targeted excision or a second stage is performed around the residual tumor, repeating the process until clear margins are achieved. A recent systematic review including 27 articles showed a 1.35% recurrence rate with MMS.111 Another potential benefit of MMS is that LMM can be detected prior to reconstruction, allowing accurate sentinel lymph node biopsy (SNLB) prior to complex reconstructions that would otherwise compromise SNLB. In addition, some authors advise that the central tumor debulking specimen should be sent for formalin-fixed paraffin embedded sections, to detect possible invasion and eventual LMM upstaging.112
Staged Excision
This technique is a variation of MMS and allows confirmation of negative margins with delayed repair of the surgical site. The usual procedure involves vertical excision with initial narrow margins examined by formalin-fixed permanent sections. Further excision is guided by histology over subsequent days. A variation of this technique which only assesses a rim of peripheral tissue has also been described and named “spaghetti technique”. A correlation with increased lesion size and with recurrent lesions and increasing surgical margin required for tumor clearance has been found with staged excision.102–104,112
Margin-controlled surgical techniques require an expert multidisciplinary team formed by expert dermatologists, Mohs micrographic surgeons and pathologists with close communication and clinical-pathological correlation that is only achievable in referral centers.100
Non-Surgical Treatment
As stated before, LM common location on the head and neck region, big size and advanced age of the patients frequently represent obstacles to surgical treatment. Alternative minimally-invasive treatments have been proposed in LM in order to decrease morbidity and recurrence rates. The greatest limitation of all these techniques is that LMM cannot be completely excluded, as the whole specimen is not examined histologically.
Cryotherapy with liquid nitrogen (cryosurgery) is an alternative approach in select patients, with recurrence rates ranging between 0% and 40%.113,114 Melanocytes are highly sensitive to low temperatures, being destroyed between −4 and −7ºC.115 A depth of at least 3 mm must be achieved in order to destroy atypical melanocytes that may have extended into the hair follicles.116 Study comparison is limited by variability in treatment regimens and duration of follow-up. A study with 18 patients with LM used two freezing cycles of 1 minute each, separated by a thaw cycle of at least 2 minutes, with frozen areas extending 1 cm beyond the borders of the lesions.117 No recurrences or metastasis were observed during a mean follow up of 75.5 months.117
Immunocryosurgery
Combination of cryosurgery sessions with imiquimod application has been successfully used for LM. This method increases local inflammation and therefore the effectiveness of the treatment in those patients who do not respond with vivid inflammation to imiquimod alone.118 Different regimes have been applied with variable results. Matas-Nadal et al119 used topical imiquimod 5% cream once daily on the lesion plus 1 cm along the lesion edge for 3 weeks, followed by one session of cryosurgery (2 cycles of 20 seconds) on the lesion plus 1 cm along the lesion edge. This was followed by 6 months of topical imiquimod cream 3 times per week. Clinical clearance was observed in 3/3 patients.
Imiquimod
Imiquimod is a topical immune-response modifier that acts through binding Toll-like receptors 7 and 8 on dendritic cells, macrophages and neutrophils, recruiting CD68+ macrophages and cells involved in cytotoxic T cell responses.120 In addition, it may interfere with adenosine receptor pathways, increasing adenylyl cyclase activity; and may have proapoptotic activity against tumor cells.13,121 Although it is off label, topical 5% imiquimod cream for LM has been extensively used, with variable results.122–125 Complete histopathologic responses (based on targeted biopsies) range from 50% to 93%, recurrence rates range from 7.1% up to 50% and mean follow-up durations range from 6 months to 4.5 years.122–126 Once again, variability in treatment regimes, assessment of outcome and duration of follow-up make comparison between studies difficult. The inability to eliminate atypical melanocytes within hair follicles due to limited penetration has been proposed as an important limiting factor for imiquimod’s higher recurrence rates.81 A recent study has investigated the role of the host immune response in order to identify responders to imiquimod.127 Samples were taken prior to starting imiquimod 5 days per week for 12 weeks. At 16 weeks, lesions were excised and assessed histologically. No differences between immune cell subset densities were observed between responders and non-responders pre-treatment. Post-treatment, responders showed up-regulated antigen presentation, type I interferon, T cell activation and tolerance induction signaling pathways.127 The optimal regimen treatment remains to be determined, but a typical regimen is an application for 5–7 days per week for 12 weeks, within a 1–2cm margin of clinically normal-appearing skin.124 A systematic review with 471 patients treated with imiquimod in different treatment schedules observed that a treatment schedule using 6–7 applications per week, with at least 60 applications total, had the greatest odds of complete clinical and histological clearance of LM.128
Imiquimod as a neoadjuvant therapy has also been used in LM, demonstrating to be useful for decreasing the necessary margin for complete clearance.121,129 In a study with 334 patients treated with imiquimod 5% cream 5 nights per week for 2 to 3 months followed by staged excisions, there was a median final margin of 2 mm with a recurrence rate of 3.9%, a mean time to recurrence of 4.3 years and a mean length of follow-up of 5.5 years.121 Addition of tazarotene 0.1% gel to imiquimod cream 5% prior to staged excision has been assessed in a randomized trial of 47 patients in order to improve imiquimod’s response.129 Complete response after 12 weeks was observed histologically in 78% of lesions treated with combined therapy versus 64% treated with monotherapy (p = 0.17). Residual LM was observed in 22% of the lesions treated with combined therapy versus 36% with monotherapy on subsequent staged excision, without recurrences at a mean follow-up of 42 months.129
Assessment of treatment response is complicated although RCM may be useful (as discussed later in the section “management”). Clinical inflammation is generally associated with adequate histologic response, although there are cases in which this was not observed and actually showed the opposite results (inflammatory reaction without histopathologic clearance).129,130 In addition, the development of LMM after imiquimod treatment for LM has been described.129,131 Whether imiquimod had a role in tumor progression or focal microinvasion was already present in these lesions is unknown.
Advantages of imiquimod include superior cosmetic result, home application, avoidance of surgical morbidity, and treatment of potential atypical melanocytes in surrounding skin. However, the main limitation of this treatment is the potential severe inflammatory reactions observed. However, these adverse effects are usually limited to dose-dependent local skin reactions that subside after stopping of treatment, and post-inflammatory pigmentation. Systemic side effects, although rare, have been described and include flu-like symptoms. The latest American guidelines for the treatment of melanoma consider imiquimod as a second-line treatment for LM when surgery is not possible (primary setting) or when optimal surgery has been performed (adjuvant setting), after carefully discussing of its limitations and associated risks with the patient and family.81
Radiotherapy
Radiotherapy (RT) can be used in LM as primary treatment or as adjuvant treatment when positive margins are found after surgery. The latest American guidelines for the treatment of melanoma support the use of radiotherapy for LM as a primary treatment when complete surgical excision is not possible81 and the European Society of Medical Oncology states that “RT for local tumor control should be considered in cases of inadequate resection margins of LMM”. Superficial RT (Grenz ray or Miescher technique) was initially used, with recurrence rates ranging from 7% to 12% in 2 large retrospective studies.132,133 However, this modality is not preferred currently due to insufficient dermal penetration of Grenz rays (1 mm into the dermis) and therefore lack of adnexal treatment, where residual atypical melanocytes can remain.134 A different approach with a slightly deeper penetration has been increasingly used, with reported cure rates of 86–91%.135 A concern with deeper penetration is scarring and pigmentary changes, which may be difficult to differentiate from recurrence. A literature review including 349 irradiated patients found a crude 5% LM recurrence rate, with a median study follow-up of 3 years.136 Resolution of pigmentation occurred after 2–24 months in the same review. Differing treatments, parameters, and dosages limit specific recommendations. The field should be expanded 1 cm beyond the area marked as involved by RCM.136 Another systematic review examined 10 studies with a total cohort of 454 patients treated with RT. Complete response rates varied between 80% and 100%, and mean recurrence rate was of 11.5%.137 Regarding electronic surface brachytherapy, it is not recommended by the latest American guidelines for the treatment of melanoma.81 A multicenter-randomized trial of imiquimod versus definitive RT is currently active and will provide results in the following years (NCT02394132).134
Photodynamic Therapy
Photodynamic therapy (PDT) is an effective, non-invasive treatment for precancerous and cancerous skin lesions. Studies have shown that PDT with 5-aminolaevulinic acid induces in vitro lysis of melanocytes.138 Karam et al139 treated 15 LM with PDT with methyl-aminolaevulinate, achieving a cure rate of 80% (12/15). They used a higher light dose (average 60 J/cm 2) than the dose for non-pigmented carcinomas (37 J/cm 2), as melanin restricts the diffusion of red light into deep layers of the epidermis. The response was assessed by clinical follow-up (18–50 months) and multiple biopsies.139 A recent study investigated the efficacy of ablative fractional laser-assisted PDT (average dose 90 J/cm 2) with 5-aminolaevulinic acid nanoemulsion for treating LM.140 The rationale for using ablative fractional laser is that it allows for the photosensitizer precursor to penetrate deep enough to reach all the atypical melanocytes. Seven out of ten lesions (70%) were histologically completely cleared after three sessions. These promising results suggest that PDT could be an alternative therapy for inoperable LM.140
Laser Therapy
Besides the use of fractional lasers to increase the penetration of PDT, described in the previous section, various forms of laser therapy have been used for the treatment of LM, with the majority of evidence derived from anecdotal reports and case studies. Carbon dioxide, Q-switched ruby, argon, neodymium-doped yttrium aluminium garnet, alexandrite laser or combinations of the above have shown poor results, with high recurrence rates.137 Lee et al reported a recurrence rate of 6.7% for carbon dioxide laser (1/15), with an initial response of 100%.141 Further studies with accurate treatment protocols are needed to confirm these results.
Ingenol Mebutate
Although ingenol mebutate has shown to induce apoptosis in melanoma cells in vitro,142 clinical data are nonconsistent. One case reported its efficacy in a 91-year-old patient with a recurrent LMM.143 Recently, a prospective multicenter study with 12 patients treated with ingenol mebutate at 150 μg/g daily for 3 days has been published, with disappointing results. Only 2 achieved complete response and no correlation between ingenolol mebutate-induced inflammation and clinical/histological clearance was observed.144
Other Non-Surgical Treatments
Other treatments such as azelaic acid, intralesional interferon alpha, fluorouracil and retinoids have proven to be ineffective for the treatment of LM and therefore are not currently recommended.27
Monitoring
Monitoring is crucial in order to detect persistence, recurrence or progression in treated LM/LMM patients. The mean time to local recurrence of primary LM and thin LMM was identified to be at least 57.5 months.145 The surgical cure rate is considered after 5 years; therefore, a long term follow-up is required. Surgery and other nonsurgical treatments can cause inflammation, pigmentation and scarring which should be taken into account when assessing treated LM/LMM. Also, identifying local recurrence can be challenging in patients with chronically sun damaged skin.
Clinically, recurrent/persistent LM/LMM can be identified by the onset of a papule, nodule or change over a treated area, which typically tends to be amelanotic or lightly pigmented with nonspecific clinical and dermoscopic features.146 Dermoscopy can sometimes improve the detection of recurrence/persistence and assist in post-treatment monitoring. The presence of “dust-like dots” is a clue for treatment failure in cases of LMM treated with imiquimod or radiotherapy.146
Follow-up after treatment can also be assisted by RCM. It is recommended to wait at least 3 months after surgery to assess the treated area with RCM to avoid inflammation and early scarring changes.147 HRCM has been reported to be useful for detecting persistence or recurrence of LM/LMM, and also for monitoring after surgical and nonsurgical treatment,148 especially in large, amelanotic or poorly pigmented lesions.68
Alarcon et al149 found no statistically significant differences between RCM and histopathology when evaluating the response of LM to imiquimod. RCM identified 70% of patients as responders to imiquimod without false-negative results, whereas response was overestimated with clinical evaluation and underestimated when using dermoscopy.
Clinical follow-up and examinations frequency depend on the AJCC primary tumor stage and the presence of additional risk factors, such as family history of melanoma, history of sunburns, dysplastic nevi, among others. LM tends to have a slower growth compared to other subtypes of melanoma; however, when LM transforms into LMM and therefore becomes invasive, its prognosis and management are the same as the other melanoma subtypes.
According to the latest 2019 European melanoma guidelines,150 follow-up for recurrence detection should be performed for at least 5 years, and follow-up for the detection of new melanomas and other skin cancers for at least 10 years. In stage 0 and IA melanoma a dermatological examination should be performed every 6 months during the first 3 years. In stage IB-IIB melanoma clinical examination is recommended every 3–6 months, together with lymph node sonography every 6 months, during the first 3 years. In stage IIC-IV with no evidence of disease, dermatological examination is advisable every 3 months during the first 3 years, together with lymph node sonography and laboratory examination (including LDH and if possible S-100) every 3–6 months. Moreover, a CT scan including neck, thorax, abdomen and pelvis or a PET-CT, and head MRI should be performed every 6 months in stage IIC-IIIC melanoma, every 3–6 months in stage IIID melanoma and every 3 months in stage IV with no evidence of disease melanoma during the first three years.150
After 4 years an annual clinical examination for stage IA melanoma seems to be adequate. For stage IB-IV with no evidence of disease from year 4 to 10 clinical examination is advisable every 6 months, and annually thereafter. Stage IV melanoma with distant metastasis follow-up should be individualized depending on symptoms, therapy and examinations.150
Conclusion
LM/LMM incidence has been increasing in the last decades. Since LM has the potential to progress into invasive LMM, careful diagnosis, treatment and management is required. Diagnosis can be difficult but non-invasive tools may help increase diagnostic accuracy. The gold-standard treatment is still surgical excision with histologically-negative margins, although alternative minimally-invasive treatments in selected patients with limited disease (LM) may help reduce morbidity and recurrence rates. However, in all cases careful monitoring after treatment with ancillary tests such as dermoscopy of RCM is fundamental, to exclude and identify persistent or recurrent disease. Randomized clinical trials comparing the efficacy of different treatments, including surgical and minimally-invasive, are needed to provide evidence-based data to improve the management of this challenging skin cancer.
Disclosure
Dr Oriol Yélamos reports personal fees from Leo Pharma and Almirall, outside the submitted work. The authors report no other potential conflicts of interest for this work.
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7332 | https://www.youtube.com/watch?v=zZ39eFo8JNA | Homework 11: #7 (Open box optimization)
Michaela Stone
2930 subscribers
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Posted: 19 Oct 2013
If an open box is made from a tin sheet 9 in. square by cutting out identical squares from each corner and bending up the resulting flaps, determine the dimensions of the largest box that can be made.
Transcript:
let's take a look at question number seven from homework 11 basically what I've done is I made a PDF file you can download the PDF file off of the homework help site you can use that you can follow along while I'm going through on the video this is the file that I've got saved on there is the exact same file that you're looking at on your screen right now so once you've watched the video it might be helpful for you to to take a look at that so that you can follow along while you're working your own question alright so what does the question say they are telling us that we want to make an open box an open box just means that there's no top on the box we're just gonna have a base and four sides and it's going to be made from a sheet of tin that's nine inches square and we're gonna make it by cutting out identical squares from each corner and bending up the flaps they want us to determine the dimensions of the largest box that can be made and we need to round our answers to two decimal places I'm gonna stop right there some of you may have come up with correct answers you're sure you've got the right answer for this but maybe you forgot to round to two decimal places say you had the same number that I have here so you had nine and you came up with one point five for the height you have to enter one point five zero and you came up with six for the length and width you have to enter six point zero zero so there's a very good chance that if you've got this correct that you just forgot to round it to two decimal places so if that's the case plug it in in the correct format and you should be all set don't have to watch the rest of the video you've got this piece of cake now if you haven't gotten the answers yet you don't know how to get them stick with me and that will walk you through it so first thing we need to do because they didn't do it for us is make it wrong it is almost impossible to solve these kinds of questions without making a draw because we don't even know what we're talking about so here's my my drawing I've got a 9-inch square sheet of ten and we're cutting identical squares out of the corner so it meets of these we don't know how big it's going to be so we'll just say that they're X by X squares we'll figure out what X is that's the whole point of this question but for now since we don't know how big they are we'll just say that they're X so if I cut an X by X square out of one corner I have to cut that same X by X square out of all the other corners well what does that look like that should probably be another picture at this point so let's take a look at what happens if we cut out the corners and if we draw the right picture we're gonna figure out what the dimensions of the base of our box are okay so here's my original nine by nine square but I've cut out an X by X square on each side so that means the width of the base of my box because I'm gonna be folding up these sides they're just like flaps once I fold them that's gonna be the sides of my box but that means that the the base is going to be 9 minus 2x I started with nine and I took X away here and X away here and the length is going to be 9 minus 2x because I started with a full nine and I'm taking away X here and executors so that's 9 minus 2x for both the length and the width so that tells me that my base the base of my box because we know what what length and width are the base of my box is gonna be 9 minus 2x times 9 minus 2x we don't know what the X is yet but it doesn't matter because it's gonna be the same for every time you see that X so this formula says the base is going to be 9 minus 2x times 9 minus 2x so we want to optimize volume they're asking us to find the best volume that we can get out of this box so we need the formula for volume that formula that's the length width times height so we have our length we found that up here that was the 9 minus 2x we have our width that's the 9 minus 2x but now think about it for a second what's our height well if I fold one of these flaps up the height is just going to be X I fold these flaps up my box is going to be X inches high because that's the length of these flaps so I need in length this 9 minus 2x times width is 9 minus 2x times height is X and all I did here was multiply it out I did the foiling that gave me this here and then I distributed the X here and just kind of changed the order around so that my highest power of X is at the beginning so you know 9 times 9 9 times negative 2/9 times negative 2x and then negative 2x times negative 2x and I end up basically once you simplified it with 4x cubed minus 36 x squared plus 81 X so now we need to optimize how do we optimize we take the derivative so my volume formula that I came up with and it's the same formula I came up with on the last page here for X cubed minus 36 x squared plus 81 X I take the derivative that's how we optimize that's the first step in figuring out what the biggest volume is going to be so I take my derivative 3 times 4 is going to give me 12x squared because I have to lower the exponent 2 times 36 is going to give me 72 X again I have to lower the exponent and then the derivative of 81 X is 81 so here is derivative that's my B Prime that's the thing I need to set equal to zero and solve for x so 12x squared minus 72 X plus 81 equals zero you want to work with those big numbers I don't want to work with those big numbers I'm going to make that as small as I possibly can so I'm just gonna divide both sides of this by three that's what I did here I divided that by three that gave me four x squared I divided this by three that gave me 24 X I divided our negative 24x and I divided this by three and that gave me 27 and if I divide 0 by three I get zero so I've kind of simplified this and now I have to factor I the way I factored it is I knew that for this 4x squared that I was gonna need a 2x times of 2x I see that I have a positive and a negative here I've got minus and plus so I knew that I needed something that added up to 24 in the middle negative 24 in the middle so I was gonna need negative numbers and that multiplied to positive 27 I need two negative numbers to be able to get a positive here and the negative here alright and the numbers that I came up with the numbers that work in this case are three and nine if you're not sure how I came up with those okay you're gonna for your question you're going to have 2x is the first part of this and you're gonna have a minus and a minus I want you to look at your middle number and divide it by two in my case it was negative 24 divided by 2 is going to give negative twelve I needed two numbers that multiply or then added up to equal negative twelve that multiply to equal 27 and those two numbers if I add negative 3 to negative 9 I'm gonna get negative 12 I multiply the two of them together I'm gonna get positive 27 so that's my thought process I'm not good with the factoring you're gonna need to see me sometime because we really need to work on that because you need to be able to do it so anyway once you've got your two tongues here you know that if we've got this equal to 0 that either 2x minus 3 equals 0 or 2x minus 9 equals 0 so we solve those we get 2x minus 3 that gives us x equals 3 halves this 2x minus 9 gives us x equals 9 hours so I've got some news for you 9 halves that's four and a half if I have my X between four and a half let's go back and look at that box really quickly if this is four and a half and this is four and a half that's my entire nine inches there and we can't have that that's not a box it's not gonna have any there's no way to fold it because my flap would go to here it would go to here and it would go down to here and down here it's not a box that's impossible so we know that we can't use this nine and a half so we need x equals three halves well what do we do with X X is the height of our box so we've got one of our answers right now remember when we looked at the picture up here X was the height of the box when we fold those flaps up X tells us the height they're asking us for the height here my height is three halves so I'm gonna put it in decimal form round two two places two decimal places that's one point five and now I know the formula for my length and width is the same we found that formula before it was nine minus 2x I'm just gonna plug in what I found for X 9 minus two times one half is nine minus three which is six but we need those two decimal places for it to be correct so both my length and width because it's a square they're the same they're going to be six or 6.00 when we enter them in and I hope this gives you step in the right direction on this like I said you can print out the PDF or take a look at the PDF to help you as you work through this with your own number |
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Glossary » True Positive Rate
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True Positive Rate
The True Positive Rate (TPR), also known as sensitivity, recall, or hit rate, is a fundamental concept in the field of machine learning, particularly in the context of classification problems. It is a statistical measure that provides insights into the performance of a classification model. The TPR is the proportion of actual positive cases that are correctly identified by the model.
What is True Positive Rate?
The True Positive Rate is derived from the concepts of True Positives and False Negatives. True Positives (TP) are the cases where the machine learning model correctly predicts the positive class. For instance, if a model is designed to predict whether an email is spam or not, a TP would be a spam email correctly identified as spam by the model.
On the other hand, False Negatives (FN) are the cases where the model incorrectly predicts the negative class. In the spam email example, a FN would be a spam email that the model incorrectly identifies as not spam. The True Positive Rate is calculated as TP / (TP + FN). This formula shows that the TPR is the proportion of actual positives (TP + FN) that are correctly identified (TP).
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Importance of True Positive Rate
The True Positive Rate is a critical measure in machine learning as it provides insights into the model’s ability to correctly identify positive cases. A high TPR indicates that the model is good at detecting positive cases, while a low TPR suggests that the model often misses positive cases.
However, the TPR should not be viewed in isolation. It should be considered alongside other performance measures such as the False Positive Rate (FPR) and Precision. This is because a model that simply classifies all cases as positive will have a high TPR but may also have a high FPR, indicating a large number of false alarms.
True Positive Rate in Different Domains
The importance of the True Positive Rate can vary depending on the domain. In some domains, such as medical diagnosis, a high TPR is crucial as the cost of missing a positive case (a patient with the disease) can be very high. In such cases, a model with a high TPR is preferred even if it has a high FPR, as false alarms (healthy patients incorrectly identified as having the disease) can be managed with further tests.
In other domains, such as spam email detection, a balance between the TPR and the FPR may be more desirable. A model with a high TPR but also a high FPR may result in many legitimate emails being incorrectly classified as spam, which can be annoying for the user. Therefore, the ideal model would have a high TPR and a low FPR.
Calculating the True Positive Rate
The True Positive Rate is calculated using the formula TP / (TP + FN). This formula shows that the TPR is the proportion of actual positives (TP + FN) that are correctly identified (TP). The TPR can be calculated directly from the confusion matrix, which is a table that summarizes the performance of a classification model.
The confusion matrix consists of four elements: True Positives (TP), False Positives (FP), True Negatives (TN), and False Negatives (FN). The TPR can be calculated by dividing the number of TP by the sum of TP and FN. This gives the proportion of actual positive cases that are correctly identified by the model.
Source:
Example of Calculating True Positive Rate
Let’s consider an example where a model is used to predict whether a patient has a certain disease. The model is tested on 100 patients, of which 50 actually have the disease. The model correctly identifies 40 of these patients as having the disease (TP), but misses 10 (FN). Therefore, the TPR is calculated as 40 / (40 + 10) = 0.8 or 80%.
This means that the model correctly identifies 80% of the patients who actually have the disease. However, this does not mean that the model is 80% accurate, as the model’s performance on the negative cases is not considered in the TPR. The model’s overall accuracy would be calculated as (TP + TN) / (TP + FP + TN + FN).
Limitations of True Positive Rate
While the True Positive Rate is a useful measure of a model’s performance on positive cases, it has its limitations. One limitation is that the TPR does not consider the model’s performance on negative cases. A model that simply classifies all cases as positive will have a high TPR, but may also have a high False Positive Rate (FPR), indicating a large number of false alarms.
Another limitation is that the TPR can be misleading if the data is imbalanced. If the number of positive cases is much smaller than the number of negative cases (called data imbalance issue), a model that simply classifies all cases as negative will have a low TPR but may still have a high overall accuracy. Therefore, the TPR should be considered alongside other performance measures such as the FPR, Precision, and Recall.
True Positive Rate vs. Other Performance Measures
The True Positive Rate is just one of many performance measures used in machine learning. Other common measures include the False Positive Rate (FPR), Precision, Recall, F1 Score, and Area Under the Receiver Operating Characteristic Curve (AUC-ROC). Each of these measures provides different insights into the model’s performance and is suitable for different scenarios.
The FPR, for instance, is the proportion of actual negative cases that are incorrectly identified by the model. It is calculated as FP / (FP + TN). The FPR is a measure of the model’s false alarms and is particularly important in scenarios where the cost of a false positive is high.
True Positive Rate vs. Precision
The True Positive Rate and Precision are both measures of a model’s performance on positive cases, but they provide different insights. The TPR is the proportion of actual positive cases that are correctly identified by the model, while Precision is the proportion of predicted positive cases that are actually positive. It is calculated as TP / (TP + FP).
A model with a high TPR but low Precision has many true positives but also many false positives. This could be a model that is overly optimistic, classifying many cases as positive. On the other hand, a model with a low TPR but high Precision has few true positives but also few false positives. This could be a model that is overly conservative, classifying few cases as positive.
True Positive Rate vs. F1 Score
The F1 Score is a measure that combines the TPR and Precision. It is the harmonic mean of the TPR and Precision, calculated as 2 (TPR Precision) / (TPR + Precision). The F1 Score provides a single measure of a model’s performance on positive cases, taking into account both the model’s ability to correctly identify positive cases (TPR) and its ability to avoid false positives (Precision).
A model with a high F1 Score has both a high TPR and a high Precision. This is a model that is good at detecting positive cases and also good at avoiding false positives. The F1 Score is particularly useful in scenarios where both false positives and false negatives are costly.
Improving the True Positive Rate
Improving the True Positive Rate is often a goal in machine learning, especially in scenarios where the cost of missing a positive case is high. There are several strategies that can be used to improve the TPR, including using a more complex model, adjusting the classification threshold, and using techniques such as oversampling or cost-sensitive learning.
Using a Complex Model
Using a more complex model can potentially improve the TPR by capturing more complex patterns in the data. However, this can also lead to overfitting, where the model performs well on the training data but poorly on new data. Therefore, it is important to validate the model’s performance on a separate test set.
Adjusting the Classification Threshold
The classification threshold is the point at which a case is classified as positive or negative. By default, this threshold is often set at 0.5, meaning that a case is classified as positive if the model’s predicted probability of the positive class is greater than 0.5. However, this threshold can be adjusted to improve the TPR.
Lowering the classification threshold will increase the TPR, as more cases will be classified as positive. However, this will also increase the FPR, as more negative cases will be incorrectly classified as positive. Therefore, adjusting the classification threshold is a trade-off between the TPR and FPR.
Using Oversampling or Cost-Sensitive Learning
Oversampling and cost-sensitive learning are techniques that can be used to improve the TPR in scenarios where the data is imbalanced. Oversampling involves replicating the positive cases in the training data to increase their prevalence, while cost-sensitive learning involves assigning a higher cost to misclassifying positive cases.
These techniques can improve the TPR by making the model more sensitive to the positive cases. However, they can also increase the FPR, as the model may become more likely to classify a case as positive. Therefore, these techniques should be used with caution and the model’s performance should be validated on a separate test set.
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7334 | https://www.youtube.com/watch?v=85xGI1lMSTU | If 3a-2b + 5c = 0. then family of straight lines ax + by + c =0 are always concurrent at a ...
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To ask Unlimited Maths doubts download Doubtnut from - If 3a-2b + 5c = 0. then family of straight lines ax + by + c =0 are always concurrent at a point whose co-ordinates is (alpha, beta), then the values of 5(alpha-beta)
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[Music] in this question if three a minus 2b plus five C equal to zero then simply Oh straight line ax plus B y plus C equal to zero are always congruent at point whose coordinates is alpha comma bit and then value of 5 times of alpha minus beta equals 2 so you get in the 3a minus 2 B plus 5 C equals to 0 and same though straight lines is a X plus b y plus C is always congruent at a point alpha comma beta so Hossam as a constant here sees constant of Isis sykov sykov Centro one over nine you're really using canvas 5/3 minus 2 B plus 5 C equals to 0 this way coefficients 1 Raonic Liscomb 5 c divided going through 3 by 5 a minus 2 by 5 b plus c equals to 0 now these two lines are congruent to each other these lines represents these same lines so you can say cap we're developing means in seconds coefficients come compare carrying it to xq gallic x equal to 3 by 5 and y values minus 2 by 5 yes y equals 2 minus 2 by 5 so these two lines are congruent at the point 3 comma 3 by 5 comma minus 2 by 5 which is given that whose coordinate is alpha beta 2 as alpha attack the coordinates of our bus guy alpha comma B type war so 3 by 5 comma minus 2 by 5 okay we find the value of 5 times of alpha minus beta so value 5 times of alpha minus beta key over there you 5 times of alpha is 3 by 5 minus of beta is minus 2 by 5 so this is plus 2 by 5 so 5 times of else M is 5 3 plus 2 which is equals to 5 so 5 so 5 cancel out of this K value gallium it was 5 5 times of alpha minus beta equal to 5 for my purpose when given that alpha beta the coordinates of the coordinates of the point oh we're family of straight lines are congruent to each other so value of 5 times of alpha minus beta equals to 5 so the idioms report this question [Music] |
7335 | https://web.williams.edu/Mathematics/sjmiller/public_html/409Fa22/addcomments.htm | Additional comments (Math 331) related to material from the
class. These additional
remarks are for your enjoyment, and will not be on homeworks or exams. These are
just meant to suggest additional topics worth considering, and I am happy to
discuss any of these further.
Friday, May 12. Video:
Harmonic Sums, Teller Problem, Grid Game:
+ Teller problem: how many people can you
see:
. We talked about what size we expected the answer to be; could be (cn,
c \sqrt{n}, cn^{\delta}, c \log n, c). We used one of my
favorite ideas:
linearity of expectation (this is a link to notes I've written on the
subject).
Another similar problem is the Marriage
Problem; here is a link to a lecture I gave on this in probability:
Marriage Problem, Poisson Random Variables:
Similar to the teller problem, it's
fascinating seeing what size one gets for an answer.
There's a really nice wikipedia
article on the secretary / marriage problem.
Somewhat related to this is the German
tank problem. What I like most about
this problem is that it's related to a lot of concepts we've studied
(conditional probabilities, breaking a complicated problem into a lot of
simpler ones), as well as the need to understand what a formula says and
re-express it in a more meaningful way. We saw the
harmonic numbers
hiding in our expression, which we can approximate using the
integral test. In a
better approximation one meets the
Euler-Mascheroni constant.
It is absolutely shocking that we can do so well in the marriage / secretary
problem (ok, mathematically, not necessarily in practice). While assuming we
know the number of applicants might sound overly restrictive, in some
situations it's actually not so unreasonable. For example, the Math/Stats
department is hiring a mathematician this year. Based on previous hiring
searches in the past few years, I expect there'll be around 700 applications,
and almost surely between 600 and 800. It's shocking that our final winning
percentage is positive and not decaying to zero with n. As a
nice exercise, try to compute the probability we end up with one of the top
two candidates. Try to come up with a strategy that will have you `settle' as
you get older (ie, start running out of candidates!).
Key input in the analysis was the sum of the harmonic series:
See here for the growth of the partial sums:
India bride walks out when groom
doesn't know math:
Harder version of
the tic-tac-toe game we discussed:
The version we did in class gives you the grid....
Yet another version of
tic-tac-toe:
Integral test:
better
version:
Stirling's formula:
We ended with the (N
\times M) lattice game; on your turn choose an ((i,j)) and remove all
((x,y)) with (x \ge i) and (y \ge j). Last one to remove a point
loses. What I love is we can prove a winning strategy exists without being able to say what it is! This just felt like a
wonderful way to end the semester and highlight the power and
limitations of pure mathematics.
Wednesday, May 10. Video:
Egg Drop Recurrence:
+ The big idea today was establishing a recurrence
relation, allowing us to reduce to previously solved cases. We gave a
similar example with our proof of the geometric series formula through a
memoryless process.
+ Hoops Problem:
- The proof we gave today of the geometric series formula (by shooting baskets)
uses many great techniques in mathematics.
The basketball example is a great in that it highlights many of the axioms
/ main theorems of probability. What events can we assign probabilities to? Is
the probability of a union the sum of the probabilities? The probability of A
is one minus the probability of not A.
It is thus well worth it to study
and ponder the proof.
- Memoryless process: once both
people miss, it is as if we've just started the game fresh. This is the key
observation to the problem, and converts an infinite sum to a finite one --
this is a major leap forward in the analysis!
- There is a real need to have good variable names. It's
easy to get confused and forget that the probability of winning depends on
the probabilities of both people and who
starts with the ball!
- We then discussed the geometric
series formula. The standard proof is nice; however, for our course the
`basketball' proof is very important, as it illustrates a key concept in
probability. Specifically, if we have a memoryless
game, then frequently after some number of moves it is as if the game
began again. This is how we were able to quickly calculate the probability
that the first shooter wins, as after both miss it is as if the game just
started.
- The geometric series formula only makes sense when \(|r| < 1\), in which case
\(1 + r + r^2 + \cdots = 1/(1-r)\); however, the right hand side makes sense for all
r other than 1. We say the function \(1/(1-r)\) is a(meromorphic)
continuation of
\(1+r+r^2+\cdots.\)
This means that they are equal when both are defined; however, \(1/(1-r)\) makes
sense for additional values of \(r\). Interpreting \(1+2+4+8+\cdots\) as \(-1\) or
\(1+2+3+4+5+ \cdots = -1/12\) actually DOES make sense, and arises in modern physics
and number theory (the latter is \(\zeta(1)\), where \(zeta(s)\) is the Riemann
zeta function)!
- We have to be careful as we cannot have \(r\) negative as it's a
probability. The way around was to split into even and odd terms, and then
see it works. Thus \(1 + r + r^2 + r^3 + \cdots = (1 + r^2 + r^4 + \cdots) +
r(1 + r^2 + r^4 + \cdots)\). We can then use the geometric series for each
factor with ratio \(r^2\), and get the sum is \(1/(1-r^2) + r/(1-r^2) =
(r+1)/(1-r^2) = 1/(1-r)\).
Monday, May 8. Video:
Dominoes and the Harmonic Series:
+ Stacking dominoes:
+ Harmonic series:
+ Euler-Mascheroni constant:
+ Big items from today: it is not always clear what the best notation is.
What do you want your unit to be? Where do you want your origin? How do
you reduce to a previous problem? The domino stacking problem is
excellent for this, as it truly highlights how challenging it can be to
find the best notation. Frequently there is a trade-off; what works well
for some aspects of the problem works less well for others. Here it was
nice to have the dominoes of size 1 and have the coordinates start at
the bottom left, and have the initial domino be labelled zero.
+ A great example of what is locally best is not globally best
are the two papers on whether or not dogs (or one dog in particular,
Elvis) know calculus. Looking at the the
path the dog takes to get the stick in the water, if the dog is in the
water close to the stick the optimal choice is to swim straight for the
stick; however, if the stick is far away it's best to swim to water, run
on water, then jump back in. In other words, if you only can go for a
few seconds the closest you can get to the stick is to swim for it;
however, if you're in it for the long haul it's worthwhile to do
something which is not locally optimal but which will lead to optimal.
For interesting articles related to this, see the two papers below by
Pennings on whether or not dogs know calculus. Click
here for
a picture of his dog, Elvis, who doesknow
calculus.
- Do dogs know calculus?
- Do dogs know bifurcations? This
paper discusses bifurcations.
An additional aside is that the bifurcation in the dog paper leads
to a nice proof of the arithmetic
/ geometric mean inequality, one of the more important ones in
math. For other proofs, see
This is a great example of a phase
transition.
Friday, May 5. Video:
Egg Problem Resolution:
+ Egg drop problem:
+ Big item today was figuring out what to 'guess'. The method of divine
inspiration is great, but it can be fickle. This is why we used the
method of free parameters. We then got a nice function (k n^{1/k}) to
optimize, then passed to the continuous function (f(x) = x n^{1/x}).
Below is a plot of the function (h(x) = x e^(1/x)); we see that its
minimum value is at (x=1).
Wednesday, May 3.
Video:
Egg Drop Problem and Combining Algorithms:
Main idea: sometimes combine algorithms and
dynamically adjust....
Watermellon problem:
Monday, May 1. Video:
Bridge and Egg Riddles:
+ The main idea was ways to convert problems to things we
have seen before (graph theory) so as not to miss things, and to be
aware of assumptions we are making....
Friday, April
28.
Video:
Theory of Bidding Tic-Tac-Toe:
+ Bidding tic tac toe:
-
-
Wednesday, April
26.
Video:
Game Theory: Rectangle Game
and Bidding Tic-Tac-Toe:
(Game Theory: Screen Capture:
+ Breaking chocolate bars:
+ Biggest loss ever:
+ Stratagema on Star Trek NextGen:
(sometimes you play for
the tie!)
- Other good clips from that episode (Peak
Performance):
and
Wednesday, April 19.
Video:
Student Presentations: Power Sums, Integrals, Squares:
Monday, April 17. Video:
Solving Rubik's cube:
(Math
313: Solving the 2x2x2 Rubik's Cube:
Solving the Rubik's cube (thanks Alan Chang and Umang Varma!):
2x2x2:
3x3x3:
4x4x4: parity issue:
Rubik's cube
(and the
G-d number), and homepage :
Friday, April 14. Video:
Contour Integration III:
+ Gaussian integral:
(see here for notes by Keith Conrad on more ways to do:
Wednesday, April 12. Video:
Contour Integration II:
+ Contour integration:
+ Key ideas: using differentiation to pull off the term we need in a
series expansion, finding best way to complexify.
Tuesday, April 11. Video: Sperner's Lemma and Fixed Point
Theorems:
+ The main idea in our proof of Sperner's lemma is a powerful one: a set
whose cardinality is odd must be non-empty! This parity knowledge turns out to
be exceptionally powerful. I know another situation where this is used: in the
theory of elliptic curves one can create a function whose order of vanishing
(conjecturally) encodes the order of the infinite part of the associated group
of rational solutions to the elliptic curves. This has applications in terms
of bounding the size of the class number. As these topics are a bit off the
standard path for this course, I'll leave it at the following brief remark for
now (we show a function (f(x)) is odd and that it has at least two zeros at
(x=0), and thus it must have a third zero!); I am
happy
to write more; if you're interested just email me (either directly or through
ephsmath@gmail.com) and I'll write
back with additional details or post them here.
+ Analysis review
handout.
+ One of the most important applications of contraction maps and fixed
point theorems is to solving first order differential equations. There is a
lot of great mathematics below the surface here.
- Detailed notes here:
(these go through the analysis preliminaries and the proof).
Lecture notes on the proof are pages 9-11 of this pdf file:
Wikipedia page:
- Example from previous year: I particularly liked seeing how quickly we converged to the
solution to (y'(t) = 2t(1 + y(t))) with (y(0) = 0). We rewrote this as an
integral equation: (y(\tau) = \int_0^\tau 2t(1 + y(t))dt), and notice that
the solution to our differential equation is a fixed point to this integral
equation! Here we have a map from functions to functions; this is a little
different than what you've seen before, and serves as a nice introduction to
functional analysis.
Integral equations:
Functional analysis:
- Video online here (from 2014):
+ Applications of Sperner's Lemma:
- Barycentric Coordinates, Fixed Point Theorem Preliminaries: Lecture from
Operations Research Math 377:
Main items:
We talked about conjugating maps to go
from one space to another, this is a very important concept and allows us to
reduce the analysis of arbitrary cases often to just one. See
for more. A
particularly nice example of this is the
Riemann Mapping Theorem.
We almost proved the
Brouwer
Fixed Point Theorem. We came close. I want to go slowly through the
argument and highlight the ideas and why we approach the proof the way we
do. This is a monumental result, and worth spending the time and doing it
right, and seeing how one would choose to introduce the various concepts.
The most important is the idea of the index of the mapping. It's similar to
a lot of things we've seen throughout the semester. We're looking for the
first occurrence of something that must happen. Notice how important it was
that the sum of the components is always 1, and that they are non-negative.
It suggests a similar result might fail if our space were....
- Brouwer Fixed Point Theorem, Linear Programming Games: Lecture from
Operations Research Math 377:
We saw earlier why we needed at least continuity. Today we saw how the
index worked, and how the Bolzano-Weierstrass theorem worked. Notice that
this is a non-constructive proof, and that at each stage of the
Spernerization we get a new triangle, not necessarily in the old.
+ Good catch that I was a bit careless and we are really doing Bolzano-Weierstrass
THREE times with three sub-sequences....
We played, or discussed playing,
IcoSoKu.
- Application: Nash equilibrium:
(his thesis is here:
scan:
, slides on talk here:
).
Monday, April 10: Video: Complex Analysis
in a day: The Path to Cauchy's Residue Formula:
Complex numbers: complex numbers are of the form (z = x + iy), with the complex conjugate
(z =
x - iy) with (i^2 = -1). While it is impossible to
totally order the
complex numbers (unlike the reals, which can), they still have many nice
properties. They are associative and commutative. It is possible to generalize
the complex numbers further. The first is the four dimensional space of the
quaternions, which are
associative but not commutative. The next generalization is the
octonians, which are eight
dimensional (over the reals) and now even associativity is lost! What's next?
The sedenions!
We showed the definition of the derivative can be rewritten and
interpreted as the first order
Taylor series does an
excellent job approximating the function. Interestingly, a function can have
continuous partial derivatives without being differentiable. The big theorem
is that if the partial derivatives are continuous then the function is
differentiable. It is possible for a function to have a limit along some paths
but not others, or to have different limits along different paths. For a
function to be differentiable, the quotient must tend to the same value no
matter what path you take. In one variable there is essentially just two
choices (approach from the left or the right); it's much more interesting in
several variables, as Cam and Kayla's artwork show.
Here is a nice example: it has the same limit along any straight line or
parabola, but a different limit along some cubic paths! Consider f(x,y) = (x8
y8)/(x2 + y8) - (x10 + y10)/(x4
y10) and take the limit as (x,y) approaches (0,0).
While the definition of complex differentiable looks innocuously similar
to that for one variable, remember that the limit must hold for any path. This
implies a variety of strange facts. A good way of viewing this is that
functions of a complex variable (z) can be written as functions of (x) and
(y).
Using (z = x + iy) and (\overline{z} = x - iy), we have (x = (z+\overline{x})/2) and
(y = (z-\overline{z})/2i).
Essentially the complex differentiable functions are those which depend on (z)
and not (\overline{z}). Thus only special combinations of x and y are allowed. Later
we'll see that complex differentiability is equivalent to our function
satisfying the Cauchy-Riemann differential equations. One way of interpreting this is
that we have a function whose corresponding function of two real variables x
and y has zero curl.
This means we have restricted ourselves to a very nice subset of real valued
functions, in particular ones where
Green's Theorem (or
Stokes' Theorem) is
particularly easy to apply.
In comparing results in real analysis to complex analysis, we met some
very interesting functions. One is
sin(1/x); though this
function isn't differentiable,
x2 sin(1/x)
is (but not infinitely often!).
Another great example is the function f(x) = exp(-1/x2) for x
not zero and 0 otherwise. To take the derivatives requires
L'Hopital's rule.
This function is extremely important in probability, as it shows that a Taylor
series does not uniquely determine a function. In probability this arises as
the moments of
a probability
distribution do not always uniquely determine the probability
distribution. This is covered in great depth in some
notes I wrote up for Math 341 (probability).
If you are unfamiliar
with Green's theorem,
watch my lecture on
Green's Theorem in a day from Calc III (you can watch it before then if
you'd like; this quickly goes over path integrals and explains the
calculations we've done).
One of the gems of complex analysis is Cauchy's
Integral Theorem, A complex differentiable function satisfies what is
called the Cauchy-Riemann
equations, and these are essentially the combination of partial
derivatives one sees in Green's theorem. In other words, the mathematics
used for Green's theorem is crucial in understanding functions of a complex
variable.
For me, I consider it one of the most beautiful gems in mathematics that
we can in some sense move the derivative of the function we're integrating
to act on the region of integration! This allows us to exchange a double
integral for a single integral for Green's theorem (or a triple integral for
a double integral in the divergence theorem). As we've seen constantly
throughout the year, often one computation is easier than another, and thus
many difficult area or volume integrals are reduced to simpler, lower
dimensional integrals. This subject is done properly in
differential forms.
Recall the exponential
function exp is defined by
(e^z
= \exp(z) = \sum_{n = 0}^{\infty} z^n/n!). This series converges for all (z). The
notation suggests that (e^z e^w = e^(z+w)); this is true, but it needs to be
proved. (What we have is an equality of three infinite sums; the proof uses
the binomial theorem.)
Using the Taylor series expansions for cosine
and sine, we find e^(iθ) = cos θ + i sin θ. From this we find |e^(iθ)| =
1; in fact, we can use these ideas to prove all trigonometric identities! For
example:
Inputs: e^(iθ) = cos θ + i sin θ and e^(iθ)
e^(iφ) = e^(i (θ+φ))
Identity: from e^(iθ) e^(iφ) = e^(i (θ+φ))
we get, upon substituting in the first identity, that (cos θ + i sin θ) (cos
φ + i sin φ) = cos(θ+φ) + i sin(θ+φ). Expanding the left hand side gives (cos
θ cos φ - sin θ sin φ) + i (sin θ cos φ + cos θ sin φ) = cos(θ+φ) + i
sin(θ+φ). Equating the real parts and the imaginary parts gives the
identities
cos(θ+φ) = cos θ cos φ - sin θ sin φ
sin(θ+φ) = sin θ cos φ + cos θ sin φ
Here's a picture that shows the standard proof with a convoluted diagram.
One can prove other identities along these
lines....
Friday, April 7: Video:
Differentiating Identities and Probability:
Differentiating identities:
Here is adifferentiating identies handout
(this became the nucleus for the section in a book I wrote). If the sums are finite
there is no difficulty justifying the method, but for infinite sums
it is very important to check to make sure we can do this interchange
(interchanging the sum and the derivative); it is frequently referred to as differentiating
under the integral sign. Differentiating
identities is a powerful technique; it creates infinitely many more identities
from a given one.
- The idea is that identities are hard to find, and if you can create more from
one that's good! We started with the geometric series formula (with a
hand-waving and then a rigorous proof), and talked about what we can get if we
can interchange the infinite sum and the derivative. Later in class we showed
that we can justify the interchange here as the tail of a geometric series is
also a geometric series; this is a fortitous situation which we can exploit.
In general one has to work harder.
Details of interchanging the sum and the derivative:
(Lecture from
my
Math 389 class).
- We talked about some of the important
standard distributions. We did the Geometric Distribution in class.
The first was the
geometric
distribution. If you thought the correct definition was to keep (p) the
probability of success and have the density function ((1-p)^{n-1}p), you're
right! This is the accepted normalization. We keep (p) for sucess, ALWAYS.
Also this way we talk about a success on toss (n), not on (n+1). Again, it
doesn't matter which way you define it, but you want to be consistent with
others.
Another is the
exponential
distribution. There are two `natural'
definitions, depending on what we want the scale parameter to be. Personally,
I prefer ((1/\lambda) \exp(-x/\lamba)) to (w exp(-xw)) (this is one of the few times
I disagree with Wikipedia); the first has mean (\lambda) while the second has
mean (1/w). You always have to be careful when looking at different sources, as
they could use different normalizations.
Binomial Distribution:
+ Wikipedia link:
+ Algebra: One of the hardest aspects of using the method of differentiating
identities is to find the right identity and the simplest path through the
algebra; fortunately it's also possible to reach the correct answer if you
take a longer route! While we could have started with (\sum_{k=0}\left({n
\atop k}\right) p^k (1-p)^{n-k} = 1), if we apply (\frac{d}{dp}) or
(p\frac{d}{dp}) we have to use a product rule, and we'd like to avoid that.
We can't just multiply through by (k) as that's our summation variable, and
whatever we do to one side must be done to the other. A nice trick is to start
with the Binomial Theorem: (\sum_{k=0}\left({n \atop k}\right)
x^k y^{n-k} = (x+y)^n), and now apply (x\frac{d}{dx}). The advantage is
clear: no product rule! All we need to do is then set (x = p) and (y =
1-p). What's nice is that, even though at the end of the day we have two
dependent terms ((p, 1-p)), we start with two independent quantities ((x,
y)) for differentiation.
Next is the
Poisson
distribution. Wikipedia redeems itself and has a
nice discussion of how it arises here. Didn't do this one in class, did it
in 2015:
.
In looking at the two distributions above,
our first though was to find the mean. We saw both were computable via
differentiating identities. For the Poisson, we saw how to compute the mean
(and even harder, the variance) by being clever about the algebra. For the
mean we needed to notice that (n / n!) equals (1/(n-1)!). For the
variance, the trick was to write (n^2) and (n(n-1 + 1)), and then when we
divide by (n!) we get (1/(n-2)! + 1/(n-1)!), and we can handle each piece.
It's
worth thinking about the variance of the Poisson. A major theme of the course
is the need to be able to look at a lengthy equation and get a feel for what
it's saying. The mean and the standard deviation are supposed to be in the
same units, so if the mean is λ then shouldn't the standard deviation be λ,
because if the variance were λ then the standard deviation would be λ1/2 and
that would have the wrong units, right? Wrong. For an exponential with density
f(x) = λ exp(-λx) the mean and standard deviation are both 1/λ, and we can see
that this is the correct λ dependence by scale issues: we exponentiate λx, so
λx must be unitless so if x is in meters say then λ is in 1/meters, and thus
this is the correct λ dependence for the mean and standard deviaton. What goes
wrong for the Poisson? Remember the density there is f(n) = λn eλ/n!;
here λ is alone in the exponential and is thus unitless! This means we can't
use the unit analysis to say that the standard deviation and the mean have the
same λ dependence.
The Poisson random variable often models the number of events in a window of
time. Also, frequently normalized spacings between events converge to
Poissonian (a great example is to look at the primes). Another is the spacings
between the ordered fractional parts of (n^k \alpha) (click here
for more).
- General advice: to differentiate an identity, you need an identity.
Seems silly to state but it's essential. Often the hardest part of these
problems is figuring out how to do the algebra in a clean way. For us, we saw
that frequently we want to move the normalization constant over to the other
side; it allows us to avoid a product or quotient rule. We also saw sometimes
it's easier to computer (E[X(X-1)]) than (E[X^2]), and then do algebra. It
all comes down to whether or not it's easier to apply (d/dx) or (x d/dx).
Wednesday, April 5: Video:
Integration by Differentiating:
We evaluated the integral of (\sin x / x) by using the
Laplace Transform,
and then specializing to a specific value. This is a very common technique:
we have a free parameter, find an answer as a function of that parameter,
and then specialize to the value we need.
Big theorems on interchanging limits and integrals:
Monotone convergence theorem:
Dominated convergence theorem:
Fatou's lemma:
Differentiating under the integral sign:
Differentiating identities and probability theory: See the Monday,
October 17, 2016 additional comment from Probability:
Lecture proving the Fundamental Theorem of Calculus through the Mean
Value Theorem: (52 minutes)
Lecture proving Green's Theorem in a day:
(51 minutes)
Monday, April 3: Video:
Pythgoras at the Bat: Introduction to Modeling:
(paper
here, slides here)
+ General links:
- Price Is Right Link:
- Williams vs Dimaggio Streak:
- Consecutive times reaching first safely (without causing an out or by
error):
- Interesting arXiv post -- the first author is a computer (his 'master'
chooses where to put him based on whom he believes did the most work on the
paper!):
+ Today's lecture serves two purposes (slides
here). Most importantly it introduces many of the key ideas and
challenges of mathematical modeling. I give this lecture Calc III and
Probability. Most students
there won't be taking
partial derivatives or integrals later in life (though you never know!);
however, almost surely you'll have a need to model, to try and describe a
complex phenomena in a tractable manner.
- Sabermetrics is
the `science' of applying math/stats reasoning to baseball. The formula I
mentioned at the start of the semester is known as the log-5
method; a better formula is the Pythagorean
Won - Loss formula (someone
linked my
paper deriving this from a reasonable model to
the wikipedia page), the topic of today's lecture. ESPN, MLB.com and all
sites like this use the Pythagorean win expectation in their expanded
series. My derivation is a nice exercise in multivariable calculus and
probability
- In general, it is sadly the case that most functions do not have a
simple closed form expression for their anti-derivative. Thus integration is
magnitudes harder than differentiation. One of the most famous that cannot
be integrated in closed form is (\exp(-x^2)), which is related to
calculating areas under the normal (or bell or Gaussian) curve. We do at
least have good series expansions to approximate it; see the entry on the erf
(or error) function.
The anti-derivative of (\ln(x)) is (x \ln(x) - x); it is a nice
exercise to compute the anti-derivative for ((\ln(x))^2) for any integer
(n). For example, if (n=4) we get (24
x - 24 x \ln(x) + 12 x (\ln x)^2
- 4 x (\ln x)^3 + x (\ln x)^4).
- Another good distribution to study for sabermetrics would be a Beta
Distribution. A nice example is the Laffer
curve from economics. I would
like to try to modify the Weibull analysis from today's lecture to Beta
distributions. The resulting integrals are harder -- if you're interested
please let me know.
- Today we discussed modeling, in particular, the interplay between finding a
model that captures the key features and one that is mathematically
tractable. While we used a problem from baseball as an example, the general
situation is frequently quite similar. Often one makes simplifying
assumptions in a model that we know are wrong, but lead to doable math (for
us, it was using continuous probability distributions in general, and in
particular the three
parameter Weibull). For more on these and related models, my
baseball paper is available here; another interesting read might be my
marketing paper for the movie industry (which
is a nice mix of modeling and linear programming, which is the linear
algebra generalization of Lagrange multipliers).
One of the most important applications of finding areas under curves
is in probability, where we may interpret these areas as the probability
that certain events happen. Key concepts are:
+ Probability distribution
+ Mean or
Expected Value
+ Standard
Deviation
+ Independence
+ Skewness and kurtosis (for
the hypercompetitive students who really want to compare themselves to
the class)
The more distributions you know, the better chance you have of finding
one that models your system of interest. Weibulls are frequently used in
survival analysis. The exponential
distribution occurs in
waiting times in lines as well as prime numbers.
In seeing whether or not data supports a theoretical contention, one
needs a way to check and see how good of a fit we have. Chi-square
tests are one of many
methods.
Much of the theory of probability was derived from people interested
in games of chance and gambling. Remember that when the house sets the
odds, the goal is to try and get half the money bet on one team and half
the money on the other. Not surprisingly, certain organizations are very
interested in these computations. Click
here for some of the details on the Bulger case (the
bookie I mentioned in class is Chico Krantz, and is referenced briefly).
Any lecture on multivariable calculus and probabilities would be
remiss if it did not mention how unlikely it is to be able to derive
closed form expressions; this is why we will study Monte
Carlo integration later.
For example, the normal
distribution is one of the
most important in probability, but there is no nice anti-derivative. We
must resort to series expansions; that expansion is so important it is
given a name: the
error function.
I strongly urge you to read the pages where we evaluate the integrals
in closed form. The methods to get these closed form expressions occur
frequently in applications. I particularly love seeing relations such as
(1/c = 1/a + 1/b); you may have seen this in resistors
in parallel or perhaps the reduced
mass from the two
body problem (masses under
gravity). Extra credit to anyone who can give me another example of
quantities with a relation such as this.
Click here for a
clip of Plinko on the Price I$ Right, or here for a showcase
showdown.
- We discussed how website like ESPN and MLB have a very limited space to
display information, especially if it's for a smart phone. Thus one cannot
show every statistic, and we have to pick and choose which ones are worth
showing. In one section I made a joke about including the team names, but
this is actually a serious comment! The MBTA (or
MTA for us old folk!) had a contest on how to redesign their
subway map of Boston. Below are links to an interesting article on the
subject and the maps.
Friday, March 17: Video:
Integration Techniques:
We discussed how to use symmetry to evaluate some integrals.
Some nice examples:
More problems:
and
List of Putnam problems and what kinds of integration work:
For a terrific application of symmetry or splitting up integrals, see
the proof of the functional equation of the Riemann zeta function via theta
functions:
We also discussed what I'm calling the `Bring it Over' method, where one
has the same expression on both sides but with different weights. Note this
is similar to solving algebra equations such as (3x + 7 = 8 x - 9).
Wednesday, March 15: Video:
Maximizing Product, Babylonian Math, Jug Problem:
Looked at the problem maximizing a restricted
product.
We studied one of my favorite problems, given (S = a_1
\cdots + a_n) with each (a_i) a positive integer, maximize the product of the (a_i). We quickly see the optimal is when each
(a_i) is 2 or 3, and since (2\ast 2 \ast 2 < 3 \ast 3) we want (3)'s
over (2)'s. We converted to a real problem and assumed there were (n)
summands, each a real number. We got a function defined on the integers to
maximize, replaced it with a function defined on the reals so calculus would
be applicable. We then curve sketched and saw the function was increasing to
its maximum and decreasing past it, so the optimal integer soln was either
to the left or right of the optimal real soln (here optimal soln is
referring to the number of summands). It's unusual to be this fortunate.
We had to maximize (a_1 \ast \cdots \ast a_n) given (a_1
\cdots + a_n = S) and each (a_i > 0). We can do this with Lagrange
multipliers, or since each (a_i) is in ([1, S]) we can appeal to the
(n=2) case because a real
continuous function on a compact set attains its max and min. What is
nice is that this existence result from real analysis improves to being
constructive; if we were at the optimal point and all coordinates were not
equal, we could simply replace two of them with the average and improve the
product.
A nice application of this problem is that for disk storage
(see radix
economy), base
3 has advantages over base
2, though base 2 has the very fast binary
search. Another nice example of base 3 occurs with the Cantor
set.
Did Babylonian Mathematics to follow our discussion on
bases. We talked about how the Babylonians did
mathematics. In addition to the horrors of base
60 (which Wikipedia tells me
is due to the Sumerians,
which must be true if they've posted it), they did give us the look-up
table. The point is to reduce long, painful calculations to pre-computed
quantities, with perhaps some (hopefully linear) interpolation as
you can't pre-compute everything. In base 60, one would need to have tables
for about 3600/2 = 1800 multiplications to be able to do (xy); the Babylonians
noticed (xy = ((x+y)^2 - x^2 - y^2) / 2), or even better (xy = ((x+y)^2 -
(x-y)^2)/4); this reduces the problem to knowing
just squares (only 60 entries needed) and the ability to subtract and divide
by 2. It's much better to do these simpler problems than the original harder
one.
We ended with a nice presentation on the jug problem, and
the differences between
DFS (depth first
search) and BFS (breadth first search).
No discussion of the jug problem would be complete without
this clip from Die Hard III:
Monday, March 13: Video:
Conway Checker Problem, (e^\pi) vs
(\pi^e):
Conway's checker problem:
(see also links
there)
Good problem: Generalize to 3dimensions, Generalize to "double"
jump: x x x [empty] -> [empty] [empty] [empty] x
(e^\pi) vs (\pi^e).
We talked a lot about how to use logarithms to make an
analysis easier, or to exponentiate. For example, ((S/x)^x = \exp(x \log(S/x))).
Here is a warning that you may have been fooled into
believing you learned certain derivatives when you hadn't. For example:
(f(x) = x^n) has derivative (n x^{n-1}). This follows
from the definition of the derivative and the binomial theorem to expand ((x+h)^n)
when (n) is a positive integer.
(f(x) = x^{p/q}) has derivative (\frac{p}{q} x^{p/q-1}).
This follows by setting (g(x) = f(x)^q = x^p) and then differentiating,
which gives (g'(x) = q f(x)^{q-1} f'(x) = p x^{p-1}), and then
substituting and solving for (f'(x)). We cannot get it the same was as the
derivative of (x^n), as that would require knowing the binomial theorem
for non-integral exponents.
(f(x) = x^{\sqrt{2}}) has derivative (\sqrt{2}
x^{\sqrt{2}-1}). This follows from using the exponential function and the
chain rule: (x^{\sqrt{2}} = \exp(\sqrt{2} \log x)).
Thus, (x^r) does have derivative (r x^{r-1}), but the
proof for general (r) goes through the exponential function.
We also used logarithms in the elementary approach to (e^\pi) vs
(\pi^e). Here is a nice argument, using (e > \pi - e > 0).
(e^\pi) vs (\pi^e): take logarithms of both sides.
(\pi \log e = \pi) vs (e \log \pi = e \log(e + \pi - e)), where the
last follows form adding zero. We do this so we can exploit log laws (log of a
product is sum of the logs).
(\pi) vs (e \log e + e \log\left(1 + \frac{\pi - e}{e}\right)), and
now we use (\log(1+x) = x - x^2/2 + x^3/3 - x^4/4 = x) plus a negative
number (decreasing in absolute value, alternating).
(\pi) vs (e + e \frac{\pi -e}{e}) + negative number.
(\pi) vs (\pi) + negative number; thus it must be (\pi \ge \pi) +
negative number, and thus working upward we find (e^\pi \ge \pi^e).
Another approach to that problem was to look at (f(x) = e^x - x^e) and
(g(x) = e^{x/e} - x). I find the second function easier to work with. Note
(g(0) = 1 > 0) and (g(e) = 0). We can figure out what's happening by
looking at the derivative: (g'(x) = e^{-1} e^{x/e} - 1). Note (g'(x) < 0)
for (0 \le x < e) and (g'(x) > 0) for (x > e). Thus (g(x)) is
monotonically decreasing to (x=e), where it is zero, and then increases.
Note the first derivative allows us to get a good feel for the shape of the
function, and thus all other (x) have (g(x) > 0), so (e^{\pi/e} > \pi),
which after exponentiation gives (e^\pi > \pi^e).
Friday, March 10: Video:
Monovariants: Tournament, Cards, Conway's
Checker Problem:
We solved the card sort problem by induction. It can also be done
with a monovariant approach. Recall the problem: Take the numbers 1, 2, ..., (n) and
randomly sort them. If the number (k) is on top, reverse the order of the
first (k) numbers. Continue as long as the top number isn't a 1. Prove that
eventually this process terminates with a 1 at the top. (So, if we start with
the ordering 4 2 1 3 5, the next ordering is 3 1 2 4 5, then 2 1 3 4 5 and
finally 1 2 3 4 5).
I think the
following solution works. It clearly suffices to show that eventually either
(1) or (n) is at the top, or (n) is at the bottom (if it's 1 we win,
while if (n) is at the top then on the next turn it's on the bottom, and an
(n) on the bottom reduces us to the case of (n-1) and we are done by
induction).
I claim that either 1 is eventually on top, or (n) is eventually on the
bottom.
Let us assume this does not happen for some (n), and without loss of
generality let (N) be the smallest integer such that (1) is never on top
and (N) is never on the bottom. As there are only finitely many ways to
order (N-1) numbers, the ordering must be cycle (and in all the elements of
the cycle, (N) is never on the top or bottom).
We have (N) is never on the top or the bottom; thus whatever card is on
the bottom must always stay on the bottom; let's call that card (k). Since
it is only the top card that causes change to happen, note that we could
safely switch (N) and (k). If we do this, we would still have an infinite
cycle but now it would only involve the numbers 1, 2, ..., (N-1). This
contradicts the minimality of (N) and thus our assumption is wrong.
Suggestion from class (John): About the monoinvariant: I think
if you compile a list of all the elements that are in order (i.e. 5
is in the 5th place) and order them from the largest to the
smallest, this list's dictionary order (i.e. compare the first
element, if they are the same compare the second) is always
increasing.
Conway's checker problem: SPOILER ALERT!
(see also links
there)
Wednesday, March 8: Video:
Chessboard Tiling, Room Movement, Reversing Cards:
Monoinvariants are a
wonderful topic, one often not seen (sadly) in the curriculum. We talked about
how it allows us to solve the virus propagation problem, but this is just the
beginning.
See Noether's Theorem for applications of conserved quantities in physics:
Slides on talk on Invariants, Monoinvariants and Extrema:
Berkeley Math Circle Lecture Notes on Monoinvariants:
Problems on invariants and monoinvariants:
We started with a chessboard problem and creating a monoinvariant to show
we cannot cover the board with 2 opposite squares missing with 1x2 dominoes.
We could also have called this a parity argument; we'll see more of
such arguments as the semester continues.
We proved a nice theorem: if (x_i \ge 0) then the maximum value of
(x_1^2 + \cdots + x_r^2) subject to (x_1 + \cdots + x_r = N) is when one
is (N) and the rest vanish. One proof was to note without loss of
generality (N=1) and then see that if (0 < x_1, x_2 < 1) then (x_1^ +
x_2^2 < x_1 + x_2) so the sum of the squares is strictly less than 1. We
could also use the monovariant argument from class. We have to be a bit
careful, as we are often using a great result from real analysis: a
continuous function on a closed, bounded set attains its maximum and minimum
(to see why we need the interval to be closed, consider (f(x) = (x-1/2) /
(x(1-x)). As (x \to 1) we have (f(x)) approaches positive infinity,
while if (x \to 0) we approach negative infinity. For more on this see the
Extreme Value
Theorem.
More chessboard problems:
Friday, March 3: Video:
12a: Monovariants, Zeckendorf Decomposition:
(slides
HERE, start at
16minutes, then go to start); 12b: Pigeon Hole Problem, Dirichlet's
Approximation Theorem:
Readings on monovariants:
Theory and examples:
Examples and problems:
Problems:
Dirichlet's approximation theorem:
(if
(\alpha) irrational then there exists integers such that (|\alpha - p/q|
< 1/q^2).
Hurwitz' theorem (improves bound to (1 / \sqrt{5} q^2):
P:roject Euler problems:
Wednesday, March 1: Video:
Irrationality Proofs, Morley's Theorem, Pigeon Hole Problems:
Rational irrational proofs (with David Montague), Mathematics
Magazine (85 (2012),
no. 2, 110--114). pdf (expanded
version: pdf;
letter from Nerode and Tennenbaum on history of proof: pdf)
See also
Irrationality of sqrt(2):
standard proof:
geometric proof:
Morley's Theorem: John Conway's simple proof is
here (just search the pdf for
Morley; interestingly it also has the irrationality of sqrt(2)).
Pigeon hole
principle: Lots of great links for readings and problems.
Monday, February
27: Video:
Geometry Problems, Points on Spheres and Hemispheres:
Today one of the key, new ideas was
the pigeonhole
principle, or Dirichlet's Box principle. There are two main steps:
identify the pigeons, identify the boxes. Usually the pigeons are obvious,
but it can be very hard to find the boxes.
The original version says if we have N+1 pigeons sent to N boxes, at least
one box gets 2 pigeons. More generally, if M > N and M pigeons are sent to N
boxes, at least one box gets ceiling(M/N) pigeons (ceiling(x) means the
smallest integer at least as large as x). This version is often very useful.
Wikipedia entry:
Fun reading:
(more extensive
notes)
(a
friendly version)
Video online here from the 2014 iteration of the class: Pigeonhole Principle: Part 1:
Pigeonhole
Principle: Part 2:
We may have proved a new theorem; I wouldn't be surprised if it's been
done before, but I think it's worth writing up as a problem and / or short
note, say for the Pi Mu Epsilon journal: given a positive integer m, if
there are 2m-2 points on a unit circle then at least m of them are on the
same closed semi-circle, and it is possible no m satisfy such a relation if
there are 2m-3 points. Anyone want to take point in writing this up? I saw
many websites with the original problem but not with generalizatoins:
(we generalized in 2-dimensions, how hard would it be in higher)? Working on
this can replace doing standard HW problems.
Friday, February
24: Video:
Pythagorean Formula, Asking Questions:
(click
here for slides)
Pythagorean
Theorem
Fermat's Last
Theorem
Beal's conjecture
spherical coordinates,
coordinates, area and volume of the n-sphere
hyperbolic
functions, hyperbolic space
Wednesday, February
22: Video:
Catalan numbers, Generating Functions, Circle Method:
We talked about the generating function approach to
Waring's problem, which
is quite challenging. Interestingly, a slightly different version of that
problem (not surprisingly known as the Easier Waring's Problem), is
elementary and easy:
The
Circle Method allows us to attack a variety of problems.
Rubinstein paper on the Hardy-Littlewood Constant and Twin Primes:
Direct link to Rubinstein's paper (not sure if this will work):
Note from one of my students:
Rubinstein and Sarnak: Chebyshev's bias:
Hardy-Littlewood Conjectures:
Twin prime conjecture:
k-tuple conjecture: (notice
the integral!).
Prime constellation:
For Star Trek fans: USS Constellation: NCC 1017:
From MemoryAlpha: The Constellation studio model was
constructed from a 1966 first edition AMT Enterprise model
kit , no. S921. (Star Trek Encyclopedia, 3rd. ed., p.
85) That particular edition sported a decal sheet with only
the "NCC-1701" decal, and, with very limited options, its
numbers had to be rearranged to create the unusually low
"NCC-1017" Constellation registry number, the first new
registry actually seen on a ship and, as it turned out, also
the only time in the original airing of the Original Series
, though
considered somewhat incongruous by many, due to the
perceived discrepancy in the numbering system.
Monday, February
20: Video:
Generating Functions, Binet's Formula, Waring's Problem, Catalan Numbers:
Generating functions: There are many ways to prove Binet's
formula
for an explicit, closed form expression for the n-th Fibonacci number. One is
through divine inspiration, the second through generating
functions and partial
fractions.
Generating functions occur in a variety of problems; there are many
applications near and dear to me in number theory (such as attacking the Goldbach or Twin
Prime Problem via
the Circle
Method).
The great utility of Binet's formula is we can jump to any Fibonacci number
without having to compute all the intermediate ones. Even though it might be
hard to work with such large numbers, we can jump to the trillionth (and if we
take logarithms then we can specify it quite well).
Generating Functions
Handout: This is from a book I'm writing on
probability. The first section is motivation, feel free to skim. Section 19.2 is
the most important. Section 19.3 is more technical and included for
completeness; we won't cover. Section 19.4 talks about convolutions -- we'll
need the very beginning to analyze the
Catalan numbers.
Catalan numbers:
A nice generalization of these occur in a
paper I wrote with some students and my kids.
Waring's problem:
We saw how to use generating functions to solve Waring's problem when
(k=1); remember Waring's problem is whether or not for a fixed (s, k) there is a
solution in non-negative integers to (x_1^k + \cdots + x_s^k = n) for
either all (n) or perhaps all (n) sufficiently large. When (k=1) this
is the Cookie Problem, and we could also solve combinatorially. If (k > 1)
we no longer have a combinatorial approach, but generating functions do
help.
For more on generating functions you can
see some videos of mine on Zeckendorf decompositions:
Cookie Monster Meets the Fibonacci Numbers. Mmmmmm -- Theorems!: (slides here).
Wednesday, February
15. Video:
Applications of the AM-GM Inequality, Zeckendorf and kilometers:
More on Arithmetic Mean - Geometric Mean inequality. This is
one of the most important inequalities in mathematics, and is a major tool in
many contest problems.
Nice page on the AM-GM and its use in problem solving:
Wikipedia page:
My notes:
Some applications in math contests:
Some more straightforward examples:
Heron's formula:
Olympiad inequalities:
Generalization of AM-GM: I. Ben-Ari and K. Conrad:
Big item in the problems we studied is symmetry, especially if not
symmetric how to symmetrize.
Key ingredient is Heron's formula:
Isoperimetric inequality
Monday, February
13. Video:
AM-GM inequality, Games
(Triangle, Rectangle, SpotIt):
Did more on AM-GM; see links from Friday, Feb 10
, especially my
notes:
When we did the proof by Lagrange multipliers, it was important to
know that there was a max/min. The fact that there are such points is
the Extreme Value Theorem. Technically the conditions are not met for us
unless we allow (a_i = 0), but we can always do by taking the (a_i \ge
\epsilon) for some very small and positive (\epsilon).+ We proved it using 'hopping' induction. Lots of types of induction.
I think this is cool: do powers of 2 and then fall back. Another
interesting technique is
transfinite induction.
Fibonacci spiral:
Mathematics of SpotIt:
(spoiler alert -- solution here).
The triangle game is related to Sperner's lemma and the Brouwer
fixed point theorem....
We didn't get to it, but the last item would have been fun versions
of tic-tac-toe. We'll hit these later, but if you're interested here's
some reading.
Earlier we talked about tic-tac-toe today
as a counting problem: how many `distinct' games are there. We are willing to
consider games that are the same under rotation or reflection as the same game;
see for
a nice analysis, or see the image
here for optimal strategy.
Probably the most famous movie occurrence of tic-tac-toe is
from Wargames; the clip
is here (the entire movie is
online here,
start around 1:44:17; this was a classic movie from my childhood).
Develin and
Payne: bidding tic-tac-toe analysis
Gobble tic-tac-toe:
Friday, February
10. Video: Recurrences and Arithmetic Mean - Geometric Mean
inequality:�
The first main item today was
the Fibonacci
series (an example of a difference
or recurrence equation).
Recurrence relations:
Recurrence relations with multiple indices:
Binet's formula:
Application of Fibonacci numbers in Roulette! See
The second main item
we studied today was the Arithmetic Mean - Geometric Mean inequality. This is
one of the most important inequalities in mathematics, and is a major tool in
many contest problems.
Nice page on the AM-GM and its use in problem solving:
Wikipedia page:
My notes:
Some applications in math contests:
Some more straightforward examples:
Heron's formula:
Olympiad inequalities:
Generalization of AM-GM: I. Ben-Ari and K. Conrad:
Wednesday, February 8. Video: Basic coding in Mathematica:
Purpose of today was to do basic coding.
Below are lectures from various classes
2017 Problem Solving (331) / Number Theory (313):
Mathematica code
(the pdf)
Video: 313:
Video 331:
2016 Probability Iteration:
Click here
for the Mathematica code
Click here for a pdf
Video:
2015 Probability Iteration:
Click here for the
Mathematica code
Click here for a pdf
Video:
some suggested problems you should try (click
here for Mathematica notebook, and
click here for a pdf).
We
discussed the cookie problem, counting the number of ways to divide 10
identical cookies among 5 distinct people (here's the classic clip where
Cookie Monster meets the
Count, whose full name is Count von Count -- they changed how he
appears!). Counting is very important; we discussed how we could (and SHOULD) do simpler variants to build intuition. It's usually called the
stars
and bars problem. What I love here is the power of changing your
perspective -- we go from a very painful brute force approach to being able to
solve it in one line.
I have posted the cookie problem on my
math riddles page (email me if
you want to contribute); someone with far more patience than I solved it by brute force. Here's their
solution. The final number, tabbed from the others, are how many distinct
rearrangements we have of this basic configuration. The total is
1001, or (10+5-1 choose 5-1). We saw this problem lead to a discussion of
multinomial coefficients. When there are two people getting cookies, say 8 and
2, there aren't (5 choose 2) ways to assign people, but (5 choose 2) 2! (we
choose the 2 people, then there are 2! ways to choose which gets the 8 and
which gets the 2). If we have 8 1 1 it's more involved. In that case it's (5
choose 3) to choose the three people, 3! ways to order which of the people
gets which number, but then we must divide by 2! (as the two people getting 1
are indistinguishable). A better way to view 3!/2! is 3! / (2! 1!) (note the
numbers on the bottom sum to the top). This is an example of a
multinomial
coefficient, a generalization of binomial coefficients. For example, if we
have MISSISSIPPI, there would be 11! ways to order the letters (order matters)
if the letters are distinguishable, but they're not. So let's put
subscripts on the letters: MI1S1S2I2S3S4I3P1P2I4.
We then have 4! ways of placing the four marked S's in the four S positions,
and so on, giving 11! / (4! 4! 2! 1!) (I like including the final 1 so that
the bottom sums to the top).
Below are the person's solutions, arranged differently than we did in
class.
0 0 0 0 10
5
0 0 0 1 9 20
0 0 0 2 8 20
0 0 0 3 7 20
0 0 0 4 6 20
0 0 0 5 5 10
0 0 1 1 8 30
0 0 1 2 7 60
0 0 1 3 6 60
0 0 1 4 5 60
0 0 2 2 6 30
0 0 2 3 5 60
0 0 2 4 4 30
0 0 3 3 4 30
0 1 1 1 7 20
0 1 1 2 6 60
0 1 1 3 5 60
0 1 1 4 4 30
0 1 2 2 5 60
0 1 2 3 4 120
0 1 3 3 3 20
0 2 2 2 4 20
0 2 2 3 3 30
1 1 1 1 6 5
1 1 1 2 5 20
1 1 1 3 4 20
1 1 2 2 4 30
1 1 2 3 3 30
1 2 2 2 3 20
2 2 2 2 2 1
- What we're really doing is solving the equation (x_1 + \cdots + x_5 = 10) in
non-negative integers. This is a very special type of
Diophantine
equation. It's actually a special case of
Waring's problem,
which looks at solving (x_1^k + \cdots + x_s^k = n) for fixed (s) and (k). These problems
are in general not accessible through combinatorics; the case (k=1) is special.
The general approach proceeds via
generating
functions, which we will cover in great detail later in the semester (it's
one of the key concepts of the class).
- Our solution to the
cookie problem is quite elegant, and in some respects reminiscent of geometry
class (remember all those proofs where the teacher cleverly adds
auxiliary lines; the difference here is we just add more cookies). While
it is possible to solve many combinatorial problems by brute force in
principle, in practice this is not a good way to go -- it is time consuming,
and quite likely that one makes a mistake. Typically one finds a way to
interpret a given quantity two ways; we can compute one of them and thus we
obtain a formula for the other. For example, we showed the number of ways of
dividing C cookies among P people is (C + P - 1 choose P-1); here all the
identical cookies are divided. What if we don't assume all the cookies are
divided -- what is the answer now? It is just Sum_{c = 0 to C} (c + P - 1
choose P - 1); this is because we are just going through all the cases (we
give out no cookies, 1 cookie, ...). What does this sum equal? Imagine now we
have another person, say the Cookie
Monster (this is one of
Cameron's favorite clips), who gets all the remaining cookies. Then dividing
at most C cookies among P people is the same as dividing exactly C cookies
among P+1 people, and hence our sum equals (C + P+1 - 1 choose P+1 - 1).
- Related to the cookie
problem is the partition problem: for the cookie problem we consider 2+3+3+1+1
different from 1+2+3+3+1 as the five people are distinct; if we don't consider
these distinct then we have a
partition
problem. It's a lot more complicated to count these, but some great
mathematics (such as Young tableau).
Monday, February 6. We began our
course in earnest today with a study of induction, one of the most common and
important proof techniques.
Video:
Analyzing inequalities, power sums, tiling and Fibonacci numbers:
We talked about how to analyze expressions such as ((a+b)(b+c)(c+a)
\ge 8abc) for (a, b, c > 0). Moments like this are the reason for this
class, talking about how to view. There's symmetry. Dimensional analysis
shows we can rescale and assume either (abc = 1) or (a+b+c = 1); turns
out the second is more useful and sets the stage to investigate it with
Lagrange
Multipliers.
A common image for Mathematical
Induction is
that of following
dominoes.
We did a bit on
Power Sums.
Lot of great math behind finding the formulas, though if we assume it's a
polynomial we can guess the coefficients by checking a few values, and then
prove by induction. Note the constant term should be zero and calculus
suggests the leading term of the sum of (n^{p-1}) for (n = 1) to (N)
should be (N^p/p).
Online Encyclopedia of Integer Sequences (homepage is )
is a tremendous resource. You can enter the first few terms of an integer
sequence, and it will list whatever sequences it knows that start this way,
provide history, generating functions, connections to parts of mathematics,
.... This is a GREAT website to know if you want to continue in mathematics.
There have been several times I've computed the first few terms of a problem,
looked up what the future terms could be (and thus had a formula to start the
induction).
Lots of good webpages with
induction problems:
We will see the
Fibonacci numbers many
times. We saw them today in covering a (2 \times n) rectangle with dominoes.
Here is a nice occurrence of a related recurrence relation in gambling: the
Double Plus One problem:
Friday, February 3. First lecture:
slides here
handout here
video here:
+ We talked about tic-tac-toe today
as a counting problem: how many `distinct' games are there. We are willing
to consider games that are the same under rotation or reflection as the same
game; see for
a nice analysis, or see the image
here for optimal strategy.
Chirality:
- more on tic-tac-toe:
How many distinct games of tic-tac-toe are there? Do
both in the case when we consider mirror images / flips to be the same
and when we don't. Remember as soon as there are three in a row the game
is over!
Redo the problem above, but now do it on a \(3 \times 3 \times 3\)
tic-tac-toe board.
Redo the above problem but on a \(3 \times 3 \times \cdots \times
3\) board, where we have a total of \(n\) dimensions!
Interesting variants:
+ Ultimate tic-tac-toe:
+ Quantum tic-tac-toe:
+ Bidding tic-tac-toe:
-
-
- Discrete bidding games:
+ Gobblet tic-tac-toe:
and
+ Probably the most famous movie occurrence of
tic-tac-toe is from Wargames; the clip
is here (the entire
movie is online here,
start around 1:44:17; this was a classic movie from my childhood).
- Clip (joint with students from OIT) on Duality: |
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by Rosen, Kenneth
Chapter 6 - Section 6.3 - Permutations and Combinations - Exercises - Page 414: 28
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5.7 The tangent function
| | | |
---
| 5.6 The cosine function | | 5.8 Summary |
5.7 The tangent function (EMBH8)
Revision (EMBH9)
Functions of the form y=tanθ for 0°≤θ≤360°
The dashed vertical lines are called the asymptotes. The asymptotes are at the values
of θ where tanθ is not defined.
Period: 180°
Domain: {θ:0°≤θ≤360°,θ≠90°;270°}
Range: \left{f(\theta):f(\theta)\in ℝ\right}
x-intercepts: \left(\text{0}\text{°};0\right),
\left(\text{180}\text{°};0\right),
\left(\text{360}\text{°};0\right)
y-intercept: \left(\text{0}\text{°};0\right)
Asymptotes: the lines \theta =\text{90}\text{°} and \theta
=\text{270}\text{°}
Functions of the form y = a \tan \theta + q
Tangent functions of the general form y = a \tan \theta + q, where a and
q are constants.
The effects of a and q on f(\theta) = a \tan \theta + q:
The effect of q on vertical shift
For q>0, f(\theta) is shifted vertically
upwards by q units.
For q<0, f(\theta) is shifted vertically
downwards by q units.
The effect of a on shape
For a>1, branches of f(\theta) are
steeper.
For 0<a<1, branches of f(\theta) are
less steep and curve more.
For a<0, there is a reflection about the
x-axis.
For -1 < a < 0, there is a reflection about
the x-axis and the branches of the graph are
less steep.
For a < -1, there is a reflection about the
x-axis and the branches of the graph are
steeper.
| | | |
---
| | a<0 | a>0 |
| q>0 | | |
| q=0 | | |
| q<0 | | |
Revision
Textbook Exercise 5.28
On separate axes, accurately draw each of the following functions for
(\text{0}\text{°} \leq \theta \leq \text{360}\text{°}):
Use tables of values if necessary.
Use graph paper if available.
For each function determine the following:
Period
Domain and range
(x)- and (y)-intercepts
Asymptotes
(y_1 = \tan \theta - \frac{1}{2})
(y_2 = - 3 \tan \theta)
(y_3 = \tan \theta + 2)
(y_4 = 2 \tan \theta - 1)
Functions of the form y=\tan (k\theta) (EMBHB)
The effects of k on a tangent graph
Complete the following table for y_1 = \tan \theta for
-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}:
| | | | | | | | |
--- --- --- --- |
| θ | -\text{360}\text{°} | -\text{300}\text{°} | -\text{240}\text{°} | -\text{180}\text{°} | -\text{120}\text{°} | -\text{60}\text{°} | \text{0}\text{°} |
| \tan \theta | | | | | | | |
| θ | \text{60}\text{°} | \text{120}\text{°} | \text{180}\text{°} | \text{240}\text{°} | \text{300}\text{°} | \text{360}\text{°} | |
| \tan \theta | | | | | | | |
2. Use the table of values to plot the graph of y_1 = \tan \theta
for -\text{360}\text{°} \leq \theta \leq
\text{360}\text{°}.
3. On the same system of axes, plot the following graphs:
y_2 = \tan (-\theta)
y_3 = \tan 3\theta
y_4 = \tan \frac{\theta}{2}
Use your sketches of the functions above to complete the following
table:
| | | | | |
--- ---
| | y_1 | y_2 | y_3 | y_4 |
| period | | | | |
| domain | | | | |
| range | | | | |
| y-intercept(s) | | | | |
| x-intercept(s) | | | | |
| asymptotes | | | | |
| effect of k | | | | |
5. What do you notice about y_1 = \tan \theta and y_2 = \tan
(-\theta)?
6. Is \tan (-\theta) = -\tan \theta a true statement? Explain your
answer.
7. Can you deduce a formula for determining the period of y = \tan k\theta?
The effect of the parameter on y = \tan k\theta
The value of k affects the period of the tangent function. If k is negative, then the
graph is reflected about the y-axis.
For k > 0:
For k > 1, the period of the tangent function decreases.
For 0 < k < 1, the period of the tangent function increases.
For k < 0:
For -1 < k < 0, the graph is reflected about the y-axis and the
period increases.
For k < -1, the graph is reflected about the y-axis and the period
decreases.
Negative angles:
\tan (-\theta) = -\tan \theta
Calculating the period:
To determine the period of y = \tan k\theta we use,
\text{Period} = \frac{\text{180}\text{°}}{|k|}
where |k| is the absolute value of k.
| | |
--- |
| k > 0 | k < 0 |
| | |
Worked example 26: Tangent function
Sketch the following functions on the same set of axes for
-\text{180}\text{°} \leq \theta \leq
\text{180}\text{°}.
y_1 = \tan \theta
y_2 = \tan \frac{3\theta}{2}
For each function determine the following:
Period
Domain and range
x- and y-intercepts
Asymptotes
Examine the equations of the form y = \tan k\theta
Notice that k > 1 for y_2 = \tan \frac{3\theta}{2}, therefore the
period of the graph decreases.
Complete a table of values
| | | | | | | | | | |
--- --- --- --- --- |
| θ | -\text{180}\text{°} | -\text{135}\text{°} | -\text{90}\text{°} | -\text{45}\text{°} | \text{0}\text{°} | \text{45}\text{°} | \text{90}\text{°} | \text{135}\text{°} | \text{180}\text{°} |
| \tan \theta | \text{0} | \text{1} | undef | -\text{1} | \text{0} | \text{1} | undef | -\text{1} | \text{0} |
| \tan \frac{3\theta}{2} | undef | -\text{0,41} | \text{1} | -\text{2,41} | \text{0} | \text{2,41} | -\text{1} | \text{0,41} | undef |
Sketch the tangent graphs
Complete the table
| | | |
---
| | y_1 = \tan \theta | y_2 = \tan \frac{3\theta}{2} |
| period | \text{180}\text{°} | \text{120}\text{°} |
| domain | {\theta: -\text{180}\text{°} \leq \theta \leq \text{180}\text{°}, \theta \ne -\text{90}\text{°}; \text{90}\text{°}} | {\theta: -\text{180}\text{°} < \theta < \text{180}\text{°}, \theta \ne -\text{60}\text{°}; \text{60}\text{°}} |
| range | {f(\theta): f(\theta) \in \mathbb{R}} | {f(\theta): f(\theta) \in \mathbb{R}} |
| y-intercept(s) | (\text{0}\text{°};0) | (\text{0}\text{°};0) |
| x-intercept(s) | (-\text{180}\text{°};0), (\text{0}\text{°};0) and (\text{180}\text{°};0) | (-\text{120}\text{°};0), (\text{0}\text{°};0) and (\text{120}\text{°};0) |
| asymptotes | \theta = -\text{90}\text{°} and \theta = \text{90}\text{°} | \theta = -\text{180}\text{°}; -\text{60}\text{°} and \text{180}\text{°} |
Discovering the characteristics
For functions of the general form: f(\theta) = y =\tan k\theta:
Domain and range
The domain of one branch is { \theta: -\frac{\text{90}\text{°}}{k} <
\theta < \frac{\text{90}\text{°}}{k}, \theta \in \mathbb{R}} because
f(\theta) is undefined for \theta = -\frac{\text{90}\text{°}}{k}
and \theta = \frac{\text{90}\text{°}}{k}.
The range is { f(\theta): f(\theta) \in \mathbb{R} } or (-\infty; \infty).
Intercepts
The x-intercepts are determined by letting f(\theta) = 0 and solving for
\theta.
The y-intercept is calculated by letting \theta = 0 and solving for
f(\theta).
\begin{align}
y &= \tan k\theta \
&= \tan \text{0}\text{°} \
&= 0
\end{align}
This gives the point (\text{0}\text{°};0).
Asymptotes
These are the values of k\theta for which \tan k\theta is undefined.
Tangent functions of the form (y = \tan k\theta)
Textbook Exercise 5.29
Sketch the following functions for (-\text{180}\text{°} \leq
\theta \leq \text{180}\text{°}). For each graph determine:
Period
Domain and range
(x)- and (y)-intercepts
Asymptotes
(f(\theta) =\tan 2\theta)
(g(\theta) =\tan \frac{3\theta}{4})
(h(\theta) =\tan (-2\theta))
(k(\theta) =\tan \frac{2\theta}{3})
Functions of the form y=\tan\left(\theta +p\right) (EMBHC)
We now consider tangent functions of the form y = \tan(\theta + p) and the effects of
parameter p.
The effects of p on a tangent graph
On the same system of axes, plot the following graphs for
-\text{360}\text{°} \leq \theta \leq
\text{360}\text{°}:
y_1 = \tan \theta
y_2 = \tan (\theta - \text{60}\text{°})
y_3 = \tan (\theta - \text{90}\text{°})
y_4 = \tan (\theta + \text{60}\text{°})
y_5 = \tan (\theta + \text{180}\text{°})
Use your sketches of the functions above to complete the following
table:
| | | | | | |
--- --- --- |
| | y_1 | y_2 | y_3 | y_4 | y_5 |
| period | | | | | |
| domain | | | | | |
| range | | | | | |
| y-intercept(s) | | | | | |
| x-intercept(s) | | | | | |
| asymptotes | | | | | |
| effect of p | | | | | |
The effect of the parameter on y = \tan(\theta + p)
The effect of p on the tangent function is a horizontal shift (or phase shift); the
entire graph slides to the left or to the right.
For p > 0, the graph of the tangent function shifts to the left by
p.
For p < 0, the graph of the tangent function shifts to the right by
p.
| | |
--- |
| p > 0 | p < 0 |
| | |
Worked example 27: Tangent function
Sketch the following functions on the same set of axes for
-\text{180}\text{°} \leq \theta \leq
\text{180}\text{°}.
y_1 = \tan \theta
y_2 = \tan (\theta + \text{30}\text{°})
For each function determine the following:
Period
Domain and range
x- and y-intercepts
Asymptotes
Examine the equations of the form y = \tan (\theta + p)
Notice that for y_1 = \tan \theta we have p = \text{0}\text{°}
(no phase shift) and for y_2 = \tan (\theta +
\text{30}\text{°}) we have p = \text{30}\text{°} > 0
and therefore the graph shifts to the left by
\text{30}\text{°}.
Complete a table of values
| | | | | | | | | | |
--- --- --- --- --- |
| θ | -\text{180}\text{°} | -\text{135}\text{°} | -\text{90}\text{°} | -\text{45}\text{°} | \text{0}\text{°} | \text{45}\text{°} | \text{90}\text{°} | \text{135}\text{°} | \text{180}\text{°} |
| \tan \theta | \text{0} | \text{1} | undef | -\text{1} | \text{0} | \text{1} | undef | -\text{1} | \text{0} |
| \tan (\theta + \text{30}\text{°}) | \text{0,58} | \text{3,73} | -\text{1,73} | -\text{0,27} | \text{0,58} | \text{3,73} | -\text{1,73} | -\text{0,27} | \text{0,58} |
Sketch the tangent graphs
Complete the table
| | | |
---
| | y_1 = \tan \theta | y_2 = \tan (\theta + \text{30}\text{°}) |
| period | \text{180}\text{°} | \text{180}\text{°} |
| domain | { \theta: -\text{180}\text{°} \leq \theta \leq \text{180}\text{°}, \theta \ne -\text{90}\text{°}; \text{90}\text{°} } | { \theta: -\text{180}\text{°} \leq \theta \leq \text{180}\text{°}, \theta \ne -\text{120}\text{°}; \text{60}\text{°} } |
| range | (-\infty;\infty) | (-\infty;\infty) |
| y-intercept(s) | (\text{0}\text{°};0) | (\text{0}\text{°};\text{0,58}) |
| x-intercept(s) | (-\text{180}\text{°};0), (\text{0}\text{°};0) and (\text{180}\text{°};0) | (-\text{30}\text{°};0) \text{ and } (\text{150}\text{°};0) |
| asymptotes | \theta = -\text{90}\text{°} \text{ and } \theta = \text{90}\text{°} | \theta = -\text{120}\text{°} \text{ and } \theta = \text{60}\text{°} |
Discovering the characteristics
For functions of the general form: f(\theta) = y =\tan (\theta + p):
Domain and range
The domain of one branch is { \theta: \theta \in (-\text{90}\text{°} - p;
\text{90}\text{°} - p) } because the function is undefined for \theta
= -\text{90}\text{°} - p and \theta = \text{90}\text{°} - p.
The range is { f(\theta): f(\theta) \in \mathbb{R} }.
Intercepts
The x-intercepts are determined by letting f(\theta) = 0 and solving for
\theta.
The y-intercept is calculated by letting \theta = \text{0}\text{°} and
solving for f(\theta).
\begin{align}
y &= \tan (\theta + p) \
&= \tan (\text{0}\text{°} + p) \
&= \tan p
\end{align}
This gives the point (\text{0}\text{°};\tan p).
Tangent functions of the form (y = \tan (\theta + p))
Textbook Exercise 5.30
Sketch the following functions for (-\text{360}\text{°} \leq
\theta \leq \text{360}\text{°}).
For each function, determine the following:
Period
Domain and range
(x)- and (y)-intercepts
Asymptotes
(f(\theta) =\tan (\theta + \text{45}\text{°}))
(g(\theta) =\tan (\theta - \text{30}\text{°}))
(h(\theta) =\tan (\theta + \text{60}\text{°}))
Sketching tangent graphs (EMBHD)
Worked example 28: Sketching a tangent graph
Sketch the graph of f(\theta) = \tan \frac{1}{2}(\theta -
\text{30}\text{°}) for -\text{180}\text{°} \leq \theta
\leq \text{180}\text{°}.
Examine the form of the equation
From the equation we see that 0 < k < 1, therefore the branches of
the graph will be less steep than the standard tangent graph y = \tan
\theta. We also notice that p < 0 so the graph will be shifted
to the right on the x-axis.
Determine the period
The period for f(\theta) = \tan \frac{1}{2}(\theta -
\text{30}\text{°}) is:
\begin{align}
\text{Period} &= \frac{\text{180}\text{°}}{|k|} \
&= \dfrac{\text{180}\text{°}}{\frac{1}{2}} \
&= \text{360}\text{°}
\end{align}
Determine the asymptotes
The standard tangent graph, y = \tan \theta, for
-\text{180}\text{°} \leq \theta \leq \text{180}\text{°} is
undefined at \theta = -\text{90}\text{°} and \theta =
\text{90}\text{°}. Therefore we can determine the asymptotes of
f(\theta) = \tan \frac{1}{2}(\theta - \text{30}\text{°}):
\frac{-\text{90}\text{°}}{\text{0,5}} +
\text{30}\text{°} = -\text{150}\text{°}
\frac{\text{90}\text{°}}{\text{0,5}} + \text{30}\text{°}
= \text{210}\text{°}
The asymptote at \theta = \text{210}\text{°} lies outside the
required interval.
Plot the points and join with a smooth curve
Period: \text{360}\text{°}
Domain: { \theta: -\text{180}\text{°} \leq \theta \leq
\text{180}\text{°}, \theta \ne -\text{150}\text{°} }
Range: (-\infty;\infty)
y-intercepts: (\text{0}\text{°};-\text{0,27})
x-intercept: (\text{30}\text{°};0)
Asymptotes: \theta = -\text{150}\text{°}
The tangent function
Textbook Exercise 5.31
Sketch the following graphs on separate axes:
(y = \tan \theta - 1) for (-\text{90}\text{°} \leq
\theta \leq \text{90}\text{°})
(f(\theta) = -\tan 2\theta) for (\text{0}\text{°}
\leq \theta \leq \text{90}\text{°})
(y = \frac{1}{2} \tan (\theta + \text{45}\text{°}))
for (\text{0}\text{°} \leq \theta \leq
\text{360}\text{°})
(y = \tan (30° - \theta)) for (-180° \leq \theta
\leq 180° )
Given the graph of (y = a \tan k\theta), determine the values of
(a) and (k).
(a = -1); (k = \frac{1}{2})
Mixed exercises
Textbook Exercise 5.32
Determine the equation for each of the following:
(f(\theta) = a \sin k\theta) and (g(\theta) = a \tan
\theta)
(f(\theta) = \frac{3}{2} \sin 2\theta) and (g(\theta) =
-\frac{3}{2} \tan \theta)
(f(\theta) = a \sin k\theta) and (g(\theta) = a \cos (
\theta + p))
(f(\theta) = -2 \sin \theta) and (g(\theta) = 2 \cos (\theta
+ \text{360}\text{°}))
(y = a \tan k\theta)
(y = 3 \tan \frac{\theta}{2})
(y = a \cos \theta + q)
(y = y = 2 \cos \theta + 2)
Given the functions (f(\theta) = 2 \sin \theta) and (g(\theta) =
\cos \theta + 1):
Sketch the graphs of both functions on the same system of
axes, for (\text{0}\text{°} \leq \theta \leq
\text{360}\text{°}). Indicate the turning points
and intercepts on the diagram.
What is the period of (f)?
(\text{360})(\text{°})
What is the amplitude of (g)?
Use your sketch to determine how many solutions there are for
the equation (2 \sin \theta - \cos \theta = 1). Give
one of the solutions.
At (\theta = \text{180}\text{°})
Indicate on your sketch where on the graph the solution to
(2 \sin \theta = -1) is found.
Determine the values of (a), (b) and (c).
(a = 2), (b = -1) and (c =
\text{240}\text{°})
What is the period of (g)?
Solve the equation (\cos \theta = \frac{1}{2}) graphically
and show your answer(s) on the diagram.
(\theta = \text{60}\text{°};
\text{300}\text{°})
Determine the equation of the new graph if (g) is reflected
about the (x)-axis and shifted to the right by
(\text{45}\text{°}).
(y = - \tan (\theta - \text{45}\text{°}))
Sketch the graphs of (y_1 = -\frac{1}{2} \sin (\theta +
\text{30}\text{°})) and (y_2 = \cos (\theta -
\text{60}\text{°})), on the same system of axes for
(\text{0}\text{°} \leq \theta \leq
\text{360}\text{°}).
| | | |
---
| Previous 5.6 The cosine function | Table of Contents | Next 5.8 Summary | |
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Abstract
Introduction
Section snippets
References (175)
Brain Research
Volume 1863, 15 September 2025, 149754
The role of NOTCH3 in CADASIL pathogenesis: insights into novel therapies
Author links open overlay panel, , ,
rights and content
Highlights
¢
CADASIL is a genetic small vessel disease caused by mutations in the NOTCH3 gene.
¢
Mutant NOTCH3 disrupts vascular smooth muscle cells, leading to white matter damage.
¢
Toxic NOTCH3 aggregation and GOM deposits are central to CADASIL pathogenesis.·
¢
New therapies like base editing and immunotherapy show promise for targeting NOTCH3.
¢
This review links disease mechanisms to emerging therapies in a translational framework.
Abstract
Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL) is a monogenetic hereditary small-vessel disorder characterised by recurrent subcortical ischemic strokes, cognitive deterioration, and other neurological symptoms. Single nucleotide mutations within the NOTCH3 gene can impair NOTCH3 processing and/or signalling, resulting in the accumulation of granular osmiophilic material (GOM) in blood vessel walls, and consequently CADASIL. Despite its significant clinical impact, there is currently no definitive treatment for CADASIL. This review provides a comprehensive analysis of the pathophysiological mechanisms underlying CADASIL, focusing on NOTCH3 mutations and their effects on protein processing and signalling. The review proposes a hypothesis that explains how NOTCH3 mutations may alter the signalling process and result in GOMs. Additionally, the review explores published therapy strategies aimed at restoring normal NOTCH3 function.
Graphical abstract
Introduction
The acronym CADASIL for cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy was suggested for the disease formerly termed hereditary multi-infarct dementia by Sourander in 1977 (Bousser and Tournier-Lasserve, 1993). CADASIL was coined by Professor Marie-Germaine Bousser in 1993 to describe a hereditary cerebral small-vessel disease with distinctive clinical and pathological features (Joutel et al., 1993, Joutel and Tournier-Lasserve, 2002). CADASIL is characterised by recurrent subcortical ischemic strokes and sometimes leads to pseudobulbar palsy and dementia (Chabriat and Bousser, 2007).
Importantly, the symptoms and prognosis of small vessel ischemic strokes, such as those seen in CADASIL, differ from other types of strokes (Arboix et al., 2010). These findings highlight that small vessel strokes have distinct clinical and risk factor profiles, which should be considered when managing CADASIL patients.
Other symptoms include migraine headaches (Tan and Markus, 2016), cognitive deterioration (Buffon et al., 2006), seizures (Oh et al., 2016), vision problems (Pretegiani et al., 2013), slow movements and tremors (parkinsonism) (Ragno et al., 2013), acute vestibular syndrome (Rufa et al., 2008), apathy and depression (Silva et al., 2023). Magnetic resonance imaging (MRI) of the brain displays signs of small deep strokes (infarcts) due to the blockage of blood vessels (Chabriat and Bousser, 2007, Caplan, 2015). MRI also displays leukoencephalopathy, characterised by structural changes in the white matter of the brain where myelin suffers significant damage (Bousser MG, Tournier-Lasserve E. Summary of the proceedings of the First International Workshop on CADASIL. Paris, May 19-21, 1993).
In addition to CADASIL, other disorders may present with overlapping clinical and radiological features, which can mimic cerebral small vessel disease through their effects on cerebral vasculature and stroke risk. These include CARASIL (a recessive hereditary small vessel disease caused by HTRA1 mutations) (Fukutake, 2011), mitochondrial encephalopathy (a group of disorders caused by mitochondrial DNA mutations leading to energy metabolism defects) (Na and Lee, 2024), lactic acidosis (a metabolic condition characterised by elevated lactic acid often linked to mitochondrial dysfunction) (Na and Lee, 2024), stroke-like episodes (transient neurological deficits mimicking stroke but with metabolic origins) (Na and Lee, 2024), and Fabrys disease (an X-linked lysosomal storage disorder causing vascular and systemic complications) (Ezgu et al., 2022).
CADASIL is inherited in an autosomal dominant pattern. Initial genetic linkage analysis in two unrelated families demonstrated that the disease locus is located on chromosome 19 (Joutel et al., 1993, Tournier-Lasserve et al., 1993). Further work investigated 8 unrelated European families and prompted the First International Workshop on CADASIL in 1993 (Bousser MG, Tournier-Lasserve E. Summary of the proceedings of the First International Workshop on CADASIL. Paris, May 19-21, 1993). The workshop aimed to establish the state of research into CADASIL and establish a basis for a European study group (Bousser MG, Tournier-Lasserve E. Summary of the proceedings of the First International Workshop on CADASIL. Paris, May 19-21, 1993). Since the first international workshop in 1993, 1702 papers have been published in CADASIL (PubMed search date: 25/01/2024). However, a definitive treatment for the disease remains elusive. CADASIL is the most common small vessel disease (SVD) resulting from mutations in a single gene (Cramer et al., 2024). The prevalence of CADASIL has been estimated at approximately 5:100,000, although it is likely higher due to under-diagnosis (Cramer et al., 2024). There is no gender or ethnic predilection (Cramer et al., 2024). Further genetic studies have demonstrated that single nucleotide mutations within NOTCH3 (19p13.2-p13.1) cause CADASIL (Chabriat et al., 1997, Salloway and Hong, 1998). This article aims to review the pathophysiology of CADASIL and efforts to ameliorate or potentially treat the disease.
Section snippets
NOTCH3 variants in CADASIL
We used NCBI (ClinVar) to review current knowledge on NOTCH3 variants in CADASIL using the following Boolean connectors (search date: 20 Feb 2024) (Landrum et al., 2018): (Cerebral arteriopathy, autosomal dominant, with subcortical infarcts and leukoencephalopathy) AND NOTCH3[Gene Name]. While ClinVar provides a traceable and curated snapshot of clinically annotated NOTCH3 variants, it is not comprehensive. Several pathogenic variants, particularly those reported in the literature but not
Overview of NOTCH proteins
NOTCH1-4 proteins are intramembrane signalling receptors found in mammals. They mediate cellcell communication to regulate various processes including stem cell differentiation (Liu et al., 2010), tissue development (Siebel and Lendahl, 2017), apoptosis (Dang, 2012), and proliferation (Khoramjoo et al., 2022). These proteins are also involved in maintaining and repairing cells (Engler et al., 2018, Gude and Sussman, 2012). NOTCH proteins have two main parts: the extracellular domain and the
Physiology of NOTCH signalling
Before describing how mutations within the NOTCH3 gene lead to CADASIL, we review how NOTCH signalling normally works. NOTCH biosynthesis has been well characterised. See (Zhou et al., 2022, van Tetering and Vooijs, 2011) for review. NOTCH is synthesised in the endoplasmic reticulum (ER) and cleaved in the Golgi (S1 cleavage) by a Furin-like protease to separate the NECD from the remaining component of NOTCH (Blaumueller et al., 1997). The remaining component is termed Notch Extracellular
Granular osmophilic material
Granular osmiophilic material (GOM) is a pathological phenotype found in the blood vessel walls (Bou-Gharios et al., 2004, Tucker et al., 2024) of individuals with CADASIL as shown in Fig. 4 (LaPoint et al., 2000). GOM deposition around vascular smooth muscle cells (VSMCs) (Lewandowska et al., 2011) and pericytes (Ruchoux et al., 2021) are one of the earliest histological signs observed in CADASIL patients, making it an important diagnostic marker (Lorenzi et al., 2017).
In 2009, a retrospective
Impaired NOTCH3 signalling
EGF-like repeats are protein domains that contain six cysteine residues. These residues form three disulfide bonds and play an important role in establishing the three-dimensional structure of the NECD (Kopan and Ilagan, 2009, Chillakuri et al., 2012). Lee and colleagues have demonstrated that mutations affecting cysteine residues can cause abnormal mobility of NOTCH3 on non-reducing gels, which indicates a shift in its conformational configuration (Lee et al., 2023). Both the loss and an
Gene therapy treatment targeting NOTCH3
Despite the current advances in CADASIL research, there is still no definitive treatment for the disease (Bersano et al., 2017). The current treatments only address the underlying symptoms, such as anti-depressant, anti-hypertensive, or anti-convulsant medications to prevent migraines in CADASIL patients (Glover et al., 2020). Since CADASIL is a monogenetic hereditary disorder, base editing approaches can be used to replace the mutant SNV with the wild-type SNV.
Wang and colleagues used human
Other therapy treatments targeting NOTCH3
There are other strategies for treating CADASIL, including immunotherapy, growth factors, and peptides. Active immunisation of transgenic mice carrying R182C mutation (TgNotchR182C) with aggregates of certain compounds can decrease the deposition of NOTCH3 around brain capillaries and lower serum levels of NOTCH3 ECD (Oliveira et al., 2023). The results demonstrated that treated mice did not show signs of inflammation, kidney toxicity, neurodegeneration, or loss of VSMCs in the vasculature.
Limitations
While this review provides a detailed overview of NOTCH3 variants and their involvement in CADASIL pathogenesis, a few limitations warrant consideration.
First, relying on publicly available databases such as NCBI ClinVar to identify disease-causing NOTCH3 variants may not capture the full spectrum of NOTCH3 variations, particularly those not submitted to ClinVar. However, ClinVar was selected because it offers a publicly accessible, standardised, and traceable dataset of clinical variant
Conclusion
CADASIL is a challenging disorder with substantial clinical implications, necessitating the development of targeted therapeutic interventions. The pathogenesis of CADASIL involves the failure of NOTCH3 protein elimination, resulting in the accumulation of GOM and vascular dysfunction. Gene therapy holds promise for restoring normal NOTCH3 function. Using a single-AAV base-editing system, adenine base editors represent a potential avenue of interest as most pathogenic mutations are C > T.
Method
All images were made using BioRender. Clinical variance data from NCBI was plotted using ggplot2 (version 3.4.4, in R (version 4.3.2). For section 4.2, Pearson's Chi-squared test with Yates' continuity correction was calculated in R using chisq.test function.
CRediT authorship contribution statement
Favour Felix-Ilemhenbhio: Writing review & editing, Writing original draft, Visualization, Methodology, Investigation, Formal analysis, Data curation, Conceptualization. Klaudia Kocsy: Writing review & editing. Mimoun Azzouz: Writing review & editing, Resources, Supervision. Arshad Majid: Writing review & editing, Resources, Supervision, Funding acquisition, Conceptualization.
Funding
F.F-I was funded by the National Institute for Health and Care Research (NIHR) Sheffield Biomedical Research Centre (NIHR203321). The views expressed are those of the author(s) and not necessarily those of the NIHR.
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
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7339 | https://www.learnenglish.com/grammar/simple-present-tense/simple-present-tense-explanation/ | GRAMMARSIMPLE PRESENT TENSE
Simple Present Tense (do-does) – With Usage, Pictures and Example Sentences
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Simple present tense (present simple tense) is a verb tense that describes the events and situations that do not change over time. We use simple present tense for the actions that happen regualarly, and that is why we use some frequency adverbs to express these repititive actions. “Do” and “does” are the auxiliary verbs of present simple tense. However, “do” and “does” are not used in positive sentences. They are used only in negative and question sentences. The auxiliary verb “does” is used for third person singular (He, She, It). In other subjects (I, You, We, They), the auxiliary verb “do” is used.
The following animated sentences are examples of present simple tense:
Does Why doesWhere does
Emily
likeseatseatdoes not eateatbuy
ice-cream.
QUICK LINKS IN THIS PAGE
⬤ Formation of simple present tense
For affirmative sentences we use the infinite form of the verb, but for negative sentences we use “not” after the auxiliary “do” and “does”. For questions we put “do/does” before the subject.
⬤ Which auxiliary (helping verb) to use for simple present tense?
The auxiliary verbs in simple present tense are “do / does“. However we use “am, is, are” to talk about a general state or condition.
Examples:
I work in the office.
I don’t work in the office.
Do you work in the office?
I am in the office.
I am in the office.
Are you in the office?
SIMILAR PAGES:
❯❯ Learn verb to be here
❯❯ Learn present continuous tense here
❯❯ Learn future simple tense (will) here
❯❯ Learn be going to future tense here
❯❯ Learn simple past tense here
❯❯ Learn past continuous tense here
❯❯ Learn present perfect tense here
⬤ Positive (Affirmative) sentences
For the formation of positive sentences in simple present tense we do not use “do” or “does” in front of the verb. This may sound strange. Because we know that the auxiliary verbs that precede the verbs help us understand the tense of the sentence. However, the verb is alone here. In addition, for the subjects “He, She, It”, the suffix “-s” is added at the end of the verb.
I like pizza.
We go abroad every summer.
She speaks three languages.
Lions eat meat.
Oliver walks to school.
They play computer games.
Maria prefers action movies.
✎ NOTE:
The verb “have” changes into “has” when it gets “-s” at the end.
She has breakfast.
She doesn’t have breakfast.
Does she have breakfast?
✎ NOTE:
In positive sentences, when the subject is “He, She, It”, we change “-s” into “-es” or “-ies“. The reason is as follows:
For the verbs ending with “-s, -ss, -sh, -ch, -x, -o” we add “-es” at the end of the verb.
brushes, kisses, catches, fixes, goes, does etc.
If the verb has “-y” at the end and it precedes with a consonant letter, we drop the “-y” and add “-ies”
try> tries, fly> flies, carry> carries etc.
⬤ Negative sentences
For the formation of negative sentences in simple present tense we use “not” together with “do / does“. The short forms are “don’t / doesn’t”
I don’t like cigarette.
Sarah doesn’t need help.
He doesn’t forget names.
They don’t go out alone.
⬤ Interrogative sentences (questions)
For the formation of question sentences (interrogative) in simple present tense we put “do / does” before the subject. This also applies to the “Wh- questions” which we call “information questions” as well.
Do you like lemonade?
Does she keep secret?
Where do you live?
Why does Molly wear pink dresses?
⬤ Sentence forms in simple present tense
Simple present tense Formula with example sentences
| (+) Affirmative sentences | (-) Negative sentences | (?) Interrogative sentences |
| I speak | I don’t speak | Do you speak? |
| You speak | You don’t speak | Do you speak? |
| He speaks | He doesn’t speak | Does he speak? |
| She speaks | She doesn’t speak | Does she speak? |
| It speaks | It doesn’t speak | Does it speak? |
| We speak | We don’t speak | Do we speak? |
| They speak | They don’t speak | Do they speak? |
⬤ Example Sentences
(+) They like basketball.
(-) They don’t like basketball.
(?) Do they like basketball?
(?) What do they like?
(+) He likes basketball.
(-) He doesn’t like basketball.
(?) Does he like basketball?
(?) What does he like?
⬤ Explanations and usages of simple present tense
Lets go on with the explanations, usages and time adverbs of simple present tense
⬤ 1- Facts or generalizations.
It is used for events and situations that never change.
Examples:
People need food in this village.
The wind blows a lot here.
Trains carry many passengers.
Smart phones cause some health problems.
Water boils at 100C degrees.
The earth revolves around the Sun.
Lions don’t eat grass.
Plants give us oxygen.
Pandas live in China.
⬤ 2- Habits, routines or repeated actions.
We use simple present tense to talk about habits and routines.
Examples:
I get up at 8 o’clock every day.
Do you drink coffee every day?
She always brushes her teeth.
Leo plays his guitar in his room.
⬤ 3- Likes and dislikes
We can also use simple present tense to talk about likes and dislikes. Some commonly used verbs are as follows: “like”,”love”, “hate”, “dislike”, “enjoy” etc.
Examples:
I love ice-cream.
Do you like playing chess?
She hates lies.
Sandra doesn’t like tea with sugar.
We like to swim.
I dislike cold weather.
⬤ 4- Scheduled events in near future
Examples:
The train arrives at 9 AM.
The bus arrives at 3 o’clock in the afternoon.
When do we board the plane?
When does the wedding ceremony start?
The films starts at 8.30
The bus arrives at 6PM.
The English lesson starts at 10.30.
⬤ 5- With non-progressive verbs
Some verbs in English doesn’t have continuous forms. These verbs are called non-progressive or non-continuous verbs. We use simple present tense with these verbs.
Some non-progressive verbs are: believe, know, remember, understand, need, hate, like, love, prefer, want, feel, mean etc.
Examples:
I know the answer. – CORRECT
I am knowing the answer. – INCORRECT
I want some sugar. – CORRECT
I am wanting some sugar. – INCORRECT
Brian feels cold. – CORRECT
Brian isn’t feeling cold. – INCORRECT
I don’t remember her name. – CORRECT
I am not remembering her name. – INCORRECT
Do you understand? – CORRECT
Are you understanding? – INCORRECT
⬤ 6- Narrating events
Example:The man opens the door and goes out slowly. He looks around carefully. Then he sees a little cat under the tree. He grabs it and says “Oh. Are you hungry?”
⬤ Adverbs of Frequency
What are Adverbs of frequency?
Simple present tense indicates repetitive actions, so it is good to say the frequency of these actions. In this case, we use some words called “Frequency adverbs”. These words are used just before the verb. This list of frequency adverbs is as follows:
⬤ List of frequency adverbs
always
usually
generally
often
normally
frequently
sometimes
occasionally
seldom
rarely
hardly ever
never
➔ Examples of frequency adverbs
Read the example sentences with the frequency adverbs and try to make similar sentences.
Examples:
I often eat eggs for the breakfast.
I never smoke.
Lisa always walks to school.
Do you usually get up early?
You hardly ever say “Thanks”.
They don’t normally go out for dinner.
My father often forgets my birthday.
I generally have breakfast before I go out.
⬤ Memory cards to learn the adverbs of frequency
The cards below have adverbs of frequency. Click on them and try to say the meaning of them in your native language.
always
often
generally
sometimes
normally
occasionally
usually
seldom
rarely
hardly ever
never
frequently
⬤ Time adverbs to use in simple present tense.
every ….
I play football ever weekend.
We go holiday ever summer.
She gets up late every day.
once, twice, three times, ten times etc.
I go to cinema once a month.
She goes out with her friends twice a week.
I call my son at least 3 times a day.
on Mondays, at the weekends, in the mornings
I call my grandparents on Saturdays.
In the evenings I take a taxi to go back home.
⬤ Verb to be (am, is, are)
We use am, is, are when we talk about a state rather than an action.
⬤ I work in London. (Action verb)
⬤ I am in London. (State verb)
➔ Action verbs examples (do, does)
Frank works at the hospital.
Frank doesn’t work at the hospital.
Does Frank work at the hospital?
Where does Frank work?
➔ Verb to be examples (am,is,are)
Frank is at the hospital.
Frank isn’t at the hospital.
Is Frank at the hospital?
Where is Frank?
⬤ Images and example sentences
Look at the images below and read the simple present tense sentences. Then try to understand the usage.
1 / 22
He loves his wife.
❮❯
⬤ A dialogue example
Here is a dialogue to learn simple present tense. Read and try to make similar dialogues.
Do you go holiday every summer?
Yes. I do.
I like summer holidays very much.
3. Where do you go for holiday?
4. In fact, I don’t go to seaside.
I like camping in the mountains.
5. Really?
Where do you go for camping?
6. It is up to us.
We usually decide before we go.
If you want, you can join us.
7. Why not?
⬤ Translate the sentences about simple present tense.
You will see some examples of simple present tense below. Translate them into your native language.
Who knows the answer?
I want some water, please.
He doesn' keep his room tidy.
We like to go to the park and feed the birds.
Steven never checks the emails at home.
I don't understand anything.
⬤ Sentence scramble game
You will see scrambled words of simple present tense sentences. Click on the words in the correct order to make meaningful sentences.
1/15
Where
do
you
work?
3
tries
❮❯
⬤ Example sentences to learn simple present tense
Here are examples of simple present tense in context below. Some are affirmative some sentences are negative and some are interrogative.
➔ 10 example sentences :
I always get up early.
I don’t like hot weather.
Mr. Anderson usually forgets to lock the door.
She keeps secrets.
Cats don’t like swimming.
She rarely writes emails.
Steven looks happy.
I like reading poems a lot.
The children brush their teeth every day.
When I buy something, I read the instructions.
⬤ Questions and answers
You will see 10 questions with their answers below. Try to understand the formation of the sentences.
➔ 10 questions and answers
How do you go to school?
I go to school by bus.
Do you like ice-cream?
Yes, I like ice-cream.
Why do you always wear sunglasses?
Because I have a problem with my eyes.
Does your father help your mother at home?
Yes, he does.
What time do you go to bed?
I go to bed at about 11 PM.
Do you make noise in the classroom?
No, I don’t.
Where are you from?
I am from Canada.
How often do you watch TV?
I sometimes watch TV.
Do you agree with me?
No, I don’t agree with you.
Are you OK?
Yes, I am OK.
⬤ Reading passage – Daily routine
My daily routine
My name is Lydia Collins. I live in a flat. My day daily routine starts very early. Every morning I wake up at six o’clock and wear my school uniform. Then I have breakfast with my father and mother. My little brother doesn’t have breakfast with us. Because it is too early for him. After breakfast I go out and wait for the school bus. At about 7 AM I get on the bus. I come back home at 3 PM. I feel tired when I come back. I have a rest and play with Dody, our cat. Then I start doing my homework. I try to finish it before 7 PM. So that I can go out and play with my friends. We have dinner at 8 PM. We often have chicken for dinner but I hate chicken. I eat it because mum gets angry. After dinner I watch TV for an hour. I go to bed early because I’m always very tired at the end of the day.
External resources:
You can also visit Wikibooks page to learn simple present tense, or watch a video for example sentences.
4.8/5 - (17 votes)
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Fill in the Blanks Quiz for Simple Present Tense
Sentence Scramble Game for Simple Present Tense |
7340 | https://www.albert.io/blog/the-best-ap-physics-1-review-guides/ | Skip to content
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AP® Physics 1
The Best AP® Physics 1 Review: Topic Summaries, Examples, and Free Practice
Welcome to Albert’s collection of science topic reviews for teaching and reviewing AP® Physics. Teachers and students can explore our easy-to-follow guides below for use at home or in the classroom.
Explore Albert's AP® Physics 1 Practice
Contents:
Unit 1 | Kinematics
Unit 2 | Force and Translational Dynamics
Unit 3 | Work, Energy, and Power
Unit 4 | Linear Momentum
Unit 5 | Torque and Rotational Dynamics
Unit 6 | Energy and Momentum of Rotating Systems
Unit 7 | Oscillations
Unit 8 | Fluids
Unit 1 | Kinematics
How do position, velocity, and acceleration relate to one another? How can kinematic equations be used to predict an object’s motion?
The articles below break down key kinematics concepts, from motion graphs to projectile motion. Whether you're studying one-dimensional movement or two-dimensional motion with vectors, these resources will help you master the fundamentals of motion.
1.1 | Scalars and Vectors in One Dimension
Review topic 1.2 | Displacement, Velocity, and Acceleration
Review topic 1.3 | Representing Motion
Review topic 1.4 | Reference Frames and Relative Motion
Review topic 1.5 | Vectors and Motion in Two Dimensions
Review topic
Unit 2 | Force and Translational Dynamics
How do forces affect the motion of objects? What role does Newton’s Laws play in understanding dynamics?
The review articles below cover fundamental force concepts, from free-body diagrams to Newton’s Laws and friction. Whether analyzing forces in equilibrium or exploring circular motion and gravity, these resources will help you understand the interactions governing motion.
2.1 | Systems and Center of Mass
Review topic 2.2 | Forces and Free-Body Diagrams
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Review topic 2.5 | Newton's Second Law
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Unit 3 | Work, Energy, and Power
How does energy transfer affect an object's motion? What is the relationship between work, energy, and power?
The review articles below explore key energy concepts, from kinetic and potential energy to the work-energy theorem and power. Whether you're analyzing forces doing work or understanding conservation of energy, these resources will help you master fundamental principles of energy transfer.
3.1 | Translational Kinetic Energy
Review topic 3.2 | Work: Conservative and Non-Conservative Forces
Review topic 3.2 | Work: Work-Energy Theorem
Review topic 3.3 | Potential Energy
Review topic 3.4 | Conservation of Energy: Law of Conservation of Energy
Review topic 3.4 | Conservation of Energy: Mechanical Energy
Review topic 3.5 | Power
Review topic
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How is momentum conserved in collisions and explosions? What is the relationship between impulse and momentum?
The review articles below explore momentum, impulse, and collisions, including how forces change an object’s motion over time. Whether analyzing elastic or inelastic collisions, these resources will help you understand the principles of momentum conservation and impulse-momentum relationships.
4.1 | Linear Momentum
Review topic 4.3 | Change in Momentum and Impulse
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How does torque influence rotational motion? What factors affect an object's rotational inertia?
The review articles below explore rotational motion, including torque, angular acceleration, and rotational inertia. Whether analyzing rotational equilibrium or applying Newton’s Laws to rotation, these resources will help you understand the principles governing rotational dynamics.
5.1 | Rotational Kinematics
Review topic 5.2 | Connecting Linear and Rotational Motion
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Review topic 5.4 | Rotational Inertia
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How is rotational energy related to linear energy? What role does angular momentum play in rotational motion?
The review articles below explore rotational energy, angular momentum, and rolling motion, including how torque and impulse affect rotating systems. Whether analyzing rotational kinetic energy or conservation of angular momentum, these resources will help you understand the fundamental principles of rotational motion.
6.1 | Rotational Kinetic Energy
Review topic 6.2 | Torque and Work
Review topic 6.3 | Angular Momentum and Angular Impulse
Review topic 6.4 | Conservation of Angular Momentum
Review topic 6.5 | Rolling
Review topic 6.6 | Motion of Orbiting Satellites
Review topic
Unit 7 | Oscillations
What causes simple harmonic motion (SHM), and how is it characterized? How do energy and frequency relate to oscillating systems?
The review articles below explore simple harmonic motion, including restoring forces, oscillations, and energy conservation in periodic motion. Whether analyzing pendulums, springs, or energy transformations in SHM, these resources will help you understand the fundamental principles of oscillatory motion.
7.1 | Defining Simple Harmonic Motion (SHM)
Review topic 7.2 | Frequency and Period of SHM
Review topic 7.3 | Representing and Analyzing SHM
Review topic 7.4 | Energy and Simple Harmonic Oscillators
Review topic
Unit 8 | Fluids
How do fluids behave under different forces and pressures? What principles govern fluid flow and buoyancy?
The review articles below explore fluid mechanics, including density, pressure, buoyancy, and fluid dynamics. Whether analyzing Archimedes' principle, Bernoulli’s equation, or the continuity equation, these resources will help you understand the fundamental behavior of fluids.
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Review topic 8.2 | Pressure
Review topic 8.3 | Fluids and Newton's Laws
Review topic 8.4 | Fluids and Conservation Laws
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A box of light bulbs contains exactly 3 light bulbs that are defective
gmatt1476
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A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs.
DS76851.01
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Post URL21 Sep 2019, 16:49
from the given, we know the number of defective bulbs,
but we don't know:
a- the number of non-defective bulbs
b- the number of the sample to be picked.
c- the method of picking (with or without replacement)
statement (1) provides us with info (c) only -->insufficient
statement (2) provides us with info (b) only -->insufficient
so both (1) and (2) are insufficient because they can't provide info (a)
for example, if the number of non-defective bulbs = 17, so the probability will be 1
and as the number of non-defective bulbs increase, the probability starts to decrease below 1
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Post URL10 Oct 2019, 06:53
MahmoudFawzy
from the given, we know the number of defective bulbs,
but we don't know:
a- the number of non-defective bulbs
b- the number of the sample to be picked.
c- the method of picking (with or without replacement)
statement (1) provides us with info (c) only -->insufficient
statement (2) provides us with info (b) only -->insufficient
so both (1) and (2) are insufficient because they can't provide info (a)
for example, if the number of non-defective bulbs = 17, so the probability will be 1
and as the number of non-defective bulbs increase, the probability starts to decrease below 1
Show more
How could not we know the info a? As the stimulus provides us with "exactly 3 defective bulbs" so i assume the rest are non-defective bulbs. What am i wrong here? Please shed some lights. Thank you
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nick1816
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Post URL15 Oct 2019, 08:04
We can't find the total number of bulbs in the box from statement 1 or 2. Hence, it's impossible to tell the probability that a sample of light bulbs picked will contain atleast 1 defective light bulb
If total number of bulbs are 21, probability that probability that a sample of 20 light bulbs picked will contain atleast 1 defective light bulb is equal to 1.
If total number of bulbs are 23, probability that probability that a sample of 20 light bulbs picked will contain atleast 1 defective light bulb is less than 1.
gmatt1476
A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs.
DS76851.01
Show more
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Post URL15 Oct 2019, 08:19
gmatt1476
A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs.
DS76851.01
Show more
Question stem- No.of defective light bulbs are 3.
Total no.of bulbs= unknown
We are asked to find -the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb.
We need to know total no.of bulbs present in the box and picking methods.
statement 1- Insufficient
Total no.of bulbs not known
Statement 2-insufficient
Picking method not known. We don't know whether 1 bulb is picked at a time, 2 picked at a time or with/without replacement.
.
Statement 1 + Statement 2
Insufficient.
Reason is we cannot arrive at a unique answer because question stem asks to find probability that atleast 1 defective bulb is picked.
There are 3 cases possible here-
one is defective, 2 defective, 3 are defective. The probability of these 3 cases gives different answer. Hence correct option is E.
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Post URL24 May 2020, 07:52
Straight forward: Since we do not know the sample size (total number of bulbs to be picked), we cannot know the probability of getting atleast 1 defective bulb
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Post URL24 May 2020, 08:08
gmatt1476
A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs.
DS76851.01
Show more
Given :-
i) There is a box of bulbs that has 3 defective bulbs. But we do not know the total number of bulbs, say n.
ii) We pick up certain bulbs, and are looking for the probability of at least 1 in them to be defective. Again we do not know the size of sample, say x.
Statements :-
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
Ok, but we are looking for the total bulbs and number of bulbs picked up as sample.
Insuff
(2) The sample will contain exactly 20 light bulbs.
x=20, but what about n. Insuff
Combined.
If n=x=20... Probability =1
But if n= 40, and x=20, probability <1.
Insuff
E
Chetan Sharma, Rimcollian
Absolute modulus 2. Fractions 3. Percentage increase 4. Combinations
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newspapersalesman
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Post URL03 Jun 2020, 22:20
MahmoudFawzy
from the given, we know the number of defective bulbs,
but we don't know:
a- the number of non-defective bulbs
b- the number of the sample to be picked.
c- the method of picking (with or without replacement)
statement (1) provides us with info (c) only -->insufficient
statement (2) provides us with info (b) only -->insufficient
so both (1) and (2) are insufficient because they can't provide info (a)
for example, if the number of non-defective bulbs = 17, so the probability will be 1
and as the number of non-defective bulbs increase, the probability starts to decrease below 1
Show more
You don't know what you are talking about bro. The reason why 1+2 is insfficient is that we don't know how many bulbs we pick. If we pick only one, the probability will be 3/20; If we pick two, the probability will be 27/95...
The reason why (2) is insufficient is also that we don't know how many bulbs we pick. it doesn't matter if we put the bulbs back or not because we are picking bulbs, not picking letters. In other words, we are looking at combination not arrangement!!
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AdiBir
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Post URL26 Jul 2020, 11:01
gmatt1476
A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs.
DS76851.01
Show more
Let's have a go at it in a structured manner.
Given:
Box has some number of bulbs(let that number be x)
Out of x bulbs, 3 are defective.
What we have to find:
Probability that out of a sample(let size of sample be y) of bulbs, at least 1 bulb is defective.
Process above info:
So we are given x bulbs, out of which 3 bulbs are defective and we need to choose a sample of y bulbs.
We have to find the probability that at least one of the bulbs in sample of size y is defective.
P(at least 1 defective) = 1 - P(no bulb defective)
Now moving on to the statements:
1)It tells that we pick bulbs one by one without replacement. This is similar to picking all y bulbs at once. But statement 1) does not provide any info about total bulbs(x) or sample size(y). So not sufficient to find probability.
Not Sufficient
2) This tells us that y=20. But still no info about x.
Not sufficient
1) and 2) combined:
Combined they tell us that we pick bulbs one by one w/o replacement and y=20. Still no information about x.
Hence Not Sufficient
Answer is E
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hemantbafna
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Post URL09 Aug 2020, 09:26
Hungluu92vn
MahmoudFawzy
from the given, we know the number of defective bulbs,
but we don't know:
a- the number of non-defective bulbs
b- the number of the sample to be picked.
c- the method of picking (with or without replacement)
statement (1) provides us with info (c) only -->insufficient
statement (2) provides us with info (b) only -->insufficient
so both (1) and (2) are insufficient because they can't provide info (a)
for example, if the number of non-defective bulbs = 17, so the probability will be 1
and as the number of non-defective bulbs increase, the probability starts to decrease below 1
How could not we know the info a? As the stimulus provides us with "exactly 3 defective bulbs" so i assume the rest are non-defective bulbs. What am i wrong here? Please shed some lights. Thank you
Show more
Statement 2 states The sample will contain exactly 20 light bulbs, it does not give us the total number of bulbs.
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DeeptiManyaExpert
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Post URL08 Apr 2021, 23:59
To Find: Probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb
Know: A box of light bulbs contains exactly 3 light bulbs that are defective.
Missing information:
1) no. of non-defective bulbs
2) the number of bulbs in the sample
3) the method of picking (with or without replacement)
St (1) tells us about the method of picking (without replacement). We don't know about the other 2 missing pieces.
INSUFFICIENT
St (2) tells us about the number of bulbs in the sample. We don't know about the other 2 missing pieces.
INSUFFICIENT
Combining the information from 2 statements give us 2 missing information but do not tell us about the number of defective bulbs.
INSUFFICIENT
Answer - E
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ueh55406
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Post URL04 Jul 2021, 00:43
1+2
So if the box had 500 box and we're picking a sample of 20 bulbs from the 500 and checking the P(of atleast one defective)
1c20, 2c20, 3c20/ 20c500
1 c 20,2 c 20,3 c 20 1 c 20,2 c 20,3 c 20 :- 1, 2 ,3 out of the 20 picked are defective.
but in the denominator, we have no idea what's the total number of bulbs, here we have taken 500 so we will get some results.
but if we took n= 1000, that denominator will become, 20 C 1000 20 C 1000. A world of difference.
hence, E.
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100mitra
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Post URL02 Sep 2021, 09:38
Correct option : E
We need concreate 4 information to solve
Total Bulb quantity sample - variable 1 (Available)
Distinct things occuring at set interval times - variable 2 (Not available)
Selection range sample - variable - 3 (Not available)
Events defective no's - variable 4 (Available)
based on below information:
A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs
A, B, D rejected
C rejected
Winner E
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Tanchat
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Post URL04 Jun 2022, 03:16
gmatt1476
A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs.
DS76851.01
Show more
Dear Expert,
From the probability formula P(E) = n(E)/n(S)
the first statement, we know the picking method.
The second statement, we only know that we picked 20 bulbs from the total bulbs that we don't know how many bulbs we have.
Combine (1) and (2), we don't know n(s) - which is total bulbs. We do know that we will pick 1 bulb 20 times without replacement
So, we cannot calculate the P(E)
Am I correct?
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Post URL28 Nov 2023, 02:26
For those of you confused about this question, read the question carefully. We are not picking one light bulb from a box. We are picking a sample of X light bulbs from a box.
(B) provides that we are picking 20 light bulbs from a box.
However, we still do not know how many non-defective light bulbs there are in the box and hence how many light bulbs, regardless of defects, there are in the box. Hence, it is E.
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gmatchile1
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Post URL17 Apr 2024, 17:13
Question stem- No.of defective light bulbs are 3.
Total no.of bulbs= unknown
We are asked to find -the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb.
We need to know total no.of bulbs present in the box and picking methods.
statement 1- Insufficient
Total no.of bulbs not known
Statement 2-insufficient
Picking method not known. We don’t know whether 1 bulb is picked at a time, 2 picked at a time or with/without replacement.
.
Statement 1 + Statement 2
Insufficient.
Reason is we cannot arrive at a unique answer because question stem asks to find probability that atleast 1 defective bulb is picked.
There are 3 cases possible here-
one is defective, 2 defective, 3 are defective. The probability of these 3 cases gives different answer.
Hence correct option is E.
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Post URL19 Sep 2024, 00:30
gmatt1476
A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs.
DS76851.01
Show more
No. of defective bulbs =3
Condition - I. No. of total bulbs or non-defective bulbs ==> missing
Condition II - No. of bulbs to be picked in said sample.
we know bulbs has to be picked 1 at a time without replacement. However, condition I&II still to be fulfilled to arrive at solution==> Insuff..
Condition II given but I still missing===>Insuf.
Combining above 2... Condition I still to be fulfilled hence--> Insufficient --Option E
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7343 | https://hamstudy.org/browse/E4_2012/E9A | Amateur Extra pool, section E9A
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Subel. E9
E1 73 questions
E2 68 questions
E3 35 questions
E4 70 questions
E5 71 questions
E6 83 questions
E7 124 questions
E8 56 questions
E9 109 questions
E0 11 questions
15 questionsE9AE9BE9CE9DE9EE9FE9GE9HHide Distractors
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Amateur Extra Class (2012-2016)
This is an old (expired) question pool. Click here to go to the current version
Subelement E9
ANTENNAS AND TRANSMISSION LINES
Section E9A
Isotropic and gain antennas: definitions; uses; radiation patterns; Basic antenna parameters: radiation resistance and reactance, gain, beamwidth, efficiency
E9 A01
Which of the following describes an isotropic antenna?
A. A grounded antenna used to measure earth conductivity
B. A horizontally polarized antenna used to compare Yagi antennas
Correct Answer C. A theoretical antenna used as a reference for antenna gain
D. A spacecraft antenna used to direct signals toward the earth
E9 A01
An isotropic radiator is a theoretical point source of electromagnetic waves which radiates the same intensity of radiation in all directions. It has no preferred direction of radiation. It radiates uniformly in all directions over a sphere centred on the source.
Isotropic radiators are used as reference radiators with which other sources are compared.
Source: Wikipedia - Isotropic Radiator
One Word Key "theoretical"
HINT: The question has the word "RADIATor" and the correct answer has the word "RADIATion" ~ (K2BHS)
Last edited by bsolov. Register to edit
Tags: none
E9 A02
How much gain does a 1/2-wavelength dipole in free space have compared to an isotropic antenna?
A. 1.55 dB
Correct Answer B. 2.15 dB
C. 3.05 dB
D. 4.30 dB
E9 A02
A reference dipole antenna is defined to have 2.15dBi gain.
A useful conversion between dBd and dBi is as follows:
dBi = dBd + 2.15
dBd = dBi - 2.15
Last edited by mschweiner. Register to edit
Tags: none
E9 A03
Which of the following antennas has no gain in any direction?
A. Quarter-wave vertical
B. Yagi
C. Half-wave dipole
Correct Answer D. Isotropic antenna
E9 A03
Isotropic antennas are ideal (theoretical) antennas that have equal power in all directions. They are used as references for antenna gain.
The word "isotropic" means "uniform in all orientations/directions".
Last edited by kk4uwv. Register to edit
Tags: none
E9 A04
Why would one need to know the feed point impedance of an antenna?
Correct Answer A. To match impedances in order to minimize standing wave ratio on the transmission line
B. To measure the near-field radiation density from a transmitting antenna
C. To calculate the front-to-side ratio of the antenna
D. To calculate the front-to-back ratio of the antenna
E9 A04
This question does not yet have an explanation! Register to add one
Tags: none
E9 A05
Which of the following factors may affect the feed point impedance of an antenna?
A. Transmission-line length
Correct Answer B. Antenna height, conductor length/diameter ratio and location of nearby conductive objects
C. Constant feed point impedance
D. Sunspot activity and time of day
E9 A05
The feed-point impedance of an antenna can be influenced by near-by conductive objects, including proximity to the ground.
Antenna height is the only answer that has anything to do with the antenna and its surrounding environment. The other answers do not affect the antenna. A dipole mounted 1/2 wavelength high over perfect ground has a feed point impedance around 72 ohms. As it is lowered, the impedance will go up to close to 100 ohms and then start dropping. By 0.1 wavelength in height, the feed point impedance approaches 20 ohms.
Another point of view - the question is about what affects the feed point of the antenna. The transmission line length, settings of an antenna tuner, and input power are not part of the impedance at (and after) the feed point of the antenna.
Last edited by gregor. Register to edit
Tags: none
E9 A06
What is included in the total resistance of an antenna system?
A. Radiation resistance plus space impedance
B. Radiation resistance plus transmission resistance
C. Transmission-line resistance plus radiation resistance
Correct Answer D. Radiation resistance plus ohmic resistance
E9 A06
Radiation resistance is that part of an antenna's feedpoint resistance that is caused by the radiation of electromagnetic waves from the antenna, as opposed to loss resistance (also called ohmic resistance) which is caused by ordinary electrical resistance in the antenna, or energy lost to nearby objects, such as the earth, which dissipate RF energy as heat.
The radiation resistance is determined by the geometry of the antenna, whereas the ohmic resistance is primarily determined by the materials of which it is made and its distance from and alignment with other nearby conductors or semi-conductors, and what those are made of.
Both radiation and ohmic resistance depend on the distribution of current in the antenna.
Eliminate the answer referring to "transmission line" because that is not what they are referring to as an "antenna system".
Last edited by bytescream. Register to edit
Tags: none
E9 A07
What is a folded dipole antenna?
A. A dipole one-quarter wavelength long
B. A type of ground-plane antenna
Correct Answer C. A dipole constructed from one wavelength of wire forming a very thin loop
D. A dipole configured to provide forward gain
E9 A07
The key is to remember that this antenna is one wavelength long "folded" in to a thin loop one-half wavelength long.
For more information see wikipedia: Folded Dipole Antenna
Silly memory trick: It's folded in half.
Last edited by radio1984. Register to edit
Tags: none
E9 A08
What is meant by antenna gain?
Correct Answer A. The ratio relating the radiated signal strength of an antenna in the direction of maximum radiation to that of a reference antenna
B. The ratio of the signal in the forward direction to that in the opposite direction
C. The ratio of the amount of power radiated by an antenna compared to the transmitter output power
D. The final amplifier gain minus the transmission-line losses, including any phasing lines present
E9 A08
This question does not yet have an explanation! Register to add one
Tags: none
E9 A09
What is meant by antenna bandwidth?
A. Antenna length divided by the number of elements
Correct Answer B. The frequency range over which an antenna satisfies a performance requirement
C. The angle between the half-power radiation points
D. The angle formed between two imaginary lines drawn through the element ends
E9 A09
Bandwidth is a measurement of frequencies, as in the width of a range of frequencies.
An antenna's bandwidth is the range of frequencies it works best on, though it's impossible to give an exact formula without the exact performance requirement.
For example the performance requirement might be "SWR less than 1.7" in which case the antenna could be modeled or tested and the exact range of frequencies for that requirement specified.
Hint: A Band "performs". The answer contains the word "performance".
Last edited by nielsenj. Register to edit
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E9 A10
How is antenna efficiency calculated?
A. (radiation resistance / transmission resistance) x 100%
Correct Answer B. (radiation resistance / total resistance) x 100%
C. (total resistance / radiation resistance) x 100%
D. (effective radiated power / transmitter output) x 100%
E9 A10
This question does not yet have an explanation! Register to add one
Tags: none
E9 A11
Which of the following choices is a way to improve the efficiency of a ground-mounted quarter-wave vertical antenna?
Correct Answer A. Install a good radial system
B. Isolate the coax shield from ground
C. Shorten the radiating element
D. Reduce the diameter of the radiating element
E9 A11
This question does not yet have an explanation! Register to add one
Tags: none
E9 A12
Which of the following factors determines ground losses for a ground-mounted vertical antenna operating in the 3-30 MHz range?
A. The standing-wave ratio
B. Distance from the transmitter
Correct Answer C. Soil conductivity
D. Take-off angle
E9 A12
Hint: GROUND losses is in the SOIL. The only answer talking about the ground. m
Last edited by fcu2fg13e5k0u4cojw9wjbpcyqu=. Register to edit
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E9 A13
How much gain does an antenna have compared to a 1/2-wavelength dipole when it has 6 dB gain over an isotropic antenna?
Correct Answer A. 3.85 dB
B. 6.0 dB
C. 8.15 dB
D. 2.79 dB
E9 A13
A 1/2-wave dipole antenna has approximately 2.15 dB of gain over an isotropic antenna. So
6 dB−2.15 dB=3.85 dB
(The directive gain factor of a half-wave dipole is 1.64. It has a 2.15 dB gain over an isotropic antenna or 10 log 10(1.64)≈2.15 dBi)
The distractor of 8.15dB relies on you remembering the 2.15dB difference between isotopic and dipole antennas, so pay careful attention to whether you're adding or subtracting 2.15dB!
Think of dBi as an inflated gain rating (which is why it's sometimes used in antenna specs instead of dBd). If an antenna has 6 dBi gain, then it will be 2.15 less when compared to a dipole.
Memory aids:
What is Half of 6 (dB)? That is the first digit in the answer, 3.85.
Last edited by gregor. Register to edit
Tags: none
E9 A14
How much gain does an antenna have compared to a 1/2-wavelength dipole when it has 12 dB gain over an isotropic antenna?
A. 6.17 dB
Correct Answer B. 9.85 dB
C. 12.5 dB
D. 14.15 dB
E9 A14
Given: It's well-known that a half-wave dipole has 2.15 dB gain over an ideal isotropic radiator.
Let H= the gain of an ideal isotropic radiator
Let W= the gain of a half-wave dipole
Therefore W=H+2.15 dB, or solving for H, H=W−2.15 dB
Now, let X= the gain of the antenna in question
If that antenna in question exhibits 12 dB over an isotropic antenna, then
X=H+12 dB
Substituting, we get
X=W−2.15 dB+12 dB=W+9.85 dB
Therefore, the antenna in question (X) exhibits a gain of 9.85 dB over that (W) of a half-wave dipole.
Remember that a half-wavength dipole has 2.15 dB gain over an ideal isotropic radiator. Then take the difference between 12 dB and 2.15 dB to arrive at the answer of 9.85 dB.
Last edited by wileyj2956. Register to edit
Tags: none
E9 A15
What is meant by the radiation resistance of an antenna?
A. The combined losses of the antenna elements and feed line
B. The specific impedance of the antenna
Correct Answer C. The value of a resistance that would dissipate the same amount of power as that radiated from an antenna
D. The resistance in the atmosphere that an antenna must overcome to be able to radiate a signal
E9 A15
The electrical resistance of an antenna is composed of its ohmic resistance plus its radiation resistance. The energy lost due to radiation resistance is the energy that is converted to electromagnetic radiation.
It isn't the combined losses of antenna elements and feed line because radiation resistance is not related to feed line losses.
It isn't the specific impedance of the antenna because the radiation resistance is only a portion of the antenna's impedance.
It isn't the resistance in the atmosphere that an antenna must overcome because the radiation resistance is not a property of the atmosphere. It is determined by the geometry of the antenna.
The correct answer is therefore the value of a resistance that would dissipate the same amount of power as that radiated from an antenna
Last edited by cactus121. Register to edit
Tags: none
Go to E8DGo to E9B
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Distance from a point to circle's closest point
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So let's assume I have a point P P in 3 D 3 D space (x 0,y 0,z 0)(x 0,y 0,z 0). And I have a circle C C that is centered at (x 1,y 1,z 1)(x 1,y 1,z 1) with a radius r r. I need to find the distance from P P to the nearest point of C C. I'm not totally sure how to define a circle in 3 D 3 D space, so suggestions there would help too :D
I really have very little idea where to begin with this (and I only have a very basic understanding of how to do the same thing with a point and a line). I haven't taken a math class in a number of years, but this concept will help tremendously in some 3 D 3 D programming I'm working on.
geometry
trigonometry
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edited Jun 20, 2015 at 0:00
Harish Chandra Rajpoot
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asked Apr 5, 2011 at 6:42
gregghzgregghz
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Do you mean a sphere? If you do mean a circle, how is its orientation given?joriki –joriki 2011-04-05 06:45:29 +00:00 Commented Apr 5, 2011 at 6:45
I do mean a circle. The orientation is so that it's surrounding the z axis ... but I'd also need to know the distance with an arbitrary rotation about the y axis.gregghz –gregghz 2011-04-05 06:47:47 +00:00 Commented Apr 5, 2011 at 6:47
We will need more information about the circle: in 3d, center and radius are not enough.André Nicolas –André Nicolas 2011-04-05 06:50:33 +00:00 Commented Apr 5, 2011 at 6:50
@user6312, I thought that might be the case. Unfortunately I don't know how to define a circle unambiguously in 3d. In my program, I'm able to define the circle with respect to the x and y axes and then rotate it as needed.gregghz –gregghz 2011-04-05 06:53:36 +00:00 Commented Apr 5, 2011 at 6:53
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Project the point onto the plane in which the circle lies. Then take the distance to the circle's centre, subtract the radius and take the absolute value to get the distance within the plane. Then you get the total distance from the distance to the plane and the distance within the plane using Pythagoras.
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answered Apr 5, 2011 at 6:50
jorikijoriki
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2 This is assuming that you mean a circle in the mathematical sense. If you actually meant a disc (a circle and its interior), then of course instead of |d−r||d−r| you need min(0,d−r)min(0,d−r) (where d d is the distance from the projected point to the centre).joriki –joriki 2011-04-05 07:03:10 +00:00 Commented Apr 5, 2011 at 7:03
That should be max(0,d−r)max(0,d−r).joriki –joriki 2011-04-05 10:39:11 +00:00 Commented Apr 5, 2011 at 10:39
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You need 3 features to define a 3D circle:
center (point)
orientation (direction vector perpendicular to plane of circle3D)
radius (positive scalar)
You need to be clear whether you're interested in the nearest point to P on the disk or the circle edge.
If you know how to set up 3D coordinate rotations and translations, you can greatly simplify the problem by adopting transforms that place the circle at the origin and flat in the z == 0 plane. You apply these desired coordinate translate and coordinate rotate transforms to point P ==>> P' ++> P". Now, with this idealized case, you should be able to solve the problem of the closest point on the circle.
Don't forget to inverse-transform your solution point Psol" ==> Psol' ==> Psol so that it comes out in starting coordinates.
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answered Aug 2, 2017 at 2:38
pbierrepbierre
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7345 | https://www.cuemath.com/calculus/homogeneous-differential-equation/ | LearnPracticeDownload
Homogeneous Differential Equation
A homogeneous differential equation is an equation containing a differentiation and a function, with a set of variables. The function f(x, y) in a homogeneous differential equation is a homogeneous function such that f(λx, λy) = λnf(x, y), for any non zero constant λ. The general form of a homogeneous differential equation is f(x, y).dy + g(x, y).dx = 0.
Let us learn more about the homogeneous differential equation, the method to solve a homogeneous differential equation, examples, faqs.
| | |
--- |
| 1. | What Is A Homogeneous Differential Equation? |
| 2. | How to Solve a Homogeneous Differential Equation? |
| 3. | Examples on Homogeneous Differential Equation |
| 4. | Practice Questions on Homogeneous Differential Equations |
| 5. | FAQs on Homogeneous Differential Equation |
What Is A Homogeneous Differential Equation?
A differential equation containing a homogeneous function is called a homogeneous differential equation. The function f(x, y) is called a homogeneous function if f(λx, λy) = λnf(x, y), for any non zero constant λ. The general form of the homogeneous differential equation is of the form f(x, y).dy + g(x, y).dx = 0. The homogeneous differential equation has the same degree for the variables x, y within the equation.
The homogeneous differential equation does not have a constant term within the equation. The linear differential equation has a constant term. The solution of a linear differential equation is possible if we are able to remove the constant term from the linear differential equation and transform it into a homogeneous differential equation. Also, the homogeneous differential equation does not have the variables x, y within any special functions such as logarithmic, or trigonometric functions.
Examples of Homogeneous Differential equations.
dy/dx = (x + y)/(x - y)
dy/dx = x(x - y)/y2
dy/dx = (x2 + y2)/xy
dy/dx = (3x + y)/(x - y)
dy/dx = (x3 + y3)/(xy2 + yx2)
In these above examples, we can substitute x = λx, and y = λy, to prove it for the homogeneous differential equation. Also, if the homogeneous differential equation is of the form dx/dy = f(x, y), and f(x, y) is a homogeneous function, then we substitute x/y = v, or x = vy. This on further integration, and substituting back the variables x, y, gives the general solution of the homogeneous differential equation.
How To Solve a Homogeneous Differential Equation?
The general solution of the homogeneous differential equation can be obtained by the integration of the given differential equation. A homogeneous differential equation of the form dy/dx = f(x, y), is solved by first separating the variable and the derivative of the particular variable on either side and then integrating it with respect to the variable.
To solve a homogeneous differential equation of the form dy/dx = f(x, y), we make the substitution y = v.x. Here it is easy to integrate and solve with this substitution. Further the differentiation of y = vx, with respect to x we get dy/dx = v + x.dv/dx. We can substitute the value of dy/dx in the expression dy/dx = f(x, y) = g(y/x) to get the below expression.
v + x.dv/dx = g(v)
xdv/dx = g(v) - v
Separating the variables x and v, we have:
(\dfrac{dv}{g(v) - v} = \dfrac{dx}{x})
Here we integrate it on both sides, which results in the following expression.
(\int \dfrac{1}{g(v) - v}.dv = \int \dfrac{1}{x}.dx)
The above expression gives the following solution, which is the general solution of the differential equation.
(\int \dfrac{1}{g(v) - v}.dv = Logx + C)
Here we substitute back the value of v = y/x, to obtain the general solution of the homogeneous differential equation. The presence of +C in the solution, refers it as a general solution, and further solving and substituting the value of +C, we can obtain the particular solution of the given homogeneous differential equation.
Related Topics
The following topics help in a better understanding of the homogeneous differential equations.
Differential Equations
Linear Differential Equation
UV Differentiation Formula
Differentiation of Trigonometric Functions
Application of Derivatives
Examples on Homogeneous Differential Equation
Example 1: Show that the differential equation (x - y).dy/dx = (x + 2y) is a homogeneous differential equation.
Solution:
(x - y).dy/dx = (x + 2y) is the given differential equation.
To prove that the above differential equation is a homogeneous differential equation, let us substitute x = λx, and y = λy.
Here we have F(x, y) = (\dfrac{(x + 2y)}{(x - y)})
F(λx, λy) = (\dfrac{(λx + 2λy)}{(λx - λy)})
F(λx, λy) = (\dfrac{λ(x + 2y)}{λ(x - y)}) = λ0f(x, y)
Therefore, the given differential equation is a homogeneous differential equation.
2. Example 2: Find the solution of the homogeneous differential equation xCos(y/x).dy/dx = yCos(y/x) + x.
Solution:
The given differential equation is xCos(y/x).dy/dx = yCos(y/x) + x
dy/dx = (\dfrac{yCos(y/x) + x}{xCos(y/x)})
dy/dx = (\dfrac{x((y/x).Cos(y/x) + 1)}{xCos(y/x)})
dy/dx = (\dfrac{((y/x).Cos(y/x) + 1)}{Cos(y/x)})
Here let us substitute y/x =v in the above expression.
dy/dx = (\dfrac{vCosv + 1}{Cosv})
Here we write y/x = v as y = vx. Differentiating y = vx on both sides we obtain dy/dx = v + x.dv/dx, which is substituted in the above expression.
v + x.dv/dx = (\dfrac{vCosv + 1}{Cosv})
x.dv/dx = (\dfrac{vCosv + 1}{Cosv}) - v
Here we separate the variables on either side of the equal to symbol.
x.dv/dx = 1/Cosv
Cosv.dv = dx/x
Integrating this expression on both sides, we have the below expression.
(\int Cosv.dv = \int \dfrac{1}{x}.dx)
Sinv = Logx + C
Here we substitute back y/x = v.
Sin y/x = Logx + C
Therefore, the solution of the homogeneous differential equation is Sin y/x = Logx + C.
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Practice Questions on Homogeneous Differential Equations
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FAQs on Homogeneous Differential Equation
What Is Homogeneous Differential Equation?
The homogeneous differential equation containing is a differential equation containing a homogeneous function. The homogeneous differential equation of the form dy/dx = f(x, y), has a homogeneous function f(x, y) such that f(λx, λy) = λnf(x, y), for any non zero constant λ. The general form of the homogeneous differential equation is of the form f(x, y).dy + g(x, y).dx = 0.
What Is the Difference Between Homogeneous and Non-Homogeneous Differential Equation?
The homogeneous differential equation consists of a homogeneous function f(x, y), such that f(λx, λy) = λnf(x, y), for any non zero constant λ. A non-homogeneous differential equation does not contain a homogeneous function. The example of a non homogeneous differential equation is a linear differential equation of the form dy/dx + Py = Q.
What Are the Examples of Homogeneous Differential Equation?
A few of the examples of homogeneous differential equations are as follows.
dy/dx = (x + y)/(x - y)
dy/dx = (x2 + y2)/xy
dy/dx = (3x + y)/(x - y)
dy/dx = (x3 + y3)/(xy2 + yx2)
What Is the Formula for Homogeneous Differential Equation?
The general form of a homogeneous differential equation is f(x, y).dy + g(x, y).dx = 0. Here the function f(x, y) is a homogeneous function such that f(λx, λy) = λnf(x, y), for any non zero constant λ. Also, there is no defined formula for a homogeneous differential equation, or to find the solution of it. The homogeneous differential equation is solved through a sequence of steps.
What Are the Steps to Solve Homogeneous Differential Equation?
The homogeneous differential equation of the form dy/dx = f(x, y), can be solved through the following sequence of steps.
Step - 1: Substitute y = vx in the given differential equation.
Step - 2: Separate the variables and the differentiation of the variables on either side of the equals to symbol.
Step - 3: Find the integration of the variables and to find the general solution containing v and x.
Step - 4: Substitute back the value of v to get the general solution in the variables x and y.
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7346 | https://www.electrical4u.com/high-pass-filter/ | Skip to content
High Pass Filter: Circuit, Transfer Function & Bode Plot
Key learnings:
High Pass Filter Definition: A high pass filter allows frequencies higher than a certain cutoff and blocks lower ones, useful in electronic filtering.
Circuit Components: These filters can use simple components like resistors and capacitors or include operational amplifiers for more complex applications.
High Pass Filter Transfer Function: The transfer function mathematically represents how the filter processes signals, emphasizing frequencies above the cutoff.
Bode Plot Overview: A Bode plot visualizes the frequency response of the filter, showing changes in signal magnitude and phase with frequency.
Practical Applications: High pass filters are used in diverse fields like audio technology and image processing to enhance signal quality.
A filter removes unwanted elements, much like a water filter cleans impurities from water. Similarly, an electric filter works to cleanse signals in electronic circuits.
An electric filter, made up of resistors, inductors, capacitors, and amplifiers, passes signals of a certain frequency while attenuating others that are too high or too low.
The frequency at which filter operates, that frequency is known as cut-off frequency. The cut-off frequency is set while designing the filter.
What is a High Pass Filter?
A high pass filter (also known as a low-cut filter or bass-cut filter) is an electronic filter that permits signals with a frequency higher than a certain cutoff frequency and attenuates signals with frequencies lower than that cutoff frequency.
The inverse of a high-pass filter is a low-pass filter, which allows signals with frequencies lower than the cut-off frequency and blocks all frequencies above this cut-off frequency. There are also band pass filters, which combine the functionality of high pass filters and low pass filters to only allow frequencies within a specific frequency range.
High Pass vs Low Pass Filters
The characteristics of a high pass filter are exactly the opposite characteristics of a low pass filter. The difference include:
| | | |
---
| | High Pass Filter (HPF) | Low Pass Filter (LPF) |
| Definition | HPF is an electric filter that allows signals with higher frequency than the cut-off frequency. It is known as the low-cut filter. | LPF is an electric filter that allows signals with a lower frequency than the cut-off frequency. It is known as the high-cut filter. |
| Circuit Diagram | In HPF, the capacitor followed by the resister. | In LPF, the resister followed by the capacitor. |
| RC Filter | | |
| Operating frequency | Higher than the cut-off frequency | Lower than the cut-off frequency. |
| Importance | It is important to cancel the low-frequency noise from the input signal. | It is important to cancel the aliasing effect. |
| Applications | It is used in amplifiers like an audio amplifier, low-noise amplifier. | It used in communication circuits as an anti-aliasing filter. |
Types of High Pass Filters
There are many types of high-pass filters according to the circuit design and components used to make a filter. The various types of high-pass filters include:
Passive High Pass Filter
The passive filter consists of only passive elements like resistor, inductor, and capacitor. It will not use any external power source or amplification components.
A passive high pass filter consists of a combination of resistor and capacitor (RC) or resistor and inductor (RL).
Active High Pass Filter
The Active filter is a combination of a passive filter with an operational amplifier (OP-AMP) or it includes an amplifier with gain control.
It is made by connecting an inverting or non-inverting component of OP-AMP with a passive filter.
RC High Pass Filter
The RC filter is one type of passive filter because it consists only of a capacitor is in series with the resistor.
The circuit diagram of high pass and low pass filter is the same, just interchange the capacitor and resistor. The circuit diagram of the RC high pass filter is as shown in the below figure.
The capacitor offers very high reactance for the signal with a frequency lower than the cut-off frequency. In this case, the capacitor act as an open switch.
The capacitor offers low reactance for the signal with a frequency higher than the cut-off frequency. In this case, the capacitor act as a close switch.
First Order High Pass Filter
First Order High pass filter consists of only one capacitor or inductor. This type of filter has a transfer function of the first order.
It means if you derive an equation in s-domain, the maximum power of ‘s’ is one. This is only possible if you use only one energy storage element like inductor and capacitor.
The first order filter can be active or passive, depending on the use of elements. If it uses only active elements, it can be a first-order filter. RC high pass filter is a first-order passive high pass filter.
Second Order High Pass Filter
Second-Order high pass filter can derive by cascading two first-order high pass filters. Therefore, it consists of two reactive components and makes a second-order circuit.
The main difference in the first order and second-order filter slope in the stop band. The slop of the second-order filter is twice of the first-order filter.
For example, if we consider a first-order Butterworth filter, the slop is +20 db/decade and for second-order Butterworth filter, the slop is +40 db/decade.
Butterworth High Pass Filter
The Butterworth filter is designed to have a flat frequency response in the pass band. So, in the pass band, there is no ripple in the frequency response. The below figure shows the circuit diagram of the first order and second-order Butterworth high pass filter with frequency response.
Chebyshev High Pass Filter
In all ranges of filters, the Chebyshev filter minimizes the error between the actual filter and the ideal filter. There are two types of filters; type-I and type-II. The type-I filter is known as “Chebyshev Filter” and the type-II filter is known as “Inverse Chebyshev Filter”.
This filter response is optimal trade between ripple and slope. If the ripple is set to 0%, the filter response is the same as the Butterworth Filter. But a ripple of 0.5% is a good choice for digital filters which make sharp slop. Below figure shows difference in frequency response for Butterworth and Chebyshev filter.
If the ripple present at the pass band, the filter is known as type-I Chebyshev Filter and if the ripple present in the stop band, the filter is known as type-II Inverse Chebyshev Filter.
There is a very fast transition between the pass band and stop band. But for this condition, the ripple will present in pass band and stop band. This type of filter is known as Elliptical Filter.
Bessel Filter
The Butterworth filter has good transient and amplitude behavior. Chebyshev filter has a good amplitude response than Butterworth filter with the expense of transient behavior.
The Bessel filter has a good transient response. But the amplitude behavior is poor. The Bessel filter is designed to get a constant group delay in the pass band.
Passive vs Active High Pass Filter
According to the components used in the circuit, filters are classified into two types; Active Filter and Passive Filter.
| | | |
---
| | Active Filter | Passive Filter |
| Circuit elements | Active Filter uses active elements like OP-AMP and Transistor. | Passive Filter uses passive elements like Capacitor and Inductor. |
| Extra power supply | It requires an extra power supply. | It operates on the signal input and not need an extra power supply. |
| Frequency limitation | It has frequency limitations. | It has no frequency limitations. |
| Cost | High | Cheaper in cost. |
| Stability | Inferior stability | Better stability |
| Weight | Low | High (Because the weight of inductor is very high) |
| Sensitivity | More sensitive | Less sensitive |
| Q Factor | High | Very Low |
| Design | It needs a complex control system. So, the design of this filter is complex. | It is easy to design. |
| Efficiency | High | Low |
| Frequency Response Characteristic | Frequency Response Characteristic is sharp | Frequency Response Characteristic is not sharp |
High Pass Filter Transfer Function Equation
The transfer function gives mathematical representation of filters. This mathematical expression gives the input to output behavior of the filter.
The transfer function of a first order high pass filter is derived in the below equations.
Output Impedance is equal to:
Input Impedance is equal to:
The transfer function is defined as the ratio of Output voltage to input voltage.
The standard form of transfer function is as:
Where:
According to this transfer function for higher frequency
And for Lower frequency
Therefore, it shows zero magnitude for lower frequency and Maximum magnitude for higher frequency.
Cutoff Frequency High Pass Filter
The cutoff frequency is defined as a frequency that creates a boundary between pass band and stop band.
For a high pass filter, if the signal frequency is more than the cutoff frequency, then it will allow passing the signal. And if the signal frequency is less than the cutoff frequency, then it will attenuate the signal.
The cutoff frequency is defined by the user at the time of designing a filter. For the first-order RC high pass filter, it is expressed as below equation. This equation is the same for a high pass as well as a low pass filter.
The cutoff frequency for second-order high pass RC filter is determined by both the resistors and capacitors. And it expressed as;
From the above equation, if the value of R1 and R2 is equal and the value of C1 and C2 is equal than the equation is expressed as;
High Pass Filter Bode Plot or Frequency Response
The frequency response or bode plot of the high pass filter is totally opposite compared to the frequency response of the low pass filter.
Using the transfer function, we can plot a frequency response of the filter circuit. The magnitude curve and phase curve of the bode plot for high pass filter is as shown in the below figure.
The magnitude curve can be obtained by the magnitude of the transfer function.
The phase curve can be obtained by the phase equation of the transfer function.
Magnitude Plot
As shown in the magnitude curve, it will attenuate the low frequency at the slope of +20 db/decade. The region from an initial point to the cutoff frequency is known as stop band.
When it crosses the cutoff frequency, it will allow the signal to pass. And the region above the cutoff frequency point is known as a pass band.
At cutoff frequency point the output voltage amplitude is 70.7% of the input voltage.
Phase Plot
At cutoff frequency, the phase angle of the output signal is +45 degree. From the phase plot, the output response of the filter shows that it can pass to infinite frequency. But in practice, the output response does not extend to infinity.
By proper selection of components, the frequency range of filter is limited.
Ideal High Pass Filter
The ideal high pass filter blocks all the signal which has frequencies lower than the cutoff frequency. It will take an immediate transition between pass band and stop band.
The magnitude response of the ideal high pass filter is as shown in the below figure. The amplitude will remain as original amplitude for signals which have a higher frequency than the cutoff frequency. And the amplitude will completely zero for signals which have a lower frequency than the cutoff frequency. Therefore, an ideal high pass filter has a flat magnitude characteristic.
The transfer function of ideal high pass filter is as shown in the equation below:
The frequency response characteristics of an ideal high pass filter is as shown in below figure.
This type of ideal characteristic of a high pass filter is not possible for practical filters. But the Butterworth filter characteristic is very close to the ideal filter.
Applications of High Pass Filters
The applications of high pass filters include:
It is used in amplifiers, equalizers, and speakers to reduce the low-frequency noise.
For sharpening the image, high pass filters are used in image processing.
It is used in various control systems.
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7347 | https://projecteuclid.org/ebooks/institute-of-mathematical-statistics-lecture-notes-monograph-series/State-of-the-art-in-probability-and-statistics/chapter/Statistical-problems-involving-permutations-with-restricted-positions/10.1214/lnms/1215090070.pdf | STATISTICAL PROBLEMS INVOLVING PERMUTATIONS WITH RESTRICTED POSITIONS PERSI DIACONIS, RONALD GRAHAM AND SUSAN P. HOLMES Stanford University, University of California and ATT, Stanford University and INRA-Biornetrie The rich world of permutation tests can be supplemented by a variety of applications where only some permutations are permitted. We consider two examples: testing in-dependence with truncated data and testing extra-sensory perception with feedback. We review relevant literature on permanents, rook polynomials and complexity. The statistical applications call for new limit theorems. We prove a few of these and offer an approach to the rest via Stein's method. Tools from the proof of van der Waerden's permanent conjecture are applied to prove a natural monotonicity conjecture.
AMS subject classiήcations: 62G09, 62G10.
Keywords and phrases: Permanents, rook polynomials, complexity, statistical test, Stein's method.
1 Introduction Definitive work on permutation testing by Willem van Zwet, his students and collaborators, has given us a rich collection of tools for probability and statistics. We have come upon a series of variations where randomization naturally takes place over a subset of all permutations. The present paper gives two examples of sets of permutations defined by restricting positions.
Throughout, a permutation π is represented in two-line notation 1 2 3 ... n π(l) π(2) π(3) • • • τr(n) with π(i) referred to as the label at position i. The restrictions are specified by a zero-one matrix Aij of dimension n with Aij equal to one if and only if label j is permitted in position i. Let SA be the set of all permitted permutations. Succinctly put: (1.1) SA = {π : UUAiπ{i) = 1} Thus if A is a matrix of all ones, SA consists of all n! permutations.
Setting the diagonal of this A equal to zero results in derangement, permu-tations with no fixed points, i.e., no points i such that π(i) = i.
196 Persi Diaconis, Ronald Graham and Susan P. Holmes The literature on the enumerative aspects of such sets of permutations is reviewed in Section 2, which makes connections to permanents, rook poly-nomials and computational complexity.
Section 3 describes statistical problems where such restricted sets arise naturally. Consider a test of independence based on paired data (Xi, Yi), (X2, Y2) {Xn, Yn)> Suppose the data is truncated in the following way: For each x there is a known set I(x) such that the pair (X, Y) can be observed if and only if Y E I(X). For example, a motion detector might only be able to detect a velocity Y which is neither too slow nor too fast.
Once movement is detected the object can be measured yielding X. Of course, such truncation usually induces dependence. Independence may be tested in the following form: Does there exist a probability measure μ on the space where Y is observed such that {1.2) P{YieBi,l<i<n\Xi,Yn for all B{ c I{Xi). Under assumption (1.2), given the unpaired data {X{ any permutation π with {Yπ(j) G /(Xi), 1 < ΐ < n} is equally likely for the paired data (X;, Yπ(i)), 1 < i < n. This allows any standard test of indepen-dence to be quantified by its permutation distribution using SA of (1.1) with A defined by if A i i ~ Λ 0 else This example raises the problem of developing a theory of the distribution of rank statistics such as Kendall's tau or 2 = 1 when π is chosen uniformly in SA In Section 3 we discuss natural examples where the restriction matrix A has an interval structure (the ones in each row are contiguous). For one-sided intervals some of the standard limit theory can be pushed through, although much remains to be done. A classical contingency table with structural zeros of Karl Pearson is treated as an example.
For two-sided intervals, we offer an exchangeable pair so that Stein's method can be used. The exchangeable pair gives a Monte Carlo Markov chain for calibrating the permutation distribution. A red-shift data set of Efron and Petrosian (1998) is treated as an example.
Permutations with restricted position appear in a different guise in skill scoring, a technique for evaluation of taste testing and extra sensory per-ception experiments with feedback to subjects permitted. This application Independence with Truncation 197 is reviewed in Section 4. Some of the tools developed to prove the van der Waerden permanent conjecture are applied here to give a simple proof of a natural monotonicity conjecture.
We hope that these two examples of permutation testing with restricted positions will interest Bill. This part of the subject can certainly use his help.
2 Permanents Let A be a zero-one square matrix of dimension n. The number of elements in SA of (1.1) is determined by the permanent of A: yZi.oj \&A\ ~ The sum in (2.1) is over all permutations in the symmetric group. Thus the permanent is like the determinant without signs. There is a large math-ematical literature on permanents. We review some of this pertaining to matching theory (Section 2.1), rook theory (Section 2.2) and complexity theory (Section 2.3).
We believe that many of the nice developments in permutation enu-meration and testing will work out nicely for the case of interval restricted permutations. An example, Fibonacci permutations, is developed in Section 2.4. It may be consulted now for motivation.
2.1 Bipartite Matchings Let [n] = {1,2,..., n} and [n 1] = {I 7,2',..., n'} be two disjoint n element sets(n — n'). A bipartite graph G is specified by giving a set of undirected edges ε = {(ii,ii), (^2^2) (^e^e)} For example, when n = 3 the graph might appear as: A matching in G is a set of vertex disjoint edges. Thus (1, l')(2,3') is a matching in the figure above. A perfect matching is a matching containing n edges. There are three perfect matchings in the figure above.
There is a one-to-one correspondence between perfect matchings and the set SA of (1.1) when A is taken as the adjacency matrix of the graph G.
198 Persi Όmconis, Ronald Graham and Susan P. Holmes Thus for the figure above: 1' 2' 3' 1 1 1 0 A = 2 1 1 1 3 0 1 1 and Per (A) = 3. This correspondence allows standard graph theory algo-rithms to be used for permutations with restricted positions. For example, given a restriction matrix A there is a polynomial time algorithm (order n 2 5/y/logn) for finding if there exists a perfect matchings using the widely available algorithm for solving the assignment problem of combinatorial op-timization (see Cook (1998)).
Naively computing the permanent from its definition (2.3) takes n.nl steps. Ryser's algorithm (see van Lint and Wilson (1992); theorem 11.2) improves this to order n.2 71. As discussed in Section 2.3 below, no substantial improvement can be expected for general restriction matrices (the problem is #-P complete). For some special cases (Sections 2.2,3.3,4.1) enumeration is feasible. For matrices A with row sums ri, Γ2,..., rn, van Lint and Wilson (1992 theorem 11.3) gives the bound Per{A) <Π? = 1(ri)!^.
Bipartite matching is the easiest case of matching theory in general graphs.
Many further results applications and references are collected in the splendid book by Lovasz and Plummer (1986).
2.2 Rook Theory This is an algebraic technique for enumerating SA The classical setting is a set of squares B of an n x n chess board. Let r(B,k) be the number of ways of placing k non-attacking rooks on the squares of B (that is, choosing k squares in B no two in the same row or column). A board B is identified with a bipartite graph G with edge (ΐ,i') if and only if {i,i') is in B. Thus r(B, k) is the number of matchings with k edges.
The rook polynomial r(G,x) is defined as (2.4) r(G,x) = Thus the number of perfect matchings is the value r(G,0). Rook poly-nomials have been extensively studied. Stanley (1986, Chapter 2) gives a useful treatment of the essentials including extensive references to the work of Goldman, Joichi, White.
Independence with Truncation 199 Riordan (1958) reviews the extensive classical literature. Godsil (1981) and Lovasz/Plummer (1986) treat the generalization to matching polynomi-als of a general graph due to Heilmann-Lieb.
Rook and matching polynomials satisfy useful recurrences, have real ze-roes and, for neat graphs, give rise to orthogonal polynomials.
2.3 Complexity Theory Evaluation of the permanent of a square matrix is a celebrated problem in modern complexity theory. Indeed, Valiant (1979) used this as the first ex-ample of a #P-complete problem. Recall (Garey and Johnson, 1978) that NP-complete problems have "yes" or "no" answers, e.g.,"Is there some sub-set sum of this list of integers equal to 137?" The class of #P-complete problems are counting problems "How many subset sums are equal to 137?" For bipartite matching there is a fast way to check existence, but the count-ing problem is #P-complete; if a polynomial time algorithm exists then thousands of other intractable problems (e.g., computing the volume of a convex polyhedra) can be solved in polynomial time. A review of the work on the permanent from a complexity viewpoint can be found in Jerrum and Sinclair(1989) or Sinclair (1993).
Modern computer science has produced efficient randomized algorithms for approximating the permanent of a dense bipartite graph (every vertex having degree at least | ) . These algorithms perform a random walk on the set of perfect matchings and almost matchings (at most one edge missing).
It has been proved that these walks converge rapidly and allow efficient selection of an essentially random perfect matching. This makes Monte Carlo quantification of the tests outlined in the introduction feasible for dense graphs.(Jerrum and Sinclair (1989), Sinclair (1993)) Unfortunately, arguments used by Jerrum and Sinclair and later workers really seem to depend on the denseness assumption. Despite extensive work over the past ten years the rate of convergence of natural random walks on the set of permutations consistent with a general restriction matrix remains open.
Among interesting recent developments we mention: Kendall, Randall and Sinclair (1996) show that the random walk algorithm works in poly-nomial time for the perfect matchings in bipartite graphs which are vertex transitive (e.g. d-dimensional rectangular grids with toroidal boundaries).
Jerrum and Vazirani (1992) give an algorithm that approximates the perma-nent of any n x n zero-one matrix in time e cnΊ( ι°9 n) . This is superpolyno-mial but better than the n2 n of Ryser's algorithm. Rasmussen (1994) gives a simple greedy approximation algorithm for the permanent which is shown to run in polynomial time for almost every graph for the usual G(n,p) model of random graphs. Finally, Karmarkar et al.(1993) followed by Barvinok 200 Persi Diaconis, Ronald Graham and Susan P. Holmes (1998) and Rasmussen (1998) show how to approximate the permanent by a stochastic algorithm which replaces the ones by cube roots of unity and takes the squared modulus of the determinant. This algorithm has good average case behavior but exponential worst case behavior.
In Section 3 below we describe walks with interval restrictions where some things can be proved.
2.4 The Example of Fibonacci Permutations This Section treats a simple example which shows that elegant theory can be developed for enumeration, random generation and the study of cycle structure. Let An be the n x n matrix with ones on, just above and just below the diagonal. Thus when n = 4 1 1 0 0 1 1 1 0 0 1 1 1 0 0 1 1 From the definition (2.1), Per (A) is multilinear. Expanding by the first row, we derive a result first noted by Lehmer (1970): PerAn — PerAn-ι + PerAn-2-From direct computation, Per(A) = 1, Per(A2) = 2, so Per(An) = F n +i is the (n + l)st Fibonacci number. The Fibonacci numbers are defined as ΓQ — U, r i = 1, Γ2 = 1, Γ3 = ^, Γ4 = o, Γ5 = O,. . . , ^n+1 = ^n "T ^n-1?
We will call the elements of SAU Fibonacci permutations.
We first develop several bijections with combinatorial objects well known to be counted by Fibonacci numbers. Permutations counted by Per(An) can be described in their cycle form. They are all permutations consisting of fixed points and pairwise adjacent transpositions. To see this, observe that permutations can be constructed as follows.
Place symbols 1,2,3,..., n in a line, put a left parenthesis at the start and a right parenthesis at the end. Proceeding sequentially, decide to pass on or to place parentheses ) ( between i and i + 1. This results in the following five permutations consistent with A\ From this description it is easy to see that the permutations enumerated by An are in one to one correspondence with the following well-known Fibonacci equivalents.
Independence with Truncation 201 Proposition 2.1 The set of Fibonacci permutations on n letters is in one-to-one correspondence with: • Subsets of [n — 1] with no consecutive elements: φ, {1}, {2}, {3}, {1,3} • binary (n-l)-tuples without two consecutive ones: 000; 100; 010; 001; 101 • Matchings in an n-path: o o o o J o—o o o ! o o—o o ' o o o—e ' o — — o o—o • Compositions ofn with all parts equal to one or two 1111,211,121,112, 22 The next proposition gives an easy, direct method for uniformly choosing a random Fibonacci permutation. It is based on a theorem of Zeckendorf (1972) and the Fibonacci numbering system.
Proposition 2.2 Any positive integer n can be uniquely expressed as n = Fki + Fk2 H + Fkt with F^ distinct Fibonacci numbers starting with F2, no two adjacent.
Thus 1 = F 2,2 = F 3,3 - F 4,4 = F 2 + F 4,5 = F 5,6 = JF5 + F 2 , 7 -F 5 + F 3,8 = F 6,9 = F 6 + F 2,10 6 - F 3 0 + F26 + F 2 4 + F12 + F 1 0, these topics are covered in Graham, Knuth, Patashnik(1989, pp.281-283). They show that the representation is easy to find, each time substracting off the largest possible Fibonacci number.
For our purposes, consider a path with n vertices and edges labeled with consecutive Fibonacci numbers starting with F2;eg for n=8: F2 F3 F4 F5 FQ FJ F% 1 2 3 5 8 13 21 The weight of a matching in this graph is the sum of the weights of the edges. The largest possible sum comes from choosing the unique maximal matching. It is easy to prove that this has weight F n+i — 1. Thus in the example above, 21 + 8 + 3 + 1 = 33 = Fg — 1. With this preparation, the algorithm is simple.
Proposition 2.3 The following algorithm produces a randomly chosen Fi-bonacci permutation on n letters.
Choose an integer U, uniformly between zero and F n+i — 1. Express U in the Fibonacci numbering system and use the edges as a matching. Using the correspondence between matchings and Fibonacci permutations completes the construction.
202 Persi Diaconis, Ronald Graham and Susan P. Holmes For a randomly chosen permutation π 6 Sn the number of fixed points F(π) has an approximate Poisson(l) distribution, the number of transposi-tions T(π) has an approximate Poisson(^) distribution, the number of cycles C(π) has an approximate Normal distribution with mean logn and variance log n, and finally the number of inversions has an approximate Normal dis-tribution with mean \ and variance |g.
The following proposition shows how these results carry over to Fibonacci permutations. In this case, the four results coalesce since T(π) = /(π), C(π) = T(τr) + 7(τr) and n = F(π) + 2T(τr).
Proposition 2.4 Let π be α randomly chosen Fibonacci permutation on n letters. Let T(π) be the number of transpositions in π. Then /n-k\ (2.5) DίΦ __ M __ V • ' ~ - . . tn (2.6) with φ = (2.7) (2.8) Proof A bijection between Fibonacci permutations with T(π) = k and fe sets of an n — k set is easy to see; Put n — k dots down in a row. Circle each element in a subset of size fc. Now working from left to right, all encircled points are expanded to two points and correspond to transpositions. Thus ® ® < > (12)(34)5 The generating function for ( n^ fc) is a classical result: ' m - Σ( n; k) (2.9) Setting z = 1 gives the classical expression (2.10) Independence with Truncation 203 , l + \/5 Λ 1-Vb where φ = — - — , ψ = 2 ' " 2 Differentiating (2.9) at z = 1 gives 2 Φ n) - 5\/5 25 Dividing by Fn+i and using (2.10), routine simplifications give (2.6,2.7) .
An interesting proof of the liming normality uses the identification of Fibonacci permutations with the set of all matchings in an n-path. In this identification, the number of transpositions corresponds to the number of edges in the associated matching. Godsil(1981) has shown that the num-ber of edges in a random matching of a general graph tends to a Normal distribution provided only that the variance tends to infinity. • Remark 2.1 The limiting normality can be proved directly from (2.5) using Stirling's formula to give a local central limit theorem. Alternatively, the generating function (2.9) can be used. Godsil's proof used above itself uses Harper's method and depends on the fact that the zeros of the matching polynomial are all real. Pitman(1997) develops Harper's method and gives easily computed error bounds to the central limit theorem.
Remark 2.2 Similar results can be developed for the case where the re-striction matrix has ones on the diagonal and k to the left and right of the diagonal in each row. The Fibonacci example has k = 1.
2.5 Other applications of permanents The above surveys only mention features of the permanent literature directly related to the present project. There are many further applications. Mallows (1957) shows how permanents appear in computing normalizing constants in non-null ranking models. Bapat (1990) surveys a variety of appearances of permanents in statistics and further applications appear in Sections 3 and 4 below. Daley and Vere-Jones (1988) show how permanents occur for point-process theory and computing moments of complex normal variables.
One of the most active recent developments is the immanants. These are expansions of the form: (2.11) 204 Persi Diaconis, Ronald Graham and Susan P. Holmes with Ξ a character of the symmetric group. Taking Ξ = 1 gives the per-manent, taking Ξ(τr) = sgn(π) gives the determinant. There is active work giving inequalities for other immanants (see Lieb (1966) and Stembridge (1991, 1992) for surveys).
3 Testing for Independence 3.1 Introduction Consider the classical problem of testing for independence without trun-cation. One observes pairs (AΊ, YΊ), (X2? ^2), > {X n, Yn) drawn indepen-dently from a joint distribution V with Xi e X,Yi £ y, suppose that V has margins V 1 and V 2. A test of the null hypothesis of independence: V — φ λ x φ 2 may be based on the empirical measure Vn. Let δ be a metric for probabilities on X x y. One class of test statistics is (3.12) Tn = δ{Vn,V ι nxV 2 n) Extending classical work of Hoeffding (1948), Blum, Kiefer and Rosenblatt (1961), and Bickel (1969), Romano (1989) show that under very mild regu-larity assumptions, the permutation distribution of the test statistic Tn gives an asymptotically consistent locally most powerful test of independence.
Consider next the truncated case explained in Section 1. The hypothesis (1.2) may be called quasi-independence in direct analogy with the similar problem of testing independence in a contingency table with structural zeros.
Clogg (1986) and Stigler (1992) review the literature and history of tests for quasi-independence with references to the work of Caussinus and Goodman.
While optimality results are not presently available in the truncated case, it is natural to consider the permutation distribution of statistics such as (3.12). This leads to a host of open problems in the world of permutations with restricted position.
We were led to present considerations by a series of papers in astrophysics literature dealing with the expanding universe. The red shift data that is collected for these problems suffers from heavy truncation problems.
For example, Figure 1 from Efron, Petrosian (1998) shows a scatterplot of 210 x — y pairs subject to interval truncation, the x coordinate corresponds to red-shift, the y coordinate corresponds to log-luminosity. A suggested theory of 'luminosity evolution' says that early quasars were brighter. This suggests that points on the right side of the picture are higher because the high redshift corresponds to high age.
Astronomers beginning with Lynden-Bell (1971, 1992) have developed permutation type tests based on Kendall's tau for dealing with these prob-lems. There is a growing statistical literature on regression in the presence of truncation; see Tsui et al.(1988) for a survey.
Independence with Truncation 210 observed data points and their boundaries 205 0 0.5 1 1.5 2 2.5 3 3.5 Figure 1. Quasar Data, 210 points, upper and lower limits.
Most previous work deals with one-sided truncation of real-valued obser-vations. The theory and practice is easier here as explained in Section 3.2 .
Efron and Petrosian (1998) have recently developed tests and estimates for the case of two-sided truncation. We develop some theory for their setup in Section 3.3. The following preliminary lemma shows that interval truncation of real valued observations leads to restriction matrices with intervals of ones in each row.
Lemma 3.1 Let a?i,... ,x n take values in an arbitrary set. Let I(xt ) be a real interval. Let yi,2/2, ,2/n be real numbers with yι G I(xχ). Suppose the ordering is chosen so that y\ < y<ι < 2 / 3 < < yn Finally, define a zero-one matrix A of dimension n by ^ ~ \ 0 e f Vj € I(xi) else Then each row in A has its ones in consecutive positions with a one in position (i,z), 1 < i < n.
Proof The intervals may be pictured as in Figure 2, which is translated 206 Persi Diaconis, Ronald Graham and Susan P. Holmes 3 -1 -0 --2 --3 < < < > ( > 0 1 2 3 4 5 6 7 Figure 2. Seven permitted intervals and interior points 10 to the matrix whose transpose is (3.13) 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 Consider row i of A. By definition, An = 1. Suppose that Aij = 1, for some i < j . Thus yj G I(xi) and of course y\ G I(xi). By monotonicity, yi G I(xi) for i < ί < j . A similar argument for j <i completes the proof. • 3.2 One-sided restriction Paired data with {y^Xi) observable, if and only if, yi G (αi,oo) give rise to one sided restriction matrices (yι G (—oo, αi) is similarly treated). This Section shows that a neat theory emerges for one-sided truncation. Sets of permutations consistent with one-sided truncation can be described as follows. Let & = (δi, & 2 ? i & n ) with 1 < b{ < n be positive integers. Let (3.14) Sb = {π : τr(t) > h, 1 < i < n} Independence with Truncation 207 Equivalently, the matrix A has Aij = 1 iff j > bi. Without loss of generality, we take b\ < 62 < h < < bn in the sequel.
Some examples: • If all bi = 1, Sb contains all permutations.
• If bi = 1,1 < i < α, b{ = 2, α + 1 < i < n, Sb contains all permutations with 1 in the first α places.
• If bi = ΐ, 5b contains only the identity permutation.
• If bi = b2 = I, bi = i — 1,1 <i <n Sb contains 2 n~ 1 permutations.
The restriction matrices arising from sets Sb give rise to what are called Ferrers boards in the rook polynomial literature. The following lemma, due essentially to Karl Pearson in 1913 shows it is easy to choose from the uniform distribution on S&.
Lemma 3.2 Let b\ < b<ι < < bi < < bn be positive integers with bi < i,l < i < n. The following algorithm results in a uniform choice from Sb Begin with a list containing 1,2,..., n.
• Choose τr(n) uniformly from J = {j : j > bn}. Delete π(n) from the list of choices.
• Choose π(n—1) from the elements j in the current list J with j > bn-.
Delete π(n — 1) from the list • : and so on...
Proof The algorithm produces an element of Sb without getting stuck because of the restriction bi πW)? with π uniformly chosen in S & . Efron and Petrosian (1992) have introduced an apparently different notion of rank test with a simple distribution theory based on lemma 3.2. The relation between these rank tests and the distribution theory above are open problems.
While Corollary 3.1 is well known in the combinatorial literature, it is often rediscovered by statisticians, see Tsai (1990).
3.3 An example with historical insights Karl Pearson considered a natural source of censored observations in his work on what is now called quasi independence. He considered families with one or more imbecilic children, cross tabulating the family size versus the birth order of the first such child. Clearly a family of j children can only have its first born special child in a position between one and j and consequently, T{j = 0 for i > j. Pearson carried out a test of independence with this truncated dataset in 1913! A historical report on Pearson's work and its later impact is given by Stigler (1992).
It is worth beginning with an exact quote of Pearson's procedure from the article by Elderton et al. (1913).
"Lastly we considered the correlation between the imbecile's place in the family and the gross size of that family. Clearly the size of the family must always be as great or greater than the imbecile's place in it, and the correlation table is accordingly one cut off at the diagonal, and there would certainly be correlation, if we proceeded to find it by the usual product Independence with Truncation 209 moment method, but such correlation is, or clearly may be, wholly spurious.
Such tables often occur and are of considerable interest for a number of reasons. They have been treated in the Laboratory recently by the following method: one variate x is greater than or equal to the other y; let us construct a table with the same marginal tables, such that y is always equal to or less than x, but let its value be distributed according to an "urn-drawing" law, i.e. purely at random. This can be done. We now have two tables, one the actual table, the other one with the same marginal frequencies, would arrive if x and y were distributed by pure chance but subject to the condition that y is equal or less than x, this table we call the independent probability table.
Now assume it to be the theoretical table, which is to be sampled to obtain the observed table, and to measure by χ 2 and P the probability that the observed result should arise as a sample from the independent probability table." We find this paragraph remarkable as an early clear example of the con-ditional permutation interpretation of the chi-square test. A careful reading reveals that Pearson is not explicit about the "urn drawing" commenting only that this can be done. In the rest of this Section we give an explicit algorithm by translating the problem into that of generating a random per-mutation with restrictions of the one-sided type and showing that lemma 3.2 achieves a particularly simple form.
To begin with, it may be useful to give the classical justification for Fisher's exact test of independence in an uncensored table. Let T{j be a table with row sums T{ and column sums Cj . Under independence the conditional distribution of Tij given r^, Cj is the multiple hypergeometric.
This may be obtained and motivated as a permutation test as follows. Sup-pose the n individuals counted by the table have row and column indicators (Xi,Yi)ι<i<n with 1 < Xi < /, 1 < Yi < J. The usual permutation test chooses π G Sn at random forming a new data set (Xi,Yπφ). The I x J table formed by this dataset has the multiple hypergeometric distribution.
Said another way, here is a simple algorithm for generating a random table drawn from the multiple hypergeometric distribution : • Place T V balls in an urn with T{ of color i.
• Draw c\ balls without replacement and set Tn to be the number of balls of color i among these, 1 < i < I.
• Draw C 2 balls from the remainder without replacement and set 22 to be the number of balls color i among these...
210 Persi Όiaconis, Ronald Graham and Susan P. Holmes We can now mimic these computations for the triangular table considered by Pearson. Call the table entries Πij with Πij = 0 for j > i. Suppose n = Σt>j n%j τhe original data can be regarded as (X^ Yk)ι<k<n with Yk < Xk This falls into the truncated data pattern with I(xi) = [l,Xi]. Following the prescription of Lemma 3.1, choose the labels i so that Y\ < Y<ι < < Yn and let Aj = < j ι . The algorithm of Lemma 3.2 translates to the following algorithm to generate the triangular table will row sums ri,... 77, column sums ci,... cj, and Tij =0 if j > i.
• Place c\ balls labeled 1 in an urn and sample τ\ of these without replacement. Let T\ be the number of balls in the sample labeled 1.
• Add C 2 balls labeled 2 to the urn. Sample r<ι from the urn without replacement. Let T21 be the number of balls labeled i in the sample ί = 1,2.
Remark 3.3 It is clear from Pearson's discussion following the quote above that he was aware of essentially this algorithm. He used it to give a closed form expression for the maximum likelihood estimates.
Very similar upper triangular tables arise in genetics in testing good-ness of fit of the Hardy-Weinberg equilibrium model. The analogous exact sampling scheme is well-known. Recently, Markov chain Monte Carlo tech-niques have been used to to do the sampling; Guo and Thompson (1992) derive such an algorithm which is a further studied in Diaconis, Graham and Holmes(1999). Lazzeroni and Lange(1997) give a stopping time ap-proach which is an early example of exact sampling. All of these ideas can be extended to the one-sided censoring case.
3.4 Two-sided restrictions All of the neat factorizations and sampling schemes in Section 3.3 disappear in the case of two-sided truncation. In this Section we introduce a graph structure on permutations in SA This gives a reversible Markov chain on SA which can be run to calibrate permutation tests. Further, the graph structure gives an exchangeable pair so that Stein's method may be used to approximate the distribution of rank statistics as in Stein (1986), Bolthausen (1984), and Bai, Chao, Liang and Zhao(1997).
Lemma 3.3 Let A be the zero-one matrix of dimension n. Suppose that for all i, An = 1 and that the ones in each row of A lie in an interval Define a graph G with vertex set SA and edges between σ and r for σ and r that differ by a transposition of labels. This graph G is connected.
Independence with Truncation 211 Proof By construction, Id € SA Connectedness is proved by showing that any σ € 5^ is connected to Id. This in turn is proved by regarding σ as a product of disjoint cycles and showing that any cycle can be broken into two resulting in another permutation still in SA Towards this end, it is useful to picture the permutations superimposed on the matrix A coloring the places in A corresponding to the permutation representation of σ. For . / 1 2 3 4 5 \ example σ — I I appears as / 1 (3.15) 0 0 \ o 1 / The rows correspond to positions, the columns to labels. An allowable tran-sition can be pictured on A as well. For example, transposing labels 2 and 5 gives I I G SA The transition is pictured via the paren-y 2 1 4 o 5 J 1 Γ] 0 0 0 E 1 1 1 i 1) 1 1 1 Ξ 1 1 0 B 1 1 (1) 0 0 1 E y 2 1 4 o 5 theses in display (3.16): (3.16) In general, transposing two labels is admissible if and only if there are two ones in the two available places in A. In the example, σ is a product of two cycles (1,5,2) and (3,4). It will be useful to picture moving along the cycle on the picture of σ on A. Prom a boxed square at (i, j) move to diagonal (j,j) and then to the unique box in row j . Finally, observe that given a cycle (ή,i2, . ,H) w i t h h smallest, the submatrix A of A formed by rows {ϊi, %2, . , k} and columns {ή, Z2,.. , π} has the row interval property with ones on the diagonal. For example, the cycle (l,5,2)above the gives A as / 1 1 1 \ (3.17) 1 0 0 1 Beginning the cycle with its smallest elements results in A having the left-most boxed element in the first column.
212 Persi Diaconis, Ronald Graham and Susan P. Holmes We may thus study the submatrix corresponding to the cycle (ii,i2j > it) in a matrix A with ones on the diagonal and having the ones in each row in an interval use {ή,22? 5^} to label the rows and columns. Thus the matrix appears as (3.18) 1 1 1 1 1 1 1 ϊ] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 . . .
. . .
1 1 1 1 1 1 1 1 1 1 1 X 1 -I 1 X 1 1 1 1 X 1 1 The proof proceeds in two cases: Case 1 i£ > i2 Then, by the row interval property there is a 1 in positions (ii,ii) and (^,22)- Thus labels 12 and it can be transposed.
Case 2 it <i2 Following around the cycle starting with the box in the first row leads to the diagonal (^2,^2)- Suppose the box at position (22? ^3) in row 12 is to the left of (^2, i<ι) then by the row interval property labels i% and i$ can be transposed. Finally, consider the case where the box in row %2 is to the right of (22,^2) the position is thus pictured: (3.19) 1 (2,2) 1 1 ... 1 1 1 1 1 1 1 1 1 1 The reader may picture the horizontal £JJ and the vertical line ly through (Ϊ2,Ϊ2) The path along the cycle next goes down to (13,12) and continues. Eventually, the cycle path must hit the box in position (it,ii). To do this, it must cross the vertical line £y. Consider the first time this happens, the path must cross the line from right to left winding up in a box at position (iΓ,ic) with ir >'i2,ic < h By the row interval property there are ones between (ir,ic) and (ir,ir) Thus AIM = 1. Further, in the first row A^ = 1. It follows that labels %i and ic can be transposed.
Independence with Truncation 213 This covers all cases and completes the proof. • Remark 3.4 The graph of Lemma 3.3 can be used to run a Markov chain on SA Prom σ G SA choose one of (£) transpositions uniformly at random and transform σ by switching the two chosen labels. If this new permutation is in SA, the walk moves there. If not, the walk stays at σ. This results in a symmetric connected Markov chain which has uniform stationary distri-bution on SA- The chain is aperiodic whenever A is not the matrix entirely filled with ones.
This chain allows calibration of any test statistic. Its proper use needs an estimate of the relaxation time. We have carried out some preliminary work which suggests that order π 2logn steps are sufficient in the case of interval structure for A, this remains a conjecture. We remark that Hanlon(1996) has determined all the eigenvalues of this Markov chain for the case of one-sided truncation.
Remark 3.5 For more general zero patterns, the graph based on transposi-tions need not be connected. For example, consider the derangements in 53.
/0 1 1\ The matrix is 1 0 1 . The two derangements are (1,3,2) and (1,2,3) \l 1 0/ in cycle notation. These are even permutations and cannot be connected by transpositions.
The 3 x 3 matrix for derangements (call it D) can be used to construct larger examples which show the difficulty of making a general theory. For example, construct a 3n x 3n matrix A by placing copies of D down the diag-onal, zeros in the remaining upper triangular part and ones in the remaining lower triangular part. It is easy to see that there are 2 n permutations in SA and that none of these can be connected to any others by transpositions.
As a second example, construct an n x n matrix A with a single copy of D in the upper left hand corner. Place zeros in the remaining places of the first three rows and ones everywhere else. Here, there are two giant components in SA which cannot be connected by transpositions.
We have shown that for n > 4 the set of all derangements is connected by transpositions. The argument proceeds by showing that any derangement can be brought to the n-cycle (1,2,3,4,..., n) in at most n—1 transpositions.
While not exploited in the current paper, there is a large class of examples of sets of permutations which are connected by transpositions; these are the linear extensions of a given partial order. See Brightwell and Winkler (1991) for an overview and references.
Remark 3.6 The Markov chain of Remark 1 above gives an exchangeable pair of random permutations (X, X')> w ^ ^ X chosen from the uniform distri-214 Persi Diaconis, Ronald Graham and Susan P. Holmes but ion on SA and X 1 one step of the chain away from X. Such an exchange-able pair forms the basis of Stein's approach to the study of Hoeίfding's combinatorial limit theorem. Bolthausen (1984) and Schneller (1989) used extensions of Stein's method to get the right Berry-Esseen bound and Edge-worth corrections. Zhao, Bai, Chao and Liang (1997) give limit theorems for double indexed statistics (a la Daniels) of form Σ α(iiJi π(i)> πU)) using Stein's method. Finally, Mann(1995) and Reinert(1998) have used Stein's method of exchangeable pairs to show that the chi-square test for indepen-dence in contingency tables has an approximate χ 2 distribution, conditional on the margins.
We have used the exchangeable pair described above to prove normal and Poisson limit theorems for the number of fixed points in a permutation chosen randomly from the set SA- There is a lot more work to be done. We note in particular that the limiting distribution of linear rank statistics is an open problem with even one-sided truncation. The distribution of Kendall's tau is an open problem in the case of two-sided truncation.
We close this Section with a statistical comment and a useful lemma. The widely used non parametric measure of association Kendall's tau applied to paired data {{xi,Vi)} can be described combinatorially as follows: Sort the pairs by increasing values of X{. Then calculate the minimum number of pairwise adjacent transpositions required to bring {yι\ into the same order as {xi}. When working with restricted positions, it is natural to ask if any admissible permutation can be brought to any other by pairwise adjacent transpositions.
The following example shows that this is not so. For n = 3 consider the / 1 1 1\ / 1 2 3\ matrix I 0 1 0 I. There are two admissible permutations ( I Vi i i/ V ; / 1 2 3 \ and I I. No pairwise adjacent transpositions of the labels is al-\ 3 2 I J lowable. The matrix has the row interval property and all transpositions connect.
It is not hard to see that pairwise adjacent transpositions connect all admissible permutations in the one-sided case.The following lemma proves connectedness in the monotone case: The intervals I(xi) = (αi,&i) can be arranged so that α\ < α<ι < as < < an;bι < 62 < 63 < < K- For the monotone case, order i by a{ increasing. Then, it is easy to see that the restriction matrix Aij = < . j ^ ' has both row interval property and column interval property, and An = 1. Call such a restriction matrix monotone.
Independence with Truncation 215 Lemma 3.4 Let A be α n-dimensionαl monotone restriction matrix. Form a graph with vertex set SA and an edge between two permutations if and only if they differ by a pairwise adjacent transposition of labels. This graph is connected.
Proof By construction, SA contains the identity. The argument proceeds by showing that any permutation π G SA can be brought to the identity by pairwise adjacent transpositions. It is useful to picture π on A by boxing entry (i,j) if π(i) = j.
(3.20) 1 1 0 o \ DD 1 1 1 / 1 2 3 4 represents I 3 4 j 2 A pairwise adjacent transposition corresponds to a basic move in two adjacent rows Recall that π has an inversion at (i,j) if i < j and π(z) > π(j). Only the identity has no inversions. Consider two consecutive rows z,z + 1 , if π(i) > π(i + 1) the picture appears as Consider the first row for which π(i) > ΐ, if π Φ identity, such a row must exist. By the row and column interval property a basic move can be made.
This reduces the number of inversions by one. Continue until no pair of adjacent rows has an inversion. This gives the identity. • Remark 3.7 While lemma 3.3 shows the graph is connected we have found simple examples of monotone restriction matrices where the distance be-tween pairs and agreeing permutations is not given by Kendall's tau. For example, let A be a 4 x 4 matrix with ones everywhere except in the upper right and lower left corners. The two permutations ( 9 1 4 0 ) 2 3 4 \ . I have Kendall's distance 5, but their graph distance is 6.
1 J. 4c Δ J and 216 Persi Diaconis, Ronald Graham and Susan P. Holmes Remark 3.8 In the case of discrete data(contingency tables) arbitrary pat-terns of truncation can be handled using the moves in Diaconis and Sturmfels (1998). For this case, the moves are easy to specify. Suppose the x-variable takes on / levels and the y variable takes on J levels. Form the complete bi-partite graph on the set {1,2,..., 1} x {1,2,..., J}. If cell (i, j) is unob-servable, delete this edge. The circuits in the remaining graph form a con-necting set of moves successively adding and deleting one while traversing the circuit. This suggests two lines of generalization. First, continuous data can be treated by discretization. Second, truncated multivariate data can be approached using the multiway table moves from Diaconis and Sturmfels (1998).
4 An application to ESP guessing experiments This final Section gives a different set of applications for permutations with restricted positions.
A classical test of parapsychology involves a deck of 25 cards made up of the following five symbols of # + « • o Each repeated five times. Under ideal conditions the deck is well shuffled and a guessing subject attempts to guess at the cards in order. Under the natural chance model, each guess has chance | of being correct and so the expected number of correct guesses is five. Of course the distribution of the number of correct guesses depends on the guessing sequence. If the subject always guesses the same symbol, there is no variability. It is not hard to show that the variance of the number of correct guesses is largest if the subject guesses some permutation of the values. In Diaconis and Graham(1981), we studied variations where feedback was given to the subject after each guess.
For example, suppose the subset is shown the card at position i after guess i (complete feedback). Then the optimal strategy is to guess a card with highest frequency among those remaining. Read (1962) shows in this case the expected number of guesses is 8.65. This is of some practical interest since many early experiments were done with feedback and 8.5+ is reported as the highest of average trials in actual trials.
The most interesting type of feedback is yes/no feedback: if a subject guesses correctly, they are told so. If they are incorrect they are only given that information. Now, the subjects optimal strategy is not obvious. We have shown in Diaconis and Graham (1981), that the greedy strategy (guess the most likely value), is only close to optimal. We also determined that the Independence with Truncation 217 expected number of correct guesses under the optimal strategy is 6.63.
The reason for discussing these matters here is twofold. First; the evaluation of the probabilities involved uses permanents. Second; a nat-ural monotonicity conjecture proved by a longish combinatorial argument in Chung, Diaconis, Graham and Mallows (1981) follows from one tool developed to prove the van der Waerden conjecture. To define things, let iV(αi, α2,..., α r; 6χ, b2, , K) be the number of arrangements of a deck of α\ + α2 + αr = n cards with αι of type i, such that symbol one does not appear in the first b\ places, symbol 2 does not appear in places b\ + 1, b\ + 2,..., & i + & 2 > and so on.
This quantity allows evaluation of the probabilities of events like: The next card is type i given bj "no" responses on type j . From the definition, iV(a, b) = Per(M) where M is the n x n zero/one matrix of the form: M = [mi] 1 1 1 1 1 1 1 [m2] 1 1 1 1 1 1 1 [m3] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 N with mi an αι x bi block of zeros and the rest all ones. The inequality to be proved is: Proposition 4.1 For any a, b with JV(a, b) φ 0, and any i, 1 < i < r N{a,b) with e{ = (0,0,..., 1,0,..., 0) the usual ith basis vector.
Remark 4.1 In the card guessing context, the inequality has the following interpretation: in a yes-no feedback experiment; the chance that the guess at the next card is of type i cannot decrease if the next card is of type i and is incorrect.
Proof The argument uses a quadratic form defined on R n. Given positive vectors Vi, V2,..., VU-Ί in RΛ define < x\y >= Per[Vu V2,..., Vn-2, ^ v] This is a symmetric bilinear form on R n. It is used in a crucial way in Egorychev's proof of the permanent conjecture. We follow the account in 218 Persi Diaconis, Ronald Graham and Susan P. Holmes van Lint and Wilson (1992). They show (Theorem 12.6) that this form is Lorentzian having n — 1 negative eigenvalues and one positive eigenvalue.
Such forms are easily seen to satisfy a "reverse Cauchy-Schwarz inequality" for x positive and y arbitrary.
(4.21) 2 > By continuity, (4.21) also holds if some of the entries in V or x are allowed to be zero. Proposition 4.1 follows from (4.21). By symmetry of the permanent, it is enough to prove it for er. Consider the three matrices corresponding to αr x 6r, αr x br + er, αr x br + 2eΓ. Move these rth blocks to the right of the full matrix. The last two columns of the full matrices with these blocks appear as αr ' 1 1 1 0 0 1 " 1 1 1 1 ' 1 1 1 0 0 1 " 1 1 0 : 0 " 1 1 1 1 1 1 1 ' 1 1 1 1 1 All other columns axe the same, matrix x and y and apply (4.21). • Call the two columns from the first Remark 4.2 In Chung, Diaconis, Graham and Mallows (1981) it was in fact shown that Πk = AΓ(α, b + kei) is log-concave: n\ > n/-+infc_i. Their proof was combinatorial and only worked for zero-one matrices. It is not clear if there is an analog of log-concavity for more general Lorentzian forms.
Acknowledgements. We thank Brad Efron for providing the original motivation and examples of truncated data, as well as many ideas on the relation to quasi-independence, Steve Stigler for the explanation of Pearson's work, Alistair Sinclair for help with the permanent literature, Steve Fienberg for pointers to the literature on structural zeros and Marc Coram, Mark Huber and Jim Fill for reading the manuscript carefully.
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PERSI DIACONIS MATHEMATICS AND STATISTICS SEQUOIA HALL STANFORD UNIVERSITY CA 94305-4065 RONALD GRAHAM COMPUTER SCIENCE UNIVERSITY OF CALIFORNIA AT SAN DIEGO AND ATT, FLORHAM PARK, NJ.
graham @ucsd. edu SUSAN HOLMES STATISTICS STANFORD UNIVERSITY AND UNITE DE BIOMETRIE INRA-MONTPELLIER, FRANCE susan @stat. Stanford, edu |
7348 | https://www.nagwa.com/en/videos/972184951295/ | Question Video: Properties of Reciprocal Functions | Nagwa
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Question Video: Properties of Reciprocal Functions
The figure shows the graph of 𝑦 = 1/𝑥. Write down the two asymptotes of 𝑦 = 1/𝑥. What is the domain of the function? What is the range of the function?
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The figure shows the graph of 𝑦 equals one divided by 𝑥. Write down the two asymptotes of 𝑦 equals one divided by 𝑥. What is the domain of the function? What is the range of the function?
Asymptotes will be lines that the graph approaches but never touches, which can be found here at 𝑥 equals zero and here at 𝑦 equals zero. Again, these are lines that the graph approaches but never actually touches.
And if we think about the equation itself, 𝑦 equals one divided by 𝑥, if we would plug in zero for 𝑥, one divided by zero is undefined. It cannot happen. So 𝑥 cannot be zero, which is why the graph never actually touches that.
And then if we would switch it, if we would make 𝑦 be zero, one divided by what gives us zero? You can’t take one and divide it by any number at all and get zero. It’s not possible. So 𝑦 cannot be zero.
Next we want to know what is the domain of the function. These are all of the values that 𝑥 can take on. Well, looking at our graph, any number for 𝑥 will work except for one. 𝑥 cannot be zero. Therefore, 𝑥 is in the real numbers, except 𝑥 cannot be zero.
And next we need to find what is the range of the function. And the range deal with the values for 𝑦. And any value of 𝑦 will work except for 𝑦 equals zero. Therefore, 𝑦 is in the real numbers, except 𝑦 cannot be zero. So these will be our final answers.
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7349 | https://artofproblemsolving.com/wiki/index.php/System_of_equations?srsltid=AfmBOooVS2-1TnGtcICD0pxqX8lp2hANnJHcQqvj7HAWES-VawJeejeN | Art of Problem Solving
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System of equations
A system of equations is a set of equations which share the same variables. Below is an example of a system of equations.
Contents
[hide]
1 Solve 2 variable equations in less than 5 seconds!!!
2 Solving Linear Systems
2.1 Gaussian Elimination
2.1.1 Problem
2.1.2 Solution
2.2 Substitution
2.2.1 Problem
2.2.2 Solution
2.3 Graphing
2.3.1 Problem
2.3.2 Solution
2.4 Advanced Methods
3 Convenient Systems
3.1 Symmetry
3.2 Clever Substitution
4 Problems
4.1 Introductory
4.2 Intermediate
5 See Also
Solve 2 variable equations in less than 5 seconds!!!
Video Link:
Solving Linear Systems
A system of linear equations is where all of the variables are to the power 1. There are three elementary ways to solve a system of linear equations.
Gaussian Elimination
Gaussian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together. Let's look at an example:
Problem
Find the ordered pair for which
Solution
We can eliminate by adding twice the second equation to the first:
Thus . We can then plug in for in either of the equations:
Thus, the solution to the system is .
Substitution
Main article: Substitution
The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We'll show how to solve the same problem from the elimination section using substitution.
Problem
Find the ordered pair for which
Solution
The first equation can be solved for :
Plugging this into the second equation yields
Thus . Plugging this into either of the equations and solving for yields .
Graphing
The third method for solving a system of linear equations is to graph them in the plane and observe where they intersect. We'll go back to our same example to illustrate this.
Problem
Find the ordered pair for which
Solution
We graph the two lines as follows:
From the graph, we can see that the solution to the system is .
Advanced Methods
Matrices can also be used to solve systems of linear equations. In fact, they provide a way to make much broader statements about systems of linear equations.
There is a whole field of mathematics devoted to the study of linear equations called linear algebra.
Convenient Systems
Some systems can be solved by taking advantage of specific forms. Such systems can often seem tough to solve at first, however.
Symmetry
Consider the below system.
The key here is to take advantage of the symmetry. If we add up all 5 equations we will have a total of 4 of each variable on the LHS. On the RHS we will have . Thus
So then subtracting the first equation from this leaves on the LHS and on the RHS. Subtracting this equation from the second equation leaves on the LHS and on the RHS. And thus we continue on in this way to find that
Clever Substitution
Consider the below system.
We can let and to get the two-variable linear system below.
Solving the system results in and . Substituting that back results in and . We can do another substitution by letting and substituting to get . Rearranging results in , so . Finally, by substituting back in, we get . Plugging back satisfies the system.
Problems
Introductory
2002 AMC 8 Problems/Problem 17
2007 iTest Problems/Problem 2
Intermediate
1989 AIME Problems/Problem 8
1993 AIME Problems/Problem 3
See Also
Algebra
Substitution
Retrieved from "
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7350 | https://www.youtube.com/watch?v=FTnDRdHFlUo | Kirchhoff's Current Law explained
James Electronics Tutorials
141 subscribers
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Posted: 4 Feb 2024
This video provides a basic explanation of Kirchhoff's Current Law and how it may be used in the form of nodal analysis for a circuit.
0:00 Intro
0:31 KCL explained
3:12 Problem one
5:49 Problem two
8:42 Outro
3 comments
Transcript:
Intro hi there welcome back to the channel and in today's video we will just be complementing last week's video covering kof's voltage law by now covering kof's current law so I'll just be starting by going through an explanation of the equivalent formula for kof's current law and how to interpret that by means of a circuit and we will put this to the test and verify kof's current law by solving some circuit problems KCL explained so let's start by discussing kof's current LA statement K's current no states that the current flowing into a node or a junction which we can write as the summation of the absolute value of the input current must be equal to the current flowing out of it which we can write as the summation of the absolute value of the output current now what do I mean mean by absolute value for those of you who don't know what absolute does is takes any negative value and turns it into a positive value and I can show you a sine wave to represent this so this is a typical sine wave here and if I was to take the absolute value of this sine wave this is what it would look like see it takes any negative part and turns it into a positive part so now we've covered that let's try to interpret interpret this formula I mean for a circuit because that's what we want to do right so let's start by drawing one here we go this is R1 R2 R3 and this is V1 voltage source and this is also node a so we have this V this current that flows out from the voltage source through this resistor which decreases the voltage as it reaches this node and at this point the current flowing into this node must be equal to the current that flows out this way and out this way I3 I2 and this is i1 so it can be said therefore that i1 equal I2 + I3 and rearranged i1 - I2 - I3 = 0 and that's how you verif ver ify KCl for a circuit so now I've covered that Basics I think we're ready to solve some Problem one problems okay let's start with problem number one what I want you to do is verify KCl for this circuit take your time if you need to and pause the video to work on the solution to start by solving this problem let's find the equival resistance so since we have just two parallel resistors we can use product over sum 1 2 over 1 + 2 which will give you a value of 2 over3 kiloohms now take note this is different from what we've been dealing with in previous videos what kilo means is time 10 10 to the^ of 3 so now we found the equivalent resistance let's find the equivalent current flowing through the circuit so 12 / 2 over 3 10 3 would give you 18 milliamps again take note the small m means M and that means 10us 3 and that's your total current so now we found the total total current flowing through the circuit we can start by verifying K by finding the current that flows across R1 and R2 so let's start with R1 use Ohm's law 12 / by by 1 10 to the 3 would give you 12 milliamps likewise for I2 12 / 2 10 3 will give you 6 milliamps so let's take a closer look at the circuit 18 milliamps flows into this node and 6 milliamps will flow that way 12 milliamps will flow that way therefore this verifies KCl for the circuit and the problem is Problem two complete same style of problem once again what I want you to do for this circuit is to verify kof's current law take your time if you need to and pause the video to work on the solution okay the first step would be to find the equivalent resistance so this time R1 is in series but however R2 and R3 are in parallel so 1 + 3 6 over 3 + 6 that's the same as 18 over 9 which is equal to 2 so our answer is 3 kils that's our total resistance so now we have the total resistance we can find the the total current flowing through the circuit so 9 volts over 3 kilms will give you 3 milliamps that's the total current so now we found the total current let's find the voltage drop across R1 because it's in series with the with the power source so voltage drop across R1 is 3 milliamps Time 1 kilohm which will give you 3vt drop so cuz 3 Vols is dropped here across R1 six vol alts will make it to node a so this is important because when we find the current that flows across R2 and R3 you will see that it sums up to 3 milliamps when we use when we account for the voltage drop I should say so six over 6 No 3 kilohms will give you 2 milliamps and for R3 6 over 6 kiloohms will give you 1 milliamp which is correct because 3 milliamps minus 2 milliamps - 1 mamp equals z that verifies KL for this circuit so Outro when thank you very much for watching I hope you enjoyed the video I learned something new about cra's current law and next week I will be introducing you to complex numbers and why we use them in electrical engineering |
7351 | https://www.teachoo.com/1725/524/Ex-7.1--7---Find-point-on-x-axis-which-is-equidistant-from/category/Ex-7.1/ | Ex 7.1, 7 - Find point on x−axis which is equidistant from
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Chapter 7 Class 10 Coordinate Geometry
Serial order wise
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Chapter 7 Class 10 Coordinate Geometry
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Ex 7.1, 7 - Chapter 7 Class 10 Coordinate Geometry
Last updated at December 13, 2024 by Teachoo
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Transcript
Ex 7.1, 7 Find the point on the x−axis which is equidistant from (2, –5) and (–2, 9). Let the given points be P (2, −5) , Q (−2, 9) And the point required be R (a, 0) As per question, point R is equidistant from P & Q Hence, PR = QR Note: Since the point is on the x−axis, y = 0 And assuming x = a Hence the point is (a, 0) Finding PR x1 = 2, y1 = −5 x2 = a, y2 = 0 PR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 𝑎 −2)2+(0−(−5))2) = √((𝑎−2)2+(5)2) = √(𝑎2+22 −2(2)(𝑎)+(5)2) = √(𝑎2+4 −4𝑎+25) = √(𝑎2 −4𝑎+29) Finding QR x1 = −2, y1 = 9 x2 = a, y2 = 0 QR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 𝑎 −(−2))2+(0−(9))2) = √((𝑎+2)2+(−9)2) = √(𝑎2+22+2(2)(𝑎)+(9)2) = √(𝑎2+4+4𝑎+81) = √(𝑎2+4𝑎+85) From the question PR = QR √(𝑎2 −4𝑎+29) = √(𝑎2+4𝑎+85) Squaring both sides (√(𝑎2 −4𝑎+29) )2 = (√(𝑎2+4𝑎+85))2 a2 – 4a + 29 = a2 + 4a + 85 a2 – 4a − a2 − 4a = 85 – 29 −8a = 56 a = 56/(−8) a = −7 Hence the required point is R(a, 0) i.e. (−7, 0)
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Chapter 7 Class 10 Coordinate Geometry
Serial order wise
Ex 7.1
Ex 7.1, 1 (i)
Ex 7.1, 1 (ii)
Ex 7.1, 1 (iii) Important
Ex 7.1, 2
Ex 7.1, 3 Important
Ex 7.1, 4
Ex 7.1, 5 Important
Ex 7.1, 6 (i)
Ex 7.1, 6 (ii) Important
Ex 7.1, 6 (iii)
Ex 7.1, 7 Important You are here
Ex 7.1, 8
Ex 7.1, 9 Important
Ex 7.1, 10
Ex 7.2→
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7352 | https://news.ycombinator.com/item?id=24287622 | | | | |
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| | | | --- | | Hacker Newsnew | past | comments | | | | | login | |
| | | | Problem Solving with Algorithms and Data Structures [pdf] (auckland.ac.nz) | | 180 points by jeremylevy on Aug 26, 2020 | hide | past | favorite | 17 comments | | | | --- | | | | | throwingawaymat on Aug 27, 2020 | next [] I like the careful review of python at the start. The "stories" in the writing seem good. If you want to learn by reading somebody explain how to solve a problem, I think this is a good text for you. I don't think there are enough exercises... with a title of "problem solving..." I was expecting more focus on (challenging?) problems in the exercises. If you want to learn by thinking about how to solve problems, this is not the text for you. I could not find a comparison of the orders of various operations on ADT's. I thought I would find something like that in Chapter 3. I think that omission makes it hard to understand the big picture. Details seem raw: The table on page 9 shows the Pythonic not equals operator is =!. 2.7 Programming Exercise 4 should really be specifying linear _in what parameter(s)_ for clarity. The operation is O(nk) presumably. Exercise 5 assumes that n = len(data) from Exercise 4. The claim that O(n log(n)) < O(n k) depends on k and n, but that isn't clearly stated. 3.14 Programming Exercise 7: this exercise has in mind only one particular solution. Does not a double ended queue have O(1) enqueue and dequeue? | | | | | | i-am-curious on Aug 27, 2020 | parent | next [] > Does not a double ended queue have O(1) enqueue and dequeue? Yes. It's a basic data structure implementable using vector or linked lists. When using the linked list, you can even do erase in O(1) assuming you have the pointer already. This also turns out to be super useful | | | | | | joeclef on Aug 27, 2020 | prev | next [] There is an updated and interactive version of this book here | | | | | | Gehinnn on Aug 27, 2020 | prev | next [] You might find the Debug Visualizer extension for VS Code helpful to visualize algorithms and datastructures . It kind of works with python, but works best with Javascript/TypeScript. | | | | | | snaveen on Aug 27, 2020 | parent | next [] Thanks for the mentioning this! | | | | | | mettamage on Aug 27, 2020 | prev | next [] I'm not impressed with the dynamic programming part, which is what I'm trying to learn right now. It simply lists it under recursion and after skimming, I see no mention of bottom up DP (iteratively) and there's only one problem about it. I'm still looking for a good resource, but this one gave me quite a bit of insight . Since it reuses the same problem over and over again and tweaks it, which gave me a better view of DP. | | | | | | dragontamer on Aug 27, 2020 | parent | next [] "Algorithm Design" by Kleinberg & Tardos is pretty good. Chapter 6 is its Dynamic Programming chapter, and there are 9 examples in the chapter with pseudocode and analysis. Some of its dynamic programming examples: Weighted Interval Scheduling, Segmented Least Squares, Subset Sums / Knapsacks, RNA Secondary Structure, Sequence Alignment (aka: Diff), Shortest Path in Graphs. All in just a ~50 page chapter. The other chapters on Network Flow, Greedy Algorithms are also excellent. "Algorithm Design" is an advanced text, but that's how it covers more material than easier textbooks since it won't waste your time on the beginner level stuff. ------------- Honestly, given the beginner-nature of the textbook here, I think they're doing a decent job. Dynamic programming isn't as fundamental as the stack / queue / sorting / searching etc. etc. that's being discussed in "Problem Solving with Algorithms and Data Structures". Dynamic programming is definitely something that should be reserved for more advanced textbooks (with maybe, at best, an introduction to the subject at this level). | | | | | | mettamage on Aug 27, 2020 | root | parent | next [] > Honestly, given the beginner-nature of the textbook here, I think they're doing a decent job. Dynamic programming isn't as fundamental as the stack / queue / sorting / searching etc. etc. that's being discussed in "Problem Solving with Algorithms and Data Structures". Ah, I think I set my expectations wrong with regards to the level. My bad. | | | | | | tsumnia on Aug 27, 2020 | prev | next [] What I love the most about this is the explicit showing of how to build the various data structures. Many DS courses will make implementing the structure the homework assignment, but the reality is the algorithms have been implemented to death online and the real benefit comes from analyzing which is most optimal. It restricts the possibilities for automatic grading, but students at this level should be progressing towards the "Evaluate" phase of Bloom's hierarchy any way. They should be getting practice assessing the benefits between data structures, not trying to reinvent the wheel. That said, many job interviews expect people to be able to implement data structures on the fly. However, I'd say that's more a limitation of the interview process rather than an issue with the textbook. | | | | | | betimsl on Aug 27, 2020 | prev | next [] I don't want to be sarcastic, but, with what else can you solve problems if not with DS and Algos? Is there any other methods that we do not know of? | | | | | | codetrotter on Aug 27, 2020 | parent | next [] You can solve them with pen and paper. And yes, you are using algorithms then too but not in the same way as on a computer. But anyways I think you read the title kind of opposite. Instead I think it is meant to reflect that algorithms and data structures, which in general have many uses, are here being used for the purpose of problem solving. I.e. to let readers know that it is specific. | | | | | | neoplatonian on Aug 27, 2020 | parent | prev | next [] Lol. Like, the rest of math? Geometry? Constructions? Counter examples? Probabilistic methods? Non-computable theorems? | | | | | | rat9988 on Aug 27, 2020 | root | parent | next [] I always felt Math use data structures and algorithms. Properties of whatever you manipulate let you have some sort of known structure with proven theorems, then you use these theorems one after an other until you get to where you maybe "want" to be . | | | | | | qntty on Aug 27, 2020 | root | parent | next [] It seems to me that mathematical objects are often more abstract than data structures are required to be. For example, you can reason about a triangle without specifying how you would represent it. You could represent a triangle as a list of three points, or as a list of side lengths and angles or even a list of angles which forces you to reason about the class of similar triangles with those angles. In fact, to represent certain algebraic structures is an entire subfield of algebra (representation theory) distinct from the parts of algebra that reason about those structures without trying to represent them. | | | | | | howling on Aug 27, 2020 | root | parent | next [] Representation theory is about study of functors between category of vector spaces and category of interest, which is quite different from the problem of different representations of a triangle. | | | | | | qntty on Aug 31, 2020 | root | parent | next [] Different how? They are both ways of making abstract descriptions of an object more concrete, which is analogous to the job of a programmer designing a data structure to concretely describe an abstractly specified computational object. My point is just that in math, you can reason about an object even if you can't concretely represent it. For example, you can figure out properties of a number or function that you don't know enough about to write down. But in programming you can't use an object unless you have a concrete representation of it, and often getting that concrete representation (in both math and programming) requires a kind of work which is quite different than the work of abstractly describing or specifying it. So it's maybe not always appropriate to call an abstract description of a mathematical object a "data structure", since often it's more analogous to (say) the interface of an object than it's implementation. | | | | | | billforsternz on Aug 26, 2020 | prev [] I am going to try and find time to read more of this. I really enjoyed a quick session with it. It felt like a model of clarity and good organization. | | |
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7354 | https://www.purplemath.com/modules/ineqrtnl.htm | Home
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Solving Rational Inequalities
Examples
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Solving rational inequalities is very similar to solving polynomial inequalities.
But because rational expressions have denominators (and therefore may have places where they're not defined), you have to be a little more careful in finding your solutions.
Content Continues Below
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How do you solve rational inequalities?
To solve a rational inequality, use these steps:
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If needed, move all the terms to one side of the inequality symbol, with zero on the other side.
If needed, combine all rational expressions into one polynomial fraction.
Factor the numerator and denominator completely.
Solve the factors for their zeroes; keep in mind that the denominator's zeroes would cause division by zero, so they cannot be included in your solution.
Use the zeroes to divide the number line into intervals.
Make a table of factors, showing where each factor is less than and greater than zero.
Multiply the factors' signs on each interval (that is, down the table's columns) to find the sign of the rational expression on that interval.
Select the intervals that match the inequality they gave you; remember to discard any endpoints that would cause division by zero.
Let's see how these instructions work in practice:
Solve the following:
They've already put this inequality into (one rational expression) with (zero) on the other side. So I can start with factoring everything:
This polynomial fraction will be zero wherever the numerator is zero, so I'll set the numerator equal to zero and solve:
(x + 2)(x + 1) = 0
x + 2 = 0 or x + 1 = 0
x = −2 or x = −1
The fraction will be undefined wherever the denominator is zero, so I'll set the denominator equal to zero and solve:
(x + 4)(x − 4) = 0
x + 4 = 0 or x − 4 = 0
x = −4 or x = 4
These four values, −4, −2, −1, and +4, divide the number line into five intervals, namely:
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(−∞, −4)
(−4, −2)
(−2, −1)
(−1, 4)
(4, +∞)
I could use "test points" to find the solution to the inequality, by picking an x-value in each interval, plugging it into the original rational expression, simplifying to get a numerical answer, and then checking the sign, but that process gets long and annoying (and is prone to errors), so I'll use the easier and faster factor-table method instead.
My factor table looks like this:
My table has one row for each factor, a row for the number line, and a row for the rational expression. Each row is split into columns, with each column corresponding to one of the intervals on the number line.
Content Continues Below
The sign of the overall rational expression is a result of the signs of its various factors, so I need to find where each factor is positive:
x + 4 > 0 for x > −4 x + 2 > 0 for x > −2 x + 1 > 0 for x > −1 x − 4 > 0 for x > 4
Now I can put "plus" signs on the intervals in each row where that row's factor is positive:
Wherever a factor isn't positive, it's negative, so I'll put "minus" signs in the remaining entries of each row:
I know that the product of an even number of "minus" signs is a plus; the product of an odd number of "minus" signs is a minus. So, by multiplying the signs down the columns (or just counting up the minusses), I get the overall sign of the rational expression on each interval:
Then the rational is positive on the intervals (−∞, −4), (−2, −1), and (4, +∞).
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Looking back at the original exercise, this is an "or equal to" inequality, so I need to consider the interval endpoints, too.
If this were a polynomial inequality, I could just throw all the interval endpoints into the solution, and I'd be done. For rational expressions, though, I have to be careful not to include any x-values that would cause division by zero.
The intervals' endpoints are −4, −2, −1, and 4. I can include −2 and −1 in the solution, because they just make the expression equal to zero by making the numerator zero. But plugging −4 or 4 into the rational expression would cause division by zero, making the rational expression undefined, so I can't include these values in the solution.
Then my full solution is:
I wrote my solution above in "interval" notation. If you have to write your solution in "inequality" notation, it would look like this:
x < −4, −2 ≤ x ≤ −1, and x > 4
Don't forget: "Infinity" is not a "number" in the way that, say, "2" is. "Infinity" cannot be "included" in your solution, so never draw a square bracket next to an "infinity" "endpoint".
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You can use the Mathway widget below to practice solving rational inequalities. Try the entered exercise, or type in your own exercise. Then click the button and select "Solve the Inequality for x" (or just "Solve for x") to compare your answer to Mathway's. (Or skip the widget,and continue with the lesson.
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7355 | https://advance.sagepub.com/users/627556/articles/648473/master/file/data/quadratics/quadratics.pdf?inline=true | Received: Added at production Revised: Added at production Accepted: Added at production DOI: xxx/xxxx
SURVEY PAPER
The Ins and Outs of Solving Quadratic Equations with Floating-Point Arithmetic
Frédéric Goualard
1Nantes Université, École Centrale Nantes, CNRS, LS2N, UMR 6004, Nantes, France
Correspondence
Frédéric Goualard. LS2N, 2, rue de la Houssinière, BP 92208, F-44322 NANTES CEDEX 3. Email: frederic.goualard@ls2n.fr
Summary
Solving quadratic equations with radicals on a computer with floating-point arith-metic requires great care to handle correctly all possible parameters. Literature on the subject glosses over the details, often considered as important but tedious to present. As a consequence, most implementations are flawed in one way or another. After having reviewed both the literature and the actual implementations in several programming languages and applications, we present an algorithm inspired from an exposition by Pat Sterbenz from 1974, adapted to take advantage of more recent re-searches in the field, which leads to a robust quadratic equation solver.
KEYWORDS:
Floating-point arithmetic, numerical errors, robustness, quadratic equation
1 INTRODUCTION
How many solutions to the quadratic equation:
𝑝 (𝑥 ) = 𝑥 2 + (1 + 2 −52 ) 𝑥 + 251 + 1 253 = 0? (1) Only one solution, as GNU GSL 2.7, Scilab 2023.0.0, and the Rust 1.69.0 mathematical library would have us believe? Two real solutions, as the Boost C++ libraries 1.82.0 and the Racket mathematical library v.8.8[cs] report? Or, maybe, two complex solutions, as found by Octave 8.2.0, the Python Numpy library 1.22.3, and MATLAB R2023a? What about 2600 𝑝 (𝑥 ) = 0 ? It definitely should have the same solutions as 𝑝 (𝑥 ) = 0 , yet GNU GSL, the Rust mathematical library, the Racket mathematical library and the Boost C++ libraries now consider that the equation has no real solution at all! Equation (1) can be solved easily by hand, and there are indeed two real solutions very close to one another. On devising an up to par quadratic equation solver, Forsythe 1 would say in 1966: “ I venture to guess that no more than five quadratic solvers exist anywhere that meet the general level of the specifications. ” More than half a century later, it seems things have not changed, or maybe for the worst. The implementers of mathematical libraries are not the only ones to blame, however. Since Forsythe’s 1966 article, aptly titled “How do you solve a quadratic equation? ”, several authors have written about the dangers lurking in the implementation of a robust quadratic equation solver, but very few, if any, have described in all the excruciating details the steps to take in order to avoid them. As Forsythe 1 put it, “ They [The details] are extremely important to actual computing, but carry less general interest than ideas just presented .” But, as the saying goes, the devil is in the details. For want of a precise exposition of a reference algorithm to robustly handle all possible inputs, actual implementations seem all to be flawed in one way or another. A robust algorithm to solve quadratic equations with radicals is the cornerstone of many more elaborate algorithms, like Muller’s method 2 for example. As Forsythe quips, “ School examples do factor with a frequency bewildering to anyone who has done mathematics outside of school! ”, but real life problems tend to exert a solver to the utmost, requiring it to deliver 2 F. Goualard
correct solutions for inputs large and small, and to handle gracefully exceptional values such as IEEE 754 3 Not-a-Number s and infinities, as well as quadratics degenerating into linear equations. We present in Section 5 a complete algorithm to solve quadratic equations with radicals, for real solutions only—finding complex solutions does not bring new problems, anyway. Contrary to our predecessors, no detail was considered too puny to be exposed. Readers who only wish to implement a robust solver can jump straight to that section after having read Section 2 for the necessary notations. An implementation of the algorithm in the Julia language is given in Program 7; it is also available on github. Most of the ideas underlying the algorithm were sketched by Sterbenz 4 in 1974 for a computer using hexadecimal pre-IEEE 754 floating-point arithmetic, in a book that appears to have been long out-of-print. It is rather distressing to observe that many writers have tried since to reimplement, often less correctly, such an algorithm. In Section 3, we take a look at algorithms presented in the literature since 1974, pointing out some of their flaws in the process; Section 4 does the same for actual implementations in current software.
2 FLOATING-POINT ARITHMETIC
The IEEE 754 standard 3 defines the representation and the properties of floating-point arithmetic on most modern computers. We only present in this section the elements that are relevant to the understanding of the algorithms presented in Sections 3 to 5. A more in-depth exposition can be found by the interested reader in Muller et al. ’s book. 5
The IEEE 754 standard defines a floating-point number 𝑣 (“ float ”, for short) as a binary 1 value of the form:
𝑣 = (−1) 𝑠 × 𝜎 × 2 𝐸 ,
where:
• 𝑠 ∈ {0 , 1} is the sign bit ;
• 𝜎 = 𝑏 0.𝑏 −1 𝑏 −2 ⋯ 𝑏 1− 𝑝 (𝑏 𝑖 ∈ {0 , 1} ), is the significand , with one bit for the integer part and 𝑝 − 1 bits for the fractional part 𝑓 ;
• 𝐸 ∈ [ 𝐸 min , 𝐸 max ] is the exponent .We note 𝑚 = (−1) 𝑠 × 𝜎 the signed significand. Given 𝑣 a floating-point number, let sign ( 𝑣 ) be defined as:
sign ( 𝑣 ) =
{ −1 if 𝑣 < 0; 1 if 𝑣 ⩾ 0. (2) A floating-point number is stored as a binary string in memory, with 𝑠 , 𝐸 , and 𝑓 stored contiguously; the integer part of the significand is not stored as it is inferred from the value of the stored exponent. The size of the string determines the precision of the floating-point format. Two formats of note are:
• The single precision format, stored as a 32-bit string;
• The double precision format, stored as a 64-bit string. To ensure continuity of the computation, the IEEE 754 standard defines special floating-point values:
• −∞ and +∞ , to obtain an affine extension of the real number system. When a result is too large in magnitude to be represented in the floating-point format, it is replaced by an infinity. This is an overflow situation;
• When some computation has no meaning over the reals (e.g., √−1 , or 0∕0 ), its result is represented by an Not-a-Number
(NaN). Some configurations of the binary string used to store the exponent 𝐸 are reserved to encode these special values. A normal float is a float whose integer part of the significand is “ 1”; a subnormal float (with an integer part equal to “ 0”) has an exponent equal to 𝐸 min . An underflow occurs when the result of a computation is so small that it must be coded by a subnormal float. This is a situation we usually try to avoid as subnormals may lead to a loss of precision. For the normal floats, the
1The latest revisions of the standard define both decimal and binary floating-point arithmetic. The binary version is still the one used the most, though. F. Goualard 3
multiplication and division by a power of 2 is errorless in the absence of underflow and overflow, as it only requires manipulating their exponent; the division by a power of 2 that underflows may not be errorless as we cannot decrement the exponent past
𝐸 min , and the binary point of the significand needs to be right-shifted in order to complete the operation, potentially losing bits in the process. Figure 1 shows the position on the real line of the floats from a format with 2 bits for the exponent and 4 bits for the significand. As can be seen on the figure, the distance ulp ( 𝑣 ) from a float 𝑣 = 𝜎 × 2 𝐸 to the next is not uniform, and equal to 21− 𝑝 × 2 𝐸 . The quantity 21− 𝑝 is called the epsilon “𝜀 ” of the format. It is sometimes defined as the distance from 1 to the smallest float greater than 1.-3.75
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-1.875
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-1.625
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-1.0
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-0.375
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0
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0.875
1.0
1.125
1.25
1.375
1.5
1.625
1.75
1.875
2.0
2.25
2.5
2.75
3.0
3.25
3.5
3.75
- +
Figure 1 Floating-point numbers on the real line (2 bits for the exponent and 4 bits for the significand). Normal number are in blue below the line, and subnormals are in magenta above it. The set of floating-point numbers is not closed for arithmetic operations (the sum of two floats may not be a float, for example). The results may, therefore, need rounding to be represented as floats. For the arithmetic operations, the IEEE 754 standard mandates that the floating-point result be the closest float to the real result ( correct rounding ). For a real value 𝑤 , we note f l ( 𝑤 )
its floating-point representation. Given two floating-point values 𝑣 1 and 𝑣 2, we have:
⎧⎪⎨⎪⎩|||f l (𝑣 1⊤𝑣 2
) − ( 𝑣 1⊤𝑣 2)||| ⩽ ulp (f l (𝑣 1⊤𝑣 2))
2
, for ⊤ ∈ {+ , −, ×, ÷};
||||f l
(√𝑣 1
)
− √𝑣 1
|||| ⩽ ulp
(
f l
(√𝑣 1
))
2
.
thanks to the correct rounding. As a shorthand, we will write f l ⟨expr ⟩ to express the fact that each step of the evaluation of the numerical expression “expr” is rounded according to the floating-point format in use. Example:
f l ⟨𝑥 2 − 𝑦 2⟩ ≡ f l (f l (𝑥 2) − f l (𝑦 2)) .
When solving quadratic equations, there are two phenomena that will be of importance because they may induce very large errors in the resulting solutions, or even in the counting of the number of solutions:
Absorption. Summing or subtracting two floats requires equating their exponents. This is done by incrementing the exponent of the smallest value in magnitude, and right-shifting the binary point of its significand accordingly. As such, we may lose some bits from the significand that cannot be represented anymore. Consider, for example the addition with 4-bit significands 1.010 × 2 10 + 1 .101 × 2 2:
1.010 ×2 10
1 .101 ×2 2 ←→ 1.010 ×2 10
0 .000 00001101 ×2 10
The underlined red bits cannot be represented in the registers of the machine and are lost. The actual computation is then
1.010 × 2 10 + 0 .000 × 2 10 . The larger value absorbed the smaller one;
Cancellation. When subtracting very close values, the only bits kept are the significands’ rightmost ones. If the operands of the subtraction originate from a previous computation, these bits may be the result of successive roundings. Consider, for example the subtraction with 8-bit significands 1.0010 111 × 2 0 − 1 .0010 010 × 2 0, where the underlined red bits are the results of roundings in previous computations:
1.0010 111 ×2 0
− 1 .0010 010 ×2 0
0.0000 101 ×2 04 F. Goualard
After renormalizing the result, we get a value 1.01 × 2 −5 whose bits are all the uncertain ones. The subtraction of close operands does not add new errors but it reveals and magnifies the past ones. The cancellation becomes a catastrophic cancellation when most or all correct bits disappear in the subtraction. The code and the examples presented in the next sections consider double precision only, even though the exposition of the main algorithm in Section 5 is done for an arbitrary size. Table 1 synthesizes the various parameters of the single and double formats.
Table 1 Parameters of single and double precision IEEE 754 floating-point formats. Name 𝑝 𝐸 min 𝐸 max 𝜀
Single precision 24 -126 127 2−23
Double precision 53 -1022 1023 2−52
3 A REVIEW OF THE LITERATURE
The method to solve quadratic equations of the form:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 , (𝑎, 𝑏, 𝑐 ) ∈ ℝ3
has been known for a very long time.According to Smith, 6 the ninth century indian mathematician Sr¯ ıdhara presented it in its current form in his book Ganita-Sara : First, compute the discriminant Δ:
Δ = 𝑏 2 − 4 𝑎𝑐 (3) Then, depending on the sign of Δ, there are either no real solution, one, or two real solutions:
⎧⎪⎨⎪⎩
Δ < 0 ∶ No real solution;
Δ = 0 ∶ 𝑥 = − 𝑏
2𝑎
;Δ > 0 ∶ 𝑥 1,2 = −𝑏 ±√Δ2𝑎 .
(4) Even though the naive algorithm based on Equation (4) was used without modification on early machines, 7 computer scientists quickly understood that it was not a good idea to use an algorithm devised for real numbers to solve quadratics on computers with floating-point arithmetic. As summarized by Metropolis (of Monte Carlo method fame) in a 1973 article, 8 there are mainly three pitfalls: 1. The values 𝑎 , 𝑏 , and 𝑐 may be so small or so large that an underflow or an overflow arises when computing Δ, even though the solutions are perfectly representable; 2. When 𝑏 2 ≫ |4𝑎𝑐 |, the formula for one of the solutions—depending on the sign of 𝑏 —suffers from catastrophic cancellation, 2 leading to a very inaccurate result; 3. When 𝑏 2 ≈ 4 𝑎𝑐 , the computation of the discriminant itself suffers from catastrophic cancellation. A large error in Δ may lead to inferring the wrong number of solutions. Solutions to the first problem were already pretty well understood and practiced early on, even though they were not always implemented systematically: to compute a complex division—a problem close to the computation of a discriminant—, Smith 9
proposed in 1962 a scaled formula in order to avoid underflows and overflows. Forsythe, 1 again, had presented in 1966 some examples of equations that would lead to overflows or underflows, and he had suggested some scaling strategy to avoid them.
2Or “ fantastic cancellation ,” as Metropolis puts it. F. Goualard 5
Sadly, as many others after him, he did not go through with it in that article or in any of his later ones, alluding to the fact that it was important but too wearisome to present. In 1956, Muller 2 used an alternative formula—amusingly called the “ citardauq ,10 ” for “quadratic” in reverse—originating from Fagnano’s XVIIIth century work, 11 to avoid the second problem:
𝑥 = −𝑏 ± √Δ2𝑎 = 2𝑐
−𝑏 ∓ √Δ
,
where “ ∓” is to be understood as the opposite sign to the one chosen for “ ±.” Since the sign between −𝑏 and √Δ is different in both formulas, it is possible to avoid catastrophic cancellation by using one formula for a solution and the other formula for the other solution, depending on the sign of 𝑏 .For the third problem, it seems that Kahan had already suggested to compute the discriminant with a precision twice the size of the working precision during a lecture at Stanford University 12 in 1966. In 2004, he introduced an algorithm 13 that avoided the need for a larger precision by simulating it. The major problems in the implementation of quadratic solvers were cleared up, at least theoretically, very early in the history of modern computers. What has been lacking ever since is an exposition of all the steps to take in order to obtain a robust solver. Authors have highlighted the necessity of handling all possible cases: large and small parameters, quadratics degenerating into linear or constant equations, cancellation problems. . . but, despite our best efforts, we could not find a presentation summarizing all these pitfalls and presenting their solutions in a clear algorithm. Details are always dismissed as trivial or tedious. This absence of an accessible reference algorithm has led to some naive or flawed algorithms benefiting from undeserved attention. Search on the web how to solve a quadratic equation and you will find over and over again the same naive algorithm implementing the mathematical formulas cited above. Even more troubling, this is also the algorithm presented in many “educational settings”, such as the CASIO manual 14 or the Numworks calculator documentation. 15
Consider such a naive algorithm, as implemented in Program 1. Program 1: Naive implementation in Julia of a quadratic equation solver.
function naive(a,b,c) a=Float64(a);b=Float64(b);c=Float64(c) delta = bb - 4ac if delta < 0 return (NaN64,NaN64) elseif delta == 0 return -b/(2a) else x1 = (-b + sqrt(delta))/(2a) x2 = (-b - sqrt(delta))/(2a) if x1 > x2 return (x2,x1) else return (x1,x2) end end end
According to Forsythe, 12 a satisfactory quadratic equation solver should handle correctly the cases where either 𝑎 , 𝑏 , or 𝑐 are zero, and for each representable solution 𝑥 𝑖 , it should compute an approximation ̂𝑥 𝑖 of it with a precision such that:
||𝑥 𝑖 −̂𝑥 𝑖 ||||𝑥 𝑖 ||
⩽ 32 𝜀.
In addition, Forsythe requires that extended precision be used only to compute the discriminant, if necessary. Program 1 fails on all these requirements. If 𝑎 is zero, we get a division by zero. In addition, consider, for example, the equation:
𝑝 1(𝑥 ) = 𝑥 2 + 2 27 𝑥 + 34 = 0 .6 F. Goualard
It has two solutions: ⎧⎪⎨⎪⎩
𝑥 1 = −2 27 +
√
254 −3 2
≈ −5 .59 × 10 −9 ;
𝑥 2 = −2 27 −
√
254 −3 2
≈ −134217728 .
Catastrophic cancellation prevents to compute 𝑥 1 accurately, and we get with Program 1:
{̂𝑥 1 = −7 .450580596923828 × 10 −9 ;̂𝑥 2 = −134217728 .
Scale the equation to 2500 𝑝 1(𝑥 ) = 0 , and Program 1 will return one solution. . . NaN. Scale the equation in the other direction to 2−1000 𝑝 (𝑥 ) = 0 , and there is only one solution again: -67108864, this time. Evidently, such lack of scale invariance is not acceptable. Some authors have been more diligent than others, however, and there are some works that still deserve some attention. The rest of this section highlights some of these references in chronological order, pointing out where they are lacking. The Julia code for all of the methods described below is available in the same Julia package on github as the final procedure in Section 5.
3.1 Nonweiler, 1968
In “Roots of low-order polynomial equations,” 16 Nonweiler presents procedures to solve quartic, cubic, and quadratic equations. He explicitly states at the beginning of the quadratic solving procedure that it will fail on overflow when 𝑎 is zero “ and in other cases ”, which does not bode well for its overall robustness. The polynomial:
𝑝 (𝑥 ) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
is rewritten as the monic polynomial:
𝑞 (𝑥 ) = 𝑥 2 − 2 𝑏 ′𝑥 + 𝑐 ′, with
{
𝑏 ′ =
−𝑏 𝑎
2
𝑐 ′ = 𝑐 𝑎
.
We can then compute Δ:
Δ = 𝑏 ′2 − 𝑐 ′.
If Δ is strictly negative, there are no real solutions. If Δ = 0 , we return 𝑏 ′; otherwise:
⎧⎪⎨⎪⎩
𝑥 1 =
{ √ Δ + 𝑏 ′ if 𝑏 ′ > 0;
𝑏 ′ − √Δ if 𝑏 ′ ⩽ 0;
𝑥 2 = 𝑐 𝑥 1
.
The discriminant is evaluated as badly as in the naive procedure and is therefore plagued by the same cancellation problem. On the other hand, the author avoids cancellation in computing the solutions by using the standard formula discerningly, according to the sign of 𝑏 , for one solution, and by using one of Viète’s formulas: 17
{ 𝑥 1 + 𝑥 2 = − 𝑏 𝑎
𝑥 1𝑥 2 = 𝑐 𝑎
for the other solution. Apart from the transformation into a monic polynomial, no effort is made towards avoiding overflows and underflows.
3.2 Jenkins, 1975
Jenkins 18 presented in 1975 a generic algorithm to find the roots of a polynomial, which is based on a quadratic solver. This solver handles separately the degenerate quadratics: 1. If 𝑎 = 0 and 𝑏 = 0 , it returns (0 , 0) ;2. If 𝑎 = 0 and 𝑏 ≠ 0, it returns (− 𝑐 𝑏 , 0) ;3. If 𝑐 = 0 , it returns (− 𝑏 𝑎 , 0) .F. Goualard 7
In the general case, it tries to avoid overflows when computing Δ by first computing the quantity 𝑒 :
𝑒 =
{ 1 − 2𝑎 𝑏 × 2𝑐 𝑏 , if |𝑏 |
2
⩾ |𝑐 |;
𝑏
2
× 𝑏
2|𝑐 |
− sign( 𝑐 ) ⋅ 𝑎, otherwise. When 𝑒 ⩾ 0 (there is no real solution otherwise), it computes
√Δ2
as:
√Δ2 =
{ √ 𝑒 × |𝑏 |
2
, if |𝑏 |
2
⩾ |𝑐 |;
√𝑒 × √|𝑐 |, otherwise. Lastly, the solutions are computed as:
⎧⎪⎪⎨⎪⎪⎩
𝑥 1 = − 𝑏
2−sign( 𝑏 )⋅
√Δ2
𝑎
;
𝑥 2 =
{ 𝑐 𝑥 1
𝑎
if 𝑥 1 ≠ 0,
0 otherwise .
in order to avoid cancellation. The formulas used lead to inaccuracies for large or small parameters. Consider for example the equation 2−1073 (𝑥 2 −𝑥 −1) = 0 ,which has the same two solutions as 𝑥 2 −𝑥 −1 = 0 : 𝑥 1 ≈ −0 .6180339887498948 and 𝑥 2 ≈ 1 .618033988749895 . This algorithm will compute the solutions −0 .5 and 1.5.
3.3 Franklin, 1977
In Chapter 1 of “Fundamental formulas of physics,” 19 edited by Menzel in 1977, Philip Franklin presents the solving of a quadratic equation as the very first formula. He does not consider the case 𝑎 = 0 , which is explicitly excluded; in addition, he only takes care of the possibility of cancellation in computing the solutions by using the citardauq when appropriate in the
modified Fagnano’s formulas :
⎧⎪⎨⎪⎩
𝑥 1 = −𝑏 −sign( 𝑏 )
√
𝑏 2−4 𝑎𝑐
2𝑎
;
𝑥 2 = −2 𝑐 𝑏 +𝑠𝑖𝑔𝑛 (𝑏 )
√
𝑏 2−4 𝑎𝑐
.
Overflows and underflows are not handled. The case 𝑏 2 ≈ 4 𝑎𝑐 , which could lead to an inaccurate computation of Δ, is not considered either.
3.4 Hamming, 1986
The method presented in “Numerical methods for scientists and engineers,” 20 authored by Hamming in 1986, does not offer much that is new: Hamming draws the attention to the risk of cancellation when computing one of the solutions when 𝑏 2 ≫ |4𝑎𝑐 |.His solutions is to compute the solution that is not affected by cancellation with the usual formula and to use Viète’s formula for the other, as Nonweiler does: {
𝑥 1 = −𝑏 −sign( 𝑏 )
√
𝑏 2−4 𝑎𝑐
2𝑎
;
𝑥 2 = 𝑐 𝑎𝑥 1
.
What is interesting is that, though he is aware of all the other problems that can plague a quadratic equation solver, he dismisses them with typical offhandedness: “ This is not the complete answer on how to evaluate the formula; we still need to worry about (1) underflow, (2) overflow, and (3) 𝑏 2 − 4 𝑎𝑐 < 0, but these are not relevant here 3.”
3.5 Young & Gregory, 1988
The procedure proposed by Young and Gregory in 1988 in their book “A survey of numerical mathematics” 21 is particularly well thought out, but with peculiar features, and it uses floating-point arithmetic from a made-up computer. If one of the parameters
3Our emphasis. 8F. Goualard
𝑎 , 𝑏 , or 𝑐 is “sufficiently small”—that is, smaller than some user-defined threshold for each parameter—, it is flushed to zero before proceeding. Their procedure is quite convoluted, because, as they put it “ [i]f we wish to develop a computer program which can solve [the quadratic] for any given value of [ 𝑎 ], [ 𝑏 ], and [ 𝑐 ], then we must first analyze certain special cases. ”All special cases for either 𝑎 , 𝑏 , and 𝑐 being zero are taken into account separately. For the general case, the algorithm rewrites the formulas to avoid overflows and underflows as much as possible. Cancellation in computing the solutions is handled as in Section 3.4. Overall, Young and Gregory’s algorithm handles correctly almost of the traps, except for the accurate computation of the discriminant. The authors are aware of the necessity to compute it with a precision that is double the working one—and write as much—but they do not offer any algorithm to do so on a computer without the proper format. In addition, being defined for a non IEEE 754 arithmetic, some care needs to be put into adapting the procedure for actual modern computers.
3.6 Baker, 1998
The 1998 article “You could learn a lot from a quadratic: I. Overloading considered harmful,” 22 by Baker, is particularly in-teresting as it starts by castigating the authors of the “Ada Language Reference Manual” from 1983 for presenting a flawed procedure (the naive one) for solving quadratic equations. Baker proceeds by recalling the properties of floating-point arithmetic and by presenting two procedures to avoid underflows and overflows. We will only consider the second one, as the first one is less efficient and precise. Baker consciously avoids considering the cases where either 𝑎 or 𝑐 are zero. To avoid underflows and overflows, he takes advantage of the representation of floating-point numbers to scale the parameters. His scaling is not error-free, however. Given the quadratic polynomial:
𝑝 (𝑥 ) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐,
we may express 𝑎 as:
𝑎 = 𝑚 𝑎 2𝑟 𝑎 2𝑒 𝑎 ,
with 𝑒 𝑎 an even integer ( 𝑒 𝑎 = 2 𝑘 𝑎 ) and 𝑟 𝑎 ∈ {0 , 1} .Let us divide 𝑝 (𝑥 ) by 𝑚 𝑎 2𝑟 𝑎 :
𝑝 𝑠 (𝑥 ) = 𝑝 (𝑥 )
𝑚 𝑎 2𝑟 𝑎
= 2 𝑒 𝑎 𝑥 2 + 𝑏 𝑚 𝑎 2𝑟 𝑎
⏟⏟⏟
𝑏 1
𝑥 + 𝑐 𝑚 𝑎 2𝑟 𝑎
⏟⏟⏟
𝑐 1
.
As Baker puts it, “ we can divide the equation through [...] without much risk of underflow or overflow ”, which is not the same thing as no risk at all: if either 𝑏 or 𝑐 are very large, for example, the division may be enough to trigger an overflow. In the same way as for 𝑎 , let 𝑐 1 = 𝑚 𝑐 1 2𝑟 𝑐 1 2𝑒 𝑐 1 , with 𝑟 𝑐 1 ∈ {0 , 1} and 𝑒 𝑐 1 = 2 𝑘 𝑐 1 . Let us make a change of variable:
𝑥 = − 𝑦 sign( 𝑏 1)2 𝑘 𝑐 1 −𝑘 𝑎 .
We get:
𝑞 (𝑦 ) = 2 2𝑘 𝑐 1 𝑦 2 − |𝑏 1|2𝑘 𝑐 1 −𝑘 𝑎 𝑦 + 𝑚 𝑐 1 2𝑟 𝑐 1 22𝑘 𝑐 1 .
Divide 𝑞 (𝑦 ) by 22𝑘 𝑐 1 :
𝑞 𝑠 (𝑦 ) = 𝑞 (𝑦 )22𝑘 𝑐 1
= 𝑦 2 − 2 |𝑏 1|2−𝑘 𝑐 1 −𝑘 𝑎 −1
⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟
𝑏 2
𝑦 + 𝑚 𝑐 1 2𝑟 𝑐 1
⏟⏟⏟
𝑐 2
.
If 𝑏 2 ⩾
√||𝑐 2||, we have two solutions:
⎧⎪⎨⎪⎩
𝑦 1 = 𝑏 2 +
√
𝑏 22 − 𝑐 2
𝑦 2 = 𝑐 2
𝑦 1
.
Note that 𝑐 2 is obtained from a computation. Hence 𝑏 22 − 𝑐 2 may lead to some cancellation. Baker proposes to compute 𝑦 1 as:
𝑦 1 = 𝑏 2
⎛⎜⎜⎜⎝
1 +
√√√√1 −
𝑐 2
𝑏 2
𝑏 2
⎞⎟⎟⎟⎠
.F. Goualard 9
Eventually, we can resubstitute 𝑥 for 𝑦 to get the solutions of the original equation:
{ 𝑥 1 = − sign (𝑏 1
) 2𝑘 𝑐 1 −𝑘 𝑎 𝑦 1;
𝑥 2 = − sign (𝑏 1
) 2𝑘 𝑐 1 −𝑘 𝑎 𝑦 2.
Baker avoids cancellation when computing solutions in the same way as Hamming in Section 3.4 and Young & Gregory in Section 3.5. He does not take any step to accurately compute the discriminant, however, which leads to errors in the number of solutions reported. Additionnally, his scaling procedure is slightly flawed when considering subnormal parameters: since the largest exponent ( 1023 , in double precision) is smaller than the absolute value of the smallest exponent ( −1074 , in double precision), we may get an overflow when computing the exponent of 𝑏 2, as −𝑘 𝑐 1 − 𝑘 𝑎 − 1 may be larger than 𝐸 max . This is, for example, the case with the equation:
2−1073 (𝑥 2 − 𝑥 − 1 ) = 0 ,
which has the two solutions:
𝑥 1,2 = −1 ± √52 .
Baker’s procedure will return the two solutions 0 and ∞.
3.7 Higham, 2002
“Accuracy and stability of numerical algorithms,” 23 by Higham, is a great book on computing with floating-point arithmetic. Concerning quadratic equations, the author has not much to say, however: he warns against the problem of computing the discriminant accurately, pointing to Kahan’s work on the subject, and he suggests to use the method already seen several times to avoid cancellation when computing the solutions:
{
𝑥 1 = −𝑏 −sign( 𝑏 )
√
𝑏 2−4 𝑎𝑐
2𝑎
;
𝑥 2 = 𝑐 𝑎𝑥 1
.
As many authors, Higham evokes the problem of underflows and overflows without elaborating, stating that “ [t]hese ideas can be built into a general strategy [...] but the details are nontrivial .”
3.8 Nievergelt, 2003
True to its title, “How (not) to solve quadratic equations,” 24 by Nievergelt, explores various algorithms, solving with radicals being only of them. In particular, Nievergelt notes that several softwares solve quadratics by computing the eigenvalues of the companion matrix, 25 which leads to inaccuracies, something we witnessed with Numpy, MATLAB and Octave for the example in the introduction of this article. To avoid cancellation when computing the solutions, Nievergelt suggests to use the modified Fagnano’s formulas:
⎧⎪⎨⎪⎩
𝑥 1 = −𝑐
(𝑏 ∕2)+sign( 𝑏 )∗ √(𝑏 ∕2) 2−4 𝑎𝑐
;
𝑥 2 = (𝑏 ∕2)+sign( 𝑏 )√(𝑏 ∕2) 2−4 𝑎𝑐
−𝑎
.
The value 𝑏 ∕2 is used instead of 𝑏 , which raises the possibility of some additional rounding error, should the value of 𝑏 be subnormal. He also suggests to consider the cases where either 𝑎 , 𝑏 , or 𝑐 are zero, without elaborating. Lastly, he advises to use a precision that is double the working one to avoid inaccuracies in computing the discriminant. Yet again, no method is proposed for systems that do not offer such a precision.
3.9 Kahan, 2004
In the 2004 article “On the cost of floating-point computation without extra-precise arithmetic,” 13 Kahan presents an algorithm to simulate double precision in order to compute the discriminant accurately. He also points out the necessity to handle exceptional parameters and underflows/overflows in a quadratic solver, but without providing details. A short MATLAB program to compute the solutions of a quadratic equation is provided in the article, without any provision to handle degenerate equations, or underflows/overflows, as the main focus seems to be on the accurate computation of the 10 F. Goualard
discriminant, for which a very length MATLAB program is also provided. The necessary algorithm is actually quite short but Kahan had to consider various environments, whose arithmetic had different properties. In section 5, we will present a short algorithm to compute the discriminant according to Kahan’s method. It is interesting to note that the properties of this algorithm have only been proved two years later by Boldo and others. 26,27,28
3.10 Einarsson, 2005
Despite its name, the book “Accuracy and reliability in scientific computing,” 29 by Einarsson, seems only concerned, when considering the solving of a quadratic equation, by the potential cancellation when computing the solutions. His method is the same as Menzel’s or Nievergelt’s, relying on the citardauq for one of the solutions. As the author says himself, “ Even a simple problem such as computing the roots of a quadratic equation needs great care. ” Nevertheless, nothing more is said to that effect.
3.11 Press et al., 2007
The often-cited book “Numerical recipes: the art of scientific computing,” 30 by Press et al. , has something to say about many numerical problems, the solving of quadratic equations being one of them. Sadly, it is disappointingly light on the matter, simply considering the cancellation problem when computing the solutions. Its solution is the one already seen several times: use the naive formula to compute the solution unaffected by cancellation and obtain the other solution through one of Viète’s formulas. Nothing is said about exceptional parameters, or underflows and overflows. The problem of accurately computing the discriminant is not considered either.
3.12 Dahlquist & Björck, 2008
The book “Numerical Methods in Scientific Computing,” 31 by Dahlquist and Björck, presents the same method as Press et al. ,with the same shortcomings. Apart from computing the solutions with the naive formula and one of Viète’s formulas, nothing is said about the computation the discriminant or the possibility of underflows and overflows.
3.13 McNamee & Pan, 2013
The book “Numerical methods for roots of polynomials,” 32 part II, by McNamee and Pan, devotes its section 12.4 to the errors encountered when solving quadratic equations. The authors warn about the case where 𝑎 is zero or near zero, remarking that it may happen often with some algorithms such as Muller’s method, which use a quadratic equation solver as one of their subroutine. The cancellation in computing solutions is addressed by using modified Fagnano’s formulas, as seen in Section 3.8, for example. The authors allude to the problem of underflows and overflows without addressing it in any way. Nothing is said about the accurate computation of the discriminant and the handling of exceptional parameters.
3.14 Panchekha et al., 2015
The article “Automatically improving accuracy for floating point expressions,” 33 by Panchekha et al. is particular in that it uses the Herbie tool to rewrite the naive formula in order to limit errors when evaluating it. For the purpose of demonstrating Herbie’s abilities, the authors only consider the formula:
𝑥 = −𝑏 − √𝑏 2 − 4 𝑎𝑐
2𝑎 .
The tool is able to rewrite the expression to avoid both catastrophic cancellation and overflows for the double precision format:
−𝑏 − √𝑏 2 − 4 𝑎𝑐
2𝑎 ⇝
⎧⎪⎪⎨⎪⎪⎩
4𝑎𝑐
−𝑏 +
√
𝑏 2−4 𝑎𝑐
2𝑎
if 𝑏 < 0;
−𝑏 −
√
𝑏 2−4 𝑎𝑐
2𝑎
if 0 ⩽ 𝑏 ⩽ 10 127 ;− 𝑏 𝑎 + 𝑐 𝑏 if 𝑏 > 10 127 .F. Goualard 11
Though interesting, the article does not present a full-fledged algorithm to solve quadratic equations and some of the problems evoked at the beginning of this section are not addressed.
3.15 Mastronardi & van Dooren, 2015
In “Revisiting the stability of computing the roots of a quadratic polynomial,” 34 Mastronardi and van Dooren present a scheme to solve quadratic equations accurately. They exclude the case 𝑎 = 0 , but consider the posssibilities that 𝑏 or 𝑐 be zero. If 𝑐 = 0 , we have: {
𝑥 1 = −𝑏 𝑎 ;
𝑥 2 = 0 .
If 𝑏 = 0 , we have: {
𝑥 1 =
√ −𝑐 𝑎 ;
𝑥 2 = − 𝑥 1.
The authors consider both complex and real roots, so they do not have to handle differently the case 𝑎𝑐 > 0. On the other hand, they do not consider the possibility of underflow or overflow when computing √−𝑐 ∕𝑎 .If 𝑎 , 𝑏 , and 𝑐 are all different from 0, we start by computing the coefficients (1 , 𝑏 1, 𝑐 1) of the monic quadratic polynomial:
𝑝 (𝑥 ) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ⇝ 𝑝 𝑚 (𝑥 ) = 𝑥 2 + 𝑏 𝑎 ⏟⏟⏟
𝑏 1
𝑥 + 𝑐 𝑎 ⏟⏟⏟
𝑐 1
,
and we then make a change of variable:
𝑦 = −𝑥 𝛼 , with 𝛼 = sign( 𝑏 1)√|𝑐 1|,
and consider the polynomial 𝑞 (𝑦 ) = 𝑝 𝑚 (− 𝛼𝑦 )∕ 𝛼 2:
𝑞 (𝑦 ) = 𝑦 2 − 2 𝛽𝑦 + 𝑒,
with:
𝛽 = |𝑏 1|
2√|𝑐 1| , 𝑒 = sign( 𝑐 1).
The roots of 𝑞 (𝑦 ) are then: {
𝑦 1 = 𝛽 + √𝛽 2 − 𝑒 ;
𝑦 2 = 𝛽 − √𝛽 2 − 𝑒.
They must be computed as follows:
⎧⎪⎪⎪⎨⎪⎪⎪⎩
𝑦 1 = 𝛽 + √𝛽 2 + 1
𝑦 2 = − 1
𝑦 1
}
if 𝑒 = −1;
𝑦 1 = 𝛽 + √(𝛽 + 1)( 𝛽 − 1)
𝑦 2 = − 1
𝑦 1
}
if 𝑒 = 1 ∧ 𝑏 ⩾ 1;
No real solution if 𝑒 = 1 ∧ 𝑏 < 1.
The roots of the original polynomial are then computed as:
{ 𝑥 1 = − 𝛼𝑦 1;
𝑥 2 = − 𝛼𝑦 2.
No special care is taken in computing the discriminant accurately. Note also that the transformations performed are not error-free and, therefore, introduce more rounding errors in the process.
3.16 Beebe, 2017
The book “Mathematical-Function Computation Handbook: Programming Using the MathCW Portable Software Library,” 35
by Beebe, is a very large book that is a treasure trove of numerical algorithms. The entire chapter 16 is devoted to quadratic equations. 12 F. Goualard
Beebe’s exposition is one of the few that explicitly states that the solver must handle all inputs, even NaNs, infinities and zeros, and that presents the formulas to use in those cases. He also presents the three major problems when solving a quadratic, noting that we need extra precision for computing the discriminant. He gives an algorithm to simulate the double precision needed, which is akin to Kahan’s. 13
Given an equation of the form:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 = 𝑚 𝑎 2𝑒 𝑎 𝑥 2 + 𝑚 𝑏 2𝑒 𝑏 𝑥 + 𝑚 𝑐 2𝑒 𝑐 ,
the algorithm starts by scaling all the coefficients by dividing them by 𝑒 = max( 𝑒 𝑎 , 𝑒 𝑏 , 𝑒 𝑐 ):
𝑚 𝑎 2𝑒 𝑎 𝑥 2 + 𝑚 𝑏 2𝑒 𝑏 𝑥 + 𝑚 𝑐 2𝑒 𝑐 = 0 ⇝ 𝑚 𝑎 2𝑒 𝑎 −𝑒 𝑥 2 + 𝑚 𝑏 2𝑒 𝑏 −𝑒 𝑥 + 𝑚 𝑐 2𝑒 𝑐 −𝑒 = 0 .
This reduces the possibilities of underflows and overflows without averting them entirely (The equation 2−1073 (𝑥 2 − 𝑥 − 1) = 0
is not correctly solved, for example, and does not give the same answers as 𝑥 2 − 𝑥 − 1 = 0 ). The solutions are then computed with the modified Fagnano’s formulas.
4 A REVIEW OF THE IMPLEMENTATIONS IN SOFTWARE
An article may dismissively consider some of the problems that can plague a quadratic equation solver, referring the reader to other publications or to “common sense” for details. A piece of software does not have that luxury, and must implement each and every strategy that is necessary to handle correctly all inputs. Not all programming languages offer standard facilities to solve quadratic equations, maybe because it is considered so simple that any programmer can implement his/her own when needed. A look at the information available on websites for programmers, such as StackOverflow, shows that the suggestions for the implementation of a quadratic solver usually range from the downright flawed to the quite incomplete. Is the code implemented in “standard” libraries better?
4.1 C++ Boost libraries
The C++ Boost libraries are a huge set of code that is oftentimes the antechamber for code that will be included into a future standard of C++ . As such, it is usually of high quality. Boost implements the boost::math::tools::quadratic_roots() function to solve quadratic equations with radicals. The comment in the header of the source code for that function refers to a discussion on StackOverflow about a numerically stable method for solving quadratic equations. The algorithm used considers separately the various cases where either 𝑎 , 𝑏 , or 𝑐 are zero. Nothing is done about the possibility of underflow or overflow, however. To avoid cancellation in computing the solutions, it uses the modified Fagnano’s formulas. The discriminant is accurately computed using Kahan’s algorithm, 13 with an fma instruction when available. Overall, the algorithm used is quite good, apart for the absence of underflow/overflow management.
4.2 GNU Scientific Library
The GNU Scientific Library is a project with a long history starting in 1996 to provide the C language with a “standard” library for numerical code. It offers the function gsl_poly_solve_quadratic() to solve quadratic equations with radicals. The algorithm handles separately the case of a quadratic degenerating into a linear equation. Strangely, when 𝑎 and 𝑏 are zero, the function returns that the equation has no solution, without testing the value of 𝑐 (if 𝑐 is zero, it should report that any value is a solution). Nothing is done to avoid underflows and overflows. In addition, the discriminant is computed with the textbook formula (Δ = 𝑏 2 − 4 𝑎𝑐 ) with no extra precision. The documentation for the function acknowledges that by stating that “[the] errors may cause a discrete change in the number of roots. ”. Cancellation in computing the solutions is avoided by using the modified Fagnano’s formulas, as the Boost libraries do. F. Goualard 13
4.3 Numworks calculator
The Numworks calculator is a French scientific calculator extensively used in France high schools. It is also on the list of permitted calculators for national exams in many other countries. It is an interesting device as it offers the possibility to program it using the Python language, and its source code is fully available. There are two ways to solve quadratic equations with the calculator: either use the module polynomial in the Python emulator, or use the "Equations” application. The polynomial module implements the naive algorithm from Program 1, plain and simple. It has, therefore, all its flaws. The “Equations” application is embedded in an environment that provides it with symbolic facilities. As such, some underflows and overflows may be averted through behind-the-scene scaling of the arguments. This is not always the case, however: for the equation 2−1073 (𝑥 2 − 𝑥 − 1) = 0 , it finds 0.5 as the only solution, while it has two:
−1 ± √52 .
According to the source code, the discriminant is computed without any extra precision, and the two solutions are obtained with the naive formulas. As a consequence, the application falls prey to all the traps laid on the path of quadratic equation solvers. Though a fantastic tool in its own right, with an unusual policy regarding its source code, it is rather sad that a teaching device such as the Numworks calculator should not offer any robust means to solve a “simple” quadratic equation.
4.4 Racket “math” library
The math module of the Racket language offers the function quadratic-solutions to solve quadratic equations. The function uses an algorithm devised by Pavel Panchekha in 2021, and presented in an entry on his blog. As the documentation states, the algorithm tries to handle cancellations and overflows correctly. Degenerate cases are not handled correctly however, and the function may return incorrect results when either 𝑎 , 𝑏 , or 𝑐 are zero (Issue #81, raised by the author). Cancellations in computing the solutions are avoided by using the modified Fagnano’s formulas. Panchekha did put some thought in the implementation of the code computing the discriminant to avoid losing accuracy and raising overflows. However, his algorithm relies on modified expressions to avoid overflows and underflows instead of scaling. As a consequence, some accuracy may be lost for very large and very small parameters (for example, the function reports that the equation 2−1073 (𝑥 2 − 𝑥 − 1) = 0 has the two solutions −2∕3 and 1.5, while the true solutions are (−1 ± √5)∕2 ). There is also an error in the computation of the discriminant, whereby some accuracy may be lost, misdiagnosing the number of solutions of an equation (for example, the equation −312499999999 𝑥 2 + 707106781186 𝑥 − 400000000000 = 0 , taken from Nievergelt’s article, 24 has the two very close solutions ≈ 1 .131369396027 and ≈ 1 .131372303775 , while Racket’s algorithm considers that it has no real solution (Issue #83, raised by the author).
4.5 Rust “roots” module
The roots module of the Rust language offers the function roots::find_roots_quadratic() to solve quadratic equations. According to its source code, the algorithm used handles separately the linear case. The discriminant is computed with the naive formula without any extra precision, and no provision is made to avoid underflows and overflows. On the other hand, cancellation is avoided in computing the solutions by using the modified Fagnano’s formulas.
4.6 Scilab
The primary way to solve a quadratic equation in Scilab seems to be with the roots() function, which, by default, computes the eigenvalues of the companion matrix. There is, however, a report titled “Scilab is not naive” 36 on the Scilab website, which presents an algorithm to solve quadratics with radicals. That algorithm seems to be used for quadratic equations in roots()
when we pass the right parameter. To avoid cancellation when computing the solutions, the algorithm uses the modified Fagnano’s formulas as seen already in Section 3.8. The report does not consider the problem of accurately computing the discriminant, and restricts its handling of 14 F. Goualard
overflows and underflows to the computation of the discriminant only. Two different expressions are used to compute 𝑠 = √Δ
in order to limit the possibility of overflow. Given 𝑏 ′ = 𝑏 ∕2 :
• If |𝑏 ′| > |𝑐 |, we compute: { 𝑒 = 1 − 𝑎 𝑏 ′
𝑐 𝑏 ′
,𝑠 = sign ( 𝑒 ) × |𝑏 ′| √|𝑒 |;
• If |𝑐 | > |𝑏 ′|, we compute: {
𝑒 = 𝑏 ′ 𝑏 ′
𝑐
− 𝑎, 𝑠 = sign ( 𝑒 ) × √|𝑐 |√|𝑒 |.
Strangely, that strategy for averting—but not completely eliminating—overflows is neither included in the algorithm presented in the report, nor present in the code of the C++ function FindQuadraticPolynomialRoots() of the Scilab source code on github.
5 A COMPLETE AND ROBUST ALGORITHM
We have witnessed in recent articles and current implementations the consequences of the very traps we were warned against in the literature from the sixties. A robust quadratic equation solver should: 1. Handle exceptional parameters (NaNs and infinities); 2. Handle degenerate cases where either 𝑎 , 𝑏 or 𝑐 are zero; 3. Avoid cancellation when computing Δ with 𝑏 2 ≈ 4 𝑎𝑐 ;4. Avoid underflows and overflows when the solutions are representable; 5. Avoid cancellation when computing the solutions when 𝑏 2 ≫ |4𝑎𝑐 |.In his book “Floating-point computation” from 1974, Sterbenz had already considered in details four of the five problems for a computer using hexadecimal pre-IEEE 754 arithmetic. The fifth problem, the accurate computation of Δ, he had dismissed by advocating to perform the computation with twice the working precision, which may not always be possible. Our presentation will address this problem by using Kahan’s algorithm from 2004. 13 We will also flesh out some of his remarks about underflows or overflows affecting intermediate computations. We will consider the five pitfalls in order, before giving the complete algorithm. Pitfall 5 will be considered together with Pitfall 4.
5.1 Pitfall 1: Exceptional parameters
Handling exceptional parameters simply requires to detect them at the beginning of the function. Programming languages usually provide methods to determine whether a float is an NaN or an infinity (e.g., isnan() and isinf() , in C and C++ ). We also have to decide what we will return in that case. In our Julia implementation, we have chosen to return a different value than when there are no real solutions ( Δ < 0).
5.2 Pitfall 2: Degenerate cases
Table 2 shows all possible degenerate cases. When all three parameters are zero, all reals are solution. What should we return then? We have decided to return the pair (∞ , −∞) in this situation. Since we will always return two solutions sorted in ascending order, this pair cannot be mistaken for the case where two solutions are not representable and their computation overflowed. When there are no solutions, real or complex, we will return one NaN. When there are only two non-real solutions, we will return a pair of NaNs. This is of course possible because the Julia language is flexible enough to permit it. With a more rigid language, we would have to use a different convention. F. Goualard 15
Table 2 Degenerate cases when solving a quadratic equation. Case Real solutions
𝑎 = 0 ∧ 𝑏 = 0 ∧ 𝑐 = 0 ℝ
𝑎 = 0 ∧ 𝑏 = 0 ∧ 𝑐 ≠ 0 ∅3. 𝑎 = 0 ∧ 𝑏 ≠ 0 ∧ 𝑐 = 0 {0} 4. 𝑎 = 0 ∧ 𝑏 ≠ 0 ∧ 𝑐 ≠ 0 {− 𝑐 𝑏 }5. 𝑎 ≠ 0 ∧ 𝑏 = 0 ∧ 𝑐 = 0 {0} 6. 𝑎𝑐 > 0 ∧ 𝑏 = 0 𝑥 ∈ ℂ
𝑎𝑐 < 0 ∧ 𝑏 = 0 {−
√
− 𝑐 𝑎 ,
√
− 𝑐 𝑎 }8. 𝑎 ≠ 0 ∧ 𝑏 ≠ 0 ∧ 𝑐 = 0 {− 𝑏 𝑎 , 0}
In Cases 4 and 8 in Table 2, the computation of −𝑐 ∕𝑏 and −𝑏 ∕𝑎 may underflow or overflow. That would mean that the solution is not representable with the floating-point format used, and such an event is therefore unavoidable. The solution returned will then be 0 or ±∞ , depending on the event. In a private communication with Gardiner and Metropolis, 37 Householder suggested to return in such a situation an unevaluated product as a pair of floats; We did not consider that solution as it would have complexified the kind of output managed by the function. Case 7 is different: we may get an underflow or an overflow when computing −𝑐 ∕𝑎 while the square root would have brought back the result into the representable domain. To avoid that, we will exploit our knowledge of the representation of IEEE 754 floating-point numbers. We have: √
− 𝑐 𝑎 =
√
− 𝑚 𝑐 2𝐸 𝑐
𝑚 𝑎 2𝐸 𝑎
=
√
− 𝑚 𝑐 2𝐸 𝑐 −𝐸 𝑎
𝑚 𝑎
Given 𝑀 and 𝐸 two integers (with 𝐸 ∈ {0 , 1} ) such that:
𝐸 𝑐 − 𝐸 𝑎 = 2 𝑀 + 𝐸
We may then write: √
− 𝑚 𝑐 2𝐸 𝑐 −𝐸 𝑎
𝑚 𝑎
=
√
− 𝑚 𝑐 22𝑀 +𝐸
𝑚 𝑎
= 2 𝑀
√
− 𝑚 𝑐 2𝐸
𝑚 𝑎
(5) This is an error-free transformation. The fraction below the square root cannot trigger any underflow or overflow anymore. In order to implement that expression, the programming language must allow isolating the signed significand and the exponent of a floating-point number, which is the case for many languages, particularly those that inherit most of their library from C. However, we have to be wary of the actual value of the exponent reported by the dedicated function. For subnormal numbers, the exponent() function in Julia will report a value smaller than 𝐸 min as it always considers the significand as normalized. As a consequence, the value of 𝑀 may be outside the domain [𝐸 min , 𝐸 max ], which would lead 2𝑀 to underflow or overflow; in the case of an overflow, for example, if the square root is less than 1, the result might still be representable. In order to avoid that situation, we set 𝑀 = 𝑀 1 + 𝑀 2 with
(𝑀 1, 𝑀 2) ∈ [ 𝐸 min , 𝐸 max ]2, and we compute Expression 5 as:
2𝑀
√
− 𝑚 𝑐 2𝐸
𝑚 𝑎
= 2 𝑀 1
⎛⎜⎜⎝
2𝑀 2
√
− 𝑚 𝑐 2𝐸
𝑚 𝑎
⎞⎟⎟⎠
(6) using the function keep_exponent_in_check() in Program 2.
5.3 Pitfall 3: Cancellation in computing Δ
The computation of the discriminant Δ according to the textbook formula:
Δ = 𝑏 2 − 4 𝑎𝑐 16 F. Goualard
Program 2: Ensuring that an exponent is always in the domain [𝐸 min , 𝐸 max ].
function keep_exponent_in_check(Kin) if -1022 <= Kin <= 1023 return (Kin, 0) elseif Kin < -1022 return (-1022, Kin+1022) else return (1023, Kin-1023) end end
entails the subtraction of two computed values, which leads to a possibility of catastrophic cancellation. Consider, for example, our introductory example:
𝑥 2 + (1 + 2 −52 )𝑥 + ( 14 + 2 −53 ) = 0
We get:
Δ = (1 + 2 −52 )2 − 4( 14 + 2 −53 ) = 2 −104
Unfortunately, 𝑏 2 = 1 + 2 −104 + 2 −51 cannot be represented with double precision, and will be rounded to f l (𝑏 2) = 1 + 2 −51 .Hence:
f l ⟨Δ⟩ = (1 + 2 −51 ) − (1 + 2 −51 ) = 0
That small rounding error leads us to misidentify the number of solutions. In 2004, Kahan 13 proposed an algorithm to accurately compute the discriminant by taking advantage of the fact that the error performed when multiplying two floats is itself a representable float. 38 It is therefore possible to collect those errors and to reinject them into the computation. Program 5 presents a version of Kahan’s algorithm that takes advantage of that when no
fma 4 instruction is available. That version requires two auxiliary functions:
• A function veltkamp_split(x) (Program 3) to separate a float x into two non-overlapping floats xhi and xlo (with
x = xhi + xlo );
• A function exactmult(x,y,pxy) (Program 4), which computes the error e = x × y − f l ( x × y).Program 3: Splitting a double precision float into two non-overlapping parts.
function veltkamp_split(x) gamma = 134217729x # 134217729 = 2^((52/2+1))+1 delta = x - gamma xhi = gamma + delta xlo = x - xhi return (xhi,xlo) end
When an fma instruction is available, a shorter algorithm exists, shown in Program 6. Note also that, when 𝑏 2 ≉ 4 𝑎𝑐 , the straight computation f l ⟨𝑏 2 − 4 𝑎𝑐 ⟩ is accurate enough. Given ̂ Δ, the value of the discriminant computed by Kahan’s algorithm, Boldo 27 proved that:
|||Δ − ̂ Δ||| ⩽ 2 ulp (̂Δ)
provided no underflow or overflow occurs. Jeannerod et al. 28 refined that bound further. We will see in the next section that we will always perform the computation of the discriminant in the absence of underflow and overflow.
4An fma instruction is an instruction on three values 𝑥 ,𝑦 ,𝑧 that can perform the operation 𝑥𝑦 +𝑧 with only one rounding error. F. Goualard 17
Program 4: Computation of the error in the multiplication of two floats.
function exactmult(x,y,pxy) (xhi,xlo) = veltkamp_split(x) (yhi,ylo) = veltkamp_split(y) t1 = -pxy + xhiyhi t2 = t1 + xhiylo t3 = t2 + xloyhi e = t3 + xloylo return e end
Program 5: Accurate computation of the discriminant with Kahan’s method 13 without an fma instruction.
function kahan_discriminant(a,b,c) d = bb - 4ac if 3abs(d) >= bb + 4ac # b^2 and 4ac are different enough? return d end p = bb q = 4ac dp = exactmult(b,b,p) dq = exactmult(4a,c,q) d = (p-q) + (dp-dq) return d end
Program 6: Accurate computation of the discriminant with Kahan’s method 13 with an fma instruction.
function kahan_discriminant_fma(a,b,c) d = bb - 4ac if 3abs(d) >= bb + 4ac # b^2 and 4ac are different enough? return d end p = bb dp = fma(b,b,-p) q = 4ac dq = fma(4a,c,-q) d = (p-q) + (dp-dq) return d end
5.4 Pitfalls 4 & 5: Underflows/overflows and cancellation in computing the solutions
When computing the discriminant, 𝑏 2, 4𝑎𝑐 , or 𝑏 2 − 4 𝑎𝑐 , even, when 𝑎𝑐 < 0, may underflow or overflow while the solutions are perfectly representable. In order to avoid it, we take advantage of the fact that scaling by a factor does not change the solutions. Starting from:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 ,
we use, once again, our knowledge of the representation of floats to express it as:
𝑚 𝑎 2𝐸 𝑎 𝑥 2 + 𝑚 𝑏 2𝐸 𝑏 𝑥 + 𝑚 𝑐 2𝐸 𝑐 = 0 .
The following scaled equation has the same solutions:
2𝐸 𝑎 −2 𝐸 𝑏
(𝑚 𝑎 2𝐸 𝑎 𝑥 2 + 𝑚 𝑏 2𝐸 𝑏 𝑥 + 𝑚 𝑐 2𝐸 𝑐
) = 0 .18 F. Goualard
We perform a change of variable:
𝑥 = 2 𝐸 𝑏 −𝐸 𝑎 𝑦.
We get:
2𝐸 𝑎 −2 𝐸 𝑏
(𝑚 𝑎 2𝐸 𝑎 𝑥 2 + 𝑚 𝑏 2𝐸 𝑏 𝑥 + 𝑚 𝑐 2𝐸 𝑐
) = 0
⇐⇒ 2𝐸 𝑎 −2 𝐸 𝑏
(𝑚 𝑎 22𝐸 𝑏 −𝐸 𝑎 𝑦 2 + 𝑚 𝑏 22𝐸 𝑏 −𝐸 𝑎 𝑦 + 𝑚 𝑐 2𝐸 𝑐
) = 0
⇐⇒ 𝑚 𝑎 𝑦 2 + 𝑚 𝑏 𝑦 + 𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏 = 0
Let us now consider the new equation:
𝑚 𝑎 𝑦 2 + 𝑚 𝑏 𝑦 + 𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏 = 0 ,
where only the constant term may be the source of an underflow or overflow. Once we get the solutions for that equation, we will perform a new change of variable from 𝑦 to 𝑥 to get the solutions sought. When computing the discriminant, the term 4𝑚 𝑎 𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏 is the only one that can underflow or overflow. Let us consider the three separate cases:
• If 4𝑚 𝑎 𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏 ∈ [2 𝐸 min , 2𝐸 max +1 ), we have neither underflow nor overflow;
• If 4𝑚 𝑎 𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏 < 2𝐸 min , we have an underflow;
• If 4𝑚 𝑎 𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏 ⩾ 2𝐸 max +1 , we have an overflow. Knowing that ||𝑚 𝑎 || and ||𝑚 𝑐 || take their value in [2 1− 𝑝 , 2) , we deduce:
• 𝐸 𝑐 + 𝐸 𝑎 − 2 𝐸 𝑏 ∈ [ 𝐸 min + 2 𝑝 − 4 , 𝐸 max − 3) : neither underflow, nor overflow;
• 𝐸 𝑐 + 𝐸 𝑎 − 2 𝐸 𝑏 < 𝐸 min + 2 𝑝 − 4 : underflow when computing the discriminant;
• 𝐸 𝑐 + 𝐸 𝑎 − 2 𝐸 𝑏 ⩾ 𝐸 max − 3 : overflow when computing the discriminant. However, if we use the function veltkamp_split() from Program 3 to compute the discriminant, we have to take into account the multiplication by 2⌊𝑝 ∕2 ⌋+1 + 1 it performs. To avoid an overflow, we then also require:
(2 ⌊𝑝 ∕2 ⌋+1 + 1) 𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏 < 2𝐸 max +1 .
To simplify, we choose a tighter bound and replace 2⌊𝑝 ∕2 ⌋+1 + 1 by 2⌊𝑝 ∕2 ⌋+2 . We get:
(2 ⌊𝑝 ∕2 ⌋+2 )𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏 < 2𝐸 max +1 ,
which gives:
𝐸 𝑐 + 𝐸 𝑎 − 2 𝐸 𝑏 < 𝐸 max − 2 −
⌊ 𝑝
2
⌋
as our new upper bound to avoid an overflow.
Case 𝐸 𝑐 + 𝐸 𝑎 − 2 𝐸 𝑏 ∈ [ 𝐸 min + 2 𝑝 − 4 , 𝐸 max − 2 −
⌊ 𝑝
2
⌋
).
The discriminant of the equation:
𝑚 𝑎
⏟⏟⏟
𝑎 ′
𝑦 2 + 𝑚 𝑏
⏟⏟⏟
𝑏 ′
𝑦 + 𝑚 𝑐 2𝐸 𝑐 +𝐸 𝑎 −2 𝐸 𝑏
⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟
𝑐 ′
= 0 ,
can be computed without any risk of underflow or overflow using Kahan’s algorithm. If it is negative—no real solution—we return the pair (NaN,NaN) ; if it is zero, we return (− 𝑏 ′∕(2 𝑎 ′)) × 2 𝐸 𝑏 −𝐸 𝑎 . The range of 𝑎 ′ and 𝑏 ′ ensure that we cannot have any spurious overflow or underflow from the division. If the discriminant is positive, we have a risk of cancellation when computing one of the solutions. As seen in the previous section, we can avoid it by using the modified Fagnano’s formulas:
⎧⎪⎪⎨⎪⎪⎩
𝑦 1 = −𝑏 ′ − sign (𝑏 ′) √
𝑏 ′2 − 4 𝑎 ′𝑐 ′
2𝑎 ′
𝑦 2 = − 2𝑐 ′
𝑏 ′ + sign ( 𝑏 ′)
√
𝑏 ′2 − 4 𝑎 ′𝑐 ′F. Goualard 19
Then, we get: { 𝑥 1 = 𝑦 12𝐸 𝑏 −𝐸 𝑎
𝑥 2 = 𝑦 22𝐸 𝑏 −𝐸 𝑎 .
As we have seen for Case 7 in Table 2, 𝐸 𝑏 − 𝐸 𝑎 may be outside the domain [𝐸 min , 𝐸 max ]. We use the same workaround as in Equation 6 and use the expressions: { 𝑥 1 = ( 𝑦 12Ω)2 Γ
𝑥 2 = ( 𝑦 22Ω)2 Γ ,
with Ω + Γ = 𝐸 𝑏 − 𝐸 𝑎 and (Ω , Γ) ∈ [ 𝐸 min , 𝐸 max ]2. The same must also be done when Δ = 0 .
Case 𝐸 𝑐 + 𝐸 𝑎 − 2 𝐸 𝑏 < 𝐸 min + 2 𝑝 − 4 .
The computation of 4𝑎 ′𝑐 ′ underflows. Consequently, we have √Δ ≈ 𝑏 ′. We may compute the first solution as:
𝑦 1 = − 𝑏 ′
𝑎 ′
To get the second solution, we may reuse the scaling as in Equation (5) and express 𝑐 ′ as:
𝑐 ′ = 𝑚 𝑐 22𝑀 +𝐸 = 2 2𝑀 𝑚 𝑐 2𝐸
⏟⏟⏟
𝑐 ′′
(7) with 𝑀 and 𝐸 two integers such that 𝐸 𝑐 + 𝐸 𝑎 − 2 𝐸 𝑏 = 2 𝑀 + 𝐸 and 𝐸 ∈ {0 , 1} . Using Viète’s formula, we obtain:
𝑦 2 = 𝑐 ′′
𝑎 ′𝑦 1
Once again, we get our original solutions by rescaling:
{ 𝑥 1 = 𝑦 12𝐸 𝑏 −𝐸 𝑎
𝑥 2 = 𝑦 222𝑀 +𝐸 𝑏 −𝐸 𝑎
Again, the exponent 𝐸 𝑏 − 𝐸 𝑎 and 2𝑀 + 𝐸 𝑏 − 𝐸 𝑎 may be outside [𝐸 min , 𝐸 max ]. We ensure the change of variable is done without spurious underflow or overflow by splitting these exponents into two if necessary, as above.
Case 𝐸 𝑐 + 𝐸 𝑎 − 2 𝐸 𝑏 ⩾ 𝐸 max − 2 −
⌊ 𝑝
2
⌋
.
We cannot solve the equation:
𝑎 ′𝑦 2 + 𝑏 ′𝑦 + 𝑐 ′ = 0 ,
directly since the discriminant will overflow on computing 4𝑎 ′𝑐 ′ or when calling veltkamp_split() on 𝑐 ′. If 𝑎𝑐 > 0, we have
𝑏 ′2 − 4 𝑎 ′𝑐 ′ < 0 and there are no real solutions. If 𝑎𝑐 < 0, the value of 4𝑎 ′𝑐 ′ is so large that we may ignore the contribution of
𝑏 ′2 and 𝑏 ′. Therefore:
𝑦 1,2 = ±√−4 𝑎 ′𝑐 ′
2𝑎 ′ = ±
√||||
𝑐 ′
𝑎 ′
|||| = ±2 𝑀
√||||
𝑐 ′′
𝑎 ′
||||
The values of 𝑐 ′′ and 𝑐 ′′ ∕𝑎 ′ may be computed without any risk of overflow, which was not necessarily the case for 𝑐 ′ and 𝑐 ′∕𝑎 ′.We rescale to get our original solutions: { 𝑥 1 = 𝑦 12𝐸 𝑏 −𝐸 𝑎
𝑥 2 = − 𝑥 1
As previously, the exponent 𝐸 𝑏 − 𝐸 𝑎 is split into two if necessary to ensure no spurious underflow or overflow. Program 7 is the complete Julia function that embodies all the formulas from this section. We call it “ sterbenz() ” as it is largely inspired from Sterbenz’s notes with Kahan’s algorithm used to accurately compute the discriminant. The sign1()
function that appears in its code implements the sign () function from Equation (2).
6 TESTS
Kahan 13 proposed to test quadratic equation solvers by considering equations of the form:
𝑀𝐹 𝑛 𝑥 2 − 2 𝑀𝐹 𝑛 −1 𝑥 + 𝑀𝐹 𝑛 −2 = 0 ,20 F. Goualard
Program 7: A robust quadratic equation solver using Sterbenz’s notes 4, adapted to double precision IEEE 754 floats, and Kahan’s algorithm 13 to compute the discriminant.
function sterbenz(a,b,c) try a=Float64(a);b=Float64(b);c=Float64(c) catch e println(stderr,"Parameters should be numbers!") return nothing end if isnan(a) || isnan(b) || isnan(c) || isinf(a) || isinf(b) || isinf(c) return (NaN64) end # Degenerate cases if a == 0 if b == 0 if c == 0 # a==0, b==0, c==0 return (Inf64, -Inf64) else # a==0, b==0, c!=0 return (NaN64) end else if c == 0 # a==0, b!=0, c==0 return (0.0) else # a==0, b!=0, c!=0 return (-c/b) end end else if b == 0 if c == 0 # a!=0, b==0, c==0 return (0.0) else # a!=0, b==0, c!=0 if sign1(a) == sign1(c) return (NaN64, NaN64) else ea = exponent(a) ec = exponent(c) ecp = ec - ea a2 = significand(a) dM = ecp & ~1 # dM = floor(ecp/2)2 M = dM>>1 # M = dM/2 E = ecp & 1 # E = odd(ecp) ? 1 : 0 c3 = significand(c)2.0^(E) S = sqrt(-c3/a2) (M1,M2) = keep_exponent_in_check(M) x1 = (S2.0^M2)2.0^M1 return (-x1,x1) end end else if c == 0 # a!=0, b!=0, c==0 if sign1(a) == sign1(b) return (-b/a, 0.0) else return (0.0, -b/a) end else # a!=0, b!=0, c!=0 ea = exponent(a) eb = exponent(b) ec = exponent(c) K = eb - ea L = ea - 2eb ecp = ec + L a2 = significand(a) b2 = significand(b) if ecp >= -920 && ecp < 995 c2 = significand(c)2.0^ecp delta = kahan_discriminant_fma(a2,b2,c2) if delta < 0 return (NaN64, NaN64) end (K1,K2) = keep_exponent_in_check(K) if delta > 0 y1 = -(2c2)/(b2+sign1(b)sqrt(delta)) y2 = -(b2+sign1(b)sqrt(delta))/(2a2) x1 = (y12.0^K2)2.0^K1 x2 = (y22.0^K2)2.0^K1 return (min(x1,x2),max(x1,x2)) end return ((-b2/(2a2))2.0^K2)2.0^K1 end dM = ecp & ~1 # dM = floor(ecp/2)2 M = dM>>1 # M = dM/2 E = ecp & 1 # E = odd(ecp) ? 1 : 0 c3 = significand(c)2.0^(E) S = sqrt(abs(c3/a2)) if ecp < -920 y1 = -b2/a2 y2 = c3/(a2y1) (dMK1, dMK2) = keep_exponent_in_check(dM+K) (K1,K2) = keep_exponent_in_check(K) x1 = (y12.0^K2)2.0^K1 x2 = (y22.0^dMK2)2.0^(dMK1) return (min(x1,x2),max(x1,x2)) end # ecp >= 995 if sign1(a) == sign1(c) return (NaN64, NaN64) else (MK1,MK2) = keep_exponent_in_check(M+K) x1 = (S2.0^MK2)2.0^(MK1) x2 = -x1 return (x2, x1) end end end end end
where 𝐹 𝑛 is the 𝑛 -th element of the Fibonacci sequence (starting from 𝐹 0), and 𝑀 = ⌊𝑅 ∕𝐹 𝑛 ⌋, with 𝑅 an integer drawn uniformly at random from [2 𝑝 −1 , 2𝑝 − 1] . The solutions are:
𝐹 𝑛 −1 ± √(−1) 𝑛
𝐹 𝑛
.
In order to consider only equations with parameters that are representable with double precision floats, we cannot exceed 𝐹 76 .To get real solutions, we also consider even values of 𝑛 only. Table 3 summarizes the results for all the methods presented in Sections 3 to 5. Out of twenty algorithms, almost half of them either compute the wrong number of solutions, or compute the wrong solutions. When a method computes only one result, it F. Goualard 21
Table 3 Results on the 38 randomized Fibonacci-based quadratic equations from 𝐹 2 to 𝐹 76 .Name Wrong Number of solutions Wrong solutions Baker 0 1Beebe 0 0Boost 0 0Dahlquist/Björck 0 0Einarsson 0 0GSL 0 0Higham 0 0Jenkins 0 16 Kahan 0 0Mastronardi/van Dooren 0 0Naive 18 0Nievergelt 0 1Nonweiler 0 0Numerical recipes 0 0Panchekha (PLDI15) 0 0Panchekha (racket) 0 37 Rust 18 0Scilab 18 0Sterbenz 0 0Young/Gregory 0 19 is counted as a wrong solution only if the value does not correspond to any of the correct solutions. In our implementation, a result ̂𝑥 is considered correct with respect to a solution 𝑥 if:
||̂𝑥 − 𝑥 || ⩽ √𝜀 max (||̂𝑥 || , |𝑥 |) (8) The randomized Fibonacci-based tests advocated by Kahan 13 are difficult to solve correctly but not so challenging after all. We have also defined a set of difficult quadratic equations that exercise the algorithms at their weakest points. All these tests, and more, are present in our Julia QuadraticEquation package available on github. There are five categories:
• Fifteen quadratics with no solution, not even complex ones (an NaN or an infinity is among the parameters);
• Four quadratics with no real solution;
• Seventeen degenerate quadratics (at least one parameter is zero);
• One quadratic with one solution;
• Eighteen quadratics with two real solutions. Table 4 presents the results for all categories with the format x/y/z, where “x” is the number of correct results, “y” the number of cases where the solver found the wrong number of solutions, and “z” the number of cases where the solutions were wrong. If a solver finds the wrong number of solutions but at least one result is correct, it is counted in the first category of error but not in the second. A result is considered correct with respect to a solution following the same definition as Equation (8). Notice that, now, only Function sterbenz() keeps a perfect score. Lastly, we generated one million quadratics randomly. The parameters are drawn among all finite floating-point numbers— even subnormals—in such a way that the resulting quadratic has two real solutions. We computed the solutions with arbitrary precision in order to be able to determine the size of the error for the results from each of our twenty methods. Table 5 reports the largest relative error overall. Unfortunately, since many methods do not handle underflows and overflows correctly, most results are distressingly large (when the result was an infinite value, the error was not considered, however). We may note that 22 F. Goualard
Table 4 Correctness of the various algorithms on challenging quadratics Name No sol. No real sol. Degenerate One sol. Two sol. Baker 0/0/15 4/0/0 2/0/15 0/1/0 16/0/2 Beebe 14/0/1 4/0/0 6/10/1 0/1/0 14/0/4 Boost 1/0/14 4/0/0 13/3/2 0/1/1 13/0/5 Dahlquist/Björck 0/0/15 4/0/0 3/10/11 0/1/1 14/0/4 Einarsson 0/0/15 4/0/0 4/10/5 0/1/1 14/0/4 GSL 1/0/14 4/0/0 13/3/2 0/1/1 13/0/5 Higham 0/0/15 4/0/0 3/10/11 0/1/1 14/0/4 Jenkins 0/0/15 4/0/0 6/10/1 0/1/0 17/0/1 Kahan 0/0/15 4/0/0 3/10/7 0/1/1 13/0/5 Mastronardi/van Dooren 0/0/15 1/0/3 6/2/9 0/1/1 14/0/4 Naive 1/0/14 4/0/0 5/9/11 0/1/1 10/2/7 Nievergelt 0/0/15 1/0/3 0/0/17 0/1/0 13/1/5 Nonweiler 0/0/15 4/0/0 7/1/9 1/0/0 13/3/4 Numerical recipes 0/0/15 4/0/0 4/10/5 0/1/1 14/0/4 Panchekha (PLDI15) 0/0/15 4/0/0 5/10/6 0/1/1 13/0/5 Panchekha (racket) 0/0/15 4/0/0 6/10/6 0/1/1 15/0/3 Rust 1/0/14 4/0/0 12/2/4 0/1/1 13/2/4 Scilab 0/0/15 4/0/0 7/9/1 1/0/0 16/2/1 Sterbenz 15/0/0 4/0/0 17/0/0 1/0/0 18/0/0 Young/Gregory 1/0/14 1/0/3 14/2/1 0/1/1 13/1/5 the algorithm used in the Racket language and the sterbenz() function exhibit both a good worst relative error and solve correctly all quadratics. This also shows that testing randomly a procedure may not be the best way to ascertain its qualities as it is still possible to devise tests that show flaws in Racket’s algorithm (See Tables 3 and 4, for example). To better assess the precision achieved by the sterbenz() procedure, we also tracked the worst error for 200 000 000 random quadratics. The largest relative error was around 1.76 𝜀 , a bit more than Forsythe’s wishes, 12 but quite satisfactory nonetheless.
7 CONCLUSION
Compare Program 1 and Program 7: one is 17 lines long and the other, 128 lines (if we count the ancillary func-tions keep_exponent_in_check() and kahan_discriminant_fma() ); one must be used with great care and a restricted set of parameters—lest it return false results—, the other can take any parameter thrown at it and return precise solutions when they can be represented as floating-point numbers. Incredibly elaborate numerical software is built from layer upon layer of ever simpler code. The simplest code, such as a quadratic equation solver, should be flawless so that, as Forsythe 1 puts it, “[...] when a quadratic equation occurs in the midst of a complex and imperfectly understood computation, one can be sure that the quadratic equation solver can be relied upon to do its part well and permit us to concentrate attention on the rest of the computation. ”The devising of Program 7 requires quite a good knowledge of the properties of IEEE 754 floating-point numbers. Should we expect that from all the programmers who implement numerical algorithms into programming language libraries and appli-cations? Probably not, and this is why those who devise these algorithms in the first place should not leave any detail out when presenting them. It is rather disquieting to compare what has been available in the literature for more than half a century—some of the algorithms published being quite good, if not perfect—with what is currently implemented in software. It is possible that the format chosen for the exposition of their algorithm by experts in numerical methods is often not the most suited to appeal to those who have to F. Goualard 23
Table 5 Worst precision for 1 000 000 random quadratics with two real solutions. Name Wrong solutions Overflows Worst relative precision Baker 36 11 14 .43 × 𝜀
Beebe 0 93804 4.5 × 10 15 × 𝜀
Boost 0 85388 1.8 × 10 161 × 𝜀
Dahlquist/Björck 0 244574 1.8 × 10 161 × 𝜀
Einarsson 0 244683 1.8 × 10 161 × 𝜀
GSL 0 244869 3.87 × 10 302 × 𝜀
Higham 0 244574 1.8 × 10 161 × 𝜀
Jenkins 0 26019 4.5 × 10 15 × 𝜀
Kahan 0 295388 7.7 × 10 302 × 𝜀
Mastronardi/van Dooren 0 196961 4.4 × 10 15 × 𝜀
Naive 33640 244574 5.5 × 10 296 × 𝜀
Nievergelt 0 243559 4.7 × 10 161 × 𝜀
Nonweiler 79013 223242 4.5 × 10 15 × 𝜀
Numerical recipes 0 244869 1.8 × 10 161 × 𝜀
Panchekha (PLDI15) 0 178856 3.8 × 10 302 × 𝜀
Panchekha (racket) 0 0 2.29 × 𝜀
Rust 33640 109 3.1 × 10 155 × 𝜀
Scilab 33839 71717 8.4 × 10 306 × 𝜀
Sterbenz 0 0 1.52 × 𝜀
Young/Gregory 194605 30596 4.5 × 10 15 × 𝜀
implement them. This rift has to be mended somehow if we expect the best numerical methods to drive out the bad ones from numerical software eventually. This is not the end of the story, though. There is one problem that Program 7 cannot prevent, and it happens when the parameters of the quadratic equation are approximations themselves. This can occur because they are obtained from a previous computation, or because they are provided by the user as decimal values or expressions. Consider for example the equation:
𝑥 2 +
√ 45 𝑥 + 15 = 0
Its discriminant is clearly zero, and the only solution should be −√1∕5 . Unfortunately, the parameters √4∕5 and 1∕5 have to be rounded to be represented as floating-point numbers. Therefore, the actual equation that is being solved is not the one the user intended, and even Program 7 will not get the expected solutions: it will return that there are no real solutions, which is true of the equation really considered:
𝑥 2 + f l
⟨√ 45
⟩
𝑥 + f l
( 15
)
= 0
One solution to this problem may be to use interval arithmetic 39 to handle such cases by enclosing the real parameters into intervals with floating-point bounds. The code for all the algorithms presented in this article has been assembled into a Julia package, QuadraticEquation , which is available on github. The package also contains challenging tests for the various solvers.
ACKNOWLEDGMENTS
Discussions with Dr. Christophe Jermann helped shape this article as it currently is. An invitation to present computer arithmetic pitfalls to high school French mathematics teachers by Pr. Magali Hersant and Dr. Emmanuel Desmontils, co-directors of the 24 F. Goualard
Institute for Research on Teaching Mathematics (IREM des Pays de la Loire), was the initial impetus for the research presented therein.
References
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What's The Difference Between Transparent & Translucent?
April 2, 2024
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A question that comes up very frequently is the specific difference between transparent and translucent material, and it’s a great question since it can make a huge impact on what sort of acrylic you end up purchasing.
Luckily, the answer is straightforward:
Transparentmaterial has the physical properties of allowing light to pass through without a large scattering of light - this means that you can actually see through the material. The exact amount of light that can pass through depends on the material, but remember that you'll always be able to clearly see through the material.
On the other hand,Translucentmaterial lets light pass through but objects on the other side can’t be seen clearly. Think stain glass windows, which allow light to come through, but you won’t be able to clearly see anything behind the material.
In the example below, the furthest right sheet is transparent, while the next sheet over is translucent
Of course, it’s easiest to see a few real-life examples using the material we have. The following video shows a few of our colors in both transparent, and translucent.
Both types of material have endless practical uses, but depending on your application it’s important to know the difference. For example, many people use translucent material for sign making, because it is excellent for backlighting.
Another common option when choosing material isopacity, which lacks the characteristics of transparent and translucent material - opaque materials allowzerolight to pass through. Here's an example of one of our opaque materials compared to translucent. You can see that the opaque material stops any light from passing through, while the transparent material allows light, and the shape of the object, to pass through.
What about acrylic rod?
The translucent and transparent properties apply to all of the products we carry, including the cast, and extruded colored rod. Again, the usefulness of each will depend on your project. We have a wide range of both transparent and translucent rods - it’s just a matter of choosing the product that fits your vision!
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7357 | https://artofproblemsolving.com/wiki/index.php/Functional_equation?srsltid=AfmBOor6ZrrWYrd6ihUrMaiVL7diF0Z65G3LwQMQaVYw1sVn54Ox8oL- | Art of Problem Solving
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Functional equation
A functional equation, roughly speaking, is an equation in which some of the unknowns to be solved for are functions. For example, the following are functional equations:
Contents
1 Introductory Topics
1.1 The Inverse of a Function
2 Intermediate Topics
2.1 Cyclic Functions
2.2 Problem Examples
3 Advanced Topics
3.1 Functions and Relations
3.2 Injectivity and Surjectivity
4 See Also
Introductory Topics
The Inverse of a Function
The inverse of a function is a function that "undoes" a function. For an example, consider the function: . The function has the property that . In this case, is called the (right) inverse function. (Similarly, a function so that is called the left inverse function. Typically the right and left inverses coincide on a suitable domain, and in this case we simply call the right and left inverse function the inverse function.) Often the inverse of a function is denoted by .
Intermediate Topics
Cyclic Functions
A cyclic function is a function that has the property that:
A classic example of such a function is because . Cyclic functions can significantly help in solving functional identities. Consider this problem:
Find such that . Let and in this functional equation. This yields two new equations:
Now, if we multiply the first equation by 3 and the second equation by 4, and add the two equations, we have:
So, clearly,
Problem Examples
2006 AMC 12A Problem 18
2007 AIME II Problem 14
Advanced Topics
Functions and Relations
Given a set and , the Cartesian Product of these sets (denoted ) gives all ordered pairs with and . Symbolically,
A relation is a subset of . A function is a special time of relation where for every in the ordered pair , there exists a unique .
Injectivity and Surjectivity
Consider a function be a function from the set to the set , i.e., is the domain of and is the codomain of .
The function is injective (or one-to-one) if for all in the domain , if and only if . Symbolically, f(x)is injective⟺(∀a,b∈X,f(a)=f(b)⟹a=b).
The function is surjective (or onto) if for all in the codomain there exists a in the domain such that . Symbolically, f(x)is surjective⟺∀a∈Y,∃b∈X:f(b)=a.
The function is bijective (or one-to-one and onto) if it is both injective and subjective. Symbolically, f(x)is bijective⟺∀a∈Y,∃!b∈X:f(b)=a.
The function has an inverse function , where , if and only if it is a bijective function.
See Also
Functions
Cauchy Functional Equation
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7358 | https://link.springer.com/article/10.1007/s13361-017-1741-9 | Log in
Mass Defect from Nuclear Physics to Mass Spectral Analysis
Critical Insight
Published:
Volume 28, pages 1836–1843, (2017)
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Journal of The American Society for Mass Spectrometry
Mass Defect from Nuclear Physics to Mass Spectral Analysis
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Soheil Pourshahian ORCID: orcid.org/0000-0002-4138-94411
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Abstract
Mass defect is associated with the binding energy of the nucleus. It is a fundamental property of the nucleus and the principle behind nuclear energy. Mass defect has also entered into the mass spectrometry terminology with the availability of high resolution mass spectrometry and has found application in mass spectral analysis. In this application, isobaric masses are differentiated and identified by their mass defect. What is the relationship between nuclear mass defect and mass defect used in mass spectral analysis, and are they the same?
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Introduction
Mass defect and binding energy of the nucleus are two related fundamental properties of atoms. Even though they are often discussed in the context of nuclear energy, mass defect and binding energy are concepts with wider applications. Mass defect exists universally in bound systems of all sizes in which the components are bound together by force. It is applicable to small systems such as the nucleus of an atom as well as large systems such as the solar system . A bound system has a lower potential energy and mass than its components in an unbound state. The difference between the mass of a bound system and its constituents in an unbound state is referred to as mass defect. Binding energy is the energy equivalent of mass defect according to Einstein’s theory of mass-energy equivalence, and is more pronounced in the atomic nucleus than the solar system due to its small size and enormous amount of energy involved. Nuclear binding energy is the source of energy of the sun and nuclear power plants.
The emergence of the concepts of nuclear mass defect and binding energy goes back to the early 20th century after atomic weights were determined accurately by chemical methods and the deviation of atomic masses from whole numbers was investigated by mass spectrograph [2,3,4,5,6,7,8]. Early research in mass spectrometry was primarily focused on determining the accurate mass and isotopic composition of elements. However, by the 1940s this work was largely complete and mass spectrometry moved from academic laboratories into research and development facilities in the petroleum and chemical industry . Resolution and accuracy of the instruments increased over the years and the accurate mass of molecules made determination of their empirical formula possible. Mass defect re-entered scientific literature this time for mass spectral analysis and applied to the identification of molecules rather than analysis of atoms [10, 11]. In order to make the following discussion clear, applications of mass defect in nuclear physics and mass spectral analysis are referred to as nuclear and chemical mass defect, respectively, even though this distinction does not exist in the literature. Chemical mass defect, defined as the difference between the monoisotopic mass and the nominal mass, became a useful criterion for sorting through a crowded mass spectrum from a complex sample in order to identify compounds of interest among many unrelated ion peaks. It was first utilized to visualize and identify different classes of compounds in petroleum samples, and later found applications in drug metabolism and pharmacokinetics studies and identification of endogenous compounds in complex biological samples .
Nuclear mass defect and binding energy are often discussed in connection with chemical mass defect and mass spectral analysis and are incorrectly considered to be the same [12,13,14,15]. Referring to both nuclear and chemical mass defect simply as mass defect can cause confusion, especially when discussed among a broader audience from different disciplines. For example, the nuclear mass defect for carbon-12 (12C) is 0.1 mass unit (u), which when converted to energy is equal to the binding energy per nucleon (protons and neutrons in the nucleus) of 7.7 mega-electron volt (MeV) for a carbon atom. On the other hand, 12C with the atomic mass of 12.0000 (selected as an integer value by convention to define atomic mass scale) has a chemical mass defect of zero when it comes to mass spectral analysis and this could be incorrectly interpreted as the equivalent of zero binding energy for carbon. One would only be able to differentiate the two usages of the term mass defect based on the context in which it is discussed.
The goal of this discussion is not to review different applications of mass defect in mass spectrometry, which can be found elsewhere , but to discuss mass defect in general. The importance of mass defect in nuclear physics is reviewed and its significance in mass spectral analysis is discussed. It is argued that nuclear and chemical mass defects are not the same and their difference is highlighted at the end, by looking at the mass defect plots of the elements in the periodic table. While nuclear mass defect reflects a physical property, chemical mass defect does not, and it is based on a convention that carbon has a mass defect of zero. It is proposed to refer to chemical mass defect as mass excess in order to eliminate confusion surrounding the usage of the term “mass defect.”
Nuclear Mass Defect and Binding Energy
Mass of the nucleus is slightly less than the added masses of its constituent protons and neutrons and this mass difference is called nuclear mass defect :
$$ \mathrm{Nuclear}\;\mathrm{Mass}\;\mathrm{Defect}=m-\left[\left(\mathrm{Z}\times {m}_H\right)+\left(\mathrm{N}\times {m}_n\right)\right] $$
(1)
where m is the atomic mass, m H is the mass of hydrogen, m n is the mass of neutron, Z is the number of protons, and N is the number of neutrons. The energy equivalent of the nuclear mass defect is known as the nuclear binding energy. In other words, binding energy is the energy released with the formation of a nucleus from its nucleons, or is the energy required to break a nucleus into its individual components.
The nuclear mass defect is a fundamental property of a nucleus and is a fixed value corresponding to a certain amount of binding energy for that nucleus. Mass defect and binding energy are important factors in the energy involved in nuclear reactions. Looking at how mass defect and binding energy change from one element to another will make the relationship between mass defect, binding energy, and nuclear energy more apparent. A plot of nuclear mass defect versus mass number for different elements is shown in Figure 1.
The nuclear mass defect changes with mass number from zero for hydrogen to a value close to –2 for uranium. The nuclear mass defect per nucleon, which is mass defect divided by mass number, is a more useful value. It provides a more meaningful way of comparison between different elements and is plotted against mass number in Figure 2a. The corresponding energy, the binding energy per nucleon, is the amount of energy that is released per nucleon upon the formation of a nucleus and is an indication of its stability (Figure 2b).
If we start from hydrogen and move to heavier atoms on the curve in Figure 2b, we find that binding energy per nucleon increases and reaches a maximum around 56, the mass number for iron. If we continue past iron, the binding energy per nucleon decreases gradually. Therefore, the medium mass nuclei are the most stable. Iron has the highest binding energy per nucleon and is the most stable nucleus (62Ni is more stable than 56Fe but it is not the most abundant isotope of nickel). The difference in the binding energy between elements provides an opportunity for producing energy if elements with lower binding energy per nucleon are converted into more stable elements with higher binding energy per nucleon. Figure 2b shows that heavy nuclei gain stability and therefore give off energy if they are fragmented into two mid-sized nuclei. The process of splitting a nucleus into smaller nuclei is known as fission. The breakdown of the uranium nucleus into more stable nuclei as a result of collisions with neutrons releases energy and is an example of a fission reaction :
$$ {}_0{}^1n+{}_{92}{}^{235}U\to {}_{56}{}^{142}Ba+{}_{36}{}^{91}Kr+3{}_0{}^1n $$
(2)
where the subscript shows atomic number Z, superscript shows mass number A, and n represents neutrons. This is the process by which nuclear energy is produced in nuclear power plants. Energy is also released if light nuclei are combined or fused together to form more massive nuclei with greater binding energy per nucleon than that of reacting species. This too is a change towards a greater stability. The process that is called fusion is exothermic only for the nuclei of mass number below 56. The reaction of deuterium and tritium to form helium is an example of a fusion reaction :
$$ {}_1{}^2H+{}_1{}^3H\to {}_2{}^4He+{}_0{}^1n $$
(3)
Nuclear fusion is the source of energy of the sun and other stars. Combination of hydrogen nuclei to form more complex nuclei was first proposed as the mechanism of production of stellar energy in 1920 after the publication of masses of isotopes by Aston . The difference in the binding energy per nucleon between hydrogen and helium is much more than between uranium and a mid-mass element such as iron and as a result hydrogen fusion can produce more energy, kilogram for kilogram, than the nuclear fission of uranium.
Mass Excess and Q-Value
The released energy as a result of the formation of nuclei can be compared to the heat of formation of molecules. The heat of formation of a molecule is the energy released with the formation of a molecule from its elements and it is a measure of the stability of the molecule. A large heat of formation is an indication of a stable molecule since a large amount of energy is required to decompose the molecule into its constituent atoms. The energy released with the formation of a nucleus from its constituent protons and neutrons is a measure of the stability of the nucleus in a similar way. The heat of formation of molecules is considerably less than the energy released with the formation of nuclei. The standard heat of formation of CO2 from carbon and oxygen is 394 KJ/mole, which is the amount of heat released under standard conditions per mole of CO2. (For comparison, the energy released as a result of the formation of a carbon nucleus from protons and neutrons is 8.9 × 109 KJ/mole). This corresponds to a mass loss of 4.4 × 10–9 g for each mole of CO2 formed. Unlike nuclear mass defect, the mass loss is too minute to be measured and is largely ignored. The energy change in chemical reactions is calculated from the heat of formation of the reactants and products. Similarly, we can calculate the amount of energy released or consumed in a nuclear reaction. The Q-value is the energy involved in a nuclear reaction and is defined as:
$$ Q\hbox{-} value={\displaystyle \sum_{initial}m{c}^2}-{\displaystyle \sum_{final}m{c}^2} $$
(4)
where m is the atomic mass and c is the speed of light. The Q-values are usually reported in units of MeV (c 2 = 931.5 MeV/u). A positive Q-value is an indication of an energetically favored reaction. The Q-value for the reaction 3, for example, is calculated using atomic masses of reactants and products:
$$ Q\hbox{-} \mathrm{value}=\left[\left(2.014102+3.016049\right)-\left(4.002603+1.008665\right)\right]\times 931.5=17.6\;\mathrm{M}\mathrm{e}\mathrm{V} $$
The neutral isotopic mass is not always given in isotope tables, but the mass excess is listed instead in units of mass or energy (MeV). It is defined as the difference between the measured atomic mass (m) and the mass number (A):
$$ {\varDelta}_A=m-A $$
(5)
where Δ A is the mass excess. Since the sum of the mass numbers on either side of a nuclear reaction is the same, A is conserved. If (A + Δ A ) is substituted for m in Equation 4, A cancels out because the total number of protons and neutrons between reactants and products are unchanged (charge and mass conservation). We are left with an equation for the Q-value that depends only on the mass excess. For the above example, one can use the mass excess in mass unit (u) to find the Q-value:
$$ Q\hbox{-} \mathrm{value}=\left[\left(0.014102+0.016049\right)-\left(0.002603+0.008665\right)\right]\times 931.5=17.6\;\mathrm{M}\mathrm{e}\mathrm{V} $$
However, it is easier to use the energy equivalent of the mass excess, which is commonly listed in tables :
$$ Q\hbox{-} \mathrm{value}=\left(13.136+14.950\right)-\left(2.425+8.071\right)=17.6\;\mathrm{M}\mathrm{e}\mathrm{V} $$
The value that is reported in the tables of nuclear properties is the mass excess (in units of energy) rather than the mass. The above example shows that the use of mass excess makes prediction of the Q-values straightforward and simplifies the calculation of energy involved in nuclear reactions. Definition of mass defect in general chemistry and physics textbooks is consistent with Equation 1 and what has been discussed so far [17, 20]. The definition used in mass spectrometry is discussed next.
Chemical Mass Defect
Application of mass spectrometry was extended from mainly atomic to molecular analysis as commercial instruments became available to meet the demands of the chemical and petroleum industry. Resolution of molecular ions with the same nominal mass but arising from different combination of elements became possible with the increasing resolution of the instruments. Petroleum samples contain many compounds with the same nominal mass but with different elemental compositions that could be identified by high resolution instruments. A goal of petroleum analysis was to identify compounds based on their class, type, and the degree of alkylation. Compound class is collectively defined as all of elemental compositions with the same heteroatom content. Compounds with the same heteroatom but with various numbers of hydrogen in their empirical formula belong to different compound types. Compound type arises from different number of double bonds or rings in the molecule. Within the same class and type, there are many compounds with varying degrees of alkylation, which differ only in the numbers of methylene (CH2) groups in their formula. They are commonly known as homologous series. Presence of a large number of different molecules made the interpretation of the mass spectra of petroleum samples a difficult task. The high-resolution spectra of such samples with many resolved peaks at the same nominal mass demanded a new approach for data interpretation. A data interpretation strategy was developed based on the chemical mass defect. Mass defect was defined as the difference between the accurate mass of the ion in question and a reference hydrocarbon ion with the same nominal mass . This approach was used to identify several new compound classes and types not reported before.
Chemical mass defect was later defined as the difference between the nominal mass and the measured monoisotopic mass by Kendrick and was utilized to facilitate analysis of petroleum samples :
$$ \mathrm{Chemical}\;\mathrm{Mass}\;\mathrm{Defect}=A-m $$
(6)
where A is the nominal mass and m is the monoisotopic mass. Nominal mass is the mass calculated using the integer mass of the most abundant isotopes of each element. It is equivalent to mass number expressed in mass unit and they are both represented by the same symbol, A, here. Therefore, chemical mass defect is the difference between the monoisotopic mass and a whole number mass, which may not be the closest integer mass. Nominal mass is the closest integer mass to the monoisotopic mass for low molecular weight compounds but this is not necessarily true at higher masses where the difference between monoisotopic and nominal masses can be quite large . For example, polystyrene, C4H9(C8H8)100H, has a nominal mass of 10458 u and monoisotopic mass of 10464.338 u. Chemical mass defect as defined in Equation 6 is what is currently used in mass spectral analysis. Because homologous series constitute a large portion of the compounds found in petroleum samples, Kendrick introduced a new mass scale based on CH2 = 14.0000 (12C scale based masses are multiplied by 14.0000/14.01565 in order to be converted to the Kendrick mass scale). On this mass scale, the repeating mass of methylene does not change the mass defect and all of the compounds in a homologous series that belong to the same class and type will have the same chemical mass defect. Plotting chemical mass defect versus nominal mass will help visualize all of the compounds present in the spectrum in a way that would not be possible by just viewing the spectrum. Figure 3 shows an example of such a plot for peaks from the high-resolution mass spectrum of a crude oil sample . Compounds belonging to the same class and type but with different number of CH2 groups will fall on a horizontal line on this plot. Similarly, compounds of the same class but different type differ by two hydrogens and will fall on horizontal lines separated by the mass defect of H2. Compounds belonging to different classes are now readily identified because their chemical mass defect will be displaced vertically from each other. Visualization of a complex mass spectrum is simplified by using a simple two-dimensional graphical display of the data based on chemical mass defect. Patterns are recognizable on the plot, and the outlier data are easily identified. Identification of a few compounds on the plot, at least one from each class, is the key to identifying the majority of the compounds. Such a plot has been used for analyzing data from a single high resolution mass spectrum of a crude oil sample containing several thousand ion peaks . Class and type assignment for so many compounds in the sample was accomplished by taking advantage of their chemical mass defect, a task that would be difficult to achieve in the absence of such a powerful data interpretation strategy. A CH2 based mass scale is historically the first one used for the analysis of crude oil samples by mass defect. Other mass scales (16O- and H2-based for example) have also been used to plot data on two-dimensional plots similar to the one shown in Figure 3, and are useful for environmental samples . The use of more than two mass scales for graphical visualization of data on higher-order plots makes data interpretation easier and increases the number of assigned chemical formulas .
Applications of Chemical Mass Defect in Mass Spectrometry
Applications of chemical mass defect in mass spectrometry have been the subject of a recent review and can be divided into two general categories. In both cases, chemical mass defect is used to facilitate the identification of compounds of interest in a complex sample in the presence of many other mass spectral peaks.
In one category, the existing chemical mass defect in the compound of interest is used to identify or track it. The characterization of crude oil samples [23, 24] or identification of metabolites [27,28,29,30], proteins , lipids , and natural organic matter falls in this category. Mass defect filters have become common in metabolite identification in the pharmaceutical industry . The central structure of drugs does not change drastically after metabolism. Therefore, the chemical mass defect of the parent drug and its metabolites usually remain similar and fall into a narrow range. In mass defect filtering, a limited chemical mass defect range is defined and ions with chemical mass defects outside of this range are removed from the spectrum through post-acquisition data processing. The result is a clean mass spectrum with less background noise or signals unrelated to the compounds of interest.
In the second category, a chemical mass defect tag is introduced into the compound of interest, which will help its differentiation from chemical noise or other compounds. In a typical complex mass spectrum such as the spectrum of a whole cell protein digest, ion signals are clustered around certain spots and there are gaps in the mass spectrum where no signal is detected because no peptides have masses at those values . Mass tags contain atoms with large chemical mass defects per nucleon. The presence of a mass tag in a compound will shift its ion signal from the crowded areas of the spectrum to the gaps where it can be easily detected and identified . Tagging proteins, for example, has been used to improve protein sequencing and identification . Fluorinated compounds have long been popular for mass calibration and as internal standards because they have chemical mass defects that are different from naturally occurring compounds and less likely to interfere with the analysis .
Comparison between Nuclear and Chemical Mass Defect
Chemical mass defect as defined by Equation 6 was originally developed for the mass spectral analysis of molecules in petroleum samples. Because chemical mass defect in a molecule is due to contribution from its constituent atoms, Equation 6 can also be used to find the chemical mass defect of atoms. This provides an equal platform for a comparison between the nuclear and chemical mass defects. In applications other than Kendrick mass defect, it is common to calculate chemical mass defect such that elements with a lower mass than carbon, such as hydrogen, have a positive mass defect (mass defect = m – A). This definition is used for plots in Figures 4 and 5. Figure 4 shows a plot of chemical mass defect versus mass number for different elements in the periodic table. One of the curves in Figure 4 is based on the 12C mass scale, whereas the other is based on the 16O mass scale. Apart from the obvious differences between the nuclear and chemical mass defect plots in Figures 1 and 4, a comparison between the two reveals a few points:
1.
Nuclear mass defect in Figure 1 changes to increasingly negative values with increasing mass number. This means that the binding energy increases with mass number and is higher for heavier atoms. The last element on the plot in Figure 1, uranium, has the highest total binding energy. Such a relationship does not exist for the chemical mass defect in Figure 4.
2. 2.
Nuclear mass defect is a negative value and has the same sign for all elements and therefore binding energies as the energy that keeps the nucleus together will all have the same sign as expected. On the other hand, chemical mass defect is positive for some and negative for others.
3. 3.
The chemical mass defect curve in Figure 4 changes by changing the mass scale from 12C to 16O, which shows that chemical mass defect is dependent on the mass scale. This is in contrast to the mass scale dependency of the nuclear mass defect. The nuclear mass defect values in Figure 1 (which is based on the 12C mass scale) will be 0.03% higher when calculated based on the 16O mass scale, a change that will not be noticeable on the graph. This dependency is small and can be ignored compared with the mass scale dependency of the chemical mass defect. Even though the nuclear mass defect for a certain nucleus is slightly higher on the 16O mass scale, the corresponding binding energy is the same whether it is calculated based on the 16O or 12C mass scale. The conversion factors from the nuclear mass defect to the binding energy are different for the 16O and 12C mass scales such that the binding energies calculated based on different mass scales will have the same value. The nuclear mass defect and binding energy are intrinsic properties and represent physical values in absolute units of mass or energy (grams or joules). Chemical mass defect is defined relative to the atomic mass unit scale, which is a matter of convention.
In addition, nuclear mass defect by definition is only concerned with the nucleus and does not include electrons, wheras chemical mass defect includes electrons. Mass defect per nucleon is another value that one can compare between the nuclear and chemical mass defects. This is the same concept reported by Aston as the packing fraction . Chemical mass defects are divided by the corresponding mass numbers and plotted as a function of mass number in Figure 5. One of the curves in Figure 5 is based on the 12C mass scale, wheras the other one is based on 16O. A comparison between Figure 5 and the nuclear mass defect per nucleon in Figure 2a shows that the position of the curve along the vertical axis is changed. This is because chemical mass defect is a relative value, with its reference zero point depending on the mass scale. The position of the curve along the vertical axis changes so that in the case of 12C mass scale, carbon-12 has a chemical mass defect of zero. On the 16O mass scale, the curve shifts along the y-axis so that oxygen-16 has a zero chemical mass defect. Consequently, both nuclear and chemical mass defects per nucleon show similar trends moving on the curves from hydrogen to uranium, but a certain element will have two different values for the nuclear and chemical mass defects. On a mass scale based on hydrogen (H = 1.0000), the nuclear and chemical mass defects will have values that are close to each other.
It is interesting to see that despite the differences between the nuclear and chemical mass defect plots in Figures 1 and 4, the nuclear and chemical mass defect per nucleon curves in Figures 2a and 5 have similar trends. This is because the underlying phenomenon behind them, the strong nuclear force, is the same. The strong nuclear force is one of the four basic forces in nature (along with the weak nuclear force, electromagnetic force and gravity), that holds the subatomic particles of the nucleus together. It pulls the protons and neutrons in the nucleus tightly together, slightly more or less tight in different nuclei. It is useful to compare an atom in the presence and absence of the nuclear force to understand its effect. In the absence of the nuclear force, an atom can be considered as a collection of protons, neutrons, and electrons. The mass of such a hypothetical atom can be calculated from the mass and number of electrons, protons, and neutrons present in the atom. This calculated mass divided by mass number for different elements, called mass per nucleon, is plotted against mass number in the top section of the plot in Figure 6 (CV-1). The mass per nucleon in the presence of the nuclear force is calculated by dividing 12C-based masses by mass numbers and is plotted against mass number in the bottom section of the plot in Figure 6 (CV-2). The effect of the nuclear force on atomic masses is clear on the plot. The almost straight line trend (CV-1), in the presence of the nuclear force, is converted to the familiar trend seen in the mass defect curves (CV-2). Therefore, the trend seen in Figures 2a and 5 is the result of the nuclear force acting inside the nucleus. We can get a hint of the strength of the nuclear force in different nuclei by measuring nuclear mass defect, a higher nuclear mass defect per nucleon indicating a stronger force per nucleon. The nuclear mass defect is the difference between the mass of a nucleus in the presence and absence of the nuclear force. Therefore, it can be calculated by subtracting the two curves (CV-1 and CV-2) in Figure 6 (electrons are present in both curves and will cancel out after subtraction). Subtracting a straight line from a function such as a sine wave will not change its trend and will produce another sine wave. Similarly, because CV-1 is close to a straight line, the resulting nuclear mass defect curve from the subtraction of CV-1 and CV-2 will have a trend similar to CV-2. Figure 6 is in fact a graphical representation of Equation 1 divided by mass number. CV-2 is a plot of m/A versus mass number and CV-1 is the second part of Equation 1 divided by mass number ([(Z × m H) + (N × m n)]/A) plotted against mass number. CV-2 has the exact same trend as the chemical mass defect plot in Figure 5. However, there is a slight difference between CV-2 and the nuclear mass defect curve in Figure 2a. Because CV-1 in Figure 6 is not a straight line (it has slight variations from one element to the next), its deviation from a straight line will cause the nuclear mass defect curve resulting from the subtraction of CV-1 and CV-2 to have a slightly different trend than CV-2. Therefore, nuclear and chemical mass defect will have slightly different trends. Superimposing the nuclear and chemical mass defect curves will reveal their differences and show that despite their similar trends, they do not follow the exact same track.
Mass Defect or Mass Excess?
The discussion so far concludes that nuclear and chemical mass defects are different. On the other hand, chemical mass defect has the same definition as mass excess, a concept introduced to simplify the calculations of energy change involved in nuclear reactions. They are both defined as the difference between the measured monoisotopic mass and nominal mass. It is therefore logical to restrict the use of mass defect to discussions of binding energy and nuclear science and use mass excess in mass spectral analysis. Looking back at the example discussed in the Introduction, 12C has a mass defect of 0.1 u and binding energy per nucleon of 7.7 MeV, but as far as mass spectral analysis is concerned, it has a mass excess of zero. However, this is against the recommendation of the International Union of Pure and Applied Chemistry (IUPAC), which supports the use of mass defect and defines it as the difference between the nominal mass and the monoisotopic mass of an atom, molecule, or ion . In applications where the decimal value of a reported mass is of interest , the use of fractional mass is more appropriate (polystyrene with the monoisotopic mass of 10464.338 u has a mass defect of 6.338 and a fractional mass of 0.338).
Conclusion
Why are atomic masses not whole numbers? The quest to answer this question and to evaluate the divergence of atomic masses from whole numbers led to the concepts of nuclear mass defect and binding energy. Later on, chemical mass defect played an important role in mass spectral analysis. Nuclear and chemical mass defects are both caused by the strong nuclear force. However, despite their common origin and close relationship, they are different concepts. Nuclear mass defect is an absolute parameter whereas chemical mass defect is a relative value. Nuclear mass defect and binding energy are intrinsic properties and are fixed values for a certain atom. On the other hand, chemical mass defect is not a fixed value and depends on the mass scale. Nuclear mass defect reflects a physical property whereas chemical mass defect is not a physical property and is based on a convention that carbon has a mass defect of zero. Considering the differences between the nuclear and chemical mass defects, it is proposed to refer to the latter as mass excess. This will eliminate confusion surrounding the use of the term mass defect especially among different disciplines and will harmonize the terminologies used in nuclear physics and mass spectrometry.
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Soheil Pourshahian
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Pourshahian, S. Mass Defect from Nuclear Physics to Mass Spectral Analysis. J. Am. Soc. Mass Spectrom. 28, 1836–1843 (2017).
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Keywords
Mass defect
Mass excess
Delta mass
Fractional mass
Mass defect filter
Packing fraction
Nuclear binding energy
High resolution mass spectrometry |
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What is the difference between refraction and dispersion?
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Originally Answered: What is the difference between wave refraction and dispersion? ·
There's just a slight confusion in terminology here.
Refraction refers to any bending of waves due to a change in speed. When water waves move through different depths, the wave is said to be refracted.
Dispersion in your context refers to the frequency dependence of refraction. In the case of light being refracted by a prism, dispersion means that the higher frequency light bends more.
In short, refraction is the bending of light, and dispersion is the frequency dependence of this behavior.
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There's just a slight confusion in terminology here.
Refraction refers to any bending of waves due to a change in speed. When water waves move through different depths, the wave is said to be refracted.
Dispersion in your context refers to the frequency dependence of refraction. In the case of light being refracted by a prism, dispersion means that the higher frequency light bends more.
In short, refraction is the bending of light, and dispersion is the frequency dependence of this behavior.
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Brad Moffat
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Refraction is the general topic referring to the bending of light as moves from a medium of one refractive index to another.
Dispersion refers to the separation of frequencies that might happen as refraction occurs. Dispersion happens if the index of refraction value changes with changing frequency.
(Please note that to the left and right of these colours that we can see there are other frequencies of light that we can’t see. The spectrum covers most of the area to the right of the prism.)
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Refraction is the general topic referring to the bending of light as moves from a medium of one refractive index to another.
Dispersion refers to the separation of frequencies that might happen as refraction occurs. Dispersion happens if the index of refraction value changes with changing frequency.
(Please note that to the left and right of these colours that we can see there are other frequencies of light that we can’t see. The spectrum covers most of the area to the right of the prism.)
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What is the difference between wave refraction and dispersion?
What is the refraction and dispersion of light?
How is light dispersion different from light refraction?
What is dispersion, and how is it related to diffraction and refraction?
What is the difference between diffraction, refraction, dispersion, and division of light? What are some examples?
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Refraction and dispersion are both optical phenomena involving the behavior of light as it passes through different media, but they refer to different processes:
Refraction
Definition: Refraction is the bending of light as it passes from one medium to another with a different density (e.g., from air to water).
Cause: This bending occurs due to a change in the speed of light in different materials. When light enters a denser medium, it slows down and bends toward the normal (an imaginary line perpendicular to the surface). When it enters a less dense medium, it speeds up and bends away from the no
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Refraction and dispersion are both optical phenomena involving the behavior of light as it passes through different media, but they refer to different processes:
Refraction
Definition: Refraction is the bending of light as it passes from one medium to another with a different density (e.g., from air to water).
Cause: This bending occurs due to a change in the speed of light in different materials. When light enters a denser medium, it slows down and bends toward the normal (an imaginary line perpendicular to the surface). When it enters a less dense medium, it speeds up and bends away from the normal.
Example: A common example of refraction is a straw appearing bent when placed in a glass of water.
Dispersion
Definition: Dispersion is the separation of light into its constituent colors (spectrum) as it passes through a medium, typically a prism.
Cause: This occurs because different wavelengths (colors) of light travel at different speeds in a medium, leading to varying degrees of refraction. Shorter wavelengths (blue light) bend more than longer wavelengths (red light).
Example: A classic example of dispersion is a rainbow or the spectrum produced when white light passes through a prism, resulting in a range of colors from red to violet.
Summary
Refraction involves the bending of light due to changes in speed when entering different media.
Dispersion involves the splitting of light into different colors due to varying degrees of refraction for different wavelengths.
Both phenomena are fundamental to understanding optics and are utilized in various applications, such as lenses and optical instruments.
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Peter Hauge
Studied optical and microwave properties of materials · Author has 2.7K answers and 1.6M answer views
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Refraction is the angular deviation of a ray of light obliquely incident on an interface of dissimilar optical media. It’s what makes underwater objects appear closer to the surface than they are. It’s fair to include rays at normal incidence, as well, so that there is always an incident, reflected and refracted ray (except for the case of TIR.)
Dispersion refers to cases where the refractive index of an optical medium varies with wavelength. It’s why prisms (doubly) refract light of differing wavelengths into differing angular regions, thereby dispersing it.
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Originally Answered: What is the difference between these terms: Scattering, Dispersion and Diffraction? ·
Scattering: is a general physical phenomenon that means “change the path of object as a result from collision”. Usually because of some impurity or inperfection. Scattering happens on object smaller or equivalent to wavelength.
Dispersion: is a phenomenon that is a result of fact, that different wavelengths have different phase speeds. Usually it means that upon entry to a different medium with different index of refracion different “colors” (wavelengths) bend with a little different angels.
Diffraction: is “bending” of light near material. At the edge, while going through apperture, while diffr
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Scattering: is a general physical phenomenon that means “change the path of object as a result from collision”. Usually because of some impurity or inperfection. Scattering happens on object smaller or equivalent to wavelength.
Dispersion: is a phenomenon that is a result of fact, that different wavelengths have different phase speeds. Usually it means that upon entry to a different medium with different index of refracion different “colors” (wavelengths) bend with a little different angels.
Diffraction: is “bending” of light near material. At the edge, while going through apperture, while diffracting on grid (both in reflection and transition.)
Scattering - light interracts with mater in “chaotic” manner.
Dispersion - light can go through material, is not absorbed, only changes phase velocity
Diffraction - light interracts with matter in well described manner.
The main difference between scattering and dispersion is, that scattering happens on objects that are in the size of wavelength whereas diffraction happens on macroscopic object with “details” (slit, apperture, edge) in the size of wavelength.
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In physics, refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium. Refraction of light is the most commonly observed phenomenon, but other waves such as sound waves and water waves also experience refraction.
In optics, dispersion is the phenomenon in which the phase velocity of a wave depends on its frequency. Media having this common property may be termed dispersive media. Sometimes the term chromatic dispersion is used for specificity. Although the term is used in the field of optics to describe light and other electromagne
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In physics, refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium. Refraction of light is the most commonly observed phenomenon, but other waves such as sound waves and water waves also experience refraction.
In optics, dispersion is the phenomenon in which the phase velocity of a wave depends on its frequency. Media having this common property may be termed dispersive media. Sometimes the term chromatic dispersion is used for specificity. Although the term is used in the field of optics to describe light and other electromagnetic waves, dispersion in the same sense can apply to any sort of wave motion such as acoustic dispersion in the case of sound and seismic waves, in gravity waves (ocean waves), and for telecommunication signals along transmission lines (such as coaxial cable) or optical fiber.
Refractio n is the change in direction of the wave. While dispersion is separation by frequency of the wave
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passage of light from one medium to another is refraction
if it is incident on the interface separating two media at an angle other than normal, it will deviate from its original direction
…………….
a ray of light (consisting of several wavelengths. in physics, such a radiation is called ‘white’ - not necessarily the colour white) travelling initially in vacuum strikes the surface of a medium at an angle (which is NOT normal to the surface)
inside the medium, each of the different constituent wavelengths will travel along different directions
this phenomenon is called ‘dispersion’
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Gadi Lenz
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Originally Answered: What is the difference between these terms: Scattering, Dispersion and Diffraction? ·
As applied to light (or electromagnetic waves, in general):
Scattering is caused by the interaction of light with material such that the light is not absorbed and momentum is conserved. One can talk about the scattering of light from a single atom and there are well known formulas to calculate the scattering in great detail. Similarly, when light scatters off a crystal (a highly “organized” structure) the it is also well-understood, predicted and measured (including the behavior as a function of wavelength). Where things become less deterministic is when the scattering media is more complex – n
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As applied to light (or electromagnetic waves, in general):
Scattering is caused by the interaction of light with material such that the light is not absorbed and momentum is conserved. One can talk about the scattering of light from a single atom and there are well known formulas to calculate the scattering in great detail. Similarly, when light scatters off a crystal (a highly “organized” structure) the it is also well-understood, predicted and measured (including the behavior as a function of wavelength). Where things become less deterministic is when the scattering media is more complex – non-crystalline solids such as glasses, turbulent liquids or gases – where the scattering is from random media. Indeed some techniques have been devised to characterize the random medium based on the scattering of coherent light by that medium (note that in turbulent media the material scattering also changes in time and space).
Dispersion, as mentioned by others, is the result of the fact that the refractive index of any material changes with wavelength. Since the phase velocity of light inside a medium is given by c/n (c speed of light in vacuum and n the refractive index), the velocity will change as its based on the light’s wavelength. This becomes very important in optical communication, where the transmitted signal has finite bandwidth and each of its constituent wavelengths travels at almightily different speeds causing the light pulses (that encode the transmitted information) to spread in time, interfere with their neighboring pulses and ultimately causing errors at the receiver.
Diffraction, is somewhat harder to understand and is relevant to a few optical phenomena. One of these has to do with focusing of laser light beams – the tighter you focus them the more they spread out (or diffract) after the focal point. Another one, mentioned by others, is what happens when you pass light through slit or hole (especially when the dimension of the hole or slit approach the wavelength of the light). Here, the light that is close to the slit/hole edge, will change direction and interfere with other parts of the light in the far field to create well-known diffraction patterns. So diffraction affects the angular direction of the propagating light.
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Author has 4.7K answers and 2.3M answer views
·5y
Originally Answered: What is the refraction and dispersion of light? ·
a ray of light is travelling in a certain medium
it strikes an interface that separates this medium from another
it enters ‘that’ medium
this event is called refraction
the next step can be talked about only if you bring in the notion of the speed of light in the two media
…………..
repeat the above experiment, albeit with the caveat that the ray of light is polychromatic - mixture of several wavelengths
the speed of light in a medium is different for different colours - ‘wavelengths’ to a person of physics
in the other medium, the constituent wavelengths will be travelling along different directions,
the
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a ray of light is travelling in a certain medium
it strikes an interface that separates this medium from another
it enters ‘that’ medium
this event is called refraction
the next step can be talked about only if you bring in the notion of the speed of light in the two media
…………..
repeat the above experiment, albeit with the caveat that the ray of light is polychromatic - mixture of several wavelengths
the speed of light in a medium is different for different colours - ‘wavelengths’ to a person of physics
in the other medium, the constituent wavelengths will be travelling along different directions,
the angular separation between different wavelengths will be quite small, though
this phenomenon is called dispersion of light
a composite ray has ‘dispersed’ into its constituents
the above statements have, however, presumed that the light ray while travelling in the first medium was not incident normally on the interface
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Jay Wacker
physicist, phd+postdoc+faculty · Author has 4.4K answers and 26.9M answer views
·13y
Originally Answered: What is the difference between wave refraction and dispersion? ·
Hongwan Liu's answer is correct.
Refraction is the bending of light, frequently due to a change in the propagation velocity of light.
Dispersion results from different frequencies of light having different velocities.
In practice, all materials that refract light will disperse it to some degree. All dispersive materials will refract light.
The key difference between the two is that dispersion is referring to the frequency dependence (which is the ability to separate different colors of light) while refrac
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Hongwan Liu's answer is correct.
Refraction is the bending of light, frequently due to a change in the propagation velocity of light.
Dispersion results from different frequencies of light having different velocities.
In practice, all materials that refract light will disperse it to some degree. All dispersive materials will refract light.
The key difference between the two is that dispersion is referring to the frequency dependence (which is the ability to separate different colors of light) while refraction is referring to the bending of light in a material.
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Steph G
Former Structural Engineer and Project Manager (1980–2010) · Author has 25.7K answers and 14.8M answer views
·5y
“Refraction refers to any bending of waves due to a change in speed. In the case of light being refracted by a prism, dispersionmeans that the higher frequency light bends more. In short, refraction is the bending of light, and dispersion is the frequency dependence of this behavior.”
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“Refraction refers to any bending of waves due to a change in speed. In the case of light being refracted by a prism, dispersionmeans that the higher frequency light bends more. In short, refraction is the bending of light, and dispersion is the frequency dependence of this behavior.”
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Nikhil Panikkar
Communications & Signal Processing Engineer · Author has 1.1K answers and 2.8M answer views
·7y
Originally Answered: What is the difference between wave refraction and dispersion? ·
Refraction refers to the change in group velocity of the wave. When this change is accompanied by a change in the phase velocity of the wave, it leads to dispersion. See this answer:
Nikhil Panikkar's answer to Why dispersion of light do not occur in vacuum or air medium?
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Will wavelength and frequency of a wave change after reflection from denser medium?
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Christian Gingras
General scientific formation · Author has 3.2K answers and 2.9M answer views
·5y
Originally Answered: Do wavelength and frequency change during reflection? ·
In case of standard mirrors with a layer of glass on top of the reflective material, the photons effectively slow down first while entering the glass. The typical speed of photons in glass is about 66% the speed of light.
The photon reach the shiny metallic layer and every part of the sinusoidal wave change direction instantaneously. This is a common theme for any waves. They do things like reflecting without needing any time to decelerate or accelerate.
In fact, every tiny section of a single oscillation, from 0 to 360 degrees, change angle right away when encountering the first atomic layer of
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In case of standard mirrors with a layer of glass on top of the reflective material, the photons effectively slow down first while entering the glass. The typical speed of photons in glass is about 66% the speed of light.
The photon reach the shiny metallic layer and every part of the sinusoidal wave change direction instantaneously. This is a common theme for any waves. They do things like reflecting without needing any time to decelerate or accelerate.
In fact, every tiny section of a single oscillation, from 0 to 360 degrees, change angle right away when encountering the first atomic layer of the shiny metal. The reflected photon then proceed to travel at the new angle still at about 66% the speed of light.
Finally, upon reaching the last atomic layer of glass, each part of the sinusoidal wave immediately return to 100% the speed of light (same in air or deep vacuum of empty space).
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What happens to the speed, frequency and wavelength of light after reflection?
Do wavelength and frequency change during reflection?
The frequency of waves does not change with reflection. Why?
What happens to the amplitude, velocity and frequency of a wave when it is reflected back from a denser medium?
Out of speed, frequency and wavelength, which parameters remain the same after reflection?
Alfred Dominic Vella
50 years teaching maths, science and computing at all levels · Author has 4.7K answers and 4.4M answer views
·Updated 6y
Although a reflection involves an absorption followed by an emission and so the reflection is not the same photon, its properties are almost the same. The phase is changed an I wonder if the frequency is not changed a small amount by the recoil of the absorbing atom.
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Although a reflection involves an absorption followed by an emission and so the reflection is not the same photon, its properties are almost the same. The phase is changed an I wonder if the frequency is not changed a small amount by the recoil of the absorbing atom.
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Lalit Patel
PhD Physics, MBA · Author has 11.9K answers and 7.2M answer views
·7y
Wavelength and frequency are intrinsic characteristics of a wave. In almost all cases, they do not change during reflection. They may change if the medium is nonlinear and/or moving.
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When a wave reflects from a boundary with a denser medium, the behavior of wavelength and frequency depends on whether you consider the incident medium (where the source and observer sit) or the transmitted medium. Key points:
Frequency remains unchanged at the boundary.
Continuity of phase at the interface requires the oscillation rate be the same for incident, reflected, and transmitted waves at the same location and time. Thus the reflected wave has the same frequency as the incident wave.
Wavelength can change only if the wave speed in the region where the wave propagates changes.
If refle
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When a wave reflects from a boundary with a denser medium, the behavior of wavelength and frequency depends on whether you consider the incident medium (where the source and observer sit) or the transmitted medium. Key points:
Frequency remains unchanged at the boundary.
Continuity of phase at the interface requires the oscillation rate be the same for incident, reflected, and transmitted waves at the same location and time. Thus the reflected wave has the same frequency as the incident wave.
Wavelength can change only if the wave speed in the region where the wave propagates changes.
If reflection occurs and the reflected wave propagates back through the same medium as the incident wave (typical case when a wave in medium 1 reflects off a denser medium 2), the reflected wave travels in medium 1 with the same speed as the incident wave → wavelength of the reflected wave equals the incident wavelength.
If part of the wave enters medium 2 (transmitted) and propagates there, the transmitted wave's speed usually differs, so its wavelength in medium 2 is λ2 = v2 / f (different from λ1 = v1 / f). Frequency f is common to both.
Phase inversion vs. no inversion affects phase but not frequency or magnitude of wavelength in medium 1:
Reflection from a denser medium typically produces a phase inversion (π phase shift), but that does not change frequency or the wavelength in the incident medium.
Examples/summary:
Sound in air reflecting off a rigid wall: reflected sound in air has same frequency and same wavelength as incident sound; transmitted (if any) into the wall would have different wavelength.
Light reflecting from a higher-index (denser) medium: reflected light in the original medium retains the same frequency and wavelength as the incident light in that medium; transmitted light inside the denser medium has reduced speed and shorter wavelength (λ = λ0 / n).
Bottom line: frequency never changes at reflection; wavelength changes only where wave speed changes — so the reflected wave in the original medium has the same wavelength as the incident wave.
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Do the frequency and wavelength change when light passes from a rarer to a denser medium?
Does the speed of a wave change when the frequency or wavelength increase?
What is the reflection at the boundary of sound waves of denser medium to rarer medium?
Does wavelength change during reflection?
For waves in any medium, as the wavelength decreases, what happens to the frequency?
Edward Ponderer
CareerTechn.Ed., Engr./arch, & Energy/env./util. · Author has 681 answers and 1.1M answer views
·6y
No, but since not all of the wave's energy will have been reflected, its intensity will have diminished, as well as the amplitude (proportional to the square root of the intensity). In terms of quantum mechanics—say of phonons (vibrational wave quanta in solids), or photons (electromagnetic wave quanta), it is a matter of a certain percentage of the quanta not being reflected, but again, the result is diminishing intensity, not change of frequency and wavelength.
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Viktor T. Toth
IT pro, part-time physicist -- patreon.com/vttoth · Author has 10.1K answers and 172.7M answer views
·7y
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Do the frequency and wavelength change when light passes from a rarer to a denser medium?
The frequency of a signal cannot change.
Don’t think of frequency as some abstract number. Think of it as the electromagnetic field being yanked back-and-forth, back-and-forth, so many times a second.
This cannot change on the boundary between two mediums. Otherwise, the electromagnetic field would be out of sync. Back-and-forth on one side becomes forth-and-back on the other side only to become back-and-forth again a short time later; this is a clear impossibility.
What does change is the wavelength, along with the speed of propagation. So a signal of a given frequency will correspond to a short
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The frequency of a signal cannot change.
Don’t think of frequency as some abstract number. Think of it as the electromagnetic field being yanked back-and-forth, back-and-forth, so many times a second.
This cannot change on the boundary between two mediums. Otherwise, the electromagnetic field would be out of sync. Back-and-forth on one side becomes forth-and-back on the other side only to become back-and-forth again a short time later; this is a clear impossibility.
What does change is the wavelength, along with the speed of propagation. So a signal of a given frequency will correspond to a shorter wavelength as it travels through a medium that is… well, never mind denser, a medium that has a higher index of refraction. In fact, this is exactly what the index of refraction is: the ratio between the vacuum speed of light and the speed (phase velocity, to be precise) of light in the medium.
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Aravindh Vasu
B.E. in Electrical and Electronics Engineering, College of Engineering Guindy, Tamil Nadu, India · Author has 191 answers and 372.3K answer views
·7y
Originally Answered: Do wavelength and frequency change during reflection? ·
Nothing happens to wavelength or the frequency of the light after reflection.What happens is that the incident light is it’s partially reflected and partially refracted if its a transparent medium and rough surfaces or dark surfaces absorbs light
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Omar Luis Curetti
Electronic & Electrical Engineer · Author has 5.5K answers and 7.2M answer views
·Updated 6y
No, it won´t. Wavelength depends only on frequency and medium wave speed, and neither of those changes by just reflecting the wave.
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Gabriel Kroeger
Former Sneakerhead at Physics Place (2011–2014)
·7y
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When a sound wave is reflected from the boundary of denser medium, what is the phase change of the wave?
Phases are quite fluid actually, just like my gender! Sometimes I’m feeling like a girl, and sometimes I’m feeling like a train. It just depends on my mood. Waves are very similar to me, or at least that’s what I tell myself to save me from my suicidal thoughts because no one is like me and no one likes me. So, just like momma always told me from the day I was born with 3 legs, “get out of my house you foul demon baby.”
The wave actually changes to a shorter wave with a higher frequency. However, some of the energy is usually lost. Just like my puppy, who was one of my only friends, was lost af
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Phases are quite fluid actually, just like my gender! Sometimes I’m feeling like a girl, and sometimes I’m feeling like a train. It just depends on my mood. Waves are very similar to me, or at least that’s what I tell myself to save me from my suicidal thoughts because no one is like me and no one likes me. So, just like momma always told me from the day I was born with 3 legs, “get out of my house you foul demon baby.”
The wave actually changes to a shorter wave with a higher frequency. However, some of the energy is usually lost. Just like my puppy, who was one of my only friends, was lost after he committed suicide because he did not want to be around me. This was difficult for me, and you could say I lost some of my wavelength. However, now, quite like a wave, the frequency of people (and animals actually) who want to die around me is higher than ever!
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David W. Vogel
optical engineer · Author has 14.5K answers and 9.9M answer views
·7y
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Do the frequency and wavelength change when light passes from a rarer to a denser medium?
The wavelength and speed change; the frequency does not change.
You have heard that the speed of light is constant, but that is only true in a vacuum. The speed is slower in a transparent medium such as glass.
The frequency cannot change, as that would produce a discontinuity at the surface. The only way to maintain continuity where the speed changes is for the wavelength to change.
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Let me share a little story from my morning routine that might help—or at least make you nod along in recognition.
Morning Chaos = Skipping Sunscreen?
One day, I was running late, trying to multitask—skincare, foundation, eyeshadow—all while sipping coffee. Makeup felt like enough protection from the sun, right? Until a friend asked, “Are you wearing sunscreen?” I paused. Makeup… sun protection? Not exactly.
My Own Learning Curve
Here’s what I discovered:
Makeup isn't sunscreen. Even foundations or BB creams that say “SPF” often don’t offer enough coverage—either the SPF number is low, or it's inconsistently applied.
A good sunscreen gives reliable, even protection. Makeup alone is like wearing an umbrella with holes—not too helpful.
My Routine Now (and Why It Works)
Cleanse and hydrate as usual.
Apply SunVex Ultra Light Fluid Sunscreen as the final step in skincare. It absorbs within seconds.
Wait 30 seconds, then start with foundation or your makeup of choice.
Reapply occasionally if you're out in the sun all day—especially if you’re sweating or outdoors.
Yes, You Still Need Sunscreen Under Makeup
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Christian Gingras
Former Software/hardware Engineer (1997–1999) · Author has 3.2K answers and 2.9M answer views
·7y
Frequency never change as this would violate the law of conservation of energy. That law is the most important to modern science. It is the final blow to superstition, magic trick, religious belief, life after death, ghost, etc All those stories that make child smile are easy to classify as non scientific since no object can appear magically and no task can be done without a source of energy.
So, if anything can change when electromagnetic wave cross anything, it is only wavelength.
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Robert James Newton
BSc in Chemistry, King's College London (KCL) (Graduated 1969) · Author has 3.4K answers and 4.6M answer views
·9y
There are no changes to wavelength and frequency with reflection because the speed of the light is not changed .
There is a change of wavelength (but not frequency) with refraction, because the speed of light is slower when travelling in a denser medium.
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Jess H. Brewer
Physics professor since 1977. · Author has 22.9K answers and 59.6M answer views
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What happens to the speed, frequency and wavelength of light after reflection?
Nothing.
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Related questions
What happens to the speed, frequency and wavelength of light after reflection?
Do wavelength and frequency change during reflection?
The frequency of waves does not change with reflection. Why?
What happens to the amplitude, velocity and frequency of a wave when it is reflected back from a denser medium?
Out of speed, frequency and wavelength, which parameters remain the same after reflection?
Do the frequency and wavelength change when light passes from a rarer to a denser medium?
Does the speed of a wave change when the frequency or wavelength increase?
What is the reflection at the boundary of sound waves of denser medium to rarer medium?
Does wavelength change during reflection?
For waves in any medium, as the wavelength decreases, what happens to the frequency?
Why does the wavelength of a wave change in a new medium?
Do you get a phase change of pi on reflection with denser medium in case of longitudinal waves?
When light travels from a rarer medium to a denser medium the wavelength decreases which means the frequency increase so why doesn't the energy of light change?
Why does the wavelength increase as light waves enter an optically denser medium?
When a wave undergoes reflection at a denser medium, what happens to its phase?
Related questions
What happens to the speed, frequency and wavelength of light after reflection?
Do wavelength and frequency change during reflection?
The frequency of waves does not change with reflection. Why?
What happens to the amplitude, velocity and frequency of a wave when it is reflected back from a denser medium?
Out of speed, frequency and wavelength, which parameters remain the same after reflection?
Do the frequency and wavelength change when light passes from a rarer to a denser medium?
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7362 | https://www.youtube.com/watch?v=Dq3eP8MtOLI | Palindromes
Complete Mathematics
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Posted: 17 Feb 2023
Is every 4-digit palindromic number divisible by 11? How do you know?
What about 5, 6, or 7 digit palindromes? Can you generalise?
2 comments
Transcript:
look at these examples of palindromic numbers every four digit palindrome is divisible by 11. is this always sometimes or never true what about five digits or six digits can you generalize for palindromes of any length okay so it's very tempting with this task to jump straight towards an algebraic proof but I think in doing so you miss out an opportunity to talk about players value a little bit so in the classroom I'll get the students to try a few four digit palindromes and check that they all seem to be divisible by 11. so nudging towards this always been true but then we have an opportunity to work systematically so I'll say what is the smallest four digit palindrome and hopefully they'll come back at me saying it's one thousand and one and when we divide that by 11 we get 91. and then I say okay what's the next smallest palindrome and what do you physically have to do to this to this number well the thousands and the units stays the same and the tens and the hundreds increase by one that's what happens as you go from the smallest palindrome to the next smallest now how do I know without calculating that this is still going to be divisible by 11. well it's the tens and the hundreds which have both increased by one we've increased it by 100 and 10. and 110 is divisible by 11. it's 10 lots of 11. so without calculating I know that this answer when I divide it by 11 would be 101. an extra 10 11. and this role this rule is going to hold for the next one as well you can see it's just these middle two digits the the tens and the hundreds which are increasing by one each time so we just did add in 110 each time so that one would be 111. now that would be nice if it worked all the way but it doesn't because it falls down when we get to one nine nine one because the the next palindrome after that um well we obviously can't just add one to these anymore it's the thousands in the units that change the next one after this is 2002 but we don't need to panic here as well because all we need to do is check the difference between these two numbers and it's quicker to just do this in your head you can see we've just added on 11 there what if you really want to stick with this idea of investigating place value you can say okay what's physically changed with this number compared to this number well the thousands and the units have increased by one so we've added on a thousand and we've added on a one and then what's happened to the hundreds and the tens well we've subtracted 900 and we've subtracted m a 90. and when you work out the result of all that calculation you just get plus 11. so as we Bridge these thousands we're only adding on 11 so it's still going to be divisible by 11. and this Rule now will hold all the way up until I get to the largest four digit palindrome which is nine thousand nine hundred and ninety nine so now we've done all that work on place value we're in a position to make more sense of the algebraic proof so what I like to do is write down four letters on the board A B C D and say to pupils okay if this is a four digit number is this a palindrome and quite quickly students will say things like it's only a palindrome if B is equal to C and A is equal to D so I say okay let's rub those out and make that happen so now we've got a B B A or Abba and is this a palindrome and pupils will help to say yes it is and now I like to say how many A's and how many B's can you see there and their instinctive response is going to be I can see two A's and two B's but here's the opportunity to link it with what we've just done with the place value stuff and actually writing the the place value heading so we've got our units tens hundreds and thousands and actually how many A's and actually how many B's are there well there are one thousand and one A's and how many B's are there there's 100 and 110 there's a hundred and ten bees in total and what do we know about these two numbers from the previous work well we know 1001 is 99a is 99 divided by 11 is 99 sorry and 110 is 10 lots of 11. so we're just one step away from the formal algebraic proof here where we factorize we take our factor of 11 out and we know that's going to be 91 a plus 10. B so we've arrived at the formal algebraic proof to why any four digit palindrome is always divisible by 11. okay so hopefully you like that nice link between the place value work and the formal algebraic proof and one thing we've not looked at yet is different length palindrome so I'll leave that for you to have a go at what about five digit or six digit palindromes are you gonna jump back and test three digit and two digit palan Jones first to try and get him try and specialize with smaller numbers have a play with the task and let me know your thoughts |
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7364 | https://www.youtube.com/watch?v=xLr0ElDc48I | How to find density, mass, and volume
MooMooMath and Science
574000 subscribers
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Description
555419 views
Posted: 17 Apr 2018
Learn to calculate density, mass, and volume.
The formula for density equals mass divided by volume.
Join this channel to get access to perks:
A cake has a mass of 500 g and volume of 2500 cm3. What is the density of the cake?
2.A solid block of wood has a mass of 6.0 g and a volume of 12.0 cm3. What is the density of the block of wood?
3.The density of a substance is 8.0 g/cm3. If a sample of the substance has a volume of 25 cm3, then what is its mass?
4.You have a weight with a mass of 820 g. The density of the weight is 5 g/cm3. What is the volume of the ball?
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Transcript:
science in this video I'd like to go over how to calculate density mass or volume first you can see I have a triangle here and the formula for density is equal to density is equal to mass divided by volume and you can use this triangle and remember you're going to the DMV so it's going to be a D em V and this is density so if you need density its mass over volume if you need volume its mass divided by density and then if you need mass its density by volume and I'll work an example for each of these okay so the first example is this let me move this up just a little bit triangle will go out of the way but I'll draw smaller ones is this a cake has a mass of 500 grams and volume of 250 centimeters q what is the density of the cake so in the first example I'm going to draw very small remember it's D em V and I know that's a little small but I think it'll work so let's see what we have yeah we're looking for density is equal to we have the mass of 500 grams so the mass is here so it's gonna go on top 500 grams and then volume which is the the V right here is 2500 and that is centimeters cubed and so when you divide 500 by 2500 that equals 0.2 and you're left with grams per centimeters cubed okay that's the first example of density okay the next next example again I'm gonna draw a little triangle d m/v it has we're looking for a solid block of wood has a mass of six grams and a volume of 12 centimeters cubed what is the density of the block of wood so again I'm looking for density the mass which goes here is 6 grams the volume it's 12 centimeters cubed when you divide 6 by 12 that gives you point 5 and again you have grams per centimeters Q okay so now we'll see how this triangle can come in to business when we work a couple other type of examples and so me pull this down and it says this the density of a substance is 8 grams per centimeters cubed if a sample of the substance has a volume of 25 centimeters cubed what is its mass so again I'm gonna go to the DMV the m/v okay and I am looking for mass mass is here so it's just density times the volume so I'm gonna put mass is equal to D times B now all I have to do is plug in the numbers well I know that it says the density is 8 grams per centimeters cubed times and the volume is 25 centimeters cubed okay and so when you multiply that that gives you 8 times 25 is 200 now what do we do with our units okay well if you want to look at this this is grams over centimeters cubed and this would be centimeters cubed over one these cancelled and you're left with grams over one or grams so it becomes 200 grams and make sure that the units are done properly and finally you have a weight with a mass of 820 grams the density of the weight is 5 grams per centimeters cube what is the volume again we'll go back to this we're trying to find volume and it's mass divided by density so V is equal to M over D so let's just plug in the numbers okay I have a mass of 820 grams have a density of 5 grams per centimeters cube okay and then when I divide 820 by 5 that equals 160 for the grams cancel and I'm left with centimeters cubed so the volume is 164 centimeters cube so there we go remember you go to the DMV the D em V and that's density mass volume and you can calculate all three of those using the triangle thanks for watching and moomoomath uploads a new math and science video every day please subscribe and share |
7365 | https://openstax.org/books/algebra-1/pages/1-12-5-writing-an-equation-given-two-points | 1.12.5 Writing an Equation Given Two Points - Algebra 1 | OpenStax
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Algebra 1
1.12.5 Writing an Equation Given Two Points
Algebra 11.12.5 Writing an Equation Given Two Points
Contents Contents
Highlights
Table of contents
Getting Started
Supporting All Learners
1 Linear Equations
Unit 1 Overview and Readiness
1.1 Exploring Expressions and Equations
1.2 Writing Equations to Model Relationships, Part 1
1.3 Writing Equations to Model Relationships, Part 2
1.4 Equations and Their Solutions
1.5 Equations and Their Graphs
1.6 Equivalent Equations
1.7 Explaining Steps for Rewriting Equations
1.8 Choosing the Correct Variable to Solve For, Part 1
1.9 Choosing the Correct Variable to Solve For, Part 2
1.10 Connecting Equations to Graphs, Part 1
1.11 Connecting Equations to Graphs, Part 2
1.12 Writing the Equation of a Line
1.12.0 Lesson Overview
1.12.1 Using the Slope Formula
1.12.2 Writing an Equation Given the Slope and y -intercept
1.12.3 Writing an Equation Given the Slope and a Point
1.12.4 Writing Equations in Different Forms
1.12.5 Writing an Equation Given Two Points
1.12.6 Choosing a Method to Write an Equation
1.12.7 Practice
1.12.8 Lesson Summary
1.13 Lines from Tables and Graphs
1.14 Writing Equations of Parallel and Perpendicular Lines
1.15 Direct Variation
Project 1: Slopes and Intercepts
Unit 1 Wrap Up
2 Linear Inequalities and Systems
3 Two-Variable Statistics
4 Functions
5 Introduction to Exponential Functions
6 Working with Polynomials
7 Introduction to Quadratic Functions
8 Quadratic Equations
9 More Quadratic Equations
Research in Practice
Appendix
Index
Search for key terms or text.
Close
1.12.5 • Writing an Equation Given Two Points
Activity
Answer the following questions.
Use the graph below to pick two points and find the slope between the two points.
Compare your answer:
The slope of the line is −1 3−1 3−1 3.
Use your two points in the slope formula to confirm the slope is correct.
Compare your answer:
Your answer may vary, but here is a sample.
m=y 2−y 1 x 2−x 1=1−3 4−(−2)=−2 6=−1 3 m=y 2−y 1 x 2−x 1=1−3 4−(−2)=−2 6=−1 3 m=y 2−y 1 x 2−x 1=1−3 4−(−2)=−2 6=−1 3
Using the slope and a point, write your equation in point-slope form.
Compare your answer:
Your answer may vary, but here is a sample.
m=−1 3 m=−1 3 m=−1 3 and (4,1)(4,1)(4,1)
y−y 1=m(x−x 1)y−y 1=m(x−x 1)y−y 1=m(x−x 1) becomes: y−1=−1 3(x−4)y−1=−1 3(x−4)y−1=−1 3(x−4).
Now write your equation in slope-intercept form.
Compare your answer:
y=−1 3 x+7 3 y=−1 3 x+7 3 y=−1 3 x+7 3
Convert slope-intercept form to standard form.
Compare your answer:
(x+3 y)=7(x+3 y)=7(x+3 y)=7
Look at your equation and the original graph. What do you notice? How can you use a graph to check your equation?
Compare your answer:
Your answer may vary, but here is a sample.
The y y y-intercept of the graph is 7 3 7 3 7 3 and the slope is −1 3−1 3−1 3.
Self Check
Which equation represents the slope-intercept form of the equation of a line containing the points(−2,−4)(−2,−4) and (1,−3)(1,−3)?
y=3 x−6 y=3 x−6
y=7 x+10 y=7 x+10
y=1 3 x−7 3 y=1 3 x−7 3
y=1 3 x−10 3 y=1 3 x−10 3
Additional Resources
Find an Equation of the Line Given Two Points
So far, you have two options for finding an equation of a line: slope-intercept or point-slope. When you start with two points, it makes more sense to use the point-slope form.
But then you need the slope. You can find the slope with just two points and then use it and one of the given points to find the equation.
Example
Write the equation of a line that contains the points (−3,−1)(−3,−1)(−3,−1) and (2,−2)(2,−2)(2,−2). Write the equation in slope-intercept form and standard form.
Step 1 - Find the slope using the given points.
Find the slope of the line through (−3,−1)(−3,−1)(−3,−1) and (2,−2)(2,−2)(2,−2).
m=y 2−y 1 x 2−x 1 m=y 2−y 1 x 2−x 1 m=y 2−y 1 x 2−x 1
m=−2−(−1)2−(−3)m=−2−(−1)2−(−3)m=−2−(−1)2−(−3)
m=−1 5 m=−1 5 m=−1 5
m=−1 5 m=−1 5 m=−1 5
Step 2 - Choose one point.
Choose either point.
(x 1,y 1)(x 1,y 1)(x 1,y 1)
(2,−2)(2,−2)(2,−2)
Step 3 - Substitute the values into the point-slope form, y−y 1=m(x−x 1)y−y 1=m(x−x 1)y−y 1=m(x−x 1).
Simplify
y−y 1 y−(−2)y+2===m(x−x 1)−1 5(x−2)−1 5 x+2 5 y−y 1=m(x−x 1)y−(−2)=−1 5(x−2)y+2=−1 5 x+2 5 y−y 1=m(x−x 1)y−(−2)=−1 5(x−2)y+2=−1 5 x+2 5
Step 4 - Write the equation in slope-intercept form.
y=−1 5 x−8 5 y=−1 5 x−8 5 y=−1 5 x−8 5
Step 5 - Convert slope-intercept form to standard form.
y=−1 5 x−8 5 y=−1 5 x−8 5 y=−1 5 x−8 5
1 5 x+y=−8 5 1 5 x+y=−8 5 1 5 x+y=−8 5
5(1 5 x+y)=5(−8 5)5(1 5 x+y)=5(−8 5)5(1 5 x+y)=5(−8 5)
(1 x+5 y)=−8(1 x+5 y)=−8(1 x+5 y)=−8
Use this table for easy reference to find an equation of a line given two points.
Step 1 - Find the slope using the given points.
m=y 2−y 1 x 2−x 1 m=y 2−y 1 x 2−x 1 m=y 2−y 1 x 2−x 1
Step 2 - Choose one point.
Step 3 - Substitute the values into the point-slope form: y−y 1=m(x−x 1)y−y 1=m(x−x 1)y−y 1=m(x−x 1).
Step 4 - Write the equation in slope-intercept form.
Step 5 - Convert slope-intercept form to standard form.
To find an equation of a line given the slope and a point, follow these steps:
Step 1 - Identify the slope.
Step 2 - Identify the point.
Step 3 - Substitute the values into the point-slope form, y−y 1=m(x−x 1)y−y 1=m(x−x 1)y−y 1=m(x−x 1).
Step 4 - Write the equation in slope-intercept form.
Step 5 - Convert slope-intercept form to standard form.
Try it
Writing the Equation of a Line Given Two Points
Write the equation of a line that contains the points(−3,5)(−3,5)(−3,5) and (−3,4)(−3,4)(−3,4).
Here is how to find an equation of a line that contains the points (−3,5)(−3,5)(−3,5) and (−3,4)(−3,4)(−3,4).
Again, the first step will be to find the slope.
Step 1 - Find the slope of the line through (−3,5)(−3,5)(−3,5) and (−3,4)(−3,4)(−3,4).
Step 2 - m=y 2−y 1 x 2−x 1 m=y 2−y 1 x 2−x 1 m=y 2−y 1 x 2−x 1
Step 3 - m=4−5−3−(−3)m=4−5−3−(−3)m=4−5−3−(−3)
Step 4 - m=−1 0 m=−1 0 m=−1 0
A line with undefined slope is a vertical line. Both given points have an x x x-coordinate of −3−3−3. So, the equation of the line is x=−3 x=−3 x=−3. There is no y y y, so the equation cannot be written in slope-intercept form.
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Book title: Algebra 1
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7366 | https://www.cs.tufts.edu/r/geometry/pdf/convexify-cccg11.pdf | CCCG 2011, Toronto ON, August 10–12, 2011 Convexifying Polygons Without Losing Visibilities Oswin Aichholzer∗ Greg Aloupis† Erik D. Demaine‡ Martin L. Demaine‡ Vida Dujmovi´ c§ Ferran Hurtado¶ Anna Lubiw∥ G¨ unter Rote∗∗ Andr´ e Schulz†† Diane L. Souvaine‡‡ Andrew Winslow‡‡ Abstract We show that any simple n-vertex polygon can be made convex, without losing internal visibilities between ver-tices, using n moves. Each move translates a vertex of the current polygon along an edge to a neighbouring vertex. In general, a vertex of the current polygon rep-resents a set of vertices of the original polygon that have become co-incident.
We also show how to modify the method so that ver-tices become very close but not co-incident, in which case we need O(n2) moves, where each move translates a single vertex.
The proof involves a new visibility property of poly-gons, namely that every simple polygon has a visibility-increasing edge where, as a point travels from one end-point of the edge to the other, the visibility region of the point increases.
1 Introduction There are many interesting problems about reconfigur-ing geometric structures while maintaining some proper-ties. Examples include: flips in triangulations , push-ing and sliding block puzzles , morphing of poly-gons and planar graphs [18, 21], and linkage reconfig-∗Institute for Software Technology, University of Technology Graz, Austria, oaich@ist.tugraz.at. Partially supported by the FWF under grant S9205-N12 NRN Industrial Geoemtry.
†D´ epartement d’Informatique, Universit´ e Libre de Bruxelles, aloupis.greg@gmail.com ‡MIT Computer Science and Artificial Intelligence Labora-tory, 32 Vassar St., Cambridge, MA 02139, USA, {edemaine, mdemaine}@mit.edu §School of Computer Science, Carleton University, vida@scs.carleton.ca ¶Departament de Matematica Aplicada II, Universitat Polit ecnica de Catalunya, ferran.hurtado@upc.edu.
Par-tially supported by projects MICINN MTM2009-07242 and Gen. Cat. DGR 2009SGR1040 ∥School of Computer Science, University of Waterloo, alubiw@uwaterloo.ca ∗∗Institut f¨ ur Informatik, Freie Universit¨ at Berlin, Taku-straße 9, 14195 Berlin, Germany. rote@inf.fu-berlin.de ††Institut f¨ ur Mathematische Logik und Grundlagenforschung, Universit¨ at M¨ unster, Germany, andre.schulz@uni-muenster.de ‡‡Department of Computer Science, Tufts University, {dls, awinslow}@cs.tufts.edu.
Research supported in part by NSF grants CCF-0830734 and CBET-0941538.
uration [7, 23]. Reconfiguration has also been studied outside the geometric domain .
This paper is about convexifying a simple polygon, i.e., continuously transforming the polygon to a con-vex polygon while maintaining simplicity. If no other structure must be maintained, this can be done in a trivial way, moving only one vertex at a time. When edge lengths must be maintained, this is a major re-sult, namely the Carpenter’s Rule Theorem [7, 23], and the reconfiguration process involves moving all vertices simultaneously.
In the Open Problem session at CCCG 2008 , Satyan Devadoss asked whether a polygon can be con-vexified without losing internal visibility between any pair of vertices, and in particular, whether this can be done by moving only one vertex at a time . We give a positive answer. We first consider a version of the prob-lem where vertices may become co-incident during the transformation, so one vertex of the polygon in general represents a set of vertices of the original polygon. We show that any polygon can be convexified by a sequence of n moves, where each move strictly increases the set of pairs of vertices that are internally visible, and each move translates one vertex along an edge of the polygon to a neighbour. In terms of the original polygon, each move translates a set of vertices along a straight line to join another set of vertices.
In Section 3 we modify our method so that vertices become very close but not coincident. In this case, we need O(n2) moves, each moving one vertex.
Internal vertex visibilities are never lost, but a single move does not necessarily add any internal vertex visibilities.
Our main tool, which may be of independent interest, is to show that every polygon has a visibility-increasing edge where, as a point travels from one endpoint of the edge to the other, the visibility region of the point in-creases.
Previous Work In the original model where coincident vertices are not allowed, Aichholzer et al. showed that any monotone polygon can be convexified without losing vertex vis-ibilities.
Their transformation moves one vertex at a time, but the number of vertex moves is not polynomi-ally bounded. If all vertices may move simultaneously, 23rd Canadian Conference on Computational Geometry, 2011 they observe that a monotone polygon can be convexi-fied in one move. They also show that, even for mono-tone polygons, it is not always possible to move just one vertex and strictly increase the set of vertex visibil-ities. Note that such an example depends crucially on prohibiting coincident vertices! If vertices are allowed to be coincident, our result shows that for any simple polygon, it is possible to move one vertex until it gains a new neighbour in the visibility graph.
The issue of allowing/disallowing coincident vertices has arisen before in problems of transforming (or “mor-phing”) polygons and straight-line graph drawings.
Cairns showed how to transform between any two straight-line planar triangulations that are combinato-rially the same, using a sequence of moves each of which translates one vertex onto another (or the reverse). He then comments that it is possible to avoid coincident vertices by keeping them a small distance apart.
A somewhat similar issue comes up in the result of Guibas and Hershberger who show that for any two simple polygons on vertices 1, 2, . . . , n such that edge (i, i + 1) has the same direction vector in both polygons, there is a morph between the polygons that preserves simplicity and the direction vectors of edges. Their method moves vertices infinitesimally close together and operates on the infinitesimal structures.
The question of moving only one vertex at a time has recently been settled in independent work by ´ Abrego et al. , who show that if a there is a transformation that convexifies a polygon without losing vertex visibilities, then the transformation can be accomplished by moving only one vertex at a time.
Although not directly relevant to this paper, we note that there is a considerable body of work on making polygons convex by means of “pivot” operations, such as flips [13, 8, 15, 25] and flipturns [3, 4].
For background on visibility graphs of polygons, see the books by Ghosh and O’Rourke .
Definitions Two points inside a polygon P are visible if the line segment between them is contained in the closed poly-gon. Given this definition, we will now use “visibility” rather than “internal visibility”. We will assume that the input polygon does not have three or more collinear vertices. It is possible to perturb a polygon to achieve this without losing internal vertex visibilities. Note the consequence that if two vertices are visible, then the line segment between them does not go through another ver-tex. For point p in P, the visibility region of p, denoted V (p), is the set of points in P visible from p.
Let a be a reflex vertex with neighbours b and b′ on the polygon boundary. Extend a line segment from b to a and beyond, until it first hits the polygon boundary at p. Define Pocket(b, a) to be the region bounded by the chain along the polygon boundary from a to p go-ing through b′, together with the line segment pa. We consider points along the line segment pa to be outside the pocket (i.e., the pocket is open along its “mouth”).
In particular, a is outside Pocket(b, a). See the shaded region in Figure 1(a).
2 Convexifying polygons Theorem 1 An n-vertex polygon can be convexified in n moves, where each move strictly increases the set of pairs of visible vertices, and each move translates one vertex of the current polygon along an incident edge to a neighbour on the polygon boundary.
The main tool in proving the theorem is the following.
We prove that if a polygon is not convex then it has an edge along which visibility increases.
More precisely, define an edge (u, v) to be a visibility-increasing edge if for every point p along the edge (u, v) we have V (u) ⊆ V (p) ⊆V (v), and there is a vertex in V (v) −V (u).
We will use a stronger induction hypothesis to prove that every non-convex polygon has a visibility-increasing edge (u, v) where v is a reflex vertex. Note that the fact that v is reflex implies that there is a vertex in V (v) −V (u).
Lemma 2 Let P be a simple polygon with reflex vertex a and edge (b, a). Then there is a visibility-increasing edge (u, v) with v reflex and u, v exterior to Pocket(b, a) such that u does not see into Pocket(b, a).
Proof. We prove the result by induction on the number of reflex vertices of the polygon exterior to the pocket.
If (b, a) is a visibility-increasing edge, then it satisfies the lemma, since b does not see into Pocket(b, a). See Figure 1(a). This takes care of the base case where every vertex v ̸= a exterior to the pocket is convex.
a a b b c (a) (b) b' p Figure 1: Visibility-increasing edges: (a) the edge (b, a) is a visibility-increasing edge; (b) vertex b is reflex, so we apply induction on (c, b).
If b is a reflex vertex then let c be the other neigh-bour of b (i.e., the neighbour not equal to a). See Fig-ure 1(b). Then Pocket(c, b) ⊇Pocket(b, a). Also, note that the reflex vertex a is exterior to Pocket(b, a) and CCCG 2011, Toronto ON, August 10–12, 2011 not exterior to Pocket(c, b). Therefore we can apply in-duction to conclude that there is a visibility-increasing edge (u, v) exterior to Pocket(c, b) such that v is reflex and u does not see into Pocket(c, b). Then u cannot see into Pocket(b, a), so (u, v) satisfies the lemma.
a b t p q x y Figure 2: Visibility-increasing edges in the general case, where we apply induction on (y, x).
We are left with the case where b is a convex vertex but (b, a) is not a visibility-increasing edge. Note that because a is a reflex vertex, V (a) contains a vertex not in V (b). Therefore, the only way that (b, a) can fail to be visibility-increasing is that there is a point p on (b, a) and a point t on the boundary of P such that t sees p, but t does not see a.
See Figure 2.
Now we rotate the line through t and p about t until it hits the poly-gon boundary. More precisely, consider the first point q along the line segment pa such that the line segment qt does not lie in the interior of P. Then some vertex x lies on the line segment qt. Note that x must be a reflex vertex. There are two paths on the polygon boundary from x to t. Take the path that does not contain a, and let y be the neighbour of x on this path. (It may happen that y = t.) We will apply induction on the edge (y, x).
Observe that Pocket(y, x) ⊇Pocket(b, a).
Also, note that the reflex vertex a is exterior to Pocket(b, a) and not exterior to Pocket(y, x). Therefore we can apply in-duction to conclude that there is a visibility-increasing edge (u, v) exterior to Pocket(y, x) such that v is reflex and u does not see into Pocket(y, x). Then u cannot see into Pocket(b, a), so (u, v) satisfies the lemma.
□ Proof. [of Theorem 1] The proof is by induction on the number of vertices. If the polygon has three ver-tices then it is already convex. For the general case, if the polygon is convex then there is nothing to prove, so suppose there is a reflex vertex. Then by Lemma 2, there is a visibility-increasing edge (u, v). The plan is to move vertex u to vertex v, resulting in a simple polygon with fewer vertices on which we apply induction. See Figure 5. Let w be the other neighbour of u on the poly-gon boundary. We have V (u) ⊆V (v) and w ∈V (u), so w must be visible to v. In particular, u is a con-vex vertex and the line segment wv does not intersect the polygon boundary except at its endpoints. There-fore moving u to v results in a simple polygon. Observe that no vertex visibilities are affected by the move, ex-cept that u gains visibilities once it reaches v (if not before).
Note that u may become collinear with two other vertices of the polygon at an intermediate point of the move, but this causes no problems.
□ 3 Avoiding coincident vertices In the previous section we showed how to convexify any polygon without losing internal visibilities, provided that vertices are allowed to become coincident. In this section we show how to avoid coincident vertices. Each set of coincident vertices from the previous method is replaced by a cluster of vertices that are close together but not coincident. One move of the previous method becomes O(n) moves, each moving a single vertex. The total number of moves is therefore O(n2). Vertex visibil-ities are never lost, but a single move might not increase vertex visibilities.
Figure 3: Cluster vertices along a single edge (top); a reflex cluster (left); and a convex cluster (right). Shaded areas indicate the interior of the polygon.
The basic idea is to replace an edge uv by a slightly outward-bent convex chain, with some points on a shal-low convex curve close to u, and other points on a shal-low convex curve close to v, see Figure 3 (top). In gen-eral, a cluster will consist of a representative vertex v, together with the vertices that have been moved to join v, and now lie on two convex curves incident to v. The representative vertex v will be at the same point in the plane as it was in the original polygon. If C is a clus-ter with representative vertex v, we will say that C is the cluster of v. Figure 3 depicts a reflex and a con-vex cluster.
In a convex cluster all vertices see each other; in a reflex cluster only vertices in the same arc see each other, and the representative vertex sees the whole cluster.
All vertices of a cluster lie in the ε-neighbourhood of 23rd Canadian Conference on Computational Geometry, 2011 the representative vertex for some sufficiently small ε.
In addition, all vertices of a cluster lie within some angle α of the original edge. See Figure 4(a).
We define values for ϵ and α that will work through-out the algorithm.
As the convexification proceeds, edges between representative vertices of the intermedi-ate polygons are always chords of the original polygon.
We will take all the chords into account when we define ε and α. We choose ε small enough that visibility be-tween two points in the ε-neighbourhoods of two vertices x and y behaves like visibility between x and y. Thus ε should be smaller than the distance between any ver-tex and a (non-coincident) chord or edge extension—see Figure 4(b). We choose α small enough that a represen-tative vertex x does not block visibilities of vertices in its cluster, and that a convex vertex remains convex— see Figure 4(c).
Apart from the constraints imposed by ϵ and α we are free to place the cluster vertices on any convex chain, and we will have occasion to alter the chain.
α ε x (a) x y z x y (b) x y (c) x Figure 4: (a) Cluster vertices are located in the shaded region determined by ε and α; (b) Constraints on ε, which must be small enough that visibility from a point within an ϵ-neighbourhood of a vertex acts like visibility from the vertex; (c) Constraints on α, which must be small enough that a vertex x does not block visibility to its cluster.
We now consider the move operation from the previ-ous section as it operates on clusters. The move oper-ation always moves a convex vertex u to join a reflex vertex v. See Figure 5. The only other vertex affected by the move is w, the other neighbour of u, which forms a triangle with u and v. Suppose without loss of gen-erality that v, u, w are in clockwise order around the polygon. When vertices are replaced by clusters, the vertices affected by the move are: all of u’s cluster; the left part of v’s cluster; and the right part of w’s clus-ter.
See Figure 6.
Note that, although the original move always increased the set of vertices visible from u, the modified move will not necessarily increase visibility from u or any of its cluster, since we do not move any vertex all the way to v.
w w v v u u Figure 5: Moving vertex u along the visibility-increasing edge (u, v) affects vertices u, v, and w, which form a triangle. Vertex v may remain reflex (left) or become convex (right).
u v w v w (a) (b) Figure 6: The operation from Figure 5 in the presence of cluster vertices: (a) the initial configuration, the final configuration shown with dashed lines, and the vertex correspondence shown with thin lines; (b) the interme-diate configuration after moving u and its cluster close to v.
We show that the transformation of clusters as shown in Figure 6 can be accomplished by moving one vertex at a time. The first phase is to move u and its cluster close to v, in a configuration congruent to their final configuration.
Move the vertices one by one starting with the vertex closest to v along the chain. Note that u loses its status as a representative vertex.
The re-sult of the first phase is shown in Figure 6(b).
Note that the union of the initial and final positions of all the vertices that are moved in the first phase is in con-vex position.
Therefore, convexity of the cluster and visibility within the cluster are maintained. Globally, as each cluster vertex moves from the neighbourhood of u’s initial position to v’s neighbourhood, its visibility changes exactly as u’s visibility changed in the original non-cluster move (stopped just before reaching v).
In the second phase (from Figure 6(b) to the final CCCG 2011, Toronto ON, August 10–12, 2011 configuration) the transformation we wish to realize is a counterclockwise rotation of w’s right cluster and a clockwise rotation of v’s left cluster to their final posi-tions. We describe how to do this for v’s left cluster. In the first step, move the vertices of v’s left cluster (one by one) close enough to v that their new positions and their final positions are in convex position, as shown in Figure 7. In the second step, move the vertices one by one to their final positions, starting with the vertex far-thest from v along the chain. Convexity of the cluster (and hence visibility within the cluster) is maintained during the second step because the union of the initial and final positions of all moved vertices is in convex po-sition. Global visibilities may be gained but are never lost.
v Figure 7: Adjusting the position of v’s left cluster vertices.
All movement takes place within the ϵ-neighbourhood of v.
The first vertex move is shown with a thin directed line. Note that this figure is not to scale since the angle α should be much smaller.
From the above ideas, we obtain the following result.
Theorem 3 An n-vertex polygon can be convexified in O(n2) moves, so that visibilities between vertices are never lost, and vertices never become coincident. Each move is a translation of a single vertex.
4 Discussion and Open Problems We have shown that any simple n-vertex polygon can be convexified in O(n2) single-vertex moves without ever decreasing the visibility graph, answering a question posed by Devadoss et al. . If coincident vertices are allowed, then n moves suffice, and each move strictly increases the visibility graph.
In the same paper, Devados et al. ask about trans-forming a polygon to decrease the visibility graph: can any simple polygon be transformed to a polygon whose visibility graph is a triangulation without ever increas-ing the visibility graph? This question remains open.
For orthogonal polygons, it would be desirable to maintain orthogonality. We conjecture than every sim-ple orthogonal polygon can be convexified (i.e., trans-formed to a rectangle) without losing visibilities, while maintaining orthogonality.
A minimal motion that maintains orthogonality is to move one edge orthogo-nal to itself (i.e., a horizontal edge moves vertically, and vice versa). However, Figure 8 shows an example where no edge can be moved orthogonally to gain visibilities.
It is possible that the current result can be generalized to straight line drawings of planar graphs: Given a pla-nar graph embedded in the plane as a straight-line draw-ing, is it possible to transform the drawing so that every internal face becomes convex, while remaining straight-line planar, and without losing internal visibilities? Our result is the special case where the drawing has only one internal face. The fact that such a transformation is possible, ignoring visibility constraints, is not at all obvious, but follows from the result by Thomassen , who showed (based on a result of Cairns ) that there is a transformation between any two straight-line planar drawings of the same embedded graph that preserves straight-line planarity. Vertices become coincident dur-ing this transformation, although that can be avoided by keeping them close but distinct. The number of vertex movements is not polynomially bounded. For further discussion on morphing of graph drawings, see [20, 21].
Finally, we make two remarks about our result on the existence of a visibility-increasing edge in any sim-ple polygon. Since good things (like ears of polygons) come in pairs, it is natural to ask whether every simple polygon has two visibility-increasing edges.
Visibility-increasing edges may have other uses in the study of visibility graphs.
A major open question is whether visibility graphs of polygons can be recognized in polynomial time (with or without the information about which edges form the polygon boundary). This is Problem 17 in the Open Problems Project .
Figure 8: An orthogonal polygon where no single edge can be moved orthogonally to gain visibilities.
23rd Canadian Conference on Computational Geometry, 2011 5 Acknowledgments This work was begun at the 26th Bellairs Workshop on Computational Geometry, co-organized by Erik De-maine and Godfried Toussaint. We thank the other par-ticipants of the workshop for stimulating discussions.
References B. M. ´ Abrego, M. Cetina, J. Lea˜ nos, and G. Salazar.
Visibility-preserving convexifications using single-vertex moves.
O. Aichholzer, M. Cetina, R. Fabila-Monroy, J. Leanos, G. Salazar, and J. Urrutia.
Convexifying monotone polygons maintaining internal visibility. Extended ab-stract, XIV Spanish Meeting on Computational Geom-etry, Alcal´ a de Henares, Spain, pages 35–38, 2011.
O. Aichholzer, C. Cort´ es, E. D. Demaine, V. Dujmovic, J. Erickson, H. Meijer, M. Overmars, B. Palop, S. Ra-maswami, and G. Toussaint. Flipturning polygons. Dis-crete & Computational Geometry, 28:231–253, 2002.
T. Biedl. Polygons Needing Many Flipturns. Discrete & Computational Geometry, 35:131–141, 2006.
P. Bose and F. Hurtado. Flips in planar graphs. Com-putational Geometry, 42:60–80, 2009.
S. S. Cairns.
Deformations of plane rectilinear com-plexes. Amer. Math. Monthly, 51(5):247–252, 1944.
R. Connelly, E. D. Demaine, and G. Rote. Straighten-ing polygonal arcs and convexifying polygonal cycles.
Discrete & Computational Geometry, 30:205–239, 2003.
B. de Sz.-Nagy. Solution of problem 3763. Amer. Math.
Monthly, 49:176–177, 1939.
E. D. Demaine, J. Mitchell, and J. O’Rourke. The open problems project. May 2010.
E. D. Demaine and J. O’Rourke.
Geometric Folding Algorithms: Linkages, Origami, Polyhedra. Cambridge University Press, 2007.
E. D. Demaine and J. O’Rourke. Open problems from CCCG 2008. In Proceedings of the 21st Canadian Con-ference on Computational Geometry (CCCG), pages 75–78, 2009.
S. L. Devadoss, R. Shah, X. Shao, and E. Winston. Vis-ibility graphs and deformations of associahedra. http: //arxiv.org/abs/0903.2848.
P. Erd˝ os. Problem 3763. Amer. Math. Monthly, 42:42, 1935.
S. Ghosh.
Visibility Algorithms in the Plane.
Cam-bridge University Press, 2007.
B. Gr¨ unbaum. How to convexify a polygon. Geocombi-natorics, 5:24–30, 1995.
R. A. Hearn and E. D. Demaine Games, puzzles, and computation. A. K. Peters, Ltd., 2009.
L. Guibas and J. Hershberger. Morphing simple poly-gons.
In Proceedings of the 10th Annual Symposium on Computational Geometry, SCG ’94, pages 267–276, New York, NY, USA, 1994. ACM.
H. N. Iben, J. F. O’Brien and E. D. Demaine. Refolding Planar Polygons.
In Proceedings of the 22nd Annual Symposium on Computational Geometry, pages 71–79, 2006. ACM.
T. Ito, E.
D. Demaine, N. J. A. Harvey, C. H. Pa-padimitriou, M. Sideri, R. Uehara and Y. Uno On the Complexity of Reconfiguration Problems. Theoretical Computer Science, in press.
A. Lubiw and M. Petrick.
Morphing planar graph drawings with bent edges. Electronic Notes in Discrete Mathematics, 31:45 – 48, 2008. The International Con-ference on Topological and Geometric Graph Theory.
A. Lubiw, M. Petrick, and M. Spriggs. Morphing or-thogonal planar graph drawings. In Proceedings of the 17th Annual ACM-SIAM Symposium on Discrete Algo-rithms, SODA ’06, pages 222–230, 2006. ACM.
J. O’Rourke. Art gallery theorems and algorithms. Ox-ford University Press, Inc., 1987.
I. Streinu. Pseudo-triangulations, rigidity and motion planning. Discrete & Computational Geometry, 34:587– 635, 2005.
C. Thomassen. Deformations of plane graphs. Jour-nal of Combinatorial Theory, Series B, 34(3):244 – 257, 1983.
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7367 | https://www.engineeringtoolbox.com/density-specific-weight-gravity-d_290.html | Engineering ToolBox
- Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
Density, Specific Weight, and Specific Gravity – Definitions & Calculator
The difference between density, specific weight, and specific gravity. Including formulas, definitions, and reference values for common substances.
Density
Density is defined as mass per unit volume . Mass is a property and the SI unit for density is kg/m3.
Density can be expressed as
ρ = m / V
= 1 / ν (1)
ρ = density (kg/m3), (slugs/ft3)
m = mass (kg), (slugs)
V = volume (m3), (ft3)
ν = specific volume (m3/kg), (ft3/slug)
What is weight and what is mass? - the difference between weight and mass
The Imperial (U.S.) units for density are slugs/ft3 but pound-mass per cubic foot - lbm/ft3 - is often used. Note that there is a difference between pound-force (lbf) and pound-mass (lbm) . Slugs can be multiplied with 32.2 for a rough value in pound-mass (lbm).
1 slug = 32.174 lbm = 14.594 kg
1 kg = 2.2046 lbm = 6.8521×10-2 slugs
density of water: 1000 kg/m3 = 1.938 slugs/ft3
See also Unit converter - mass and Unit converter - density
On atomic level - particles are packed tighter inside a substance with higher density. Density is a physical property - constant at a given temperature and pressure - and may be helpful for identification of substances.
Below on this page: Specific gravity (relative density), Specific gravity for gases, Specific weight, Calculation examples
Se also:
Densities for some common materials
Water - Density, Specific Weight and Thermal Expantion Coefficient - variation with temperature at 1, 68 and 680 atm, SI and Imperial units
Air - Density, Specific Weight and Thermal Expantion Coefficient - variation with temperature and pressure, SI and Imperial units
How to measure density of liquid petroleum products
Specific Gravity
Specific Gravity (Relative Density) - SG - is a dimensionless unit defined as the ratio of the density of a substance to the density of water - at a specified temperature and can be expressed as
SG = ρsubstance / ρH2O (2)
where
SG = Specific Gravity of the substance
ρsubstance = density of the fluid or substance (kg/m3)
ρH2O = density of water - normally at temperature 4 oC (kg/m3)
It is common to use the density of water at 4 oC (39 oF) as a reference since water at this point has its highest density of 1000 kg/m3 or 1.940 slugs/ft3 .
Since Specific Gravity - SG - is dimensionless, it has the same value in the SI system and the imperial English system (BG). SG of a fluid has the same numerical value as its density expressed in g/mL or Mg/m3. Water is normally also used as reference when calculating the specific gravity for solids.
See also Thermophysical Properties of Water - Density, Freezing temperature, Boiling temperature, Latent heat of melting, Latent heat of evaporation, Critical temperature ...
Specific Gravity for some common Materials
Specific Gravities common Substances
| | |
| Acetylene | 0.0017 |
| Air, dry | 0.0013 |
| Alcohol | 0.82 |
| Aluminum | 2.72 |
| Brass | 8.48 |
| Cadmium | 8.57 |
| Chromium | 7.03 |
| Copper | 8.79 |
| Carbon dioxide | 0.00198 |
| Carbon monoxide | 0.00126 |
| Cast iron | 7.20 |
| Hydrogen | 0.00009 |
| Lead | 11.35 |
| Mercury | 13.59 |
| Nickel | 8.73 |
| Nitrogen | 0.00125 |
| Nylon | 1.12 |
| Oxygen | 0.00143 |
| Paraffin | 0.80 |
| Petrol | 0.72 |
| PVC | 1.36 |
| Rubber | 0.96 |
| Steel | 7.82 |
| Tin | 7.28 |
| Zinc | 7.12 |
| Water (4 oC) | 1.00 |
| Water, sea | 1.027 |
| Wood, Oak | 0.77 |
Specific Gravity of gases is normally calculated with reference to air - and defined as the ratio of the density of the gas to the density of the air - at a specified temperature and pressure.
The Specific Gravity can be calculated as
SG = ρgas / ρair (3)
where
SG = specific gravity of gas
ρgas = density of gas (kg/m3)
ρair = density of air (normally at NTP - 1.204 (kg/m3))
NTP - Normal Temperature and Pressure - defined as 20 oC (293.15 K, 68 oF) and 1 atm (101.325 kN/m2, 101.325 kPa, 14.7 psia, 0 psig, 30 in Hg, 760 torr)
Molecular weights can be used to calculate Specific Gravity if the densities of the gas and the air are evaluated at the same pressure and temperature.
See also Thermophysical Properties of Air - density, viscosity, critical temperature and pressure, triple point, enthalpi and entropi, thermal conductivity and diffusicity,......
Specific Weight
Specific Weight is defined as weight per unit volume . Weight is a force . The SI unit for specific weight is (N/m3). The imperial unit is (lb/ft3).
Specific Weight (or force per unit volume) can be expressed as
γ = ρ ag (4)
where
γ = specific weight (N/m3), (lb/ft3)
ρ = density (kg/m3), (slugs/ft3)
ag = acceleration of gravity (9.807 m/s2, 32.174 ft/s2 under normal conditions)
What is weight and what is mass? - the difference between weight and mass
Specific Weight for Some common Materials
Specific Weight common Materials
| Product | Specific Weight - γ - | |
| Imperial Units (lb/ft3) | SI Units (kN/m3) |
| Aluminum | 172 | 27 |
| Brass | 540 | 84.5 |
| Carbon tetrachloride | 99.4 | 15.6 |
| Copper | 570 | 89 |
| Ethyl Alcohol | 49.3 | 7.74 |
| Gasoline | 42.5 | 6.67 |
| Glycerin | 78.6 | 12.4 |
| Kerosene | 50 | 7.9 |
| Mercury | 847 | 133.7 |
| SAE 20 Motor Oil | 57 | 8.95 |
| Seawater | 63.9 | 10.03 |
| Stainless Steel | 499 - 512 | 78 - 80 |
| Water | 62.4 | 9.81 |
| Wrought Iron | 474 - 499 | 74 - 78 |
Material Properties
Example 1: Density of a Golf ball
A golf ball has a diameter of 42 mm and a mass of 45 g. The volume of the golf ball can be calculated as
V = (4/3) π ((42 mm) (0.001 m/mm) / 2)3
= 3.8×10-5 m3
The density of the golf ball can then be calculated as
ρ = (45 g) (0.001 kg/g) / (3.8×10-5 m3)
= 1184 kg/m3
Example 2: Using Density to Identify a Material
An unknown liquid substance has a mass of 18.5 g and occupies a volume of 23.4 ml (milliliter).
The density of the substance can be calculated as
ρ = ((18.5 g) / (1000 g/kg)) / ((23.4 ml) / ((1000 ml/l) (1000 l/m3)))
= (18.5×10-3 kg) / (23.4×10-6 m3)
= 790 kg/m3
If we look up the densities of some common liquids we find that ethyl alcohol - or ethanol - has a density of 789 kg/m3. The liquid may be ethyl alcohol!
Example 3: Density to Calculate Volume Mass
The density of titanium is 4507 kg/m3. The mass of 0.17 m3 volume titanium can be calculated as
m = (0.17 m3) (4507 kg/m3)
= 766.2 kg
Note! - be aware that there is a difference between " bulk density" and actual "solid or material density". This may not be clear in the description of products. Always double check values with other sources before important calculations.
Example 4: Specific Gravity of Iron
The density of iron is 7850 kg/m3. The specific gravity of iron related to water with density 1000 kg/m3 is
SGiron = (7850 kg/m3) / (1000 kg/m3)
= 7.85
Example 5: Specific Weight of Water
The density of water is 1000 kg/m3 at 4 °C (39 °F).
The specific weight in SI units is
γ = (1000 kg/m3) (9.81 m/s2)
= 9810 N/m3
= 9.81 kN/m3
The density of water is 1.940 slugs/ft3 at 39 °F (4 °C).
The specific weight in Imperial units is
γ = (1.940 slugs/ft3) (32.174 ft/s2)
= 62.4 lb/ft3
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From Wikipedia, the free encyclopedia
Academic title
| | |
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| | This article may need to be rewritten to comply with Wikipedia's quality standards. You can help. The talk page may contain suggestions. (November 2018) |
An adjunct professor is a type of academic appointment in higher education who does not work at the establishment full-time. The terms of this appointment and the job security of the tenure vary in different parts of the world, but the term is generally agreed to mean a bona-fide part-time faculty member in an adjunct position at an institution of higher education.
Terminology
[edit]
An adjunct professor may also be called an adjunct lecturer, an adjunct instructor, or adjunct faculty. Collectively, they may be referred to as contingent academic labor. The rank of sessional lecturer in Canadian universities is similar to the US concept.
Americas
[edit]
Main article: Adjunct professors in North America
In the United States, an adjunct is, in most cases, a non-tenure-track faculty member. However, it can also be a scholar or teacher whose primary employer is not the school or department with which they have adjunct status.
Adjunct professors make up the majority of instructors in higher education (post-secondary) institutions. As with other part-time workers, they are paid less than full-time professors and do not receive employee benefits such as health insurance or an office.
In most cases, adjunct professors need a master's degree, but in some cases only require a bachelor's degree and relevant experience. However, over a third have a doctoral degree. In many universities, the title "adjunct professor" (or variations thereof, such as "adjunct associate professor") implies a PhD or other terminal degree; those with a master's or bachelor's degree may receive the title of "adjunct lecturer". In 2018 the American Association of University Professors (AAUP) expressed concern that only a quarter of university positions are tenure-track, with implications for job security and academic freedom. The AAUP analysis determined that 73% of university teaching positions in the United States are non-tenure track.
In Canada, adjunct professors are often nominated in recognition of active involvement with the appointing institution. At the same time, they are employed by the government, industry, a profession, or another institution. The term "course lecturer," rather than "adjunct," is used if the appointment is strictly to teach one or more courses. In contrast, the US uses this title for all instructors.
In Argentina and Brazil, the designation professor adjunto implies stable employment.
Europe
[edit]
In Portugal, the designation professor adjunto implies stable full-time employment in a polytechnic university.
In parts of Spain, the term professor adjunto refers to a non-tenured position.
In Hungary, there exists a similar term adjunktus, as well as adiunkt in Poland, although only the term is similar, as adjunktus in Hungarian means full-time employed assistant professor, not a bona fide lecturer.
In Finland, the Docents' Union of Finland recommend the term adjunct professor or associate professor in English as a translation of the title of docent. However, the official translation used by the universities granting the title is "Title of Docent". A docentship should be regarded as an educational title not connected with the employment rank as such, rather an assurance of the level of expertise, to enable the person to advance further in their academic career. The rank of a docent entitles scientists to be principal investigators, lead research groups, and act as the supervisors of doctoral students.
Some universities in The Netherlands have adjunct professors, where the title applies to the highest ranking variant of associate professor, thus having quite a distinct interpretation from the American use of the term.
In Italy, the term adjunct professor is used to translate the title of Professore a contratto.
In France, the term adjunct professor refers to “Chargé de cours”.
In Germany, the term adjunct professor translates to the title of außerplanmäßiger Professor and is abbreviated apl. Prof.
South Asia
[edit]
In Bangladesh, private universities follow the title adjunct professor or adjunct associate professor to imply non-tenure faculty members.
In Pakistan, adjunct (assistant/associate) professors are also considered as non-regular faculty members, and usually, posts are given to Pakistani overseas scientists under a faculty development program.
Southeast Asia and Oceania
[edit]
In Australia, the term adjunct is reserved for academics and researchers from outside the university who have a close association with the university, e.g., through supervision of PhD students, recognized by an honorary title reflective of their rank and standing (adjunct lecturer, senior lecturer, associate professor or professor).
In Thailand, adjunct (assistant/associate) professors are considered "non-regular officers".
References
[edit]
^ "What is an Adjunct Professor? Job Description & Salary". 1 September 2015. Archived from the original on 18 June 2018. Retrieved 16 April 2018.
^ Hall, Lee (22 June 2015). "I am an adjunct professor who teaches five classes. I earn less than a pet-sitter - Lee Hall". the Guardian.
^ "Classification of Ranks and Titles » Faculty Handbook - Boston University". Boston University.
^ "What is an Adjunct Professor? Job Description & Salary | Resilient Educator". September 2015.
^ "How to Become an Adjunct Professor: Job, Education, Salary". Archived from the original on 2018-06-18. Retrieved 2018-04-16.
^ Colleen Flaherty, 'New Data on Adjunct Instructors,' Inside Higher Ed, November 2, 2018]
^ "About three-quarters of all faculty positions are off the tenure track, according to a new AAUP analysis".
^ "About three-quarters of all faculty positions are off the tenure track, according to a new AAUP analysis". www.insidehighered.com. Retrieved 2021-01-07.
^ Davies, Gwen (2001-11-05). "Policy and Procedures for Appointment Of Adjunct Professors" (PDF). Retrieved 2022-04-13.
^ "Finnish Academic Job Titles Explained". Academic Positions.com.
^ Australia, The University of Western. "Adjunct and clinical titles". www.hr.uwa.edu.au.
^ "Policies and Procedures Library - The University of Queensland, Australia". ppl.app.uq.edu.au.
^ "Adjunct Appointments". www.jcu.edu.au. Archived from the original on 2021-05-10. Retrieved 2018-04-16.
^ www.tweek.com.au, (TWEEK!). "SCU Policy Library - Adjunct, Visiting and Conjoint Appointments Policy". policies.scu.edu.au.
| Academic ranks overview | |
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| Overview | Faculty Ranks and titles Professorship Tenure track |
| North American system | Tenure track Assistant professor Associate professor Professor (full) Endowed chair and Distinguished professor Non-tenure track Adjunct professor Instructor Lecturer |
| Commonwealth system | Research and teaching track Associate lecturer Lecturer Senior lecturer Reader or associate professor Professor Research track Research assistant Research fellow Senior research fellow Professor or research professor |
| Junior positions | Teaching track Teaching assistant Teaching fellow Associate instructor Research track Research assistant Research associate Research fellow Postdoctoral researcher |
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7369 | https://www.reddit.com/r/askmath/comments/180v9wv/using_the_cauchyschwarz_inequality_for/ | Using the Cauchy-Schwarz inequality for minimization/maximization (specific problem, plus general questions) : r/askmath
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•2 yr. ago
bobotheking
Using the Cauchy-Schwarz inequality for minimization/maximization (specific problem, plus general questions)
Linear Algebra
There is a special digression within David Poole's outstanding (by my assessment) textbook, Linear Algebra: A Modern Introduction. I'm using the second edition and the topic begins on page 556, with the example problem I'm interested in on page 560. This isn't homework; I'm just studying it because I think it's neat.
Poole points out that certain optimization problems can be solved without using calculus. You express the quantity to be extremized as an inner product (or product of vector moduli), then apply the Cauchy-Schwarz inequality, || ≤ |A||B|. (It's often helpful in this context to square both sides of the inequality). Equality is satisfied when the vectors are scalar multiples of one another and so depending which side of the inequality your expression resides, this proves its minimum or maximum value. One nifty feature of this approach is that if your function is f(x, y), you first obtain the maximum/minimum value of this function. This is in contrast to calculus, where setting the derivatives to zero first gets you the x and y values, which then need to be plugged in to find the value of the function. Furthermore, constraints can be handled pretty easily if they can be expressed in terms of the vectors you introduce.
Anyway, there are fifteen problems in the digression and I basically waltzed through fourteen of them. Several of the problems are some restatement of inequalities involving various means: quadratic mean ≥ arithmetic mean ≥ geometric mean ≥ harmonic mean (explicitly in problem 12, but implicitly throughout the exercises). Other problems include:
Prove that among all rectangles with area 100 square units, the square has the smallest perimeter.
What is the minimum value of f(x) = x + 1/x for x>0?
Find the minimum value of f(x, y, z) = (x+y)(y+z)(z+x) if x, y, and z are positive real numbers such that xyz=1.
Find the maximum value of f(x,y,z) = x + 2y + 4z subject to x 2 + 2y 2 + z 2 = 1.
Find the minimum value of f(x,y,z) = x 2 + y 2 + 1/2z 2 subject to x+y+z=10. [Note how the role of the linear and vector relations has been swapped in comparison to the previous problem and correspondingly, we seek a minimum here as opposed to a maximum in problem 8.]
Etc., but this gives you an idea of the types of problems this technique is suited for, plus you can practice it if you're not overly familiar with the procedure.
Problem 15 is a different beast, however:
15. Let x and y be positive real numbers with x+y=1. Show that the minimum value of f(x,y) = (x + 1/x)2+ (y + 1/y)2is 25/2, and determine the values of x and y for which it occurs.
I must have spent hours on this problem throwing everything I could at it. Because the constraint is linear and the function is quadratic and we're minimizing the function, it's apparent that we want our inner product to involve the constraint while f(x,y) should work its way in on the right hand side. Unfortunately, the constraint and quadratic functions don't really "communicate well" with each other and I can't come up with an inequality that involves both expressions.
For example, let's try a fairly straightforward approach: A = [x+1/x, y+1/y], B = [1, 1]. Applying the Cauchy-Schwarz inequality (squared):
2 ≤ |A|2|B|2
(x+1/x + y+1/y)2 ≤ 2[(x+1/x)2 + (y+1/y)2]
(x+1/x)2 + (y+1/y)2 + 2(x+1/x)(y+1/y) ≤ 2[(x+1/x)2 + (y+1/y)2]
2xy + 2x/y + 2y/x + 2/xy ≤ (x+1/x)2 + (y+1/y)2
From here, we can rewrite the constraint by dividing through by xy as 1/x + 1/y = 1/xy, moving us maybe marginally forward:
2xy + 2x/y + 2y/x + 2/x + 2/y ≤ (x+1/x)2 + (y+1/y)2
and at this point I'm basically stuck. I've tried three or four other definitions of vector functions to no avail, but can't help but feel I'm overlooking something obvious. Does anyone have any ideas on how to solve this? Just giving me suitable definitions for our vectors should be enough to point me in the right direction.
Also, I'm quite interested in the general theory of what's going on here. I toyed around with more generalized versions, such as A = [x p y q, x q y p], B = [x r y s, x s y r], but I don't recall that leading anywhere interesting. Is the utility of this approach limited to things that "look like" various means and/or ellipsoids? How does one quickly determine whether the Cauchy-Schwarz inequality can be applied in this way and furthermore, how do you then determine appropriate vector functions such that you won't spend 10 minutes going down a dead end path? Are there perhaps some resources that delve into the general theory?
Thanks in advance!
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How to Convert Seconds to Hours Quickly with Examples
We all have seen three hands on a clock face. The minute hand indicates the time in minutes, the hour hand indicates the time in hours, and the second hand shows the time in seconds. There are 12 central divisions on the clock, each lasting one hour, and five minor divisions spaced equally between them.
Five minutes are passed when the minute hand transitions from one large division to the subsequent large division. Each of the 12 significant divisions lasts five minutes, making the equation for an hour: 12 x 5 = 60 minutes.
Time is the eternal progression of past, present, and future events, indicating that time comprises every series of events that have occurred or will occur in the future. It is measured in minutes, seconds, and hours. The clock is used to measure time more precisely and accurately.
What Is a Clock?
The clock is a device for measuring the time that displays the passage of hours, minutes, and seconds.
Clock
The units used to measure time are the hour, minute, and second. The minute hand is the largest and fastest hand on the clock, followed by the second hand and the Hour hand.
Time
Time is defined as the duration of all events or the specific instant that something occurs. Time includes all events that have occurred or will occur in the future.
Time is written as- Hour: Minutes: Seconds.
Example: 02: 48:28
Time keeper
Two Different Types of Clocks
Clock with a 12-hour format: As the name implies, this clock displays time from 1 to 12. These clocks lack digital displays, and the wall clocks are an example.
Clock in 24-hour format: This clock displays the time from 1 to 24 as its name says. These clocks are digital, and some wrist-watches, railway’s digital clocks, etc. are examples of such clock format.
In a 12-hour clock, the entire 24-hour period is divided into two segments, from midnight to noon and from midday to midnight. Ante Meridiem, or before noon, is what "A.M." stands for. As a result, we add the A.M. suffix to the time from midnight to noon. For instance, we might say that it is 9 am. Post Meridiem, or "afternoon," is what "P.M." stands for. As a result, we add the P.M. suffix to the end of the time from noon to midnight, for instance, 4 p.m.
Hour, Minute, and Seconds Hands
Seconds to Hours Formula
To convert from seconds to hours, divide the number of seconds by 3600.
For instance:
Convert 60 seconds in an hour.
As we know the seconds to hours formula, we need to divide the given time by 3600. Since, here we have 60 seconds, 60 divided by 3600 will give us 0.016 hours.
Conversions:
1 Hour = 60 Minutes
1 Minute =$\dfrac{1}{60}$ Hour
1 Minute = 60 Seconds
1 Second = $\dfrac{1}{60}$ Minutes
How to Convert Seconds to Minutes
To change seconds to minutes, multiply the digit by 60. Simply reverse the process of converting minutes to seconds if you start from seconds and need to convert to minutes. To change secs to mins, divide by 60.
For instance:
30 secs divided by 60 mins equals 0.5 mins.
One minute and 30 seconds are identical to 90 seconds.
How to Convert Minutes to Seconds
Keep in mind that one minute consists of 60 seconds. Remember the magic number: 60 seconds for every minute before you begin your conversion. It will be much simpler to convert if you keep this in mind.
For instance:
Change 5 min 37 sec into seconds.
Solution: 1 min = 60 sec
Thus, 5 min = 5 x 60 = 300 sec.
5 min 37 sec = 300 sec + 37 sec = 337 seconds
Solved Examples
Example 1: Convert 270 minutes into seconds.
Ans: We know that one hour is equal to 60 minutes.
Additionally, we know that one minute equals 60 seconds.
Thus, 270 minutes equals 270 x 60, or 16,200 seconds.
Example 2: Convert 4800 seconds into hours.
Ans: First, we need to convert seconds into minutes and then minutes into hours.
$= 4800 \text{ seconds }= \left(4800 \div 60)\text{ minutes }(1\text{ second }=\dfrac{1}{60}\text{ minute}\right)$
$= 80$minutes
$= 80 \div 60$ hours $\left(\text{ minute }=\dfrac{1}{60}\text{ hour}\right)$
= 1 hour 20 minutes
Example 3: Convert 36000 seconds into hours.
Ans: First, we have to convert seconds into minutes and then minutes into hours.
$= 36000 \text{ seconds }= \left(3600 \div 60)\text{ minutes }(1\text{ second }=\dfrac{1}{60}\text{ minute}\right)$
$= 600$minutes
$= 600 \div 60$ hours $\left(\text{ minute }=\dfrac{1}{60}\text{ hour}\right)$
= 10 hours
Hands of a Clock
Try on Your Own
Q1. How many minutes are in 600 seconds?
Ans: 10 minutes are in 600 seconds
Q2. Convert 7300 seconds to minutes to hours.
Ans: 7,300 Seconds = 2.0277 Hours = 2 Hours, 1 Minute and 39 Seconds.
Summary
Time is the continued ongoing growth of past, present, and future scenes. It simply means every sequence of events that has happened, which is occurring or will happen, is created by time. It is estimated in hours, minutes, and seconds. The device used to measure time is known as a clock. The hour hand is the smallest and slowest in the watch, then comes the second hand, and the largest is the minute hand.
FAQs on Convert Seconds to Hours – Step-by-Step Guide
What is the equation to convert seconds to hours?
To convert seconds to hours, use the equation:$$ \text{Hours} = \frac{\text{Seconds}}{3600} $$This is because 1 hour = 3600 seconds. For example, to convert 7200 seconds to hours:$$ \text{Hours} = \frac{7200}{3600} = 2 \text{ hours} $$Vedantu provides interactive sessions to help you understand and practice such conversions easily in mathematics and science subjects at all academic levels.
What is the equation for converting hours to seconds?
To convert hours to seconds, multiply the number of hours by 3600. The equation is:$$ \text{Seconds} = \text{Hours} \times 3600 $$For example, converting 3 hours to seconds:$$ \text{Seconds} = 3 \times 3600 = 10,800 \text{ seconds} $$Vedantu's expert tutors can guide you through more such unit conversions and help improve your problem-solving skills.
How to convert 1 s to 1 h?
To convert 1 second to hours, divide by 3600 since there are 3600 seconds in an hour:$$ 1 \text{ s} = \frac{1}{3600} \text{ h} \approx 0.00027778 \text{ h} $$This small value reflects how many hours just one second makes up. Vedantu provides step-by-step guidance for such math and science conversions, helping students build conceptual understanding.
Is 7200 seconds 2 hours?
Yes, 7200 seconds is equal to 2 hours. You can check this by dividing 7200 by the number of seconds in an hour (3600):$$ \frac{7200}{3600} = 2 \text{ hours} $$Vedantu’s study materials make it easy to learn and practice time conversions and other foundational math concepts.
How do you convert minutes and seconds into hours?
To convert minutes and seconds into hours:
Convert minutes to seconds: $\text{Minutes} \times 60$
Add all seconds together
Divide the total by 3600
Formula:$$ \text{Hours} = \frac{(\text{Minutes} \times 60) + \text{Seconds}}{3600} $$Vedantu’s interactive sessions include numerous examples and practice problems that help reinforce these conversion skills.
What is an easy trick to remember seconds to hours conversion?
A simple trick to remember seconds to hours conversion is:
There are 3600 seconds in 1 hour.
To find hours, divide the number of seconds by 3600.
For quick estimation, you can also remember:3600 seconds = 1 hour, 1800 seconds = 0.5 hour, and 7200 seconds = 2 hours. Vedantu teachers use similar memory aids to make unit conversions effortless for students.
How can you convert decimal hours to seconds?
To convert decimal hours to seconds, multiply the decimal hour value by 3600:$$ \text{Seconds} = \text{Decimal\ Hours} \times 3600 $$For example, 1.5 hours would be:$$ 1.5 \times 3600 = 5400 \text{ seconds} $$Vedantu’s curriculum covers both decimal and fractional time conversions for comprehensive learning.
Why is it important to understand time conversions in math and science?
Understanding time conversions is essential in math and science because:
Many calculations in physics, chemistry, and everyday life require conversion between units of time.
Accurate conversions are necessary for solving problems related to speed, rate, and scheduling.
It enhances logical thinking and foundation for advanced topics.
Vedantu integrates such real-world applications into lessons for better comprehension and exam readiness.
Can Vedantu help with practice problems on converting seconds to hours?
Absolutely! Vedantu offers a wide variety of practice problems and interactive quizzes that focus on converting seconds to hours and other unit conversions. These resources help students master the concept through guided examples and immediate feedback, building confidence and proficiency in mathematics.
What are some real-life examples where converting seconds to hours is used?
Converting seconds to hours is useful in various scenarios, such as:
Calculating total travel time for journeys measured in seconds.
Understanding durations in scientific experiments or data logging.
Time tracking for exercise, sports, or project tasks.
Vedantu’s teachers often incorporate such real-life applications in lessons to make learning meaningful and practical for students.
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1 Construction of Right Triangles
2 Using Right Triangles to find Unknown SidesToggle Using Right Triangles to find Unknown Sides subsection
2.1 Finding unknown sides from two other sides
2.2 Finding unknown sides from one side and one (non-right) angle
2.3 Cosine example
3 Properties of the cosine and sine functionsToggle Properties of the cosine and sine functions subsection
3.1 Period
3.2 Half Angle and Double Angle Formulas
Trigonometry/Thales Theorem
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Construction of Right Triangles
[edit | edit source]
Provided AC is a diameter, angle B is always a constant right angle
Right triangles are easily constructed. Recall that a diameter is a straight line which starts at one point on the circle and goes through the center to the other side. Using this property of a circle:
Construct a diameter of a circle. Call the points where the diameter touches the circle a and c.
Choose a distinct point b on the circle (any point that is neither a nor c).
Construct line segments connecting points a, b and c.
Provided the above three directions are followed, the resulting triangle Δ abc will be a right triangle. This result is known as Thales' theorem.
This right triangle can be further divided into two isosceles triangles by adding a line segment from b to the center of the circle.
To simplify the following discussion, we specify that the circle has diameter 1 and is oriented such that the diameter drawn above runs left to right. We shall denote the angle of the triangle at the right θ and that at the left φ. The sides of the right triangle will be labeled, starting on the right, a, b, c, so that a is the rightmost side, c the leftmost, and b the diameter. We know from earlier that side b is opposite the right angle and is called the hypotenuse. Side a is opposite angle φ, while side c is opposite angle θ. We reiterate that the diameter, now called b, shall be assumed to have length 1 except where otherwise noted.
By constructing a right angle to the diameter at one of the points where it crosses the circle and then using the method outlined earlier for producing binary fractions of the right angle, we can construct one of the angles, say θ, as an angle of known measure between 0 and π / 2. The measure of the other angle φ is then π - π/2 - θ = π/2 - θ, the complement of angle θ. Likewise, we can bisect the diameter of the circle to produce lengths which are binary fractions of the length of the diameter. Using a compass, a binary fractional length of the diameter can be used to construct side a (or c) having a known size (with regard to the diameter b) from which side c can be constructed.
Using Right Triangles to find Unknown Sides
[edit | edit source]
Finding unknown sides from two other sides
[edit | edit source]
From the Pythagorean Theorem, we know that:
a 2+b 2=c 2.{\displaystyle a^{2}+b^{2}=c^{2}.\,}
Recall that c (the diameter/hypotenuse in the above example) has been defined as having a length of one, therefore:
b 2=1−a 2.{\displaystyle b^{2}=1-a^{2}.\,}
permits us to calculate the length of b squared. It may happen that b squared is a fraction such as 1/4 for which a rational square root can be found, in this case as 1/2, alternatively, we could use Newton's Method to find an approximate value for b.
Finding unknown sides from one side and one (non-right) angle
[edit | edit source]
As any triangle could be compared to our basic triangle (formed from a circle with a diameter of one), a table enumerating the relationships between angles and side lengths would be very useful to understanding the properties of any triangle. However, such a table would be unwieldy in practice and it is often not necessary to know the exact value.
Of course, given an angle θ{\displaystyle \theta }, we could construct a right angled triangle using ruler and compass that had θ{\displaystyle \theta \,} as one of its angles, we could then measure the length of the side that corresponds to a to evaluate the cos{\displaystyle \cos } function. Such measurements would necessarily be inexact; it would be a problem in physics to see how accurately such measurements can be made; using trigonometry we can make precise predictions with which the results of these physical measurements can be compared.
The most common way of communicating the idea that relationships exist without providing exact details takes the form of a 'function'. A function is like a machine that takes some simple input and produces some simple output. Usually a function defines some kind of rule ([Function]) and provides us a handy notation useful in trigonometry. That way, we know that we're using a certain relationship without needing to know the exact numerical values. Basic trigonometric functions are simply stand-ins for the relationship between angles and sides of a triangle.
One such function, which allows us to know the relationship between any value of θ{\displaystyle \theta \,} and the corresponding value of a{\displaystyle a\,} is called the cosine or cos{\displaystyle \cos \,}. This universal relationship is represented as: a c=cosθ{\displaystyle {\frac {a}{c}}=\cos \theta } which would save us the work of constructing angles and lengths and making difficult deductions from them.
This means that if you know the cosine of an angle, you also know the relationship between the lengths of the sides. The actual size of the triangle can be bigger or smaller, but the mathematical relationship represented by the cosine does not change so long as the size of the angle remains constant.
Cosine example
[edit | edit source]
Some explicit values for the cos function are known. For θ=0{\displaystyle \theta \,=0}, sides a{\displaystyle a\,} and c{\displaystyle c\,} coincide: a=c=a c=1{\displaystyle a=c={\frac {a}{c}}=1}, so cos(0)=1{\displaystyle \cos \left(0\right)=1}. For θ=π 2{\displaystyle \theta ={\frac {\pi }{2}}}, sides b{\displaystyle b\,} and c{\displaystyle c\,} coincide and are of length 1, and side a{\displaystyle a\,} is of zero length, consequently a=0{\displaystyle a\,=0}, c=1{\displaystyle c\,=1}. a c=0{\displaystyle {\frac {a}{c}}=0} and cosπ 2=0{\displaystyle \cos {\frac {\pi }{2}}=0}.
The simplest right angle triangle we can draw is the isosceles right angled triangle, it has a pair of angles of size π 4{\displaystyle {\frac {\pi }{4}}} radians, and if its hypotenuse is considered to be of length one, then the sides a{\displaystyle a\,} and b{\displaystyle b} are of length 1 2{\displaystyle {\sqrt {\frac {1}{2}}}} as can be verified by the theorem of Pythagoras. If the side a{\displaystyle a\,} is chosen to be the same length as the radius of the circle containing the right angled triangle, then the right hand isosceles triangle obtained by splitting the right angled triangle from the circumference of the circle to its center is an equilateral triangle, so θ{\displaystyle \theta } must be π 3{\displaystyle {\frac {\pi }{3}}}, and ϕ{\displaystyle \phi } must be π 6{\displaystyle {\frac {\pi }{6}}} and b 2{\displaystyle b\,^{2}} must be 1−1 2⋅1 2=3 4{\displaystyle 1-{\frac {1}{2}}\cdot {\frac {1}{2}}={\frac {3}{4}}}.
Properties of the cosine and sine functions
[edit | edit source]
Period
[edit | edit source]
A full revolution is an angle of 2π radians, so increasing an angle by this amount gets you exactly back to your starting point. Therefore, a perfect circle is maintained, along with the relationships formed by triangles within, by adding 2π to any angle θ. This is called the period, --the size of the angle or the time period over which the relationships begin to repeat (correlating the two is complex, and allows us to talk about wave theory).
Using functions, we can represent this fact in terms of the cosine function by stating that: cosθ=cos(θ+2 π n),{\displaystyle \cos \theta =\cos \left(\theta +2\pi n\right),\,}
Knowing the period of the sine and cosine functions (and by derivation, that of other functions) is useful because it means that we can substitute one angle for another when we know that the period is the same. This helps in calculations, such as when there is the need to add or subtract angles.
Half Angle and Double Angle Formulas
[edit | edit source]
We can derive a formula for cos(θ/2) in terms of cos(θ) which allows us to find the value of the cos() function for many more angles. To derive the formula, draw an isosceles triangle, draw a circle through its corners, connect the center of the circle with radii to each corner of the isoceles triangle, extend the radius through the apex of the isoceles triangle into a diameter of the circle and connect the point where the diameter crosses the other side of the circle with lines to the other corners of the isoceles triangle.
Therefore:
cos(θ/2)=1+cosθ 2{\displaystyle \cos(\theta /2)={\sqrt {\frac {1+\cos \theta }{2}}}}
which gives a method of calculating the cos() of half an angle in terms of the cos() of the original angle. For this reason is called the "Cosine Half Angle Formula".
The half angle formula can be applied to split the newly discovered angle which in turn can be split again ad infinitum. Of course, each new split involves finding the square root of a term with a square root, so this cannot be recommended as an effective procedure for computing values of the cos() function.
Equation can be inverted to find cos(θ) in terms of cos(θ)/2:
cos(θ/2)=1+cos(θ)2⇒cos 2(θ/2)=1+cos(θ)2⇒cos 2(θ/2)−1 2=cos(θ)2⇒2 cos 2(θ/2)−1=cos(θ){\displaystyle {\begin{matrix}&\cos(\theta /2)={\sqrt {\frac {1+\cos(\theta )}{2}}}\\Rightarrow &\cos ^{2}(\theta /2)={\frac {1+\cos(\theta )}{2}}\\Rightarrow &\cos ^{2}(\theta /2)-{\frac {1}{2}}={\frac {\cos(\theta )}{2}}\\Rightarrow &2\cos ^{2}(\theta /2)-1=\cos(\theta )\\end{matrix}}}
substituting δ=θ 2{\displaystyle \delta \ ={\frac {\theta }{2}}} gives:
2 cos 2(δ)−1=cos(2 δ){\displaystyle 2\cos ^{2}(\delta )-1=\cos(2\delta )\,}
that is a formula for the cos() of double an angle in terms of the cos() of the original angle, and is called the "cosine double angle sum formula".
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7374 | https://blogs.ams.org/mathgradblog/2017/04/08/ordered-fields-order/ | Ordered Fields and When You Can’t Order Them |
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Ordered Fields and When You Can’t Order Them
Posted onApril 8, 2017byTom Gannon
The real numbers have an ordering on them–given two numbers and , we can tell whether or .So as math people, we like to generalize this to other sets–when can we say that a general set is ordered? In this post, we’re going to explain the explicit of idea of what it means for a field to be ordered, and then show that the complex numbers can’t be ordered–no matter what ordering you put on it. (If you don’t know what a field is, just think of the real numbers or the complex numbers .)
In the real numbers, we say that exactly when , that is, is positive. So since we’re trying to generalize the idea of “ordering”, one way is to do that is to figure out how to generalize the idea of “positive” numbers. So (and I’m being a little loose here), let’s say we only have the idea of “adding”, which we’ll denote with +, and multiplication, which we’ll denote by “”. This is sort of what it means to be a “general field” (You might want to think about which properties of positive numbers can be defined only with +, , or if this is too vague, just keep reading.)
Here are the two that are most important about the positive numbers–one is that if you add two positive numbers, you get a positive number again. The other is that if you multiply two positive numbers, you get a positive number again. These properties say that whatever set of positive numbers we have, it must be closed under addition and closed under multiplication resepectively.
Oh–and one other point. If we pick a generic number , we want to say that either is positive, negative, or zero. Also, has to be exactly one of them (so 10 can’t also be negative, for example). It turns out that’s all we need to make our definition of an ordered field:
Definition:We say that a field is an ordered field if it has a set (of “positive numbers”) such that:
( is closed under addition) If we have two elements and , then their sum is also in , that is, .
( is closed under multiplication) If we have two elements and , then their product is also in , that is, .
(All nonzero numbers are positive or negative) For all in our field, exactly one of the following holds: or .
Now we’ll show something pretty cool.
Proposition:The complex numbers is not an ordered field.
Proof: To show this, we’re going to use a method called proof by contradiction. We’re essentially going to show that ifwas an ordered field, something bad will happen. So let’s assumewas an ordered field and see if we can find anything weird happening.
Well one special element in that’s not in the real numbers is , where . So since , either or is positive, according to (3) above.
If was positive, then is a positive number, by (2). But again by (2), this says that is positive, so and -1 are both positive. This violates (3).
Okay, so what if was positive instead? Well, a pretty similar thing happens, since will still be positive, so we’ll get the same contradiction that 1 and -1 are both positive.
So there’s no way to order the complex numbers, at least as a field. Woah! That’s pretty neat. The mathematicians reading this may argue that if you loosen up your definition of just a set ordering, instead of a field ordering, you could put an ordering on . But instead of doing that and arguing with your computer screen, you should try to prove to yourself that any finite field can’t be ordered (as a field). It’s more fun that way.
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7375 | https://badangling.com/american-bass-species-guide/ | American Bass Fish Species Guide - Types of Bass Fish
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Black Bass
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North America hosts a passionate obsession with its bass fish species and subspecies. From general recreation, huge payout tournaments, and cooking, bass have become a familiar sight across the continent’s fisheries.
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Across North America, there are four main types of bass fish species:
1. Black Bass
Black bass are freshwater fish with elongated bodies and are well-known, hard-fighting fish. Colors vary between subspecies, and they all have spiny and soft rayed dorsal fins, most of which are joined together to form a single fin. While they are technically part of the sunfish family, they grow much larger than your typical sunfish, making them appealing to anglers.
Undoubtedly the most popular bass species, America’s obsession with black bass has created a multi-billion-dollar industry focused on entertainment and tournaments.
You can find black bass in all freshwater bodies across North America, from deep lakes to shallow streams. Almost every method of fishing can be used to catch black bass, from live bait to fly fishing and everything in between!
Smallmouth Bass
Often referred to as bronzebacks, these light brown-colored fish are one of the hardest-fighting subspecies on this list. Smallmouths caught in clear and cold freshwater have very defined colors, making their vertical lines evident. Adult smallmouths often present with red eyes and are distributed throughout North America.
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Largemouth Bass
This is probably the most popular subspecies of black bass, and the one most people will have heard of. Their sheer size makes them one of North America’s most coveted fish species. The lures and tactics you can use to catch a largemouth bass are almost endless due to their diverse diet, made up of shad, crawfish, worms, and insects.
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Spotted Bass
Spotted bass are often confused for their largemouth cousins. Two of the main differences that spotties have are dark spots along the lower half of their bodies and a connected dorsal fin. Spotted bass eat and behave much like largemouth bass, being similarly hard fighters (pound-for-pound) when on the end of the line.
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Redeye Bass
Redeye bass may be a new species to many lake fishermen, but creek anglers will be familiar with them. A small member of the black bass family, redeye bass have brown-green sides that are lighter in the center and decorated with vertical stripes. As you may expect, they get their name from their piercing red eyes.
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Shoal Bass
The shoal bass could be considered a new bass species. Until 1999, it wasn’t recognized as a stand-alone species and was considered a subspecies of the redeye bass. A native only to Georgia, Alabama, and Florida, this creek-based bass species can be caught using various tactics and baits; favorites amongst anglers include fly fishing and float fishing.
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Suwannee Bass
Allowing for their average weight, suwannee bass are shorter in length than many other bass species discussed here. Found only in southern Georgia and northern Floria, these stunning fish have brown bodies with olive mottling across their sides. Their face is decorated with prominent horizontal striped markings, a common feature across the black bass family.
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2. Temperate Bass
The general coloration of temperate bass differs slightly from subspecies to subspecies. Members of the temperate bass family display stripes that run horizontally along their bodies and have pearl white undersides with light grey backs. Compared to other bass species, temperate bass have smaller mouths, thin lips, and large, rigid scales.
While temperate bass are widely distributed through North America’s freshwater lakes and rivers, they can also be caught in brackish water. Some species are actually anadromous, meaning they spend most of their life in brackish or saltwater and only enter freshwater to spawn.
Their diet is typically made up of shad and other baitfish, along with small crustaceans and insects. Temperate bass are known to form large schools, making them a popular option among anglers; being in the right place at the right time can lead to a highly productive day, plus they put up genuinely swashbuckling fights.
Watch out when handling these fish. Temperate bass carry sharp, spiny dorsal fins that were once used as a defense mechanism and are also equipped with razor-sharp gill plates.
Striped Bass
A highly desired bass species, striped bass display seven to nine prominent horizontal stripes that run from head to tail. Striped bass can reach enormous sizes, with individuals over 50 pounds not being unheard of.
Stripers migrate between saltwater and freshwater, meaning they have one of the largest ranges of any bass species. From the Gulf of Saint Lawrence to the Colorado River, ripping a large swimbait through the water is a highly effective striped bass tactic.
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White Bass
White bass are freshwater fish distributed across North America; they can be caught anywhere from small, running streams to large, stagnant lakes. Catching a white bass can be as simple as throwing a white grub (or any bright color) and working it quickly through the middle of the water column.
Often mistaken for striped bass, white bass usually come in between 12-15 inches in length and so are considerably smaller than their bigger striped bass brothers.
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Yellow Bass
The yellow bass is a small fish, probably closer in size to perch than most of the bass species on this list.Typically caught from clear rivers, yellow bass display defined horizontal lines from head to tail, similar to striped and white bass. If the conditions suit, a grub, rooster tail, or any small spinner can land you multiple yellow bass in a single session.
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3. Saltwater Bass
Many species of ocean-dwelling bass are closely related to their freshwater counterparts. While the ocean hosts many different subspecies, three of note are: spotted sand bass, calico bass, and barred sand bass.
Spotted Sand Bass
As you would expect, spotted sand bass boast a stunning pattern made from hundreds of black and brown spots displayed on their bodies. They also have a distinctive dorsal fin, where the third spine is considerably longer than the preceding and proceeding spines.
Usually between 12-16 inches in length and 3-4 lb in weight, these beautiful fish are distributed in the central-eastern range of the Pacific Ocean. As an ambush predator, spotted sand bass feed on smaller baitfish such as herring. Using a natural colored grub or live bait can land you a PB spotted bass.
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Calico Bass
Undoubtedly, the calico bass is one of the most unique-looking bass species, with varying shades of brown, black, gold, olive, and white decorating their bodies.
Also known as kelp bass, calico bass congregate around underwater kelp forests, using the vegetation for cover as they ambush prey feeding on zooplankton. A highly desired species by anglers, calico bass also make for excellent table fare and are a staple in many Mexican cities.
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Barred Sand Bass
The barred sand bass is found off the southern coast of California and northern Mexico. While not huge, this species puts up one hell of a fight, providing decent battles to anglers using light tackle.
Similarly to spotted sand bass, their third dorsal spine is also significantly elongated. As the two species share a similar distribution and diet, you’ll need to look for other distinctive characteristics to separate the two. One of which is the clearly defined, dark vertical bars running down the sides of this species.
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4. Other Bass
North America is also home to other species that are popular amongst anglers. We’ve grouped fish that are non-native or hybrids of native species in this category. While these fish offer excellent recreational opportunities, it is essential that these bass populations do not grow to a point where they negatively affect domestic bass species.
Peacock Bass
Strictly speaking, peacock bass are not technically bass, although anglers still treat them as such. Peacock bass were introduced to south Florida from South America in the 1980s to keep invasive species in check, and so are recognized as non-native to North America.
These fish are known for their explosive bite and the intense fights that ensue. As popular as peacock bass are, you will have to concentrate your focus on south Florida if you plan to fish for them.
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Hybrid Bass
In recent years, hybrid bass have changed the bass fishing game. Hybridization can occur between largemouth and smallmouth bass or, more commonly, between striped and white bass. Here we look at the striped/white hybrid.
Luckily, hybridizing fish does not affect an ecosystem as much as it does with other animals. Targeting a hybrid fish can be a great way to expand your fishing knowledge and catch fish that others rarely target.
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Frequently Asked Questions
Here we look at the most common questions that people ask when it comes to bass fish species in America. We aim to unravel many mysteries surrounding these beautiful and fascinating fish. Regardless of your current knowledge level, there’s always more to discover in the vast world of bass fishing!
When to Hit the Water? Your Guide to the Bass Fishing Season
When does the official bass fishing season start? Which season is the best for catching bass? Dive into these questions and more with our complete guide that examines the intricate dance of bass behavior throughout the seasons. From the chilly grips of winter to the vibrant warmth of summer, learn how water temperatures, sunlight, and other environmental factors shape the habits of these elusive fish.
With expert advice on how to handle each seasons, this guide is a treasure trove for both seasoned anglers and novices.
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Help: When Is the Best Time to Fish for Bass?
Here we aim to answer the question ‘What is the best time of day to fish for bass?’. Although generalized advice about the best times to fish during each season can be useful, mastering bass fishing involves more than just watching the clock. It requires a deep understanding of the rhythms of bass behavior throughout the day and across different seasons.
Delve into the specifics of early morning strategies where quieter waters and less competition lead to better catches. As the sun sets, discover the prime conditions that make late afternoon and dusk some of the best times to land a bass, with detailed tips on lure selection and fishing techniques.
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Tackle Box Truths: Can You Eat Bass Fish & Should You?
Plunge into the heated discussion that dominates fishing forums and dinner tables alike: can you eat bass? We tackle this contentious issue head-on, shedding light on the nuances that make this a hotly debated topic. Discover the taste secrets of different bass species and navigate through the tangled nets of regulations and ethical considerations.
Our comprehensive guide addresses not only one of the most frequently asked questions in the US bass fishing community, but also delves into several related topics. An essential read for every curious angler!
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Conclusion
There is no simple answer when people ask the question, ‘what is a bass fish?’ and identifying them can be challenging. This guide and the subsequent pages are designed to make identifying and catching specific bass species easier.
After catching a few different subspecies, you will undoubtedly be hooked, like much of North America. Bass fishing is a diverse, engaging sport and shouldn’t be limited to a single species, so get out there and catch as many different types of bass as possible!
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7376 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8444071/ | A new affinity matrix weighted k-nearest neighbors graph to improve spectral clustering accuracy - PMC
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PeerJ Comput Sci
. 2021 Sep 6;7:e692. doi: 10.7717/peerj-cs.692
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A new affinity matrix weighted k-nearest neighbors graph to improve spectral clustering accuracy
Muhammad Jamal Ahmed
Muhammad Jamal Ahmed
1 The School of Computer Science and Engineering, Kyungpook National University, Daegu, Daegu, South Korea
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1, Faisal Saeed
Faisal Saeed
1 The School of Computer Science and Engineering, Kyungpook National University, Daegu, Daegu, South Korea
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1, Anand Paul
Anand Paul
1 The School of Computer Science and Engineering, Kyungpook National University, Daegu, Daegu, South Korea
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1,✉, Sadeeq Jan
Sadeeq Jan
2 Department of Computer Science & IT, University of Engineering Technology Peshawar, Peshawar, Peshawar, Pakistan
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2, Hyuncheol Seo
Hyuncheol Seo
3 School of Architectural, Civil, Environmental and Energy Engineering, Kyungpook National University, Daegu, Daegu, South Korea
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3,✉
Editor: Anand Nayyar
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1 The School of Computer Science and Engineering, Kyungpook National University, Daegu, Daegu, South Korea
2 Department of Computer Science & IT, University of Engineering Technology Peshawar, Peshawar, Peshawar, Pakistan
3 School of Architectural, Civil, Environmental and Energy Engineering, Kyungpook National University, Daegu, Daegu, South Korea
✉
Corresponding author.
Received 2021 Jun 2; Accepted 2021 Aug 4; Collection date 2021.
© 2021 Ahmed et al.
This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, reproduction and adaptation in any medium and for any purpose provided that it is properly attributed. For attribution, the original author(s), title, publication source (PeerJ Computer Science) and either DOI or URL of the article must be cited.
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PMCID: PMC8444071 PMID: 34604521
Abstract
Researchers have thought about clustering approaches that incorporate traditional clustering methods and deep learning techniques. These approaches normally boost the performance of clustering. Getting knowledge from large data-sets is quite an interesting task. In this case, we use some dimensionality reduction and clustering techniques. Spectral clustering is gaining popularity recently because of its performance. Lately, numerous techniques have been introduced to boost spectral clustering performance. One of the most significant part of these techniques is to construct a similarity graph. We introduced weighted k-nearest neighbors technique for the construction of similarity graph. Using this new metric for the construction of affinity matrix, we achieved good results as we tested it both on real and artificial data-sets.
Keywords: K-nearest neighbors, Spectral clustering, Eigen decomposition, Affinity matrix
Introduction
Clustering data is a prevailing technique used in unsupervised learning; its goal is to breakdown data into clusters (Ünal, Almalaq & Ekici, 2021) in a way that representatives of the identical cluster are better identical to each other, conferring to some resemblance measure (Frate et al., 2021) than any two members from two different groups. The categorical partition of recent clustering techniques can be as follows: hierarchical clustering, partitioning clustering, grid-based clustering, and density-based clustering, correspondingly. Although the preceding clustering techniques displayed decent achievement, but those methods in its applicability to big data because of their highly computation complexity are limited (Xu & Zhang, 2004). In real-world complications diverse applications of clustering are revised e.g. in Danesh, Dorrigiv & Yaghmaee (2020). The procedure is effectively cast-off e.g. in management for hazard appraisal Bharti & Jindal (2020) and Jain (2010) or in portfolio management (Sanchez-Silva, 2009). Even though several clustering approaches have been suggested in the late periods, see e.g. Chen et al. (2003) or Zhang & Maringer (2010), a prevalent clustering approach which is worthy of dealing with any kind of clustering problem; subsequently, the real-life clusters may be of diverse densities, random complicated shapes and instable sizes. Path-based clustering and spectral clustering are dualistic developed clustering methods recently and these both have conveyed remarkable outcomes in a numeral of stimulating clustering problems, although path-based and spectral clustering both are not sufficiently robust counters to outliers and noise in the data. Regardless of the auspicious achievement of path-based and spectral clustering algorithms testified on some problematic data sets, there are some certain situations when both of these algorithms do not achieve that auspicious performance. Choosing affinity matrix is of great significance, it will be the reason whether the clustering results will be good or bad. Affinity matrix is generally defined in a similar manner to the Gaussian kernel based on inter-point Euclidean distance in the input space. Clustering data is an essential and complex problem in pattern recognition, computer vision and data mining, such as gene analysis, object classification, image segmentation and study of social networks. The intention of clustering data, also called cluster analytics or analysis, is to observe the ordinary grouping(s) of a collection of objects, points, or patterns (Jain, Murty & Flynn, 1999). Webster expresses the analysis of cluster as “a classification method using statistics for determining whether the entities of a population plunge into unlike clusters by formulating quantitative assessments of numerous features” (Xu & WunschII, 2005). There is a wide use of cluster analysis in abundant applications, inclusive of image processing, data analysis, market research, and pattern recognition (Jain, 2005; Larose, 2005). Dynamic development is going on in data clustering, Statistics, Data Mining, Biology, Spatial Database Technology, Marketing and Machine Learning are the main contributing zones of research. Cluster analysis has developed as an extremely vigorous topic in data mining research in the recent times and the reason to this is the enormous volume of data collected in databases ( Coming to clustering techniques developed in recent time, one of the utmost extensively used clustering is spectral clustering; the reason behind its extensive usage includes solid mathematical ground-works and minimal norms on data distribution. Spectral clustering, in its unique unsupervised form, utilizes pairwise similarity to cluster samples, which is further expressed and concluded through a graph Laplacian matrix, although in this system there is no means to certify that the consequential groups or clusters resemble to the other user-defined designs or semantic of categories and cadres in the data. The spectral clustering technique is smooth and straightforward to implement, while in performance it dominates conventional clustering techniques like the k-means clustering algorithm. Principally, there are three main steps in spectral clustering: pre-processing, decomposition, and grouping. A similarity graph and its adjacency matrix are forged for the data set in the first step (pre-processing) of spectral clustering. In the second step(decomposition), during the eigen vectors of the matrix presentation of data set is changed. In the third step (grouping), groups or clusters are mined from the fresh depiction.
The rest of our manuscript is organised as follows: the next section concisely briefs the existing work of constructing similarity graph and spectral clustering algorithms. In “Proposed Approach”, we discussed the proposed approach. After that we have the “Results and Discussion” section, and followed by the concluded remarks.
Related Work
Clustering of data have been considered for an extensive period of time, and a complete overview to the construction of graphs techniques can be found in Sumathi & Esakkirajan (2001). The following section comprises of brief review and numerous current approaches for spectral clustering, graph partitioning and construction (Han & Kamber, 2000). Spectral clustering literature can be categorized into two classes (Sumathi & Esakkirajan, 2001). One is the emphasis on clustering data in the presence of a similarity graph when it is given, and the second one is the emphasis on the construction of the similarity graph using a precise spectral clustering technique. Consider we have Z = (x 1, . . .x n) a set of data points and some concept of similarity s pq ≥ 0 among all subsets of data points x p and x q. The clustering instinctive objective is to categorize the data points into numerous clusters in such a way that points in the identical cluster are related and points in unlike clusters are different to individual ones (Chen, Fang & Saad, 2009). If the problem is not to have additional material (information) than similarities (Rathore et al., 2018) among data points, an appropriate technique of characterizing the data is to construct a similarity graph G = (V, E). In the graph every vertex “v i” signifies a data point “x p”. A certain threshold is set to check the connectivity between two vertices. If the similarity between two corresponding data points “x p” and “x q” is positive or larger than the certain threshold then edge is weighted by s pq accordingly. This situation is illustrated by Fig. 1.
Figure 1. Explanation of assigning a data point to a cluster.
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Now we can reformulate the complication of clustering by means of similarity graph. So, the utmost desire or need is to find segregated region (partition) (Saeed, 2018) of the graph to such a degree that edges among diverse groups have minor weights. What this means is that points in different groups are different from each other. While the edges inside a cluster have larger weights, this acknowledges that points inside the identical cluster are similar to each data point. In the primary class, there are various research that advances the performance and efficiency of clustering Macleod, Luk & Titterington (1987). In particular, Gul, Paul & Ahmad (2019), Saeed (2019) and Belkin & Niyogi (2004) find the amount of principal Eigen vectors and segregate the data using latent tree models. Bentley (1975) Used NMF (non-negative matrix factorization) with spectral clustering, and recommended non negative and sparse spectral clustering technique. Bentley (1980) proposed Eigen-vector selection algorithm with novelty in informative/relevant, which defines the total numbers of clusters. Bentley (1980) introduced a technique with the idea of divide-and-conquer style for assembling approximate the k-nearest neighbour graph. In this technique, the data-points are recursively divided into subgroups and are overlapped, then a single k-nearest neighbour graph is constructed on every individual small subgroup. The concluding graph is fabricated by integrating all those subgroup graphs organised overall by means of overlapping fragments. Analytically, it was reported by the authors that their approach had O(dn 1.22) time complexity. Lately, Bentley (1980) introduced an alternate effective method by means of the same kind of idea. But the main difference is that the datasets are without overlapping recursively divided. To escalate the k-nearest neighbour reminiscence, it assembles several elementary graphs by reiterating the division technique for numerous times period. To form a decent division, the two approaches use convention direction to segregate the data set. The k-nearest neighbour search is an instance-based learning problem, which states in the training phase we cannot promote query points. Nonetheless, construction of k-nearest neighbour graph is an exude problem which has all the query points at his hand. Therefore, construction of such a graph is uncomplicated in routine, and we could take benefits of its various characteristic to propose more effective algorithm.
Proposed Approach
In this section we introduce a new similarity graph technique for spectral clustering. We propose a weighted k-nearest neighbors approach for spectral clustering. The reason behind using weighted k-nearest neighbors approach is the use of hyper-parameter k, which influences the performance of k-nearest neighbors. Spectral approach has two significant roles that makes it very attractive. First, it provides us with a mathematically-sound formulation (Charikar, 2002). The second advantage is computation speed. Spectral methods have become standard techniques in algebraic graph theory (Wang et al., 2012). The most widely used techniques utilize eigenvalues and eigenvectors of the adjacency matrix of the graph. Further, nowadays the significance has deviated slightly to the spectrum of the intently associated Laplacian. Indeed, Mohar (Chung, 1997) wrote that the Laplacian spectrum is more essential than that of the adjacency matrix associated field where the spectral technique has been promoted comprising ordering (Chung, 1997), partitioning (Mohar, 2004) and clustering (Juvan & Mohar, 1992). Moreover, the areas like clustering, partitioning, and ordering, unlike graph drawing practice discrete quantization’s of the eigenvectors, which operates the eigenvectors deprived of any amendments.
Techniques used for clustering
There are two extensive techniques for clustering:
Compactness
Connectivity
Data points that are positioned nearby to one another are clustered in the same group and are compact everywhere the centre of cluster. Through distance between the observations, the proximity can be measured, like k-Means.
Connected data points or immediately next to each other are clustered together. Even though the remoteness between two data points is minor, if there is no connection, the data points are not clustered into a group. Spectral clustering is kind of method that pursue this technique.
Weighted-KNN based spectral clustering
The proposed method comprises of the following steps as shown in Fig. 2. In the first step, we do pre-processing, followed by building a similarity graph of the given dataset, which is the most important step in spectral clustering. Then, we introduce weighted K-NN for building the similarity graph. In the next step, we generate the graph Laplacian and for that we need to find the degree of the node and degree matrix. Decomposition is the next step in which we find the Eigen values and Eigen vectors of the Graph Laplacian. Finally, we cluster or group the data with the standard k-means technique. In Fig. 2, we can see the overall illustration of the proposed approach.
Figure 2. Proposed architecture.
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Pre-processing
The most significant and important phase in data mining is pre-processing the data. Distance-Based techniques such as K-NN, support vector machine, and k-means are probably the utmost techniques which are affected by the variety of features. For missing data, we implemented k-nearest neighbors which uses feature similarity. Values of any new data points can be predicted by using feature similarity. What we want to convey is that a new point is allotted a value based on the position of that point, how sharply it bears a resemblance to the points in the training data. It generates an elementary mean impute, and the resulting list is later used to formulate a KDTree. Next, it uses the KDTree to figure out nearest neighbours (NN). After finding the nearest neighbors, it holds the weighted average of the nearest neighbors. Consequently, we scaled our data before employing weighted-KNN in order to ensure that all attributes contribute equally to the outcome. To ensure this, we used standardization, in which it replaces the values by the z-scores by using the following Eq. (1).
(1)
This restructures the features with standard deviation of 1 (σ = 1) and a mean of ‘0’ (= 0).
Similarity graph
In this step we built a similarity graph based on weighted K-NN, in the pattern of an adjacency matrix. An adjacency matrix is a form of square matrix which is to characterize a finite graph. The features of the matrix signify the adjacency of vertices in a graph. Suppose a graph (a simple one) the adjacency matrix is a square matrix S such that:
(2)
In the matrix diagonal features are all zero because loops (edges from a vertex to itself) are not acceptable in minimal graphs. For the construction of Similarity Graph, we preferred to use an improved variant of k-nearest neighbors known as weighted k-NN proposed by Xu & WunschII (2005). Dudani in Pothen, Simon & Liou (1990) proposed the primary weighted approach for k-NN voting. In this method from the interval (0,1) weights are taken accordingly. The more the closer the more the data point will have weightage, the most nearby neighbor data point is weighted with 1, the outermost data point with 0 and as for the other data points, they are scaled among by the linear mapping illustrated in Eq. (3).
(3)
There are two additional alternatives suggested by Pothen, Simon & Liou (1990), the rank weight (Eq. (4)) and the inverse distance weight (Eq. (5)).
(4)
(5)
Inverse of the squared distance can be more reliable instead of the inverse distance (Shi & Malik, 2000; Dudani, 1976; Mitchell, 1997). Probably, in both we may have a possibility of division by zero. We can solve it by addition of a negligible constant as shown in Eq. (4).
(6)
In Eq. (1) the exclusion of the Kth neighbor by the weighting function from the voting process in the condition when dk d1, since dk = 0 for i = k. Mitchell (1997) offers a simplification of the weighting function by presenting fresh parameters, s ≥ k and a ≥ 0. By using Macleod introduced parameters, we can conquer that deficiency. After the numerous combo previous parameters, which have been examined in Lucińska & Wierzchoń (2012), we will use s = k with a = 1.
(7)
Laplacian matrix
In this step we project the data onto a lower dimensional space. This measure deals with the possibility if some elements of the identical cluster may be not nearby in the provided dimension. Thus, the reduction of dimensional space takes place here so that the data points gets closer and thus those data points can be grouped together in the same cluster by a conventional clustering technique. This whole process is done through the procedure of computing the Laplacian Matrix. For the Laplacian Matrix computation we need to define degree of a node. The definition of the degree of m th node is:
(8)
The Laplacian Matrix is defined as: L = D − A, where ‘D’ is the diagonal matrix of degrees and ‘A’ is the adjacency matrix defined by above equation. Symmetric normalized Laplacian matrix is described as following in Charikar (2002):
(9)
Decomposition
Spectral or eigen decomposition is the presentation in the form of factorization of a certain matrix into a recognised structure, therefore the matrix is characterized pertaining its eigen values and eigen vectors. Consequently, this factorization can be done only on diagonalizable matrices. A non-zero vector of dimension N is an eigen vector of a square N × N matrix E, if it fulfils the equation:
(10)
where is the eigen vector of matrix E corresponding to Lambda λ, and λ is scalar. Categorically, the eigenvectors are the set of vectors that E (linear transformation) purely lengthens or contracts, and the quantity that the linear transformation lengthens/contracts by is the eigenvalue. Equation (10) is known as the eigenvalue problem or equation. Computing the initial q eigenvectors of the Laplacian graph. The major eigenvalue of E correlate to the minimum eigenvalue of L, so we found our first eigenvector by this way. Moreover, for the next values, we recommend using the Von Mises iteration or Power Method because of its time complexity which is O(n 2).
Grouping and clustering
In this section, finally, we implement an ordinary k-means algorithm on the newly achieved set of vectors of reduced dimension for the final clustering and grouping. K-means is a technique of clustering vector quantization, formerly from signal processing, that intent to panel ‘n’ observations into ‘k’ groups in which each observation belongs to the cluster with the nearest mean cluster centres or cluster centroid, serving as a prototype of the cluster. The goal of the clustering technique ‘k-means’ is to classify concealed variables of the large amount of data. The underlying or hidden variables are the centroids of clusters of that data-set. Through the smallest distance the representatives of a cluster or group are determined of each data to the centroid of the cluster.
Results and Discussions
We made a comparison of our work with the original work (Wu et al., 2008) on two datasets. The original work uses sign-less Laplacian matrix and mutual k-nearest neighbors, whereas we use Symmetric normalized Laplacian matrix and weighted k-nearest neighbors. Our proposed algorithm uses the same properties of eigenvectors as of the original work. For the purpose of comparison, the utmost performance of our algorithm the parameter σ values were selected manually, as defined by Fischer & Poland (2004). We preferred to use their values. The algorithms are assessed on two datasets, they cap an extensive variety of complexities. One is artificial dataset and other is real world problem. Artificial dataset “blobs” is generated by diffusing the data points by means of Gaussian distribution.
In Figs. 3 and 4, we can observe the difference of using different metrics for generating similarity graphs.
Figure 3. K-NN constructed similarity graph.
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Figure 4. Weighted K-NN constructed similarity graph.
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Figures 5 and 6 illustrates all the sorted eigen values of the given dataset. Figure 5 shows the sorted eigen values resulted through k-nearest neighbors graph and Fig. 6 illustrates the sorted eigen values resulted by our proposed technique weighted k-nn. We can see a clear difference between both the methods.
Figure 5. Sorted eigen values of graph Laplacian using K-NN.
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Figure 6. Sorted eigen values of graph Laplacian using weighted K-NN.
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In Figs. 7 and 8, we observe the first ten eigenvalues of both the algorithms and then for these eigenvalues we consider their respected corresponding eigen vectors. Overall, resulting zero or less than zero eigenvalues, is too restraining when the groups or clusters are not segregated allied (connected) components. In order to deal with this, we define the number of groups or clusters we need to discover. Occasionally it is essential to place eigenvalues in lowest to highest order. But then we also may need to rearrange the eigen vectors so they still go with the same eigenvalues. We sort the eigenvalues and keep the corresponding values of eigenvalues and their vectors.
Figure 7. First 10 sorted eigen values of graph Laplacian using K-NN.
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Figure 8. First 10 sorted eigen values using weighted K-NN.
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We have used k-means algorithm for the final clustering as it can be seen in Fig. 9. The reason behind using k-means is to get more than two clusters that is in this case the numbers 0–9.
Figure 9. Clusters illustrating different digits.
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Now we will discuss the results generated from the artificial data set blobs. Artificial data set “blobs” is generated by diffusing the data points by means of Gaussian distribution. Figures 10 and 11 represents the similarity graphs produced by using k-nearest neighbors graph and the proposed affinity matrix weighted k-nearest neighbors graph.
Figure 10. K-NN constructed similarity graph.
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Figure 11. Weighted K-NN constructed similarity graph.
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The reason behind use of weighted k-nn for generating the similarity measure is one of the various problems that influence the achievement and performance of the K-nearest neighbors is the optimal value of the hyper parameter “k”. If we take k too small, the algorithm would be too susceptible to outliers. And, if “k” is very large, then there are chances that the neighborhood might comprise a lot of data points from different classes, as the problem can be seen in Figures 10 and 11. Also in Figs. 10 and 11, the difference between both the similarity graph construction is visible. Additionally, there may be an issue of combining the class labels in k- Nearest Neighbors approach. Taking the majority vote is the simplest method of all, but there is probability of a problem if the distance between nearest neighbors varies widely and the neighbouring data points more accurately point out the class of the object.
In Fig. 12, we can see all the sorted eigen values of Graph Laplacian. Using Graph Laplacian we want to make the structure of the data obvious so that we can make the clusters in the data clear and obvious.
Figure 12. All sorted eigen values of dataset 2.
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Figures 13 and 14 illustrates the first 10 sorted eigen values of Graph Laplacian, we generate the eigen values with both the techniques k-nearest neighbors graph and weighted k-nearest neighbors graph.
Figure 13. First 10 sorted eigen values of K-NN.
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Figure 14. First 10 sorted eigen values of weighted K-NNFi.
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Figure 15 illustrates the confusion matrix, by which we can observe the correlation between different attributes of the dataset.
Figure 15. Confusion matrix.
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In Fig. 16, for the final grouping or clustering we used the Fiedler vector to partition the data points. Eigen vector that corresponds to smallest eigen value (non-zero) is the Fiedler vector. Cluster one contains the indices values below zero and rest of the values are assigned to the second cluster. Association of indicator vector with each eigen value (non-zero) of a matrix is must. Indicator vector (individual one) perfectly comprises binary values to specify clusters association, as well as they are orthogonal to each other.
Figure 16. Clusters illustrating different digits.
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In Table 1, we compared different algorithms for spectral clustering. Our algorithm stands tall among all of these.
Table 1. Comparison of our algorithm with other algorithms.
| Algorithm | Homogeneity | Completeness | V Measure |
--- --- |
| LSC-R | 0.53 | 0.84 | 0.65 |
| LSC-K | 0.55 | 0.88 | 0.69 |
| Speclus (K-NN based) | 0.56 | 0.94 | 0.73 |
| W-KNN based Spectral Clustering | 0.58 | 0.98 | 0.78 |
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Conclusion
In this paper, we have introduced a new similarity metric known as weighted k-nearest neighbors for the construction of the affinity matrix. It is constructed based-on the weighted K-NN in which we give weight-age to every node based on how far or near the nodes are. If the node is near-by it is given greater weight-age and if it is far away it’s given less weight-age.
Our experimental results shows that a good similarity metric is very significant for spectral algorithm in order to achieve good results. Our results shows that our technique is good enough to cluster the data in a good way.
Supplemental Information
Supplemental Information 1. Affinity Matrix Code.
Click here for additional data file. (590.9KB, ipynb)
DOI: 10.7717/peerj-cs.692/supp-1
Funding Statement
This work was supported by the National Research Foundation of Korea (NRF) grants funded by the Korean government under reference number (2020R1A2C1012196). The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript.
Contributor Information
Anand Paul, Email: paul.editor@gmail.com.
Hyuncheol Seo, Email: charles@knu.ac.kr.
Additional Information and Declarations
Competing Interests
The authors declare that they have no competing interests.
Author Contributions
Muhammad Jamal Ahmed conceived and designed the experiments, performed the experiments, performed the computation work, authored or reviewed drafts of the paper, and approved the final draft.
Faisal Saeed analyzed the data, prepared figures and/or tables, and approved the final draft.
Anand Paul analyzed the data, authored or reviewed drafts of the paper, and approved the final draft.
Sadeeq Jan analyzed the data, prepared figures and/or tables, and approved the final draft.
Hyuncheol Seo performed the computation work, authored or reviewed drafts of the paper, proposal statement and draft and finances, and approved the final draft.
Data Availability
The following information was supplied regarding data availability:
The data is available in KAGGLE:
The affinity matrix source code is available as a Supplemental File.
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7377 | https://math.stackexchange.com/questions/1326429/intuition-of-multivariable-chain-rule | Skip to main content
Intuition of multivariable chain rule
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I was learning/reviewing the chain rule for multivariable calculus and was wondering why the multivariable calculus chain rule is a function of summation of products of derivatives rather than just product of derivatives, like its single variable counter part.
In particular I want to fix how I think of the chain rule and generalize my thoughts/intuition to multiple variables.
Usually the way I used to remember the chain rule was by the usual "trick" of "canceling out" the middle dummy variables i.e. consider y = f(x(t)), then:
dydt=dydxdxdt
and since the dx's cancel out, the chain rule works! Wohooo, super intuitive, easy to remember and even though its not mathematically rigorous, at least it sort of makes sense.
However, for multiple variables the equation looks very different. Consider z=f(x(t),y(t)), then its chain rule derivative is:
dzdt=∂f∂xdxdt+∂f∂ydydt
Even though there is some of the same "canceling" trick, the equation doesn't quite make as much intuitive sense to me or where it came from.
So what is the intuition behind this equation? Why is it a summation of products of derivatives? Anyone have a good way of generalizing the intuition from one variable to multiple variables? Or maybe we have to change our intuition in a significant way and that is fine, as long as its more useful for multiple variable calculus!
Quick comment, by intuition, I don't necessarily mean analogies to physics, but it can be to conceptual ideas in mathematics. So bringing in explanations say of real analysis and linear algebra that appeal the intuition/concepts of those areas are welcome! We all have different type of intuitions. :)
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edited Jun 16, 2015 at 5:37
Charlie Parker
asked Jun 15, 2015 at 16:48
Charlie ParkerCharlie Parker
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Think about displacement: moving in the z direction is the same as stepping first in the x direction then in the y direction.
– Demosthene
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Jun 15, 2015 at 16:51
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I've fixed your derivatives to partial ones. (Couldn't stand knowing they were wrong!)
– Chappers
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Jun 15, 2015 at 16:53
@Chappers no worries. They where definitively wrong.
– Charlie Parker
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Jun 15, 2015 at 17:10
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The problem with intuition about cancelling differentials, it isn't safe. And yet, the method of differentials is stupidly successful.
Let me give a standard example of intuitions downfall. First, since partials cancel,
∂z∂y∂y∂x∂x∂z=1
except, it doesn't. Actually, with the right interpretation,
∂z∂y∂y∂x∂x∂z=−1.
In particular, we assume x,y,z are related by some level function F(x,y,z)=0 then dF=Fxdx+Fydy+Fzdz thus
∂z∂y=dzdy∣∣∣dx=0=−FyFz
with more words, if we consider z as a function of x,y then the partial derivative of z whilst holding x fixed is −Fy/Fz. Notice, I simply take the total differential of F and solve for dz/dy while setting dx=0. This is an example of how the differential notation is naively successful (because, careful application of the implicit function theorem yields the same outcome). Likewise, intuitive calculation with dx,dy,dz yields
∂y∂x=dydx∣∣∣dz=0=−FxFy
∂x∂z=dxdz∣∣∣dy=0=−FzFx
Thus,
∂z∂y∂y∂x∂x∂z=(−FyFz)(−FxFy)(−FzFx)=−1.
Getting back to your posed question. Why are there sums of derivatives? Well, in short, because the multivariate function can change in all of its arguments. As the derivative is a linear approximation to the change in the function we have little hope except to see formulas formed from sums of all the possible things which can change the outcome. This is the multivariate chain rule. It accounts for each entry in an entirely symmetrical manner. Ok, these sort of explainations don't settle well with me. The real answer in my estimation is matrix multiplication. The chain-rules really fall out of multiplication of Jacobian matrices which in turn come from the chain-rule in its pure form D(F∘G)=DF∘DG. But, perhaps this isn't intuition. That said, it is my intuition.
I'll add a little example to explain how the matrix multiplication works together with the Jacobian matrix to capture the chain rule. Suppose X⃗ :R2uv→R3xyz and F⃗ =⟨P,Q,R⟩:R3xyz→R3. Here I use the notation R2uv to indicate u,v serve as the coordinates. Here you can think of X⃗ as a parametrization of a surface and F⃗ as a vector field in three dimensional space. The composition F⃗ ∘X⃗ is commonly considered in the calculation of flux of F⃗ through the surface parametrized by X⃗ . In this case, the Jacobian of X⃗ is given by
JX⃗ =[∂X⃗ ∂u|∂X⃗ ∂v]=⎡⎣⎢∂ux∂uy∂uz∂vx∂vy∂vz⎤⎦⎥
and the Jacobian of F⃗ is given by
JF⃗ =[∂F⃗ ∂x|∂F⃗ ∂y|∂F⃗ ∂z]=⎡⎣⎢∂xP∂xQ∂xR∂yP∂yQ∂yR∂zP∂zQ∂zR⎤⎦⎥
Setting G⃗ =F⃗ ∘X⃗ we find from the matrix form of the chain rule that: (suppressing point dependence)
JG⃗ =JF⃗ JX⃗ =⎡⎣⎢∂xP∂xQ∂xR∂yP∂yQ∂yR∂zP∂zQ∂zR⎤⎦⎥⎡⎣⎢∂ux∂uy∂uz∂vx∂vy∂vz⎤⎦⎥=⎡⎣⎢⎢∂xP∂ux+∂yP∂uy+∂zP∂uz∂xQ∂ux+∂yQ∂uy+∂zQ∂uz∂xR∂ux+∂yR∂uy+∂zR∂uz∂xP∂vx+∂yP∂vy+∂zP∂vz∂xQ∂vx+∂yQ∂vy+∂zQ∂vz∂xR∂vx+∂yR∂vy+∂zR∂vz⎤⎦⎥⎥
For example, in the (1,1) entry we read off:
∂G1∂u=∂∂u[P(x(u,v),y(u,v),z(u,v))]=∂P∂x∂x∂u+∂P∂y∂y∂u+∂P∂z∂z∂u
Notice the matrix JG⃗ contains all 6 interesting chain rules involving composition of the component functions P,Q,R of F⃗ composed with the component functions x,y,z of u,v.
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answered Jun 15, 2015 at 22:02
James S. CookJames S. Cook
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It would have been nice if you had expanded on the last part about your intuition and matrix multiplication with an example or something.
– baxx
Commented
Aug 21, 2018 at 11:32
@baxx I'm not so sure it's all that nice :)
– James S. Cook
Commented
Aug 22, 2018 at 3:06
What does this line mean? And how was it derived?
∂y∂x=dydx∣∣∣dz=0=−FxFy
– gyro
Commented
Dec 21, 2022 at 11:43
@IvanaGyro given F(x,y,z)=0 take the total differential to obtain Fxdx+Fydy+Fzdz=0 then if you can solve for dy/dx with dz=0 and obtain dy/dx=−Fx/Fy. In all honesty, the notation ∂y/∂x is ambiguous, all of this is a shorthand for a careful application of the implicit function theorem.
– James S. Cook
Commented
Dec 22, 2022 at 22:52
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I got it. Thank you for explaining.
– gyro
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Dec 23, 2022 at 11:49
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dzdt measures what changes about z when you change t a little. Since z=f(x(t),y(t)), two things to the input of z change when you change t a little: x changes a little, and y changes a little. Both of these changes affect z a little. The first term is the part of the total change in z coming from x changing, and the second term is the part of the total change in z coming from y changing.
Symbolically (writing ∂(−) for a small change in (−)):
z+δz=f(x(t+δt),y(t+δt))≈f(x(t)+δx,y(t)+δy)≈f(x(t),y(t))+∂f∂xδx+∂f∂yδy
where δx≈dxdtδt and δy≈dydtδt.
The canceling trick is a very bad idea; it sort of works for ordinary calculus but as you can see it fails very badly for multivariable calculus. The problem is that partial derivatives do not behave anything like fractions. What they actually behave like are coefficients of a matrix (and the general form of the chain rule involves matrix multiplication); this will become much clearer if you take a linear algebra course.
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answered Jun 16, 2015 at 3:28
Qiaochu YuanQiaochu Yuan
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I've taken a linear algebra course. Do you mind expanding how that is relevant? Is it because u can express the differential operator as a matrix and the coeffs of polynomials stacked in vectors? I'd love to see how linear algebra matters. Whenever there is a linear algebra explanation, things become much more intuitive for me.
– Charlie Parker
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Jun 16, 2015 at 5:35
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@Charlie: in general, the derivative of a function f:Rn→Rm at a point is its best linear approximation at that point; explicitly it takes the form of a matrix. The content of the multivariate chain rule is that taking derivatives respects composition: it sends a composition of multivariate functions to the corresponding composition of best linear approximations, which explicitly becomes matrix multiplication.
– Qiaochu Yuan
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Jun 16, 2015 at 5:52
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7378 | https://www.math.ru.nl/~bosma/students/kirkels/mscthesis.pdf | Irreducibility Certificates for Polynomials with Integer Coefficients Bart Kirkels August 2004 3 Voorwoord Dit is ’m dan: mijn doctoraalscriptie. Na vijf jaar wiskunde gestudeerd te hebben, ben ik het laatste half jaar met onderzoek bezig geweest en de resultaten daarvan zijn in dit verslag te lezen.
Deze scriptie kwam er natuurlijk met de hulp van velen, die ik nu graag wil bedanken.
In de eerste plaats mijn afstudeerbegeleider Wieb Bosma. Hij heeft het onderzoek in goede banen geleid, en maakte altijd tijd om er over te praten, ook al had hij het vaak erg druk.
Bedankt voor de fijne samenwerking Wieb!
Ook wil ik Bas Spitters, de tweede lezer, bedanken voor het herhaaldelijk doorlezen en becommentari¨ eren van dit stuk, ook al zat hij midden in een verhuizing.
De officieuze derde lezer is Roel Willems. Hem wil ik bedanken voor het lezen en corrigeren van deze scriptie en het prettige samenwerken bij (onder andere) Commutatieve Algebra, Recursietheorie en het implementeren van de MPQS.
Tijdens de studie heb ik ook (meer dan) voldoende afleiding en koffie gehad, waarvoor ik Desda graag bedank. Ook alle studenten, met name iedereen uit mijn jaar: bedankt voor de supergezellige tijd!
Ik heb de afgelopen vijf jaar zo veel prachtigs geleerd dat ik daarvoor alle docenten en de rest van de staf wil bedanken. Ook iedereen waar ik mee samengewerkt heb in commissies en besturen wil ik bedanken voor de goede samenwerking.
Ook nog veel dank aan mijn familie en vrienden die me gesteund hebben en (in hele positieve zin) afgeleid hebben naast de studie. Voor hun is er achter in deze scriptie een Nederlandse, niet-wiskundige, samenvatting.
En ten slotte wil ik Sanne bedanken, voor al het fijne dat we tot nu toe al meegemaakt hebben, en ook voor het nakijken van de samenvatting natuurlijk.
Veel plezier met het lezen van deze scriptie!
Bart Kirkels Nijmegen, augustus 2004 Contents 1 Introduction 7 2 Certificates: Definitions and Examples 9 2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 2.2 Cantor’s Certificate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 2.3 Weinberger’s Certificate . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 2.4 The LLL-algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 3 The Certificates 13 3.1 Eisenstein . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 3.1.1 The Certificate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 3.1.2 Checking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 3.1.3 Sch¨ onemann’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . .
14 3.2 Bunyakovsky’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 3.2.1 Multiple Evaluations . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 3.2.2 One Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 3.2.3 Bunyakovsky’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . .
17 3.2.4 Checking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 3.3 Modulo p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 3.3.1 Irreducibility over Finite Fields . . . . . . . . . . . . . . . . . . . . .
18 3.3.2 Modulo-combinations . . . . . . . . . . . . . . . . . . . . . . . . . .
19 3.3.3 Checking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 4 Galois Theory and Modulo-Factorisations 23 4.1 Polynomials without Certificate . . . . . . . . . . . . . . . . . . . . . . . . .
23 4.2 Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 4.2.1 Which Galois groups can Occur? . . . . . . . . . . . . . . . . . . . .
26 4.3 Results in Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 4.3.1 Polynomials with Prime Degree . . . . . . . . . . . . . . . . . . . . .
27 4.3.2 Polynomials with Composite Degree . . . . . . . . . . . . . . . . . .
27 4.3.3 Swinnerton-Dyer Polynomials . . . . . . . . . . . . . . . . . . . . . .
29 4.3.4 The Galois Inverse-problem . . . . . . . . . . . . . . . . . . . . . . .
30 4.3.5 Distribution of Galois groups . . . . . . . . . . . . . . . . . . . . . .
30 5 6 CONTENTS 5 Modulo p: Improved 31 5.1 The Hensel-algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 5.2 The Hensel-certificate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 5.3 The Improved Algorithm: Combining 3.28 and 5.3 . . . . . . . . . . . . . .
33 5.4 Checking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33 5.5 Worst Case Scenario . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34 6 Test Results 35 6.1 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 6.2 Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36 6.2.1 Modulo p-algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . .
36 6.2.2 Algorithms based on Bunyakovsky’s Conjecture . . . . . . . . . . . .
37 6.2.3 Eisenstein . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 6.2.4 A Swinnerton-Dyer Polynomial . . . . . . . . . . . . . . . . . . . . .
38 6.2.5 Final Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 7 Formalizing Mathematics 39 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 7.1.1 What is formalizing? . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 7.1.2 Computer Mathematics . . . . . . . . . . . . . . . . . . . . . . . . .
40 7.2 Coq and a Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41 7.2.1 Introducting Coq with Examples . . . . . . . . . . . . . . . . . . . .
41 7.2.2 Formalizing Theorem 3.22 . . . . . . . . . . . . . . . . . . . . . . . .
43 7.3 Combination with Magma . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 7.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46 8 Samenvatting 47 Chapter 1 Introduction Let f be an integer polynomial. We are interested in its factorisation. So we ask a computer algebra system what it is. If f is reducible then we will get its factors and can easily check the multiplication. But what if the answer is that our polynomial is irreducible? How can the system convince us? Well, usually it really does not, but there are certain criteria that can guarantee the irreducibility of polynomials. Irreducibility itself is a negative property in the following sense: there do not exist non-trivial factors. The criteria need a positive property, such as the existence of a certain number. When we know this number we only have to check whether it is really correct or not, and have proved irreducibility in that way. This proving can be done on a computer, using a so called Proof Assistant.
In this thesis some of these criteria will be investigated. I will describe an algorithm that assigns an irreducibility certificate (based on two criteria) to every irreducible polynomial in Z[X]. I will show that these certificates are almost always easy to check. I have also proved the correctness of a criterion in the Proof Assistent Coq.
To achieve this the notions of an irreducibility certificate and of an irreducibility criterion will be made more precise in Chapter 2. Some certificates, that do not suffice for our purposes, are then given as examples.
In Chapter 3 some practical certificates are given and for every certificate it is proved that its existence for a polynomial f ∈Z[X] guarantees irreducibility of f. One of these certificates is the ‘modulo p-certificate’. It is based on factorizing f modulo primes.
In Chapter 4 this certificate is investigated using Galois theory. It is proved that for a polynomial f the existence of a modulo p-certificate is equivalent to having a special condition on the Galois group of f. Now Galois theory is used to prove that for polynomials of prime degree there always exists a certificate. For polynomials of composite degree it is proved that for almost every such f there exists a certificate. We also see that these certificates are easy to check.
7 8 CHAPTER 1. INTRODUCTION In Chapter 5 we give an algorithm that assigns a certificate to every irreducible poly-nomial. I have implemented this algorithm in the Computer Algebra System Magma.
This algorithm tries to assign a modulo p-certificate, using the factorisations of the polynomial f modulo the primes below some bound (based on the degree of f). If a modulo p-certificate has not been found, we can assume that it does not exist and we now use the modulo-factorisations of f to find a ‘Hensel-certificate’. Such a certificate exists for every irreducible polynomial (but can be hard to check). An other advantage of the Hensel-part of our algorithm is that it will find a non-trivial factor of f if f is reducible.
I have also implemented algorithms that assign the other certificates of Chapter 3 to irreducible polynomials (when possible). All of these algorithms are tested in Chapter 6.
In Chapter 7 the subject of formalizing mathematics on a computer is introduced.
Programs that can be used for formalizing are called Proof Assistants. Coq is the Proof Assistant that I have used. Examples of using Coq (in the setting of this thesis) are given.
I have formalized an irreducibility criterion in Coq; I will sketch how this was done. The formalizing process can be combined with the computations of Magma. This combination can be used to formally prove irreducibility of polynomials.
In Computer Algebra Systems the factorisation of a polynomial is usually determined by using the LLL-algorithm . The mathematical theory for proving the correctness of this algorithm is at this moment too advanced to formalize in Coq. That is the reason why we want to be able to prove irreducibility in another way.
Eventually, in Chapter 8, I give a short non-mathematical summary of this thesis for everyone who is interested in what I have done.
Chapter 2 Certificates: Definitions and Examples In this chapter we shall first explain the notion of an irreducibility certificate. Then we take a look at two specific certificates. The first one seems to be quite good, but generating these certificates is very time-consuming. The second one is an example of a conditional certificate, and we shall see how it depends on a conjecture.
By Gauss’ lemma we can prove the irreducibility of a polynomial over Q, by proving the irreducibility of a corresponding polynomial over Z. That is why we will give irreducibility certificates for polynomials over Z. So in this chapter f will be a polynomial in Z[X].
2.1 Definition Definition 2.1 A criterion is an existential condition; by this we mean that it is of the form: ‘there exist mathematical objects M such that . . . ’.
Definition 2.2 More specifically: an irreducibility criterion is an existential condition for a polynomial f that guarantees the irreducibility of f.
A famous example is Eisenstein’s criterion which will be part of our investigation in the next chapter.
Definition 2.3 An irreducibility certificate for f and an irreducibility criterion, consists of mathematical objects M that satisfy the criterion for f.
Definition 2.4 A checking-algorithm for a criterion is a procedure to decide whether a certificate M satisfies the criterion or not.
The main purpose of this thesis is to give certificates together with a correctness proof.
This can be achieved by formalizing the criterion and the checking-algorithm in a so called ‘proof assistant’. Chapter 7 will be about formalizing, but for now it is enough to know that the proof of the sufficiency of the criterion should not be too advanced, so that it can be formalized.
9 10 CHAPTER 2. CERTIFICATES: DEFINITIONS AND EXAMPLES Of course we would also like a checking-algorithm to be fast. It would be nice if it would require an amount of computation polynomial in the size of f. (By the size of f we mean the number of bits required to represent all coefficients of f.) But we shall also see certificates of which we can not guarantee an easy check, but that are on average quite fast to check.
There is also another aspect, that of finding certificates. We hope to find certificates for every irreducible polynomial, but for most criteria there are irreducible polynomials that do not have a certificate, or for which it takes a lot of time to find a certificate.
2.2 Cantor’s Certificate In 1981 David Cantor showed that for every irreducible polynomial there exists an irreducibility certificate that can be checked in time polynomial in the size of f.
First some definitions: Definition 2.5 Let f ∈Z[X]. The height of f, denoted by H(f), is the maximum of the absolute values of its coefficients.
Definition 2.6 A triple of polynomials (f, g, h) ∈Z[X]3, with degrees n, k, l respectively, is cantorian if: • f and g have the same degree, say n = k ≥1 • l = n −1 • g is monic (i.e. gk = 1) and g0 ̸= 0 • f(X) divides g(h(X)) • either |gk−1| > 1 + Pk−2 j=0 |gj| or gk−1 = 0 and gk−2 > 1 + Pk−3 j=0 |gj| The following theorem, proved by Cantor, states that a cantorian triple can act as an irreducibility certificate: Theorem 2.7 If (f, g, h) is cantorian, then f is irreducible.
I will now sketch the proof (it is in [6, Lemma 1 and 2]).
First it is proved that g ∈Z[X], satisfying the cantorian conditions, is irreducible. To conclude it is proved that if f divides g(h(X)) with g irreducible, then f itself is irreducible.
And so we have the Cantor-criterion and the corresponding Cantor-certificate: Certificate 2.8 (Cantor) A triple (f, g, h) of polynomials in Z[X].
Criterion 2.9 There exists a triple (f, g, h) of polynomials in Z[X], that is cantorian.
2.2. CANTOR’S CERTIFICATE 11 Cantor has also proved that such a certificate, with the polynomials bounded in height, can always be found: Theorem 2.10 If f is an irreducible polynomial then there are polynomials g and h such that (f, g, h) is cantorian and H(g), H(h) ≤(128)2·deg(f)2H(f)3·deg(f)3.
Again, the proof is in [6, Lemma’s 4 to 7] and we shall sketch it: A huge n −1-dimensional cube, that is divided in many small cubes, is created, as well as linear forms depending on the (complex) roots of f. We let these forms act on vectors and in this way get vectors in the cube. Now the pigeonhole principle is used to find two vectors that are in the same little cube. Out of these two vectors we can form h ∈Z[X] that satisfies the cantorian conditions. From h it is easy to construct g, and eventually it is proved that g also satisfies the cantorian conditions.
Because g and h are bounded we see that we can check whether a triple (f, g, h) is cantorian or not, in time polynomial in the size of f. So now we have easy-to-check certificates for every irreducible polynomial; a checking-algorithm just has to check the (easy) cantorian conditions on bounded polynomials!
Alas, we do not really have them: in Cantor’s article g and h are constructed, but as we have seen in the sketch of the proof above this is an enormous task.
But now that we have Cantor’s theorem we may try to find certificates in an other way.
Unfortunately there is no fast method known at this moment. We know that a solution exists for all irreducible polynomials so we can try all g and h with the correct height.
Obviously the height-limit is quite high here so this criterion does not guarantee that we can find a certificate quickly.
Another problem is that the mathematics involved is quite advanced and therefore it would take a lot of time to formalize all of it. And this is what we wanted, to be absolutely certain of irreducibility!
Example Suppose we want to show that f = X3 + X + 1 is irreducible, with a Cantor-certificate. Then we can start trying all polynomials g of degree 3 and all polynomials h of degree 2 with coefficients smaller than (128)2·deg(f)2H(f)3·deg(f)3 = 12818. But if we try we can find g = X3 + 10X2 + X + 1 and h = −4X2 + 3X −6, but it is clear that we can not be certain of finding them fast. Once we have these g and h it is easy to check that (f, g, h) is cantorian.
Coming to a conclusion we can now say that there exists a Cantor-certificate for every irreducible polynomial and that such a certificate can be checked easily. But the major drawbacks are that at this moment there is no known method to find certificates fast, and that it will be an enormous task to formalize the mathematics involved.
12 CHAPTER 2. CERTIFICATES: DEFINITIONS AND EXAMPLES 2.3 Weinberger’s Certificate and the Riemann Hypothesis A certificate which has a greater disadvantage is that of Weinberger. In he proves that we can compute the number of irreducible factors of a polynomial in time polynomial in the size of f if the Riemann Hypothesis (RH) holds. So we can also quickly compute whether it is irreducible (i.e. the number of factors is 1), or not.
This is how it works: It is proved that the average number of linear factors of f modulo all primes, is the number of irreducible factors of f. Therefore we factor f modulo 2, 3, 5, 7, . . ., until the average numbers of linear factors of f modulo a prime can be determined. The question now is where we can stop to be sure of the average number. This is where the RH is needed. With this hypothesis (which says something about the distribution of prime numbers), a bound on the primes to be used can be calculated.
The certificate for an irreducible polynomial f would then consist of all modulo-factorisations which were needed according to the RH.
We now see what the problem is: if the RH does not hold then the average after a finite number of primes might not be the total average. So this algorithm could either give a too high or too low number of irreducible factors of f as an answer. So the algorithm (and thus a certificate) relies too heavy on an unproved conjecture.
Remark Even when the hypothesis would be proved, another problem is that in practice the bound on the primes would be very large, so that the checking would take a lot of time.
2.4 The LLL-algorithm In 1982 the LLL-algorithm was introduced by Lenstra, Lenstra and Lovasz . With this algorithm factorizing f can be done in time polynomial in the size of f. There have been made various improvements on this method and the algorithm used in Magma for example is based on a version by Van Hoeij .
So we could add an empty certificate to an irreducible polynomial, because using the LLL-algorithm we can check that is irreducible in polynomial time.
The problem for us is that we want to be completely sure that a polynomial is irreducible.
As, for example, there might be bugs in a computer algebra system, we want to have a formalized algorithm. The mathematics used to prove the correctness of the LLL-algorithm is in the present state of technology too far out of reach to be formalized.
Chapter 3 The Certificates In this chapter I will give practical irreducibility certificates for polynomials over Z.
The polynomial will be denoted by f; f = fnXn + fn−1Xn−1 + . . . + f1X + f0 ∈Z[X], with n ≥2, fn ̸= 0, and gcd(f0, . . . , fn) = 1.
In every section I will first give the certificate and criterion and then prove that the crite-rion guarantees irreducibility. Next comes some additional information, depending on the certificate and eventually I will describe how the certificate can be checked (and how fast).
3.1 Eisenstein The criterion of Eisenstein is well-known and, if the polynomial is in the right form, it is very easy to use, even without a computer. A disadvantage is that in practice polynomials with large degrees and coefficients have no Eisenstein-style certificate. This will be illustrated with examples in Chapter 6.
3.1.1 The Certificate Certificate 3.1 (Eisenstein) p ∈N Criterion 3.2 There exists a prime p such that p ∤fn, p2 ∤f0, and p | fi for 0 ≤i < n.
Theorem 3.3 If the criterion holds for f, then f is irreducible.
Proof Suppose we have p as in the proposition. Suppose also that f is reducible, say f = gh = (grXr + . . . + g0)(hn−rXn−r + . . . + h0), with 1 ≤r < n. Then f0 = g0h0, so p | g0h0 but p2 ∤g0h0, so let’s say that p | g0, p ∤h0.
Now use the principle of induction to prove that p divides all coefficients of g: We know that p | g0. Suppose that p divides g0, g1, . . . , gi−1 (i < n of course).
Now we let hj = 0 if j > n −r (so they are defined now). Then we have that fi = g0hi +g1hi−1 +. . .+gi−1h1 +gih0, and we know that p | fi, so p divides the right hand side of the equation. The assumption is that p divides g0, . . . , gi−1, 13 14 CHAPTER 3. THE CERTIFICATES so we see that p must divide gih0 as well. Since p ∤h0, we have that p | gi.
Hence p divides all coefficients of g, so it divides g itself.
This means that p | f, which is a contradiction because p ∤fn.
So f is irreducible.
□ Example Let f = X5 + 500X4 −15X3 + 10X −30, then f is irreducible by Eisenstein, with p = 5. So the certificate is 5 here.
Remark Let f ∈Z[X], a ∈Z. Then f(X) is irreducible ⇔f(X + a) is irreducible.
Proof If f(X) has a non-trivial factor g(X) then f(X + a) has the non-trivial factor g(X + a) and vice versa. Hence f(X) does not have a non-trivial factor ⇔f(X + a) does not have a non-trivial factor.
□ And now we have a new certificate: Certificate 3.4 (p, a), with p ∈N, a ∈Z.
Criterion 3.5 There exist p ∈N, a ∈Z such that f(X + a) is irreducible by Eisenstein, with p.
Theorem 3.6 If the criterion holds for f, then f is irreducible.
The proof is clear with the remark above.
Example Let f = X3 + 27X2 + 222X + 562, then f(X) does not have an Eisenstein-certificate, but f(X −1) = X3 + 24X2 + 171X + 366 does, with p = 3. So f is irreducible.
The certificate is (3, −1).
3.1.2 Checking To check a certificate 3.4 we have to compute f(X + a) and check all coefficients. We also need a proof that p is indeed prime. Because p can not be ‘really big’ (it divides the constant coefficient for example) the complete check is quite fast.
3.1.3 Sch¨ onemann’s Theorem The criterion of Eisenstein was formulated in 1850 and was in fact a special case of a theorem already formulated by Sch¨ onemann in 1846 .
Theorem 3.7 Let F = f n + pg ∈Z[X], with n ≥1, p ≥2 a prime, f, g ∈Z[X] such that deg(f n) > deg(g) and f is irreducible in Fp[X] and f does not divide g in Fp[X]. Then F is irreducible.
Proof Suppose that F = F1F2 is a non-trivial factorisation. Then F = F1F2 in Fp[X]. F = f n, f is irreducible in Fp[X], and so it follows that there exist u, v ∈N, u + v = n and g1, g2 ∈Z[X] such that F1 = f u + pg1, F2 = f v + pg2, 3.2. BUNYAKOVSKY’S CONJECTURE 15 with deg(g1) < u · deg(f), deg(g2) < v · deg(f). So F = f n + pg = (f u + pg1)(f v + pg2) = f n + p(f ug2 + f vg1 + pg1g2) Hence g = (f ug2 + f vg1 + pg1g2). Without loss of generality we may suppose that u ≤v, so that g = f u · h + pg1g2, with h = g2 + f v−ug1 ∈Z[X]. Viewing this equation modulo p we now have that g = f u · h. So f divides g in Fp[X]: contradiction, unless u = 0. But if u = 0, then F1 = 1 and we have a trivial factorisation of F, and thus we arrive at a contradiction now as well.
□ We can see that Eisenstein’s criterion is a special case by taking f(X) = X and g(X) = 1 p(fn−1Xn−1 + . . . + f1X + f0). As p | fi for 0 ≤i < n, we see that g ∈Z[X].
3.2 Bunyakovsky’s Conjecture In the previous chapter we have seen Weinberger’s algorithm, which needed a conjecture to be certain of correct certificates. In this section we shall discuss certificates that always imply irreducibility. But there is a conjecture (of Bunyakovsky) that guarantees the existence of such certificates for every irreducible polynomial. The certificates consist of one or more integers i for which f is a prime or a unit. First we shall describe the certificates that need multiple evaluations.
3.2.1 Multiple Evaluations These certificates need multiple evaluations. They were found in .
First two remarks that will be used in the proofs.
Remark 1 Let g ∈Z[X] be a polynomial of degree k ≥1. Then for a ∈Z there are at most k integers i such that g(i) = a. To see this suppose that there are more than k such integers. Then the polynomial g(i)−a, also of degree k ≥1 has more than k integer zeroes, which is impossible.
Remark 2 If f is reducible, say f = gh, with deg(g), deg(h) ≥1, and f(i) is prime for i ∈N, then g(i) or h(i) is a unit. As this is in Z, the units are 1 and −1.
Certificate 3.8 2n + 1 integers.
Criterion 3.9 There exist 2n + 1 integers i1, . . . , i2n+1, such that f(ij) is a prime or unit for all j.
Theorem 3.10 If the criterion holds for f, then f is irreducible.
Proof Suppose f = gh, with deg(g) = k ≥1, deg(h) = l ≥1. Then from Remark 1 we know that g(i) = 1 for at most k integers i, also g(i) = −1 for at most k integers i. Thus the maximum number of times that g(i) equals 1 or −1 is 2 · k. In the same way h(i) can equal 1 or −1 at most 2 · l times. As a result of Remark 2, the maximum possible number of distinct values of i for which f(i) can be a prime or a unit is 2(k + l) = 2n. This is a contradiction with our assumption, so f is irreducible.
□ 16 CHAPTER 3. THE CERTIFICATES Example Let f = X4 + 3X3 + 17X2 −89X + 3. Then we have the following table: X −5 −2 −1 0 2 4 5 7 8 f(X) 1123 241 107 3 −67 367 983 3643 6011 So we have 9 (= 2 · deg(f) + 1) prime values and we have that f is irreducible.
The corresponding certificate is (−5, −2, −1, 0, 2, 4, 5, 7, 8).
We can do with fewer primes, but then we need an extra proposition and some conditions on the distribution of these evaluation points.
Proposition 3.11 Let g ∈Z[X], with deg(g) = d and i1, i2 ∈Z, such that |i1 −i2| > 2 and |g(i1)| = |g(i2)| = 1. Then g(i1) = g(i2).
Proof Suppose g(i1) = 1, g(i2) = −1. Subtracting these two equations we get gd(id 1 −id 2) + . . . + g2(i2 1 −i2 2) + g1(i1 −i2) = 2.
All terms in the sum are divisible by i1−i2, so (i1−i2) | 2. This is in contradiction with our assumption that |i1 −i2| > 2. In the same way we can arrive at a contradiction if g(i1) = −1, g(i2) = 1. So g(i1) = g(i2).
□ Certificate 3.12 n + 1 integers.
Criterion 3.13 There exist n + 1 integers ij that differ pairwise by more than 2, such that f(ij) is a prime or unit for all j.
Theorem 3.14 If the criterion holds for f, then f is irreducible.
Proof Suppose f = gh, with deg(g), deg(h) ≥1. With the previous proposition and Remark 1 we know that there are at most deg(g) integers i differing pairwise by more than 2, for which |g(i)| = 1. In the same way we have that there are at most deg(h) integers i differing pairwise by more than 2, for which |h(i)| = 1.
By Remark 2 there can now only be n(= deg(g)+deg(h)) i’s (differing pairwise by more than 2) for which f(i) is a prime or a unit. This is in contradiction with our assumption, so f is irreducible.
□ Example Let f = X4 + 3X3 + 17X2 −89X + 3. Then we have the following table: X −8 −5 0 4 7 f(X) 4363 1123 3 367 3643 So we have 5 (=deg(f) + 1) prime values and we have that f is irreducible.
The certificate is (−8, −5, 0, 4, 7).
3.2. BUNYAKOVSKY’S CONJECTURE 17 3.2.2 One Evaluation For this certificate, by John Brillhart , we just need one prime evaluation.
Certificate 3.15 x0 ∈Z Criterion 3.16 There exist m ∈N∗greater than the moduli of the (complex) zeros of f, and x0 ∈Z with |x0| ≥m + 1, such that f(x0) is prime.
Theorem 3.17 If the criterion holds for f, then f is irreducible.
Proof Suppose f = gh, with deg(h) ≥1. Then m is certainly greater than the moduli of the zeros of g and h.
|f(x0)| = |g(x0)| |h(x0)| is prime.
Let deg(h) = r, then hr is its leading coefficient. Let α1, . . . , αr be the zeros of h. Then |h(x0)| = |hr| Qr i=1 |x0 −αi| > 1, because |hr| ≥1, |x0| ≥m + 1 and for all i : m > αi. So |h(x0)| is prime. If deg(g) ≥1 then (in the same way) |g(x0)| > 1, which gives rise to a contradiction, so g is constant and |g(x0)| = 1, hence g = ±1, and we have that f is irreducible.
□ Example Let f = X4 + 3X3 + 17X2 −89X + 3. The complex zeros of f were calculated and were approximately: 0.034, 2.708, −2.871 + 4.941i, −2.871 −4.941i. So the moduli are less than 1, 3, 6 and 6. Hence we can take m = 6.
We now see that f(7) = 3643 (which is prime), so f is irreducible.
The certificate simply is 7.
Example One larger example: Let f = X37 + 90823490832X19 −4082408240240000333. m = 5 suffices, and f(−150) = –327624661361185544956242665668594483126156114682006835937500004082408240240000333 completes the proof. Of course a problem now is: how do we know that the monstrous evaluation is really prime? For that purpose we might for example use Pocklington’s certificate [24, 25]. But when primes become this big this certificate will take a very long time to check.
3.2.3 Bunyakovsky’s Conjecture So we have seen the certificates, but do they always exist?
To investigate this, we first need a definition: Definition 3.18 Let f ∈Z[X]. The fixed divisor of f, denoted by df, is the largest positive integer d such that d | f(a) for all a ∈Z.
Example Let f = X2 + 9X −4. Now f(i) is always even, f(0) = −4 and f(1) = 6, so here df = 2.
18 CHAPTER 3. THE CERTIFICATES We see that if df > 1, then f(i) can be prime for only finitely many i’s. In 1857 Bunyakovksy stated the following conjecture : Conjecture 3.19 If f is irreducible, then d−1 f f(i) is prime for infinitely many i.
At this moment we only know this conjecture to be true for the special case that deg(f) = 1. Then it follows from Dirichlet’s theorem on primes in arithmetic progressions.
The conjecture is generally believed to be true.
So, if Bunyakovsky’s conjecture holds we always have certificates for irreducible polynomi-als. But what we really need is an even stronger conjecture that puts an upper bound on where to find a prime evaluation.
3.2.4 Checking A certificate consists of the i’s for which f(i) is prime or a unit. We then have to calculate the f(i)’s and prove that they are prime (or a unit). These primes may become monstruous as we have seen in an example, so we really can’t say how long it takes to check this certificate. (And finding a certificate we can be completely certain of will of course cost a lot more time).
But in practice, when we trust the primality check of Magma and try to prove irreducibility for some smaller polynomials (something like deg(f) ≤20 and coefficients ≤106), it works quite well, especially Brillhart’s certificate 3.15. For test results, see Chapter 6.
3.3 Modulo p In this section p will always be a (positive) prime number.
In practice, factoring over small finite fields is easier than over Z. Unfortunately an ir-reducible factor of f might factor over a finite field. But a modulo-factorisation can still provide us with some information.
3.3.1 Irreducibility over Finite Fields Certificate 3.20 p ∈N Criterion 3.21 There exist a prime p such that f is irreducible over Fp, and p does not divide the leading coefficient fn.
Theorem 3.22 If the criterion holds for f, then f is irreducible.
Proof If f = gh in Z[X] then f = gh in Fp[X].
So if f is irreducible over Fp and p ∤fn, then f is irreducible over Z.
□ 3.3. MODULO P 19 Example Let f = X4+3X3−2X2+X−17 in Z[X], so that in F3[X] : f = X4+X2+X+1, which is irreducible (see next proposition). We conclude that f is irreducible.
The certificate is 3.
To check a certificate we need a way to prove irreducibility in Fp[X]. For this purpose we have the next proposition: Proposition 3.23 A polynomial f ∈Fp[X] with n = deg(f) ≥1 is irreducible if and only if: (i) f | Xpn −X, and (ii) gcd(Xpn/t −X, f) = 1 for all prime divisors t of n.
Proof The proof is in [15, page 382]. We use the fact that X pd −X ∈Fp[X] is, for any d ≥1, the product of all monic irreducible polynomials in Fp[X] whose degree divides d. To prove this some algebra of finite fields is needed, like Fermat’s little theorem.
□ 3.3.2 Modulo-combinations In this subsection the ‘modulo p’-method will first be introduced, the reasons for the use of this method will be given, and we will take a look at the certificates of this method.
With certificate 3.20 we have seen that it is useful to factor our polynomial f modulo a prime p. (Remark: In Chapter 6 we prove that we only need certificates for monic polynomials. For these polynomials we automatically have that p does not divide the leading coefficient.) For this certificate we were only interested in primes p such that f modulo p is irreducible, but we can take advantage of (almost) any prime p.
We know that a factor of f remains a factor when working modulo p (and these factors may be reducible over Fp). Let Dp be the set of degrees of all factors (not necessarily irreducible) of f modulo p, and D the set of degrees of all factors of f (in Z[X]). Then we have that D ⊆Dp. This holds for any prime, so if p1, . . . , pk are all different primes then D ⊆Tk i=1 Dpi. So we have the next powerful certificate and criterion: Certificate 3.24 A set of natural numbers (p1, . . . , pk).
Criterion 3.25 There exist primes p1, . . . , pk such that Tk i=1 Dpi = {0, deg(f)}.
Example Let f = X5 −45X4 −261X3 −8404X2 −18078X −135764. For p = 3 we have that f = X5 + 2X2 + 1 = (X3 + 2X2 + 2X + 2)(X2 + X + 2), so that D3 = {0, 2, 3, 5}.
For p = 5 we have that f = X5 + 4X3 + X2 + 2X + 1 = (X + 1)(X4 + 4X3 + X + 1), so D5 = {0, 1, 4, 5}. Now D3 ∩D5 = {0, 5}, hence the primes 3 and 5 form an irreducibility certificate for f.
20 CHAPTER 3. THE CERTIFICATES Implementation I have implemented an algorithm based on the ‘modulo p’-method in the Computer Algebra System Magma .
The algorithm is available online at
It works with a related concept, that of partitions of the degrees of factors of f.
Definition 3.26 By Pp we shall denote the sequence of degrees of irreducible factors of f modulo p (in non-decreasing order, for uniqueness), so Pp is a partition of deg(f). Pp is called the decomposition type of f modulo p.
Example In the example above we have that P3 = [2, 3] and P5 = [1, 4]. For f = X5 we have that Pp = [1, 1, 1, 1, 1] for every prime p. So Pp has to be a tuple: it may contain the same number several times.
Definition 3.27 Let P1 and P2 be two partitions. We say that P1 is a parent of P2 if every element of P1 is a sum of elements of P2 and every element of P2 appears exactly once in these sums.
Example [1, 2, 3, 4] is a parent of [1, 1, 1, 2, 2, 3]: take 1+2 and 1+3 as 3 and 4. Or take 1+1 and 2+2 as 2 and 4.
And now we can formulate a certificate in the partition-style: Certificate 3.28 A set of natural numbers (p1, . . . , pk).
Criterion 3.29 Primes p1, . . . , pk such that the only common parent of all the Ppi is [deg(f)].
These partitions give us more information than all the degrees: Example We compare f = X3 and g = X3 + X. Then P3(f) = [1, 1, 1], P3(g) = [1, 2], as g = X(X2 + 1), so there is a difference. But D3(f) = D3(g) = {0, 1, 2, 3}. So the partition modulo 3 tells us that g has a factor of degree 2, while all the degrees do not give any information, because every degree is possible.
What we also know is that: (p1, . . . , pk) is an irreducibility certificate for f of type 3.28 if and only if it is an irreducibility certificate for f of type 3.24. This means that the partitions can only give more information when irreducibility can not yet be proved. The advantage of having more information will be used in the final algorithm, in Chapter 5.
Reasons for using this algorithm There are various reasons for using this algorithm: • We almost always get a certificate. This will be made precise in Chapter 4.
• Certificates are generated very fast. Test results will show this, see Chapter 6.
• We can make the certificates in such a way that they are easy to check.
We will treat that now.
3.3. MODULO P 21 3.3.3 Checking Easier-to-check Certificates Until now the certificates simply consisted of the primes for which the modulo-factorisations proved irreducibility. But in the process of computing these primes we of course compute these modulo-factorisations. And the certificate will be larger, but much easier to check if we incorporate these factorisations.
Example Let f = X5 −45X4 −261X3 −8404X2 −18078X −135764, as in the previous examples. The certificate was simply (3, 5), but now we add the factorisations, so that the certificate becomes ({3, (X3 + 2X2 + 2X + 2)(X2 + X + 2)}, {5, (X + 1)(X4 + 4X3 + X + 1)}).
Checking Certificates In proposition 3.23 we gave a simple way to prove irreducibility over a finite field. We use this to check a certificate.
When we get a certificate we do the following: 1. For every prime: check that it is prime. As we will see later primes will be quite small, so if we trust a small prime-list this check can be very fast.
2. For every prime p and the factorisation of f modulo p: check that it is a factorisation modulo the prime.
3. For every factor: check that it is irreducible using proposition 3.23.
4. Construct Dp for every prime and check that the intersection of all of them really is {0, deg(f)}.
The next chapter will give more information on how many factorisations we have to check, and whether we will always find a certificate or not. This will be accomplished by looking at the corresponding Galois theory.
22 CHAPTER 3. THE CERTIFICATES Chapter 4 Galois Theory and Modulo-Factorisations In the previous chapter we have seen several irreducibility certificates. We will take a closer look at certificate 3.28 which combines modulo p-factorisations. In this chapter that certificate is the only one that we will discuss.
In the first section we will see examples of irreducible polynomials that do not have such a certificate. Then we will discuss some Galois theory, this will make the situation clearer and afterwards we can use results from Galois theory to understand the existence of our certificate better.
4.1 Polynomials without Certificate There are irreducible polynomials that are reducible modulo every prime. This has first been observed by D. Hilbert .
The standard example of a polynomial which has this property is in the next example.
Example Let f = X4 + 1. Then f is irreducible, but reducible modulo every prime.
Proof f(X +1) = X4+4X3+6X2+4X +2, which is irreducible by Eisenstein’s criterion with p = 2. So f is irreducible over Z.
For p = 2 we have that f = (X + 1)4. For p odd there are three different cases: p ≡1 mod 4, then −1 is a square in Fp, so f = (X2−√−1)(X2+√−1) ∈Fp[X] p ≡7 mod 8, then 2 is a square in Fp, so f = (X2 + √ 2X + 1)(X2 − √ 2X + 1) p ≡3 mod 8, then −2 is a square in Fp, so f = (X2 + √−2X −1)(X2 −√−2X −1) So f is reducible modulo every prime p.
□ The information on squares in finite fields (using quadratic reciprocity) can be found in [1, page 181] for example.
23 24 CHAPTER 4. GALOIS THEORY AND MODULO-FACTORISATIONS With this example we see that we even have something stronger: the combination of modulo-factorisations will only tell us that if there exists a non-trivial factor then it must have degree 2.
As we are interested in finding certificates, we would also like to know for which polyno-mials no certificate can be found. In order to find some structure, we tried to find more polynomials of degree 4 that do not have a modulo p-certificate. Later we will see that degree 4 is the first interesting degree. We have found a big group of polynomials for which no modulo p-certificate can be found. There are other polynomials of degree 4 that do not have such a certificate, but we shall also see that they are related with the following group of polynomials: Examples Let a, b ∈Z. Then we have proved that f = X 4 + aX2 + b2 has a factor of degree 2 modulo every prime p.
Proof First we derive sufficient conditions.
Let p be a prime. From now on we work modulo p, so a, b are in Fp now.
Then we have to prove that there are numbers c, d, e and f in Fp such that X4 + aX2 + b2 = (X2 + cX + d)(X2 + eX + f) = X4 + (c + e)X3 + (d + f + ce)X2 + (cf + de)X + d f.
We take e = −c to have the term of degree 3 correct: X4 + aX2 + b2 = X4 + (d + f −c2)X2 + (cf −cd)X + d f.
Taking c = 0 or d = f fixes our linear term.
c = 0 : X4 + aX2 + b2 = X4 + (d + f)X2 + d f So we take f = a −d to get: X4 + aX2 + b2 = X4 + aX2 + (da −d2) and now we know: If there is d ∈Fp such that da −d2 = b2 then f has a factor of degree 2.
d = f : X4 + aX2 + b2 = X4 + (2d −c2)X2 + d2 Now we can either take d = b or d = −b: d = b : X4 + aX2 + b2 = X4 + (2b −c2)X2 + b2 What we need now is that 2b −c2 = a, thus 2b −a = c2.
In terms of the Legendre symbol (and taking a, b ∈Z): If (2b−a p ) = 1 then f has a factor of degree 2.
d = −b : X4 + aX2 + b2 = X4 + (−2b −c2)X2 + b2 What we need now is that −2b −c2 = a, thus −2b −a = c2.
In terms of the Legendre symbol (and taking a, b ∈Z): If (−2b−a p ) = 1 then f has a factor of degree 2.
Now we shall show that these found conditions can always be fulfilled, i.e.: For every prime p, for all a, b ∈Fp we have that (−2b−a p ) = 1 or ( 2b−a p ) = 1 or that there exists d ∈Fp such that da −d2 = b2.
For p = 2 this is trivial: if a = 0 then d = b does it.
Otherwise a = 1 (modulo p) and then ( 2b−a p ) = (2b−1 2 ) = (1 2) = 1.
4.2. GALOIS THEORY 25 So from now on p is odd.
If p | a : We must prove that ( 2b p ) = 1 or ( −2b p ) = 1 or ( −b2 p ) = 1.
If p ≡1 mod 4 we have that ( −1 p ) = 1, so then ( −b2 p ) = (−1 p )(b2 p ) = 1 · 1 = 1.
Otherwise we have that ( −1 p ) = −1, so ( 2b p ) = 1 or ( −2b p ) = 1, which had to be proved. If p | b : We can take d = 0 and are done.
So we can assume that p ∤a and p ∤b.
Suppose that ( 2b−a p ) = (−2b−a p ) = −1, then ( (2b−a)(−2b−a) p ) = 1. So we have c := p (2b −a)(−2b −a) ∈Fp. We can divide by 2 as p is odd.
So we have d := c+a 2 ∈Fp. Now da−d2 = ca+a2 2 −(c+a 2 )2 = 2ac+2a2 4 −c2+2ac+a2 4 = a2−c2 4 = a2−(2b−a)(−2b−a) 4 = a2+4b2−a2 4 = b2. So in this case we have a correct d.
The only remaining cases are ( 2b−a p ) = 0 and ( −2b−a p ) = 0. In the first case a = 2b in Fp and we can take d to be b so that da−d2 = ba−b2 = 2b2 −b2 = b2.
In the second case a = −2b in Fp and we can take d to be −b so that da −d2 = −ba −b2 = 2b2 −b2 = b2.
So we see now that f always has a factor of degree 2 modulo a prime p.
□ And now we have that every irreducible polynomial of the form f = X 4 + aX2 + b2 is reducible modulo every prime.
By trying more irreducible polynomials it seemed as if there were only examples of polyno-mials of composite degree that were reducible modulo every prime. This will become clear as we look at some Galois theory next.
4.2 Galois Theory Let f ∈Z[X] be primitive, monic and irreducible over Z, with degree n. Let G be the Galois group of the splitting field K of f over Q. Then each element σ of G (i.e. each automorphism of K) can be seen as a permutation of the n roots of f and in that way it has a unique decomposition into disjoint cycles, say of lengths λ1 ≤. . . ≤λr. Because λ1 + . . . + λr = n we have that λ = (λ1, . . . , λr) is a partition of n. We shall call λ the cycle type of σ.
Definition 4.1 Let λ be a partition of n, then we denote by Hλ(⊆G) the set of automor-phisms of K that have cycle type λ.
Definition 4.2 Again λ is a partition of n. By µ(λ) we denote the relative frequency with which the cycle type λ occurs in G. So µ(λ) = #Hλ/#G.
26 CHAPTER 4. GALOIS THEORY AND MODULO-FACTORISATIONS We have already seen partitions of n: when we discussed combining modulo-factorisations.
And now we have a beautiful theorem of Frobenius connecting these partitions: Theorem 4.3 (Frobenius) The density of the set of primes p for which f has a given decomposition type λ exists, and is equal to µ(λ).
The existence of the so called Frobenius substitution proves a stronger result, namely that all factorisations modulo a prime have their decomposition type equal to a cycle type that appears in G. Chebotar¨ ev has proved a stronger density theorem that uses this substitution. More information about Frobenius’ theorem and a proof of Chebotar¨ ev’s density theorem can be found in .
Example Let f = X4 + 1. Then f is the minimal polynomial of ζ8, so the splitting field of f is Q(ζ8) and we have the following diagram of the subfields: Q Q( √ 2) Q(√−2) Q(i) Q(ζ8) The Galois group of f is the Klein group V4. This group consists of four elements: the identity and three elements of cycle type (2, 2). So we now know that modulo one fourth of all primes f splits in four linear factors. And for all other primes f splits in two factors of degree 2.
And so we have now proved that there is no modulo p-certificate for f = X 4 + 1. In general to determine for which polynomials we can not find a certificate, it may be very helpful to consider Galois groups of polynomials.
But determining a Galois group of a polynomial is time-consuming and hard to check, so we will look at the situation in a very general way and use the Galois groups just to make general statements about the modulo p-certificates.
4.2.1 Which Galois groups can Occur?
We have already mentioned that G can be seen as a group permutating the roots of f, hence as a subgroup of Sn.
To talk about the Galois groups that can occur we need the following definition: Definition 4.4 A subgroup G of S(Z) is called transitive if for every ordered pair (x, y) of elements of Z there exists σ ∈G such that σ(x) = y.
4.3. RESULTS IN GALOIS THEORY 27 And we have the following proposition which is proved in [18, page 45], or can be found in any other standard text book in Galois theory.
Proposition 4.5 f is irreducible ⇔G is transitive.
So we see that the Galois group of an irreducible polynomial of degree n is a transitive subgroup of Sn. And our question has become more precise: which of these groups can not provide its corresponding polynomials with a certificate and how often / when do these groups occur? Results on these questions are treated in the next section.
4.3 Results in Galois Theory In this section we will deal with the following topics: 1. Polynomials with prime degree.
2. Polynomials with composite degree, especially 4 and 6.
3. Swinnerton-Dyer polynomials.
4. The Galois inverse problem.
5. The distribution of Galois groups.
4.3.1 Polynomials with Prime Degree Let f be a polynomial of prime degree q. Then the Galois group G of f acts transitively on the set of the q roots of f, and so G possesses an element σ of order q which acts cyclically on the roots of f in the splitting field K of f. So the cycle type of σ is (q) so there are infinitely many primes p for which the decomposition type of f modulo p is (q) as well, in other words: f is irreducible modulo these primes.
So we always get a certificate if f has prime degree!
4.3.2 Polynomials with Composite Degree In it is proved that for every n ∈N that is composite, there is an irreducible polynomial f of degree n that is reducible modulo every prime.
A sketch of the proof: A transitive soluble permutation group G of degree n, which does not possess an element of cycle type (n), is constructed. Then f ∈Z[X] is constructed such that G works transitively on the n roots of f. By the previous section we know that f then can never be irreducible modulo a prime.
It is possible that an irreducible polynomial is reducible modulo every prime but still has an irreducibility certificate, we shall see this for a polynomial of degree 4, with Galois group A4.
28 CHAPTER 4. GALOIS THEORY AND MODULO-FACTORISATIONS It seems likely that for every composite degree there probably are irreducible polynomials for which we can not find a certificate this way. We will now investigate the degrees 4 and 6.
Degree 4 Let f be an irreducible polynomial of degree 4. Then the Galois group G of f is a transitive subgroup of S4 as we have already seen. S4 has five transitive subgroups: S4, D4, A4, V4 and C4.
S4, D4 and C4 all contain an element of order 4, so in these cases there are primes modulo which f is irreducible, and we are guaranteed of a certificate. For example: if G ≃C4 then two elements are of order 4, one element has cycle type (2, 2) and the identity element has cycle type (1, 1, 1, 1). Hence in this case the set of primes modulo which f is irreducible has density 1 2 in the set of all primes.
If G ≃A4 then G contains 8 elements of cycle type (1, 3), 3 elements of cycle type (2, 2) and the identity element. As the only common parent of (1, 3) and (2, 2) is (4) we are sure to get a certificate.
But we have seen that there are irreducible polynomials that don’t have a certificate. And this is where Klein’s group V4 comes in. As it has just has the identity element and 3 elements with cycle type (2, 2) we see that we can’t get a certificate for a polynomial with this Galois group.
And now we can conclude that for modulo p-certificates: An irreducible polynomial of degree 4 has no certificate ⇔G ≃V4 Galois group V4 Let f have Galois group G ≃V4. We can then denote the 4 elements of G by e, a, b and c, with e the identity element and a2 = b2 = c2 = e. We then also have that ab = c. If we let K be the splitting field of f over Q, we have the following Galois diagrams: G ≃V4 < a > < b > < ab > < e > Q M1 M2 M3 K M1, M2 and M3 are quadratic intermediate fields of K : Q.
4.3. RESULTS IN GALOIS THEORY 29 As the intermediate fields are quadratic we know that there are d, e ∈Z, such that K can be written as Q( √ d, √e, √ de) = Q( √ d + √e). Now we determine the minimal polynomial of α := √ d + √e (which must be of degree 4): α2 = d + 2 √ de + e, so (α2 −(d + e))2 = 4de, thus α4 −2(d + e)α2 + (d + e)2 −4de = 0. As (d+e)2−4de = d2−2de+e2 = (d−e)2, the minimal polynomial is of the form X4+aX2+b2.
This is the form we found by trying to find polynomials without a certificate in Magma. We know that there are polynomials that have Galois group G ≃V4, that are not of this form; X4 + 4X3 + 5X2 + 2X + 1 for example. But we have now proved that the splitting field K of such a polynomial also is the splitting field of a polynomial of the form X 4 + aX2 + b2.
Degree 6 For degree 6 we can do something similar as for degree 4. There are 16 transitive subgroups of S6, of which seven do not have an element of cycle type (6). All seven do contain at least one element of cycle type (3, 3). Six out of seven in addition have an element with one of the following cycle types: (2, 4), (2, 2, 2) or (1, 5). And thus there is only one subgroup for which there are no certificates. This subgroup has order 12 and is generated by the elements (1, 4)(2, 5) and (1, 3, 5)(2, 4, 6). We shall denote this group by P6. There are polynomials which have P6 as Galois group, with Magma I have found f = X6 + 8X4 + 11X2 −4 for example.
Higher Degrees For both degrees 4 and 6 we have found just one Galois group for which no certificate can be found, but for higher degrees the total number of transitive subgroups grows quite hard and with that the number of groups for which there is no certificate grows. For example: for degree 9 there are 34 transitive subgroups of S9 of which 5 do not give a certificate and for degree 8 we already have 50 transitive subgroups of S8, of which 24 do not give a certificate. The question whether all subgroups occur as Galois groups will be discussed in subsection 4.3.4.
4.3.3 Swinnerton-Dyer Polynomials Definition 4.6 Let i ∈N∗. Then we define the ith Swinnerton-Dyer polynomial as f = Y (X ± √ 2 ± √ 3 ± √ 5 ± . . . ± √pi) ∈Z[X] where the product runs over all (2i) possible combinations of + and −signs, and where pi is the ith prime.
From Galois theory we know that f is irreducible. But for any prime p, Fp2 contains all the square roots modulo p, so √ 2 mod p, . . . , √pi mod p are all in Fp2. Hence f modulo p splits in linear factors over Fp2. And so we have that f modulo p splits in factors of degree ≤2 over Fp.
So in the best situation (to find a certificate) we have a prime p such that f splits in 2i−1 quadratic factors modulo p. And so for i ≥2 we can not find a certificate with our algorithm for the Swinnerton-Dyer polynomials.
30 CHAPTER 4. GALOIS THEORY AND MODULO-FACTORISATIONS 4.3.4 The Galois Inverse-problem We have now seen that the Galois group of an irreducible polynomial f is a transitive subgroup of Sdeg(f). A somewhat inverse question is: Let G be a transitive subgroup of Sn. Does there exist an irreducible polynomial for which the Galois group is G?
This is the so called Galois inverse-problem and until now it has not been solved. However it is conjectured to be true and for deg(f) ≤15 explicit realizations for every transitive subgroup have been constructed in . In Magma these polynomials form a special database; for every transitive group G (of order ≤15) one can ask a corresponding polynomial with the command PolynomialWithGaloisGroup.
4.3.5 Distribution of Galois groups So for the usage of our algorithm we will assume that every transitive group can occur as a Galois group. But what is known about the distribution of the Galois groups over the irreducible polynomials?
Well, in it is proved that almost every irreducible polynomial with integer coefficients has its Galois group equal to the full symmetric group. There are elements with cycle type (n) in Sn, so we can always find a certificate in this case. This is what happens in the article: Definition 4.7 By En(N) we denote the number of monic polynomials in Z[X] of degree n with all coefficients in absolute value ≤N that do not have Sn as their Galois group.
Definition 4.8 By Rn(N) we denote the number of reducible monic polynomials in Z[X] of degree n with all coefficients in absolute value ≤N.
Rn(N) can be estimated by ≪N n−1 and in the article cited above it is proved that En(N) ≪N n−1 2 (log N)1−γ where γ = γn > 0.
And so we see that almost every polynomial which does not have Sn as its Galois group, is reducible. In fact we have that the set of the irreducible polynomials that do not have the full symmetric group as Galois group has density zero in the set of all irreducible polynomials.
Having the full symmetric group as Galois group is not a necessary condition for finding certificates, so we can can conclude that almost every irreducible polynomial (namely the ones with Galois group Sn or another group that guarantees certificates) has a certificate.
Chapter 5 Modulo p: Improved In the previous chapter we have seen that there are irreducible polynomials f for which we can not find a certificate of the form 3.28 with the modulo p-algorithm.
In this chapter we shall give an algorithm that will provide a certificate in these cases. It is based on the modulo-factorisations that have been computed.
5.1 The Hensel-algorithm Suppose we have a polynomial f of which we think that it is irreducible (because that is what Magma tells us for example). Then we start looking for a certificate with the modulo p-algorithm. After trying some primes we still have not found a certificate. But at that point we do have some information: all the modulo-factorisations. An example to illustrate this and to show what to do now: Example Let f = X6 +8X4 +11X2 −4. In the previous chapter we have seen that there is no certificate for f. The factorisation of f modulo p = 3 is: f = (X 3+X2+2)(X3+2X2+1) ∈Fp[X], so we know that if f is reducible then it has a factor of degree 3. We know even more: if f is reducible, say f = gh then we can say (without loss of generality) that g = X3 + X2 + 2 and h = X3 + 2X2 + 1 ∈Fp[X].
Remark As the discriminant of f is 212314, 3 does not divide it. We can not use a prime here that divides the discriminant, because if p | disc(f) we would not have a squarefree factorisation of f modulo p. The guarantee of such a squarefree factorisation is needed for Hensel-lifting the factorisation. This will be treated later in this section.
We can now use a bound on the coefficients of possible factors by Mignotte . To state a theorem based on this bound [8, Section 3.5.1], we first need a definition: Definition 5.1 For f we define |f| = (Pn i=0 |fi|2)1/2.
So in our case |f| = √1 + 64 + 121 + 16 = √ 202.
31 32 CHAPTER 5. MODULO P: IMPROVED The theorem that bounds the coefficients of a polynomial that divides f is the following: Theorem 5.2 If g = Xk + . . . + g1X + g0 ∈Z[X] divides f then we have for all j that |gj| ≤ k −1 j |f| + k −1 j −1 |fn|.
So in our case we have a possible monic factor of degree 3, say g = X 3 + g2X2 + g1X + g0.
Then the theorem tells us that: |g2| ≤ 2 2 |f| + 2 1 < 17 |g1| ≤ 2 1 |f| + 2 0 < 30 |g0| ≤|f| + 1 < 16 So the coefficients of g must all be < 30 in absolute value. We call this bound on the maximum of the absolute values of the coefficient the Mignotte bound and denote it by B.
What we do now is Hensel-lift our factorisation in Fp[X] to a power of p that is greater than 2B. (More information on Hensel-lifting, and the proof of uniqueness can be found in [15, Section 15.4].) We now know that g = g, as all coefficients must be less than B, in absolute value.
In our case we lift the factorisation into F81[X]. The factorisation then becomes: f = (X3 −14X2 + 21X + 2)(X3 + 14X2 + 21X −2). And now we know that if f is reducible then its factors have to be in this factorisation. But we can easily see that these factors do not divide f and so we are certain that f is irreducible!
In the unlikely case that we would have found a factor (for instance because of a bug in the system), we now would have known that f is reducible! So this algorithm actually prevents us from errors.
5.2 The Hensel-certificate Certificate 5.3 q ∈N Criterion 5.4 There exists a prime power q that is greater than two times the Mignotte bound, and such that all factors of f modulo q do not divide f.
The proof that this criterion guarantees irreducibility of f has been given above.
5.3. THE IMPROVED ALGORITHM: COMBINING 3.28 AND 5.3 33 5.3 The Improved Algorithm: Combining 3.28 and 5.3 Let deg(f) = n. To find one of these two certificates for f we do the following: • If n is prime then we will find a modulo p-certificate, so we start searching and know by the theory of Chapter 4 that we will find such a certificate.
• Otherwise n is composite and this will be our algorithm: 1. Try to find a modulo p-certificate, with the primes ≤30 · n, while keeping the modulo-factorisations in memory. The bound has been found by testing.
2. If we have a certificate now, we are done. Otherwise we pick out a prime for which f has the fewest factors modulo p.
3. Use p and the factorisation of f modulo p to find a Hensel-certificate (which can always be found). And we are done.
We now have an algorithm that can output two kinds of certificates. But we can easily decide which of the two we have, as will be discussed next.
5.4 Checking If the certificate consists of more than one number we know that it is a modulo p-certificate (3.28). Otherwise, if the one number is not prime we know that we have a Hensel-certificate (5.3) and if the number is prime we compute the factorisation of f modulo it. If f is irreducible we are done and otherwise we know we have a Hensel-certificate.
If our modulo p-algorithm works, we now have an upper bound on the prime numbers, so the checking (3.3.3) can be done really quick now.
Otherwise checking the ‘Hensel’-certificate means: 1. Compute f’s Mignotte bound B and check that q > 2B.
2. Compute the factorisation F of f modulo q.
3. Check the irreducibility of the factors found, over Fp, with proposition 3.23.
4. Check that all possible factors are not factors.
In fact we also need to check that we have tried all possible factors, but this should be very clear from the algorithm used.
34 CHAPTER 5. MODULO P: IMPROVED 5.5 Worst Case Scenario We now know that we most of the time will get a modulo p-certificate, and else we probably have a modulo-factorisation with not so many factors. But the Swinnerton-Dyer polynomials are the worst polynomials that can be encountered: as we have seen the factors of their factorisation modulo a prime always have degree ≤2. And then we have a lot of factors to check.
Example Let f be the 4th Swinnerton-Dyer polynomial, which is of degree 24 = 16.
Suppose f splits in quadratic factors modulo some prime. Then we have 8 factors that we Hensel-lift. Then we first have to check if any of these lifted factors divide f, next all combinations of 2 and 3 factors and then half of all possible combinations of 4 factors (this is because if we have checked that such a combination is not a factor, the complementary combination can also not be a factor!). In total these are 8 1 + 8 2 + 8 3 + 1 2 8 4 = 8+28+56+ 35 = 127 possible factors, and it is clear that for larger Swinnerton-Dyer polynomials this number grows exponentially in the degree of f. For the 5th Swinnerton-Dyer polynomial we have to check 32, 767 possible factors, for example.
Chapter 6 Test Results Before we give the test results, we first state some remarks.
6.1 Remarks Polynomials we have to Certify Remark We only need certificates for monic polynomials, because we can check irre-ducibility of a non-monic polynomial by that of a monic one.
Proof Let f = fnXn + fn−1Xn−1 + . . . + f1X + f0 ∈Z[X], with n ≥2, fn ̸= 0 and gcd(f0, . . . , fn) = 1. Suppose that f is reducible, say f = gh, with deg(g) = k ≥1, deg(h) = n −k ≥1.
Claim: The monic polynomial f n−1 n f( X fn ) is reducible.
Proof fn = gkhn−k, so f n−1 n f( X fn ) = gkhn−kf n−2 n g( X fn )h( X fn ) = (f k−1 n hn−k(g( X fn ))) (f n−k−1 n gk(h( X fn ))).
□ So the irreducibility of this monic polynomial proves the irreducibility of f. □ The Modulo p-certificate In Chapter 3 we have seen Eisenstein’s criterion and we have seen that for this criterion it sometimes works to look at f(X + a) for some a ∈Z. This will not work for the mod-ulo p-algorithm, as the Galois groups corresponding to f(X) and f(X +a) are the same. So: Remark If there does not exist a certificate for f(X) then neither does there exist one for f(X + a).
Now let n ∈N∗, f = X2 + n!. Then f is irreducible in Z[X], as √ −n! / ∈Z. But we also see that for p ≤n : f = X2 ∈Fp[X], so a certificate can only be found if p > n. With the theory of Chapter 4 we know that such a certificate will be found. And so we can conclude with the final remark: Remark There is no m ∈N such that all modulo p-certificates contain only primes < m.
35 36 CHAPTER 6. TEST RESULTS 6.2 Testing For this test I have implemented the following certificates in Magma: The improved modulo p-certificate (3.28 combined with 5.3).
Two certificates based on Bunyakovsky’s conjecture: certificate 3.12 and Brillhart’s certificate (3.15).
The extended certificate of Eisenstein (3.4).
Some of the following polynomials are said to have random coefficients. By this I mean: Let f be of degree n. Then the coefficient of the term of degree i is a random number which is in absolute value ≤10n+1−i.
All tested polynomials are monic (because of the first remark in this chapter), and denounced irreducible by Magma.
We try to find certificates for the following groups of polynomials: (A3) All monic polynomials of degree 3 with coefficients in [−5, . . . , 5].
(R4) A set of 100 polynomials of degree 4 with random coefficients.
(S8) A set of 100 polynomials of degree 8 with coefficients in [−100, . . . , 100] (random).
(R8) A set of 100 polynomials of degree 8 with random coefficients.
(R17) A set of 100 polynomials of degree 17 with random coefficients.
(S100) A set of 10 polynomials of degree 100 with coefficients in [−100, . . . , 100] (random).
(R100) A set of 10 polynomials of degree 100 with random coefficients.
We will now look at the performance of the algorithms on all these sets of polynomials.
6.2.1 Modulo p-algorithm The test results are in the following table. By the average number of primes we mean the average number of primes in a certificate. The smaller the primes in a certificate are, the easier we can check this certificate. So we also give the largest prime that occurs in the certificates, as well as the average (rounded of to one decimal) of all the primes occuring.
Set of polynomials A3 R4 S8 R8 R17 S100 R100 Average number of primes 1 1.3 1.9 2.0 2.8 3.6 3.7 Largest prime 41 29 37 31 43 31 13 Average of all primes 3.9 5.3 5.5 5.7 5.9 5.4 4.9 For every polynomial we found a modulo p-certificate, so no Hensel-lifting was needed. The primes are all very small (≤43), so the certificates can be checked easy. This also means that we can find these certificates very fast.
Conclusion The modulo p-algorithm has a good performance on all sets of polynomials. Certificates are found quickly and can be checked easily.
6.2. TESTING 37 6.2.2 Algorithms based on Bunyakovsky’s Conjecture Brillhart’s certificate 3.15 is in this test always easier to check than certificate 3.12. This is because for every tested polynomial f, the certificate 3.12 contains at least one prime evaluation that is greater than the prime evaluation in Brillhart’s certificate.
Because of its better performance we only give the test results for the algorithm that assigns Brillhart’s certificate (if this is possible).
For all polynomials of degree ≤17 such a certificate is found. As a reminder: a certificate now consists of a number x0 ∈Z. The smaller the absolute value of this x0 is, the sooner we find it. These are the test results on the x0’s found: Set of polynomials A3 R4 S8 R8 R17 Largest x0 in absolute value 20 283 204 202 502 Average absolute value of x0 5.5 61 68 69 96 The primes that we find seem to grow exponentially in the degree of the polynomial, as we can see in the test results on the primes: Set Average of all primes Largest prime A3 246 6347 R4 115322138 5629706083 S8 80602014447841089 4404810068836754723 R8 37703978841734883 1893697149435378341 R17 115850744758137640950336926116401219582360968 a 46-digit prime S100 − a 238-digit prime We have used our algorithm for one polynomial of degree 100, and it took more than 1000 seconds of processor time in Magma to find the 238-digit prime. So for large polynomials this certificate is not practical to find or check.
Conclusion For polynomials of low degree certificates are found quickly. But already for polynomials of degree 4 the primes are quite large. So if we trust the primality check in Magma then this algorithm gives good certificates in a fast way for not too large polynomials.
6.2.3 Eisenstein For the set A3 we find a certificate for 266 of the 1002 polynomials (taking |a| ≤100). For the set R4 we find a certificate for 3 out of 100 polynomials (also taking |a| ≤100). For all other sets we do not find certificates, even for large a. For large a we have the disadvantage that the coefficients of f(X + a) become huge, so that trying to find a criterion takes a lot of processor time.
38 CHAPTER 6. TEST RESULTS Conclusion If a polynomial is in the right form, then the Eisenstein-criterion works very well.
But for larger polynomials, it is not practical to try to find a certificate, as there are too few certificates to be found in this way.
6.2.4 A Swinnerton-Dyer Polynomial For all polynomials that were tested a modulo p-certificate could be found. So for these polynomials we have not needed the Hensel-part of our algorithm. To check the performance of the Hensel-part we have tried the algorithms on the 4th Swinnerton-Dyer polynomial (of degree 16), denoted here by f.
With Eisenstein, for |a| ≤100, no certificate can be found.
Brillhart’s certificate is found in 0.4 seconds processor time. It consists of x0 = 108.
f(108) = 338615958064361660344430273649169, a 33-digit prime number.
When we use the modulo p-method we do not find a suitable combination for p < 480, so we start Hensel-lifting and find q = 711 as our Hensel-certificate. This is computed in 2.08 seconds processor time.
So for this f a Brillhart-certificate is found faster, but to check it will take longer than checking our Hensel-certificate. So even in this case we prefer the modulo p-method.
6.2.5 Final Conclusion For low degree the certificates of Eisenstein or Brillhart can be faster, and in this case the Brillhart-certificate can be checked fast (as the average prime number found is low). If an Eisenstein certificate is found, it can always be checked fast. But the modulo p-method also gives good certificates quickly.
For the modulo p-algorithm we see from our tests that the prime numbers found in these certificates are in general very low (all were ≤43), the number of primes in a certificate increases slightly for higher degrees. So the conclusion is that this is a very good method, that almost always gives easy-to-check certificates.
Chapter 7 Formalizing Mathematics In this chapter I will introduce the notion of formalizing mathematics. A program for doing this on a computer, the proof assistant Coq , will be used to give examples. For this thesis I have formally proved the correctness of theorem 3.22 in Coq, which will be discussed next.
Eventually, I will say something about combining the Magma and Coq systems. Such a combination of systems would allow us to rigorously prove irreducibility of polynomials using a computer.
7.1 Introduction 7.1.1 What is formalizing?
Normally, mathematics (like the theorems proved in this thesis) is informal. This means that it is written in a sort of natural human language, and that not every small detail of a proof is completely worked out. We can formalize mathematics by formulating everything in a logical system. A proof then only consists of the reasoning steps allowed by this system, and it has to be complete, in order to be a proof.
The process of checking a proof can be automated if it is formal: all allowed reasoning steps are known and so a proof can be checked step by step. A program in which mathematics is formalized is called a Proof Assistant (PA). In a PA the human user enters the mathematical notions, theorems and proofs, while the program checks if all details are correct. This can be compared to a word processor in which a spell checker is used to check the correctness of the single words.
For checking large pieces of software and hardware these PAs are used in the industry.
A problem for mathematicians with these industrial programs is that they contain a large kernel of axioms and deduction rules. Hence there is a reasonable chance of having conflict-ing axioms, which has actually happened in such systems. So if we have checked software we can be quite sure of its correctness, but not completely, because of possible errors in the checking software. The great advantage of such a system is its great performance, which is of course needed in the industry.
39 40 CHAPTER 7. FORMALIZING MATHEMATICS We, as theoretical mathematicians, are of course more interested in the correctness of our proofs. For this purpose PAs with a small proof kernel are used. Some years ago the practice was to create complete formalized proofs (by hand) and then let the PA check it.
Most recent systems are interactive, in the following sense: the user states a lemma, and proves it step by step. The PA checks all these steps.
We can be sure of the correctness of a PA such as Coq, as its proof kernel is small. This is called the De Bruijn criterion for proof checkers, called after prof. de Bruijn who was one of the first to develop a proof checker for mathematics on a computer in the 1960s, the Automath system .
Of course a very fundamental problem is that we can never be totally sure of anything, but if we accept the (few) simple rules in the proof kernel, then we know that all formalized proofs are correct, as they consist only of cases of these rules. In Coq, the program I have used, this kernel consists of deduction rules for a so called type theory: a formal language in which mathematics can be expressed.
I shall finish this subsection by stating advantages of formalizing mathematics (using a PA): • The proofs are correct.
• The mathematics already formalized has a high accessibility. Therefore it is practical to formalize more mathematical theory by building on what has been formalized.
• It is a good way of sharing mathematics. There are also programs to represent the formalized mathematics in a nice way.
I will give examples of formalizing mathematics in Coq in the next section, but first we look at formalizing in a wider perspective: Computer Mathematics.
7.1.2 Computer Mathematics With Computer Mathematics, we mean the activities performed on a computer by mathe-maticians, these are: Presentation: This thesis, for instance, is typed in L A T EX. The typesetting of mathematics and mathematical symbols is an important activity that is done with a computer.
Computation: The computer can be used as a very powerful calculator, but in the field of Computer Algebra (CA) the computer is also used to perform symbolic computations.
This means that we can for example calculate in an exact way with algebraic numbers.
For instance, √ 2 is kept in memory by its minimal polynomial, X 2 −2, and by doing so we obtain exact answers, we do not round off.
Proving: Formalizing mathematics. We have discussed this in the previous subsection.
7.2. COQ AND A CRITERION 41 The very longterm goal of the field of Computer Mathematics is to create the ideal mathematical workspace. This is a computer system in which mathematicians can perform all of their mathematical work. That includes defining new notions, perform (symbolic) computations, proving and publishing. In this workspace it would be as easy as doing mathematics on paper, but then with all the help of a computer.
If I for instance would write this thesis in such a workspace, I could compute and prove in the document that I am writing. It is obvious that such a workspace has not yet been realized, but it is clear that such a system has advantages for mathematicians.
7.2 Coq and a Criterion 7.2.1 Introducting Coq with Examples All Coq-commands are typed in ‘this style’. In Coq there is a universe for datatypes and a universe for propositions. We start with some propositional logic: Let A and B be propositions. This is denoted by ‘Variables A B : Prop.’ in Coq. Then we can state the following lemma: ‘Lemma triv1 : (A ∧B) -> B.’ When we load this lemma the interactive proving-session starts and we are in the following environment, where we have to find a proof of (A∧B) →B: 1 subgoal A : Prop B : Prop -----------(A ∧B) -> B So the goal is under the line here, and the variables are above the line. We now need a proof of (A ∧B) →B, to do this we can give a proof of B using a proof of A ∧B. This is done with the ‘intro’ command. We want to give this proof of A ∧B the name HAB (as it becomes a Hypothesis in the proof), so we give the command ‘intro HAB.’ and get: 1 subgoal A : Prop B : Prop HAB : A ∧B -----------B 42 CHAPTER 7. FORMALIZING MATHEMATICS We want to use the proof HAB to obtain a proof of A and a proof of B. To do so we eliminate HAB by ‘elim HAB.’. By doing this we still have the proof of A ∧B; we now get a proof of A and a proof of B as well. So we have: 1 subgoal A : Prop B : Prop HAB : A ∧B -----------A -> B -> B Now we can obtain the proofs HA for A and HB for B by ‘intros HA HB.’: 1 subgoal A : Prop B : Prop HAB : A ∧B HA : A HB : B -----------B Now we have a proof of B, namely HB and this is what we need. So we can state ‘exact HB’ and the proof is completed.
During the interactive process we have gone through, Coq has created a complete proof object for our lemma in the type theory-language. The check of this proof can now be done by the small type-checking part, and this is done by giving the command ‘Qed.’. The proof object we have can now be checked by any type-checking program, so anyone interested in the correctness of our proof can test it with his or her favorite type-checking program.
The lemma we have just proved is trivial to us, but it took some steps to obtain a formal proof. As mathematicians, we really want Coq to be able to figure out such a simple, straight-forward proof by itself, by trying some rules.
For this purpose tactics have been written. They combine several commands and try to find a proof. In the case of our lemma using the simple tactic ‘trivial.’ finds a proof at once.
We shall now first define some mathematical objects, functions and properties of objects in Coq. This is done to give an impression of what is possible in Coq, and to show that the language is quite readable to mathematicians.
7.2. COQ AND A CRITERION 43 We have the set nat of natural numbers in Coq. They are defined inductively as follows: Inductive nat : Set := | O : nat | S : nat -> nat.
So ‘nat’ consists of elements of the form ‘(S (S ...(S 0)...))’. As we have given no meaning to members of the set ‘nat’, every element is automatically different from any other. This implies that there are infinitely many different elements in ‘nat’. The definition is inductive, hence all elements of ‘nat’ are of the form ‘(S (S ...(S 0)...))’.
We can think of ‘0’ as the number zero and of ‘S’ as the successor function.
Then ‘(S (S (S 0)))’ represents the number three for example.
Next functions on ‘nat’ are defined in Coq. Additition and multiplication can for instance be defined in a recursive way. The notation is improved as well, so that Coq can represent ‘(S (S (S (S 0))))’ by ‘4’, for example. We take a look at some definitions based on ‘nat’: The property that n in nat is even can be defined in the following natural way: Definition IsEven (n:nat) := exists k:nat, n = 2k.
We can now prove (IsEven 6) by explicitly giving k = 3. To have a better check for even numbers we could also prove formally that the definition is equivalent with: the last decimal digit is even. This would require defining more mathematical notions, but it would afterwards be easy to prove a number even.
This subsection is concluded with the definition of a function: n 7→2n: Definition Double (n:nat):nat := 2n.
By the second :nat we mean that the outcome (2n) is in nat. If this would not be clear to Coq we would have to prove this to finish the definition.
7.2.2 Formalizing Theorem 3.22 My objective was to formally prove theorem 3.22: If f is irreducible in Fp[X], then f is irreducible in Z[X].
To prove this the mathematical notions that are needed have to be formalized first.
These are: finite fields of prime characteristic (Fp), polynomials and some definitions like ‘(irreducible f)’.
44 CHAPTER 7. FORMALIZING MATHEMATICS For the formalisation I have used the C-CoRN-library1 [10, 12] for Coq, developed by the Foundations Group of Computer Science of the University of Nijmegen.
Polynomials have been formalized in C-CoRN. Rings have been formalized as well and the set of all rings is denoted by ‘Ring’. For a ring R (‘Variable R : Ring.’) we define a polynomial with coefficients in R as: Inductive poly : Type := | poly zero : poly | poly linear : R -> poly -> poly.
So we have the zero polynomial (‘poly zero’) and polynomials of the form c + X ∗f (‘(poly linear c f)’) with c in R, and f a polynomial. In C-CoRN, there are also lemmas for working with polynomials, and definitions of degrees and coefficients for example.
The notion of a finite field of prime characteristic, Fp, has been formalized by Vince Barany.
In his files he defines the notions of the greatest common divisor, primality of integers, and modular arithmetic and ultimately proves that Fp is a field for every prime p. He also states and proves lots of lemma’s. These files are now online on the C-CoRN-website .
Polynomials and Fp thus have been formalized and can be used. For R, we have the formal polynomial ring over R by (poly ring R).
We now define ‘Variable p : positive.’, a positive integer, and state ‘Hypothesis Hpprime : (IsPrime p).’, so that p is prime.
Now we can define Fp, Fp[X] and Z[X] by Definition fp := (Fp p Hprime).
Definition fpx := (poly ring fp).
Definition zx := (poly ring Z as Ring).
Lots of lemmas had to be proved, an example: Lemma fpx resp coef : forall (f:zx)(n:nat), (zfp (nth coeff n f)) [=] (nth coeff n (zxfpx f)).
This lemma states that for f ∈Z[X], the n-th coefficient modulo p of f equals the n-th coefficient of f modulo p. Because much has been proved already, the proof is quite short: induction f.
intuition.
induction n.
intuition.
astepl (zfp (nth coeff n f)).
astepr (nth coeff n (zxfpx f)).
apply (IHf n).
Qed.
1Most definitions in C-CoRN start with a ‘c’, for reasons that are not of interest here. For readibility we drop this ‘c’ in the definitions in this chapter.
7.3. COMBINATION WITH MAGMA 45 We use induction on f and n and later use the induction hypothesis for f, IHf. The astepl and astepr commands rewrite the left and right side of an equation, respectively.
And finally the theorem was stated and proved: Theorem irrcrit : forall f:zx, (irreducible fp (zxfpx f)) -> (irreducible f).
The complete formalisation is available online at
7.3 Combination with Magma As we have seen, formalized proofs are certainly correct. But computation in a PA is very slow. When we multiply in nat for example, we have to make a lot of calculation steps as all these natural numbers are of the form ‘(S (S ...(S 0)...))’. On the other hand we have Computer Algebra Systems (CASs), such as Magma, that can compute very fast, but that may contain bugs. So it would be very nice to be able to combine these two types of systems, in our case Coq and Magma. This combination is possible in several ways: • In subsection 7.1.2 we have already mentioned the ideal mathematical workspace.
In such a workspace both systems are embedded and equally important.
• We could also build in a small theorem prover in Magma, for proving the correctness of algorithms.
• And finally we can use Magma to help Coq with its proofs.
These last two combinations are, of course, easier to realize than the workspace. And as there will probably always be specialized CASs for specific areas of mathematics, the last kind of combination will always be of interest.
For us this last combination gives a way to formally prove irreducibility, using certificates: Suppose that we want to prove (irreducible f) in Coq, for f in Z[X]. We have seen that certificates can help us. But finding a certificate is difficult, especially in a PA. So this is why we want Magma to find it. In Coq we have to prove the theorem: ‘If there exists a certificate, then f is irreducible.’, so that if Magma gives us a certificate, Coq only has to verify the correctness of this certificate.
An example: we have proved theorem 3.22. So to prove the irreducibility of f over Z it suffices to prove that f is irreducible over Fp for some prime p. Magma can help us find this p, so that we only have to check that p really is prime and that f ∈Fp[X] is irreducible.
For this last fact a formalisation of proposition 3.23 can be used. Now Magma can again be of assistance by verifying the conditions of that proposition. For example: we have to check that f | Xpn −X. To help us Magma can perform the division and return g such that f · g = Xpn −X. Now Coq can check this multiplication (this is easy) and in this way it is proved that f indeed divides Xpn −X.
46 CHAPTER 7. FORMALIZING MATHEMATICS And so we see that to prove irreducibility with certificate 3.20 we also need a formalisation of proposition 3.23, and a program that can communicate the outcome of a computation in Magma to Coq.
The program Maplemode can be used to let Maple compute for Coq. Bas Spitters and I have made some adaptations to Maplemode, with the help of Dan Synek, such that we could let Magma do some computations for Coq. The factorizing of polynomials is not yet implemented, as more theory has to be formalized for this.
So the combination of Magma and Coq has been initiated, but there is more work to do before we can formally prove irreducibility in Coq, using computations in Magma.
7.4 Conclusions The proof assistant Coq was introduced, and used to formally prove theorem 3.22.
I have also shown how Coq can be used to formally prove the irreducibility of polynomials in Z[X], using certificates and the computer algebra system Magma for finding certificates.
At this moment, formalizing mathematics is practiced by few mathematicians. One of the reason is that formalizing is not a ‘mathematician-friendly’ activity. Some problems are: • The formalized proofs must be given in full detail. Tactics have improved this, but it still is a long road from an informal to a formal proof.
• The mathematics that has been formalized is not very advanced. There are not many areas of mathematics of which recent results can be easily formalized. In the C-CoRN-project a constructive proof of the fundamental theorem of algebra (FTA) has been formalized, but it took a full year to formalize all mathematics involved and prove everything.
• Formalizing is still disconnected from the other Computer Mathematics-activities.
The ideal workspace has not been realized, but there is more and more interaction between different systems.
Combining different systems is the pioneering work for the workspace, so it will be of use to formalize more certificates and to actually prove the irreducibility of (large) polynomials using a Proof Assistant.
Chapter 8 Samenvatting Deze scriptie heeft als titel ‘Irreducibiliteitscertificaten voor veeltermen met gehele co¨ efficienten’. Wat hiermee bedoeld wordt vertel ik zo, maar ik zal eerst een andere (meer wereldse) situatie beschrijven, die de sfeer van deze scriptie duidelijk zal maken.
Stel je voor dat je een lot koopt bij het postkantoor. De loting is een beetje raar, want de trekking blijft voor je verborgen. Na de trekking wil je weten of je iets gewonnen hebt en je gaat dus weer naar het postkantoor. Daar word je botweg verteld dat je geen prijs hebt.
Tja, wat doe je dan? Aangezien er geen trekkingslijst is kun je moeilijk in protest gaan, maar je hebt geen enkele goede reden om te geloven dat je ´ echt geen prijs hebt. . .
In deze scriptie nemen veeltermen met gehele co¨ efficienten (vanaf nu noem ik ze gewoon veeltermen) de plaats van de loten in. De vraag is of zo’n veelterm reducibel is, ofwel of je lot een prijs oplevert. Als je prijs hebt krijg je die en ben je tevreden. Net zo krijgen we voor een reducibele veelterm een niet-triviale factor.
Als een veelterm geen niet-triviale factoren heeft noemen we de veelterm irreducibel, niet reducibel dus. (Het komt dus overeen met een lot dat geen prijs oplevert.) De vraag is nu hoe we er zeker van kunnen zijn dat een veelterm irreducibel is. Bij de loten is het makkelijk: als je maar de trekkingslijst hebt, kun je nagaan of je een prijs hebt. Maar een probleem voor de veeltermen is, dat er daar oneindig veel van zijn, en daarvan zijn er ook weer oneindig veel reducibel en oneindig veel irreducibel. Dus een lijst opstellen van reducibele veeltermen heeft geen zin.
Wat er in deze scriptie eigenlijk gebeurt is dat ik naar speciale eigenschappen van irreducibele veeltermen ga kijken. Voorbeeldje bij de loten: Stel dat je weet dat alle loten die in de prijzen zijn gevallen eindigen op 37. Als je nu een lot hebt dat eindigt op 13 heb je pech, maar je hebt dan wel de zekerheid dat je geen prijs hebt. Zo een speciale eigenschap, waarmee je zeker weet dat je geen prijs hebt (in dit geval het eindigen op 13) noemen we nu een Geen-Prijs-Bewijs, ofwel een GPB. Voor irreducibele veeltermen heet dit een irreducibiliteitscertificaat.
47 48 CHAPTER 8. SAMENVATTING In deze scriptie gebeurt het volgende: Eerst maak ik duidelijk wat ik precies bedoel met een irreducibiliteitscertificaat en geef er een paar voorbeelden van, ongeveer zoals hierboven, maar dan wat moeilijker en uitgebreider. Vervolgens ga ik ´ e´ en soort certificaat beter bestuderen, met behulp van Galoistheorie. In de lotenwereld was Galois degene die als eerste precies kon zeggen hoe de loten gemaakt worden en de loting tot stand komt. Zo kom ik er achter voor hoeveel loten, zonder prijs, we een GPB kunnen vinden. En ik vind ook grote groepen loten waarvoor ik zeker weet dat ik een GPB kan vinden als ze geen prijs hebben.
Daarna breid ik de GPBs die ik onderzocht heb uit, zodat ik voor ´ elk lot dat niet in de prijzen valt een GPB kan vinden. Door uitgebreid te testen laat ik daarna zien dat deze GPBs eenvoudiger te controleren zijn dan een aantal andere soorten GPBs. De conclusie is dus dat ik een goede GPB heb gemaakt, dat voor elk lot zonder prijs gevonden kan worden.
Hoofdstuk 7 zal ik zonder loten proberen uit te leggen. Dat hoofdstuk gaat over het formaliseren van wiskunde in een computer. Hiermee bedoel ik, dat ik samen met een com-puter formele, correcte bewijzen ga geven. Ik heb van een irreducibiliteitscertificaat formeel bewezen, met een computer dus, dat het de irreducibiliteit van een veelterm impliceert. De systemen om wiskunde te formaliseren staan nog enigszins in de kinderschoenen, maar ik denk dat ze steeds vaker gebruikt zullen worden.
Nu, zoals beloofd, een uitleg van de titel. Een veelterm is een functie van de vorm fnXn + fn−1Xn−1 + . . . + f1X + f0. Een voorbeeld is X4 + 3X3 −6X2 + 8.
De co¨ efficienten zijn de getallen voor de verschillende X-machten, hier dus 1, 3, −6, 0 en 8.
De graad is de hoogste macht van X, hier dus 4.
We bekijken veeltermen met gehele co¨ efficienten, dus 1 2X2 doet bijvoorbeeld niet mee.
Je kunt met veeltermen ongeveer rekenen zoals met getallen, zo hebben we bijvoorbeeld (X2 + 7X −6) + (3X2 −8X + 1) = 4X2 −X −5 en (X2 + 2X + 3) × (3X2 + 5X −7) = 3X4 + 11X3 + 12X2 + X −21. Het rekenen met veeltermen wordt voor een deel op de middelbare school geleerd.
Met de vermenigvuldiging van net zien we dat X 2 + 2X + 3 een deler van 3X4 + 11X3 + 12X2 + X −21 is. Zo’n deler noemen we nu een factor. We noemen een factor niet-triviaal als de graad 1 of hoger is, oftewel: als er een X in voorkomt. Als een veelterm geen niet-triviale factoren heeft, noemen we die veelterm irreducibel (dat betekent: niet te ontbinden).
Ik hoop hiermee het onderwerp van mijn scriptie begrijpelijk uit te hebben gelegd.
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Springer-Verlag (1993) The Coq Development Team. The Coq Proof Assistent Reference Manual, Version 8.0, available online: c ⃝INRIA (2004) Lu´ ıs Cruz-Filipe, Herman Geuvers and Freek Wiedijk. C-Corn: the Constructive Coq Repository at Nijmegen, to appear in MKM 2004, Springer-Verlag (2004) David Delahaye and Micaela Mayero. A Maple Mode for Coq, available online at (2002) Foundations Group, Computer Science, University of Nijmegen. Constructive Coq Repository at Nijmegen. Available online at
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H.C. Pocklington. The Determination of the Prime or Composite Nature of Large Numbers by Fermat’s Theorem, in Proc. Cambridge Phil. Soc. 18, pp. 29–30, (1914) V.R. Pratt. Every prime has a succinct certificate, in SIAM J. Comput. Vol. 4, pp. 214–220 (1974) T. Sch¨ onemann. Grundz¨ uge einer allgemeinen Theorie der h¨ ohern Congruenzen, deren Modul eine reelle Primzahl ist, in J. reine angew. Math. 31, pp. 269–325 (1846) P. Stevenhagen and H.W. Lenstra, Jr. Chebotar¨ ev and his Density Theorem, in The Mathematical Intelligencer, Vol. 18, No. 2, pp. 26–37 (1996) P.J. Weinberger. Finding the Number of Factors of a Polynomial, in Journal of Algorithms, 5, pp. 180–186 (1984) |
7379 | https://www.cuemath.com/algebra/interval-notation/ | LearnPracticeDownload
Interval Notation
Interval notation is a method to represent an interval on a number line. In other words, it is a way of writing subsets of the real number line. An interval comprises the numbers lying between two specific given numbers. For example, the set of numbers x satisfying 0 ≤ x ≤ 5 is an interval that contains 0, 5, and all numbers between 0 and 5.
Let us understand the interval notation and different types of intervals in detail using solved examples.
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| 1. | What is Interval Notation? |
| 2. | Different Types Of Intervals |
| 3. | Notations For Different Types of Intervals |
| 4. | Number Line Representation of Different Types of Intervals |
| 5. | Converting Inequality To Interval Notation |
| 6. | FAQs on Interval Notation |
What is Interval Notation?
Interval Notation is a way of expressing a subset of real numbers by the numbers that bound them. We can use this notation to represent inequalities. We know an interval expressed as 1 < x < 5 denotes a set of numbers lying between 1 and 5.
Examples of Interval Notation
Suppose we want to express the set of real numbers {x |-2 < x < 5} using an interval. This can be expressed as interval notation (-2, 5).
The set of real numbers can be expressed as (-∞, ∞).
Different Types Of Intervals
Intervals can be classified based on the numbers the set comprises. Some sets include the endpoints specified in the notation, while some might partially or not include the endpoints. In general, there are three types of intervals given as,
Open Interval
Closed Interval
Half-Open Interval
Open Interval
This type of interval does not include the endpoints of the inequality. For example, the set {x | -3 < x < 1} does not include the endpoints, -3 and 1. This is expressed using open interval notation: (-3, 1).
Closed Interval
This type of interval includes the endpoints of the inequality. For example, the set {x | -3 ≤ x ≤ 1} include the endpoints, -3 and 1. This is expressed using closed interval notation: [-3,1].
Half-Open Interval
This type of interval includes only one of the endpoints of the inequality. For example, the set {x | -3 ≤ x < 1} include the endpoint -3. This is expressed using half-open interval notation: [-3,1)
Notations For Different Types of Intervals
We can follow certain rules and symbols to represent the interval notation for different types of intervals. Let us understand different symbols that can be used to write a particular type of interval.
Symbol for Interval Notation
The notations we use for different intervals are:
[ ]: This is a square bracket that is used when both the endpoints are included in the set.
( ): This is a round bracket that is used when both the endpoints are excluded in the set.
( ]: This is a semi-open bracket that is used when the left endpoint is excluded and the right endpoint is included in the set.
[ ): This is also a semi-open bracket that is used when the left endpoint is included and the right endpoint is excluded in the set.
Number Line Representation of Different Types Of Intervals
Different types of interval notation can be represented on a number line. Look at the handy table that distinguishes between all the types of intervals using their representation on a number line.
| Interval Notation | Inequality | Number Line | Type of Interval |
--- --- |
| (a, b) | {x | a < x < b} | | Open Interval |
| [a, b] | {x | a ≤ x ≤ b} | | Closed interval |
| [a, ∞) | {x | x ≥ a} | | Half-Open Interval |
| (a, ∞) | {x | x > a} | | Half-Open Interval |
| (-∞, a) | {x | x < a} | | Half-Open Interval |
| (-∞, a] | {x | x ≤ a} | | Half-Open Interval |
Converting Inequality to Interval Notation
Follow the steps mentioned below to convert an inequality to interval notation.
Graph the solution set of the interval on a number line.
Write the numbers in the interval notation with a smaller number appearing first on the number line on the left.
If the set is unbounded on the left, use the symbol "-∞" and if it is unbounded on right, use the symbol "∞".
Let's take a few examples of inequality and convert them to interval notation.
| Inequality | Number Line | Interval Notation |
---
| x ≤ 3 | | (-∞, 3] |
| x < 5 | | (-∞, 5) |
| x ≥ 2 | | (2,∞] |
Important Notes on Interval Notation:
Interval notation is used to express the set of inequalities.
There are 3 types of interval notation: open interval closed interval, and half-open interval.
The interval with no infinity symbol is called a bounded interval.
The interval containing the infinity symbol is called an unbounded interval.
Related Articles:
Sets
Number Line
Real Numbers
greater than or equal to
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Interval Notation Worksheet
Worksheet on Number Lines
Examples on Interval Notation
Example 1: For a person to be elected as the president, he should be a minimum of 35 years old. Give an interval representing this information.
Solution:
Let's represent the age of the president by A.
It is given that the person's age A should be a minimum of 35
This means that A should be either greater than or equal to 35
Hence, this is represented by the inequality, A ≥ 35
So, the required interval is [35, ∞).
2. Example 2: For a school to participate in an Olympiad exam, the number of students from the school should be a minimum of 10. Represent this using an interval notation.
Solution:
Let x represent the number of students participating in a class, then x should be greater than or equal to 10
This means x ≥ 10
The interval notation is [10, ∞).
So, the number of students is represented by [10, ∞).
3. Example 3: Lola needs at least 1500 calories a day but the calorie intake should not exceed 1800 calories. Represent the possible amount of calories she could eat using interval notation.
Solution:
Let's represent the number of calories by x.
The inequality representing the possible amount of calories is given by 1500 ≤ x ≤ 1800
The interval notation for this inequality is [1500, 1800]
The required interval notation is [1500, 1800].
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Practice Questions on Interval Notation
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FAQs on Interval Notation
What do you Mean by Interval Notation?
Interval notation is a method to represent any subset of the real number line. We use different symbols based on the type of interval to write its notation. For example, the set of numbers x satisfying 1 ≤ x ≤ 6 is an interval that contains 1, 6, and all numbers between 1 and 6.
What is Interval Notation on a Graph?
When we represent the solution set of an interval on a number line, that is a graph for the interval notation.
How to Graph Interval Notation using Number Line?
We can graph the interval notation for a given set of numbers based on the type of number and specific symbols for the brackets used to enclose the set for the given particular type.
What is the ∪ Symbol for Interval Notation?
The union "∪" symbol is used to denote the union of two or more intervals in any Interval notation. Interval notation is defined as the method used to represent any subset of the real number line.
What are the Types of Intervals?
There are different types of intervals that can be represented by following a different set of rules for interval notation. These types of interval notation can be given as,
Open Interval
Closed Interval
Half-Open Interval
How to Convert Inequality to Interval Notation?
We can convert inequality to interval notation using the below-given steps,
Firstly, we need to graph the solution set of the interval on a number line.
Then write the numbers in the interval notation with a smaller number appearing first on the number line on the left.
Use the symbol "-∞" for the unbounded set on left and if it is unbounded on right, use the symbol "∞".
How do you Exclude Numbers in Interval Notation?
We use the round brackets to exclude numbers in interval notation. These numbers are generally the endpoints of the given set. To exclude a set of numbers in between, we can use two different sets and club them together using the union symbol '∪'.
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7380 | https://www.piping-designer.com/index.php/properties/fluid-mechanics/2500-hazen-williams-coefficient | Hazen-Williams Coefficient
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Hazen-Williams Coefficient
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Physics
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Hazen-Williams Coefficient
Tags:
Coefficient
Hazen-Williams
Flow Rate
Hazen-Williams Coefficient for Flow Velocity Formula
C=v 1.318⋅r 0.63 h⋅m 0.54 C=v 1.318⋅r h 0.63⋅m 0.54 (Hazen-Williams Coefficient)
v=C⋅1.318⋅r 0.63 h⋅m 0.54 v=C⋅1.318⋅r h 0.63⋅m 0.54
r h=(v C⋅1.318⋅m 0.54)1/0.63 r h=(v C⋅1.318⋅m 0.54)1/0.63
m=(v C⋅1.318⋅r 0.63 h)1/0.54 m=(v C⋅1.318⋅r h 0.63)1/0.54
SymbolEnglishMetric
C C = Hazen-Williams Coefficient, see below for values d i m e n s i o n l e s s d i m e n s i o n l e s s-
v v = Fluid Mean Flow Velocityf t/s e c f t/s e c-
r h r h = Hydraulic Radiusf t f t-
m m = Hydraulic Gradd i m e n s i o n l e s s d i m e n s i o n l e s s-
Hazen-Williams coefficient, abbreviated as C,also called Hazen-Williams friction coefficient or, Hazen-Williams roughness coefficient, a dimensionless number, is used in the Hazen-Williams Equation. The coefficient is used in fluid dynamics to calculate the resistance of water flow in a pipe network. The lower the coefficient, the smoother the pipe is. The higher the coefficient, the less fluid flow is restricted. By using pipe materials with improved flow characteristics, energy costs for pumping can be reduced or smaller pipes can be used.
The Hazen-Williams coefficient represents the internal roughness of the pipe, taking into account factors such as pipe material, age, and condition. It is used to incorporate the effect of pipe roughness on the flow characteristics. The higher the coefficient, the smoother the pipe surface, resulting in a higher flow rate for a given pressure drop. The Hazen-Williams coefficient varies depending on the pipe material, and it is typically determined through empirical testing. The coefficient values for different pipe materials are usually available in engineering references and design manuals.
Hazen-Williams Coefficient for Flow Rate Formula
C=Q 0.285⋅d 2.63⋅m 0.54 C=Q 0.285⋅d 2.63⋅m 0.54 (Hazen-Williams Coefficient)
Q=C⋅0.285⋅d 2.63⋅m 0.54 Q=C⋅0.285⋅d 2.63⋅m 0.54
d=(Q C⋅0.285⋅m 0.54)1/2.63 d=(Q C⋅0.285⋅m 0.54)1/2.63
m=(Q C⋅0.285⋅d 2.63)1/0.54 m=(Q C⋅0.285⋅d 2.63)1/0.54
SymbolEnglishMetric
C C = Hazen-Williams Coefficient, see below for values d i m e n s i o n l e s s d i m e n s i o n l e s s-
Q Q = Fluid Flow Ratef t 3/s e c f t 3/s e c-
d d = Pipe Inside Diameteri n i n-
m m = Hydraulic Graded i m e n s i o n l e s s d i m e n s i o n l e s s-
It's important to note that the Hazen-Williams equation is an empirical approximation and is most accurate for steady, uniform flow conditions in water supply systems. For more complex or non-uniform flow situations, other equations, such as the Darcy-Weisbach equation, may be more appropriate.
Note that the Hazen-Williams Coefficient is '''not''' the same as the Darcy-Weibach-Colebrook friction factor, f. These are not in any way related to each other.
Hazen-Williams Coefficient
MaterialCoefficient
Aluminum 130 - 150
Asbestos Cement 120 - 150
Asphalt-lined iron or steel 140
Brass 130
Cast Iron, cement lined 140
Cast Iron, coated 110 - 140
Cast Iron, new unlined 130
Cast Iron, old unlined 40 - 120
Cast Iron, uncoated 100 - 140
Cast Iron, 10 years old 107 - 113
Cast Iron, 20 years old 89 - 100
Cast Iron, 30 years old 75 - 90
Cast Iron, 40 years old 64 - 83
Cement lining 140
Concrete 100 - 140
Concrete, old 100 - 110
Copper 130 - 140
Corrugated Metal Pipe 60
Corrugated Steel 60
Deteriorated old pipes 60 - 80
Ductile Iron 120 - 145
Fiberglass 150
Galvanized Iron 100 - 120
Glass 130
Lead 130
Polyethylene 140
PVC, PE, GRP 120 - 150
Steel, new unlined 120
Steel, 15 years 200
Steel, riveted joints 95 - 110
Steel, welded joints 100 - 140
Steel, welded joints, lined 110 - 140
Steel, welded or steamless 100 - 120
Tin 130
Wood Stave 110
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How To Convert A GPA To N/MM2
ScienceMathTechnologyFeatures
Math
Algebra
How To Convert A GPA To N/MM2
ByMelissa MayerUpdated Mar 24, 2022
krisanapong detraphiphat/Moment/GettyImages
Gigapascals, atmospheres, millimeters of mercury – when you read these common units for measuring pressure, your head might start spinning. It can feel especially overwhelming if you must convert between units. However, with a basic understanding of units and prefixes, the concept of pressure and unit conversion is straightforward and easy to master.
What is Pressure?
What is Pressure?
When gas or fluid fills a container, the individual atoms and molecules in that substance don't sit still. Instead, they move around inside the container, bouncing off its walls. This movement creates force or stress pushing against the walls of the container. This is pressure, and it's measured in units of force (or stress) per unit of square area.
The concept of physical pressure is all around you in the real world. You must understand pressure when checking or filling a bicycle or automobile tire to the manufacturer's specifications. When it comes to weather, you hear about atmospheric pressure, or the pressure the atmosphere exerts on the planet. In terms of personal health, many people measure blood pressure every day; this is the measurement of the pressure your blood cells exert on the walls of your blood vessels during and between heartbeats.
Units and Prefixes
Units and Prefixes
Common units for measuring pressure include pounds per square inch (psi), atmospheres (atm), bars, millimeters of mercury (mmHg) and pascals (Pa). This last unit—pascals—is part of the International System of Units and therefore uses prefixes to indicate larger or smaller values. For example, a megapascal (MPa) comprises a million pascals because "mega" indicates "million" while a gigapascal (GPa) comprises a billion pascals because "giga" indicates "billion." One of the most important things to remember when it comes to pressure is that one pascal is equal to one newton per square meter (N/m 2).
Converting from GPa to N/mm2
Converting from GPa to N/mm2
In order to convert from gigapascals (GPa) to newtons per square millimeter (N/mm 2), you must perform step-wise conversions. First, notice the prefix attached to GPa and convert to the base unit Pa. To do this, multiply the value—for instance, 3 GPa—by the value of the prefix "giga," or 1 billion. 3 GPa is the same as 3 billion pascals.
Next, remember that one pascal equals one newton per square meter (N/m 2). This means you can directly substitute N/m 2 so that your value now reads 3 billion N/m 2. Finally, notice the prefix attached to your goal unit, N/mm 2. When converting from m 2 to mm 2, your prefix is "milli," or 1 thousand. To move from the larger m 2 to the smaller mm 2, divide your value by 1 thousand. In the example, you end up with 3 million N/mm 2. Therefore, 3 GPa is the same as 3 million N/mm 2.
References
Encyclopaedia Britannica: Pressure
Khan Academy: Pressure
Cite This Article
MLA
Mayer, Melissa. "How To Convert A GPA To N/MM2" sciencing.com, 25 June 2018.
APA
Mayer, Melissa. (2018, June 25). How To Convert A GPA To N/MM2. sciencing.com. Retrieved from
Chicago
Mayer, Melissa. How To Convert A GPA To N/MM2 last modified March 24, 2022.
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7382 | https://www.hilotutor.com/archives_brindle.html | Make Your Point > Archived Issues > BRINDLE Send Make Your Point issues straight to your inbox. | | | | | | | | | | | --- --- --- --- --- | | | | | --- | | | | | | | pronounce BRINDLE: | Say it "BRIN dull." To hear it, click here. | | | connect this word to others: | | | | | how to use it: | "Brindle" is one of those rare, sophisticated words that helps you get really precise. mottledinstead.
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The words brindle and brindledoften describe the fur coats of certain animals.
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The English word "brindle" probably came from a Middle English or Old English word meaning "brown, as if burnt or branded." (Sources disagree a bit on this.)
Today, "brindle" and "brindled" both mean "streaked with brown."
The English word "brindle" probably came from a Middle English or Old English word meaning "brown, as if burnt or branded." (Sources disagree a bit on this.)
Today, "brindle" and "brindled" both mean "streaked with brown."
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And a brindle cow:
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And some brindle brick:
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Part of speech:
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And some brindle brick:
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Part of speech:
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Part of speech:
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"Brindle" is one of those rare, sophisticated words that helps you get really precise.
mottledinstead.
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"A sleek, white, and brindled [dog] thrusts himself into the conversation... Its manners are disgusting, and it gobbles over its food."
— Rudyard Kipling, From Sea to Sea: Letters of Travel, 1913
"[The director Todd Haynes], a trim, boyish fifty-eight, with dishevelled brindle hair, was standing at the epicenter of his newest drama..."
— John Lahr, The New Yorker, 4 November 2019
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Explain the meaning of "brindle" without saying "streaked" or "streaky."
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When you're reading fiction and you meet a character described as "brindled," it's often someone big, strong, rugged, and full of authority--someone who calls to mind the power of some brindled animal, like a dog, wolf, cow, horse, bear, or lion.
Here's John Trotwood Moore, making an especially clear comparison between man and beast:
A large dog, brindled and lean, walked complacently and condescendingly in, followed by his master. At a glance, the least imaginative could see that Jud Carpenter, the Whipper-in of the Acme Cotton Mills, and Bonaparte, his dog, were well mated.
The man was large, raw-boned and brindled, and he, also, walked in, complacently and condescendingly.
With this example in mind, talk about another character from a book or movie who you could describe as brindled. What personality traits seem to match this character's dark, streaky appearance? What animal(s) does this character remind you of?
Need help thinking of someone? How about Hagrid from Harry Potter, Chewbacca from Star Wars, or Ezekiel from The Walking Dead?
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| Awesome, I'm glad it helped! |
Thanks for letting me know!
If you have any questions about this word, please message me at Liesl@HiloTutor.com.
| |
| study it: |
Explain the meaning of "brindle" without saying "streaked" or "streaky."
| |
| try it out: |
When you're reading fiction and you meet a character described as "brindled," it's often someone big, strong, rugged, and full of authority--someone who calls to mind the power of some brindled animal, like a dog, wolf, cow, horse, bear, or lion.
Here's John Trotwood Moore, making an especially clear comparison between man and beast:
A large dog, brindled and lean, walked complacently and condescendingly in, followed by his master. At a glance, the least imaginative could see that Jud Carpenter, the Whipper-in of the Acme Cotton Mills, and Bonaparte, his dog, were well mated.
The man was large, raw-boned and brindled, and he, also, walked in, complacently and condescendingly.
With this example in mind, talk about another character from a book or movie who you could describe as brindled. What personality traits seem to match this character's dark, streaky appearance? What animal(s) does this character remind you of?
Need help thinking of someone? How about Hagrid from Harry Potter, Chewbacca from Star Wars, or Ezekiel from The Walking Dead?
| |
| If you like, use this space to write. (To save your work, copy and paste it into an email or a document.) |
| | | |
---
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| |
| Awesome, I'm glad it helped! |
Awesome, I'm glad it helped!
Thanks for letting me know!
If you have any questions about this word, please message me at Liesl@HiloTutor.com.
| |
| study it: |
Explain the meaning of "brindle" without saying "streaked" or "streaky."
| |
| try it out: |
When you're reading fiction and you meet a character described as "brindled," it's often someone big, strong, rugged, and full of authority--someone who calls to mind the power of some brindled animal, like a dog, wolf, cow, horse, bear, or lion.
Here's John Trotwood Moore, making an especially clear comparison between man and beast:
A large dog, brindled and lean, walked complacently and condescendingly in, followed by his master. At a glance, the least imaginative could see that Jud Carpenter, the Whipper-in of the Acme Cotton Mills, and Bonaparte, his dog, were well mated.
The man was large, raw-boned and brindled, and he, also, walked in, complacently and condescendingly.
With this example in mind, talk about another character from a book or movie who you could describe as brindled. What personality traits seem to match this character's dark, streaky appearance? What animal(s) does this character remind you of?
Need help thinking of someone? How about Hagrid from Harry Potter, Chewbacca from Star Wars, or Ezekiel from The Walking Dead?
| |
| If you like, use this space to write. (To save your work, copy and paste it into an email or a document.) |
| |
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When you're reading fiction and you meet a character described as "brindled," it's often someone big, strong, rugged, and full of authority--someone who calls to mind the power of some brindled animal, like a dog, wolf, cow, horse, bear, or lion.
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Scipy: Centroid of convex hull
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Asked 10 years, 2 months ago
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how can I calculate the centroid of a convex hull using python and scipy? All I found are methods for computing Area and Volume.
regards,frank.
python
scipy
convex
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asked Jul 22, 2015 at 12:02
OD IUMOD IUM
1,634 3 3 gold badges 19 19 silver badges 27 27 bronze badges
1
4 If you are looking for the actual barycenter of the volume circumscribed by the convex hull, assuming uniform density, do not use the answers below. The just give you the mean of the coordinates of the points describing the hull.Raketenolli –Raketenolli 2017-02-07 17:40:38 +00:00 Commented Feb 7, 2017 at 17:40
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Assuming you have constructed the convex hull using scipy.spatial.ConvexHull, the returned object should then have the positions of the points, so the centroid may be as simple as,
```python
import numpy as np
from scipy.spatial import ConvexHull
points = np.random.rand(30, 2) # 30 random points in 2-D
hull = ConvexHull(points)
Get centoid
cx = np.mean(hull.points[hull.vertices,0])
cy = np.mean(hull.points[hull.vertices,1])
```
Which you can plot as follows,
```python
import matplotlib.pyplot as plt
Plot convex hull
for simplex in hull.simplices:
plt.plot(points[simplex, 0], points[simplex, 1], 'k-')
Plot centroid
plt.plot(cx, cy,'x',ms=20)
plt.show()
```
The scipy convex hull is based on Qhull which should have method centrum, from the Qhull docs,
A centrum is a point on a facet's hyperplane. A centrum is the average of a facet's vertices. Neighboring facets are convex if each centrum is below the neighbor facet's hyperplane.
where the centrum is the same as a centroid for simple facets,
For simplicial facets with d vertices, the centrum is equivalent to the centroid or center of gravity.
As scipy doesn't seem to provide this, you could define your own in a child class to hull,
```python
class CHull(ConvexHull):
def __init__(self, points):
ConvexHull.__init__(self, points)
def centrum(self):
c = []
for i in range(self.points.shape):
c.append(np.mean(self.points[self.vertices,i]))
return c
hull = CHull(points)
c = hull.centrum()
```
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edited Jul 22, 2015 at 12:53
answered Jul 22, 2015 at 12:27
Ed SmithEd Smith
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OD IUM
OD IUMOver a year ago
Thx for helping. I will check if the calculated centroid is always inside the hull.... BTW: do you know how to plot the hull in 3D? I'm using Axes3D and scatter. Simply scatter the simplices?
2015-07-22T12:51:57.93Z+00:00
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Ed Smith
Ed SmithOver a year ago
I've just added a method with append so should work in N dimensions. I guess you can plot the vertices as a scatter using matplotlib as in matplotlib.org/mpl_toolkits/mplot3d/tutorial.html#scatter-plots or a line plot using the simplices as above but in 3D...
2015-07-22T12:56:43.867Z+00:00
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The simple 'mean' method is not correct if the hull has points that are irregularly spaced around the perimeter, or at least will give a skewed answer. The best approach that I use is to calculate the centroids of the delaunay triangles of the hull points. This will weight the calculation to calculate the centroid as the COM of the shape, not just the average of vertices:
```python
from scipy.spatial import ConvexHull, Delaunay
def _centroid_poly(poly):
T = Delaunay(poly).simplices
n = T.shape
W = np.zeros(n)
C = 0
for m in range(n):
sp = poly[T[m, :], :]
W[m] = ConvexHull(sp).volume
C += W[m] np.mean(sp, axis=0)
return C / np.sum(W)
poly = np.random.rand(30, 2)
will calculate the centroid of the convex hull of poly
centroid_hull = _centroid_poly(poly)
```
Something like this should work.
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edited Jul 18, 2021 at 15:15
klwire
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answered Sep 11, 2019 at 4:41
Alec DayAlec Day
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To find the geometric centre of the hull's vertices simply use,
```python
Calculate geometric centroid of convex hull
hull = ConvexHull(points)
centroid = np.mean(points[hull.vertices, :], axis=0)
```
To plot the hull try,
```python
import numpy as np
import pylab as pl
import scipy as sp
import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d as a3
Plot polyhedra
ax = a3.Axes3D(pl.figure())
facetCol = sp.rand(3)
for simplex in hull.simplices:
vtx = [points[simplex,:], points[simplex,:], points[simplex,:]]
tri = a3.art3d.Poly3DCollection([vtx], linewidths = 2, alpha = 0.8)
tri.set_color(facetCol)
tri.set_edgecolor('k')
ax.add_collection3d(tri)
Plot centroid
ax.scatter(centroid0], centroid, centroid)
plt.axis('equal')
plt.axis('off')
plt.show()
```
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answered Jul 22, 2015 at 15:27
DrBwtsDrBwts
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This solution is correct also if the hull has points that are irregularly spaced around the perimeter.
```python
import numpy as np
from scipy.spatial import ConvexHull
def areaPoly(points):
area = 0
nPoints = len(points)
j = nPoints - 1
i = 0
for point in points:
p1 = points[i]
p2 = points[j]
area += (p1p2)
area -= (p1p2)
j = i
i += 1
area /= 2
return area
def centroidPoly(points):
nPoints = len(points)
x = 0
y = 0
j = nPoints - 1
i = 0
for point in points:
p1 = points[i]
p2 = points[j]
f = p1p2 - p2p1
x += (p1 + p2)f
y += (p1 + p2)f
j = i
i += 1
area = areaPoly(hull_points)
f = area6
return [x/f, y/f]
'points' is an array of tuples (x, y)
points = np.array(points)
hull = ConvexHull(points)
hull_points = points[hull.vertices]
centroid = centroidPoly(hull_points)
```
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edited Mar 14, 2021 at 9:38
answered Mar 13, 2021 at 12:02
exsurge-domineexsurge-domine
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DaveL17
DaveL17Over a year ago
Thank you for taking the time to provide an answer to the question. Since questions often receive multiple answers, you are encouraged to provide some context to your answer -- such as explaining why your solution may be better than other solutions.
2021-03-13T15:36:31.95Z+00:00
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7384 | https://courses.cs.vt.edu/cs2104/Fall13/notes/T16_Algorithms.pdf | Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Algorithms 1 Algorithms are the threads that tie together most of the subfields of computer science.
Something magically beautiful happens when a sequence of commands and decisions is able to marshal a collection of data into organized patterns or to discover hidden structure.
Donald Knuth Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain 2 Definition algorithm an effective method expressed as a finite list of well-defined instructions for calculating a function effective method (or procedure) a procedure that reduces the solution of some class of problems to a series of rote steps which, if followed to the letter, and as far as may be necessary, is bound to: -always give some answer rather than ever give no answer; -always give the right answer and never give a wrong answer; -always be completed in a finite number of steps, rather than in an infinite number; -work for all instances of problems of the class.
Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Properties of an Algorithm 3 An algorithm must possess the following properties: finiteness: The algorithm must always terminate after a finite number of steps.
definiteness: Each step must be precisely defined; the actions to be carried out must be rigorously and unambiguously specified for each case.
input: An algorithm has zero or more inputs, taken from a specified set of objects.
output: An algorithm has one or more outputs, which have a specified relation to the inputs.
effectiveness: All operations to be performed must be sufficiently basic that they can be done exactly and in finite length. Knuth Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Problems vs Algorithms vs Programs 4 For each problem or class of problems, there may be many different algorithms.
For each algorithm, there may be many different implementations (programs).
. .
. .
. .
p r o b l e m algorithm 1 algorithm 2 algorithm k . . .
. .
. .
. .
. .
. .
. .
program 1 program 2 program n Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Expressing Algorithms 5 An algorithm may be expressed in a number of ways, including: natural language: usually verbose and ambiguous flow charts: avoid most (if not all) issues of ambiguity; difficult to modify w/o specialized tools; largely standardized pseudo-code: also avoids most issues of ambiguity; vaguely resembles common elements of programming languages; no particular agreement on syntax programming language: tend to require expressing low-level details that are not necessary for a high-level understanding Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Common Elements of Algorithms 6 acquire data (input) some means of reading values from an external source; most algorithms require data values to define the specific problem (e.g., coefficients of a polynomial) computation some means of performing arithmetic computations, comparisons, testing logical conditions, and so forth...
selection some means of choosing among two or more possible courses of action, based upon initial data, user input and/or computed results iteration some means of repeatedly executing a collection of instructions, for a fixed number of times or until some logical condition holds report results (output) some means of reporting computed results to the user, or requesting additional data from the user Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain pseudo-Language 7 See the posted notes on pseudo-language notation.
Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Simple Example: Area of a Trapezoid 8 algorithm AreaOfTrapezoid takes number Height, number lowerBase, number upperBase # Computes the area of a trapezoid.
Pre: input values must be non-negative real numbers.
number averageWidth, areaOfTrapezoid averageWidth := ( upperBase + lowerBase ) / 2 areaOfTrapezoid := averageWidth Height display areaOfTrapezoid halt Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Simple Example: N Factorial 9 algorithm Factorial takes number N # Computes N! = 1 2 . . . N-1 N, for N >= 1 # Pre: input value must be a non-negative integer.
number nFactorial nFactorial := 1 while N > 1 nFactorial := nFactorial N N := N - 1 endwhile display nFactorial halt Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Example: Finding Longest Run 10 algorithm LongestRun takes list number List, number Sz # Given a list of values, finds the length of the longest sequence # of values that are in strictly increasing order.
Pre: input List must contain Sz elements.
number currentPosition # specifies list element currently # being examined number maxRunLength # stores length of longest run seen # so far number thisRunLength # stores length of current run if Sz <= 0 # if list is empty, no runs...
display 0 halt endif currentPosition := 1 # start with first element in list maxRunLength := 1 # it forms a run of length 1 thisRunLength := 1 # continues on next slide...
Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Example: Finding Longest Run 11 # ...continued from previous slide while currentPosition < Sz if ( List[currentPosition] < List[currentPosition + 1] ) thisRunLength := thisRunLength + 1 else if ( thisRunLength > maxRunLength ) maxRunLength := thisRunLength endif thisRunLength := 1 endif currentPosition := currentPosition + 1 endwhile display maxRunLength halt QTP: is this algorithm correct?
Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Testing Correctness 12 How do we know whether an algorithm is actually correct?
First, the logical analysis of the problem we performed in order to design the algorithm should give us confidence that we have identified a valid procedure for finding a solution.
Second, we can test the algorithm by choosing different sets of input values, carrying out the algorithm, and checking to see if the resulting solution does, in fact, work.
BUT… no matter how much testing we do, unless there are only a finite number of possible input values for the algorithm to consider, testing can never prove that the algorithm produces correct results in all cases.
Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Proving Correctness 13 We can attempt to construct a formal, mathematical proof that, if the algorithm is given valid input values then the results obtained from the algorithm must be a solution to the problem.
We should expect that such a proof be provided for every algorithm.
In the absence of such a proof, we should view the purported algorithm as nothing more than a heuristic procedure, which may or may not yield the expected results.
Algorithms Intro Problem Solving in Computer Science CS@VT ©2011-12 McQuain Measuring Performance 14 How can we talk precisely about the "cost" of running an algorithm?
What does "cost" mean? Time? Space? Both? Something else?
And, if we settle on one thing to measure, how do we actually obtain a measurement that makes sense?
This is primarily a topic for a course in algorithms, like CS 3114 or CS 4104. |
7385 | https://www.quora.com/What-is-the-most-effective-round-robin-tournament-format-for-all-sports-and-what-makes-it-the-best | Something went wrong. Wait a moment and try again.
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What is the most effective 'round-robin' tournament format for all sports and what makes it the best?
Patrick Loudon
Former life in beer, now part time teacher, writer
·
Author has 11K answers and 1.1M answer views
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1y
Seems UEFA has it right allowing for upstart teams to beat powerhouses but to allow the quality sides to float to the top in group play before the 16 team knockout. With American Pro, Collegiate, and Scholastic Sport, a loss in a tournament and you are gone, with the single exception of College baseball alone, which takes a a thoughtful a unique approach to elimination. with Conference, Regionals, Super Regionals and concluding with a winner/losers double elimination before a 3 game series for the title. It gives teams an opportunity to redeem and return to form.
Related questions
What is a round-robin tournament, and why is it called so?
How do I organise a round-robin tournament for 18 players, each having 4 opponents?
What are the rules of a round-robin tournament?
What are the disadvantages of a round-robin tournament?
What are the advantages and disadvantages of a league or round-robin tournament?
Raghuraman R
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6y
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What are some of the best formats (e.g., round robin) you have come across in sports tournaments?
FIFA world cup has quite a popular, steady and least controversial format for such a widely followed tournament.
Anonymous
5y
Related
What is a round-robin tournament, and why is it called so?
In a round-robin tournament, each player (or team) plays each other competitor (usually once, sometimes twice in a double round-robin tournament). In some competitions (e.g., World Cup in soccer and cricket), there is a preliminary round-robin phase where members of a pool play each other to determine which go through to the later elimination stage of the competition.
"The term round-robin is derived from the French term 'ruban', meaning 'ribbon'. Over a long period of time, the term was corrupted and idiomized to 'robin'."
link Round-robin tournament - Wikipedia
The phrase round-robin had earlie
In a round-robin tournament, each player (or team) plays each other competitor (usually once, sometimes twice in a double round-robin tournament). In some competitions (e.g., World Cup in soccer and cricket), there is a preliminary round-robin phase where members of a pool play each other to determine which go through to the later elimination stage of the competition.
"The term round-robin is derived from the French term 'ruban', meaning 'ribbon'. Over a long period of time, the term was corrupted and idiomized to 'robin'."
link Round-robin tournament - Wikipedia
The phrase round-robin had earlier been used to describe "correspondence to a single address authored or signed by numerous individuals (as found in a petition). ... The modern use of the term dates from the 17th Century French ruban rond (round ribbon). This described the practice of signatories to petitions against authority (usually Government officials petitioning the Crown) appending their names on a document in a non-hierarchical circle or ribbon pattern (and so disguising the order in which they have signed) in order that none may be identified as a ring leader."
link Round-robin - Wikipedia
Larry Thoman
27+ years working in the industry, 46+ years as a player
·
Author has 241 answers and 953.2K answer views
·
Jul 24
Related
What is the best way to run a tournament in sports where each team plays every other team once?
The format you would use for such a tournament would be called round robin. In team sports, like football, soccer, baseball, etc. it would not be the ideal choice due to the factors that James Shane and MarkB452 state. Unless you wanted a limited number of teams.
To figure out if you have the playing field capacity to accommodate such a tournament, you would first need to figure out the number of games it will take to complete such a tournament. So, as an example, let’s say you have 7 teams in a round robin tournament. How many games does that entail? The formula that I’ve always used is take o
The format you would use for such a tournament would be called round robin. In team sports, like football, soccer, baseball, etc. it would not be the ideal choice due to the factors that James Shane and MarkB452 state. Unless you wanted a limited number of teams.
To figure out if you have the playing field capacity to accommodate such a tournament, you would first need to figure out the number of games it will take to complete such a tournament. So, as an example, let’s say you have 7 teams in a round robin tournament. How many games does that entail? The formula that I’ve always used is take one less than the number of teams (since a team can’t play itself), and then start counting down by 1. In this example that would be 6+5+4+3+2+1=21.
And since there are 2 teams playing in each match, you also know that you can only play 3 games at a time (provided you have that field capacity). In the above example, that means 1 of the 7 teams sits out per round (gets a bye). So how many rounds does it take to complete a 7-team round robin? Well, you take the total number of matches (21) and divide by the number of games that can be played in one round (3) to get 7 rounds needed to complete a 7-team round robin.
Now that you have some basic figures as to what is required for such a tournament, you then need to figure if you have the field capacity and time to run it. Each sport will vary with the average time it takes to complete a game. This requires some knowledge of the sport as even timed sports like football and basketball will vary widely due to time needed for umpire and referee calls, introductions, halftime (and perhaps quarter) intervals, time-outs, etc.. Let’s say in this example that the average game requires 3 hours to complete.
Assuming you have 3 fields available for play, it would take a minimum 21 hours (7 rounds times 3 hours each) to complete all the games. But you also have to consider time between games needed to prepare the field for the next game, restrictions on the number of games that a team can play in one day, possible weather delays (for outdoor sports), and other such things.
Hopefully this will give you a better idea of what all would be entailed to run a round robin tournament.
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Related questions
What is the difference between a round-robin tournament format and a knockout tournament format?
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Which tournament is also known as a round-robin tournament?
James Shand
Former Retired manufacturing supervisor
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Author has 703 answers and 405.6K answer views
·
2y
Related
What is the best way to run a tournament in sports where each team plays every other team once?
I agree with Mark below. Too many teams make for a very long tournament and isn’t really feasible. You should probably limit the number to about 10 teams. That’s doable as many HS football teams play opponents one time. But, if you are talking about a short baseball or softball tournament, then ten is too many. A three day tournament would have all games on multiple fields the first day.
Using team #1 as the example, 1 plays 2, 1 plays 3 and 1 plays 4. But, three games a day is the limit. Three fields would allow seven teams to play. Three games would be completed the first day. The next day 1
I agree with Mark below. Too many teams make for a very long tournament and isn’t really feasible. You should probably limit the number to about 10 teams. That’s doable as many HS football teams play opponents one time. But, if you are talking about a short baseball or softball tournament, then ten is too many. A three day tournament would have all games on multiple fields the first day.
Using team #1 as the example, 1 plays 2, 1 plays 3 and 1 plays 4. But, three games a day is the limit. Three fields would allow seven teams to play. Three games would be completed the first day. The next day 1 plays 5, 1 plays 6 and 1 plays 7. Final day, the top two teams play for the championship. But, this is really pushing it. That’s why double elimination tournaments are preferred.
Philip
Fan for over 40 years
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Author has 5K answers and 3.2M answer views
·
3y
Related
Why is the round-robin tournament format so prevalent in professional football leagues?
The round robin tournament shows which team is the best over a longer period of time than a knockout competition.
A knockout tournament only tells you which team was best on that particular day. If we look at the FA cup for example we quite often get teams from a lower league beating premier league opposition (Nottingham Forest defeating Arsenal from the 2021–22 FA Cup for example).
The best leagues operate a dual round robin so each team play every other team at both home and away, this gives everybody the same advantage meaning that nobody has any advantage over anybody else.
Some smaller leagu
The round robin tournament shows which team is the best over a longer period of time than a knockout competition.
A knockout tournament only tells you which team was best on that particular day. If we look at the FA cup for example we quite often get teams from a lower league beating premier league opposition (Nottingham Forest defeating Arsenal from the 2021–22 FA Cup for example).
The best leagues operate a dual round robin so each team play every other team at both home and away, this gives everybody the same advantage meaning that nobody has any advantage over anybody else.
Some smaller leagues (SPFL for example) have and additional set of round robin fixtures after the dual round robin to increase the number of fixtures to around 33 giving the season the same length as other European Leagues
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Malcolm McLeod
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How can I create a round robin tournament with 10 teams, each playing another team only once, only playing 5 rounds, and playing each of 5 games only once?
While the other answer is technically correct in terms of the definition of a ‘round-robin’ tournament I am going to assume that you are simply looking for a way to schedule a tournament so that everyone plays five games and doesn’t play anyone twice.
My approach would be to assign the ten teams into two sets of five (either randomly, or using some seeding if the teams are ranked in some way).
(NOTE: If you are seeding then it is important to observe that being in the same ‘set’ means that you DON’T play anyone else in that set. You could go with 1–5 and 6–10 but that might be a bit harsh on the
While the other answer is technically correct in terms of the definition of a ‘round-robin’ tournament I am going to assume that you are simply looking for a way to schedule a tournament so that everyone plays five games and doesn’t play anyone twice.
My approach would be to assign the ten teams into two sets of five (either randomly, or using some seeding if the teams are ranked in some way).
(NOTE: If you are seeding then it is important to observe that being in the same ‘set’ means that you DON’T play anyone else in that set. You could go with 1–5 and 6–10 but that might be a bit harsh on the lower teams. Something like 1,2,5,7,9 and 3,4,8,8,10 is probably better.)
Then to create a schedule you can simple create a 5 x 5 table as below and place teams from one set along the columns and the other set down the rows:
So in Round 1 you have the following games:
Team 1 v 3; 2 v 4; 5 v 6; 7 v 8; 9 v 10 and then Round 2 games are the pink, round 3 the grey etc.
Add a diagonal stripe through each box and you can use the chart to record the results as well:
You just have to make sure you are careful to remember the the W or L in the bottom left relates to the team on the left hand side and the W or L in the top right relates to the team at the top of the column. You can double check as the sum of all the wins should come to 25.
Arda
NBA I EuroLeague I Basketball Enthusiast
·
Author has 176 answers and 615.7K answer views
·
5y
Related
What's the best return to play format for the NBA, and why?
Just recently, NBA discusses the potential ways to resume the NBA 2019–20 Season. There are 4 Potential Formants right now. Let’s start to discuss these:
16 Teams…
Jumping immediately to the NBA PlayOffs?
By applying this format, risk of spreading infection is going to be less. However, it won’t be so fair to some teams: Portland Trail Blazers, New Orleans Pelicans, Sacramento Kings, and San Antonio Spurs.
20 Teams…
Similar to World Cup
. Probably 5 Groups, which is basing on regular-season records. Everyone playing with each other twice, and just the top 2 Teams advancing to the Final.
22 Teams…
Just recently, NBA discusses the potential ways to resume the NBA 2019–20 Season. There are 4 Potential Formants right now. Let’s start to discuss these:
16 Teams… Jumping immediately to the NBA PlayOffs? By applying this format, risk of spreading infection is going to be less. However, it won’t be so fair to some teams: Portland Trail Blazers, New Orleans Pelicans, Sacramento Kings, and San Antonio Spurs.
20 Teams… Similar to World Cup. Probably 5 Groups, which is basing on regular-season records. Everyone playing with each other twice, and just the top 2 Teams advancing to the Final.
22 Teams… Bringing 6 teams more? “NBA can build a simple eight-team bracket with either a single-elimination format or short series.” Similar to a Regular Season Games, and a play-in Tournament. Scenario doesn’t so much convenient for the safety.
30 Teams… Holding a 72-Game Season? After that a play-in Tournament. By applying that this TV Deals can be beneficial to the Companies.
All these Scenarios are carrying some advantages and disadvantages. By 16 Teams, some teams can’t join so their seasons are going to end up with a disappointment. However, by inviting 30 Teams, there won’t be enough time to complete all of these in an adequate time, which can lead some problems to the 2020–21 Season.
NBA is currently saying that resuming can be done on July, 31 in Orlando, however, it can change too.
Source: NBA discussed four potential formats for return to play during Board of Governors call Friday, per report
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Robert Pratt
Former Division II College Mascot at Missouri Western State University (1999–2000)
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6y
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How can I create a round robin tournament with 10 teams, each playing another team only once, only playing 5 rounds, and playing each of 5 games only once?
How can I create a round robin tournament with 10 teams, each playing another team only once, only playing 5 rounds, and playing each of 5 games only once?
My solution, based on the parameters listed, is to do the following:
Pot 1: Teams A,B,C,D,E
Pot 2: Teams F,G,H,I,J
Pot 1’s schedule would look like this. (Similar pattern for Pot 2)
Round 1: AB, CD [E off]
Round 2: AC, DE [B off]
Round 3: AD, BE [C off]
Round 4: AE, BC [D off]
Round 5: BD, CE [A off]
So, a team’s first four games are a round-robin tournament within each pot.
The final round of the tournament would be
Pot 1 1st place v Pot 2 1st pl
How can I create a round robin tournament with 10 teams, each playing another team only once, only playing 5 rounds, and playing each of 5 games only once?
My solution, based on the parameters listed, is to do the following:
Pot 1: Teams A,B,C,D,E
Pot 2: Teams F,G,H,I,J
Pot 1’s schedule would look like this. (Similar pattern for Pot 2)
Round 1: AB, CD [E off]
Round 2: AC, DE [B off]
Round 3: AD, BE [C off]
Round 4: AE, BC [D off]
Round 5: BD, CE [A off]
So, a team’s first four games are a round-robin tournament within each pot.
The final round of the tournament would be
Pot 1 1st place v Pot 2 1st place for champion/2nd place
Pot 1 2nd place v Pot 2 2nd place for 3rd place/4th place
and so on, though I may not label 5th - 10th place, i might just call them “final games”.
To make this work, there would be six total “game days”, and each team would have one “off day” in this tournament.
Depending on the sport, a tie-breaker may need to be determined to select rankings, i.e.
A: 3–1, beat C, D, E, lost to B
B: 3–1, beat A, D, E, lost to C
C: 3–1, beat B, D, E, lost to A
D: 1–3, beat E, lost to A, B, C
E: 0–4, needs more practice.
Mark Johnson
MS in Economics (college major), Southern Illinois University Edwardsville
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Author has 1K answers and 2.8M answer views
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5y
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How do you schedule a round-robin tournament?
How do you schedule a round-robin tournament?
The only thing tricky about scheduling a round-robin tournament is that the number of participants has to be limited to the number of rounds plus one. A round robin of 12 people has 11 rounds. Even a Game/30 tournament can only get 4–5 rounds in a day. While 2 or 3 day long tournaments used to be the rule, they are now the exception.
You can make the tournament invitational, or limit participation to the first 7 players to register.
Another alternative is to play Quads. A Quad is a group of 4 players who play round robin against each other…for a total
How do you schedule a round-robin tournament?
The only thing tricky about scheduling a round-robin tournament is that the number of participants has to be limited to the number of rounds plus one. A round robin of 12 people has 11 rounds. Even a Game/30 tournament can only get 4–5 rounds in a day. While 2 or 3 day long tournaments used to be the rule, they are now the exception.
You can make the tournament invitational, or limit participation to the first 7 players to register.
Another alternative is to play Quads. A Quad is a group of 4 players who play round robin against each other…for a total of 3 rounds. Then you can just cut off registrations to 4, or 8, or 12…etc. If you get 3 players who don’t play in a quad, you can just let them play one another, with the odd player sitting out while the first two play.
When you plan the tournament, you need to be able to describe the tournament so you can advertise it.
Pairing a Round Robin sounds easy, but there are a number of problems. What do you do when one player (Bob) drops out after playing his first 3 games (lets say there were supposed to be 7 rounds). You then have to adjust the pairings, and decide whether those who already played Bob keep the points they won (or lost) or whether those games don’t count in the final. If you fought hard to beat Bob, you don’t want to have that loss eliminated from your results. But if you haven’t played him yet, you don’t want those who played him to get credit for a game you never got a chance to play.
Fortunately there are instructions in the Official Rule Book on how to change the pairings and how to count the games. Tournament Directors hate situations like happened with Bob, because trying to explain what just happened to all the other players is a total pain in the neck.
There are also rules about how to arrange the pairings to say which player gets white in each round in order to balance colors fairly by the end of the tournament. Again, this is in the official rule book. One way around this is to play double RR tournaments, where each player plays all the others twice…once as white and once as black. But that means you have to have time for twice as many rounds. That works nicely in a blitz tournament, but for longer time controls or for larger tournaments it is tough. A round robin of 12 players like the US Championship has to go from 11 games to 22. That’s a major commitment of time and resources on everybody’s part.
Another solution for Round Robin format tournaments is if you’re running a chess club and decide to run one. You just play one or two games a week against the other participants. But if Bob doesn’t show one week because his boss called him in or his anniversary was that night, then it puts things “off-cycle”. So you would need a week or two for folks to play rescheduled games.
When you’re expecting a large number of players, that’s one of the reasons most directors revert to Swiss System tournaments. A 5 round tournament is usually enough to guarantee a single winner among 32 participants, though draws can mess that up a little. A round robin of that crowd would require 31 rounds. No thank you.
Studied Commerce at Arwachin Bharti Bhawan Senior Secondary School, Vivek Vihar (Graduated 2021)
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Author has 212 answers and 1M answer views
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7y
Related
What are the "Round Robin" and "Knock Out" formats of sports tournaments?
Round Robin:
A
round-robin tournament
(or
all-play-all tournament
) is a competition in which each contestant meets all other contestants in turn. It contrasts with an elimination tournament.
Knockout:
A single-elimination,
knockout
, or sudden death
tournament
is a type of elimination
tournament
where the loser of each match-up is immediately eliminated from winning the
tournament
.
Round-robin tournament - Wikipedia
Single-elimination tournament - Wikipedia
Round Robin:
A
round-robin tournament
(or
all-play-all tournament
) is a competition in which each contestant meets all other contestants in turn. It contrasts with an elimination tournament.
Knockout:
A single-elimination,
knockout
, or sudden death
tournament
is a type of elimination
tournament
where the loser of each match-up is immediately eliminated from winning the
tournament
.
Footnotes
Round-robin tournament - Wikipedia
Single-elimination tournament - Wikipedia
Daniel Oakes
United State Chess Federation rating 2100+
·
Author has 423 answers and 157.3K answer views
·
3y
Related
How many games are in a 6-team round-robin tournament?
I’m assuming you mean each game is a single match, like if it were soccer, a team plays a single game against another team. If multiple games are played (Like in a chess Olympiad, if Poland plays China, 4 players on each team play in the team match, so the one pairing (China vs. Poland) requires 4 games).
‘if 1 pairing = 1 match, then the answer is 15, and there are 2 pretty good ways to get there. You have 6 teams, and each team will play 5 games, so 6x5 = 30. But we’re double counting; we’ve counted China vs. Poland twice - as one of China’s 5 games, and also as one of Poland’s 5 games. Same
I’m assuming you mean each game is a single match, like if it were soccer, a team plays a single game against another team. If multiple games are played (Like in a chess Olympiad, if Poland plays China, 4 players on each team play in the team match, so the one pairing (China vs. Poland) requires 4 games).
‘if 1 pairing = 1 match, then the answer is 15, and there are 2 pretty good ways to get there. You have 6 teams, and each team will play 5 games, so 6x5 = 30. But we’re double counting; we’ve counted China vs. Poland twice - as one of China’s 5 games, and also as one of Poland’s 5 games. Same with all of the other games. So we have to divide by 2, and that gives us 15.
the other way to get there is, how many rounds will there be? 5, because each team plays in every round until they’ve played all 5 of the other teams. So there will be 5 rounds. How many games will there be in each round? 3 - If you pair up 6 teams as opponents, you get 3 games. So 5 Rounds x 3 games in each round = 15/
Jack Rudd
BSc in Mathematics, The Open University (Graduated 2009)
·
Author has 278 answers and 89.6K answer views
·
2y
Related
What is a round-robin tournament? Is it possible for a round-robin tournament to end without having any winner, or with more than one winner? How?
A round-robin, also known as an all-play-all, also (in team sports) known as a league, is a tournament where every player plays every other player a fixed number of times (most commonly 1 or 2).
It can end with a tie for first place in a number of ways - suppose A beats B, B beats C, C beats A, and all three of them beat all the other competitors. Then A, B and C will tie for first place.
Related questions
What is a round-robin tournament, and why is it called so?
How do I organise a round-robin tournament for 18 players, each having 4 opponents?
What are the rules of a round-robin tournament?
What are the disadvantages of a round-robin tournament?
What are the advantages and disadvantages of a league or round-robin tournament?
What is the difference between a round-robin tournament format and a knockout tournament format?
Why is the round-robin tournament format so prevalent in professional football leagues?
What is a league or round-robin tournament?
How do you schedule a round-robin tournament?
Which tournament is also known as a round-robin tournament?
What is a round-robin tournament?
What is the difference between a round-robin tournament and a single league tournament?
How is a round-robin tournament winner determined?
What purpose does a round-robin tournament serve?
What is the other name of round robin tournament?
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7386 | https://www.youtube.com/watch?v=X6S-Bl0dbjs | Simplifying Algebraic Expressions with Parentheses
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Transcript:
Hello! Today we're going to take a look at simplifying algebraic expressions with parentheses. Before we do that, let's review what like terms are. Like terms are any two terms that have the same variable raised to the same exponent. So, some examples of like terms are: x squared and 7x squared. 3xy and 4xy are also examples of like terms, and so are 2x and 5x. Now, an example of two terms that are not like terms would be 6x and 6x squared. Even though they have the same coefficient and the same variable, the variables are not raised to the same power, so these two terms are not like terms. The coefficients for like terms can be different, and in fact, they usually are. So, like here: 2x and 5x are like terms, even though the coefficient 2 is different from the coefficient 5. So, now that we've reviewed what like terms are, let's use this information to simplify some expressions. So, we're going to start with the expression, 3x squared plus 2x minus 5, and add that to, 7x squared minus 3x plus 14. Since all our terms are being added or subtracted, we can use the associative and commutative properties to regroup our terms so our like terms are next to one another. Remember, subtracting can be thought of as adding a negative, so these properties will still work. So, let's combine our like terms. Our like terms are 3x squared and 7x squared, 2x and negative 3x, and negative 5 and 14. So we're going to regroup and write them this way: We're going to do, 3x squared plus 7x squared. Then we'll add our next set of like terms: 2x minus 3x. Plus our last set of like terms: negative 5 plus 14. Now we can simplify each set of parentheses. So, 3x squared plus 7x squared, is 10x squared. 2x minus 3x, is negative x, so we have plus negative x. Plus, negative 5 plus 14 is 9. Now we can simplify this just a little bit further by changing this "adding a negative number" to just subtracting. So our final answer is going to be, 10x squared minus x plus 9. Since this isn't an equation, we can't solve for anything. The most we can do is simplify, so this is our final answer. Let's try another example. x plus 2xy minus 3y, plus negative 17x minus 9xy plus 27y. So first, we're going to regroup our terms so our like terms are together. So let's start by underlining our like terms. We have x and negative 17x as like terms, 2xy and negative 9xy are like terms, and negative 3y and 27y are also like terms. So now we can write them in parentheses. We have x minus 17x, plus 2xy minus 9xy, plus negative 3y plus 27y. And now all we have to do is simplify each set of parentheses. So, x minus 17x is negative 16x. 2xy minus 9xy is negative 7xy, so we're just going to do minus 7xy. And negative 3y plus 27y is 24y, so we're going to add 24y. So, our final answer is, negative 16x minus 7xy plus 24y. Before we go, I want to try one more example that's slightly more difficult. 6x squared plus 4x minus 2, minus, 2x squared minus 18x plus 5. So, notice that this expression has a minus sign between the two sets of parentheses. Remember how I said earlier that subtracting is the same thing as adding a negative? This will help us in simplifying this expression. So, let's rewrite this as, 6x squared plus 4x minus 2, plus negative 1 times, 2x squared minus 18x plus 5. So, this is where the adding a negative number comes in. But the negative number has to be 1 because that doesn't change the value of the actual expression; 1 times anything is the original expression. So, this doesn't change the value of anything, it just makes it look slightly different so we can understand where our next steps come from. So, now that we have a negative 1 here, we need to distribute this negative 1 into each term in the second set of parentheses. So, we're going to leave this first set of parentheses exactly the same: 6x squared plus 4x minus 2. Plus, negative 1 times 2x squared, is negative 2x squared. Negative 1 times negative 18x, is plus 18x. And negative 1 times positive 5, is negative 5. So, notice that we have the exact same set of parentheses, just the sign has changed in front of each term. So, now we're going to follow the exact same steps as we have before. We're going to underline our like terms, so we have 6x squared and negative 2x squared. 4x and 18x are also like terms, and so are negative 2 and negative 5. So now we're going to regroup these like terms into their own sets of parentheses. So we'll have, 6x squared minus 2x squared, plus 4x plus 18x, plus negative 2 minus 5. Now we'll simply simplify each set of parentheses. So, 6x squared minus 2x squared is 4x squared. 4x plus 18x is 22x. And negative 2 minus 5 is negative 7. So our final answer is, 4x squared plus 22x minus 7. And there you have it! I hope that this video made this topic more understandable for you. Thanks for watching and happy studying. |
7387 | https://personalpages.manchester.ac.uk/staff/david.d.apsley/lectures/turbbl/integral.pdf | Turbulent Boundary Layers 4 - 1 David Apsley h δ boundary-layer profile equivalent displaced profile δ h+ h streamline boundary 4. INTEGRAL ANALYSIS OF THE BOUNDARY LAYER SPRING 2009 4.1 Characteristic integral lengths 4.2 The momentum-integral relation 4.3 Application to laminar flow 4.4 Application to turbulent flow Examples The integral analysis of boundary layers was originally proposed by Theodore von Kármán and K. Pohlhausen in separate papers in 1921. 4.1 Characteristic Integral Lengths Displacement Thickness The flow near the surface is retarded, so that the streamlines must be displaced outwards to satisfy continuity. To reduce the total mass flow rate of a frictionless fluid by the same amount, the surface would have to be displaced outward by a distance , called the displacement thickness: deficit flux mass y U U U e e e e = ⌡ ⌠ − = ∞ 0 d ) ( Momentum Thickness The loss of momentum flux for the mass flux U dy between adjacent streamlines is y U U U e d ) ( × − . Hence the total loss of momentum flux is equivalent to the removal of momentum through a distance , called the momentum thickness: t lux defici momentum f y U U U U e e e d ) ( 0 2 = ⌡ ⌠ − = ∞ For incompressible flow (with uniform free-stream density) these take particularly simple forms: Displacement thickness: ⌡ ⌠ − = ∞ 0 d ) 1 ( y U U e (1) Momentum thickness: ⌡ ⌠ − = ∞ 0 d ) 1 ( y U U U U e e (2) y U equal areas y U δ e e Turbulent Boundary Layers 4 - 2 David Apsley Shape Factor The ratio of displacement thickness to momentum thickness is called the shape factor H: Shape factor: ≡ H (3) Since ) 1 ( 1 e e e U U U U U U − > − the shape factor is always greater than 1. A large shape factor is an indicator of a boundary layer near separation. y U y small H large H 4.2 The Momentum-Integral Relation 4.2.1 Zero-Pressure-Gradient Boundary Layer 0 x τ w CONTROL VOLUME streamline For a zero-pressure-gradient boundary layer ( e and Ue constant) the rate of loss of momentum equals the total drag on the surface up to that point; i.e., per unit width: ⌡ ⌠ = = x w e e x x D U 0 force drag total deficit flux momentum 2 d ) (3 2 1 3 2 1 Differentiating: w e eU x ) ( d d 2 = Dividing by 2 e eU we can also express it in a simpler non-dimensional form: 2 d d f c x = Integrating, gives the total drag coefficient (= average skin-friction coefficient) over length L: L L cD ) ( 2 = The drag coefficient can then be deduced entirely from the downstream velocity profile. Turbulent Boundary Layers 4 - 3 David Apsley 4.2.2 General Case The following derivation applies to a steady, compressible or incompressible, 2-dimensional boundary layer with arbitrary free-stream pressure gradient. For complete generality we will allow for a wall transpiration velocity Vw as well as a wall shear stress w. Start with the integral momentum flux deficit: ⌡ ⌠ − = ∞ 0 2 d ) ( y U U U U e e e Differentiate with respect to x, taking the differential operator through the integral sign and using the product rule, { } ⌡ ⌠ ∂ ∂ − + − ∂ ∂ = ⌡ ⌠ − ∂ ∂ = ∞ ∞ 0 0 2 d d d ) ( ) ( d ) ( ) ( d d y x U U x U U U U x U y U U U x U x e e e e e In the first underlined term substitute from the continuity equation: y V x U ∂ ∂ − = ∂ ∂ ) ( ) ( In the second underlined term substitute from the boundary-layer momentum equation: y x U U y U V x U U e e e ∂ ∂ + + ∂ ∂ − = ∂ ∂ d d Then ⌡ ⌠ ∂ ∂ − − − ∂ ∂ + − ∂ ∂ − = ⌡ ⌠ ∂ ∂ − − ∂ ∂ + + − ∂ ∂ − = ∞ ∞ 0 0 2 d d d ) ( ) ( ) ( d d d d d ) ( ) ( ) ( d d y y x U U U y U V U U y V y y x U U y U V x U U U U y V U x e e e e e e e e e e e Collecting y derivatives together, and simplifying using the product rule and the fact that Ue is independent of y, { } ⌡ ⌠ − − ⌡ ⌠ + − ∂ ∂ − = ∞ ∞ 0 0 2 d ) ( d d d ) ( ) ( d d y U U x U y U U V y U x e e e e e e ⇒ ) ( d d ) ( ) ( d d 0 2 e e e e e e U x U U U V U x − + − − = ∞ Since Ue – U and both vanish at y = ∞, whilst U = 0, V = Vw and = w at y = 0, ) ( d d ) ( d d 2 e e e w e w w e e U x U U V U x − + = i.e. e w w w e e e e e U V x U U U x d d ) ( d d 2 + = + Turbulent Boundary Layers 4 - 4 David Apsley Kármán’s integral relation 4 3 4 2 1 3 2 1 4 43 4 42 1 4 3 4 2 1 ion transpirat wall stress wall gradient pressure stream -free of effect momentum of loss of rate 2 d d ) ( d d e w w w e e e e e U V x U U U x + = + (4) For constant-density flow the most compact form comes from expanding the first differential and dividing by 2 e U : e w f e e U V c x U U H x + = + + 2 d d ) 2 ( d d (5) where 2 2 1 e w f U c = skin-friction coefficient = H shape factor Notes. (1) The results also hold along a curved-wall boundary layer, provided that the radius of curvature is much greater than the thickness of the boundary layer. (2) The integral relations hold whether the boundary layer is laminar, turbulent or transitional. (3) The integral relation is exact only for the correct mean velocity profile. However, the power of the scheme is that it provides an excellent estimate of drag with only a half-decent approximation for U(y). (4) For a flat-plate boundary layer with zero pressure gradient and no wall transpiration, Kármán’s integral relation takes the particularly simple form: Zero pressure-gradient boundary layer: 2 d d f c x = (6) In this case the total-drag coefficient L U L D L c e D 2 2 1 ) ( ) ( = where ⌡ ⌠ = L w x L D 0 d ) ( is given by L L L cD ) ( 2 ) ( = (7) The total drag is entirely determined by the downstream momentum thickness. (5) The shape factor H is only relevant when there is a free-stream pressure gradient (dUe/dx ≠ 0). A large shape factor indicates a boundary layer approaching separation. (6) An alternative proof of Kármán’s integral relation – starting from the differential boundary-layer equations – is given in the Examples Section. Turbulent Boundary Layers 4 - 5 David Apsley 4.3 Application to Laminar Flow A fully-developed profile may be approximated by: > ≤ = ), ( y U y y f U U e e (8) where 0 ) 1 ( , 1 ) 1 ( , or 0 ) 0 ( , 0 ) 0 ( = ′ = ∞ ≠ ′ = f f f f . (Why is the second condition important?) The integral boundary-layer depths are proportional to : , B A = = (9) and the skin-friction coefficient 2 2 1 0 Re ) 0 ( 2 d d C U f U y U c e e y f = ′ = = = (10) where A, B and C are constants determined by the particular profile adopted: ) 0 ( 2 , d ) 1 ( , d ) 1 ( 1 0 1 0 f C f f B f A ′ = ⌡ ⌠ − = ⌡ ⌠ − = (11) Substituting the expressions for and cf into the integral relation: = ⇒ = ⇒ = e e f U B C x U C x B c x d d 2 d d 2 d d 2 Hence: 2 / 1 2 / 1 = e U x B C or 2 / 1 2 / 1 Re− = x B C x (12) (x is measured from some virtual origin; in practice, this can be taken as the leading edge). Since 2 / 1 x ∝ we have x x 2 1 d d = and hence x x c f d d 2 = = Summary of Results For a Zero-Pressure-Gradient Laminar Boundary Layer B A H B A = = = , , ) ( 2 ) ( , L c L c x c f D f = = where ) 0 ( 2 , d ) 1 ( , d ) 1 ( 1 0 1 0 f C f f B f A ′ = ⌡ ⌠ − = ⌡ ⌠ − = e U x B C 2 / 1 = Turbulent Boundary Layers 4 - 6 David Apsley Thus, assuming only that the profile is self-similar, but without any knowledge about its precise form, we can confirm that 2 / 1 x ∝ (or 2 / 1 Re− ∝ x x ) and 2 / 1 Re ) ( − ∝ L D L c The simplest profile satisfying on 0 d d , 0 on 0 = = = = = y y U U U y U e is (the channel-flow profile) ) 2 ( y y U U e − = or 2 2 ) ( − = f This has 4 , , 15 2 3 1 = = = C B A and gives the following approximations. Approximation Blasius solution e U x 83 .
1 e U x 72 .
1 e U x 730 .
0 e U x 664 .
0 H 2.5 2.59 In both cases, x c f = ) ( 2 ) ( L c L c f D = Thus, the drag is correct to within 10% and the shape factor to within 3.5%, even for a simple approximation. Much better approximations are given in the Examples section. Turbulent Boundary Layers 4 - 7 David Apsley 4.4 Application to Turbulent Flow The momentum integral relation for a zero-pressure-gradient boundary layer is 2 d d f c x = To make this tractable we adopt the power-law approximations that we have met before: Velocity profile: 7 / 1 ) ( y U U e = (13) Skin-friction coefficient: 6 / 1 Re 0205 .
0 − = f c (where Re e U = ) (14) From the first of these we can compute displacement and momentum thickness: / , 7 / 1 y U U e = = Displacement thickness: 8 1 d ) 1 ( d ) 1 ( 1 0 7 / 1 0 = ⌡ ⌠ − = ⌡ ⌠ − ≡ y U U e (15) Momentum thickness: 72 7 d ) 1 ( d ) 1 ( 1 0 7 / 1 7 / 1 0 = ⌡ ⌠ − = ⌡ ⌠ − ≡ y U U U U e e (16) Shape factor: 7 9 = ≡ H (17) Substituting for skin-friction coefficient cf and momentum thickness in the momentum integral relation, 6 / 1 ) ( 01025 .
0 d d 72 7 e U x = ⇒ x U e d ) ( 01025 .
0 d 72 7 6 / 1 6 / 1 = ⇒ x U e 6 / 1 6 / 7 ) ( 01025 .
0 7 6 72 7 = × ⇒ 6 / 1 6 / 7 ) ( 123 .
0 ) ( x U x e = Hence 7 / 1 Re 166 .
0 − = x x or 7 / 6 Re 166 .
0 Re x = (18) Thus, the boundary layer grows as 7 / 6 x ∝ Since 7 / 6 x ∝ we have x x x 12 1 7 6 d d = = , and hence: Turbulent Boundary Layers 4 - 8 David Apsley 7 / 1 Re 0277 .
0 12 1 2 d d 2 − = × = = x f x x c (19) (Alternatively, use 6 / 1 Re 0205 .
0 − = f c and substitute for Re .) 7 / 1 Re 032 .
0 ) ( 6 7 2 ) ( − = = = L f D L c L L c (20) i.e. the total drag coefficient is only 7/6 times the local skin friction coefficient at the end of the plate. Summary of Results for a Zero-Pressure-Gradient Turbulent Boundary Layer Boundary-layer depth: 7 9 , 72 7 , 8 1 = = = H where 7 / 1 Re 166 .
0 − = x x or 7 / 6 x ∝ Frictional drag: 7 / 1 Re 0277 .
0 − = x f c 7 / 1 Re 032 .
0 ) ( 6 7 ) ( − = = L f D L c L c Note. An earlier and alternative correlation by Prandtl (see Q3 in the Examples) and based on pipe-flow data gave a skin friction coefficient proportional to 5 / 1 − x and resulted in a drag coefficient 5 / 1 Re 072 .
0 ) ( − = L D L c . This is still widely used today. 0 0.001 0.002 0.003 0.004 0.005 0.006 1.00E+05 1.00E+06 1.00E+07 1.00E+08 ReL cD Prandtl Modern Turbulent Boundary Layers 4 - 9 David Apsley Examples Question 1. Show that, for a power-law velocity profile of the form ) ( / 1 < = y y U U n e the displacement thickness, momentum thickness and shape factor are given by 1 1 + = n , ) 2 )( 1 ( + + = n n n , n n H 2 + = Question 2. Laminar boundary layer profiles may be assumed to be approximately of the form ), ( y f U U e = = where 0 ) 1 ( , 1 ) 1 ( , 0 ) 0 ( = ′ = = f f f . Use an integral analysis with the following profiles to find expressions for , , H, cf and cD and compare with Blasius’ solution. (a) 3 2 1 2 3 − (b) 4 3 2 2 + − (c) ) 2 sin( Question 3. (White) An alternative (but less accurate) analysis of turbulent flat-plate flow was given by Prandtl in 1927, using a wall shear-stress formula from pipe flow: 4 / 1 2 0225 .
0 = e e w U U Show that an integral analysis with this formula and a 1/7 power law profile for velocity leads to the following relations for turbulent flat-plate flow: 5 / 1 Re 37 .
0 x x = , 5 / 1 Re 058 .
0 x f c = , 5 / 1 Re 072 .
0 ) ( L D L c = Question 4. Show that, in general, if the skin-friction coefficient and momentum thickness for a zero-pressure-gradient flat-plate boundary layer are related by n f A c / 1 Re− = then 1 1 1 Re 2 ) 1 ( + − + + = n x n n A n n x , x n n c f 1 2 + = , ) ( ) 1 ( ) ( L c n n L c f D + = Use this to confirm the results in Section 4.4 and Question 3 above. Turbulent Boundary Layers 4 - 10 David Apsley Question 5. Why would a simple power law be a bad approximation to a laminar boundary-layer velocity profile? Question 6. For purely laminar or purely turbulent flat-plate boundary layers the momentum thickness θ at distance x from the leading edge is given by: = θ − − ) turbulent ( , Re 0158 .
0 ) laminar ( , Re 664 .
0 7 / 1 2 / 1 x x x where / Re x U e x = and Ue = free-stream velocity, = kinematic viscosity. In practice, a turbulent boundary layer is preceded by an initial laminar region. Since f c x 2 1 /d d = and the skin-friction coefficient cf is finite, the momentum thickness must be continuous at the transition point. Irrespective of any intermediate development, the total drag coefficient for one side of a plate of length L depends only upon the momentum thickness at the end: L L L cD ) ( 2 ) ( θ = (a) Develop a procedure, illustrated by a flowchart showing calculation steps, or by a computer program, to find the momentum thickness at the end of, and total drag coefficient for, a flat plate of arbitrary length L. (Rex,tr is assumed to be given.) Using this procedure, find: (b) the transition length; (c) the momentum thickness at the downstream end; (d) the drag coefficient cD; (e) the total drag on one side of a smooth rectangular plate of size 4 m × 2 m in an air flow of 8 m s–1 parallel to the long side. Take Rex,tr = 5×105 and, for air, = 1.2 kg m–3, = 1.5×10–5 m2 s–1. Question 7. A wind tunnel has a test section 0.5 m square and 6 m long. To preserve a constant free-stream velocity it is sometimes desirable to slant the walls outward. (a) Explain the purpose of this wall adjustment. (b) If only the roof is to be adjusted, estimate the angle at which it should be slanted in order to preserve a free-stream velocity of 40 m s–1 over a working section from x = 2 m to x = 4 m, stating any assumptions made. Question 8. A square-section duct of side 30 cm and length 12 m carries a throughflow of air at flow rate 3 m3 s–1. Assuming that turbulent boundary layers grow along the side walls from the front edge, estimate: (a) the pressure drop along the duct; (b) the drag on the side walls. laminar turbulent transition point x U e Turbulent Boundary Layers 4 - 11 David Apsley Question 9. Starting from the boundary-layer equations in differential form: y x U U y U V x U U e e e ∂ ∂ + = ∂ ∂ + ∂ ∂ d d (momentum) 0 ) ( ) ( = ∂ ∂ + ∂ ∂ y V x U (continuity) derive Kármán’s integral relation. Question 10. A boundary-layer mean-velocity profile is approximated by > < = ) ( ) ( ln y U y y u E u U e where U is continuous at y = . Find and in terms of u /Ue and . Question 11. Assume that the following mean-velocity profile holds across a turbulent boundary layer. ) ( 2 ) ln( 1 y f B y u u U + + = where 3 2 2 3 ) ( η − η = η f (a) Find an expression for U/Ue as a function of and the quantity , where e U u = , ) (δ ≡U U e (b) Hence, or otherwise, show that the displacement thickness δ is given by: ) 1 ( + = (c) Find a similar expression for / where is the momentum thickness. Question 12. In a 2-dimensional, incompressible boundary layer the energy thickness E may be defined by the integral energy flux deficit: ⌡ ⌠ − = ∞ 0 2 2 3 d ) ( y U U U U e E e where Ue is the free-stream velocity and U is the wall-parallel mean-velocity component in the boundary layer. Using the boundary-layer equations with shear stress , derive the following relation for incompressible boundary-layer flow over a plane surface with wall transpiration velocity Vw: 2 0 3 d 2 ) ( d d e w E e U V y y U U x + ⌡ ⌠ ∂ ∂ = ∞ Give an interpretation of all the terms. Turbulent Boundary Layers 4 - 12 David Apsley Answers (2) Profile x U e x U e H (a) 3 2 1 2 3 − 74 .
1 13 70 4 3 2 / 1 = 646 .
0 70 117 2 1 2 / 1 = 69 .
2 13 35 = (b) 4 3 2 2 + − 75 .
1 37 35 5 9 2 / 1 = 685 .
0 35 37 3 2 2 / 1 = 55 .
2 74 189 = (c) ) 2 sin( 74 .
1 ) 2 / 2 ( 2 2 / 1 = − − 655 .
0 ) 2 / 2 ( 2 / 1 = − 66 .
2 2 / 2 2 = − − (d) Blasius 1.72 0.664 2.59 In all cases, x c f = , ) ( 2 ) ( L c L c f D = (6) (b) xtr = 0.94 m (c) L = 6.8×10–3 m (d) cD = 0.0034 (e) D = 1.05 N (7) (b) 0.41 degrees (8) (a) 560 Pa (b) 56 N (10) e U u = ) 2 1 ( e e U u U u − = (11) (a) { } ) 1 2 3 ( 2 ln 1 3 2 − − + + = e U U (c) + + − + = 2 2 35 52 6 19 2 ) 1 ( |
7388 | https://mapscaping.com/what-is-a-thematic-map/ | Understanding Thematic Maps - September 29, 2025
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Understanding Thematic Maps
March 6, 2023Uncategorized
Understanding Thematic Maps
In this blog, we will dig into the two main classes of maps, focusing specifically on thematic maps and their various types.
TWO Classes of Maps: General Reference Maps and Thematic Maps
Maps can be categorized into two primary classes: general reference maps and thematic maps. Understanding the distinction between these classes is crucial for interpreting spatial information effectively.
General Purpose Maps
General purpose maps, also known as reference maps, provide a broad overview of geographical features without delving into analysis. They display natural and human-made features, allowing users to locate places and understand basic geography.
Features: Includes bodies of water, roads, parks, political boundaries, and elevation.
Examples: Street maps and highway maps.
No Analysis: These maps do not interpret data or provide analytical insights.
Here’s a table summarizing the differences between thematic maps and general purpose maps:
| Aspect | Thematic Maps | Topographic Maps |
---
| Focus | Represent a specific theme or subject matter | Represent physical features and landmarks |
| Data Representation | Use symbols, colors, and patterns for theme representation | Use contour lines, symbols, and colors for terrain features |
| Variability | Various types (e.g., choropleth, dot density, isoline) | Follow a standard format and scale |
| Purpose | Visualize the spatial distribution of chosen variables | Depict elevation, terrain, and natural/human-made features |
| Common Uses | Analyzing demographic, environmental, or economic data | Navigation, planning, and engineering purposes |
The table highlights the key differences in focus, data representation, variability, purpose, and common uses between thematic and topographic maps.
There are two types of Thematic Maps: Qualitative and Quantitative
Qualitative Thematic Maps
Qualitative thematic maps focus on specific themes or subjects, emphasizing the “where” aspect of data rather than the “how much.” They illustrate spatial distributions without providing numerical values.
Purpose: To show the distribution of characteristics, such as language families or cultural regions.
Examples: Maps indicating the locations of different language speakers worldwide.
Types of Quantitative Thematic Maps
Choropleth maps
Choropleth maps use predefined boundaries to present data through varying shades or colours. They are popular for visualizing demographic data across regions.
Strengths: Easy to create and interpret; widely used in media.
Weaknesses: Generalization may obscure local variations; multiple regions may appear identical despite differences.
Advantages of Choropleth Maps:
Easy to read: Choropleth maps are easy to read and understand, even for people with little or no background in geography or data analysis.
Good for displaying data by geographic area: Choropleth maps are an effective way to display data that is aggregated by geographic areas, such as states, counties, or cities.
Can show patterns and trends: Choropleth maps can be used to show patterns and trends in data that vary across different regions or sub-regions, allowing for comparisons between different areas.
Useful for making spatial comparisons: Choropleth maps can be useful for making spatial comparisons between different areas, highlighting differences or similarities in the data being mapped.
Disadvantages of Choropleth Maps:
Data limitations: Choropleth maps are limited by the availability and quality of data for the geographic areas being mapped. Small or sparsely populated areas may have limited data available, which can make it difficult to accurately map the data.
Misleading results: Choropleth maps can be subject to the modifiable areal unit problem, which can lead to misleading results if the choice of sub-regions affects the analysis. For example, different ways of dividing a region into sub-regions can produce different results.
Data classification issues: The choice of data classification method used in choropleth maps can also affect the results of the analysis. Different methods can produce different maps with different patterns and trends.
Lack of detail: Choropleth maps can lack detail about the actual data values being mapped within each sub-region. The use of colors or shading can make it difficult to discern differences between values within a given sub-region.
Dot density maps
Dot density or dot distribution maps are powerful tools for visualizing spatial distributions and patterns. Each dot on these maps represents a specific quantity, allowing viewers to understand how a phenomenon is spread across a given area.
Representation: Each dot can symbolize a fixed number of units, such as one individual or a specific quantity of resources.
Spatial Patterns: By analyzing the concentration of dots, one can identify trends, clusters, or areas of dispersion.
Advantages of Dot Density Maps:
Easy to read: Dot density maps are easy to read and understand, even for people with little or no background in geography or data analysis.
Good for showing relative values: Dot density maps are an effective way to show relative values of data between different geographic areas, such as states or counties.
Can show patterns and trends: Dot density maps can be used to show patterns and trends in data that vary across different regions or sub-regions, allowing for comparisons between different areas.
Can display detailed information: Dot density maps can display detailed information about the actual data values being mapped within each sub-region, by showing the number or density of dots within each area.
Disadvantages of Dot Density Maps:
Limited by data: Dot density maps are limited by the availability and quality of data for the geographic areas being mapped. Small or sparsely populated areas may have limited data available, which can make it difficult to accurately map the data.
Limited by dot size: The size of the dots used in dot density maps can be limited by the scale of the map or the size of the geographic areas being mapped. If the dots are too small, they may be difficult to see or count. If they are too large, they can overlap or obscure other dots, making it difficult to accurately represent the data.
Potential for bias: The placement of dots within each sub-region can be subject to bias or manipulation, depending on the data source or mapping method used.
Lack of detail: Dot density maps may not provide as much detail about the data values within each sub-region as other types of thematic maps, such as choropleth maps or graduated symbol maps.
Cartograms
Cartograms are a unique type of graphic that distorts land area based on thematic variables like population or economic data. While not technically maps, they provide a different perspective on spatial data.
Characteristics: Areas are resized according to the variable being represented, which can lead to unconventional shapes and layouts.
Usage: Effective for visualizing relative sizes, such as countries sized by population rather than land area.
Advantages of Cartograms:
Visually impactful: Cartograms are visually striking and can quickly draw attention to the patterns and trends in the data being mapped.
Good for showing relative values: Cartograms are an effective way to show relative values of data between different geographic areas, such as states or countries.
Can show patterns and trends: Cartograms can be used to show patterns and trends in data that vary across different regions or sub-regions, allowing for comparisons between different areas.
Can display detailed information: Cartograms can display detailed information about the actual data values being mapped within each sub-region, by showing the size or shape of each area.
Disadvantages of Cartograms:
Limited by data: Cartograms are limited by the availability and quality of data for the geographic areas being mapped. Small or sparsely populated areas may have limited data available, which can make it difficult to accurately map the data.
Potential for distortion: The use of different shapes or sizes to represent data values can lead to distortions in the overall shape or size of the geographic areas being mapped, depending on the mapping method used.
Can be difficult to interpret: The interpretation of cartograms can be subjective, as the data values are represented by different shapes or sizes rather than discrete symbols or colors. This can make it difficult to discern differences between values within a given sub-region.
Potential for bias: The choice of shapes or sizes used to represent data values can be subject to bias or manipulation, depending on the data source or mapping method used.
Heat maps
Heat maps use colors to represent the intensity or concentration of data values over a geographic area. The colors used typically range from cool to warm colors to represent low to high values of the data variable. Heat maps are often used to show patterns of activity or behavior, such as traffic density or social media usage.
Advantages of Heat Maps:
Easy to read: Heat maps are easy to read and understand, even for people with little or no background in geography or data analysis.
Good for displaying density data: Heat maps are an effective way to display density data, showing the concentration or intensity of data values over a geographic area.
Can show patterns and trends: Heat maps can be used to show patterns and trends in data that vary across different regions or sub-regions, allowing for comparisons between different areas.
Useful for making spatial comparisons: Heat maps can be useful for making spatial comparisons between different areas, highlighting differences or similarities in the data being mapped.
Disadvantages of Heat Maps:
Data limitations: Heat maps are limited by the availability and quality of data for the geographic areas being mapped. Small or sparsely populated areas may have limited data available, which can make it difficult to accurately map the data.
Limited by color scale: The color scale used in heat maps can affect the visual impact of the map, and different color scales can produce different results.
Potential for bias: The placement of data points within each sub-region can be subject to bias or manipulation, depending on the data source or mapping method used.
Can be difficult to interpret: The interpretation of heat maps can be subjective, as the data values are represented by colors rather than discrete symbols or shapes. This can make it difficult to discern differences between values within a given sub-region.
Graduated symbols
Graduated symbol or proportional symbol maps use symbols of different sizes to represent different values of a data variable. The symbols are usually scaled proportionally to the values of the variable they represent, with larger symbols representing higher values. Graduated symbols are commonly used to show data related to population, income, or other quantitative variables.
Proportional symbol maps employ varying sizes of symbols to represent quantities at specific locations. This method effectively combines both location and magnitude in one visual representation.
Customization: Symbols can be adjusted in size, color, and pattern to enhance readability and convey data effectively.
Applications: Often used to illustrate demographic data, economic statistics, or resource distributions.
Advantages of Graduated Symbol Maps:
Easy to read: Graduated symbol maps are easy to read and understand, even for people with little or no background in geography or data analysis.
Good for showing quantitative data: Graduated symbol maps are an effective way to show quantitative data, such as population, income, or other numerical values.
Can show patterns and trends: Graduated symbol maps can be used to show patterns and trends in data that vary across different regions or sub-regions, allowing for comparisons between different areas.
Useful for making spatial comparisons: Graduated symbol maps can be useful for making spatial comparisons between different areas, highlighting differences or similarities in the data being mapped.
Disadvantages of Graduated Symbol Maps:
Limited by data: Graduated symbol maps are limited by the availability and quality of data for the geographic areas being mapped. Small or sparsely populated areas may have limited data available, which can make it difficult to accurately map the data.
Limited by symbol size: The size of the symbols used in graduated symbol maps can be limited by the scale of the map or the size of the geographic areas being mapped. If the symbols are too small, they may be difficult to see or count. If they are too large, they can overlap or obscure other symbols, making it difficult to accurately represent the data.
Potential for bias: The placement of symbols within each sub-region can be subject to bias or manipulation, depending on the data source or mapping method used.
Lack of detail: Graduated symbol maps may not provide as much detail about the data values within each sub-region as other types of thematic maps, such as choropleth maps or dot density maps. The use of symbols can make it difficult to discern differences between values within a given sub-region.
Isopleth maps
Isopleth maps, also known as contour maps or isoline maps. The contour lines connect points of equal value, and the density or spacing of the lines indicates the degree of change between the values. Isopleth maps are commonly used to show data related to elevation, temperature, precipitation, and other continuous variables.
Examples: Commonly used in weather forecasting and topographical maps.
Terminology: Lines on these maps have specific names based on the data they represent, such as isobars for air pressure and isotherms for temperature.
Advantages of Isopleth Maps:
Good for displaying continuous data: Isopleth maps are an effective way to display continuous data, such as elevation, temperature, or precipitation.
Can show patterns and trends: Isopleth maps can be used to show patterns and trends in data that vary across different regions or sub-regions, allowing for comparisons between different areas.
Can display detailed information: Isopleth maps can display detailed information about the actual data values being mapped within each sub-region, by showing the contour lines that connect points of equal value.
Useful for making spatial comparisons: Isopleth maps can be useful for making spatial comparisons between different areas, highlighting differences or similarities in the data being mapped.
Disadvantages of Isopleth Maps:
Limited by data: Isopleth maps are limited by the availability and quality of data for the geographic areas being mapped. Small or sparsely populated areas may have limited data available, which can make it difficult to accurately map the data.
Limited by contour interval: The contour interval used in isopleth maps can affect the visual impact of the map, and different contour intervals can produce different results.
Can be difficult to interpret: The interpretation of isopleth maps can be subjective, as the contour lines are representing continuous data values rather than discrete symbols or shapes.
Potential for bias: The interpolation of data values between sampled points to create the contour lines can be subject to bias or manipulation, depending on the data source or mapping method used.
In conclusion
Thematic maps are a powerful tool for visualizing data related to geographic areas. Each type of thematic map, including choropleth maps, dot density maps, graduated symbol maps, isopleth maps, and cartograms, has its own unique advantages and disadvantages.
The choice of map type will depend on the specific data being mapped, the geographic scale of the analysis, and the goals of the analysis.
Frequently asked questions about thematic maps and thematic mapping:
What is a Thematic Map?
A thematic map is a type of map specifically designed to show the spatial distribution of one or more themes or variables, such as population density, climate patterns, or economic trends. Unlike general reference maps, which display a variety of physical and political features, thematic maps focus on specific data related to a theme.
2. What are the Different Types of Thematic Maps?
Choropleth Maps: Use variations in shading, coloring, or the placing of symbols within predefined areas (like countries or states) to indicate the average values of a property or quantity in those areas.
Isopleth Maps: Depict lines that connect points of equal value, commonly used for weather maps (e.g., isobars, isotherms).
Dot Distribution Maps: Represent the presence of a feature with a dot symbol. The density of dots in an area helps to perceive the distribution pattern.
Proportional Symbol Maps: Use symbols of different sizes to represent data associated with different areas or locations.
Flow Maps: Illustrate the movement of objects between different areas, commonly used for migration or transportation data.
How are Thematic Maps Used?
Thematic maps are used in various fields for visualizing and analyzing spatial patterns and relationships. For instance, in environmental science, they can depict pollution levels or deforestation; in economics, they can show income levels or employment rates across regions; in public health, they can represent disease prevalence or healthcare access.
2. What are the Key Elements of a Thematic Map?
Essential components include:
Title: Describes what the map is about.
Legend: Explains the symbols and color codes used.
Scale: Indicates the ratio of a distance on the map to the corresponding distance on the ground.
Data Source: Cites where the thematic data came from.
Orientation (Compass Rose): Shows map direction (North, South, East, West).
How Do You Create a Thematic Map?
To create a thematic map, start by defining the purpose and selecting appropriate data. Use GIS software or mapping tools to visualize this data. Choose the type of thematic map that best represents your data. Customize elements like colors, symbols, and scales for clarity and impact.
2. What are the Advantages and Limitations of Thematic Maps?
Advantages include their ability to simplify complex data and highlight spatial patterns and relationships. However, they have limitations, such as potential misinterpretation due to symbol choices or color scales, and they can be misleading if data is not accurately represented.
3. How to Interpret Thematic Maps Accurately?
Accurate interpretation requires understanding the map’s legend, scale, and the nature of the data presented. It’s important to consider the data source and methodology used in data collection and to be aware of any potential biases or limitations in the data.
4. What’s the Difference Between Thematic Maps and Reference Maps?
Thematic maps focus on specific themes or data, conveying a particular message or showcasing patterns. Reference maps, on the other hand, provide general information about physical and political features, like roads, rivers, city locations, and political boundaries.
5. What are Examples of Thematic Maps?
Examples include a map showing average household income by county, a global map of earthquake occurrences, or a map depicting the prevalence of a certain language or dialect in different regions.
6. How Has Technology Changed Thematic Mapping?
Advancements in GIS, remote sensing, and computer-aided design have revolutionized thematic mapping. These technologies allow for more accurate data collection, easier manipulation and analysis of spatial data, and the creation of more dynamic, interactive maps.
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I'm Daniel O'Donohue, the voice and creator behind The MapScaping Podcast ( A podcast for the geospatial community ). With a professional background as a geospatial specialist, I've spent years harnessing the power of spatial to unravel the complexities of our world, one layer at a time.
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7389 | https://omgcenter.com/2012/02/21/tenuousness-and-tenacity-and-the-commencement-of-lent/ | Tenuousness and Tenacity and the Commencement of Lent | OMG Center
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Posted on February 21, 2012 in Ash Wednesday, Christianity, Church, Cross, Hope, Joy, Lent, Suffering
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Tenuousness and Tenacity and the Commencement of Lent
It’s 8:04 on Tuesday morning, and I’m sitting in the waiting room at the hospital after just sending my son off to yet another surgery.
I am somewhat ashamed that I don’t remember how many surgeries he’s had since the accident in 2004; maybe that’s because I hope that each one is the last one.
Today is also Fat Tuesday, the day before Ash Wednesday.
And on this surgery/Mardi Gras day, naturally I’m thinking etymologies.
It might be an odd thing, to some, that word histories are swirling in my mind at this moment, but it is a comforting habit of mine (and a fine antidote to the blaring “Law and Order: Special Victims” showing in the waiting room for our viewing pleasure). This idiosyncrasy of mine helps me sort through realities from new angles, deepening how I think through the narrow and through the broad.
The words I’m considering are relatives to each other: tenuous and tenacious.
The fusion of Karl’s surgery day, Fat Tuesday, and Ash Wednesday fuse tenuousness and tenacity too.
Tenuous comes from the Latin word meaning “thin,” (tenuis), which has as its root the prefix meaning “to stretch.”
Tenacious comes from the Latin meaning “the act of holding fast,” (tenacitas)“tough, holding fast,” (tenax)“to hold,” (tenere).
Every time that Karl’s little body is cloaked in surgical scrubs, and that little plastic mask is placed over his little face, I make a little sign of the cross on his forehead and am reminded, again, that life is tenuous. There is such a thin (tenuis) line between life and death.
But every time that Karl’s little body is cloaked in surgical scrubs, and that little plastic mask is placed over his little face, and I make a little sign of the cross on his forehead, I am reminded, again, that he is tenacious. He is tough (tenax) and holds (tenere) forth with mischievous and contagious “joyful defiance.”
I think that Mardi Gras is our tenacious defiance of the tenuousness of life.
And that isn’t all bad.
Traditionally, Lent has been cloaked too, not in surgical scrubs but in dreariness and in the subjugation of joy.
I’ve come to think that sometimes one needs tenacity just to get through Lent.
I’m not opposed to setting aside 40 days of one’s year to reflect on the tenuousness of life, a tenuousness that is starkly marked by many with the imposition of an ashen cross on the forehead.
Marking life’s tenuousness also marks its sacredness. The cinders smear into forehead grease that life has no business being squandered.
It’s to be savored. And stewarded.
And one can steward the savoring.
And one can do that with some upward heel-kicking, pancake stuffing, music enjoying indulgence (metaphorically and literally speaking), and by inviting others to join into the festive fray.
At WomanKind in Richmond, a woman announced her church’s new tradition of giving up negativity during Lent. The congregation is St. James Episcopal Church in Leesburg, VA, and you can read about their plot to abandon cranky thoughts here.
They caught the idea of this new kind of Lenten discipline here. The point of a Lenten negativity fast is that one can develop of habit of moving away from pessimism, criticism, contempt, and impatience (not only toward others, but also toward oneself), and moving toward constructive naming and solving of problems, and hope, and celebration.
That takes tenacity.
It is now 9:14 in the evening. Poor Karl came through the surgery well, but had deep and awful pain during recovery. Those were tenuous moments, I tell you what, seeing his little face wrinkled in pain and fear.
I smeared his forehead and his cheeks with tears while I sung “Hush Little Baby” and tenaciously insisted that he get more morphine, and then more, and then more.
And by the time he was wheeled into his room, he was cracking his why-did-the-chicken-cross-the-road joke and picking his nose like he was picking a banjo, grinning all the while, and I don’t even think it was due to the morphine.
And then he laughed some more, and napped a bit (as did the mama), and then the grandma came and brought a dino squeeze toy that pelts balls out of its jaws, and then he ate, and then he puked, and then he pretended to puke just to flip out the mama, and now he’s asleep.
And I am finally writing the first OMG blog in a month, the between-two-houses action of moving and painting caught in midstream, the fabulous WomanKind conference and this surgery now behind me, and I am now subsumed by a deep exhaustion, and sipping smuggled red wine from a styrofoam cup.
And I find that I too am caught in the tension of tenacity and tenuousness, of Fat Tuesday and Ash Wednesday, of knowing that, Fat Tuesday aside, life is slender, and so let us be joyful in it.
Sign up atwww.omgcenter.comto have OMG blogs delivered straight to your inbox!
Do you have questions or issues about theology, vocation, or life matters? Anna invites you to ask your questions and engage in mutual theological mulling: theological direction, if you will.
Contact Anna atanna@omgcenter.comto visit about personal or congregational consultations, as well as to speak about booking her to present at your next event.
She also runs The Spent Dandelion Theological Retreat Center, where you can come to Retreat, Reflect, and Restore at her North Shore home. Visitwww.spentdandelion.comto learn more and book a stay.
Check out @omgcenter and @spentdandelion on Twitter too.
And last, but not least, check out Anna’s first book “I Can Do No Other: The Church’s New Here We Stand Moment,”along with her newest book, Joyful Defiance: Death Does Not Win The Day, both published by Fortress Press.
~~~~~
10 Comments
Debbie Domeier on February 21, 2012 God bless your family especially Karl. After going through all kinds of crap with my previous congregation over the homosexuality issue, you put the world into perspective Anna … ! Thank you. I don’t understand why some people can not find love in their hearts for their neighbor, not to mention, wasting so much time on negativity when there is true suffering in this world. I like the negativity fast. May God hold you in the palm of God’s hand. Peace. Reply
OMG on February 21, 2012 Thank you, Debbie! I’m sorry that you struggled in your previous congregation. That notion of a negativity fast is yet one more way to help letting go of past angers, angers that really only feed on the present and do nothing about the past. Here’s to hoping for a restorative Lent for you then! Reply
Brooke Taylor on February 22, 2012 God bless Karl on this first day of Lent … and all his days. Thank you for this blog and for coming to WomanKind, especially knowing you would be in surgery with Karl just a few days hence. (I am the grandmother who talked to you about stem cells.) Your workshop was a highlight of an event with many highlights. I think everyone sitting with me Saturday morning is starting a negativity fast today, even though we all know how difficult it will be. Thanks for passing the idea along to so many more people. Reply
OMG on February 22, 2012 Thank you for your words! That “negativity fast” notion is such a good one, too good to hoard! Peace to you and to your family, and thanks for being in my workshop! Reply
Anne Ryan on February 22, 2012 Sending many hugs and prayers your way, Anna. What a blessing you continue to be to those of us who’ve finally found you:) Godspeed as you deal with the challenges on your journey. Reply
OMG on February 28, 2012 Thank you so much, Anne! His first day of school was yesterday, and he had a good day. The students were glad to see him as well! Reply
Char on February 22, 2012 Great message and I love the idea of a negativity fast for Lent! That whole thing about giving something up seems to have lost its pallor, at least for me for now, but this seems like a beautiful idea. Peace, love, healing, and a bunch of others good things are wished you’re way. Reply
OMG on February 28, 2012 So glad you are in my world, Char! Reply
Ruth Modlin Ellett on February 26, 2012 Anna- Being with you and your sweet family last weekend was a miracle for me. You, Karl, Else, your mom and family are in my prayers. With love and admiration. Reply
OMG on February 28, 2012 Ruth, you defined hospitality while we were there. I thank you again for that, and now also for this comment. Peace! Reply
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7390 | https://r-coder.com/binomial-distribution-r/ | R CODER
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The binomial distribution
The dbinom function
Plot of the binomial probability function in R
The pbinom function
pbinom example
Plot of the binomial cumulative distribution in R
The qbinom function
Plot of the binomial quantile function in R
The rbinom function
Binomial distribution in R
Statistics with R
Distributions
The binomial distribution is a discrete distribution that counts the number of successes in Bernoulli experiments or trials. In this tutorial we will explain how to work with the binomial distribution in R with the dbinom, pbinom, qbinom, and rbinom functions and how to create the plots of the probability mass, distribution and quantile functions.
The binomial distribution
Denote a Bernoulli process as the repetition of a random experiment (a Bernoulli trial) where each independent observation is classified as success if the event occurs or failure otherwise and the proportion of successes in the population is constant and it doesn’t depend on its size.
Let X∼B(n,p), this is, a random variable that follows a binomial distribution, being n the number of Bernoulli trials, p the probability of success and q=1−p the probability of failure:
The probability mass function (PMF) is P(X=x)=(xn)pxqn−x if x=0,1,2,…,n.
The cumulative distribution function (CDF) is F(x)=Iq(1−x,n−x).
The quantile function is Q(p)=F−1(p).
The expected mean and variance of X are E(X)=np and Var(X)=npq, respectively.
The functions of the previous lists can be computed in R for a set of values with the dbinom (probability), pbinom (distribution) and qbinom (quantile) functions. In addition, the rbinom function allows drawing n random samples from a binomial distribution in R. The following table describes briefly these R functions.
| Function | Description |
--- |
| dbinom | Binomial probability mass function (Probability function) |
| pbinom | Binomial distribution (Cumulative distribution function) |
| qbinom | Binomial quantile function |
| rbinom | Binomial pseudorandom number generation |
In the following sections we will review each of these functions in detail.
The dbinom function
In order to calculate the binomial probability function for a set of values x, a number of trials n and a probability of success p you can make use of the dbinom function, which has the following syntax:
```
dbinom syntax
dbinom(x, # X-axis values (x = 0, 1, 2, ..., n)
size, # Number of trials (n > = 0)
prob, # The probability of success on each trial
log = FALSE) # If TRUE, probabilities are given as log
```
For instance, if you want to calculate the binomial probability mass function for x=1,2,…,10 and a probability of success in each trial of 0.2, you can type:
dbinom(= 1: 10, = 10, =0.2)
```
Output
0.2684354560 0.3019898880 0.2013265920 0.0880803840 0.0264241152
0.0055050240 0.0007864320 0.0000737280 0.0000040960 0.0000001024
```
Plot of the binomial probability function in R
The binomial probability mass function can be plotted in R making use of the plot function, passing the output of the dbinom function of a set of values to the first argument of the function and setting type = “h” as follows:
```
Grid of X-axis values
x <- 1:80
size = 80, prob = 0.2
plot(dbinom(x, size = 80, prob = 0.2), type = "h", lwd = 2,
main = "Binomial probability function",
ylab = "P(X = x)", xlab = "Number of successes")
size = 80, prob = 0.3
lines(dbinom(x, size = 80, prob = 0.3), type = "h",
lwd = 2, col = rgb(1,0,0, 0.7))
size = 80, prob = 0.4
lines(dbinom(x, size = 80, prob = 0.4), type = "h",
lwd = 2, col = rgb(0, 1, 0, 0.7))
Add a legend
legend("topright", legend = c("80 0.2", "80 0.3", "80 0.4"),
title = "size prob", title.adj = 0.95,
lty = 1, col = 1:3, lwd = 2, box.lty = 0)
```
The pbinom function
In order to calculate the probability of a variable X following a binomial distribution taking values lower than or equal to x you can use the pbinom function, which arguments are described below:
```
pbinom syntax
pbinom(q, # Quantile or vector of quantiles
size, # Number of trials (n > = 0)
prob, # The probability of success on each trial
lower.tail = TRUE, # If TRUE, probabilities are P(X <= x), or P(X > x) otherwise
log.p = FALSE) # If TRUE, probabilities are given as log
```
By ways of illustration, the probability of the success occurring less than 3 times if the number of trials is 10 and the probability of success is 0.3 is:
```
pbinom(3, = 10, =0.3)
0.6496107 or 64.96%
```
As the binomial distribution is discrete, the previous probability could also be calculated adding each value of the probability function up to three:
```
sum(dbinom(0: 3, = 10, =0.3))
0.6496107 or 64.96%
```
As the binomial distribution is discrete, the cumulative probability can be calculated adding the corresponding probabilities of the probability function. The following R function allows visualizing the probabilities that are added based on a lower bound and an upper bound.
```
size: number of trials (n > = 0)
prob: probability of success on each trial
lb: lower bound of the sum
ub: upper bound of the sum
col: color
lwd: line width
binom_sum <- function(size, prob, lb, ub, col = 4, lwd = 1, ...) {
x <- 0:size
if (missing(lb)) {
lb <- min(x)
}
if (missing(ub)) {
ub <- max(x)
}
plot(dbinom(x, size = size, prob = prob), type = "h", lwd = lwd, ...)
if(lb == min(x) & ub == max(x)) {
color <- col
} else {
color <- rep(1, length(x))
color[(lb + 1):ub ] <- col
}
lines(dbinom(x, size = size, prob = prob), type = "h",
col = color, lwd = lwd, ...)
}
```
As an example, you can represent the probabilities that are added to calculate the probability of a binomial variable taking values equal or lower than 5 if the number of trials is 20 and the probability of success is 0.2 with the following code:
binom_sum(size = 20, prob = 0.2, lwd = 2, col = 2, ub = 5,
ylab = "P(X = x)", xlab = "Number of successes")
The sum of the probabilities displayed in red is equal to pbinom(5, size = 20, prob = 0.2).
pbinom example
In this section we will review a more complete example to understand how to calculate binomial probabilities in several scenarios. Consider that a basketball player scores 4 out of 10 baskets (p=0.4). If the player thows 20 baskets (20 trials):
The probability of scoring 6 or less baskets, P(X≤6), is:
pbinom(6, size = 20, prob = 0.4) # 0.2500107 or 25%
1 - pbinom(6, size = 20, prob = 0.4, lower.tail = FALSE) # Equivalent
This probability can also be calculated adding the corresponding elements of the binomial probability function, as we pointed out in the previous section:
```
sum(dbinom(0: 6, = 20, =0.4))
0.2500107 or 25%
```
Using the function that we defined before we can represent the calculated probability:
binom_sum(size = 20, prob = 0.4, ub = 6, lwd = 2,
ylab = "P(X = x)", xlab = "Number of successes")
The probability of scoring less than 6 baskets, P(X<6), is:
pbinom(5, size = 20, prob = 0.4) # 0.125599 or 12.56%
1 - pbinom(5, size = 20, prob = 0.4, lower.tail = FALSE) # Equivalent
sum(dbinom(0:5, size = 20, prob = 0.4)) # Equivalent
Note that we set 5 on the first argument of the function instead of 6 because the binomial distribution is discrete, so P(X<6)=P(X≤5). The calculated probability can be represented with the sum of the following probabilities of the probability mass function:
binom_sum(size = 20, prob = 0.4, ub = 5, lwd = 2,
ylab = "P(X = x)", xlab = "Number of successes")
The probability of scoring more than 12 baskets, P(X>12), is:
pbinom(12, size = 20, prob = 0.4, lower.tail = FALSE) # 0.02102893 or 2.1%
1 - pbinom(12, size = 20, prob = 0.4) # Equivalent
sum(dbinom(13:20, size = 20, prob = 0.4)) # Equivalent
This probability corresponds to:
binom_sum(size = 20, prob = 0.4, lb = 12, lwd = 2,
ylab = "P(X = x)", xlab = "Number of successes")
The probability of scoring between 7 and 11 baskets, P(7≤X≤11), is:
pbinom(11, size = 20, prob = 0.4) - pbinom(7, size = 20, prob = 0.4) # 0.5275807 or 52.8%
sum(dbinom(8:11, size = 20, prob = 0.4)) # Equivalent
The corresponding plot can be created with the following code:
binom_sum(size = 20, prob = 0.4, lb = 7, ub = 11, lwd = 2,
ylab = "P(X = x)", xlab = "Number of successes")
As the binomial distribution is a discrete distribution P(X=x)=0, so P(X≥x)=P(X>x) and P(X≤x)=P(X<x).
Plot of the binomial cumulative distribution in R
The binomial distribution function can be plotted in R with the plot function, setting type = “s” and passing the output of the pbinom function for a specific number of experiments and a probability of success.
The following block of code can be used to plot the binomial cumulative distribution functions for 80 trials and different probabilities.
```
Grid of X-axis values
x <- 1:80
size = 80, prob = 0.2
plot(pbinom(x, size = 80, prob = 0.2), type = "s", lwd = 2,
main = "Binomial distribution function",
xlab = "Number of successes", ylab = "F(x)")
size = 80, prob = 0.3
lines(pbinom(x, size = 80, prob = 0.3), type = "s",
lwd = 2, col = 2)
size = 80, prob = 0.4
lines(pbinom(x, size = 80, prob = 0.4), type = "s",
lwd = 2, col = 3)
Add a legend
legend("bottomright", legend = c("80 0.2", "80 0.3", "80 0.4"),
title = "size prob", title.adj = 0.95,
lty = 1, col = 1:3, lwd = 2, box.lty = 0)
```
The qbinom function
Given a probability or a set of probabilities, the qbinom function allows you to obtain the corresponding binomial quantile. The following block of code describes briefly the arguments of the function:
```
qbinom syntax
qbinom(p, # Probability or vector of probabilities
size, # Number of trials (n > = 0)
prob, # The probability of success on each trial
lower.tail = TRUE, # If TRUE, probabilities are P(X <= x), or P(X > x) otherwise
log.p = FALSE) # If TRUE, probabilities are given as log
```
As an example, the binomial quantile for the probability 0.4 if n=5 and p=0.7 is:
```
qbinom(=0.4, = 5, =0.7)
3
```
Plot of the binomial quantile function in R
The binomial quantile function can be plotted in R for a set of probabilities, a number of trials and a probability of success with the following code:
```
Grid of X-axis values
x <- 1:80
size = 80, prob = 0.2
plot(qbinom(seq(0, 1, 0.001), size = 80, prob = 0.2),
main = "Binomial quantile function",
ylab = "Q(p)", xlab = "p",
type = "s", col = 3, xaxt = "n")
axis(1, labels = seq(0, 1, 0.1), at = 0:10 100)
size = 80, prob = 0.3
lines(qbinom(seq(0, 1, 0.001), size = 80, prob = 0.3), type = "s", col = 2)
size = 80, prob = 0.4
lines(qbinom(seq(0, 1, 0.001), size = 80, prob = 0.4), type = "s", col = 1)
Add a legend
legend("topleft", legend = c("80 0.2", "80 0.3", "80 0.4"),
title = "size prob", title.adj = 0.95,
lty = 1, col = 1:3, lwd = 2, box.lty = 0)
```
The rbinom function
The rbinom function allows you to draw n random observations from a binomial distribution in R. The arguments of the function are described below:
```
rbinom syntax
rbinom(n, # Number of random observations to be generated
size, # Number of trials (> = 0)
prob) # The probability of success on each trial
```
If you want to obtain, for instance, 15 random observations from a binomial distribution if the number of trials is 30 and the probability of success on each trial is 0.1 you can type:
rbinom(= 15, = 30, =0.1)
7 3 2 1 5 1 1 4 4 3 1 5 2 4 1
Nonetheless, if you don’t specify a seed before executing the function you will obtain a different set of random observations. If you want to make the output reproducible you can set a seed as follows:
set.seed(2)
rbinom(n = 15, size = 30, prob = 0.1)
2 4 3 1 6 6 1 5 3 3 3 2 4 1 2
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7391 | https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-inscribed-angles/v/inscribed-angles-exercise-example | Inscribed angles (video) | Circles | Khan Academy
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High school geometry
Course: High school geometry>Unit 8
Lesson 7: Inscribed angles
Inscribed angles
Inscribed angles
Challenge problems: Inscribed angles
Inscribed angle theorem proof
Inscribed angle theorem proof
Math>
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Circles>
Inscribed angles
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Inscribed angles
TEKS.Math: G.12A
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About About this video Transcript
Sal finds a missing inscribed angle using the inscribed angle theorem.
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hasitak11170 7 years ago Posted 7 years ago. Direct link to hasitak11170's post “At 0:46, Sal says that "w...” more At 0:46 , Sal says that "we know from the inscribed angle theorem ...."
What exactly is the inscribed angle theorem? Is there another video somewhere that I missed, because I am doing this mission from the beginning? If not, is there a link somewhere that explains this concept?
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Glorfindel 7 years ago Posted 7 years ago. Direct link to Glorfindel's post “The inscribed angle theor...” more The inscribed angle theorem states that the inscribed angle has one half the degree of the central angle that shares the same arc with the inscribed angle. The theorem is explained later in the video.
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josh 10 years ago Posted 10 years ago. Direct link to josh's post “Can someone please explai...” more Can someone please explain? I think I need some help on this.
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mrsdanielgray 2 years ago Posted 2 years ago. Direct link to mrsdanielgray's post “Just take half of the ang...” more Just take half of the angle(divide by 2) and you get an outcome.
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sewerurchin 8 months ago Posted 8 months ago. Direct link to sewerurchin's post “I got the same problem 3 ...” more I got the same problem 3 times in a row on the exercise ;)
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vihaan 2 years ago Posted 2 years ago. Direct link to vihaan's post “This might be a dumb ques...” more This might be a dumb question but what are inscribed angles?
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kubleeka 2 years ago Posted 2 years ago. Direct link to kubleeka's post “We say an angle is inscri...” more We say an angle is inscribed in a circle if the vertex is on the edge of the circle, and the legs go through the interior of the circle.
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lived4adream 5 years ago Posted 5 years ago. Direct link to lived4adream's post “Don't we actually calcula...” more Don't we actually calculate the angle using Θ=arc length/radius? As the radius(distance) is doubled (=diameter in that case), initial Θ is multiplied by 1/2.
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HZWang 5 years ago Posted 5 years ago. Direct link to HZWang's post “Hi lived4adream, the answ...” more Hi lived4adream, the answer is no, we don't. The ratio you are talking about is the radian measurement(arc length/radius). Radians are not used for inscribed angles; their purpose is to resemble and serve as a unit of measurement for the central angle derived from the ratio of the arc length of a central angle and the radius of the circle. Besides, in this case, AD and CD are not diameters of circle B. The basis of the inscribed angle theorem is a bit more complicated and different from what you are thinking of.
Overall, great question!
Hope you found this helpful and feel free to ask if you have any more questions!
~Hannah
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Caleb Allen 2 years ago Posted 2 years ago. Direct link to Caleb Allen's post “this has nothing to do wi...” more this has nothing to do with the questions in the exercise?
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noormohamed1616 7 years ago Posted 7 years ago. Direct link to noormohamed1616's post “when he says <ABC he take...” more when he says <ABC he takes it the way show in the video. my question is, why should we not take the other angle i.e., the greater angles more than 180 one?
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Khaled Fayed Ghaleb 7 years ago Posted 7 years ago. Direct link to Khaled Fayed Ghaleb's post “If you refer to 0:15; you...” more If you refer to 0:15 ; you could understand by other way that it is the angle of intersection between the line AB and line BC at the vertex B.
and by common thinking and stated in this course before we measure the less angle (angle is corner in latin) unless the problem define the opposite
please refer to
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PI Technology Π 6 years ago Posted 6 years ago. Direct link to PI Technology Π's post “What is the definition of...” more What is the definition of inscribed angle ?
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𝕐𝕒𝕤𝕙𝕒𝕤 𝕊 2 years ago Posted 2 years ago. Direct link to 𝕐𝕒𝕤𝕙𝕒𝕤 𝕊's post “An inscribed angle is the...” more An inscribed angle is the angle formed in the interior of a circle when two chords intersect the same arc.
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asims001 2 years ago Posted 2 years ago. Direct link to asims001's post “what are inscribed angles” more what are inscribed angles
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𝕐𝕒𝕤𝕙𝕒𝕤 𝕊 2 years ago Posted 2 years ago. Direct link to 𝕐𝕒𝕤𝕙𝕒𝕤 𝕊's post “An inscribed angle is the...” more An inscribed angle is the angle formed in the interior of a circle when two chords intersect the same arc.
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saumyadangol2010 a year ago Posted a year ago. Direct link to saumyadangol2010's post “Where is a video on the i...” more Where is a video on the inscribed angle theorem?
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Birchlight thedragon🐲 5 months ago Posted 5 months ago. Direct link to Birchlight thedragon🐲's post “ more That's the link to the video!
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Video transcript
[Voiceover] A circle is centered on point B. We see that right over there. That's the center of this big, blue circle. Points A, C, and D lie on its circumference. We see that. Points A, C, and D lie on the circumference. If angle ABC... So ABC; So that's this central angle right over here; measures 132 degrees. Alright, so this is 132 degrees. What does angle ADC measure? A, D, C. So let's think about how these are related. ABC is a central angle. ADC is an inscribed angle. And they intercept the same arc. The arc AC. They both intercept this arc right over here. And we know from the inscribed angle theorem that an inscribed angle that intercepts the same arc as a central angle is going to have half the angle measure. And it even looks that way right over here. So if ABC- if the central angle is 132 degrees, then the inscribed angle that intercepts the same arc is going to be half of that. So half of 132 degrees is what? It is 66 degrees. We can check our answer, and we got it right.
Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube
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7392 | https://www.khanacademy.org/economics-finance-domain/microeconomics/elasticity-tutorial/price-elasticity-tutorial/a/price-elasticity-of-demand-and-price-elasticity-of-supply-cnx | Price elasticity of demand and price elasticity of supply (article) | Khan Academy
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Microeconomics
Course: Microeconomics>Unit 3
Lesson 1: Price elasticity of demand
Introduction to price elasticity of demand
Price elasticity of demand using the midpoint method
More on elasticity of demand
Determinants of price elasticity of demand
Determinants of elasticity example
Price Elasticity of Demand and its Determinants
Perfect inelasticity and perfect elasticity of demand
Constant unit elasticity
Total revenue and elasticity
More on total revenue and elasticity
Elasticity and strange percent changes
Price elasticity of demand and price elasticity of supply
Elasticity in the long run and short run
Elasticity and tax revenue
Determinants of price elasticity and the total revenue rule
Economics>
Microeconomics>
Elasticity>
Price elasticity of demand
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Price elasticity of demand and price elasticity of supply
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How do quantities supplied and demanded react to changes in price?
Key points
Price elasticity measures the responsiveness of the quantity demanded or supplied of a good to a change in its price. It is computed as the percentage change in quantity demanded—or supplied—divided by the percentage change in price.
Elasticity can be described as elastic—or very responsive—unit elastic, or inelastic—not very responsive.
Elastic demand or supply curves indicate that the quantity demanded or supplied responds to price changes in a greater than proportional manner.
An inelastic demand or supply curve is one where a given percentage change in price will cause a smaller percentage change in quantity demanded or supplied.
Unitary elasticity means that a given percentage change in price leads to an equal percentage change in quantity demanded or supplied.
What is price elasticity?
Both demand and supply curves show the relationship between price and the number of units demanded or supplied. Price elasticity is the ratio between the percentage change in the quantity demanded, Q d, or supplied, Q s, and the corresponding percent change in price.
The price elasticity of demand is the percentage change in the quantity demanded of a good or service divided by the percentage change in the price. The price elasticity of supply is the percentage change in quantity supplied divided by the percentage change in price.
Elasticities can be usefully divided into five broad categories: perfectly elastic, elastic, perfectly inelastic, inelastic, and unitary. An elastic demand or elastic supply is one in which the elasticity is greater than one, indicating a high responsiveness to changes in price. An inelastic demand or inelastic supply is one in which elasticity is less than one, indicating low responsiveness to price changes. Unitary elasticities indicate proportional responsiveness of either demand or supply.
Perfectly elastic and perfectly inelastic refer to the two extremes of elasticity. Perfectly elastic means the response to price is complete and infinite: a change in price results in the quantity falling to zero. Perfectly inelastic means that there is no change in quantity at all when price changes.
| If . . . | It Is Called . . . |
--- |
| %change in quantity%change in price=∞ | Perfectly elasti |
| %change in quantity%change in price>1 | Elastic |
| %change in quantity%change in price=1 | Unitary |
| %change in quantity%change in price<1 | Inelastic |
| %change in quantity%change in price=0 | Perfectly inelastic |
Using the midpoint method to calculate elasticity
To calculate elasticity, instead of using simple percentage changes in quantity and price, economists sometimes use the average percent change in both quantity and price. This is called the Midpoint Method for Elasticity:
Midpoint method for elasticity=Q 2−Q 1(Q 2+Q 1 2)P 2−P 1(P 2+P 1 2)
The advantage of the midpoint method is that we get the same elasticity between two price points whether there is a price increase or decrease. This is because the formula uses the same base for both cases. The midpoint method is referred to as the arc elasticity in some textbooks.
Using the point elasticity of demand to calculate elasticity
A drawback of the midpoint method is that as the two points get farther apart, the elasticity value loses its meaning. For this reason, some economists prefer to use the point elasticity method. In this method, you need to know what values represent the initial values and what values represent the new values.
Point elasticity=new Q−initial Q initial Q initial P−new P initial P
Calculating price elasticity of demand
Let’s apply these formulas to a practice scenario. We'll calculate the elasticity between points A and B in the graph below.
Image credit: Figure 1 in "Price Elasticity of Demand and Price Elasticity of Supply" by OpenStaxCollege, CC BY 4.0
First, apply the formula to calculate the elasticity as price decreases from $70 at point B to $60 at point A:
%change in quantity=3,000–2,800(3,000+2,800)/2×100=200 2,900×100=6.9%change in price=60–70(60+70)/2×100=–10 65×100=–15.4 Price elasticity of demand=6.9%–15.4%=0.45
The elasticity of demand between point A and point B is 6.9%–15.4%, or 0.45. Because this amount is smaller than one, we know that the demand is inelastic in this interval.
Wait a minute! Shouldn't the price elasticity of demand be negative here?
Price elasticities of demand are always negative since price and quantity demanded always move in opposite directions on the demand curve. But, by convention, we talk about elasticities as positive numbers. So mathematically, we take the absolute value of the result. From now on, we'll ignore this detail and just remember to interpret elasticities as positive numbers.
This means that, along the demand curve between point B and point A, if the price changes by 1%, the quantity demanded will change by 0.45%. A change in the price will result in a smaller percentage change in the quantity demanded. For example, a 10% increase in the price will result in only a 4.5% decrease in quantity demanded. A 10% decrease in the price will result in only a 4.5% increase in the quantity demanded.
I'd like to do another practice problem.
Let's do another practice problem to cement this concept!
Calculate the price elasticity of demand for an increase in price from G to H in the graph above.
Step 1. Start with the formula for price elasticity of demand:
Step 2. Use the Midpoint Method formula to calculate percent change in quantity and price:
%change in quantity=Q 2–Q 1(Q 2+Q 1)/2×100%change in quantity=1,600–1,800(1,600+1,800)/2×100=–200 1,700×100=–11.76
%change in price=P 2–P 1(P 2+P 1)/2×100%change in price=130–120(130+120)/2×100=10 125×100=8.0
Step 4. Use the results of step three to fill in the formula for price elasticity of demand:
Price elasticity of demand=%change in quantity%change in price=−11.76 8=1.47
The elasticity of demand from G to H is 1.47.
Recall that the elasticity between point A point B was 0.45. So, demand was inelastic between points A and B and elastic between points G and H. This shows us that price elasticity of demand changes at different points along a straight-line demand curve.
Calculating the price elasticity of supply
Now let's try calculating the price elasticity of supply. We use the same formula as we did for price elasticity of demand:
Price elasticity of supply=%change in quantity%change in price
Assume that an apartment rents for $650 per month and, at that price, 10,000 units are rented—you can see these number represented graphically below. When the price increases to $700 per month, 13,000 units are supplied into the market.
By what percentage does apartment supply increase? What is the price sensitivity?
Image credit: Figure 2 in "Price Elasticity of Demand and Price Elasticity of Supply" by OpenStaxCollege, CC BY 4.0
We'll start by using the Midpoint Method to calculate percentage change in price and quantity:
%change in quantity=13,000–10,000(13,000+10,000)/2×100=3,000 11,500×100=26.1%change in price=$700–$650($700+$650)/2×100=50 675×100=7.4
Next, we take the results of our calculations and plug them into the formula for price elasticity of supply:
Price elasticity of supply=%change in quantity%change in price=26.1 7.4=3.53
Again, as with the elasticity of demand, the elasticity of supply is not followed by any units. Elasticity is a ratio of one percentage change to another percentage change—nothing more. It is read as an absolute value. In this case, a 1% rise in price causes an increase in quantity supplied of 3.5%. The greater than one elasticity of supply means that the percentage change in quantity supplied will be greater than a one percent price change.
Hang on! This sounds familiar. Is the elasticity the slope?
It is a common mistake to confuse the slope of either the supply or demand curve with its elasticity.
The slope is the rate of change in units along the curve, or the rise/run—change in y over the change in x. For example, in the graph above, at each point shown on the demand curve, price drops by $10 and the number of units demanded increases by 200. So the slope is −10/200 along the entire demand curve; it does not change.
The price elasticity, however, changes along the curve. Elasticity between point A and point B was 0.45 and increased to 1.47 between points point G and point H. Elasticity is the percentage change, which is a different calculation from the slope and has a different meaning.
When we are at the upper end of a demand curve, where price is high and the quantity demanded is low, a small change in the quantity demanded—even, say, one unit—is pretty big in percentage terms. A change in price of a dollar is going to be much less important in percentage terms than it would have been at the bottom of the demand curve. Likewise, at the bottom of the demand curve, that one unit change when the quantity demanded is high will be small as a percentage.
So, at one end of the demand curve, where we have a large percentage change in quantity demanded over a small percentage change in price, the elasticity value would be high, or demand would be relatively elastic. Even with the same change in the price and the same change in the quantity demanded, at the other end of the demand curve the quantity is much higher, and the price is much lower, so the percentage change in quantity demanded is smaller and the percentage change in price is much higher. That means at the bottom of the curve we have a small numerator over a large denominator, so the elasticity measure would be much lower, or inelastic.
As we move along the demand curve, the values for quantity and price go up or down, depending on which way we are moving, so the percentages for a $1 difference in price or a one unit difference in quantity will change as well, which means the ratios of those percentages will change.
Summary
Price elasticity measures the responsiveness of the quantity demanded or supplied of a good to a change in its price. It is computed as the percentage change in quantity demanded—or supplied—divided by the percentage change in price.
Elasticity can be described as elastic—or very responsive—unit elastic, or inelastic—not very responsive.
Elastic demand or supply curves indicate that the quantity demanded or supplied responds to price changes in a greater than proportional manner.
An inelastic demand or supply curve is one where a given percentage change in price will cause a smaller percentage change in quantity demanded or supplied.
Unitary elasticity means that a given percentage change in price leads to an equal percentage change in quantity demanded or supplied.
Self-check questions
Using the data shown in the table below about demand for smart phones, calculate the price elasticity of demand from point B to point C, point D to point E, and point G to point H. Classify the elasticity at each point as elastic, inelastic, or unit elastic.
| Points | P | Q |
---
| A | 60 | 3,000 |
| B | 70 | 2,800 |
| C | 80 | 2,600 |
| D | 90 | 2,400 |
| E | 100 | 2,200 |
| F | 110 | 2,000 |
| G | 120 | 1,800 |
| H | 130 | 1,600 |
Show solution.
From point B to point C, price rises from $70 to $80, and Qd decreases from 2,800 to 2,600. Using these numbers, we can fill in the formulas for percent change in quantity and price:
%change in quantity=2600–2800(2600+2800)÷2×100=–200 2700×100=–7.41%change in price=80–70(80+70)÷2×100=10 75×100=13.33
Now, we can use the results of our calculations above to fill in the formula for elasticity of demand:
Elasticity of demand=–7.41%13.33%=0.56
The demand curve is inelastic in this area; that is, its elasticity value is less than one.
Now follow the same process using the data from point D to point E:
%change in quantity=2200–2400(2200+2400)÷2×100=–200 2300×100=–8.7%change in price=100–90(100+90)÷2×100=10 95×100=10.53 Elasticity of demand=–8.7%10.53%=0.83
The demand curve is inelastic in this area; that is, its elasticity value is less than one.
And from Point G to point H:
%change in quantity=1600–1800 1700×100=–200 1700×100=–11.76%change in price=130–120 125×100=10 125×100=8%Elasticity of demand=–11.76%8%=–1.47
The demand curve is elastic in this interval.
Using the data shown in in the table below about supply of alarm clocks, calculate the price elasticity of supply from: point J to point K, point L to point M, and point N to point P. Classify the elasticity at each point as elastic, inelastic, or unit elastic.
| Point | Price | Quantity Supplied |
---
| J | $8 | 50 |
| K | $9 | 70 |
| L | $10 | 80 |
| M | $11 | 88 |
| N | $12 | 95 |
| P | $13 | 100 |
Show solution.
From point J to point K, price rises from $8 to $9, and quantity rises from 50 to 70. Using these numbers, we can fill in the formulas for percent change in quantity and price:
%change in quantity=70–50(70+50)÷2×100=20 60×100=33.33%change in price=$9–$8($9+$8)÷2×100=1 8.5×100=11.76
Now, we can use the results of our calculations above to fill in the formula for elasticity of supply:
Elasticity of supply=33.33%11.76%=2.83
The supply curve is elastic in this area; that is, its elasticity value is greater than one.
Now follow the same process using the data from point L to point M; the price rises from $10 to $11, while the Qs rises from 80 to 88.
%change in quantity=88–80(88+80)÷2×100=8 84×100=9.52%change in price=$11–$10($11+$10)÷2×100=1 10.5×100=9.52 Elasticity of supply=9.52%9.52%=1.0
The supply curve has unitary elasticity in this area.
And finally, from point N to point P, the price rises from $12 to $13, and Qs rises from 95 to 100.
%change in quantity=100–95(100+95)÷2×100=5 97.5×100=5.13%change in price=$13–$12($13+$12)÷2×100=1 12.5×100=8.0 Elasticity of supply=5.13%8.0%=0.64
The supply curve is inelastic in this region of the supply curve.
Review Questions
What is the formula for calculating elasticity?
What is the price elasticity of demand? Can you explain it in your own words?
What is the price elasticity of supply? Can you explain it in your own words?
Critical-thinking questions
Transatlantic air travel in business class has an estimated elasticity of demand of 0.40 less than transatlantic air travel in economy class, which has an estimated price elasticity of 0.62. Why do you think this is the case?
What is the relationship between price elasticity and position on the demand curve? For example, as you move up the demand curve to higher prices and lower quantities, what happens to the measured elasticity? How would you explain that?
Problems
The equation for a demand curve is P=48–3 Q. What is the elasticity in moving from a quantity of 5 to a quantity of 6?
The equation for a demand curve is P=2/Q. What is the elasticity of demand as price falls from 5 to 4? What is the elasticity of demand as the price falls from 9 to 8? Would you expect these answers to be the same?
The equation for a supply curve is 4 P=Q. What is the elasticity of supply as price rises from 3 to 4? What is the elasticity of supply as the price rises from 7 to 8? Would you expect these answers to be the same?
The equation for a supply curve is P=3 Q–8. What is the elasticity in moving from a price of 4 to a price of 7?
Attribution
Attribution
This article is a modified derivative of "Price Elasticity of Demand and Price Elasticity of Supply" by OpenStaxCollege, CC BY 4.0.
The modified article is licensed under a CC BY-NC-SA 4.0 license.
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hugofetiza 8 years ago Posted 8 years ago. Direct link to hugofetiza's post “I feel as if the apartmen...” more I feel as if the apartment problem was not worded properly. In the first sentence, "rented" should be "supplied" since what is rented isn't always the full supply.
For reference, I put the problem below:
Assume that an apartment rents for $650 per month and, at that price, 10,000 units are rented—you can see these number represented graphically below. When the price increases to $700 per month, 13,000 units are supplied into the market.
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anutkalaund 4 years ago Posted 4 years ago. Direct link to anutkalaund's post “Why do we use arc elastic...” more Why do we use arc elasticity in the first task?
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Sakshi Shah 8 years ago Posted 8 years ago. Direct link to Sakshi Shah's post “In the problem two, can s...” more In the problem two, can someone help me with a justification as to why both the price falls indicate a unitary elasticity. How can we prove it mathematically?
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Akshaya Balaji 7 years ago Posted 7 years ago. Direct link to Akshaya Balaji's post “Hi. Under price elasticit...” more Hi. Under price elasticity of demand , the calculation for elasticity in the figure 1 from A to B, since the price is already low and there is a decrease from $70 to $60, I gather that the degree of responsiveness would be higher and it would be elastic i.e. greater than 1..isn't that possible? I've been having doubts of a similar nature..please help!
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Andrew M 7 years ago Posted 7 years ago. Direct link to Andrew M's post “The elasticity can vary f...” more The elasticity can vary from point to point along the demand curve. It's not constant for most demand curve shapes.
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rosemerybarrera4 2 months ago Posted 2 months ago. Direct link to rosemerybarrera4's post “how do we use the midpoin...” more how do we use the midpoint method to calculate the elasticity ?
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Jasmine a month ago Posted a month ago. Direct link to Jasmine's post “You can find it out above...” more You can find it out above in the text.
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divayanshbargotra a year ago Posted a year ago. Direct link to divayanshbargotra's post “I could work out how elas...” more I could work out how elasticity for the curves (problem 2) P=2/Q and (problem 3) P=Q/4 would always be equal to 1 at all points on the curve with the help of Qi.Z's answer in the comments.
However, I have a hunch that there is a way to judge when the curve is going to be unit elastic at all points just by looking at the equation of the curve. Is this the case, if so, how can we figure out which equations will lead to unit elasticity at all points?
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malikakbaar 5 years ago Posted 5 years ago. Direct link to malikakbaar's post “I still don't understand ...” more I still don't understand why do we need to have the midpoint method? In my intuition, it seems that the method shifts the meaning of the elasticity itself. Can anyone share the why behind it? Also, in the real case, when will economist calculate with or without midpoint?
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I am bad at maths 5 years ago Posted 5 years ago. Direct link to I am bad at maths's post “when do we use the midpoi...” more when do we use the midpoint method or the normal method to calculate the elasticity? Which one should we use to do the exercises on Khanacademy?
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7393 | https://proofwiki.org/wiki/Focal_Property_of_Ellipse | Focal Property of Ellipse
From ProofWiki
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Contents
1 Theorem
2 Proof
3 Also known as
4 Sources
Theorem
Let $\EE$ be an ellipse.
Let $P$ be an arbitrary point on $\EE$.
Let $APB$ be a tangent to $\EE$ at $P$.
Then:
: $\angle APF_1 = \angle BPF_2$
where $F_1$ and $F_2$ are the foci of $\EE$.
Proof
| |
| This theorem requires a proof.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Also known as
The focal property of an ellipse is also known as the reflection property.
In the context of physics, it is also called the optical property or acoustical property
Sources
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): ellipse
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): ellipse
Retrieved from "
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Foci of Ellipses
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7394 | https://louis.pressbooks.pub/appliedcalculus/chapter/3-1-optimization/ | Skip to content
Chapter 3: Applications of the Derivative
Section 3.1: Optimization
Learning Objectives
By the end of this section, the student should be able to:
Formulate a real-world problem as a mathematical function that requires optimization.
Utilize optimization techniques to identify the absolute extrema (minimum and/or maximum) of a given function.
Use the first derivative test to find local extremes.
Use the second derivative test to find local extremes.
Real-World Optimization Problems
In theory and applications, we often want to maximize or minimize some quantity. An engineer may want to maximize the speed of a new computer or minimize the heat produced by an appliance. A manufacturer may want to maximize profits and market share or minimize waste. A student may want to maximize a grade in calculus or minimize the hours of study needed to earn a particular grade.
Without calculus, we only know how to find the optimum points in a few specific examples (for example, we know how to find the vertex of a parabola). But what if we need to optimize an unfamiliar function?
The best way we have without calculus is to examine the graph of the function, perhaps using technology. But our view depends on the viewing window we choose – we might miss something important. In addition, we’ll probably only get an approximation this way. (In some cases, that will be good enough.)
Calculus provides ways of drastically narrowing the number of points we need to examine to find the exact locations of maximums and minimums while at the same time ensuring that we haven’t missed anything important.
Video
Local Maxima and Minima
Before we examine how calculus can help us find maximums and minimums, we need to define the concepts we will develop and use.
Definitions (Local Maxima and Minima)
[latex]f(x)[/latex] has a local maximum at [latex]x=a[/latex] if [latex]f(a) \geq f(x)[/latex] for all [latex]x[/latex] near [latex]a[/latex].
[latex]f(x)[/latex] has a local minimum at [latex]x=a[/latex] if [latex]f(a) \leq f(x)[/latex] for all [latex]x[/latex] near [latex]a[/latex].
[latex]f(x)[/latex] has a local extreme at [latex]x=a[/latex] if [latex]f(a)[/latex] is a local maximum or minimum.
The plurals of these are maxima and minima. We often simply say “max” or “min”; it saves a lot of syllables.
Some books say “relative” instead of “local.”
The process of finding maxima or minima is called optimization.
A point is a local max (or min) if it is higher (lower) than all the nearby points. These points come from the shape of the graph.
Definitions (Global Maxima and Minima)
[latex]f(x)[/latex] has a global maximum at [latex]x=a[/latex] if [latex]f(a) \geq f(x)[/latex] for all [latex]x[/latex] in the domain of [latex]f(x)[/latex].
[latex]f(x)[/latex] has a global minimum at [latex]x=a[/latex] if [latex]f(a) \leq f(x)[/latex] for all [latex]x[/latex] in the domain of [latex]f(x)[/latex].
[latex]f(x)[/latex] has a global extreme at [latex]x=a[/latex] if [latex]f(a)[/latex] is a global maximum or minimum.
Some books say “absolute” instead of “global.”
A point is a global max (or min) if it is higher (lower) than every point on the graph. These points come from the shape of the graph and the window through which we view the graph.
The local and global extremes of the function in Figure 1 are labeled. You should notice that every global extreme is also a local extreme, but there are local extremes that are not global extremes.
If [latex]h(x)[/latex] is the height of the earth above sea level at the location [latex]x[/latex], then the global maximum of [latex]h[/latex] is [latex]h(summit \ of \ Mt. \ Everest) = 29,028[/latex] feet. The local maximum of [latex]h[/latex] for the United States is [latex]h(summit \ of \ Mt. \ McKinley) = 20,320[/latex] feet. The local minimum of [latex]h[/latex] for the United States is [latex]h(Death \ Valley) = -282[/latex] feet.
Example 1
The table shows the annual calculus enrollments at a large university. Which years had local maximum or minimum calculus enrollments? What were the global maximum and minimum enrollments in calculus?
| | | | | | | | | | | | |
--- --- --- --- --- --- |
| Year | 2000 | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 |
| Enrollment | 1257 | 1324 | 1378 | 1336 | 1389 | 1450 | 1523 | 1582 | 1567 | 1545 | 1571 |
There were local maxima in [latex]2002[/latex] and [latex]2007[/latex]; the global maximum was [latex]1582[/latex] students in [latex]2007[/latex]. There were local minima in [latex]2003[/latex] and 2009; the global minimum was [latex]1257[/latex] students in [latex]2000[/latex]. We choose not to think of [latex]2000[/latex] as a local minimum or [latex]2010[/latex] as a local maximum; however, some books would include the endpoints. We are allowed to have a global maximum or global minimum at an endpoint.
Finding Maxima and Minima of a Function
What must the tangent line look like at a local max or min? Look at these two graphs again – you’ll see that at all the extreme points, the tangent line is horizontal (so [latex]f' = 0[/latex]). There is one cusp in the blue graph – the tangent line is vertical there (so [latex]f'[/latex] is undefined).
That gives us the clue how to find extreme values.
Definitions
A critical number for a function [latex]f[/latex] is a value [latex]x = a[/latex] in the domain of [latex]f[/latex] where either [latex]f'(a) = 0[/latex] or [latex]f'(a)[/latex] is undefined.
A critical point for a function f is a point (a, f(a)) where a is a critical number of f.
A local max or min of f can only occur at a critical point.
Example 2
Find the critical points of [latex]f(x) = x^3 - 6x^2 + 9x + 2[/latex].
A critical number of [latex]f[/latex] can occur only where [latex]f'(x) = 0[/latex] or where [latex]f'[/latex] does not exist.
[latex]f '(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)[/latex] So [latex]f'(x) = 0[/latex] at [latex]x = 1[/latex] and [latex]x = 3[/latex] (and no other values of [latex]x[/latex]). There are no places where [latex]f'[/latex] is undefined.
The critical numbers are [latex]x = 1[/latex] and [latex]x= 3[/latex]. So the critical points are latex[/latex] and latex[/latex].
These are the only possible locations of local extremes of [latex]f[/latex]. We haven’t discussed yet how to tell whether either of these points is actually a local extreme of [latex]f[/latex] or which kind it might be. But we can be certain that no other point is a local extreme.
The graph of [latex]f[/latex] below shows that latex = (1, 6)[/latex] is a local maximum and latex = (3, 2)[/latex] is a local minimum. This function does not have a global maximum or minimum.
Example 3
Find all local extremes of [latex]f(x) = x^3[/latex].
[latex]f(x) = x^3[/latex] is differentiable for all [latex]x[/latex], and [latex]f'(x) = 3x^2[/latex]. The only place where [latex]f'(x) = 0[/latex] is at [latex]x = 0[/latex], so the only candidate is the critical point (0,0).
If [latex]x \gt 0[/latex], then [latex]f(x) = x^3 \gt 0 = f(0)[/latex], so [latex]f(0)[/latex] is not a local maximum.
Similarly, if [latex]x \lt 0[/latex], then [latex]f(x) = x^3 \lt 0 = f(0)[/latex] so [latex]f(0)[/latex] is not a local minimum.
The critical point (0,0) is the only candidate to be a local extreme of [latex]f[/latex], and based on the graph, this candidate did not turn out to be a local extreme of [latex]f[/latex]. The function [latex]f(x) = x^3[/latex] does not have any local extremes.
Remember this example! It is not enough to find the critical points – we can only say that [latex]f[/latex] might have a local extreme at the critical points.
First and Second Derivative Tests
Is That Critical Point a Maximum or Minimum (or Neither)?
Once we have found the critical points of [latex]f[/latex], we still have the problem of determining whether these points are maxima, minima or neither.
All of the graphs in the figure below have a critical point at latex[/latex]. It is clear from the graphs that the point latex[/latex] is a local maximum in (a) and (d), latex[/latex] is a local minimum in (b) and (e), and latex[/latex] is not a local extreme in (c) and (f).
The critical numbers only give the possible locations of extremes, and some critical numbers are not the locations of extremes. The critical numbers are the candidates for the locations of maxima and minima.
[latex]f'[/latex] and Extreme Values of [latex]f[/latex]
Four possible shapes of graphs are shown here – in each graph, the point marked by an arrow is a critical point, where [latex]f'(x) = 0[/latex]. What happens to the derivative near the critical point?
At a local max, such as in the graph on the left, the function increases on the left of the local max, then decreases on the right. The derivative is first positive, then negative at a local max. At a local min, the function decreases to the left and increases to the right, so the derivative is first negative, then positive. When there isn’t a local extreme, the function continues to increase (or decrease) right past the critical point – the derivative doesn’t change sign.
The First Derivative Test
The First Derivative Test for Extremes
Find the critical points of [latex]f[/latex].
For each critical number c, examine the sign of [latex]f'[/latex] to the left and to the right of c. What happens to the sign as you move from left to right?
If [latex]f'(x)[/latex] changes from positive to negative at [latex]x = c[/latex], then [latex]f[/latex] has a local maximum at latex[/latex].
If [latex]f'(x)[/latex] changes from negative to positive at [latex]x = c[/latex], then [latex]f[/latex] has a local minimum at latex[/latex].
If [latex]f'(x)[/latex] does not change sign at [latex]x = c[/latex], then latex[/latex] is neither a local maximum nor a local minimum.
Example 4
Find the critical points of [latex]f(x) = x^3 - 6x^2 + 9x + 2[/latex] and classify them as local max, local min, or neither.
We already found the critical points; they are latex[/latex] and latex[/latex].
Now we can use the first derivative test to classify each. Recall that [latex]f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)[/latex]. The factored form is easiest to work with here, so let’s use that.
At (1, 6) we could choose a number slightly less than 1 to plug into the formula for [latex]f'[/latex] – perhaps use [latex]x = 0[/latex] or [latex]x = 0.9[/latex]. Then we could examine its sign. But we don’t care about the numerical value; all we are interested in is its sign. And for that, we don’t have to do any plugging in:
If [latex]x[/latex] is a little less than 1, then [latex]x-1[/latex] is negative, and [latex]x-3[/latex] is negative. So [latex]f' = 3(x - 1)(x - 3)[/latex] will be pos(neg)(neg) = positive.
For [latex]x[/latex] a little more than 1, we can evaluate [latex]f'[/latex] at a number more than 1 (but less than 3, we don’t want to go past the next critical point!) – perhaps [latex]x = 2[/latex]. Or we can make a quick sign argument like what we did above: for [latex]x[/latex] a little more than 1, [latex]f' = 3(x - 1)(x - 3)[/latex] will be pos(pos)(neg) = negative.
So [latex]f'[/latex] changes from positive to negative, which means there is a local max at latex[/latex].
As another approach, we could draw a number line and mark the critical numbers:
We already know the derivative is zero or undefined at the critical numbers. On each interval between these values, the derivative will stay the same sign. To determine the sign, we could pick a test value in each interval and evaluate the derivative at those points (or use the sign approach used above).
At latex \ f'[/latex] changes from negative to positive, so there is a local min at latex[/latex].This confirms what we saw before in the graph.
[latex]f''[/latex] and Extreme Values of [latex]f[/latex]
The concavity of a function can also help us determine whether a critical point is a maximum or minimum or neither. For example, if a point is at the bottom of a concave up function, then the point is a minimum.
The Second Derivative Test
The Second Derivative Test for Extremes
Find all critical points of [latex]f[/latex]. For those critical points where [latex]f'(c) = 0[/latex], find [latex]f''(c)[/latex].
If [latex]f''(c) \lt 0[/latex] (negative), then [latex]f[/latex] is concave down and has a local maximum at [latex]x = c[/latex].
If [latex]f''(c) \gt 0[/latex] (positive), then [latex]f[/latex] is concave up and has a local minimum at [latex]x = c[/latex].
If [latex]f''(c) = 0[/latex], then [latex]f[/latex] may have a local maximum, a minimum or neither at [latex]x = c[/latex].
The cartoon faces can help you remember the second derivative test.
Example 5
[latex]f(x) = 2x^3 - 15x^2 + 24x - 7[/latex] has critical numbers [latex]x =[/latex] 1 and 4. Use the second derivative test for extremes to determine whether [latex]f(1)[/latex] and [latex]f(4)[/latex] are maximums or minimums or neither.
We need to find the second derivative:[latex]\begin{align} f(x)=& 2x^3 - 15x^2 + 24x - 7\ f'(x)=& 6x^2 - 30x + 24\ f''(x)=& 12x - 30 \end{align}[/latex]
Then we just need to evaluate [latex]f''[/latex] at each critical number:
[latex]x = 1[/latex]: [latex]f''(1)=12(1)-30 \lt 0[/latex], so there is a local maximum at [latex]x = 1[/latex].
[latex]x = 4[/latex]: [latex]f''(4)=12(4)-30 \gt 0[/latex], so there is a local minimum at [latex]x = 4[/latex].
Many students like the second derivative test. The second derivative test is often easier to use than the first derivative test. You only have to find the sign of one number for each critical number rather than two. And if your function is a polynomial, its second derivative will probably be a simpler function than the derivative.
However, if you needed a product rule, quotient rule, or chain rule to find the first derivative, finding the second derivative can be a lot of work. Also, even if the second derivative is easy, the second derivative test doesn’t always give an answer. The first derivative test will always give you an answer.
Use whichever test you want to. But remember – you have to do some tests to be sure that your critical point actually is a local max or min.
Examples
Video
Global Maxima and Minima
In applications, we often want to find the global extreme; knowing that a critical point is a local extreme is not enough.
For example, if we want to make the greatest profit, we want to make the absolutely greatest profit of all. How do we find global max and min?
There are just a few additional things to think about.
Endpoint Extremes
The local extremes of a function occur at critical points – these are points in the function that we can find by thinking about the shape (and using the derivative to help us). But if we’re looking at a function on a closed interval, the endpoints could be extremes. These endpoint extremes are not related to the shape of the function; they have to do with the interval, the window through which we’re viewing the function.
In the graph above, it appears that there are three critical points – one local min, one local max, and one that is neither one. But the global max, the highest point of all, is at the left endpoint. The global min, the lowest point of all, is at the right endpoint.
How do we decide if endpoints are global max or min? It’s easier than you expected – simply plug in the endpoints, along with all the critical numbers, and compare [latex]y[/latex]-values.
Example 6
Find the global max and min of [latex]f(x) = x^3 - 3x^2 - 9x + 5[/latex] for [latex]-2 \leq x \leq 6[/latex].
[latex]f'(x) = 3x^2 - 6x - 9 = 3(x + 1)(x - 3)[/latex]. We need to find critical points, and we need to check the endpoints.
[latex]f'(x) = 3(x + 1)(x - 3) = 0[/latex] when [latex]x = -1[/latex] and [latex]x = 3[/latex]. The endpoints of the interval are [latex]x = -2[/latex] and [latex]x = 6[/latex].
Now we simply compare the values of [latex]f[/latex] at these four values of [latex]x[/latex]:
| | |
--- |
| [latex]x[/latex] | [latex]f(x)[/latex] |
| [latex]-2[/latex] | [latex]3[/latex] |
| [latex]-1[/latex] | [latex]10[/latex] |
| [latex]3[/latex] | [latex]-22[/latex] |
| [latex]6[/latex] | [latex]59[/latex] |
The global minimum of [latex]f[/latex] on [latex][ -2, 6][/latex] is -22, when [latex]x = 3[/latex], and the global maximum of [latex]f[/latex] on [latex][ -2, 6][/latex] is 59, when [latex]x = 6[/latex].
If There Is Only One Critical Point
If the function has only one critical point and it’s a local max (or min), then it must be the global max (or min). To see this, think about the geometry. Look at the graph on the left – there is a local max, and the graph goes down on either side of the critical point. Suppose there was some other point that was higher – then the graph would have to turn around. But that turning point would have shown up as another critical point. If there’s only one critical point, then the graph can never turn back around.
When in Doubt, Graph It and Look
If you are trying to find a global max or min on an open interval (or the whole real line) and there is more than one critical point, then you need to look at the graph to decide whether there is a global max or min. Be sure that all your critical points show in your graph and that you graph beyond them – that will tell you what you want to know.
Example 7
Find the global max and min of [latex]f(x) = x^3 - 6x^2 + 9x + 2[/latex].
We have previously found that latex[/latex] is a local max and latex[/latex] is a local min. This is not a closed interval, and there are two critical points, so we must turn to the graph of the function to find the global max and min.
The graph of [latex]f[/latex] shows that points to the left of [latex]x = 4[/latex] have [latex]y[/latex]-values greater than [latex]6[/latex], so latex[/latex] is not a global max. Likewise, if [latex]x[/latex] is negative, [latex]y[/latex] is less than [latex]2[/latex], so latex[/latex] is not a global min. There are no endpoints, so we’ve exhausted all the possibilities. This function does not have a global maximum or minimum.
Key Takeaways
The only places where a function can have a global extreme are critical points or endpoints.
If the function has only one critical point, and it’s a local extreme, then it is also the global extreme.
If there are endpoints, find the global extremes by comparing [latex]y[/latex]-values at all the critical points and at the endpoints.
When in doubt, graph the function to be sure. (However, unless the problem explicitly tells you otherwise, it is not enough to just use the graph to get your answer.)
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7395 | https://brighterly.com/math/coefficient/ | Math tutors
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Coefficient – Definition With Examples
Coefficient – Definition With Examples
Jo-ann Caballes
Updated on January 8, 2024
Table of Contents
Welcome to another exhilarating episode of mathematical exploration at Brighterly – a fun, interactive platform designed to nurture the budding mathematicians of tomorrow. Our topic today revolves around coefficients, a foundational concept in the world of mathematics and particularly vital in the field of algebra. A solid grasp of coefficients equips young learners with the tools to simplify and solve algebraic expressions and equations. So, what does this seemingly complex term, ‘coefficient,’ signify? Let’s delve deeper!
In the realm of algebra, a coefficient is a numerical or symbolic factor that precedes a variable (or a group of variables combined by multiplication) within an algebraic expression or equation. This numerical or symbolic multiplier is crucial as it dictates the quantity of a particular variable in the equation. For instance, in the term 7x, ‘7’ is the coefficient, while ‘x’ is the variable. In this context, ‘7’ implies that we have seven instances or ‘lots’ of the variable ‘x’. Intriguingly, coefficients are not restricted to whole numbers alone. They can also be fractions, decimals, negative numbers, or even numbers involving complex calculations.
Types of Coefficients
As we delve into the topic of coefficients, we distinguish between two primary types: numeric coefficients and variable coefficients. Both of these types carry unique characteristics and functions, influencing the way they interact within equations and expressions.
Definition of Numeric Coefficients
Numeric Coefficients are precisely what their name suggests. They are numerical values that precede a variable in an algebraic expression or equation. Acting as multipliers, they provide a fixed, predetermined quantity for the variable in question. This quantity remains constant unless the equation or expression undergoes some form of manipulation. For instance, in the algebraic term 8y, the numeric coefficient is ‘8’.
Definition of Variable Coefficients
Conversely, Variable Coefficients are variables that multiply other variables or mathematical expressions. Their uniqueness lies in their variability, unlike their numeric counterparts. The value of a variable coefficient can change, contingent on the values assigned to other variables within the equation or expression. For instance, in the algebraic expression xy, ‘x’ serves as a variable coefficient for ‘y’.
Properties of Coefficients
Grasping the properties of coefficients is paramount in achieving proficiency in manipulating algebraic expressions. These properties play a vital role in simplifying and solving complex mathematical equations.
Properties of Numeric Coefficients
A significant property of a numeric coefficient is that it exhibits the distributive property over addition or subtraction. Also known as the distributive property, this rule implies that a(b + c) equals ab + ac. In this equation, ‘a’ serves as the numeric coefficient.
Properties of Variable Coefficients
Similar to numeric coefficients, variable coefficients also adhere to the distributive property. For example, in the mathematical expression x(y + z), ‘x’ functions as a variable coefficient, and this expression can be expanded to xy + xz, following the distributive property.
Difference Between Numeric and Variable Coefficients
The fundamental distinction between numeric and variable coefficients resides in their constancy. While numeric coefficients provide a fixed value that does not change, variable coefficients do not adhere to constancy. The value of a variable coefficient can fluctuate, contingent on the values assigned to other variables in the equation.
Coefficients in Algebraic Expressions
In the context of algebra, coefficients serve as indispensable tools for writing, interpreting, and simplifying algebraic expressions and equations. They aid in understanding the quantity of variables and their distribution within an equation or expression.
Writing Algebraic Expressions With Numeric Coefficients
When it comes to formulating algebraic expressions with numeric coefficients, it primarily involves multiplying a numeric value with a variable. For example, the statement “four times a number” could be translated into the algebraic expression 4x, where ‘4’ is the numeric coefficient.
Writing Algebraic Expressions With Variable Coefficients
For variable coefficients, the process involves one variable influencing another variable. Thus, if you encounter a phrase such as “a number ‘x’ times another number ‘y’,” you would formulate it as the algebraic expression xy, where ‘x’ acts as the variable coefficient.
Practice Problems on Coefficients
Can you identify the coefficient in the term: 3xy?
How would you write an algebraic expression for “7 times a number ‘z'”?
How would you simplify the expression: 4(x + 2)?
Encourage your child to attempt these problems independently. You can use our innovative Brighterly Math Solver to verify the answers!
Conclusion
To wrap things up, coefficients represent a core concept in the field of algebra, facilitating a more profound comprehension and simplification of mathematical equations and expressions. Mastery of this concept opens doors to solving complex algebraic problems and lays a solid foundation for understanding more advanced mathematical theories. At Brighterly, we believe in the potential of every child to excel in mathematics, and our mission is to make this learning journey enjoyable and fulfilling. We encourage learners to practice regularly and to use the resources available on our platform to reinforce their understanding. Remember, every great mathematician started with the basics. Let’s keep the spirit of discovery alive, and as always, keep practicing!
Frequently Asked Questions on Coefficients
What exactly is a coefficient?
A coefficient, in the context of algebra, is a numerical or symbolic factor that multiplies a variable within an algebraic expression or equation. It determines the quantity of the variable in the equation or expression. For instance, in the term 3x, ‘3’ is the coefficient, implying we have three lots of ‘x’.
How do numeric and variable coefficients differ?
Numeric coefficients are constant numerical multipliers in an algebraic expression or equation, such as ‘7’ in 7x. They don’t change their value unless the equation or expression is manipulated. On the other hand, variable coefficients are variable multipliers of other variables or expressions, such as ‘x’ in xy. Their value can vary depending on the values of other variables in the equation or expression.
What role do coefficients play in algebraic expressions?
Coefficients play a pivotal role in the formulation, interpretation, and simplification of algebraic expressions and equations. They indicate the quantity and distribution of variables within these expressions and equations, thereby facilitating their manipulation and resolution.
Can coefficients be negative or fractions?
Yes, coefficients can indeed be negative numbers or fractions. For example, in the term -5x, ‘-5’ is a negative coefficient, and in the term (1/2)x, ‘1/2’ is a fractional coefficient. They follow the same rules as other coefficients when it comes to mathematical operations.
How does understanding coefficients help in learning algebra?
Understanding coefficients is critical to learning algebra because it allows students to simplify and manipulate algebraic expressions and equations. By knowing how coefficients function and interact with variables, students can solve for unknowns, simplify expressions, and understand the relationships between different parts of an equation.
Information Sources:
MathWorld – Coefficient
Wolfram Alpha – Coefficients in Algebraic Expressions
BBC Bitesize – Writing Algebraic Expressions
Jo-ann Caballes
14 articles
As a seasoned educator with a Bachelor’s in Secondary Education and over three years of experience, I specialize in making mathematics accessible to students of all backgrounds through Brighterly. My expertise extends beyond teaching; I blog about innovative educational strategies and have a keen interest in child psychology and curriculum development. My approach is shaped by a belief in practical, real-life application of math, making learning both impactful and enjoyable.
Table of Contents
Types of Coefficients
Definition of Numeric Coefficients
Definition of Variable Coefficients
Properties of Coefficients
Properties of Numeric Coefficients
Properties of Variable Coefficients
Difference Between Numeric and Variable Coefficients
Coefficients in Algebraic Expressions
Writing Algebraic Expressions With Numeric Coefficients
Writing Algebraic Expressions With Variable Coefficients
Practice Problems on Coefficients
Conclusion
Frequently Asked Questions on Coefficients
What exactly is a coefficient?
How do numeric and variable coefficients differ?
What role do coefficients play in algebraic expressions?
Can coefficients be negative or fractions?
How does understanding coefficients help in learning algebra?
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2018 AIME I Problems/Problem 14
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 Get solution faster in Sol 2
5 Solution 3
6 Video Solution
Problem
Let be a heptagon. A frog starts jumping at vertex . From any vertex of the heptagon except , the frog may jump to either of the two adjacent vertices. When it reaches vertex , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than jumps that end at .
Solution 1
This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line . We can count the number of left/right (L/R) paths of length that start at and end at either or .
We can visualize the paths using the common grid counting method by starting at the origin , so that a right (R) move corresponds to moving 1 in the positive direction, and a left (L) move corresponds to moving 1 in the positive direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines or . Letting be the number of such paths from to under these constraints, we have the following base cases:
and recursive step for .
The filled in grid will look something like this, where the lower-left corresponds to the origin:
The bolded numbers on the top diagonal represent the number of paths from to in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from to in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line . The total number of ways is .
(Solution by scrabbler94)
Solution 2
Let denotes the number of sequences with length that ends at . Define similarly for the other vertices. We seek for a recursive formula for . Computing a few terms we have , , , , and .
Using the formula yields , , , , , , , and .
Finally adding yields .
~ Nafer
Get solution faster in Sol 2
In the 5th to last line, we have by shifting the index in line 2.
Solution 3
For vertices the number of ways to get there after jumps is the sum of the number of ways to get to the adjacent vertices after jumps. For vertices and the number of ways to get there after jumps is he number of ways to get to and after jumps.
Video Solution
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7397 | https://www.youtube.com/watch?v=euhtXUgBEts | The Jigsaw Method
Cult of Pedagogy
66900 subscribers
7466 likes
Description
839219 views
Posted: 15 Apr 2015
This cooperative learning strategy increases student engagement, encourages collaboration, and results in better learning. Learn how to use the basic Jigsaw method, another variation called Jigsaw II, and get tips for troubleshooting, like what to do if you can't divide students evenly.
Transcript:
Introduction Hey there, this is Jennifer Gonzalez for Cult of Pedagogy. In this video, I will be showing you how to use the Jigsaw Method. First I'll show you the basic structure of
the Jigsaw, then a variation known as Jigsaw II, and finally I'll give you some tips for
troubleshooting this strategy. So what is the Jigsaw Method? Developed by Elliot Aronson in 1971, Jigsaw is a cooperative learning strategy
where each student in a group takes responsibility for one chunk of the content, then teaches
it to the other group members. Like the pieces of a jigsaw puzzle, students
fit their individual chunks together to form a complete body of knowledge. Here are the steps of a basic jigsaw (sometimes
referred to as Jigsaw I) Step 1 Divide Students Step 1 is to divide students into groups of
4 to 6 people per group. Jigsaw works best when you have the same number
of students in each team, So avoid having some groups of four, some
of five, and some of six. Later in this video, I'll show you what to do if you can't divide
students into perfectly even teams. For this example, we'll assume you're working
with a class of exactly 30 students who can be divided evenly into groups of six. We'll
call these the Jigsaw Groups. Step 2 Divide Content Step 2: Divide your content into 4 to 6 chunks. It's important to divide the content into
the same number of chunks as the number of students in each group. So if you have six
students per group, break your content into six chunks. If you're only going to have five
students in each group, then you'll only need five chunks. Suppose you're a history teacher and you're
doing an overview of different types of government. You could divide your content into these chunks:
democracy, dictatorship, monarchy, republic, totalitarianism, and theocracy. By the way: These index cards just represent
chunks of content. You don't need to use actual index cards to do jigsaw. A chunk of content can
be a section of a textbook chapter, a handout containing information, or an online resource. Step 3 Assign Content Step 3: Assign one chunk of content to each
person in the Jigsaw Group. Each group has one person responsible for
one chunk of the content. That person will be expected to teach that
chunk to the rest of the group. At this point, students don't really interact
with other members of their group; they just read and study their own chunk of content
independently. Then, their independent study is fortified by the next step... Step 4 Meet in Expert Groups Step 4: Have students meet in Expert Groups. After each student has studied his or her
chunk independently, they gather with all the other students who have been assigned
to the same chunk. These are called Expert Groups. Within each expert group, students compare
their ideas and work together to prepare some kind of presentation to give to their Jigsaw
Groups. During this time, gaps in individual students' knowledge can be filled, misconceptions
can be cleared up, and important concepts can be reinforced. Step 5 Return to Jigsaw Groups Step 5: Students return to Jigsaw Groups. Now that students have studied their chunks
in their expert groups, They return to their original jigsaw groups,
where each student takes a turn presenting their
chunk of information. Meanwhile, the other students listen carefully, take notes, and
ask lots of questions. Once the first expert has gone, the others
take their turns, As each "expert" teaches their chunk of content, The others in the group are learning it. Step 6 Assessment Step 6: Assess all students on all the content. The assessment can be a simple quiz to make
sure all students got a basic understanding of all the material. Be sure to include all
content chunks in this quiz. Jigsaw II is a variation on the basic structure
of Jigsaw. Developed by Robert Slavin in 1986, Jigsaw
II makes one significant tweak to the basic Jigsaw. The difference is in how the assessment is
treated. In Jigsaw I, students are assessed individually and receive just one score. In Jigsaw II, quiz scores are given once to
individual students, then each group's scores are averaged to generate a group score. This
builds in competition between groups and encourages students to work harder at helping each other
learn the material well. Here are a few tips for troubleshooting this
strategy. Troubleshooting One problem you might encounter is this: What
if students don't divide evenly? Now ideally, you'd have a perfectly divisible
group. But as we all know, that kind of perfection rarely happens, and even if you have perfection
in your plans, one absent student can throw your whole game off. First, remember that you can create groups
of 4, 5, or 6 (and some jigsaw advocates even allow
for groups of 2 or 3), so that should help minimize "extra" students. Still, if you end
up with a few extras, just assign two students in the same group the
same chunk. Now, what if some "experts" don't teach the
material very well? If this group is depending on this one student
to teach them about monarchies, and he's not the strongest student, that group is kind
of out of luck. You can anticipate this problem when creating
your groups. One thing you can do, if you have an uneven number of students, is pair
up two students on the same chunk who might be stronger together than they would be on
their own. Also, it's the responsibility of the expert
group to make sure that everyone is prepared to present their chunk to their respective
jigsaw groups. If one student isn't really getting it, make sure the rest of the group
gives that student extra preparation... So they'll be ready to teach the material
to their jigsaw group. A full description of this strategy and its
history can be found at the Jigsaw Classroom website. Additional information about Jigsaw
was obtained from Silver, Strong, and Perini's book, The Strategic Teacher. Additional Information Now before you go, I have some stuff for you to click. Ready? I have a ton of other teaching strategy videos.
To watch some of them, click here. To subscribe to my YouTube channel and receive a notification
every time I make a new video, click here. To read more information about the Jigsaw
strategy, click here. And to visit my website, Cult of Pedagogy, which you totally should,
click this guy right here. Thanks for watching and have a great day. |
7398 | https://stackoverflow.com/questions/7164397/find-the-min-max-excluding-zeros-in-a-numpy-array-or-a-tuple-in-python | Find the min/max excluding zeros in a numpy array (or a tuple) in python - Stack Overflow
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Find the min/max excluding zeros in a numpy array (or a tuple) in python
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Asked 14 years, 1 month ago
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I have an array. The valid values are not zero (either positive or negetive). I want to find the minimum and maximum within the array which should not take zeros into account. For example if the numbers are only negative. Zeros will be problematic.
python
numpy
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edited Apr 23, 2016 at 11:45
BrechtDeMan
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asked Aug 23, 2011 at 16:33
ShanShan
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1
4 What is your question? What have you tried? How didn't it work?hmakholm left over Monica –hmakholm left over Monica 2011-08-23 16:36:26 +00:00 Commented Aug 23, 2011 at 16:36
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How about:
python
import numpy as np
minval = np.min(a[np.nonzero(a)])
maxval = np.max(a[np.nonzero(a)])
where a is your array.
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answered Aug 23, 2011 at 16:36
JoshAdelJoshAdel
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Sven Marnach
Sven MarnachOver a year ago
@Shan: Using masked arrays can avoid the copy created by a[np.nonzero(a)] -- see my answer.
2011-08-23T16:58:55.557Z+00:00
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If you can choose the "invalid" value in your array, it is better to use nan instead of 0:
```python
a = numpy.array([1.0, numpy.nan, 2.0])
numpy.nanmax(a)
2.0
numpy.nanmin(a)
1.0
```
If this is not possible, you can use an array mask:
```python
a = numpy.array([1.0, 0.0, 2.0])
masked_a = numpy.ma.masked_equal(a, 0.0, copy=False)
masked_a.max()
2.0
masked_a.min()
1.0
```
Compared to Josh's answer using advanced indexing, this has the advantage of avoiding to create a copy of the array.
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edited Jul 17, 2020 at 18:49
answered Aug 23, 2011 at 16:57
Sven MarnachSven Marnach
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JoshAdel
JoshAdelOver a year ago
@Sven: when I do ma.base is a I get false, so it doesn't look like ma is just a view of a and there is a copy of the memory somewhere. Or am I testing this the wrong way?
2011-08-23T17:12:43.923Z+00:00
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unutbu
unutbuOver a year ago
@JoshAdel: by default, np.ma.masked_equal makes a copy of a. To get a view, use ma = np.ma.masked_equal(a, 0.0, copy=False).
2011-08-26T13:14:25.79Z+00:00
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Boycott OpenAI sellouts
Boycott OpenAI selloutsOver a year ago
This worked perfectly for me - I had a case where I needed to retrieve the index of the min/max nonzero value.
2015-09-01T19:45:04.74Z+00:00
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Brunox13
Brunox13Over a year ago
Just a side comment: I feel like naming the view (ma) same as the module used to create it (numpy.ma) is unnecessarily confusing for someone trying to understand this code, when it could have been named literally anything else more descriptive. Why not a_view, for example??
2020-07-17T16:08:09.567Z+00:00
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Sven Marnach
Sven MarnachOver a year ago
@Brunox13 It wouldn't have occurred to me that this is confusing, since it's just three lines of code, so thanks for the feedback! I went with masked_a.
2020-07-17T18:51:04.003Z+00:00
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Here's another way of masking which I think is easier to remember (although it does copy the array). For the case in point, it goes like this:
```python
import numpy
a = numpy.array([1.0, 0.0, 2.0])
ma = a[a != 0]
ma.max()
2.0
ma.min()
1.0
```
It generalizes to other expressions such as a > 0, numpy.isnan(a), ... And you can combine masks with standard operators (+ means OR, means AND, - means NOT) e.g:
```python
Identify elements that are outside interpolation domain or NaN
outside = (xi < x) + (eta < y) + (xi > x[-1]) + (eta > y[-1])
outside += numpy.isnan(xi) + numpy.isnan(eta)
inside = -outside
xi = xi[inside]
eta = eta[inside]
```
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answered Jun 6, 2012 at 8:51
uniomniuniomni
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Masked arrays in general are designed exactly for these kind of purposes. You can leverage masking zeros from an array (or ANY other kind of mask you desire, even masks that are more complicated than a simple equality) and do pretty much most of the stuff you do on regular arrays on your masked array. You can also specify an axis for which you wish to find the min along:
python
import numpy.ma as ma
mx = ma.masked_array(x, mask=x==0)
mx.min()
Example input:
python
x = np.array([1.0, 0.0, 2.0])
output:
python
1.0
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edited Aug 4, 2020 at 2:59
answered Aug 3, 2020 at 23:49
EhsanEhsan
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You could use a generator expression to filter out the zeros:
python
array = [-2, 0, -4, 0, -3, -2]
max(x for x in array if x != 0)
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answered Aug 23, 2011 at 16:37
Chris PickettChris Pickett
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JoshAdel
JoshAdelOver a year ago
I think the OP is talking about numpy arrays and not python lists. There is a difference, although your solution is correct for the later. Not going to downvote, but just so you know.
2011-08-23T16:39:09.683Z+00:00
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Chris Pickett
Chris PickettOver a year ago
Ahh, just saw array, didn't see the numpy tag.
2011-08-23T16:54:27.627Z+00:00
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endolith
endolithOver a year ago
This still works on numpy arrays, so it's a valid answer. For small arrays like this, it's significantly faster than the numpy version, too.
2019-09-01T03:48:21.547Z+00:00
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A simple way would be to use a list comprehension to exclude zeros.
```python
tup = (0, 1, 2, 5, 2)
min([x for x in tup if x !=0])
1
```
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answered Aug 23, 2011 at 16:36
WilduckWilduck
14.2k 13 13 gold badges 63 63 silver badges 91 91 bronze badges
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Wilduck
WilduckOver a year ago
The title of the question does say "in a numpy array (or a tuple)".
2011-08-23T16:41:48.96Z+00:00
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JoshAdel
JoshAdelOver a year ago
didn't see tuple in the title and only read the question which says array. I stand corrected +1. I jumped on you (and the other response), because people post solutions to numpy questions treating them as if they are numpy lists, which they aren't. It's a personal pet peeve, since the numpy solution is often wildly more efficient.
2011-08-23T16:44:24.987Z+00:00
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JoshAdel
JoshAdelOver a year ago
slight correction of what I wrote above: '....treating them as if they are python lists...'
2011-08-23T16:56:13.783Z+00:00
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NeilG
NeilGOct 9, 2024 at 5:42
No need for an additional list construction inside the min function, as demonstrated in Chris Pickett's answer
2024-10-09T05:42:11.027Z+00:00
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you can do this:
```python
yourlist = [-1, 2, 3, 4, -10, -7, -6, -5, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, -7, -10, 5, 9, 1, 0, 11, 20, -20]
minlist = min(filter(lambda x: x != 0,yourlist))
maxlist = max(filter(lambda x: x != 0,yourlist))
print(maxlist, minlist)
output is: 20 -20
```
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answered Apr 21, 2024 at 21:35
Omid asadiOmid asadi
1
1 Comment
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bfontaine
bfontaineOver a year ago
This looks like a duplicate of this answer.
2024-04-23T14:49:36.83Z+00:00
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7399 | https://gregorygundersen.com/blog/2023/02/01/estimating-square-roots/ | Estimating Square Roots in Your Head
I explore an ancient algorithm, sometimes called Heron's method, for estimating square roots without a calculator.
Published
01 February 2023
Imagine we want to compute the square root of a number n. The basic idea of
Heron’s
method,
named after the mathematician and engineer, Heron of
Alexandria, is to find a number g that is close to n and to then
average that with the number n/g, which corrects for the fact that g either
over- or underestimates n.
I like this algorithm because it is simple and works surprisingly
well. However, I first learned about it in (Benjamin & Shermer, 2006), which
did not provide a particularly deep explanation or analysis for why this method works. The goal of this post is to better
understand Heron’s method. How does it work? Why does it work? And how good are
the estimates?
The algorithm
Let’s demonstrate the method with an example. Consider computing the square root of n=33. We start by finding a number that forms a perfect square that is
close to 33. Here, let’s pick g=6, since 62=36. Then we compute a
second number, b=n/g. In practice, computing b in your head may require an approximation. Here, we can compute
it exactly as 33/6=5.5. Then our final guess is the average of these two numbers or
n≈2g+b,(1)
which in our example is
33=5.74456264654...≈26+5.5=5.75.(2)
That is pretty good. The relative error is less than 0.1%! And
furthermore, this is pretty straightforward to do in your head when n isn’t
too large.
Why does this work?
Intuitively, Heron’s method works because g is either an over- or
underestimate of n. Then n/g is an under- or overestimate,
respectively, of n. So the average of g and n/g should be closer to
n than either g or n/g is.
While this was probably Heron’s reasoning, we can offer a better explanation
using calculus: Heron’s method works because we’re performing a first-order Taylor
approximation around our initial guess. Put more specifically, we’re making a linear approximation of the nonlinear square root function at the point
g2.
To see this, recall that the general form of a Taylor expansion about a
point a is
f(x)=f(a)+k=1∑∞k!f(k)(a)(x−a)k,(3)
where the notation f(k) denotes the k-th derivative of f. If we define f(x)=x, then
f′(x)=2x1,(4)
and so the first-order Taylor approximation of x is
x≈a+2ax−a.(5)
Now choose x=n, and let h(x) denote the Taylor expansion around a=g2. Then we have
n≈h(n)=g+2gn−g2=2g+n/g=2g+b.(6)
And this is exactly what we calculated above.
Geometric interpretation
In general, the first-order Taylor expansion approximates a possibly
nonlinear function f(x) with a linear function at the point a:
f(x)≈f(a)+f′(a)(x−a)y-intercept + slope × x-shift.(7)
Thus, the Taylor approximation represents the tangent line to f(x) at the point (a,f(a)). We can see this for f(x)=x in Figure 1. This is a useful visualization because it highlights something
interesting: we expect Heron’s approximation to be worse for
smaller numbers. That’s because the square root function is “more curved”
(speaking loosely) for numbers closer to zero (Figure 2, left). As the square root function
flattens out for larger numbers, the linear approximation improves (Figure 2,
right).
Figure 1. A visualization of Heron's
method, or a first-order Taylor approximation of f(x)=x. We
construct a linear approximation h(x) (red dashed line) to the nonlinear function
f(x) (blue line). We then guess h(n) (black dot). Our
error is the absolute vertical difference between h(n) (black
dot) and n (blue dot).
How good is the approximation?
How good is this method? Did we just
get lucky with n=33 or does Heron’s method typically produce sensible
estimates of n?
To answer this question, I’m replicating a nice figure from an article from MathemAfrica, in
which the author makes a plot with the input number n on the x-axis and the absolute error
∣∣∣n−h(n)∣∣∣(8)
on the y-axis (Figure 2, blue line). (Note that when programming Heron’s
method, we must decide if we want to find g2 by searching numbers greater or
less than n; here, I’ve set g as the first integer less than
the square root of n, or as g = floor(sqrt(n)).) I like this figure because it captures
two interesting properties of Heron’s method. First, as we discussed above, the
Taylor approximation will typically be worse when n is small (when f(x)=x;
this is not true in general). And second, the error drops to zero on perfect
squares and increases roughly linearly between perfect squares.
Figure 2. The absolute (blue) and
relative (red) errors between the true value n and the estimate
using Heron's approximation h(n). The errors are zero when n is a perfect
square and are smaller on average for small n than for large n.
The MathemAfrica post focuses on lowering this absolute error
by judiciuosly picking the initial guess g. This is interesting as analysis. However, in my mind, this is
unnecessarily complicated for most practical mental math scenarios, i.e. for quick sanity checking rather than in a demonstration or competition. Why it is overly complicated? Well, the relative error,
n∣n−h(n)∣,(9)
rapidly decays to less than a percentage point or so (Figure 2, red line).
If
you’re not using a calculator to compute a square root, you’re probably just
getting a rough idea of a problem. And if we actually wanted to lower the absolute error and didn’t care about a human’s
mental limits, we should just expand the Taylor approximation to higher orders.
Benjamin, A., & Shermer, M. (2006). Secrets of mental math: the mathemagician’s guide to lightning calculation and amazing math tricks. Crown. |
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